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# Net Present Value: Micron Technology
Suppose Micron Technology (NasdaqGS: MU - http://ca.finance.yahoo.com/q?s=MU&ql=0) is considering a new project that will cost \$2,225,000 (initial cash outflow). The company has provided the following cash flow figures to you:
Year Cash Flow
0 -\$2,225,000
1 350,000
2 939,000
3 720,000
4 500,000
5 900,000
If the Micron Technology's cost of capital (discount rate) is 9%, what is the project's net present value? Based on your analysis and findings, what would you recommend to the executives and the shareholders of Micron Technology? Should the project be accepted? The shareholders of Micron Technology would also like to know the meaning of NPV concept.
You may use the following steps to calculate NPV:
1) Calculate present value (PV) of cash inflow (CF)
PV of CF = CF1 / (1+r)1 + CF2 / (1+r)2 + CF3 / (1+r)3 + CF4 / (1+r)4 + CF5 / (1+r)5
2) Calculate NPV
NPV = Total PV of CF - Initial cash outflow
or
-Initial cash outflow + Total PV of CF
r = Discount rate (9%)
If you do not know how to use calculator, please use the present value tables.
Brealey, R.A., Myers, S.C., & Allen, F. (2005). Principles of corporate finance, 8th Edition. The McGraw?Hill. Retrieved May, 2012, from http://jcooney.ba.ttu.edu/fin3322/Brealey%20Files/Appendix%20A%20-%20Present%20Value%20Tables.pdf (Please use Table 1)
#### Solution Preview
1)
PV of CF = CF1 / (1+r)1 + CF2 / (1+r)2 + CF3 / (1+r)3 + CF4 / (1+r)4 + CF5 ...
#### Solution Summary
The following posting helps calculate net present value.
\$2.19 | 478 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-22 | longest | en | 0.805264 |
https://codeforces.com/topic/61709/en4 | 1,624,513,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00522.warc.gz | 159,181,137 | 15,861 | [Tutorial] Searching Binary Indexed Tree in O(log(N))
Revision en4, by sdnr1, 2018-08-21 10:48:16
NOTE : Knowledge of Binary Indexed Trees is a prerequisite.
(If you don't wanna read the entire blog, Binary Lifting is the answer :P)
## Problem Statement
Assume we need to solve the following problem. We have an array, A of length n with only non-negative values. We want to perform the following operations on this array:
1. Update value at a given position
2. Compute prefix sum of A upto i, i ≤ n
3. Search for a prefix sum (something like a lower_bound in the prefix sums array of A)
## Basic Solution
Seeing such a problem we might think of using a Binary Indexed Tree (BIT) and implementing a binary search for type 3 operation. Its easy to see that binary search is possible here because prefix sums array is monotonic (only non-negative values in A).
The only issue with this is that binary search in a BIT has time complexity of O(log2(N)) (other operations can be done in O(log(N))). Even though this is naive,
Implementation
Most of the times this would be fast enough (because of small constant of above technique). But if the time limit is very tight, we will need something faster. Also we must note that there are other techniques like segment trees, policy based data structures, treaps, etc. which can perform operation 3 in O(log(N)). But they are harder to implement and have a high constant factor associated with their time complexity due to which they might be even slower than O(log2(N)) of BIT.
Hence we need an efficient searching method in BIT itself.
## Efficient Solution
We will make use of binary lifting to achieve O(log(N)) (well I actually do not know if this technique has a name but I am calling it binary lifting because the algorithm is similar to binary lifting in sparse tables).
#### Implementation :
int bit[N]; // BIT array
int bit_search(int v)
{
int sum = 0;
int pos = 0;
for(int i=LOGN; i>=0; i--)
{
if(pos + (1 << i) < N and sum + bit[pos + (1 << i)] < v)
{
sum += bit[pos + (1 << i)];
pos += (1 << i);
}
}
return pos + 1;
}
#### History
Revisions
Rev. Lang. By When Δ Comment
en19 sdnr1 2018-08-22 15:35:12 132
en18 sdnr1 2018-08-22 13:14:18 53
en17 sdnr1 2018-08-22 13:12:08 114
en16 sdnr1 2018-08-22 12:52:42 174
en15 sdnr1 2018-08-22 12:50:46 390 (published)
en14 sdnr1 2018-08-22 11:41:30 953 (saved to drafts)
en13 MikeMirzayanov 2018-08-22 11:10:35 7
en12 sdnr1 2018-08-22 11:05:32 20
en11 sdnr1 2018-08-22 10:48:05 25 (published)
en10 sdnr1 2018-08-22 10:46:44 36 (saved to drafts)
en9 sdnr1 2018-08-21 22:37:35 39
en8 sdnr1 2018-08-21 22:28:02 1056 Added some more insight for better understanding of the algorithm
en7 sdnr1 2018-08-21 21:57:33 96
en6 sdnr1 2018-08-21 21:25:18 988 (published)
en5 sdnr1 2018-08-21 18:19:38 867
en4 sdnr1 2018-08-21 10:48:16 321
en3 sdnr1 2018-08-21 10:12:15 228
en2 sdnr1 2018-08-21 10:07:53 214
en1 sdnr1 2018-08-21 09:55:51 1869 Initial revision (saved to drafts) | 974 | 2,978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-25 | latest | en | 0.883884 |
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# HELP WITH A RTC LOGIC
## 5 posts in this topic
Hi,
I was trying to do an logic with rtc, but I know i'm in the wrong way to do this, can someone give me a light about how can I do this on a better way?
you insert an hour and minute to start a engine, and a hour and minute to stop this engine by the hmi. The way i'm doing this is:
(ActualHour>=SetOnHour)&(ActualMinute>=SetOnMinute) == EngineON
(ActualHour=SetOffHour)&(ActualMinute=SetOffMinute) == EngineOFF
I know that are completely wrong because
1st: If i select 5AM to start and 5PM to stop, after 5PM, the system will start again.
2nd:Thinking you selected to start 5AM and stop 5PM, and know its 8AM, the system will start by hours normally, but if you have selected 20 minute and know its 15, the system will not work
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For each bound, and for the current time, convert hour and minute to minute-of-day. (zero to 1439.) Then you can do normal bounds checking on those integers.
1 person likes this
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Wow, how i didn't try this hahaha, felling like a dumbass
I will do this and i come later to show the results and possible problems... thank you for the enlightenment
```
```
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##### Share on other sites
Problem solved with the function
(SetOffHour>=ActualHour)&(ActualHour>=SetOnHour) == engine (on/off)
Btw, with this, another problem has appeared, this doesnt solve this question: When the SetOFFHour is lower than the ActualHour, so, if I select 5Am to Start and 4Am to stop, after 23:59, the system just stop and will only start at 5am again
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Posted (edited)
On 04/03/2023 at 11:38 PM, Rodrigo Balsalobre said:
Problem solved with the function
(SetOffHour>=ActualHour)&(ActualHour>=SetOnHour) == engine (on/off)
Btw, with this, another problem has appeared, this doesnt solve this question: When the SetOFFHour is lower than the ActualHour, so, if I select 5Am to Start and 4Am to stop, after 23:59, the system just stop and will only start at 5am again
i'm here again and the fuction to solve is
(SetOffHour>=ActualHour>=SetOnHour) == engine (on/off)
OR
(SetOffHour>=0 && SetOffHour<=SetOnHour<=ActualHour) == engine on/off
these two conditions solve the problem
Edited by Rodrigo Balsalobre
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A094774 Decimal expansion of Pi^(e - Pi). 0
6, 1, 5, 9, 5, 7, 9, 6, 7, 3, 9, 1, 5, 2, 2, 1, 0, 8, 0, 2, 7, 5, 8, 4, 2, 0, 8, 7, 2, 3, 0, 7, 3, 8, 9, 7, 9, 9, 6, 1, 0, 3, 3, 9, 1, 3, 9, 1, 6, 1, 1, 7, 1, 6, 2, 5, 6, 5, 5, 7, 7, 6, 5, 0, 3, 5, 1, 5, 4, 7, 3, 7, 9, 0, 5, 9, 2, 5, 5, 9, 2, 6, 8, 6, 8, 7, 3, 1, 9, 9, 2, 9, 2, 1, 7, 3, 7, 9, 2, 7, 3, 4, 2, 6, 2 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS EXAMPLE 0.6159579673915221080275842087 MATHEMATICA RealDigits[Pi^(E-Pi), 10, 120][[1]] (* Harvey P. Dale, May 08 2018 *) CROSSREFS Cf. A001113, A000796, A059742. Sequence in context: A257704 A248411 A011439 * A231925 A193239 A023406 Adjacent sequences: A094771 A094772 A094773 * A094775 A094776 A094777 KEYWORD cons,nonn AUTHOR Mohammad K. Azarian, Jun 09 2004 EXTENSIONS Offset corrected by Mohammad K. Azarian, Dec 11 2008 STATUS approved
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https://fdocuments.in/document/fe-formulations-hand-out.html | 1,620,304,830,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00589.warc.gz | 279,176,773 | 18,107 | of 24
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## Documents
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FE Formulation handout
### Transcript of Fe Formulations Hand Out
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Finite element formulations
Joel Cugnoni, LMAF / EPFL
March 13, 2013
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
1 Finite Element familiesIntroductionOverview of FE families
2 Continuum Elements3D continuum2D plane strain / stress2D axisymmetry
3 Structural ElementsShell elementsBeam elements
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
IntroductionOverview of FE families
Class of FE formulations
Existing classes of Finite Elements can be characterized by thefollowing criteria:
Geometry modeling:Modeling space: number of coordinates to describe geometry(3D, 2D, 1D)Basic Topology: basic type of topology (solid, surface, wire)
Physical modeling:Physics: the behaviour that is modelled type of DOFs,elementary matrices & resultsPhysical modeling space: 3D, 2D planar / axisymm., 1D number & meaning of DOFs
FE formulation:Element shape: hex, tetra, triangle, wedge, quad.Interpolation: FE shape functions order Number of nodesIntegration: integration scheme (type / order)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
IntroductionOverview of FE families
Main families of Finite Elements
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
3D continuum model
3D continuum model:3D geometry / 3D continuum behaviour / 3D loads, may have
symmetries !!Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
3D continuum elements
Nodal Coordinate:
xj = {x1, x2, x3}
Nodal DOF:
qj = {u1, u2, u3}
Coordinate Transform:
eT : x = x() = aH() ex
Displacement Interpolation:
euh = eH eq
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
Example: linear hexahedron finite element
Master element geometry:
supported by 8 corner nodes, coordinates: (1, 2, 3) [1, 1]3
Basis functions:
h1(1, 2, 3) =18(1 1)(1 2)(1 3)
h2(1, 2, 3) =18(1 + 1)(1 2)(1 3)
h3(1, 2, 3) =18(1 + 1)(1 + 2)(1 3)
h4(1, 2, 3) =18(1 1)(1 + 2)(1 3)
h5(1, 2, 3) =18(1 1)(1 2)(1 + 3)
h6(1, 2, 3) =18(1 + 1)(1 2)(1 + 3)
h7(1, 2, 3) =18(1 + 1)(1 + 2)(1 + 3)
h8(1, 2, 3) =18(1 1)(1 + 2)(1 + 3)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane strain model
Plane Strain: constrained in longitudinal direction2D geometry / 2D continuum behaviour / 2D loads, infinite
depthJoel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane strain
2D plane strain: definition
The base hypothesis of 2D plane strain problem is:
u1,2 = u1,2(x1, x2) & u3 = 0 33 = 23 = 13 = 0
The constitutive relationship is then: 112212
= E(1 + )(1 2)
1 0 1 00 0 122
112212
Note that 13 = 23 = 0 but 33 6= 0 (Poisson effect):
33 =E(11 + 22)
(1 + )(1 2)Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane strain elements
Nodal Coordinate:
xj = {x1, x2}
Nodal DOF:
qj = {u1, u2}
Geometry & displacement interp.:
x = aH(1,2)ex ; euh = eH eq
Stress / strain results:
= {11, 22, 12}T = {11, 22, 12}T
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane stress model
Plane Stress: no constraints in longitudinal direction2D geometry / 2D continuum behaviour / negligible depth
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane stress
2D plane stress: definition
The base hypothesis of 2D plane stress problem is:
33 = 23 = 13 = 0 & u1,2 = u1,2(x1, x2)
The constitutive relationship is then written: 112212
= E(1 2)
1 0 1 00 0 12
112212
Note that 33 6= 0 and is derived from the Poisson effects:
33 = E
(11 + 22)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D plane stress elements
Nodal Coordinate:
xj = {x1, x2}
Nodal DOF:
qj = {u1, u2}
Geometry & displacement interp.:
x = aH(1,2)ex ; euh = eH eq
Stress / strain results:
= {11, 22, 12}T = {11, 22, 12}T
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D axisymmetric model
Axisymmetric model: revolution geometry2D axisymm. geometry / 3D continuum behaviour / 2D axisymm.
loadsJoel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D axisymmetric elasticity
2D axisymmetric elasticity
2D axisymmetric models are written in cylindrical coordinates{x1,2,3} {r , z , }. The axisymmetric problem derives from thehypotheses that it is invariant with coordinate and thus thedisplacement, stress & strains fields depend only on thecoordinates r and z .
ur = ur (r , z); uz = uz(r , z); u = 0
rr =urr
; zz =uzz
; =urr
; rz =urz
+uzr
; r = z = 0
rr , zz , rz , = f (r , z); r = z = 0
Note that, even if we have reduced the dimensionnality of theproblem, the constitutive behaviour is fully 3D.
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
3D continuum2D plane strain / stress2D axisymmetry
2D axisymmetric elements
Nodal Coordinate:
xj = {x1, x2} = {r , z}
Axis of symmetry = OX2 Nodal DOF:
qj = {u1, u2} = {ur , uz}
Geometry & displacement interp.:
x = aH(1,2)ex ; euh = eH eq
Stress / strain results:
= {rr , zz , , rz}T = {rr , zz , , rz}T
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Shell elementsBeam elements
Shell elements
Shell part (3D geometry / 2.5D structural behaviour)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Shell elementsBeam elements
Shell elements
Nodal Coordinates, thickness, normal vector :
xj = {x1, x2, x3} ; nj ; t jGeometric interpolation:
x() =i
ahi (1, 2)(exi +
1
23 t
i ni)
Nodal DOF:qj = {u1, u2, u3, ur1, ur2(, ur3)}
Displacement interpolation:
euh() =ai
hi (1, 2)(e ui +
1
23 t
i[ur1 v1 + ur2 v2)
]Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Shell elementsBeam elements
Beam elements
Wire part (3D geometry / 1.5D structural behaviour)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Shell elementsBeam elements
Beam elements
Nodal Coordinates, dimensions, normal vectors :
xj = {x1, x2, x3} ; nj2; nj3; t j2; t j3Geometric interpolation:
x() =i
ahi (1)(exi +
1
22 t
i2 n
i2 +
1
23 t
i3 n
i3)
Nodal DOF:qj = {u1, u2, u3, ur1, ur2, ur3)}
(rotations / displacement expressed in the global coord. system)
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Example 1: TGV Bogie
Suitable modeling methods:
3D solids, use directly the 3D CAD model
3D shells (relatively thin plates), need to build surface model
Joel Cugnoni, LMAF / EPFL Finite element formulations
• OverviewFinite Element familiesContinuum ElementsStructural Elements
Examples
Example 2: Aircraft fuselage
Suitable modeling methods:
3D shells (thin skins), need to build surface model3D solids, | 2,634 | 8,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-21 | latest | en | 0.565586 |
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### Sticky Triangles
##### Stage: 2 Challenge Level:
Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
### Lost Books
##### Stage: 2 Challenge Level:
While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book?
### Chain of Changes
##### Stage: 1 Challenge Level:
Arrange the shapes in a line so that you change either colour or shape in the next piece along. Can you find several ways to start with a blue triangle and end with a red circle?
### Triangular Hexagons
##### Stage: 2 Challenge Level:
Investigate these hexagons drawn from different sized equilateral triangles.
### The Numbers Give the Design
##### Stage: 2 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Chairs and Tables
##### Stage: 1 Challenge Level:
Make a chair and table out of interlocking cubes, making sure that the chair fits under the table!
### Street Sequences
##### Stage: 1 and 2 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
### Play a Merry Tune
##### Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Domino Sets
##### Stage: 2 Challenge Level:
How do you know if your set of dominoes is complete?
### Calendar Patterns
##### Stage: 2 Challenge Level:
In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers?
### Mobile Numbers
##### Stage: 1 and 2 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time?
### Function Machines
##### Stage: 2 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### Cube Bricks and Daisy Chains
##### Stage: 1 Challenge Level:
Daisy and Akram were making number patterns. Daisy was using beads that looked like flowers and Akram was using cube bricks. First they were counting in twos.
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### Domino Patterns
##### Stage: 1 Challenge Level:
What patterns can you make with a set of dominoes?
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Holes
##### Stage: 1 and 2 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Magazines
##### Stage: 2 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Next Number
##### Stage: 2 Short Challenge Level:
Find the next number in this pattern: 3, 7, 19, 55 ...
### A Shapely Network
##### Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
### The Tomato and the Bean
##### Stage: 1 Challenge Level:
At the beginning of May Tom put his tomato plant outside. On the same day he sowed a bean in another pot. When will the two be the same height?
### Sounds Great!
##### Stage: 1 Challenge Level:
Investigate the different sounds you can make by putting the owls and donkeys on the wheel.
##### Stage: 2 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Pyramid Numbers
##### Stage: 2 Challenge Level:
What are the next three numbers in this sequence? Can you explain why are they called pyramid numbers?
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Exploring Wild & Wonderful Number Patterns
##### Stage: 2 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Amazing Alphabet Maze
##### Stage: 1 Challenge Level:
Can you go from A to Z right through the alphabet in the hexagonal maze?
### Light Blue - Dark Blue
##### Stage: 2 Challenge Level:
Investigate the successive areas of light blue in these diagrams.
### The Mathemagician's Seven Spells
##### Stage: 2 Challenge Level:
"Tell me the next two numbers in each of these seven minor spells", chanted the Mathemagician, "And the great spell will crumble away!" Can you help Anna and David break the spell?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### Generating Number Patterns: an Email Conversation
##### Stage: 2, 3 and 4
This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution. | 1,739 | 7,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-39 | latest | en | 0.955753 |
https://studylib.net/doc/5325843/lecture-3----astronomical-coordinate-systems | 1,547,775,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118031216-00202.warc.gz | 652,527,572 | 33,616 | # Lecture 3 -- Astronomical Coordinate Systems
```Lecture 3 -- Astronomical Coordinate
Systems
Constellation of the Day…Aquila
Look at constellation maps on course
Last time: seasonal differences in
the sky
At different times of year we see
different constellations in the
evening sky, etc. Can be
understood as the Sun moving
through different constellations
Measuring the position of the Sun against
the background stars
The path of the Sun through the stars
Note that the Sun only moves through
certain constellations
Virgo, Libra,Scorpius,
Sagittarius,
Capricornus, etc. What
is the connection here?
Question: what’s causing this?
Demo
Also look at online animation with the book
web site
Last time: Earth’s orbital motion (revolution)
Plane of Earth’s orbit around Sun
called “plane of ecliptic”
Revolution cannot explain seasonal
changes in rising and setting of Sun and
Moon
Obliquity of the Ecliptic and the
Altitude Angle of the Sun
Explanation of Seasonal
Variations: tilt of the Earth’s axis:
obliquity of the ecliptic
The celestial sphere, the celestial pole,
and the celestial equator
Two Lines on
the Sky
•The ecliptic
•The celestial
equator
•See Figure
2.11
Astronomical Scientific Terms
•
•
•
•
•
•
Meridian
Celestial sphere
Zenith
Azimuth and altitude
Ecliptic
Celestial equator
For new purposes, we need a different
coordinate system
Analogy: I am riding my bike on a dirt road near
Lone Tree, and want to describe to someone in
London the location of a radio tower I see in the
distance.
Question: what system of coordinates do I use?
A New Coordinate System: Celestial
Coordinates
• The stars “stick together” and define
their own reference system. The planets
move with respect to them
• Celestial coordinates are Right
Ascension and Declination
• Right Ascension ….. Longitude
• Declination ….latitude
• http://sohowww.nascom.nasa.gov/
Questions
```
12 cards
26 cards
17 cards
25 cards
24 cards | 500 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.808104 |
https://discussions.unity.com/t/calculate-up-distance-displacement-for-3d-infinity-runner-game/212479 | 1,716,200,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00120.warc.gz | 182,225,841 | 7,831 | # Calculate UP Distance Displacement for 3D Infinity Runner Game
Hi guys, I’m a little noob with programming and need some help with an issue.
I’m doing an infinity runner 3D space shooter where my character only moves on the Y Up axis and can rotate his spaceship 360 degrees to move left and right like a rocket.
I made a distance script counter where it shows the distance that the player traveled flying upwards when he press the fly button, but I need some help with only counts the distance if he is “going up” , because the script still counts the distance even if the ship is flying down or is flying sideways.
So the question is: how to write for the distance counter script only counts meters if the spaceship is flying up?
Here is my script
public class distanceCounter : MonoBehaviour {
``````Text mtTxt;
[SerializeField]
private float distance;
[SerializeField]
private GameObject player;
[SerializeField]
private float curPos;
void Start ()
{
mtTxt = GetComponent<Text> ();
curPos = player.transform.position.y;
}
void Update ()
{
if (TouchControls.startMeters == true)
{
distance += Time.deltaTime * curPos;
}
mtTxt.text = "Meters: " + distance + " m";
}
``````
}
Hi,
I am assuming that curPos is capturing the position of the player (ship) at the start of the game. So what you need is a sense of how much the player has flown up. I am not sure how you implement your infinite scrolling world, so I am assuming that when your spaceship goes “up”, your camera also goes “up” with it and new content keeps getting generated (higher and higher from ground level). If that is the case, then you should be calculating the distance as follows:
``````private float startPos, curPos;
void Start ()
{
mtTxt = GetComponent<Text> ();
startPos = player.transform.position.y;
distance = 0f;
}
void Update ()
{
if (TouchControls.startMeters == true)
{
curPos = player.transform.position.y;
distance = Mathf.Max(0f, curPos - startPos);
}
mtTxt.text = "Meters: " + distance + " m";
}
``````
Hi jupt001,
A simple solution then would be to add distance only if your player is pressing the forward or the up key; i.e. he is progressing further. You would then conditionally add distance as a notion of time he spends clicking the up button, and likewise deducting distance when he clicks the down button.
So lets assume you have an Update function in the Player’s rocket script which is as follows:
``````public float rocketSpeed = 10f;
private float distanceSoFar = 0f;
void Update()
{
float move = Input.GetAxis("Vertical");
if(move != 0)
distanceSoFar += rocketSpeed*Time.deltaTime*Mathf.sign(move);
}
``````
This is the closest to what I need. I just need help to figure out how to control the velocity of meters degrease when the player stop pressing the fly button and start falling.
Example: When player reaches 100 meters and stop pressing “fly btn” he starts to fall and the meters counter decreases but it does so fast like -70 meters and I want that it decreases just -20 meters from the player’s last position and not -70 meters like it is happening.
`````` private float multiplier;
private int distance;
private GameObject player;
private float curPos;
void Update ()
{
curPos = player.transform.position.y;
multiplier += Time.deltaTime * 0.06f;
if (TouchControls.startMeters == true)
{
distance = (int)Mathf.Max(0f, curPos * multiplier);
}
``````
Dear @jupt001,
Thank you for sending across the setup. It helped me wrap my head around to what you actually wanted to achieve, and how to do it. Pls find the updated package with new code here
The main logic for the movement is placed now in BGScroller script which is also responsible for scrolling the canvas. The dot product helps to project the local up vector of the rocket to the global Y axis to get a sense of whether it is headed up or down. The distance update is done only when fly button is pressed. I also added a test update function in TouchControls for testing with arrows on PC. Pls. turn the testingForNonTouchDevices variable to false before testing on your touch device.
I took the liberty of cleaning up your code and putting in lots of comments to help you write better code for future projects. A summary of some of these ideas are:
1. Keep GUI scripts simple… only related to GUI and not some calculation involving some other scene objects. Do those calculations in scripts attached to the scene objects themselves, and then pass on a reference to the script or object concerned to the GUI script
2. Avoid use of static variables to pass on information from one class to another. Use them only if these variables are going to be shared across all instances of the script, or if there is only ONE instance, example a single Player
3. Find references to objects only once in Start(), and not every frame in Update()
4. See how I am getting a handle to a script from another script (via an object reference, which can be input manually or automatically found by tag), and then accessing public variables of this script
Okay, I know this is a lot to take in, and Rome was definitely not built in one day
I hope this helps you create a super awesome game, and I look forward to you sharing it with me once you finish it! Good luck buddy! | 1,167 | 5,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.846651 |
https://www.teacherspayteachers.com/Product/Measurement-Data-Bundle-CCSS-1255028 | 1,548,237,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584328678.85/warc/CC-MAIN-20190123085337-20190123111337-00412.warc.gz | 962,302,986 | 21,792 | # Measurement & Data Bundle (CCSS)
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\$10.00 | 1,009 | 4,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-04 | latest | en | 0.867579 |
https://stats.stackexchange.com/questions/592407/combining-averages-to-improve-estimates | 1,718,855,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00546.warc.gz | 474,941,777 | 39,406 | # Combining Averages to Improve Estimates?
I am an MBA student taking courses in statistics.
Recently, I posted a question (Can "Pooled Averages" be considered as "Better" compared to "Individual Averages"?) in which I asked if averaging different averages together can improve the quality of an estimate.
Now, I have a slightly modified version of this question.
Suppose we are only given information on whether different people in the population have a specific disease (i.e. we are not provided with a dataset, only provided with summary information):
• Suppose the probability of having a specific disease in the entire population is p1
• Suppose the probability of having a specific disease for the population of men is p2
• Suppose the probability of having a specific disease for the population of women is p3
• Suppose the probability of having a specific disease for the population of young people is p4
• Suppose the probability of having a specific disease for the population of old people is p5
Now, imagine we realize that we forgot to collect the data for this one young man. What is the probability that he has this disease?
I see several different methods of estimating this probability:
• Method 1 (Overall Effect): p1
• Method 2 (Gender Effect): p2
• Method 3 (Age Effect): p4
• Method 4 (Average Age and Gender Effect): (p2 + p4)/2
• Method 5 (Average Overall, Age and Gender Effect) : (p1 + p2 + p4)/3
• Method 6 (Average Overall and Age Effect): (p1 + p4)/2
• Method 7 (Average Overall and Gender Effect): (p1 + p2)/2
I have the following question: Using statistics, do we have any analysis that can compare the validity of these methods? For example, perhaps some of these methods are meaningless as they double count and introduce noise into the estimates? Or perhaps some of these methods will result in very large standard errors?
I have been having this discussion with my classmates and different people think its either Method 1, Method 4 or Method 5. But is there any way to formally establish which of these methods is the best?
EXTRA: I spent some thinking about a simulation experiment in R to see which Method is better. Imagine you have a dataset that contains information on the disease, age and gender of patients:
### DATA FOR PROBLEM
# Disease = 1, No Disease = 0
Disease <- c(1,0)
Disease <- sample(Disease, 10000, replace=TRUE, prob=c(0.3, 0.7))
# Male = 1, Female = 0
Gender <- c(1,0)
Gender <- sample(Gender, 10000, replace=TRUE, prob=c(0.5, 0.5))
# Old = 1, Young = 0
Age <- c(1,0)
Age <- sample(Age, 10000, replace=TRUE, prob=c(0.5, 0.5))
Patient_ID = 1:1000
Simulation_Data = data.frame(Patient_ID, Disease, Gender, Age)
Now, imagine that a new patient enters with a randomly assigned gender, age and disease status. We can now compare which of these methods will provide a better estimate by running this simulation many times:
#### LOOP
results = list()
for (i in 1:1000)
{
# SIMULATE DATA FOR A NEW PATIENT
Patient_Being_Tested_Disease_i = ifelse( runif(1, 0, 1) > 0.5, 1,0)
Patient_Being_Tested_Gender_i = ifelse( runif(1, 0, 1) > 0.5, 1,0)
Patient_Being_Tested_Age_i = ifelse( runif(1, 0, 1) > 0.5, 1,0)
Overall_Prob = mean(Simulation_Data$Disease) Patient_Gender_Data_i = Simulation_Data[which( Simulation_Data$Gender == Patient_Being_Tested_Gender_i), ]
Patient_Gender_Prob_i = mean(Patient_Gender_Data_i$Disease) Patient_Age_Data_i = Simulation_Data[which( Simulation_Data$Age == Patient_Being_Tested_Age_i), ]
Patient_Age_Prob_i = mean(Patient_Age_Data_i$Disease) Method_1_i = Overall_Prob Method_2_i = mean(Patient_Gender_Data_i$Disease)
Method_3_i = mean(Patient_Age_Data_i$Disease) Method_4_i = (Patient_Gender_Prob_i + Patient_Age_Prob_i)/2 Method_5_i = (Overall_Prob + Patient_Gender_Prob_i + Patient_Age_Prob_i)/3 Method_6_i = (Overall_Prob + Method_3_i)/2 Method_7_i = (Overall_Prob + Method_2_i)/2 methods_i <- c(Method_1 = Method_1_i , Method_2 = Method_2_i, Method_3 = Method_3_i, Method_4 = Method_4_i , Method_5 = Method_5_i, Method_6 = Method_6_i , Method_7 = Method_7_i) winner_i = ifelse(Patient_Being_Tested_Disease_i == 0, names(methods_i)[which.min(methods_i)], names(methods_i)[which.max(methods_i)]) print(winner_i) results_i = data.frame(i, winner_i) results[[i]] <- results_i } final <- do.call(rbind.data.frame, results) counts <- table(final$winner_i)
barplot(counts, main="Which Method Won?",
xlab="Method", ylab = "Number of Wins")
My logic is - in the case when the patient has the disease, you want to select the Method with the largest probability. And in the case when the patient does not have the disease, you want to select the method with the smallest probability.
I hope I did this correctly!
• It is a bit unclear what you mean with by the combination of these multiple effects. You might be talking about a linear model, but given your recent post it seems like you are averaging the effects. In relation to a linear model it might be interesting to read about the piranha problem; you can not generally add up multiple effects. Commented Oct 15, 2022 at 18:18
• Would this approach that I have done be similar in theory to the spirit of a linear model? Commented Oct 15, 2022 at 18:23
• No averaging would be different. In a linear model you would be adding multiple effects and two combined effects would be ending up as bigger (adding up), instead of being averaged. Commented Oct 15, 2022 at 18:29 | 1,456 | 5,432 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.946643 |
http://www.topperlearning.com/forums/home-work-help-19/the-displacement-physics-motion-in-a-straight-line-57004/reply | 1,495,523,637,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607591.53/warc/CC-MAIN-20170523064440-20170523084440-00229.warc.gz | 706,568,230 | 38,145 | Question
Thu July 05, 2012 By:
# the displacement
Thu July 05, 2012
According to the question, displacement, s = kt3
where k is the constant of proportionality.
Now acceleration, a = d2s/dt2
so that a = 6kt
hence acclaration is proportional to the time
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# A bird moves with velocity 20m/s in a direction making an angle of 60 degree with the eastern line and 60 degree with vertical upwards. The velocity vector in rectangular form : a) 10 i + 10 j + 10 k b) 10 i +10(2)^1/2 j + 10 k c) -20 i +10(2)^1/2 j + 20 k d) 10(2)^1/2 i + 10 j + k
Mon August 29, 2016
Home Work Help | 213 | 628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-22 | latest | en | 0.858622 |
https://link.springer.com/chapter/10.1007/978-3-030-44787-8_14 | 1,686,110,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00712.warc.gz | 416,274,064 | 82,870 | There are four mechanisms that cause sound energy to be absorbed and sound waves to be attenuated as they propagate in a single-component, homogeneous fluid:
• Viscous (shear) effects in bulk fluids
• Thermal conduction in bulk fluids
• Molecular absorption in bulk fluids
• Thermoviscous boundary layer losses
The decrease in the amplitude of acoustical disturbances or in the amplitude of vibrational motion (due to dissipative mechanisms) has been a topic of interest throughout this textbook. In this chapter, we will capitalize on our investment in such analyses to develop an understanding of the attenuation of sound waves in fluids that are not influenced by proximity to solid surfaces. Such dissipation mechanisms are particularly important at very high frequencies and short distances or very low frequencies over geological distances.
The parallel addition of a mechanical resistance element to the stiffness and mass of a simple harmonic oscillator led to an exponential decay in the amplitude of vibration with time in Sect. 2.4. The (mechanically) series combination of a stiffness element and a mechanical resistance in the Maxwell model of Sect. 4.4.1, and in the Standard Linear Model of viscoelasticity in Sect. 4.4.2 introduced the concept of a relaxation time, τR, that had significant effects on the elastic (in-phase) and dissipative (quadrature) responses as a function of the nondimensional frequency, ωτR. Those response curves were “universal” in the sense that causality linked the elastic and dissipative responses through the Kramers-Kronig relations, as presented in Sect. 4.4.4.
That relaxation time perspective, along with its associated mathematical consequences, will be essential to the development of expressions for attenuation of sound in media that can be characterized by one or more relaxation times related to those internal degrees of freedom that make the equation of state a function of frequency. Examples of these relaxation time effects include the rate of collisions between different molecular species in a gas (e.g., nitrogen and water vapor in air), the pressure dependence of ionic association-dissociation of dissolved salts in seawater (e.g., MgSO4 and H3BO3), and evaporation-condensation effects when a fluid is oscillating about equilibrium with its vapor (e.g., fog droplets in air or gas bubbles in liquids).
The viscous drag on a fluid oscillating within the neck of a Helmholtz resonator, combined with the thermal relaxation of adiabatic temperature changes at the (isothermal) surface of that resonator’s compliance, led to energy dissipation in lumped-element fluidic oscillators in Sect. 9.4.4, producing damping that limited the quality factor of those resonances in exactly the same way as mechanical resistance limited the quality factor of a driven simple harmonic oscillator, which was first introduced as a consequence of similitude (i.e., dimensional analysis) in Sect. 1.7.1.
The thermoviscous boundary layer dissipation, summarized in Eq. (9.38), was used to calculate the attenuation of plane waves traveling in a waveguide in Sect. 13.5.5. As will be demonstrated explicitly in this chapter, thermoviscous boundary layer losses provide the dominant dissipation mechanism at low frequencies (i.e., lumped element systems and waveguides below cut-off) for most laboratory-sized objects (including the laboratory itself when treated as a three-dimensional enclosure). For fluid systems that are not dominated by dissipation on solid surfaces in close proximity to the fluids they contain, the dissipation due to losses within the fluid itself (i.e., bulk losses) can be calculated directly from the hydrodynamic equations of Sect. 7.3.
To reintroduce the concepts complex wavenumber or complex frequency that typically characterize the attenuation of sound over space or time, a simple solution of the Navier-Stokes equation will first be derived. That approach will not provide the correct results for attenuation of sound, even in the absence of relaxation effects, because it does not properly take into account the relationship between shear deformation and hydrostatic compression in fluids that are necessary to produce plane waves (see Fig. 14.1). That relationship was used to relate the modulus of unilateral compression for isotropic solids (aka the dilatational modulus) to other isotropic moduli in Sect. 4.2.2 and in Fig. 4.3. The complete solution for bulk losses due to a fluid’s shear viscosity, μ; thermal conductivity, κ; and the relaxation of internal degrees of freedom (i.e., “bulk viscosity), ζ, will follow that nearly correct introductory treatment and will be based upon arguments related to entropy production, like the analysis in Sect. 9.3.3.
## 1 An Almost Correct Expression for Viscous Attenuation
Because we started with a complete hydrodynamic description of homogeneous, isotropic, single-component fluids using the Navier-Stokes equation, we are now well-prepared to investigate the dissipation mechanisms that attenuate the amplitude of sound waves propagating far from the influence of any solid boundaries. A one-dimensional linearized version of the Navier-Stokes equation (9.2) is reproduced below:
$${\rho}_m\frac{\partial {v}_1\left(x,t\right)}{\partial t}+\frac{\partial {p}_1\left(x,t\right)}{\partial x}-\mu \frac{\partial^2{v}_1\left(x,t\right)}{\partial {x}^2}=0$$
(14.1)
The linearized, one-dimensional continuity equation (10.1) is not affected by the inclusion of viscosity in the Navier-Stokes equation.
$$\frac{\partial {\rho}_1\left(x,t\right)}{\partial t}+{\rho}_m\frac{\partial {v}_1\left(x,t\right)}{\partial x}=0$$
(14.2)
Since the thermal conductivity of the fluid is ignored, κ = 0, the linearized version of the adiabatic equation of state can still be invoked to eliminate ρ1(x, t) in favor of p1(x, t), allowing expression of the continuity equation in terms of the same variables used in Eq. (14.1).
$${\displaystyle \begin{array}{c}{\rho}_1\left(x,t\right)={\left(\frac{\partial \rho }{\partial p}\right)}_s{p}_1\left(x,t\right)=\frac{p_1\left(x,t\right)}{c^2}\\ {}\Rightarrow \kern1em \frac{1}{\rho_m{c}^2}\frac{\partial {p}_1\left(x,t\right)}{\partial t}+\frac{\partial {v}_1\left(x,t\right)}{\partial x}=0\end{array}}$$
(14.3)
As done so many times before, the dispersion relation, ω (k), will be calculated by assuming a right-going traveling wave to convert the homogeneous partial differential equations (14.1) and (14.3) to coupled algebraic equations.
$${\displaystyle \begin{array}{c}-{jkp}_1+\left(j{\omega \rho}_m+{k}^2\mu \right){v}_1=0\\ {} j\omega \frac{p_1}{\rho_m{c}^2}-{jkv}_1=0\end{array}}$$
(14.4)
There are now both real and imaginary terms in the coupled algebraic equations, unlike their nondissipative equivalents, such as Eqs. (10.16) and (10.17). The existence of a nontrivial solution to Eq. (14.4) requires that the determinant of the coefficients vanish.
$$\left|\begin{array}{cc}\frac{j\omega}{\rho_m{c}^2}& - jk\\ {}- jk& \left(j{\omega \rho}_m+{k}^2\mu \right)\end{array}\right|=0$$
(14.5)
The evaluation of this determinant leads to the secular equation that will specify the complex wavenumber, k, in terms of the angular frequency,Footnote 1 ω.
$${\left(\frac{\omega }{c}\right)}^2={k}^2\left(1+\frac{j\omega \mu}{\rho_m{c}^2}\right)$$
(14.6)
If ωμ/ρmc2 < < 1, the binomial expansion can be used twice to approximate the spatial attenuation coefficient, α.
$$\mathbf{k}\cong \frac{\omega }{c}-j\frac{\omega^2\mu }{2{\rho}_m{c}^3}=k-j{\alpha}_{almost}$$
(14.7)
To remind ourselves that this result is not completely correct, this spatial attenuation coefficient has been designated αalmost = ω2(μ/2ρmc3). The form of this result, specifically the fact that the attenuation is proportional to ω2/ρm, suggests that experimental results could be plotted as a function of the square of frequency divided by mean pressure, as shown in Fig. 14.4.
It is also useful to notice that μ/ρmc2 in Eq. (14.6) has the units of time, so the “small parameter” in those binominal expansions of Eq. (14.6) is of the form $$j{\omega \tau}_{\overline{\mathrm{\ell}}}$$. Equally important is the recognition that such a relaxation time, $${\tau}_{\overline{\mathrm{\ell}}}$$, is on the order of the collision time in a gas, based on the mean free path, $$\overline{\mathrm{\ell}}$$, derived from simple kinetic theory of gases in Sect. 9.5.1 [1].
$${\tau}_{\overline{\mathrm{\ell}}}=\frac{\overline{\mathrm{\ell}}}{c}\cong \frac{3{\mu}_{gas}}{\rho_m{c}^2}=\frac{3}{\gamma}\frac{\mu_{gas}}{p_m}$$
(14.8)
This is different from the relaxation times, τR, which can characterize the time dependence of the equation of state or the response of a viscoelastic medium described in Sect. 4.4, where ωτR ⪒ 1. At frequencies above $${\omega}_{\overline{\mathrm{\ell}}}\cong {\left({\tau}_{\overline{\mathrm{\ell}}}\right)}^{-1}$$, the assumptions that underlie the hydrodynamic approach are no longer valid (see Chap. 7, Problem 1). For air near room temperature and at atmospheric pressure, $${\tau}_{\overline{\mathrm{\ell}}}$$ ≅ 400 ps, so f = ω /2π ≅ 400 MHz. This is identical with the result obtained in Eq. (9.24) for the critical frequency, ωcrit, at which sound propagation in air transitions from adiabatic at low frequencies to isothermal at high frequencies. The regime where $${\omega \tau}_{\overline{\mathrm{\ell}}}$$ > 1 becomes questionable within the context of a (phenomenological) hydrodynamic theory [1].
Unlike the complex wavenumbers of exponentially decaying thermal and viscous waves near boundaries, examined in Sects. 9.3.1 and 9.4.2, where ℜe[k] = ℑm[k], most attenuation mechanisms in bulk fluids far from boundaries have ℜe[k] ≫ ℑm[k] or |k| ≫ α. If k is substituted into the expression for the pressure associated with a single-frequency one-dimensional plane wave traveling in the +x direction, p1 (x, t), it is easy to see that α leads to an exponential decay in the amplitude with propagation distance for the sound wave.
$${p}_1\left(x,t\right)=\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t-\mathbf{k}x\right)}\right]=\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\;\left[\omega\;t-\left(k- j\alpha \right)x\right]}\right]={e}^{-\alpha x}\Re \mathrm{e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t- kx\right)}\right]$$
(14.9)
Since space and time can be transformed by the sound speed, a real temporal attenuation coefficient can be defined, β = α c, to describe the rate at which the amplitude of the plane wave decays in time.
$${p}_1\left(x,t\right)={e}^{-\beta t}\mathit{\Re e}\left[\hat{\mathbf{p}}{e}^{j\left(\omega\;t- kx\right)}\right]\kern1em \mathrm{where}\kern1em \beta =\alpha c$$
(14.10)
Although the result for αalmost is not exactly correct, it does exhibit a feature of the correct result for the spatial attenuation coefficient that includes thermoviscous dissipation that is given by αclassical in Eq. (14.31) where internal relaxation effects are discussed in Sect. 14.5.Footnote 2
Unlike the spatial attenuation coefficient for dissipation of plane waves in a waveguide, given in Eq. (13.78), which is proportional to$$\sqrt{\omega }$$, Eq. (14.7) shows that αalmost is proportional to the square of the frequency, as is αclassical.
## 2 Bulk Thermoviscous Attenuation in Fluids
Although the previous results for αalmost are incomplete, it both has provided an introduction to the complex wavenumber, k, that determines the attenuation distance and has introduced a relaxation time, $${\tau}_{\overline{\mathrm{\ell}}}$$, that sets an upper limit to the frequencies above which the continuum model of a fluid is not appropriate. One reason that previous result for the viscous attenuation is not complete is that we have ignored the fact that the fluid deformation corresponding to the passage of a plane wave requires the superposition of two shear deformations and a hydrostatic compression. This superposition of shear strain and hydrostatic strain is illustrated schematically in Fig. 14.1. Of course, the result for the attenuation coefficient, αalmost, in Sect. 14.1, also does not yet include the thermal conductivity, κ, of the fluid.
To incorporate all of the dissipative effects in a fluid, it is necessary to start from the complete expression for entropy production in a single-component homogeneous fluid. The mechanical energy dissipation, Emech, is the maximum amount of work that can be done in going from a given non-equilibrium state of energy, Eo, back to equilibrium, E(S), which occurs when the transition is reversible (i.e., without a change in entropy) [2]. $${\dot{E}}_{mech}$$ is the rate at which the mechanical energy is dissipated by the periodic transitions from the non-equilibrium state to the equilibrium state as orchestrated by the wave motion.
$${\dot{\Pi}}_{mech}=-\dot{E}(S)=-\left(\frac{\partial E}{\partial S}\right)\dot{S}={T}_m\dot{S}$$
(14.11)
The right-most expression in Eq. (14.11) uses the fact that the derivative of the energy with respect to the entropy is the equilibrium value of the mean absolute temperature, Tm. The entropy equation (7.43) can be written so that the shear stresses and the hydrostatic stresses can be expressed symmetrically in Cartesian components.Footnote 3
$$\rho T\left(\frac{\partial s}{\partial t}+\overrightarrow{v}\cdotp \overrightarrow{\nabla}s\right)=\nabla \cdotp \left(\kappa\;\overrightarrow{\nabla}T\right)+\frac{1}{2}\mu {\left(\frac{\partial {v}_i}{\partial {x}_k}+\frac{\partial {v}_k}{\partial {x}_i}-\frac{2}{3}\frac{\partial {v}_i}{\partial {x}_i}\right)}^2+\zeta {\left(\nabla \cdotp \overrightarrow{v}\right)}^2$$
(14.12)
The two cross derivatives, (∂vi/∂xk) and (∂vk/∂xi), represent the two shear deformations with the hydrostatic component removed: $$\raisebox{1ex}{2}\!\left/ \!\raisebox{-1ex}{3}\right.\;\left(\partial {v}_i/\partial {x}_i\right)$$ [3]. The square of the hydrostatic deformation is represented by $${\left(\nabla \cdotp \overrightarrow{v}\right)}^2$$. The hydrostatic deformation is multiplied by a new positive scalar coefficient, ζ, that must have the same units as the shear viscosity [Pa-s].
Having the form of a conservation equation (see Sect. 10.5), the right-hand side of Eq. (14.12) represents the rate of entropy production, $$\dot{S}$$, caused by thermal conduction, viscous shear, and some possible entropy production mechanism (unspecified at this point but eventually related to the time dependence of the equation of state) associated with the hydrostatic deformation. Using Eq. (14.12), the dissipated mechanical power, Πmech, can be evaluated by integrating over a volume element that includes the plane wave disturbance, dV.
$${\Pi}_{mech}=-\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV-\frac{\mu }{2}\int {\left(\frac{\partial {v}_i}{\partial {x}_k}+\frac{\partial {v}_k}{\partial {x}_i}-\frac{2}{3}\frac{\partial {v}_i}{\partial {x}_i}\right)}^2 dV-\zeta \int {\left(\nabla \cdotp \overrightarrow{v}\right)}^2 dV$$
(14.13)
Since we are still attempting a solution in the linear limit, the lowest-order contribution to the power dissipation must be second order in the wave’s displacement from equilibrium; in this case, $${T}_1^2\kern0.5em \mathrm{and}\kern0.5em {\left|{\overrightarrow{v}}_1\right|}^2$$, hence it is positive definite (see Sect. 10.5). For that reason, the absolute temperature, T, can be taken outside the integral and represented by Tm, since allowing for acoustical variation of that temperature term would add a correction to the thermal conduction loss that is third order in displacements from equilibrium.
For a plane wave propagating in the x direction, it is convenient to express vx = v1 sin (ω t − kx), setting vy = vz = 0. Substitution into the last two terms of Eq. (14.13) produces the (nonthermal) mechanical dissipation.
$$-\left(\frac{4}{3}\mu +\zeta \right)\int {\left(\frac{\partial {v}_1}{\partial x}\right)}^2 dV=-{k}^2\left(\frac{4}{3}\mu +\zeta \right){v}_1^2\int {\cos}^2\left(\omega t- kx\right)\; dV$$
(14.14)
Since we are only interested in the time-averaged power dissipation, the contributions from the nonthermal terms in Eq. (14.13) is$$-\left({k}^2/2\right)\left[\left(4\mu /3\right)+\zeta \right]{v}_1^2\kern0.1em {V}_o$$, where Vo is the volume of the fluid under consideration through which the plane wave is propagating.
It is worth comparing the appearance of the factor, 4/3, that multiplies the shear viscosity, μ, with the corresponding expression for the modulus of unilateral compression, D (aka the dilatational modulus), introduced in Sect. 4.2.2, to the shear modulus, G, and bulk modulus, B, in Table 4.1: D = (4G/3) + B. Again, this is a direct consequence of the fact that the distortion produced by a plane wave can be decomposed into two shears (related to G) and a hydrostatic compression (related to B).
The result in Eq. (14.14), without ζ, was first produced by Stokes who expressed the result as the temporal attenuation coefficient [4]. The lack of agreement between his theoretical predictions and experimental measurements provided the starting point for the modern attempts to account for attenuation in terms of molecular relaxation [5]. The spatial attenuation coefficient, due to viscous dissipation, was first introduced by Stefan in 1866Footnote 4 [6]. The first calculation to include both the effects of thermal conductivity and shear viscosity on the absorption of sound was published by Kirchhoff in 1868 [7].
To evaluate the contribution of thermal conduction to the mechanical dissipation in Eq. (14.13), the temperature change needs to be related to the pressure change to evaluate the one-dimensional temperature gradient, (∂T/∂x). For an ideal gas, this relation should be familiar, having been derived in Eqs. (1.21) and (7.25).
$${\left(\frac{\partial T}{\partial p}\right)}_s=\left(\frac{\gamma -1}{\gamma}\right)\frac{T_m}{p_m}$$
(14.15)
Since we seek an attenuation coefficient that would be applicable to all fluids, a more general expression for (∂T/∂p)s needs to be calculated to evaluate (∂T/∂x).
$${\left(\frac{\partial T}{\partial x}\right)}_s={\left(\frac{\partial T}{\partial p}\right)}_s{\left(\frac{\partial p}{\partial v}\right)}_s{\left(\frac{\partial v}{\partial x}\right)}_s=-{\rho}_mc{\left(\frac{\partial T}{\partial p}\right)}_s{kv}_1\cos \left(\omega\;t- kx\right)$$
(14.16)
The derivative of pressure with respect to velocity for a nearly adiabatic plane wave is a direct consequence of the Euler equation: p1 = ρmcv1.
The derivative of temperature with respect to density can be evaluated using the enthalpy function, H(S, p) = U + pV, that sums the internal energy, U, introduced in Sect. 7.1.2 to calculate heat capacities, with the mechanical work, W = pV. From Eqs. (7.8, 7.9, and 7.10), the internal energy, U(S,V), can be transformed into the enthalpy, H(S, p), using the product rule for differentiation. In thermodynamics, this operation is known as a Legendre transformation [8].
$${\displaystyle \begin{array}{c} dU= TdS- pdV= TdS-d(pV)+ Vdp\\ {}\Rightarrow d\left(U+ pV\right)\equiv dH= TdS+ Vdp\end{array}}$$
(14.17)
The change in enthalpy, dH(S, p), can be expanded in a Taylor series, retaining only the linear terms.
$$dH={\left(\frac{\partial H}{\partial S}\right)}_p dS+{\left(\frac{\partial H}{\partial p}\right)}_S dp$$
(14.18)
Comparison of Eqs. (14.17) and (14.18) can be used to evaluate those derivatives.
$${\left(\frac{\partial H}{\partial S}\right)}_p=T\kern1em \mathrm{and}\kern1em {\left(\frac{\partial H}{\partial p}\right)}_S=V$$
(14.19)
Since the order of differentiation is irrelevant, the mixed partial derivatives must be equal.
$$\frac{\partial^2H}{\partial p\partial S}=\frac{\partial^2H}{\partial S\partial p}\kern1em \Rightarrow \kern1em {\left(\frac{\partial T}{\partial p}\right)}_S={\left(\frac{\partial V}{\partial S}\right)}_p$$
(14.20)
This result is one of several thermodynamic identities known as the Maxwell relations [9].
$${\left(\frac{\partial V}{\partial S}\right)}_p={\left(\frac{\partial V}{\partial T}\right)}_p{\left(\frac{\partial T}{\partial S}\right)}_p$$
(14.21)
The result in Eq. (14.21) can be expressed in terms of tabulated material properties [10] using the definition of the (extensive) heat capacity at constant pressure, Cp, or the (intensive) specific heat (per unit mass) at constant pressure, cp, from Eq. (7.14), and the definition of the isobaric (constant pressure) volume coefficient of thermal expansion, βp.
$${C}_p=T{\left(\frac{\partial S}{\partial T}\right)}_p\kern1em \mathrm{or}\kern1em {c}_p=\kern0.5em T{\left(\frac{\partial s}{\partial T}\right)}_p\kern1em \mathrm{and}\kern1em {\beta}_p=\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_p=-\frac{1}{\rho_m}{\left(\frac{\partial \rho }{\partial T}\right)}_p$$
(14.22)
These results can be combined to produce an expression for the temperature gradient required to evaluate the thermal conduction integral in Eq. (14.13) using Eq. (14.16).
$$\frac{\partial T}{\partial x}=c\frac{\beta_p{T}_m}{c_p}\frac{\partial {v}_1}{\partial x}=-c\frac{\beta_p{T}_m}{c_p}{v}_1k\cos \left(\omega\;t- kx\right)$$
(14.23)
As before, our interest will be in the time-averaged value for evaluation of the thermal conduction term in the integral expression for Πmech in Eq. (14.12).
$${\left\langle -\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV\right\rangle}_t=\frac{-\kappa {c}^2{\beta}_p^2{v}_1^2{k}^2}{2{c}_p^2}{V}_o$$
(14.24)
This result can be evaluated in terms of the difference in the specific heats that was shown by thermodynamic arguments to be CPCV = ℜ, in Eq. (7.14), for an ideal gas, or cpcv = ℜ/M, where ℜ is the universal gas constant and M is the mean molecular or atomic mass of the ideal gas or ideal gas mixture. By the same thermodynamic arguments, the general result for the specific heat difference can be expressed in terms of the fluid parameters in Eq. (14.22) [11].
$${c}_p-{c}_v=T{\beta}_p^2{\left(\frac{\partial p}{\partial \rho}\right)}_T=T{\beta}_p^2\left(\frac{c_v}{c_p}\right){\left(\frac{\partial p}{\partial \rho}\right)}_s=T{\beta}_p^2{c}^2\left(\frac{c_v}{c_p}\right)$$
(14.25)
The relationship between the square of the isothermal sound speed, (∂p/∂ρ)T, and the square of the adiabatic sound speed, (∂p/∂ρ)s = c2, should be familiar since (cv/cp) ≡ γ−1.
Substitution of Eq. (14.25) into Eq. (14.24) provides a compact expression for the time-averaged power dissipation that is valid for ideal gases as well as for all other homogeneous fluids.
$${\left\langle -\frac{\kappa }{T_m}\int {\left(\nabla T\right)}^2 dV\right\rangle}_t=-\left(\frac{1}{2}\right)\kappa {k}^2{v}_1^2{V}_o\left(\frac{1}{c_v}-\frac{1}{c_p}\right)$$
(14.26)
Combining Eq. (14.26) with Eq. (14.14) provides an expression for the time-averaged mechanical power dissipation due to all of the irreversible dissipation mechanisms.
$${\left\langle {\Pi}_{mech}\right\rangle}_t=-\left(\frac{1}{2}\right){k}^2{v}_1^2{V}_o\left[\left(\frac{4}{3}\mu +\zeta \right)+\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]$$
(14.27)
The total energy, E, of the plane wave occupying the volume, Vo, can be expressed in terms of the maximum kinetic energy density, (KE)max.
$$E={\left(\mathrm{KE}\right)}_{\mathrm{max}}{V}_o=\left(\frac{1}{2}\right){\rho}_m{v}_1^2{V}_o$$
(14.28)
Since the decay rate of the energy is twice that of the amplitude decay rate, the spatial attenuation constant that reflects thermoviscous losses (and whatever ζ represents!), αT-V, can be written in terms of the time-averaged power dissipation, 〈Πmecht, in Eq. (14.27), and the average total energy, E, in Eq. (14.28).
$${\alpha}_{T-V}=\frac{\left|{\left\langle {\Pi}_{mech}\right\rangle}_t\right|}{2 cE}=\frac{\omega^2}{2{\rho}_m{c}^3}\left[\left(\frac{4}{3}\mu +\zeta \right)+\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]$$
(14.29)
This final result for αT-V is valid for all fluids as long as the decrease in the sound wave’s amplitude over the distance of a single wavelength is relatively small, αT −V λ ≪ 1, since the stored energy was calculated for an undamped sound wave.
We also see that this result is similar to the “almost correct” result, αalmost, calculated from the Navier-Stokes equation in Sect. 14.1. The dependence on frequency, ω; mass density, ρm; and sound speed, c, is identical, but the shear viscosity, μ, is no longer the only transport property of the medium that contributes to attenuation of the sound wave; in Eq. (14.29), μ has been replaced by the term within the square bracket.
As demonstrated earlier in Eq. (14.8), the expression for attenuation given in Eq. (14.29) will always be valid for sound in gases, since the kinematic viscosity, νgas = μgas/ρm, is on the order of the product of the mean free path, $$\overline{\mathrm{\ell}}$$, times the mean thermal velocity of the gas molecules or the sound speed.
$$\frac{\nu_{gas}\omega }{c^2}\cong \overline{\mathrm{\ell}}\frac{\omega }{c}\cong \frac{\overline{\mathrm{\ell}}}{\lambda}\ll 1$$
(14.30)
## 3 Classical Thermoviscous Attenuation
Before the role of molecular relaxation was appreciated and the associated dissipative coefficient, ζ, was introduced, attenuation of sound due to thermoviscous losses was calculated by Kirchhoff [7]. That result is often called the classical absorption coefficient, αclassical.
$${\alpha}_{classical}=\frac{\omega^2}{2{\rho}_m{c}^3}\left[\frac{4}{3}\mu +\kappa \left(\frac{1}{c_v}-\frac{1}{c_p}\right)\right]=\frac{\omega^2}{2{c}^3}\left[\frac{4}{3}\frac{\mu }{\rho_m}+\frac{\kappa }{\rho_m{c}_p}\left(\frac{c_p}{c_v}-1\right)\right]$$
(14.31)
For an ideal gas, the classical attenuation coefficient can be expressed more transparently in terms of the kinematic viscosity, ν = μ/ρm; the polytropic coefficient, γ = cp/cv; and the dimensionless ratio of the thermal and viscous diffusion constants, known as the Prandtl number, Pr ≡ (μ/cpκ) = (δν/δκ)2, that was introduced in Sect. 9.5.4.
$${\alpha}_{classical}=\frac{\omega^2\nu }{2{c}^3}\left[\frac{4}{3}+\frac{\left(\gamma -1\right)}{\Pr}\right]\kern1em \Rightarrow \kern1em \frac{\alpha_{classical}}{f^2}=\frac{2{\pi}^2\nu }{c^3}\left[\frac{4}{3}+\frac{\left(\gamma -1\right)}{\Pr}\right]$$
(14.32)
Most single-component gases and many gas mixtures have Pr ≅ 2/3. For air at atmospheric pressure and 20 °C, ν = 1.51 ×10−5 m2/s, Pr = 0.709, γ = 1.402, and c = 343.2 m/s. Under those conditions, αclassical / f 2 = 1.40 × 10−11 s2/m. The accepted value of α/f2 in the high-frequency limit is 1.84 × 10−11 s2/m. This discrepancy is due to the absence of ζ in Eq. (14.32) [12].
Since the difference between the specific heats at constant pressure and constant volume is small for liquids, the viscous contribution to the classical attenuation constant is dominant. For pure water at 280 K, $${\nu}_{H_2O}$$ = 1.44 × 10−6 m2/s and$${\Pr}_{H_2O}$$ = 10.4, [13] with $${c}_{H_2O}$$ = 1500 m/s, making αclassical/f 2 = 1.1 × 10−14 s2/m for freshwater. The accepted value of α/f2 in the high-frequency limit is 2.5 × 10−14 s2/m, again due to the absence of ζ in Eq. (14.32) [14].
## 4 The Time-Dependent Equation of State
The distortion of a fluid element caused by passage of a plane wave was decomposed into shear deformations, which changed the shape of the element, and a hydrostatic deformation, which changed the volume of the element, as diagrammed schematically in Fig. 14.1. Each of those deformations introduced irreversibility that increased entropy as expressed in Eq. (14.12), leading to energy dissipation as expressed in Eq. (14.13). Based on the discussion in Sect. 9.4, the shear viscosity, μ, was introduced to relate the shear deformations to the dissipative shear stresses.
Another coefficient, ζ, was introduced to relate entropy production to the divergence of the fluid’s velocity field, $$\nabla \cdotp \overrightarrow{v}$$. The continuity equation requires that when $$\nabla \cdotp \overrightarrow{v}\ne 0$$, the density of the fluid must also be changing: (∂ρ/∂t) ≠ 0. Why should a change in the fluid’s density be related to irreversible entropy production?
When the phenomenological model was introduced, it was assumed that only five variables were required to completely specify the state of a homogeneous, isotropic, single-component fluid: one mechanical variable (p or ρ) and one thermal variable (s or T), along with the three components of velocity (e.g., vx, vy, and vz). For a static fluid, |v| = 0, only two variables were required, resulting in the laws of equilibrium thermodynamics (i.e., energy conservation and entropy increase) rather than the laws of hydrodynamics (that also incorporate thermodynamics). The evolution of those variables was determined by the imposition of five conservation equations (i.e., mass, entropy, and vector momentum). That assertion included an implicit assumption that an equation of state existed and it could be used to relate the thermodynamic variables (mechanical and thermal) to each other instantaneously.
For some fluids, the assumption of an instantaneous response of the density to changes in the pressure is not valid. (Noble gases are one notable exception, since they do not have any rotational degrees of freedom.) The microscopic models of gases that were based on the kinetic theory introduced the concept of collision times between the constituent particles (atoms and/or molecules) and the Equipartition Theorem in Eq. (7.2) that stated that through these collisions, an equilibrium could be established that distributed the total thermal energy of the system equitably (on average) among all of the available degrees of freedom. What has been neglected (to this point) was the fact that the collisions take a non-zero time to establish this equilibrium; if the conditions of the fluid element are changing during this time, the system might never reach equilibrium.
How did we get away with this “five-variable fraud” for so long? One answer is hidden in the transition from adiabatic sound speed in an ideal gas to the isothermal sound speed. Equation (9.24) defined a critical frequency, ωcrit, at which the speed of thermal diffusion was equal to the speed of sound propagation. At that frequency, the wavelength of sound corresponded to a distance, which was about 20 times the average spacing between particles, known as the mean free path between collisions, $$\overline{\mathrm{\ell}}$$. Since that collision frequency was so much higher than our frequencies of interest, the equilibration between translational and rotational degrees of freedom in gases of polyatomic molecules occurred so quickly that the equation of state appeared to act instantaneously [15]. Even though a vibrating object couples to the translational degrees of freedom in a gas, the translational and rotational degrees of freedom came into equilibrium in much less time than the period of the vibrating object’s oscillations. That “fraud” was obscured by our use of γ = 7/5 in the expression for sound speed which treated the air as instantly sharing energy between the internal translational and rotational degrees of freedom.
When there are other components in a gas or liquid, they may have relaxation times that are sufficiently close to the acoustic periods of interest that their “equilibration” to the acoustically induced changes cannot be considered to occur instantaneously. In air, for example, if there is water vapor present, it will equilibrate with the O2 and N2 over times that are comparable to the periods of sound waves of interest for human perception (i.e., 20 Hz ≲ f ≲ 20 kHz). Figure 14.2 shows the relaxation frequencies as a function of the mole fraction, h, of H2O and also relative humidity as a percentage [16].
Those equilibration times are dependent upon the gas mixture’s temperature, pressure, and mixture concentration (i.e., mole fraction of water vapor, h, or relative humidity, RH) [17]. It is apparent that the relaxation frequencies in Fig. 14.2 are in the audio range for ordinary values of temperature and humidity [17].
$${\displaystyle \begin{array}{c}{f}_{rO}=\frac{p_m}{p_{ref}}\left\{24+\left[\frac{\left(4.04x{10}^4h\right)\left(0.02+h\right)}{0.391+h}\right]\right\}\\ {}{f}_{rN}=\frac{p_m}{p_{ref}}{\left(\frac{T}{T_{ref}}\right)}^{-\frac{1}{2}}\left(9+280h\kern0.5em \exp \left\{-4.170\left[{\left(\frac{T}{T_{ref}}\right)}^{-1/3}-1\right]\right\}\right)\end{array}}$$
(14.33)
The relaxation frequencies for water vapor and nitrogen, frN, and for water vapor and oxygen, frO, assume a standard atmospheric composition with 78.1% nitrogen, 20.9% oxygen, and 314 ppm carbon dioxide at a reference pressure, pref = 101,325 Pa, and reference temperature, Tref = 293.15 K = 20.0 °C. In Eq. (14.33), the molar concentration of water vapor, h, is expressed in percent. For ordinary atmospheric conditions near sea level, 0.2% ≲ h ≲ 2.0%.
To relate RH to h (both in %), it is first necessary to calculate the saturated vapor pressure of water in air, psat, relative to ambient pressure, pref = 101.325 kPa, using the triple-point isotherm temperature, T01 = 273.16 K = +0.01 °C.
$$\frac{p_{sat}}{p_{ref}}={10}^C;\kern1em C=-6.8346{\left(\frac{T_{01}}{T}\right)}^{1.261}+4.6151$$
(14.34)
The molar concentration of water vapor, h, in percent, can then be expressed in terms of the relative humidity, RH, also in percent [17].
$$h= RH\left(\frac{p_{sat}}{p_m}\right)$$
(14.35)
A similar effect is observed in seawater where boric acid, B(OH)3, and magnesium sulfate, MgSO4, have relaxation frequencies of 1.18 kHz and 145 kHz, respectively, at 20 °C [18]. In the case of these salts, the relaxation time represents the pressure-dependent association-dissociation reaction between the dissolved salts and their ions.
## 5 Attenuation due to Internal Relaxation Times
“If a system is in stable equilibrium, then any spontaneous change of its parameters must bring about processes which tend to restore the system to equilibrium.” H. L. Le ChâtelierFootnote 5
A new positive scalar coefficient, ζ, was introduced in the entropy conservation Eq. (14.12) to scale the irreversibility of hydrostatic fluid deformations. It has the same units as the shear viscosity [Pa-s]Footnote 6 and is usually of about the same magnitude. If the medium does not possess any additional internal degrees of freedom that have to be brought into equilibrium, then its value can be identically zero. That constant is zero for the noble gases (He, Ne, Ar, Kr, Xe, and Rn) that are intrinsically monatomic with atoms that are spherically symmetrical, thus lacking rotational degrees of freedom (see Sect. 7.2). On the other hand, as suggested in Fig. 14.2, if there are processes with relaxation times that are near the frequencies of interest, the value of ζ can be orders of magnitude greater than μ near those frequencies.
With acoustical compressions or expansions, as in any rapid change of state, the fluid cannot remain in thermodynamic equilibrium. Following Le Châtlier’s Principle,5 the system will attempt to return to a new equilibrium state that is consistent with the new parameter values that moved it away from its previous state of equilibrium. In some cases, this equilibration takes place very quickly so that the medium behaves as though it were in equilibrium at all times. In other cases, the equilibration is slow, and the medium never catches up. In either case, the processes that attempt to reestablish equilibrium are irreversible and therefore create entropy and dissipate energy.
If ξ represents some physical parameter of the fluid and ξo represents the value of ξ at equilibrium, then if the fluid is not in equilibrium, ξ will vary with time. If the fluid is not too far from equilibrium, so the difference, ξ − ξo, is small (i.e., |ξ − ξo|/ξo ≪ 1), and then the rate of change of that parameter, $$\dot{\xi}$$, can be expanded in a Taylor series retaining only the first term and recognizing that any zero-order contribution to $$\dot{\xi}$$ must vanish since ξ = ξo at equilibrium.
$$\dot{\xi}=-\frac{\left(\xi -{\xi}_o\right)}{\tau_R}$$
(14.36)
This suggests an exponential relaxation of the system toward its new equilibrium state. Le Châtelier’s Principle requires that the rate must be negative and that the relaxation time, τR, must be positive.
For acoustically induced sinusoidal variations in the parameter, ξ, at frequency, ω, the sound speed will depend upon the relative values of the period of the sound, T = 2π/ω, and the relaxation time, τR. If the period of the disturbance is long compared to the exponential equilibration time, τR, so that ωτR = 2πτR/T ≪ 1, then the fluid will remain nearly in equilibrium at all times during the acoustic disturbance. In that limit, the sound speed will be the equilibrium sound speed, co. In the opposite limit, ωτR = 2πτR/T ≫ 1, the medium’s sound speed, c > co, will be determined by the fluid’s elastic response if the internal degrees of freedom cannot be excited by the disturbance. Said another way, the internal degrees of freedom are “frozen out” in that limit; they simply do not have enough time to participate before the state of the system has changed.
One way to think about this effect is to consider the sound speed in a gas of diatomic molecules that possess three translational degrees of freedom and two rotational degrees of freedom. The specific heats of monatomic and polyatomic gases were discussed in Sect. 7.2, and the relationship between the sound speed in such gases and the specific heat ratio, γ = cp/cv, is provided in Eq. (10.22). If the rotational degrees of freedom are not excited, then the gas behaves as though it were monatomic, so γ = 5/3. If the rotational and translational degrees of freedom are always in equilibrium, then γ = 7/5 < 5/3, so $${c}_{\infty }=\sqrt{25/21}\kern0.5em {c}_o$$.
The mathematical “machinery” needed to represent the attenuation and dispersion of sound waves in a homogeneous medium with an internal degree of freedom, or a “relaxing sixth variable,” has already been developed to describe viscoelastic solids in Sect. 4.4.2. Figure 4.25 could just as well describe the propagation speed (solid line) as a function of the nondimensional frequency, ωτR, with co being the limiting sound speed for ωτR = 2πτR/T ≪ 1 and c being the sound speed for ωτR = 2πτR/T ≫ 1. In addition, the Kramers-Kronig relations of Sect. 4.4.4 would still apply; the variation in sound speed with frequency requires a frequency-dependent attenuation, shown in Fig. 4.25 as the dashed line, and vice versa.
The transformation of the results derived for the stiffness and damping of a viscoelastic medium simply requires that the sound speed is proportional to the square root of the elastic modulus (i.e., stiffness) as expressed in Eq. (10.21). That substitution allows Eq. (4.67) to produce the propagation speed as a function of the nondimensional frequency, ωτR.
$${c}^2={c}_o^2+\left({c}_{\infty}^2-{c}_o^2\right)\frac{{\left({\omega \tau}_R\right)}^2}{1+{\left({\omega \tau}_R\right)}^2}$$
(14.37)
The same approach applied to Eq. (4.70) provides the attenuation per wavelength, αλ, as function of the dimensionless frequency, ωτR.
$$\left(\alpha \lambda \right)=2\pi \frac{\left({c}_{\infty}^2-{c}_o^2\right)\left({\omega \tau}_R\right)}{c_o^2\left[1+{\left({\omega \tau}_R\right)}^2\right]+\left({c}_{\infty}^2-{c}_o^2\right){\left({\omega \tau}_R\right)}^2}$$
(14.38)
Following Eq. (4.71) or Eq. (4.89), the maximum value of attenuation per wavelength in Eq. (14.38) will occur at a unique value of the nondimensional frequency, (ωτR)max.
$${\left({\omega \tau}_R\right)}_{\mathrm{max}}=\frac{c_o}{c_{\infty }}\kern1em \mathrm{and}\kern1em {\left(\alpha \lambda \right)}_{\mathrm{max}}=\pi \frac{\left({c}_{\infty}^2-{c}_o^2\right)}{c_{\infty }{c}_o}$$
(14.39)
The consequence of the Kramers-Kronig relations for such single relaxation time phenomena, as emphasized in Sect. 4.4.2, is that the attenuation is entirely determined by the dispersion, and vice versa.
Using these results, it is possible to write simple universal expressions for attenuation due to excitation of internal degrees of freedom in terms of the relaxation frequency, fR = (2πτR)−1.
$$\frac{\alpha \lambda}{{\left(\alpha \lambda \right)}_{\mathrm{max}}}=\frac{2}{\frac{f_R}{f}+\frac{f}{f_R}}\kern1em \Rightarrow \kern1em \alpha (f)=\left[\frac{2{\left(\alpha \lambda \right)}_{\mathrm{max}}}{cf_R}\right]\frac{f^2}{1+{\left(\frac{f}{f_R}\right)}^2}$$
(14.40)
The variation in the attenuation per wavelength, αλ, and the propagation speed, c, as a function of nondimensional frequency, ωτR, is plotted in Fig. 14.3 and should be compared to the plot for a viscoelastic solid in Fig. 4.25, which exhibits identical behavior. For relaxation frequencies, fR, that are much higher than the frequency of interest, f, the attenuation constant’s quadratic frequency dependence is recovered, as was derived in Eq. (14.7) for αalmost and in Eq. (14.29) for αT-V.
### 5.1 Relaxation Attenuation in Gases and Gas Mixtures
The first example of the effects of the relaxation of an internal degree of freedom on sound speed and attenuation in gas is taken from the measurements of Shields in fluorine [19]. Halogen vapors (e.g., chlorine, fluorine, bromine, iodine) are unique in that they consist of homonuclear diatomic molecules that have appreciable vibrational energy, EV = (n + 1/2) ℏωV, at room temperature (see Sect. 7.2.2). Because the diatomic molecules behave as simple (quantum mechanical) harmonic oscillators, that internal degree of freedom can be characterized by a single relaxation time corresponding to the radian period of their harmonic oscillations.
Figure 14.4 shows the measured values of attenuation per wavelength, αλ, and sound speed, c, as a function of frequency (also normalized by pressure), in units of kHz/atm., for fluorine gas at 102 ± 2 °C, after correction for boundary layer losses at the surface of the tube containing the gas [20].
The vibrational relaxation time for diatomic fluorine at 102 °C is τR = 10.7 μs, corresponding to a relaxation frequency, fR = (2πτR)−1 = 14.9 kHz. Based on the fit to the sound speed, the limiting speeds are co = 332 m/s and c = 339 m/s. The peak in the attenuation per wavelength, (αλ)max = 0.13, occurs at (ωτR)max = 0.98, based on Eq. (14.39), in excellent agreement with the data in Fig. 14.4.
The relaxation attenuation in humid air is more complicated since the two relaxation frequencies for equilibration of the water vapor with the nitrogen and with the oxygen are different, as expressed in Eq. (14.33) and plotted in Fig. 14.2. Since energy loss is cumulative, it is possible to express the attenuation constant, αtot, as the sum of the attenuation caused by the classical value, αclassical, and the contributions from the two relaxation processes.
$${\alpha}_{tot}={\alpha}_{classical}+{\alpha}_{O_2}+{\alpha}_{N_2}$$
(14.41)
The pressure, frequency, and temperature dependence for the total attenuation coefficient is provided in combination with the relaxation times of Eq. (14.33) and plotted as a function of the frequency/pressure ratio for various values of the relative humidity in Fig. 14.5 [21].
$${\displaystyle \begin{array}{l}\frac{\alpha_{Air}}{f^2}=1.84\times {10}^{-11}{\left(\frac{p_m}{p_{ref}}\right)}^{-1}{\left(\frac{T}{T_{ref}}\right)}^{\frac{1}{2}}+{\left(\frac{T}{T_{ref}}\right)}^{-5/2}\\ {}\times \left\{0.01275\kern0.5em {e}^{-2,239/T}\left[\frac{f_{rO}}{f_{rO}^2+{f}^2}\right]+0.1068{e}^{-3,352/T}\left[\frac{f_{rN}}{f_{rN}^2+{f}^2}\right]\right\}\end{array}}$$
(14.42)
The difference between the classical attenuation constant and the total is clearly very large. At 2 kHz and 1 atm., αclassical = 0.02 dB/m, but with 10% relative humidity, the attenuation is αtot = 0.80 dB/m.
Values for the attenuation in dB/km are tabulated in a standard for different values of temperature from −25 °C to +50 °C and 10% ≤ RH ≤ 100% for pure tones with frequencies from 50 Hz to 10 kHz in $$\raisebox{1ex}{1}\!\left/ \!\raisebox{-1ex}{3}\right.$$ -octave increments [17]. A small subset of that data are presented in Table 14.1.
### 5.2 Relaxation Attenuation in Fresh and Salt Water
As earlier noted in Sect. 14.3, the measured attenuation of sound in water is greater than αclassical /f2 based on the shear viscosity by more than a factor of two. In the calculation of αclassical for water, the thermal conductivity was neglected. It can be shown that neglect of the thermal conductivity is not the cause of this discrepancy. Measurements of attenuation at 4 °C, where water has its density maximum and the thermal expansion coefficient vanishes, mean that cp = cv, so according to Eq. (14.25), there are no temperature changes associated with the acoustical pressure changes [22].
The excess attenuation has been ascribed to a structural relaxation process wherein a molecular rearrangement is caused by the acoustically produced pressure changes. During acoustic compression, the water molecules are brought closer together and are rearranged by being repacked more closely. This repacking takes a non-zero amount of time and leads to relaxational attenuation that makes ζ ≠ 0 [23]. The relaxation time as a function of water temperature for this process, τR, is on the order of picoseconds and is plotted as a function of temperature in Fig. 14.6.
At 4 °C, τR ≅ 3.5 ps, corresponding to a relaxation frequency, fR = (2πτR)−1 = 45 GHz, well above experimentally accessible frequencies. For that reason, there are no “relaxation bumps” in the attenuation vs. frequency, as seen in Fig. 14.7. Nonetheless, this structural relaxation makes ζ > μ, accounting for the excess attenuation in pure water.
The attenuation of sound in seawater is similar to that in humid air where the relaxation frequencies are within a frequency range of interest. In seawater, there are two pressure-dependent ionic association-dissociation reactions due to the dissolved boric acid, B(OH)3, and the dissolved magnesium sulfate, MgSO4. Their contributions to the attenuation have the generic form introduced in Eq. (14.40).
$${\displaystyle \begin{array}{c}{\mathrm{Mg}\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}\leftrightarrow {\mathrm{Mg}}^{+3}+{\mathrm{SO}}_4^{-2}+{\mathrm{H}}_2\mathrm{O};\kern0.5em {\alpha}_{{\mathrm{Mg}\mathrm{SO}}_4}\cong \frac{4.6\times {10}^{-3}{\left(\mathrm{kHz}\right)}^2}{4100+{\left(\mathrm{kHz}\right)}^2}\\ {}\mathrm{B}{\left(\mathrm{OH}\right)}_3+{\left(\mathrm{OH}\right)}^{-1}\leftrightarrow \mathrm{B}{\left(\mathrm{OH}\right)}_4^{-1};\kern0.5em {\alpha}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}\cong \frac{1.2\times {10}^{-5}{\left(\mathrm{kHz}\right)}^2}{1+{\left(\mathrm{kHz}\right)}^2}\end{array}}$$
(14.43)
The approximate attenuation values, $${\alpha}_{{\mathrm{MgSO}}_4}\kern0.5em \mathrm{and}\kern0.5em {\alpha}_{B{\left(\mathrm{OH}\right)}_3}$$, in Eq. (14.43) are in units of [m−1] when the frequency is expressed in kHz.
Those reactions that have relaxation frequencies that depend upon absolute temperature, T, and salinity, S, that is expressed in parts per thousand, ‰, and those relaxation frequencies, $${f}_{{\mathrm{rBH}}_3{\mathrm{O}}_3}\kern0.5em \mathrm{and}\kern0.5em {f}_{{\mathrm{rMgSO}}_4}$$, are in hertz [24, 25].
$${\displaystyle \begin{array}{c}{f}_{{\mathrm{rBH}}_3{\mathrm{O}}_3}=2,800\sqrt{\mathrm{S}/35}\kern0.5em \times {10}^{\left[4-\left(1,245/T\right)\right]}\\ {}{f}_{{\mathrm{rMfSO}}_4}=\frac{8,170\times {10}^{\left[8-\left(1,990/T\right)\right]}}{1+0.008\left(\mathrm{S}-35\right)}\end{array}}$$
(14.44)
For salinity, S = 35‰, and T = 293 K, $${f}_{\mathrm{rB}{\left(\mathrm{OH}\right)}_3}$$ = 1.58 kHz and$${f}_{{\mathrm{rMgSO}}_4}$$ = 132 kHz.
The attenuation coefficient has the expected form, based on Eq. (14.40), and is plotted as a function of frequency in Fig. 14.7.
$$\frac{\alpha }{f^2}=\frac{A_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}{f}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}}{f^2+{f}_{\mathrm{B}{\left(\mathrm{OH}\right)}_3}^2}+\frac{P_{{\mathrm{MgSO}}_4}{A}_{{\mathrm{MgSO}}_4}{f}_{{\mathrm{MgSO}}_4}}{f^2+{f}_{{\mathrm{MgSO}}_4}^2}+{A}_o{P}_o$$
(14.45)
Approximate expressions for the coefficients representing the relaxation strengths, A, and pressure correction factors, P, are provided by Fisher and Simmons [18] with more accurate values provided by François and Garrison [24, 25].
## 6 Transmission Loss
The fact that the bulk attenuation of sound in fluids is quadratic in the frequency has important consequences for ultrasonics (f > 20 kHz) and for long-range sound propagation. At the extremely low frequencies, infrasound in the Earth’s atmosphere can propagate around the entire globe, and the sound of breaking waves generated by a storm on the Pacific coast of the United States has been detected by a low-frequency microphone at the Bureau of Standards in Washington, DC. An International Monitoring System with 60 infrasound monitoring stations has been deployed globally to detect violations of the Comprehensive Nuclear-Test-Ban Treaty [27].
### 6.1 Short and Very Short Wavelengths
As discussed in Sect. 12.8.1, the Rayleigh resolution criterion implies that the smallest feature that can be resolved in an ultrasonic image will be limited by the wavelength of the sound used to produce the image. Since many ultrasonic imaging systems are used in biomedical applications, we can assume a speed of sound in biological tissue that is approximately equal to the speed of sound in water, $${c}_{{\mathrm{H}}_2\mathrm{O}}$$ = 1500 m/s [28]. To resolve an object that is about a millimeter would then require sound at a frequency, f = c/λ ≅ 1.5 MHz. At that frequency, the attenuation of sound in liver tissue is over 2 dB/cm, so a roundtrip transmission loss to go to a depth of 10 cm is 40 dB.
Because the speed of sound in liquids is typically 200,000 times slower than the speed of light, it is possible to achieve optical wavelength resolution of about 5000 Å = 0.5 μm using sound at a frequency of 3 GHz. In addition, acoustical microscopy produces image “contrast” due to variation in the acoustic absorption of the specimens and scattering that arises from the acoustic impedance mismatch between the specimen and the surrounding material due density and compressibility differences (see Sects. 12.6.1 and 12.6.2). Such sources of contrast will reveal completely different information about a specimen than can be deduced due to changes in optical index of refraction or optical reflectance. In addition, sound can penetrate an optically opaque object, and staining is not required for contrast enhancement.Footnote 7
As mentioned, acoustic microscopy is limited by the fact that attenuation is such a strong function of frequency. At 1.0 GHz, the attenuation of sound in water is 200 dB/mm [29]. Despite the high attenuation loss, acoustic microscopes can image red blood cells acoustically at 1.1 GHz with a resolution equivalent to an oil-immersion optical microscope at a magnification of 1000 [30]. Subcellular details as small as 0.1–0.2 μm (e.g., nuclei, nucleoli, mitochondria, and actin cables) have been resolved due to the extraordinary contrast that can differentiate various cytoplasmic organelles [31].
The greatest resolution that has been achieved using acoustical microscopy has been accomplished in superfluid helium at temperatures near absolute zero, which has a sound speed, c1 ≅ 240 m/s, a speed that is even lower than the speed of sound in air. Since the dynamics of liquid helium at temperatures below Tλ = 2.17 K are determined by quantum mechanics, the attenuation mechanisms are different than those for classical fluids, which also gives it a much smaller attenuation. At low temperatures, T < 0.5 K, the phonon mean free path is controlled by scattering from “rotons,” which are quantized collective excitation of the superfluid [32]. Sound wavelengths in liquid helium shorter than 2000 Å = 0.2 μm, in a non-imaging experiment, at frequencies of 1.0 GHz, had been studied before 1970 by Imai and Rudnick [33].
### 6.2 Very Long Wavelengths
At the opposite extreme, at much lower frequencies, the absorption can be quite small. Although the worldwide network of infrasound monitoring sites, using electronic pressure sensors and sophisticated signal processing, that is being used to assure compliance with the Comprehensive Nuclear-Test-Ban Treaty has already been mentioned [27], the most famous measurement of long-distance infrasound propagation was made using barometers.
On 27 August 1883, the island of Krakatoa, in Indonesia east of Java, was destroyed by an immense volcanic explosion. The resulting pressure wave was recorded for days afterward at more than 50 weather stations worldwide. Several of those stations recorded as many as seven passages of the wave as it circled the globe:
“The barograph in Glasgow recorded seven passages: at 11 hours, 25 hours, 48 hours, 59 hours, 84 hours, 94 hours, and 121 hours (5 days) after the eruption.” [34]
About 4 h after the explosion, the pressure pulse appeared on a barograph in Calcutta. In 6 h, the pulse reached Tokyo; in 10 h, Vienna; and in 15 h, New York. The period of the pulse was between 100 and 200 s corresponding to a fundamental frequency of about 7 mHz. Its propagation velocity was between 300 and 325 m/s [35]. Although that is close to the speed of sound in air near room temperature, the wave was similar to a shallow water gravity wave in which the height of the atmosphere rose and fell with the passage of the wave [36].
## 7 Quantum Mechanical Manifestations in Classical Mechanics
“The major role of microscopic theory is to derive phenomenological theory.” G. E. UhlenbeckFootnote 8
Although acoustics is justifiably identified as a field of classical phenomenology, there are many acoustical effects that have their origin in the microscopic theory of atoms and therefore manifest macroscopic behaviors that can only be explained in terms of quantum mechanics. The effects of these “hidden variables” have been manifest throughout this textbook starting with the damping of simple harmonic oscillators that connects the “system” to the environment, thus producing Brownian motion [37], which was related to the more general theory coupling fluctuations and dissipation [38] in Chap. 2.Footnote 9
This theme recurred in Chap. 7 when the quantization of energy levels for molecular vibration and rotation influenced the specific heat of gases and in Chap. 9 where a simple kinetic theory of gases was used to determine the pressure and temperature variation of viscosity and thermal conductivity. Now we see in this chapter how structural relaxations in water [23], like those in Fig. 5.23 for the four crystalline structures of plutonium [39], scattering of phonons and rotons in superfluids [32], or molecular vibrations in F2 [19] and collision times in gas mixtures [15], or chemical reactions [24, 25], manifest themselves in the attenuation of sound.
These internal relaxation effects have been incorporated into our phenomenological theory through the introduction of an additional dissipative process that has been quantified by the introduction of a frequency-dependent parameter, ζ, that shares the same units with the coefficient of shear viscosity, μ. That coincidence has led to this new parameter being called the coefficient of “bulk viscosity” (or sometimes “second viscosity”), even though it is independent of the shear deformation of the fluid and is not the source of momentum transport.
### Talk Like an Acoustician
Viscoelasticity Mean free path Viscous drag Kinetic theory Thermal relaxation Einstein summation convention Thermoviscous boundary layer Bulk viscosity Spatial attenuation coefficient Second viscosity Temporal attenuation coefficient Le Châtelier’s principle Shear strain Nondimensional frequency Hydrostatic strain Vibrational relaxation time Collision time Structural relaxation Enthalpy function Association-dissociation reactions Legendre transformation Kramers-Kronig relations Maxwell relations Relaxation time
### Exercises
1. 1.
Bulk attenuation and reverberation time. An expression was provided in Eq. (13.29) to incorporate the attenuation in air contained within an enclosure into the expression for reverberation time. A “useful correlation” was provided in Eq. (13.30) that was applicable for 1500 Hz ≤ f ≤ 10,000 Hz and for relative humidity in the range 20% ≤ RH ≤ 70%.
1. (a)
Average frequency dependence. Table 14.1 (left) provides the attenuation in dB/km from the ANSI/ASA standard for the frequencies within the range specified at 20 °C for RH = 50% [17]. Plot the log10 of the spatial attenuation, α, in m−1 vs. the log10 of frequency, f, in kHz, to determine the power law dependence on frequency (see Sect. 1.9.3). Is your result proportional to f1.7 to within the statistical uncertainty of your least-squares fit? Keep in mind that attenuation expressed in [dB/m] must be multiplied by 0.1151 ≅ [10log10(e2)]−1 to convert to m−1 (sometimes including the dimensionless “Nepers” to report results in Nepers/m).
2. (b)
Humidity dependence. Table 14.1 (right) also includes the attenuation in dB/km at 4.0 kHz and 20 °C for 20% ≤ RH ≤ 70%. The “useful correlation” claims that the correction to frequency dependence for variations in relative humidity should be linear in (50%/RH). How close is that presumed humidity dependence to values in the table for variation in relative humidity at 4.0 kHz?
2. 2.
The mother of all PA systems. Shown in Fig. 14.8 is a loudspeaker that can produce 30,000 watts of acoustic power by modulating a pressurized air stream (like a siren) using a cylindrical “valve” mounted on a voice coil, like that used for an electrodynamic loudspeaker. That sound source, located at the apex of the horn, is called a “modulated airstream loudspeaker.” [40]
Such a public address system was developed to tell illiterate enemy combatants, in their native language, to put down their weapons and surrender from a distance that is greater than the distance that could be traversed by artillery shells. Although the bandwidth of telephone speech for very good intelligibility is generally 300 Hz to 3.4 kHz, for this problem, we will focus on the propagation of a 1 kHz pure tone. Since this system was deployed in desert terrain, assume that RH = 10%, Tm = +50 °C = 122 °F, and pm = 100 kPa.
1. (a)
Sound level at 100 m. Assuming hemispherical spreading and no sound absorption, what is the root-mean-square acoustic pressure amplitude if the source produces 30 kW of acoustic power and any nonlinear effects that might cause harmonic distortion (see Sect. 15.2.3) can be neglected? Also neglect any refractive effects due to sound speed gradients caused by temperature or wind as discussed in Sect. 11.3. Report your results as both in r.m.s. pressure amplitude and in dB re: 20 μParms.
2. (b)
Greater distances. Repeat part a, but determine the sound pressure at 1.0 km and 3.0 km, again neglecting attenuation.
3. (c)
Include attenuation. Determine the spatial attenuation coefficient under these conditions at 1.0 kHz and use it to reduce the sound pressure at 0.1 km, 1.0 km, and 3.0 km below that obtained due only to hemispherical spreading.
4. 3.
Pump wave attenuation for a parametric array. The generation of highly directional sound beams from the nonlinear acoustical interaction of two colinear high-frequency sound beams will be discussed in Sect. 15.3.3. Calculate the exponential attenuation length, = α−1, of a typical 40 kHz beam in dry air, RH = 0%, and moist air with RH = 60% using the graph in Fig. 14.5.
5. 4.
Siren. The siren shown in Fig. 14.9 consumed 2500 ft3/min of air at a pressure of 5 psi above ambient to produce 50 horsepower of acoustic power at 500 Hz, and did so with 72% efficiency [41].
1. (a)
Hydraulic power. How much time-averaged power, 〈Πhyt = (Δp)|U|, is available, in watts and in horsepower, from the specified volume flow rate, U, and the available pressure drop, Δp?
2. (b)
Hemispherical spreading. Assuming the mean temperature during the measurement was Tm = 20 °C and pm = 100 kPa, what would be the root-mean-square pressure a distance of 1000 ft. from the siren if it produced a sound power of 50 horsepower? Report your results both in pressure and dB re: 20 μParms.
3. (c)
Include attenuation. If RH = 50%, how much additional loss, in dB, would be produced by absorption at the same distance?
6. 5.
The SOFAR channel. One method for locating pilots who crash over the ocean is to drop an explosive sound source that is set to detonate at a depth that is equal to the axis of the deep sound channel that is created by a sound speed profile like the one shown in Fig. 11.8. In that figure, the axis of the sound channel is 1112 m below the ocean’s surface. Sound that is trapped in that channel will spread cylindrically, rather than spherically, beyond a transition distance that will be assumed to be much shorter than the distance of interest, so the sound amplitude will decrease in proportion to $$\sqrt{R}$$, where R is the distance between the source and the receiver.
1. (a)
Cylindrical spreading. What would be the loss, in dB, due to cylindrical spreading over 5000 km relative to the level 1 km from the source?
2. (b)
Include attenuation. Using the results for seawater in Fig. 14.7, what would be the spatial attenuation, in dB, that would have to multiply the cylindrical spreading loss over 5.0 km calculated in part (a) for sound with a frequency of 100 Hz? Repeat for the loss due to attenuation in seawater after 5000 km.
7. 6.
High-frequency relaxational attenuation constant for air and water.
1. (a)
Limiting frequency dependence. Using Eq. (14.40), show that for frequencies well above the highest relaxational frequency, fR, the attenuation of sound is independent of frequency.
2. (b)
Relaxational attenuation constant (bulk viscosity) of air. In Eq. (14.42), the measured high-frequency limit of the spatial attenuation constant in air is αair /f2 = 1.84 × 10−11 s2/m. Calculate αclassical/f2 for air at atmospheric pressure and 20 °C, and use both high-frequency results (i.e., with and without relaxation effects) to determine the value for ζair in the high-frequency limit.
3. (c)
Relaxational attenuation constant (bulk viscosity) of pure water. The measured high-frequency limit of the spatial attenuation constant in pure water at 4 °C is $$\underset{f\gg {f}_R}{\lim }{\alpha}_{{\mathrm{H}}_2\mathrm{O}}/{f}^2=$$ 2.5 × 10−14 s2/m. Based on αclassical / f2 for pure water at atmospheric pressure and 4 °C, determine the value for $${\zeta}_{H_2O}$$ in the high-frequency limit. | 17,091 | 62,794 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.923886 |
http://www.life123.com/technology/online-games/farkle/guide-to-farkle-rules.shtml | 1,432,727,467,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928965.67/warc/CC-MAIN-20150521113208-00218-ip-10-180-206-219.ec2.internal.warc.gz | 558,432,888 | 14,792 | # Guide to Farkle Rules
With this guide to Farkle rules, you can jump right into this digital dice game. While Farkle has been around for a long time, a recent surge in popularity of dice games online has rejuvenated its reputation. Because the rules of Farkle are anything but intuitive, you may have been intimidated to even try and play. However, a review of the rules can teach any potential players what they need to know.
Players And Materials
Before learning the rules of Farkle, it's worth understanding what's usually needed to play in a real-world Farkle game. Then, you can apply the rules to the virtual world. To play Farkle, you will need six dice and at least two people. Some players enjoy using a cup for shaking the dice in real-world Farkle. A pencil or pen and some paper are necessary for keeping score, but most online Farkle games will do it for you.
Object Of The Game
The object of Farkle is to be the first player to score 10,000 points. Certain numbers and combinations of the rolled dice have a specific value, and the roller with the best luck wins by accumulating the most points.
Procedures And Rules
Each player rolls the dice in turn. In order to get on the scoreboard, the player must roll at least 1,000. On a roll, a 1 is worth 100 points, and a 5 is worth 50 points. Two 1s equals 200, and three 1s equals 500. Two 5s equals 100 points, and three 5s equals 500 points. A straight is worth 1000, and that means the dice combine for an order of 1-2-3-4-5-6. Any set of 3 equals the number multiplied by 100. That is, three 3s equals 300, and three 5s equals 500. After you score, you can remove those dice and bank them for points, or roll the remaining dice again. The risk of rolling again is that you may roll numbers or combinations with no value. This is called a Farkle. If you roll a Farkle, you lose your points.
Scoring Variations And Options
Some players choose to add their own scoring quirks, either by changing the value of combinations or the manner in which a player can reach 10,000 points. There may be more flexible interpretations of the rules online, as there are with many Web games. Being imaginative with scoring can be fun, and also create a more customized Farkle experience.
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What is Farkle? This dice game has been popular for a long time and has recently been embraced by the social networking crowd.
The Farkle dice game is easy to learn, but, despite its simplicity, it is still an addictive game of chance that is perfect for Facebook users. | 613 | 2,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2015-22 | latest | en | 0.956373 |
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# OG SC #107
Author Message
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Manager
Joined: 28 Jan 2005
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23 Feb 2005, 03:26
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Hello again there,
OG SC # 107
Downzoning, zoning that typically results in the reduction of housing density, allows for more open space in areas where little water or services exist.
A: little water or services exist
B: little water or services exists
C: few services and little water exists
D: there is little water or services available
E: there are few services and little available water
My question)
In regards to E, correct choice, I think the correct should be “there are few services and little water availableâ€
_________________
Best regards,
If you have any questions
New!
Senior Manager
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23 Feb 2005, 07:37
E has to be chosen because the rest are grammatically wrong.
A - cannot say little services
B - 'exists' is wrong. It should be 'exist'. Also, the violation in A applies here.
C - wrong. It jas to be 'exist' because A and B refer to things.
D - The violation in A applies here.
So, it has to be E.
[quote="Taku"]Hello again there,
OG SC # 107
Downzoning, zoning that typically results in the reduction of housing density, allows for more open space in areas where little water or services exist.
A: little water or services exist
B: little water or services exists
C: few services and little water exists
D: there is little water or services available
E: there are few services and little available water
My question)
In regards to E, correct choice, I think the correct should be “there are few services and little water availableâ€
_________________
Awaiting response,
Thnx & Rgds,
Chandra
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23 Feb 2005, 08:16
water is uncountable and service is countable. u use 'little' with uncountable words whereas 'few' with countable words. so A,B and D are eliminated. C changes the meaning of the sentence by placing 'and' instead of 'or'. so, this leaves E as the best choice. i hope its good enuff expl for u.
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02 Apr 2005, 17:15
[quote="Taku"]Hello again there,
OG SC # 107
Downzoning, zoning that typically results in the reduction of housing density, allows for more open space in areas where [u]little water or services exist[/u].
A: little water or services exist
B: little water or services exists
C: few services and little water exists
D: there is little water or services available
E: there are few services and little available water
i did'nt get what u want to ask!!!!
explain again
My question)
In regards to E, correct choice, I think the correct should be “there are few services and little water availableâ€
_________________
i hate when people do'nt post the OA, it leaves in guessing!!!!
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02 Apr 2005, 18:04
faizaniftikhar83 wrote:
C changes the meaning of the sentence by placing 'and' instead of 'or'.
Moreover, in C the verb exists, which is singular, doesnot agree with the plural subjects.
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02 Apr 2005, 20:35
The question here is, what is the difference between
(i) little available water
(ii) little water available
IMO, the first suggests that there is less amount of useful water(water that can be used), whereas the second suggests that there is little water.
Downzoning is basically reducing the density of housing - this will imply that with more houses/property, the quality of water could get worse and hence I think (i) fits the sentence better than (ii). However, as others have reasoned, (E) is the best of the available options (or options available? )
02 Apr 2005, 20:35
Similar topics Replies Last post
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Topics:
SC from OG11 0 14 Nov 2008, 15:13
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Display posts from previous: Sort by | 1,413 | 5,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-04 | latest | en | 0.901889 |
https://www.physicsforums.com/threads/maxima-cas-question-re-forcing-evaluation.390828/ | 1,590,565,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00205.warc.gz | 863,810,936 | 15,311 | # Maxima CAS question re: forcing evaluation
## Main Question or Discussion Point
I'm wondering if anyone knows how, using the Maxima CAS, to force the evaluation of an expression? For example, if a function returns something like the following as a solution:
$$-\frac{{2}^{\frac{6}{4\,log\left( 2\right) -5}+2}-5\,\left( \frac{6}{4\,log\left( 2\right) -5}+2\right) }{log\left( 2\right) \,{2}^{\frac{6}{4\,log\left( 2\right) -5}+2}-5}+\frac{6}{4\,log\left( 2\right) -5}+2$$
it sure would be nice to ask Maxima to evaluate it to a certain number of digits! :uhh:
Hi bitrex,
Have you found the answer to your question ? I am quite interested in too.
Maxima give me the following result and I don't manage Maxima to evaluate where as my TI-89 succeed !
$$$\int_{0}^{0.25}\int_{-\frac{\pi }{32}}^{\frac{\pi }{32}}\frac{\mathrm{cos}\left( \phi-\frac{\pi }{8}\right) }{{\left( {\left( -z-0.05\right) }^{2}-4\,\mathrm{cos}\left( \phi-\frac{\pi }{8}\right) +5\right) }^{\frac{3}{2}}}d\phi\,\left( -z-0.05\right) dz$$$
What is the keyword to force evaluation in Maxima ?
Thanks,
binoyte | 389 | 1,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-24 | longest | en | 0.704385 |
http://www.docstoc.com/docs/76121448/Impact-of-Financial-Ratio-to-Beta-Risk | 1,386,944,795,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164946521/warc/CC-MAIN-20131204134906-00058-ip-10-33-133-15.ec2.internal.warc.gz | 316,861,639 | 33,077 | # Impact of Financial Ratio to Beta Risk
Document Sample
``` Financial Analysis of Projects for Henson Civil
Overview and Explanation
Henson Civil currently relies on Gross Margin as the main indicator of financial performance of projects. The
purpose of this analysis is to demonstrate that Gross Margin does NOT accurately measure and report value.
Therefore, a set of "value-based" metrics should compliment the current accounting metrics that are in place.
The use of value-based metrics will provide a more balanced view of financial performance.
Projects that are substantially complete have been evaluated based on their cash flows. Cash flow is the basis
for financial analysis since unlike Gross Margin, cash flow is real and not estimated or forecasted. The following
three metrics have been applied to ascertain if projects are creating value:
1. Net Cash Flow Ratio
The Net Cash Flow Ratio is the Net Cash Flow after expenses per the S11 Project Report in relation to
the project's total cost (amount paid out). For example, a 15% ratio would imply that for every
\$ 1.00 in costs paid out, we were able to pocket \$ 1.15 or 15% of the cost. The higher the
ratio, the more likely the project is to create value. A ratio below the Cost of Capital for
Henson would imply that the project may have negatively impacted shareholder value. This is
the first step for isolating projects that may not create value for Henson.
2. Rate of Return
Rate of Return is the return a project generates in relation to the cost of the investment Henson
makes in the project. The cost of the investment is considered the Cost of Capital.
3. Net Present Value
Net Present Value is the value of the project derived by discounting the cash inflows and
cash outflows at the Cost of Capital.
Why Evaluate Projects Using Cash Flow and not Gross Margin?
The value that a project creates is a function of the project's cash flow. The "Net" Cash Flow or the
residual amount that we can withdraw from the project after we pay for everything is often called
Free Cash Flow. When assigning a value to a company, business unit, or project, Analyst often
discount the Free Cash Flow to arrive at a Net Present Value to approximate the Fair Market
Value of what is being valued. Therefore, the use of "cash flow" type metrics is a preferred
way of ascertaining value-created by a project as opposed to looking at traditional accounting
metrics such as Gross Margin.
Summarize Results
Correlation Graph - Poor Relationship between Gross Margin and Rate of Return
Correlation Graph - Strong Relationship between Net Cash Flow Ratio and Rate of Return
Preliminary Review by Cash Flow Ratio
Summarize Rate of Return Analysis
Summarize Net Present Value Analysis
Detail Analysis for Project:
M5L88100 - Municipal Center East Ohio
S8K11561 - Sewer Plants - ST/ FM / LLM
S8K11562 - Water Pipelines - ST/ FM / LLM
M7L55220 - Office Parks - I 96 Corridor
S8K66100 - Cost Model for DS / LMT
S4C19901 - Stack & Block Model - PNT
M3L17001 - Central Subway - CA ODT
S4B16771 - Security Retrofit - Port Authority
M3C18901 - Design Specs for National Towers
M4B17701 - Security Upgrade - Chemical Facilities
S5M19910 - Research Lab Facility - US / CA / BZ
M7D19880 - School Upgrades - West NP
S5C81050 - Master Facility Plan - DLL SE Region
M9K10601 - Facility Support Team - ILBBC
S6X12677 - Tunnel Design for DOT
S8L72110 - CM Support for Yukon SC
S5L11850 - Surveys - Railroad West Lines
Summarize Results of Analysis
< Current Metrics >
Cost to Date
In House Project
< - - - - - - - - - - - Value Based Metric - - - - - - - - - - - - > Gross Gross
Project Name Prj No. Cash Flow Ratio Rate of Return Net Present \$ Margin Margin
Ranked Best to Worst Performing by Rate of Return
Tunnel Design for DOT S6X12677 124.76% 130% \$ 103,026 36.74% 36.26%
CM Support for Yukon SC S8L72110 99.02% 106% \$ 145,154 63.08% 62.23%
Design Specs for Nat Towers M3C18901 101.28% 101% \$ 55,299 51.31% 49.03%
Security Retrofit - Port Auth S4B16771 75.59% 80% \$ 427,993 47.61% 28.20%
School Upgrades - West NP M7D19880 77.32% 77% \$ 202,161 43.81% 41.15%
Surveys - R/R West Lines S5L11850 66.09% 68% \$ 1,819,817 41.10% 31.64%
Stack/Block Model - PNT S4C19901 63.25% 63% \$ 38,416 40.00% 38.83%
Cost Model Program - DS/LMT S8K66100 46.52% 63% \$ 53,271 41.10% 38.57%
Mun Center - East Ohio M5L88100 14.21% 57% \$ 422,711 49.62% 11.33%
Master Fac Plan - DLL SE Region S5C81050 34.17% 49% \$ 16,286 45.54% 39.74%
Central Subway - CA ODT M3L17001 23.85% 22% \$ 8,925 46.44% 44.89%
Security Upgrade - Chem Fac M4B17701 14.12% 13% \$ 41,784 43.41% 26.05%
Research Lab Fac - US/CA/BZ S5M19910 9.66% 10% \$ 102 43.97% 4.68%
Sewer Plants - ST/FM/LLM S8K11561 6.18% 9% \$ (2,509) 34.53% 11.00%
Water Pipelines - ST/FM/LLM S8K11562 -3.34% -8% \$ (53,681) 8.90% 6.54%
Office Parks - I96 Corridor M7L55220 -4.25% -41% \$ (34,929) 39.43% 27.26%
Recommendations to Management
Projects that provide marginal returns should not be developed as part of the company's growth strategy unless
there is some specific justification (such as establishing a toehold position in a new market, client is referring
other clients to Henson, client is helping Henson improve its overall processes, etc.) to warrant continued
investment in such projects. Projects that are marginal tend to be projects which have:
1. High Proportion of Direct Costs to Labor Costs. Since Direct Costs, such as Sub Contractors are not
marked up very high compared to Henson Labor, projects with cost structures that are primarily Direct
are very likely to be marginal projects for Henson. All costs must be paid and reported financially and as
such, management cannot simply focus on the "in-house service" portion of a project only, ignoring
the cash outflows that do not comprise in-house section on the S11 Report. Value is driven by ALL
cash outflows and cash inflows associated with a project.
2. Project costs (cash outflows) are too high in relation to project revenue billed (cash inflows). In order
to create positive valuations, Henson needs to consider a Toll Gate whereby Total Budgeted Project
Revenues are at least 35% higher than Total Budgeted Project Costs. This will "protect" projects from
negative valuations and ensure the entire portfolio of projects contributes to positive valuations.
3. Rates of Return and Net Cash Flow Ratios below 10% (Cost of Capital). Henson should consider
reporting the Net Cash Flow Ratio as the "approximate" project rate of return on the S11 Report.
For projects that are substantially complete (good history of both cash inflows and outflows),
there appears to be a high correlation between the Net Cash Flow Ratio and the Project's Rate of
Return. A quick and easy way to approximate a project's rate of return, is to calculate the Net Cash
Flow Ratio per the S11 Project Report.
Basic Recommendation
Going forward, Henson should consider inclusion of "valuation" type metrics for evaluating the financial
performance of projects since it is critical that projects have positive impacts on shareholder value.
The current approach of evaluating performance by looking at Gross Margin is inappropriate
for meeting the financial objective of increasing shareholder wealth. This financial analysis provides
a basic introduction to "value-based" type metrics and the graph below illustrates the "poor"
correlation between Gross Margin and Value Based Metrics.
Correlation Graph #1 - Poor Correlation between Gross Margin and Rate of Return
The following graph summarizes the poor relationship between Gross Margin and Value Based Metrics. Therefore,
Gross Margin should not be relied upon as an indication of project value.
Correlation between GM and Rate of Return
140%
120%
100%
80%
& Return %'s
80%
GM & Return %'s
60% Rate of Return
40% In House GM
Total Prj GM
20%
0%
-20% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-40%
-60%
Number of Projects Analyzed
Correlation Graph # 2
The following graph illustrates the strong relationship between Net Cash Flow Ratio and a Project's Rate of Return. Therefore,
Managers can easily approximate value creation by simply calculating the Net Cash Flow Ratio off the S11. If the Ratio is
above 10%, then the project is most likely increasing shareholder value for Henson.
Strong Relationship between Net Cash Flow Ratio and
Project Rate of Return
140.00%
120.00%
100.00%
Ratio % & Return %
80.00%
60.00%
Cash Flow Ratio
40.00%
Ratio % & Return %
40.00%
Rate of Return
20.00%
0.00%
-20.00% 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
-40.00%
-60.00%
16 Projects Analyzed - Best to Worst
Preliminary Review by Ranking Net Cash Flow Ratio
In an effort to quickly identify projects that could be destroying shareholder value, all projects that were considered substantially
complete or billed out were ranked according to the Ratio of Cash Inflows to Outflows per the S11 Project Report.
Projects that were considered marginal or suspect were selected for further analysis using Rate of Return and Net Present Value.
Net Cash Flow Ratio = Cash Collected in Excess of Cost / Total Cost per the Cash Flow Section on the S11.
Ranked from worst to best
Project Net Cash Total Net Cash
Project Description Number Flow Project Cost Flow Ratio Comments
xxxxxxxxxxxxxxxxx xxxxxxxx (9,046) 38,017 -23.79% Ouch!!!!!
xxxxxxxxxxxxxxxxx xxxxxxxx (29,417) 691,519 -4.25% Ouch!!!!!
xxxxxxxxxxxxxxxxx xxxxxxxx (31,507) 944,581 -3.34% Ouch!!!!!
xxxxxxxxxxxxxxxxx xxxxxxxx 215,235 3,485,161 6.18% :( Not creating value
xxxxxxxxxxxxxxxxx xxxxxxxx 18,501 191,594 9.66% :(
xxxxxxxxxxxxxxxxx xxxxxxxx 215,988 1,530,148 14.12% Marginal
xxxxxxxxxxxxxxxxx xxxxxxxx 1,616,181 11,373,498 14.21% :( Marginal
xxxxxxxxxxxxxxxxx xxxxxxxx 361,175 1,528,886 23.62%
xxxxxxxxxxxxxxxxx xxxxxxxx 19,606 82,205 23.85%
xxxxxxxxxxxxxxxxx xxxxxxxx 162,165 651,149 24.90%
xxxxxxxxxxxxxxxxx xxxxxxxx 93,913 354,193 26.51%
xxxxxxxxxxxxxxxxx xxxxxxxx 114,022 415,302 27.46%
xxxxxxxxxxxxxxxxx xxxxxxxx 791,671 2,423,894 32.66%
xxxxxxxxxxxxxxxxx xxxxxxxx 5,492 16,479 33.33%
xxxxxxxxxxxxxxxxx xxxxxxxx 16,862 49,345 34.17%
xxxxxxxxxxxxxxxxx xxxxxxxx 198,608 576,282 34.46%
xxxxxxxxxxxxxxxxx xxxxxxxx 140,900 386,220 36.48% Project not complete
xxxxxxxxxxxxxxxxx xxxxxxxx 138,612 377,385 36.73%
xxxxxxxxxxxxxxxxx xxxxxxxx 272,858 729,571 37.40%
xxxxxxxxxxxxxxxxx xxxxxxxx 36,496 97,262 37.52%
xxxxxxxxxxxxxxxxx xxxxxxxx 55,008 146,393 37.58%
xxxxxxxxxxxxxxxxx xxxxxxxx 73,195 189,214 38.68%
xxxxxxxxxxxxxxxxx xxxxxxxx 134,748 346,977 38.83%
xxxxxxxxxxxxxxxxx xxxxxxxx 323,794 757,549 42.74% Project not complete
xxxxxxxxxxxxxxxxx xxxxxxxx 570,602 1,320,746 43.20%
xxxxxxxxxxxxxxxxx xxxxxxxx 119,221 273,892 43.53% Project not complete
xxxxxxxxxxxxxxxxx xxxxxxxx 109,525 250,005 43.81%
xxxxxxxxxxxxxxxxx xxxxxxxx 72,442 158,139 45.81%
xxxxxxxxxxxxxxxxx xxxxxxxx 166,130 362,508 45.83%
xxxxxxxxxxxxxxxxx xxxxxxxx 55,872 120,105 46.52%
Project Net Cash Total Net Cash
Project Description Number Flow Project Cost Flow Ratio Comments
xxxxxxxxxxxxxxxxx xxxxxxxx 245,839 522,822 47.02%
xxxxxxxxxxxxxxxxx xxxxxxxx 9,681 20,358 47.55%
xxxxxxxxxxxxxxxxx xxxxxxxx 58,430 122,177 47.82%
xxxxxxxxxxxxxxxxx xxxxxxxx 154,073 320,505 48.07%
xxxxxxxxxxxxxxxxx xxxxxxxx 118,107 227,067 52.01%
xxxxxxxxxxxxxxxxx xxxxxxxx 1,492 2,764 53.98%
xxxxxxxxxxxxxxxxx xxxxxxxx 91,553 163,689 55.93%
xxxxxxxxxxxxxxxxx xxxxxxxx 25,499 45,498 56.04%
xxxxxxxxxxxxxxxxx xxxxxxxx 292,216 517,303 56.49%
xxxxxxxxxxxxxxxxx xxxxxxxx 22,107 38,179 57.90%
xxxxxxxxxxxxxxxxx xxxxxxxx 50,559 84,491 59.84%
xxxxxxxxxxxxxxxxx xxxxxxxx 2,282,078 3,793,402 60.16%
xxxxxxxxxxxxxxxxx xxxxxxxx 114,668 188,979 60.68%
xxxxxxxxxxxxxxxxx xxxxxxxx 260,077 428,329 60.72%
xxxxxxxxxxxxxxxxx xxxxxxxx 130,290 210,642 61.85%
xxxxxxxxxxxxxxxxx xxxxxxxx 54,072 85,494 63.25%
xxxxxxxxxxxxxxxxx xxxxxxxx 142,112 215,287 66.01%
xxxxxxxxxxxxxxxxx xxxxxxxx 2,495,106 3,775,128 66.09%
xxxxxxxxxxxxxxxxx xxxxxxxx 383,502 566,132 67.74%
xxxxxxxxxxxxxxxxx xxxxxxxx 37,226 54,595 68.19%
xxxxxxxxxxxxxxxxx xxxxxxxx 31,767 44,380 71.58%
xxxxxxxxxxxxxxxxx xxxxxxxx 39,636 54,508 72.72%
xxxxxxxxxxxxxxxxx xxxxxxxx 548,158 725,193 75.59%
xxxxxxxxxxxxxxxxx xxxxxxxx 73,065 96,129 76.01%
xxxxxxxxxxxxxxxxx xxxxxxxx 277,583 359,011 77.32%
xxxxxxxxxxxxxxxxx xxxxxxxx 32,049 40,991 78.19%
xxxxxxxxxxxxxxxxx xxxxxxxx 130,923 166,805 78.49%
xxxxxxxxxxxxxxxxx xxxxxxxx 14,448 18,159 79.56%
xxxxxxxxxxxxxxxxx xxxxxxxx 4,281,268 5,371,634 79.70%
xxxxxxxxxxxxxxxxx xxxxxxxx 152,058 187,327 81.17%
xxxxxxxxxxxxxxxxx xxxxxxxx 19,408 23,060 84.16%
xxxxxxxxxxxxxxxxx xxxxxxxx 150,480 177,017 85.01%
xxxxxxxxxxxxxxxxx xxxxxxxx 293,034 332,418 88.15%
xxxxxxxxxxxxxxxxx xxxxxxxx 59,681 67,658 88.21%
xxxxxxxxxxxxxxxxx xxxxxxxx 14,942 15,643 95.52%
xxxxxxxxxxxxxxxxx xxxxxxxx 208,548 218,264 95.55%
xxxxxxxxxxxxxxxxx xxxxxxxx 26,827 27,830 96.40%
xxxxxxxxxxxxxxxxx xxxxxxxx 67,530 68,432 98.68%
xxxxxxxxxxxxxxxxx xxxxxxxx 185,091 186,925 99.02% Project not complete
Project Net Cash Total Net Cash
Project Description Number Flow Project Cost Flow Ratio Comments
xxxxxxxxxxxxxxxxx xxxxxxxx 73,331 72,404 101.28%
xxxxxxxxxxxxxxxxx xxxxxxxx 60,959 59,784 101.97%
xxxxxxxxxxxxxxxxx xxxxxxxx 128,375 102,899 124.76%
xxxxxxxxxxxxxxxxx xxxxxxxx 18,041 12,011 150.20%
xxxxxxxxxxxxxxxxx xxxxxxxx 4,909 2,653 185.04%
xxxxxxxxxxxxxxxxx xxxxxxxx 16,314 5,491 297.10%
Rate of Return for Projects
Summarize Rates of Return
Basic Concept:
All investments must generate returns higher than the cost of the investment; otherwise
the investment will destroy value. The rate of return for Henson is based on the cost of
capital. When a projects's rate of return is higher than the Cost of Capital, the project
increases shareholder value. If the project's rate of return is below the Cost of Capital,
the project will negatively impact shareholder wealth. Additionally, we can add a "risk
premium" to those projects that carry higher risk. For the sake of keeping things
simple, we will simply use Henson's Cost of Capital as our discount rate.
Calculating Rate of Return:
Rate of Return is calculated as the Internal Rate of Return or IRR by discounting cash
inflows (payments made by clients) and cash outflows (amounts paid out for burdened
labor, direct costs, etc.) at a rate whereby the inflows = outflows. Another way of
understanding the calculation is to simply find the discount rate that gives a Net Present
Value of - 0 - for the project.
Required Inputs:
In order to calculate a project's rate of return, the Project Accountant will need to run:
1) OAR800 - Billing and Receipt History Report to ascertain cash inflows by month/year
2) Monthly Burdened Cost during the life of the project, summarized by month. Run Expenditure
Inquiry, export to Excel, format Expenditure Date to Month/Year, SubTotal by Expenditure Date.
3) Schedule the cash inflows (Step 1) and cash outflows (Step 2), calculate the Net Cash Flows,
and use the Excel Financial Functions to calculate Rate of Return and Net Present Value.
Summarize Rates of Returns for Projects Analyzed:
Project: M5L88100 - Municipal Center East Ohio 57%
Project: S8K11561 - Sewer Plants - ST/FM/LLM 9%
Project: S8K11562 - Water Pipelines -ST/FM/LLM -8%
Project: M7L55220 - Office Parks I-96 Corridor -41%
Project: S4C19901 - Stack & Block Model - PNT 63%
Project: S8K66100 - Cost Model Program - DS/LMT 63%
Project: M3L17001 - Central Subway - CA ODT 22%
Project: S4B16771 - Security Retrofit - Port Authority 80%
Project: M3C18901 - Design Specs for National Towers 101%
Project: M4B17701 - Security Upgrades for Chem Facilities 13%
Project: S5M19910 - Research Lab Facility - US/CA/BZ 10%
Project: M7D19880 - School Upgrades - West NP 77%
Project: S5C81050 - Master Plan Facility - DLL SE 49%
Project: S6X12677 - Tunnel Design for DOT 130%
Project: S8L72110 - CM Support for Yukon SC 106%
Project: S5L11850 - Railroad Surveys West Lines 68%
Net Present Value for Projects
Summarize Net Present Values
Basic Concept:
The Accounting Model tells us how much we invest in projects, but it tells us nothing about how well our
investment is doing over time. The Finance Model on the other hand considers the following:
1. Time Vaue of Money - \$ 1,000 a year from now is not worth a \$ 1,000 today
2. Risk - We expect to get paid \$ 1,000 when the work is done, but we're not absolutely certain
3. Opportunity Cost - If we have the use of \$ 1,000 now, we have an "opportunity" to do something with it
Because the Finance Model includes these assumptions in evaluating performance, it tends to be a more
accurate model than the Accounting Model. One of the primary metrics within the Finance Model is
Net Present Value. Net Present Value is what a project is worth today.
Calculating Net Present Value:
Net Present Value is calculated by discounting the cash inflows and cash outflows using a discount rate.
The discount rate used is the Cost of Capital (Cost of Debt + Cost of Equity). Projects that earn more than
the Cost of Capital have a positive Net Present Value and projects that earn less than the Cost of
Capital have a negative Net Present Value.
Required Inputs:
Same as Rate of Return. Within Microsoft Excel, use the financial function =NPV for annual cash flows
and the financial function =XNPV for monthly cash flows.
Summarize Amount of Value Created:
Project: S8K11561 - Sewer Plants ST/FM/LLM \$ (2,509)
Project: S8K11562 - Water Pipelines ST/FM/LLM \$ (53,681)
Project: M7L55220 - Office Parks I 96 Corridor \$ (34,929)
Project: S4C19901 - Stack & Block Model PNT \$ 53,271
Project: S8K66100 - Cost Model Program DS / LMT \$ 38,416
Project: M3L17001 - Central Subway - CA ODT \$ 8,925
Project: S4B16771 - Security Retrofit for Port Authority \$ 427,993
Project: M3C18901 - Design Specs for National Towers \$ 55,299
Project: M4B17701 - Security Upgrades - Chem Fac \$ 41,784
Project: S5M19910 - Research Lab Fac - US / CA / BZ \$ 102
Project: M7D19880 - School Upgrades West NP \$ 202,161
Project: S5C81050 - Master Fac Plan - DLL SE Region \$ 16,286
Project: S6X12677 - Tunnel Design for DOT \$ 103,026
Project: S8L72110 - CM Support for Yukon SC \$ 145,154
Project: S5L11850 - Surveys - R/R West Lines \$ 1,819,817
Calculate Project Rate of Return & Net Present Value
Project: M5L88100 - Mun Center - East Ohio
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Aug-99 0 619,306 -619,306 (614,401)
2 Sep-99 473,289 74,799 398,490 392,203
3 Oct-99 794,309 540,350 253,959 247,972
4 Nov-99 398,970 1,237,911 -838,941 (812,676)
5 Dec-99 455,191 668,203 -213,012 (204,709)
6 Jan-00 454,364 147,003 307,361 293,041
7 Feb-00 364,702 280,725 83,977 79,430
8 Mar-00 942,948 650,551 292,397 274,376
9 Apr-00 0 37,955 -37,955 (35,334)
10 May-00 429,794 424,520 5,274 4,871
11 Jun-00 985,343 212,622 772,721 708,003
12 Jul-00 0 586,256 -586,256 (532,901)
13 Aug-00 296,744 104,678 192,066 173,203
14 Sep-00 0 -323,186 323,186 289,138
15 Oct-00 0 62,720 -62,720 (55,668)
16 Nov-00 324,021 23,709 300,312 264,435
17 Dec-00 0 38,176 -38,176 (33,349)
18 Jan-01 0 35,147 -35,147 (30,460)
19 Feb-01 0 9,527 -9,527 (8,191)
20 Mar-01 0 4,058 -4,058 (3,462)
21 Apr-01 0 50,029 -50,029 (42,335)
22 May-01 0 5,014 -5,014 (4,209)
23 Jun-01 0 373 -373 (311)
24 Jul-01 0 0 0 0 Summarize Net Cash Flow on Annual Basis:
25 Aug-01 0 242 -242 (199) Year Net Flow
26 Sep-01 187,986 487 187,499 152,479 1999 -1,018,809
27 Oct-01 0 3,486 -3,486 (2,812) 2000 1,552,187
28 Nov-01 0 1,400 -1,400 (1,120) 2001 78,159
29 Dec-01 0 64 -64 (51)
TOTAL 6,107,661 5,496,124 496,964
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 113.43% 57.23% (1)
Net Present Value => 504,867 \$422,308.86 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S8K11561 - Sewer Plants - ST/FM/LLM
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Nov-99 0 10,593 -10,593 (10,509)
2 Dec-99 0 20,381 -20,381 (20,059)
3 Jan-00 0 44,924 -44,924 (43,865)
4 Feb-00 0 23,304 -23,304 (22,574)
5 Mar-00 0 4,690 -4,690 (4,507)
6 Apr-00 0 3,736 -3,736 (3,562)
7 May-00 0 311 -311 (295)
8 Jun-00 0 1,216 -1,216 (1,141)
9 Jul-00 0 151 -151 (140)
10 Aug-00 0 3,049 -3,049 (2,816)
11 Sep-00 0 1,366 -1,366 (1,251)
12 Oct-00 0 2,577 -2,577 (2,343)
13 Nov-00 0 1,273 -1,273 (1,148)
14 Dec-00 0 12,367 -12,367 (11,064)
15 Jan-01 0 17,909 -17,909 (15,895)
16 Feb-01 0 23,071 -23,071 (20,315)
17 Mar-01 0 16,591 -16,591 (14,494)
18 Apr-01 13,827 21,802 -7,975 (6,911)
19 May-01 43,995 27,283 16,712 14,368
20 Jun-01 53,120 16,750 36,370 31,023
21 Jul-01 0 32,110 -32,110 (27,171)
22 Aug-01 0 32,305 -32,305 (27,120)
23 Sep-01 96,520 50,233 46,287 38,551
24 Oct-01 0 205,285 -205,285 (169,620)
25 Nov-01 0 51,715 -51,715 (42,392)
26 Dec-01 88,315 237,089 -148,774 (120,987)
27 Jan-02 69,322 541,986 -472,664 (381,339)
28 Feb-02 -503 44,881 -45,384 (36,325)
29 Mar-02 607,831 63,899 543,932 431,912
30 Apr-02 282,362 57,002 225,360 177,532
31 May-02 186,651 242,078 -55,427 (43,318)
16 76ea7090-3e56-4669-91bb-7ba4598e27ae.xls Matt H. Evans
32 Jun-02 0 37,202 -37,202 (28,844)
33 Jul-02 338,337 187,047 151,290 116,372 Summarize Net Cash Flow on Annual Basis:
34 Aug-02 228,807 352,155 -123,348 (94,127) Year Net Flow
35 Sep-02 555,619 376,016 179,603 135,971 1999 -30,974
36 Oct-02 432,531 479,497 -46,966 (35,274) 2000 -98,963
37 Nov-02 595,904 246,225 349,679 260,551 2001 -436,365
38 Dec-02 112,941 147,855 -34,914 (25,809) 2002 633,960
TOTAL 3,705,579 3,637,922 (8,935)
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 8.59% 9.08% (1)
Net Present Value => -6,188 (\$2,634.75) (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
17 76ea7090-3e56-4669-91bb-7ba4598e27ae.xls Matt H. Evans
Calculate Project Rate of Return & Net Present Value
Project: S8K11562 - Water Pipelines - ST/FM/LLM
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Apr-00 0 7,370 -7,370 (7,312)
2 May-00 0 28,624 -28,624 (28,172)
3 Jun-00 0 28,065 -28,065 (27,404)
4 Jul-00 0 22,871 -22,871 (22,155)
5 Aug-00 74,658 19,944 54,714 52,581
6 Sep-00 31,996 68,090 -36,094 (34,413)
7 Oct-00 0 48,835 -48,835 (46,191)
8 Nov-00 72,342 34,706 37,636 35,316
9 Dec-00 0 53,189 -53,189 (49,516)
10 Jan-01 144,287 60,375 83,912 77,498
11 Feb-01 0 32,979 -32,979 (30,217)
12 Mar-01 0 93,959 -93,959 (85,408)
13 Apr-01 0 59,037 -59,037 (53,239)
14 May-01 0 67,693 -67,693 (60,561)
15 Jun-01 231,688 20,009 211,679 187,878
16 Jul-01 0 46,019 -46,019 (40,522)
17 Aug-01 0 85,140 -85,140 (74,374)
18 Sep-01 91,961 88,839 3,122 2,706
19 Oct-01 0 25,974 -25,974 (22,332)
20 Nov-01 0 8,603 -8,603 (7,338)
21 Dec-01 0 10,163 -10,163 (8,600)
22 Jan-02 0 7,212 -7,212 (6,055)
23 Feb-02 0 1,314 -1,314 (1,094)
24 Mar-02 0 817 -817 (675) Summarize Net Cash Flow on Annual Basis:
25 Apr-02 0 3 -3 (3) Year Net Flow
26 May-02 0 20,703 -20,703 (16,837) 2000 -132,699
27 Jun-02 266,140 2,285 263,855 212,874 2001 -130,855
28 Jul-02 0 281 -281 (225) 2002 232,044
29 Aug-02 0 0 0 0
30 Sep-02 0 0 0 0
31 Oct-02 0 1,469 -1,469 (1,148)
32 Nov-02 0 10 -10 (8)
TOTAL 913,072 944,581 (54,945)
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 0.00% -8.18% (1)
Net Present Value => -54,562 (\$53,722.26) (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: M7L55220 - Office Parks - I96 Corridor
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Jun-01 0 3,848 -3,848 (3,817)
2 Jul-01 0 7,701 -7,701 (7,580)
3 Aug-01 0 23,474 -23,474 (22,921)
4 Sep-01 0 33,426 -33,426 (32,380)
5 Oct-01 0 45,373 -45,373 (43,605)
6 Nov-01 0 66,787 -66,787 (63,675)
7 Dec-01 142,462 44,294 98,168 92,853
8 Jan-02 0 57,594 -57,594 (54,045)
9 Feb-02 47,254 72,093 -24,839 (23,123)
10 Mar-02 47,225 47,348 -123 (113)
11 Apr-02 94,450 34,892 59,558 54,570 Summarize Net Cash Flow on Annual Basis:
12 May-02 0 47,205 -47,205 (42,909) Year Net Flow
13 Jun-02 47,225 36,811 10,414 9,391 2001 -82,441
14 Jul-02 47,355 34,592 12,763 11,418 2002 48,396
15 Aug-02 94,450 40,292 54,158 48,068
16 Sep-02 0 30,650 -30,650 (26,989)
17 Oct-02 0 30,430 -30,430 (26,583)
18 Nov-02 141,675 39,331 102,344 88,696
TOTAL 662,096 696,141 (42,741)
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 0.00% -41.30% (1)
Net Present Value => -42,720 (\$34,929.73) (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S8K66100 - Cost Model Program/Study for DS/LMT
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Aug-01 0 1,835 -1,835 (1,820)
2 Sep-01 0 19,112 -19,112 (18,810)
3 Oct-01 0 8,613 -8,613 (8,410)
4 Nov-01 48,882 9,337 39,545 38,307
5 Dec-01 0 2,452 -2,452 (2,357)
6 Jan-02 0 15,491 -15,491 (14,769)
7 Feb-02 29,329 10,416 18,913 17,889
8 Mar-02 0 12,519 -12,519 (11,747)
9 Apr-02 0 6,362 -6,362 (5,922)
10 May-02 0 17,157 -17,157 (15,845)
11 Jun-02 0 15,653 -15,653 (14,342)
12 Jul-02 97,764 441 97,323 88,466
13 Aug-02 0 0 0 0 Summarize Net Cash Flow on Annual Basis:
14 Sep-02 0 498 -498 (445) Year Net Flow
15 Oct-02 0 201 -201 (179) OUTFLOW -120,111
16 Nov-02 19,553 25 19,528 17,195 INFLOW 195,528
TOTAL 195,528 120,111 67,210
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 956.43% 62.79%
Net Present Value => 68,044 \$53,224.09 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S4C19901 - Stack & Block Model - PNT
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Sep-01 0 23,078 -23,078 (22,895)
2 Oct-01 0 24,064 -24,064 (23,684)
3 Nov-01 0 20,892 -20,892 (20,399)
4 Dec-01 0 16,603 -16,603 (16,084)
5 Jan-02 0 674 -674 (648)
6 Feb-02 139,566 183 139,383 132,889
7 Mar-02 0 0 0 0
8 Apr-02 0 0 0 0
9 May-02 0 0 0 0
10 Jun-02 0 0 0 0
11 Jul-02 0 0 0 0
12 Aug-02 0 0 0 0 Summarize Net Cash Flow on Annual Basis:
13 Sep-02 0 0 0 0 Year Net Flow
14 Oct-02 0 0 0 0 OUTFLOW -85,494
15 Nov-02 207 0 207 184 INFLOW 139,773
TOTAL 139,773 85,494 49,363
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 386.72% 63.49%
Net Present Value => 49,906 \$38,381.94 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
Calculate Project Rate of Return & Net Present Value
Project: M3L17001 - Central Subway System - CA ODT
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Jun-02 0 4,386 -4,386 (4,352)
2 Jul-02 0 20,898 -20,898 (20,568)
3 Aug-02 0 33,354 -33,354 (32,568) Summarize Net Cash Flow on Annual Basis:
4 Sep-02 0 30,749 -30,749 (29,786) Year Net Flow
5 Oct-02 0 -10,548 10,548 10,137 OUTFLOW -83,112
6 Nov-02 101,766 4,272 97,494 92,951 INFLOW 101,766
TOTAL 101,766 83,112 15,814
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 117.09% 22.45%
Net Present Value => 16,032 \$8,904.43 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Project Rate of Return & Net Present Value
Project: S4B16771 - Security Retrofit for Port Authority
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Mar-01 0 2,457 -2,457 (2,438)
2 Apr-01 0 2,843 -2,843 (2,798)
3 May-01 0 3,316 -3,316 (3,238)
4 Jun-01 0 58,052 -58,052 (56,235)
5 Jul-01 45,760 81,482 -35,722 (34,329)
6 Aug-01 181,834 21,380 160,454 152,978
7 Sep-01 103,587 15,193 88,394 83,608
8 Oct-01 195,554 16,095 179,459 168,399
9 Nov-01 19,617 116,624 -97,007 (90,307)
10 Dec-01 456 57,250 -56,794 (52,453)
11 Jan-02 4,816 60,800 -55,984 (51,295)
12 Feb-02 33,128 24,073 9,055 8,231
13 Mar-02 159,260 51,958 107,302 96,764
14 Apr-02 32,148 20,738 11,410 10,208
15 May-02 11,224 26,340 -15,116 (13,417) Summarize Net Cash Flow on Annual Basis:
16 Jun-02 25,578 53,640 -28,062 (24,709) Year Net Flow
17 Jul-02 60,873 44,081 16,792 14,669 OUTFLOW -731,021
18 Aug-02 226,957 23,396 203,561 176,414 INFLOW 1,314,578
19 Sep-02 41,701 25,905 15,796 13,581
20 Oct-02 58,249 9,575 48,674 41,517
21 Nov-02 72,604 11,409 61,195 51,784
22 Dec-02 41,232 4,414 36,818 30,909
TOTAL 1,314,578 731,021 517,843
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 22770.32% 79.83%
Net Present Value => 524,251 427,659 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Project Rate of Return & Net Present Value
Project: M3C18901 - Design Specs for National Towers
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Mar-01 0 1,255 -1,255 (1,245)
2 Apr-01 0 14,511 -14,511 (14,282)
3 May-01 0 5,993 -5,993 (5,852)
4 Jun-01 0 21,183 -21,183 (20,520)
5 Jul-01 76,690 19,478 57,212 54,982
6 Aug-01 7,126 5,336 1,790 1,707
7 Sep-01 32,432 59 32,373 30,621
8 Oct-01 25,992 941 25,051 23,507
9 Nov-01 0 1,423 -1,423 (1,324)
10 Dec-01 0 489 -489 (452)
11 Jan-02 0 1,534 -1,534 (1,406)
12 Feb-02 0 0 0 0
13 Mar-02 671 30 641 578
14 Apr-02 0 0 0 0 Summarize Net Cash Flow on Annual Basis:
15 May-02 2,785 0 2,785 2,472 Year Net Flow
16 Jun-02 0 0 0 0 OUTFLOW -72,404
17 Jul-02 0 0 0 0 INFLOW 145,696
18 Aug-02 0 0 0 0
19 Sep-02 0 174 -174 (149)
TOTAL 145,696 72,404 68,637
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 4923.99% 101.23%
Net Present Value => 69,335 55,260 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Project Rate of Return & Net Present Value
Project: M4B17701 - Security Upgrade - Chemical Facility
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Apr-01 725 -725 (719)
2 May-01 3,638 -3,638 (3,581)
3 Jun-01 13,617 -13,617 (13,296)
4 Jul-01 12,892 -12,892 (12,488)
5 Aug-01 4,801 -4,801 (4,614)
6 Sep-01 0 10,308 -10,308 (9,828)
7 Oct-01 63,214 8,772 54,442 51,495
8 Nov-01 3,278 12,927 -9,649 (9,054)
9 Dec-01 17,043 -17,043 (15,866)
10 Jan-02 37,772 37,733 39 36
11 Feb-02 64,744 25,227 39,517 36,208
12 Mar-02 40,638 -40,638 (36,940)
13 Apr-02 54,576 -54,576 (49,217)
14 May-02 165,911 48,656 117,255 104,902
15 Jun-02 178,709 -178,709 (158,616) Summarize Net Cash Flow on Annual Basis:
16 Jul-02 541,648 227,891 313,757 276,273 Year Net Flow
17 Aug-02 288,165 -288,165 (251,730) OUTFLOW -1,548,022
18 Sep-02 461,777 136,454 325,323 281,938 INFLOW 1,746,104
19 Oct-02 201,106 153,581 47,525 40,861
20 Nov-02 206,654 271,667 -65,013 (55,454)
TOTAL 1,746,104 1,548,022 170,310
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 601.75% 12.80% (1)
Net Present Value => 174,705 \$41,455.77 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S5M19910 - Research Lab Facility - US / CA / BZ Local
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 May-01 0 191 -191 (189)
2 Jun-01 0 736 -736 (725)
3 Jul-01 0 16,779 -16,779 (16,384)
4 Aug-01 19,179 153 19,026 18,431
5 Sep-01 17,574 15,628 1,946 1,870
6 Oct-01 0 13,695 -13,695 (13,057)
7 Nov-01 33,856 16,050 17,806 16,842
8 Dec-01 17,111 29,461 -12,350 (11,589)
9 Jan-02 17,264 90 17,174 15,988
10 Feb-02 0 14,320 -14,320 (13,225)
11 Mar-02 16,833 1,900 14,933 13,682
12 Apr-02 34,019 30,405 3,614 3,285
13 May-02 17,540 17,721 -181 (164)
14 Jun-02 17,712 17,802 -90 (81)
15 Jul-02 16,491 251 16,240 14,414
16 Aug-02 0 13,888 -13,888 (12,229) Summarize Net Cash Flow on Annual Basis:
17 Sep-02 0 1 -1 (1) Year Net Flow
18 Oct-02 0 2,428 -2,428 (2,104) OUTFLOW -191,655
19 Nov-02 2,513 95 2,418 2,079 INFLOW 210,092
20 Dec-02 0 60 -60 (52)
TOTAL 210,092 191,655 16,793
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 1136.34% 9.62% (1)
Net Present Value => 16,986 \$63.61 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: M7D19880 - K12 School Upgrades - West NP Schools
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Aug-01 6,506 -6,506 (6,455)
2 Sep-01 48,253 21,412 26,841 26,418
3 Oct-01 51,253 38,670 12,583 12,286
4 Nov-01 58,145 36,998 21,147 20,485
5 Dec-01 58,561 19,697 38,864 37,349
6 Jan-02 64,254 23,879 40,375 38,494
7 Feb-02 96,254 24,767 71,487 67,617
8 Mar-02 0 33,782 -33,782 (31,700)
9 Apr-02 53,217 28,590 24,627 22,926
10 May-02 48,217 35,692 12,525 11,567
11 Jun-02 48,217 23,224 24,993 22,900
12 Jul-02 58,717 20,787 37,930 34,478 Summarize Net Cash Flow on Annual Basis:
13 Aug-02 51,319 26,944 24,375 21,981 Year Net Flow
14 Sep-02 0 13,679 -13,679 (12,238) outflow -359,417
15 Oct-02 0 3,665 -3,665 (3,253) inflow 636,407
16 Nov-02 0 1,125 -1,125 (991)
TOTAL 636,407 359,417 261,866
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 9964463440.00% <=Too Distorted , Estimate and use => 77.07%
Net Present Value => 264,438 \$202,000.15 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S5C81050 - Master Facility Plan for DLL SE Region
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Oct-01 0 2,179 -2,179 (2,161)
2 Nov-01 0 13,362 -13,362 (13,151)
3 Dec-01 4,647 702 3,945 3,852
4 Jan-02 35,218 4,637 30,581 29,624
5 Feb-02 5,362 6,597 -1,235 (1,186)
6 Mar-02 9,325 418 8,907 8,492
7 Apr-02 0 5,136 -5,136 (4,858)
8 May-02 0 5,760 -5,760 (5,405)
9 Jun-02 0 7,128 -7,128 (6,635)
10 Jul-02 11,591 1,328 10,263 9,479 Summarize Net Cash Flow on Annual Basis:
11 Aug-02 0 2,093 -2,093 (1,918) Year Net Flow
12 Sep-02 0 7 -7 (6) outflow -49,345
13 Oct-02 45 0 45 41 inflow 73,607
14 Nov-02 0 0 0 0
15 Dec-02 7,419 0 7,419 6,585
TOTAL 73,607 49,345 22,751
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 11774.14% 49.17% (1)
Net Present Value => 22,984 \$16,268.76 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: M9K10601 - Support Team for Fac Mgmt - ILBBC
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Sep-01 48,253 0 48,253 47,871
2 Oct-01 51,253 0 51,253 50,444
3 Nov-01 58,145 21,967 36,178 35,325
4 Dec-01 58,561 26,077 32,484 31,467
5 Jan-02 64,254 24,299 39,955 38,398
6 Feb-02 48,217 22,765 25,452 24,266
7 Mar-02 48,217 29,499 18,718 17,705
8 Apr-02 53,217 6,120 47,097 44,194
9 May-02 48,217 6,662 41,555 38,685
10 Jun-02 48,217 501 47,716 44,069 Summarize Net Cash Flow on Annual Basis:
11 Jul-02 58,717 -13,262 71,979 65,951 Year Net Flow
12 Aug-02 51,291 1,089 50,202 45,633 outflow -151,829
13 Sep-02 0 5,071 -5,071 (4,573) inflow 636,559
14 Oct-02 0 9,480 -9,480 (8,482)
15 Nov-02 0 6,643 -6,643 (5,896)
16 Dec-02 0 4,917 -4,917 (4,329)
TOTAL 636,559 151,829 460,727
DO NOT INCLUDE THIS PROJECT IN ANALYSIS - SOME COSTS ARE SITTING IN COMPANION PROJECT
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 0.00% 319%
Net Present Value => 465,280 \$391,566.24 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
(1): Rate of Return for this project using annualized cash flows
(2): Net Present Value of project is negative when the Rate of Return is less than the Cost of Capital
(3): Cost of Capital is used as the Cost of Investment, no risk premium has been added to project
Calculate Project Rate of Return & Net Present Value
Project: S6X12677 - Tunnel Design for DOT
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 May-02 0 13,660 -13,660 (13,552)
2 Jun-02 0 17,354 -17,354 (17,080)
3 Jul-02 0 24,180 -24,180 (23,610)
4 Aug-02 115,373 14,856 100,517 97,370
5 Sep-02 0 22,058 -22,058 (21,198) Summarize Net Cash Flow on Annual Basis:
6 Oct-02 49,539 10,300 39,239 37,411 Year Net Flow
7 Nov-02 66,362 501 65,861 62,295 outflow -102,910
8 Dec-02 5,127 1 5,126 4,810 inflow 236,401
TOTAL 236,401 102,910 126,446
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 13533.62% <=Too Distorted , Estimate and use => 129.72%
Net Present Value => 127,658 \$102,960.15 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Project Rate of Return & Net Present Value
Project: S8L72110 - CM Support for Yukon SC
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Nov-01 0 2,700 -2,700 (2,679)
2 Dec-01 0 7,998 -7,998 (7,872)
3 Jan-02 104,297 65,677 38,620 37,710
4 Feb-02 0 11,078 -11,078 (10,731)
5 Mar-02 160,040 14,606 145,434 139,765
6 Apr-02 30,407 10,233 20,174 19,234
7 May-02 38,181 9,840 28,341 26,807
8 Jun-02 19,545 9,111 10,434 9,791
9 Jul-02 19,545 10,851 8,694 8,094
10 Aug-02 0 16,368 -16,368 (15,117)
11 Sep-02 0 12,544 -12,544 (11,493) Summarize Net Cash Flow on Annual Basis:
12 Oct-02 0 8,347 -8,347 (7,587) Year Net Flow
13 Nov-02 0 10,101 -10,101 (9,109) outflow -180,540
14 Dec-02 0 -8,914 8,914 7,975 inflow 372,015
TOTAL 372,015 180,540 184,786
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 87410691.88% <=Too Distorted , Estimate and use => 106.06%
Net Present Value => 186,550 \$145,054.64 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Project Rate of Return & Net Present Value
Project: S5L11850 - Surveys for R/R West Lines
Monthly Cash Cash Net Present
Period Mo/Yr Inflows Outflows Cash Value
Payments * B-Labor/ODC ** (simple calc)
1 Mar-99 0 18,004 -18,004 (17,862)
2 Apr-99 0 19,123 -19,123 (18,822)
3 May-99 0 21,624 -21,624 (21,114)
4 Jun-99 0 29,786 -29,786 (28,853)
5 Jul-99 0 68,217 -68,217 (65,558)
6 Aug-99 0 39,922 -39,922 (38,062)
7 Sep-99 0 35,095 -35,095 (33,194)
8 Oct-99 541,112 58,543 482,569 452,827
9 Nov-99 0 36,409 -36,409 (33,894)
10 Dec-99 0 17,082 -17,082 (15,776)
11 Jan-00 53,200 53,464 -264 (242)
12 Feb-00 122,800 43,003 79,797 72,535
13 Mar-00 53,200 44,497 8,703 7,848
14 Apr-00 53,200 46,648 6,552 5,862
15 May-00 53,200 53,364 -164 (146)
16 Jun-00 63,060 90,251 -27,191 (23,943)
17 Jul-00 53,200 72,005 -18,805 (16,427)
18 Aug-00 0 87,541 -87,541 (75,867)
19 Sep-00 117,167 122,576 -5,409 (4,651)
20 Oct-00 356,980 117,905 239,075 203,923
21 Nov-00 348,039 116,370 231,669 196,041
22 Dec-00 0 110,489 -110,489 (92,756)
23 Jan-01 286,669 107,025 179,644 149,618
24 Feb-01 366,254 138,081 228,173 188,531
25 Mar-01 187,644 159,527 28,117 23,048
26 Apr-01 229,747 122,754 106,993 87,009
27 May-01 53,200 124,095 -70,895 (57,197)
28 Jun-01 617,678 132,278 485,400 388,512
29 Jul-01 53,200 89,699 -36,499 (28,982)
30 Aug-01 226,513 138,205 88,308 69,566
31 Sep-01 25,737 108,423 -82,686 (64,622)
32 Oct-01 300,711 139,030 161,681 125,357
33 Nov-01 0 150,906 -150,906 (116,077)
34 Dec-01 406,624 122,966 283,658 216,461
35 Jan-02 0 112,147 -112,147 (84,902)
36 Feb-02 454,929 94,555 360,374 270,665
37 Mar-02 275,923 96,468 179,455 133,715
38 Apr-02 0 79,595 -79,595 (58,838)
39 May-02 299,430 81,832 217,598 159,578 Summarize Net Cash Flow on Annual Basis:
40 Jun-02 0 69,690 -69,690 (50,703) Year Net Flow
41 Jul-02 0 64,688 -64,688 (46,691) outflow -3,729,372
42 Aug-02 340,864 89,689 251,175 179,860 inflow 6,269,997
43 Sep-02 0 61,758 -61,758 (43,873)
44 Oct-02 174,747 57,350 117,397 82,739
45 Nov-02 154,969 66,018 88,951 62,194
46 Dec-02 0 20,673 -20,673 (14,340)
TOTAL 6,269,997 3,729,372 2,022,495
Economic Analysis on Monthly Basis: Economic Analysis on Annual Basis:
Project Rate of Return => 1176.86% 68.12%
Net Present Value => 2,055,878 \$1,818,272.77 (2)
Monthly Cost of Investment => 0.80% (Annual Cost of Capital divided by 12) 9.58% (3)
* Per Oracle Report OAR800 - Billing and Receipt History
** Total Burdened Cost per Detail Dump of all Charges
Calculate Cost of Capital
Compare to Rate of Return
The following attributes are used to calculate Henson's Cost of Capital:
Risk free rate of return 4.50% (per Henson 2001 10-K Filing with the SEC)
Beta Factor 0.55 (per Yahoo Finance)
Market rate of return 16.50% (per review of 5 year returns from Value Line)
Interest Rate on Debt 6.25% (per Henson 2001 10-K Filing with the SEC)
Effective Tax Rate 34.50% (per Henson 2001 10-K Filing with the SEC)
Calculate Weighted Average Cost of Capital:
weighted
* B.V. ** %'s avg %
Cost of Equity 11.10% 592 78.31% 8.69%
Cost of Debt 4.09% 164 21.69% 0.89%
756 100.00%
Weighted Average Cost of Capital 9.58%
* Book Values for Total Equity and Total Debt per 10-K Filing expressed in millions
** Market Value weights are suppose to be used, but in the absence of market values
for equity and debt, book value weights were used.
Summarize Cut Off Dates for Monthly Reporting Periods
FY 2000 FY 2001 FY 2002 FY 2003
Mo/Yr Cutoff Mo/Yr Cutoff Mo/Yr Cutoff Mo/Yr Cutoff
Oct-99 10/29/1999 Oct-00 10/27/2000 Oct-01 10/26/2001 Oct-02 10/25/2002
Nov-99 11/26/1999 Nov-00 11/24/2000 Nov-01 11/23/2001 Nov-02 11/22/2002
Dec-99 12/31/1999 Dec-00 12/29/2000 Dec-01 12/28/2001 Dec-02 12/27/2002
Jan-00 1/28/2000 Jan-01 1/26/2001 Jan-02 1/25/2002 Jan-03 1/24/2003
Feb-00 2/25/2000 Feb-01 2/23/2001 Feb-02 2/22/2002 Feb-03
Mar-00 3/31/2000 Mar-01 3/30/2001 Mar-02 3/29/2002 Mar-03
Apr-00 4/28/2000 Apr-01 4/27/2001 Apr-02 4/26/2002 Apr-03
May-00 5/26/2000 May-01 5/25/2001 May-02 5/24/2002 May-03
Jun-00 6/30/2000 Jun-01 6/29/2001 Jun-02 6/28/2002 Jun-03
Jul-00 7/28/2000 Jul-01 7/27/2001 Jul-02 7/26/2002 Jul-03
Aug-00 8/25/2000 Aug-01 8/24/2001 Aug-02 8/23/2002 Aug-03
Sep-00 9/29/2000 Sep-01 9/28/2001 Sep-02 9/27/2002 Sep-03
Run Various Test to Verify Calculations
Compare Annual Calc's to Monthly Calc's - Should Not Vary Widely:
Year Inflow Outflow Net
6/1/2002 90,000 180,000 -90,000 -180,000 90,000
12/1/2002 270,000 120,000 150,000 -120,000 270,000
Total 360,000 300,000 20.00%
NPV 42,149 IRR 11%
Date XIRR 1.770035
Jan-00 7,500 15,000 -7,500
Feb-00 7,500 15,000 -7,500
Mar-00 7,500 15,000 -7,500
Apr-00 7,500 15,000 -7,500
May-00 7,500 15,000 -7,500
Jun-00 7,500 15,000 -7,500
Jul-00 7,500 15,000 -7,500
Aug-00 7,500 15,000 -7,500
Sep-00 7,500 15,000 -7,500
Oct-00 7,500 15,000 -7,500
Nov-00 7,500 15,000 -7,500
Dec-00 7,500 15,000 -7,500 -90,000
Jan-01 22,500 10,000 12,500
Feb-01 22,500 10,000 12,500
Mar-01 22,500 10,000 12,500
Apr-01 22,500 10,000 12,500
May-01 22,500 10,000 12,500
Jun-01 22,500 10,000 12,500
Jul-01 22,500 10,000 12,500
Aug-01 22,500 10,000 12,500
Sep-01 22,500 10,000 12,500
Oct-01 22,500 10,000 12,500
Nov-01 22,500 10,000 12,500 -90,000
Dec-01 22,500 10,000 12,500 150,000
NPV \$43,396
Finance Cost Mo Rate
Cost of Capital 10.00% (used as reinvestment rate) 0.83%
Annual Monthly
Regular Rate of Return 67% 4% 67%
Modified Rate of Return 5% 2%
Irregular Payment IRR n/a 66.62%
USE THE IRR CALC IN BOLD FOR ALL RATE OF RETURN
CALCULATIONS ON ALL PROJECTS
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Mar 3 comment Why does electrical current start to flow? I've looked at the question, but it is not going in the same direction. The most important part in my question is to state the EM field for all points in space and then explain how that drives the electrons. The answer there only mentions there is "voltage from the power plant", but it doesn't draw the whole picture. Mar 2 awarded Editor Mar 2 revised Why does electrical current start to flow? added 1 characters in body Mar 2 asked Why does electrical current start to flow? Feb 25 comment How are fundamental forces transmitted? Thanks. There are sort of similar questions, but my specific question how localization and distance come into play I couldn't find (it only mentions "quantum field" without further explanation which would help me). Maybe someone can elaborate on locality in particular? Feb 25 asked How are fundamental forces transmitted? Sep 9 comment How do you prove $S=-\sum p\ln p$? @Ron: I'm not saying you are wrong. I'm saying you just haven't given an answer related to the question. And I made the experience that generally people who start their answer with either "It's very easy and well known" or "I know where you are wrong" usually haven't understood the question. Feel free to check my posts word by word. Sep 9 comment How do you prove $S=-\sum p\ln p$? This is pretty interesting! Yet, it misses the question :( It does not answer why I should apply entropy to a system which is already completely determined by other laws. Why would I assume that a system is the max entropy distribution? One has to proof that. And it shouldn't be a hand-wavy argument either. Also the 2nd and 3rd paragraph of my question is left out. Many people claim that just everything follows entropy (black holes, life, ...). OK, so I have a particle circling another. What are p_i? What is the entropy? Sep 7 comment How do you prove $S=-\sum p\ln p$? The common Carnot argument wouldn't help. The question how it is supposed to be connected with all microscopic processes in general would still be as open as before. Someone must have tried that before? I know common literature but it's not in there :(; S is whatever people use for showing irreversibility. Sep 7 comment How do you prove $S=-\sum p\ln p$? Shannon entropy exists, but it's still no answer why it is used in physics. Shannon entropy probably has some presuppositions. So why does physics satisfy these presuppositions? Sep 7 asked How do you prove $S=-\sum p\ln p$? Sep 6 awarded Student Sep 5 asked Learn QM algebraic formulations and interpretations | 608 | 2,656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-18 | latest | en | 0.965506 |
https://stats.stackexchange.com/questions/635210/when-should-grouping-variables-interact-in-a-mixed-effects-model?noredirect=1 | 1,726,502,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00686.warc.gz | 506,298,640 | 41,534 | # When should grouping variables interact in a mixed-effects model?
I was reading this post which is relevant to a research project I'm working on now. I think that I understand the difference between crossed and nested random effects, e.g. as described here.
The first post I linked asks about differences between the random effects structure Ticks~(1|Year)+(1|Site) vs. Ticks~(1|Year)+(1|Site)+(1|Year:Site), i.e. allowing the number of ticks at each Site to vary across years, in addition to varying across years and across sites. Since there are multiple observations of each site during each year, this model is estimable and more flexible than the model that omits the term (1|Year:Site).
While the excellent GLMM FAQ does not mention interactions between grouping variables, this is mentioned in another writing by Ben Bolker here, which also describes this term as allowing the effect of each year to vary by site.
I'm trying to understand more about this interaction between the grouping variables and when it might be necessary, and whether this differs between terms with a 1 on the LHS or a variable on the LHS of the grouping term. Assuming such a model is estimable, under what circumstances would one want to use the model y ~ (x|f) + (x|g) instead of y ~ (x|f) + (x|g) + (x|f:g)?
From a lot of what I've read, it seems that the former model is often recommended for fully crossed designs. Does this justify leaving out the extra term? Why?
Am I just overthinking this problem, and the inclusion of this term should be based on domain knowledge and the specific data one has?
• These models are nested within one another. Have you tried using a likelihood ratio test (anova() in R) to determine whether the added complexity of the (1|f:g) intercept fits the data better than the less complex model without it? If you have substantive reasons to prefer the more complex model, then so be it. But you can also use the LRT to help you if you don't have such reasons. Commented Dec 19, 2023 at 0:02
• @ErikRuzek Yes, and it appears to benefit the model. But as we know, just because an LRT or AIC says to include the term doesn't mean it's always best from an inferential standpoint, so I was curious if something about the study design should affect this decision. Commented Dec 19, 2023 at 17:13
I am going to work from the simpler model in which you do not have a random (or varying) slope for x and instead just have a set of random (or varying) intercepts:
m1 <- lmer(dv ~ 1 + (1|f) + (1|g) + (1|f:g), d)
The random (or varying) intercepts can be interpreted as such:
1. (1|f) - the random intercept for f is shared across all groups g for a given f.
2. (1|g) - the random intercept for g is shared across all groups f for a given g.
3. (1|f:g) - the random intercept for f:g is shared among the unique groups comprised by combinations of f and g.
Keep in mind that in order to validly estimate #3, you need to have many cases in each combination. If you only have one case per combination, then this if completely confounded with the residual term.
So how do you decide whether to include (1|f:g) in the model? Substantively, you can ask yourself whether you think the combinations of f and g are distinct enough from their fellow f and/or g members that they need their own unique variance term. One may say, "of course they are!" But the problem is that one might not have enough data to reliably estimate this additional random parameter. That could mean that you get a model convergence error, a weird parameter estimate, etc.
All that said, model fit criteria is a perfectly fine justification one can use to determine whether the added complexity of the additional interaction intercept improves the fit of the model relative to the simpler additive model. This answers the question of whether the more complex model does a better job of accounting for the unexplained variance in the outcome than the simpler model. | 914 | 3,954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.956744 |
https://www.engineersadda.co/gate22-ce-daily-practice-quiz-14-july-2021 | 1,627,100,659,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00445.warc.gz | 760,794,756 | 23,800 | # GATE’22 CE:- Daily Practice Quiz 14-July-2021
Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous
Each question carries 2 mark
Negative marking: 1/3 mark
Time: 12 Minutes
Q1. The state of stresses on an element is shown in the given figure. The values of stresses are σ_x (=32MPa):σ_y (=-10MPa) and major principal stress σ_1 (=40 MPa). The minor principal stress ‘σ_2’ is
(a) -22 MPa
(b) -18 MPa
(c) 22 MPa
(d) Indeterminable due to insufficient data
Q2. What is the limiting principle tensile stress in prestress uncracked concrete member of M25 grade?
(a) 1.2 MPa
(b) 1.5 MPa
(c) 2 MPa
(d) 2.5 MPa
Q3. A steady, two-dimensional, incompressible flow field is represented by u = x + 3y + 3 and V = 2x – y – 8 in this flow field, the stagnation point is
(a) (3, 2)
(b) (-3, 2.5)
(c) (-3, -2)
(d) (3, -2)
Q4. Which one of the following represents relative density of saturated sand deposited having moisture content of 25%, if maximum and minimum void ratio of sand are 0.95 and 0.45 respectively and specific gravity of sand particles is 2.6?
(a) 40%
(b) 50%
(c) 60%
(d) 70%
Q5. The kinematic indeterminacy of the structure shown in the figure is equal to
(a) 14
(b) 15
(c) 16
(d) 17
Solutions
S1. Ans.(b)
Sol. σ_x+σ_y=σ_1+σ_2
32+(-10)=40+σ_2
▭(σ_2= -18 MPa)
S2. Ans.(a)
Sol. As per IS
Maximum principal tensile stress in prestressed concrete member = 0.24 √fck
=0.24√25
=0.24×5
=1.20 MPa
S3. Ans.(d)
Sol. Given,
u = x + 3y + 3
v = 2x – y – 8
For stagnation point u = 0 & v = 0
x+3y+3=0—-(1)
2x-y-8=0—-(2)
By solving equation (1) & (2)
▭(■(x=3@y= -2))⇒(3,-2)
S4. Ans.(c)
Sol. Given, S = 1 (Saturated sand)
e_max=0.95
e_min=0.45
w=25%=0.25
G=2.6
I_D=?
We know,
Se=WG
1×e=0.25×2.6
▭(e=0.65)
Now,
I_D=(e_max-e)/(e_max-e_min )
■(0.95-0.65 @0.95-0.45)×100
▭(I_D=60%)
S5. Ans.(a)
Sol. No.of inextensible members (m) =15
No.of external reactions (r_e) =7
No.of joints (j) =12
Internal hinged reactions (r_r) =0
Kinematic indeterminacy (D_k) =3j-r_e-m+r_r
=(3×12)-7-15+0
=36-22
▭(D_k=14)
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Are you sure you want to skip this step? | 1,269 | 3,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-31 | latest | en | 0.703398 |
https://www.physicsforums.com/threads/calculate-turbine-gross-and-net-output.869665/ | 1,660,752,047,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00748.warc.gz | 844,915,429 | 18,278 | # Calculate Turbine gross and net output
Nemo's
## Homework Statement
A steam power plant consists of a boiler, a turbine, a condenser and a pump. The temperature of the inner walls of the boiler is 350oC and the temperature of the condenser cooling water is 20oC. During a certain interval of time, the heat added to the boiler is 2.9x106 kJ and the heat rejected in the condenser is 2.1x106 kJ. If the pump work is 250000 kJ, estimate : a. The turbine gross and net output. b. The thermal efficiency of the power plant. c. The maximum possible efficiency of the power plant.
[/B]
## Homework Equations
Second law of thermodynamics.
## The Attempt at a Solution
Applying law of conservation of energy to get Wout (Turbine work)
Win-Wout+Qh-Ql=0
250M-Wout+2.9G-2.1G=0
Wout=250.8 MJ
Staff Emeritus
Homework Helper
## Homework Statement
A steam power plant consists of a boiler, a turbine, a condenser and a pump. The temperature of the inner walls of the boiler is 350oC and the temperature of the condenser cooling water is 20oC. During a certain interval of time, the heat added to the boiler is 2.9x106 kJ and the heat rejected in the condenser is 2.1x106 kJ. If the pump work is 250000 kJ, estimate : a. The turbine gross and net output. b. The thermal efficiency of the power plant. c. The maximum possible efficiency of the power plant.
[/B]
## Homework Equations
Second law of thermodynamics.
## The Attempt at a Solution
Applying law of conservation of energy to get Wout (Turbine work)
Win-Wout+Qh-Ql=0
250M-Wout+2.9G-2.1G=0
Wout=250.8 MJ
What's the gross output of the turbine? The net output? What about parts b) and c)?
Nemo's
What's the gross output of the turbine? The net output? What about parts b) and c)?
I actually don't understand the difference between the net and gross outputs.
a)I need to make sure that it's correct to get the turbine work output using the law of conservation of energy so I can proceed to b.
b) If I get the turbine output power from a then the efficiency=total power input/total power output =Wout/250M
c)Maximum efficiency =1-TL/TH =1- ((20+273)/(350+273))=0.53=53%
Please correct me if I am wrong.
Staff Emeritus
Homework Helper
I actually don't understand the difference between the net and gross outputs.
a)I need to make sure that it's correct to get the turbine work output using the law of conservation of energy so I can proceed to b.
b) If I get the turbine output power from a then the efficiency=total power input/total power output =Wout/250M
c)Maximum efficiency =1-TL/TH =1- ((20+273)/(350+273))=0.53=53%
Please correct me if I am wrong.
The turbine is the only thing between the boiler and the condenser. What does that tell you about the gross work output?
The pump is between the condenser and the boiler. Part of the gross output of the turbine must be used to drive the pump. Net work output is ...?
Nemo's
The turbine is the only thing between the boiler and the condenser. What does that tell you about the gross work output?
The pump is between the condenser and the boiler. Part of the gross output of the turbine must be used to drive the pump. Net work output is ...?
I think that makes the gross work output = turbine work output = 250.8M and the net output =turbine work output - pump work = 0.8M (according to my previous calculations).
Staff Emeritus
Homework Helper
I think that makes the gross work output = turbine work output = 250.8M and the net output =turbine work output - pump work = 0.8M (according to my previous calculations).
Your original calculation 1) doesn't make sense and 2) is arithmetically incorrect
Applying law of conservation of energy to get Wout (Turbine work)
Win-Wout+Qh-Ql=0
250M-Wout+2.9G-2.1G=0
Wout=250.8 MJ
2.9 G - 2.1 G = 800 M joules
250 M + 800 M = 1050 M joules - Wout = 0
So, how did you come up with Wout = 250.8 M joules? None of your original numbers even have that degree of precision.
Nemo's
Your original calculation 1) doesn't make sense and 2) is arithmetically incorrect
2.9 G - 2.1 G = 800 M joules
250 M + 800 M = 1050 M joules - Wout = 0
So, how did you come up with Wout = 250.8 M joules? None of your original numbers even have that degree of precision.
Yes I made an arithmetic mistake. Wout should be 1050M according to my original calculation. so net output 1050-250=800M
I think I know why my original calculation doesn't make sense. I should appply the first law of thermodynamics. However, I don't know the mass or the enthalpy.
Staff Emeritus
Homework Helper
Yes I made an arithmetic mistake. Wout should be 1050M according to my original calculation. so net output 1050-250=800M
No, this is still wrong, but it is arithmetically correct. Where does the 1050 come from?
I think I know why my original calculation doesn't make sense. I should appply the first law of thermodynamics. However, I don't know the mass or the enthalpy.
You don't need to know either. You're given the total heat quantities, which is what you would have obtained from knowing the mass flow rates and the enthalpies anyway.
Nemo's
No, this is still wrong, but it is arithmetically correct. Where does the 1050 come from?
You don't need to know either. You're given the total heat quantities, which is what you would have obtained from knowing the mass flow rates and the enthalpies anyway.
O.k what if I say turbine work output = Qh-Ql=2.9G-2.1G=800M would that be correct? I am not sure because similarly I was expecting pump work = 2.9G-2.1G to apply but it doesn't.
Staff Emeritus
Homework Helper
O.k what if I say turbine work output = Qh-Ql=2.9G-2.1G=800M would that be correct?
Gross or net output?
I am not sure because similarly I was expecting pump work = 2.9G-2.1G to apply but it doesn't.
Why would you think that? The pump work is explicitly given as 250,000 kJ, or 250 MJ, in the problem statement.
2.9 GJ - 2.1 GJ ≠ 250 MJ
Nemo's
Gross or net output?
Gross=800M
Why would you think that? The pump work is explicitly given as 250,000 kJ, or 250 MJ, in the problem statement.
2.9 GJ - 2.1 GJ ≠ 250 MJ
Net=800m-250M=550M
This makes sense to me. I hope I got it right this time.
Staff Emeritus
Homework Helper
Gross=800M
Net=800m-250M=550M
This makes sense to me. I hope I got it right this time.
Yes, that's much better.
Nemo's
Yes, that's much better.
b)thermal efficiency = Net Work /Qh = 550M/2.9G =19%
c) carnot efficiency = 53% (previusly calculated)
So is this the correct solution ?
Staff Emeritus | 1,820 | 6,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-33 | latest | en | 0.85291 |
https://deliveredin24.com/what-is-the-companys-net-working-capital/ | 1,675,418,059,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00035.warc.gz | 213,551,448 | 9,704 | # What is the company’s net working capital?
Learning Goal: I’m working on a management writing question and need an explanation and answer to help me learn.Q1. XZY has net sales of 5,320,140; net income of 2,145,700; cost of goods sold 1,300,000; and EBIT 2,200,000. Calculate the gross profit and the operating profit margin for the firm. (Show your calculations) (1 Mark)- Ch 4 Q2. Prepare a common sized Balance Sheet for the below Balance sheet? (Show your calculations) (1 Mark)- Ch 4 Cash 21,000 Acct/Rec 52,000 Inventories 200,500 Current assets 273,500 Net fixed assets 132,000 Total assets 405,500 Accts/Pay 22,800 Accrued expenses 21,000 Short-term N/P 8,700 Current liabilities 52,500 Long-term debt 150,000 Total liabilities 202,500 Owner’s equity 203,000 Total liabilities and owners’ equity 405,500 Q3. ABC company generated total sales of \$32,565,420 during fiscal 2021. Depreciation and amortization for the year totaled \$1,278,120, and cost of goods sold was \$21,400,000. Interest expense for the year was \$6,341,250 and selling, general, and administrative expenses totaled \$2,556,610 for the year. If the company’s tax rate was average 30 percent, what is its net income after taxes? (Show your calculations) (1 Mark)- Ch 3 Q4. BBB company had cash and marketable securities worth \$400,134 accounts payables worth \$2,490,357, inventory of \$1,321,500, accounts receivables of \$2,188,128, short-term notes payable worth \$120,000, other current liabilities of 200,000, and other current assets of \$521,800. What is the company’s net working capital? (Show your calculations) (1 Mark)-Ch 3 Q5. In your own words, explain the difference between Brokers and Dealers? (Show your calculations) (1 Mark)-Ch 2
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No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 911 | 3,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-06 | latest | en | 0.900343 |
http://mathhelpforum.com/math-topics/182600-uniform-acceleration.html | 1,566,692,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00473.warc.gz | 125,786,194 | 10,179 | 1. ## Uniform Acceleration
A car starts from rest with uniform acceleration of 10m/s^2 until it reaches a speed of 24m/s . It travels at this speed for 3hrs and then comes to rest with uniform retardation in 10secs.
What would be the average speed of the car the whole journey and the retardation of the car in the last 10 secs?
2. There are three phases for the movement:
- acceleration phase (when it accelerates from 0 m/s to 24 m/s at a rate of 10 m/s^2)
- traveling phase (three hours traveling at 24 m/s)
- deceleration phase (when it decelerates from 24 m/s to 0/ms in 10 seconds)
Both questions can be answered by using the physics formulas for uniform motions.
As for the average speed of the car for the whole journey, you should use the weighted mean of the three phases.... like~
((averagespd acceleration phase)*(time taken on acceleration phase) + (aspd traveling)*(time traveling) + (aspd deceleration)*(time deceleration))/(totaltime)
3. Originally Posted by MathBond
A car starts from rest with uniform acceleration of 10m/s^2 until it reaches a speed of 24m/s^2 . It travels at this speed for 3hrs and then comes to rest with uniform retardation in 10secs.
What would be the average speed of the car the whole journey and the retardation of the car in the last 10 secs? | 323 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.914269 |
https://crypto.stackexchange.com/questions/22855/diffie-hellman-application/22866 | 1,579,278,092,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589861.0/warc/CC-MAIN-20200117152059-20200117180059-00036.warc.gz | 403,558,852 | 31,287 | # Diffie-Hellman Application
Alice and Bob want to perform a Diffie-Hellman key exchange using the group $G$, a primitive root $g$, Alice’s secret key $k_A$ and Bob’s secret key $k_B$. In each case below compute the element of $G$ that Alice sends to Bob, the element that Bob sends to Alice, and the secret key that Alice and Bob will share.
1. $G=\mathbb Z/163\mathbb Z$(as an additive group), $g = 2$, $k_A = 128$, $k_B = 65$.
2. $G=\mathbb F_{163}^*$, $g = 2$, $k_A = 128$, $k_B = 65$
3. $G = [0, 1) ∩ \mathbb Q$ with group operation $$a\Diamond b = [a + b] = a + b − \left \lceil{a+b}\right \rceil$$ $g = 23/123$, $k_A = 358$ and $k_B = −44$.
4. $G = GL_2(\mathbb F_{17})$, $g= \left[\begin{array}{ c c } 2 & 3 \\ 4 & 5 \end{array} \right]$, $k_A = 13$, $k_B = 5$.
For $1)$ I did the following, Alice sends Bob $A = (128)2 \bmod p$ and then Bob sends Alice $B = (65)2 \bmod p$. Their secret key is $(128+65)2 \bmod p$.
For $2)$ I did this: Alice sends Bob $A = 2^{128}$ $mod p$ and Bob sends Alice $B = 2^{65} \bmod p$ Therefore, their secret key is $B = 2^{(65)(128)} \bmod p$.
For $3)$ Alice sends Bob $128\Diamond 23/123$ and Bob sends Alice $65\Diamond 23/123$ so their secret key is $(128\Diamond 65)\Diamond 23/123$. However I don't know whether this is right or not.
I don't know how to do the fourth one.
Any help is appreciated. Thank you.
• You don't have (3) correct; Alice sends Bob $116/123$, Bob sends Alice $95/123$, and their secret key is $62/123$. How did I get that? Well, consider the group operation, and if $a=p/q$, what is $a\Diamond a\Diamond ... \Diamond a$ – poncho Feb 10 '15 at 11:13
• The ceiling function makes any fraction in it as the highest integer. So Alice sends Bob $358+23/123 - \left \lceil{358+23/123}\right \rceil=358+23/123 - 359$ which is not the same as you provided. – mike russel Feb 10 '15 at 13:58
• No, Alice does not send $358\diamond 23/123$ (for one, $358$ is not a member of the group $G$); instead, she sends $23/123\ \diamond\ 23/123\ \diamond\ 23/123\ \diamond ... \diamond\ 23/123$ – poncho Feb 11 '15 at 4:50
• $23/123\diamond 23/123$ is $46/123 - 1$ ? – mike russel Feb 11 '15 at 5:49
• Never mind I got it. $23/123$ multiplied $358$ times mod $123 = 116/123$ Although i don't understand why it's mod 123 – mike russel Feb 11 '15 at 6:00
Eve shouldn't be able to find the shared secret easily from the messages Alice and Bob send.
In the questions 1 and 2, you said that the shared secret is the sum/the product of the two messages, but anyone can compute them. Therefore, you are wrong. The shared secret should be:
1) $B=(128)(65)2 = (65)(128)2\mod p$,
2) $B=(2^{(65)})^{(128)}=(2^{(128)})^{(65)}\mod p$.
Once you understood these two examples, I am sure you will be able to solve the 2 other questions easily by thinking a bit more about it.
• But for the second one, $2^{(65)^{(128)}} = 2^{(65)(128)}$. What confuses me about 3 is the primitive root $g$. I can't think of a way to express it other than what i came up with. – mike russel Feb 10 '15 at 13:42 | 1,054 | 3,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-05 | latest | en | 0.772796 |
https://rambamhospital.com/ratios-and-proportions-worksheet/ | 1,656,527,191,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00142.warc.gz | 532,202,090 | 15,599 | # Ratios And Proportions Worksheet
Ratios And Proportions Worksheet. It can come many forms such as 3 out of. Our proportions worksheets review whole number and decimal proportions as well as provide simple proportion word problems.
These worksheets for grade 6 mathematics ratio and proportion cover all important topics which can come in your standard 6 tests and examinations. Ratio and proportion worksheets on this page you will find: A ratio is s a fraction like 3/4.
### These Worksheets For Grade 6 Mathematics Ratio And Proportion Cover All Important Topics Which Can Come In Your Standard 6 Tests And Examinations.
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### Write Each Of The Following Rates As A.
Ratios can be written by using the word 'to', 3 to 2, by using a colon:, 3 : Some important facts about ratio and proportion worksheet for class 6 for comparing quantities of the same type we commonly use the method of taking difference between the quantities. Added specifically, it reflects the adeptness of actor disinterestedness to awning all outstanding. | 326 | 1,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.900696 |
https://byjusexamprep.com/tips-to-prepare-trigonometry-for-nda-exam-i | 1,686,435,115,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646350.59/warc/CC-MAIN-20230610200654-20230610230654-00269.warc.gz | 175,587,285 | 59,366 | # Tips to Prepare Trigonometry for NDA exam
By Dhruv Kumar|Updated : May 17th, 2021
Trigonometry section of Elementary Mathematics carries the 2nd highest weightage in the NDA exam, after Algebra. So, it is important that one is familiar with the preparation tips & tricks of trigonometry for the NDA exam, to score good marks in the mathematics section.
With the help of this post, we will help the aspirants understand how they should prepare trigonometry for the mathematics section of the NDA exam. So, read these NDA trigonometry preparation tips now to score full marks in this topic. We have also covered the important trigonometric identities for the NDA exam, which the students must remember.
The subtopics in trigonometry are angles and their measure in degree and in radians, trigonometric ratios, sum and difference formulae, trigonometric identities, multiple and sub-multiple angles, properties of a triangle, inverse trigonometric functions, and applications of trigonometry.
The candidate should be well informed about the correct syllabus, that is, the topics and sub-topics and the marks distribution or weightage, before proceeding with his study plan to avoid any confusion. Solving at least 70% of the mathematics section with 90% accuracy will be beneficial for the candidate.
## How to prepare Trigonometry for the Mathematics section of NDA exam
One can follow the under-mentioned tips for proper preparation of the trigonometry in the maths section of the Combined Defense Services exam!
### 1. Memorize certain trigonometric values
The candidate should try remembering trigonometric values such as Sin 18, Cos 18, Sin 36, Cos 36, etc. In addition to this, he should also memorize the trigonometric values of angles such as 0, 30, 45, 60, 90 degrees.
### 2. Opt for shortcuts
• Trigonometry is one of those topics wherein shortcuts are extremely handy.
• The candidate must learn various tricks and shortcuts for better time management.
• For instance, to remember the table of trigonometric values, the candidate must only memorize the values of Sin 0, Sin 30, Sin 45, Sin 60, Sin 90. All other values for those angles, that is, Cos, Tan, Cot, Sec, and Cosec can be derived from the sine value as given below:
• Tan A = Sin A / Cos A
• Cot A = 1 / Tan A
• Sec A = 1 / Cos A
• Cosec A = 1 / Sin A
### 3. Prepare a weekly schedule ahead of every week
• Mathematics is a subject that one can get a good command over by practising it daily.
• The candidate should prepare the entire week’s schedule in advance.
• He should ensure that at least one sub-topic each from all the high weightage topics, including trigonometry, is solved daily in order to ensure consistency and to avoid losing touch.
• The candidate should start with the difficult topics first if he has more time; else he should start by strengthening those topics that he finds easier and then move on to the difficult ones and look for shortcuts to understand them.
### 4. Prepare a revision sheet for formulae
• The candidate should prepare a revision sheet by writing down all the trigonometric formulae in a systematic manner to aide in quick referencing and last-minute preparations.
### 5. Solve mock papers and previous year's question papers
• Solving mock papers will give the candidates an estimate of the amount of time to be spent per question and also help the candidate decide if he needs to work on his time management.
• Previous years’ question papers will help the candidate get an idea about the paper pattern and the weightage given to each sub-topic under trigonometry.
• A regular NDA & NA paper has about 10-15 questions from trigonometry, with a majority of weightage being usually given to measurement of angles and trigonometric ratios.
• The candidate should be well informed about the context and sub-topic that these questions are derived from.
• Test simulators are another way to get a real-time experience of the NDA and NA examination. These simulations ensure that the candidates get a feel of the exam beforehand.
### 6. Approximations and calculations
• The candidate should work on improving his calculations by practising mental arithmetic and quick math.
• This includes memorizing squares and cubes of the first 10-15 digits, multiplication tables to 25, etc.
• The candidate should also use approximations wherever possible. If the values given in the options deviate to a great degree, approximation comes in handy to get the solution faster.
• However, if the values given in the options do not deviate much, the candidate should use the accurate values for calculation.
### 7. Use reference books or guides
While NCERT books give a brief but comprehensive idea about each topic, other reference books can also be used for a more immersive and in-depth information about various topics. We have already shared the list of useful books for different sections of the NDA exam. Go through it to know the sources you can refer to prepare this topic.
## Important Trigonometric Identities for NDA Exam
• Sin2 A + Cos2 A = 1
• 1 + Tan2 A = Sec2 A
• 1 + Cot2 A = Cosec2 A
• Sin (A+B) = Sin A Cos B + Cos A Sin B
• Sin (A-B) = Sin A Cos B – Cos A Sin B
• Cos (A+B) = Cos A Cos B – Sin A Sin B
• Cos (A-B) = Cos A Cos B + Sin A Sin B
• Tan (A+B) = (Tan A + Tan B) / (1 – Tan A Tan B)
• Tan (A-B) = (Tan A – Tan B) / (1 + Tan A Tan B)
• Cot (A+B) = (Cot A – Cot B) / (1 – Cot A Cot B)
• Cot (A-B) = (Cot A + Cot B) / (1 + Cot A Cot B)
• Sin 2A = 2 Sin A Cos A
• Cos 2A = Cos2 A – SinA = (2 Cos2 A – 1) = (1 - SinA)
• Tan 2A = 2 Tan A/ (1 – Tan2 A)
• Sin A Cos B = ½ (Sin (A+B) + Sin (A-B))
• Cos A Sin B = ½ (Sin (A+B) - Sin (A-B))
• Cos A Cos B = ½ (Cos (A+B) + Cos (A-B))
• Sin A Sin B = ½ (Cos (A-B) - Cos (A+B))
• Sin 3A = 3 Sin A – 4 SinA
• Cos 3A = 4 Cos3 A – 3 Cos A
• Tan 3A = (3 Tan A – Tan3 A) / (1 – 3 Tan2 A)
All the best for your exams, | 1,513 | 5,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-23 | latest | en | 0.8798 |
http://www.reddit.com/r/javahelp/comments/1kzj15/theory_only_conceptually_how_could_i_break_a/?sort=old | 1,433,333,610,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036742.18/warc/CC-MAIN-20150601214356-00084-ip-10-180-206-219.ec2.internal.warc.gz | 626,700,911 | 16,864 | [–] 1 point2 points (11 children)
sorry, this has been archived and can no longer be voted on
Well, what are the tokens? Are they meant to be abstract? I don't understand the idea of arrays discouraged either, is it just so you need to think of a new way of storing the tokens?
But with the information I'm given, I'm going to assume that a token is a set of unbroken integers - so "11a4324taau324a" would have "11","4324" and "324" as tokens.
Since I don't understand the assignment, I can't get much farther - but if you're not allowed to use any kind of bucket to store the tokens, then you have to store some kind of representation of the tokens. Look at that string - if you could use arrays, it would be a simple matter of throwing all those numbers into an array separately. but is there any way you can think of to store a representation of those numbers in a single variable?
I really can't tell how vague you were accidentally (if at all) vs how vague the professor is trying to be, but please try to post more information and we can try to help better, this assignment is far to open ended to get anywhere on right now, as everything is up to interpretation
[–][deleted] (10 children)
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[deleted]
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ah, good, I can work with this then. There's a couple of ways you can do this, I'm going to try to push you toward the way I think is easier.
I'm going to assume that getting to the part of the program where you have the single line is doable for you, and start discussing it from the point you've pulled out your "1232 2312 0080" line.
Using one of the functions above will let you know when you've hit something that signifies the end of a number. Say you've already parsed your first number, and you've just got to the end of 2312. At this point, you know that the second number ends after 9 characters into the string...if you also knew where that number started, you could use a function to pull out just those four numbers, couldn't you?
And if the program output is exactly like you say - you don't even have to store that 2312 anywhere. After you print it out, if you just add it to your current 'sum' - that's all you'll need it for.
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[deleted]
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Are you still stuck on this?
Would I need to store it as a String before using on of those operations?
Well, Integer.parseInt(String someNumberString) takes a String in as an argument. So whatever String you pass into parseInt has to be a String representation of a single number.
getting the input from scanner and printing it back out using System.out, I did made a method with char and isDigit to return a boolean, but I am not sure if that will end up working out for me in the end.
That's a good start.
Have you tried stepping through the input String one character at a time, yet? That's what you should try to code next, if you haven't already.
[–][deleted] (5 children)
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[deleted]
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Ah, I'm sorry to hear that. Are you still interested in solving the problem?
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[deleted]
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Do you want code, or do you want to step through it in english?
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[deleted]
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Based on what the end goal is (a sum of numbers) you don't really need an array of the tokens anyway.
Basically i would loop through the characters in the string and have a state flag used for telling whether the current character is a "boundary character" or an "input character". The boundary is whitespace and the input is everything else. When you reach a boundary character, everything in between the last boundary position (or the start of the line) and the current position is your current token. Parse the token as an integer and add it to your running sum.
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[deleted]
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Just a boolean condition that indicates if you are at a boundary character or not. You can just print the tokens as you find them.
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As for the no-arrays specification, you might be able to use a `StringBuilder` (or `StringBuffer`) and append what would be each part of the array to the `StringBuffer.` You might separate the sections of the array with a non-printing character (like a newline, or something else, so long as it isn't a character you might be putting in the `StringBuffer.` Sorry if that's unclear, but I didn't want to post code, which would have made it much clearer. I could post a sample `String` if OP says that's allowed.
Edit: You might even be able to write a class that handles this behavior. A class `Array` (possibly generic, if that's needed) that has methods you would normally use to modify an array. | 1,347 | 5,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2015-22 | latest | en | 0.973872 |
https://gmatclub.com/forum/last-year-a-record-number-of-new-manufacturing-jobs-were-50260.html | 1,487,954,655,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171620.77/warc/CC-MAIN-20170219104611-00099-ip-10-171-10-108.ec2.internal.warc.gz | 725,162,125 | 64,874 | Last year a record number of new manufacturing jobs were : GMAT Critical Reasoning (CR)
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# Last year a record number of new manufacturing jobs were
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Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 08:37
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Last year a record number of new manufacturing jobs were created. Will this year bring another record? Well, a new manufacturing job is created either within an existing company or by the start-up of a new company. Within existing firms, new jobs have been created this year at well below last year’s record pace. At the same time, there is considerable evidence that the number of new companies starting up will be no higher this year than it was last year, and surely the new companies starting up this year will create no more jobs per company than did last year’s start-ups. Clearly, it can be concluded that the number of new jobs created this year will fall short of last year’s record.
In the argument given, the two portions in boldface play which of the following roles?
(A) The first is a prediction that, if accurate, would provide support for the main conclusion of the argument;
the second is that main conclusion.
(B) The first is a prediction that, if accurate, would provide support for the main conclusion of the argument;
the second is a conclusion drawn in order to support that main conclusion.
(C) The first is an objection that the argument rejects; the second is the main conclusion of the argument.
(D) The first is an objection that the argument rejects; the second presents a conclusion that could be drawn
if that objection were allowed to stand.
(E) The first is a claim that has been advanced in support of a position that the argument opposes; the second
is a claim advanced in support of the main conclusion of the argument.
[Reveal] Spoiler: OA
Last edited by WaterFlowsUp on 14 Sep 2014, 05:37, edited 1 time in total.
The question has been changed to the proper OG version.
If you have any questions
New!
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 08:58
ajisha wrote:
Last year a record number of new manufacturing jobs were created. Will this year
bring another record? Well, any new manufacturing job is created either within
an existing company or by the start-up of a new company.
Within existing firms,
new jobs have been created this year at well below last year’s record pace. At the
same time, there is considerable evidence that the number of new companies starting
up this year will be no higher than it was last year and there is no reason to think
that the new companies starting up this year will create more jobs per company
than did last year’s start-ups.
So clearly, the number of new jobs created this year
will fall short of last year’s record.
In the argument given, the two portions in boldface play which of the following
roles?
A. The first provides evidence in support of the main conclusion of the argument;
the second is a claim that argument challenges.
B. The first is a generalization that the argument seeks to establish; the second is
a conclusion that the argument draws in order to support that generalization.
C. The first is a generalization that the argument seeks to establish; the second is
a judgment that has been advanced in order to challenge that generalization.
D. The first is presented as obvious truth on which the argument is based; the
second is a claim that has been advanced in support of a position that the
argument opposes.
E. The first is presented as obvious truth on which the argument is based; the second is a
judgment advanced in support of the main conclusion of the argument.
Between D and E, I feel E is more appropriate..
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 09:09
i was also down to D or E!!! picking E although I'm not 100% sure...
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 09:13
Straight E...the argument doesnt oppose the final position...it begins by asking an answer to a question and finally ends up taking a particular position after discussion.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 09:24
vineetgupta wrote:
Straight E...the argument doesnt oppose the final position...it begins by asking an answer to a question and finally ends up taking a particular position after discussion.
Well, IMO, the argument does seem to be OPPOSING an implicit assumption that because last year the jobs were great so this year too they would be great. The argument seems to be advocating a stance against this position (though this position has not been stated anywhere in the passage) - but the TONE implies it.
Anyway - E finally rules.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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09 Aug 2007, 09:35
dwivedys wrote:
vineetgupta wrote:
Straight E...the argument doesnt oppose the final position...it begins by asking an answer to a question and finally ends up taking a particular position after discussion.
Well, IMO, the argument does seem to be OPPOSING an implicit assumption that because last year the jobs were great so this year too they would be great. The argument seems to be advocating a stance against this position (though this position has not been stated anywhere in the passage) - but the TONE implies it.
Anyway - E finally rules.
You are right...the author uses a negative tone to support his conclusion.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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12 Aug 2007, 00:41
ajisha wrote:
Last year a record number of new manufacturing jobs were created. Will this year
bring another record? Well, any new manufacturing job is created either within
an existing company or by the start-up of a new company.
Within existing firms,
new jobs have been created this year at well below last year’s record pace. At the
same time, there is considerable evidence that the number of new companies starting
up this year will be no higher than it was last year and there is no reason to think
that the new companies starting up this year will create more jobs per company
than did last year’s start-ups.
So clearly, the number of new jobs created this year
will fall short of last year’s record.
In the argument given, the two portions in boldface play which of the following
roles?
A. The first provides evidence in support of the main conclusion of the argument;
the second is a claim that argument challenges.
B. The first is a generalization that the argument seeks to establish; the second is
a conclusion that the argument draws in order to support that generalization.
C. The first is a generalization that the argument seeks to establish; the second is
a judgment that has been advanced in order to challenge that generalization.
D. The first is presented as obvious truth on which the argument is based; the
second is a claim that has been advanced in support of a position that the
argument opposes.
E. The first is presented as obvious truth on which the argument is based; the second is a
judgment advanced in support of the main conclusion of the argument.
Can we please have the OA? Thanks.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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13 Aug 2007, 22:35
agree with E
First is presented as a general statement
and the second supports the argument
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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14 Aug 2007, 05:47
Tough
E
I went against D because the guy isn't opposing an argument, he's making one.
I hate these questions, after a while the letters start to float around, as they do in Alphabetti's soup...... yah...... Mental Endurance
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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14 Aug 2007, 20:03
Good question.
I can see picking D after being mentally exhausted and halfway through GMAT.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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24 Aug 2014, 04:09
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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31 Aug 2014, 04:53
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Hey guys, it's a modified question from the OG 13 --> it's not the original version. In the original version other parts of the sentence are underlined and the correct answer is A.
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Re: Last year a record number of new manufacturing jobs were [#permalink]
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14 Sep 2014, 05:38
@BrainLab : thanks for bringing this to the notice. I have made necessary changes.
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15 Feb 2015, 03:33
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to be honest I've stoped searching after choice A. It has stated precisely the structure of the bold face part.
1. Prediction: the new companies starting up this year WILL
2. Conclusion: CLEARLY... the number of new jobs created this year will fall short of last year’s record.
--> A CORRECT: First is a prediction that supports the conclusion, second is this conclusion.
And I think it's not a Sub 600 Level question. Describe a role questions can be easily attributed to 600-700 and above Level questions, taking in to account, that one has to read a text for about 40 sec.....
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Last year a record number of new manufacturing jobs were [#permalink]
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14 Jan 2016, 10:11
Hey guys,
bringing this topic back to life with one question:
how can you start a prediction with word "surely" as in "and surely the new companies starting up this year will create no more jobs per company than did last year’s start-ups". This is what threw me off in A) and I hate myself for that as it is for sure the best answer. I also know that "surely" isn't underlined - does that mean that it should not be taken into consideration in "describe the role" questions?
Please kindly advise as this is not the first time I've made a mistake because of applying a bit of logic. Surely and prediction just don't make any sense to me, however you twist it and turn it.
Happy to show my appreciation with kudos!
Thanks guys,
Jay
Intern
Joined: 15 Sep 2015
Posts: 7
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Kudos [?]: 0 [0], given: 0
Re: Last year a record number of new manufacturing jobs were [#permalink]
### Show Tags
13 Feb 2016, 08:35
Hi Jay.. its not a fact "surely" does not mean it has happened and can be taken granted. Its is predicted that "surely" it will happen. In Gmat if you see words like It "must be true", "Surely will happen" are all claims and since it has not happened yet its a prediction. Hope this helps
Re: Last year a record number of new manufacturing jobs were [#permalink] 13 Feb 2016, 08:35
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# Last year a record number of new manufacturing jobs were
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,905 | 15,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-09 | longest | en | 0.926583 |
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Maths Arithmetic Progression Sequence To Find The Term in Number Pattern To See Whether The Sequence is an A.P. or Not nth Term of an A.P. / Sum of n terms of an A.P. Problems Related to tn and Sn Word Problems Circle Circles Passing Through One, Two and Three Points Secant and Tangent Tangent Theorem and Its Converse Theorem: About Tangents Drawn From External Point Theorem of Touching Circles Arc of a Circle–Major Arc, Minor Arc Congruence of Arcs Theorems Related to Congruent Arcs and Congruent Chords Inscribed Angle, its theorem and Interrupted Arc Theorem of Cyclic Quadrilateral and Its Converse Theorem of Angle Between Tangent and Secant, Converse of Tangent Secant Theorem Theorem of Internal Division of Chords / External Division of Chords Tangent Secant Segments Theorem English Idioms & phrases Intelligence / Mental Ability Verbal Test Blood relations Puzzle test Logical sequence of words Verification of truth of the statement Analytical reasoning Problems on clocks Situation reaction test Logical reasoning Science Heat Environmental Management Towards Green Energy General Knowledge / Current Affairs History Mass media and history Nationalism in Europe and India Political Science Political parties Democracy & diversity Geography Natural vegetation and wild life Water resources & agriculture | 276 | 1,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-16 | latest | en | 0.764649 |
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# winter 2004 2nd mt - CHEM 14B" YOUR NAME Instructor Dr...
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Unformatted text preview: CHEM 14B '" YOUR; NAME Instructor: Dr. Laurence Lavelle WINTER 2004 2ND MIDTERM (Total number of pages = 7) (Total points = 50) (Total time = 50 mins) YOUR DISCUSSION SECTION .................................... .. YOUR TA is: Franklin Ow Sam Ho Carlos Hernandez Mo Chuautemoc Arellanes Write in pen. Show all your work. Check your units and significant figures. Think clearly. Good Luck. Constants and Formulas Planck constant, h = 6.63 x 10—34 J - s » Avogadro constant, NA = 6.02 x 1023 mol'1 Gas constant, R = 8.314 J.K‘1.mol'1 = 8.206 x 10'2 L.atm.K".mol'1 = 62.364 L.Torr.K'1.mol’1 Speed of light, 0 = 3.0 x 108 m.s’1 Faraday's constant, F = 96,485 C.mol‘1 Water, specific heat capacity =4.18 .J.°C".g'1 0°C=273.15K 1L=1drn3 iatm:10‘l.325 kPa 71:3.14 E=hv AE=q+w q=nCAT Wz—PXAV SszInW PV=nRT w=—V1IV2PdV = -nRTin% AS = 9% 1 ASTI T2 = Tiff? 99% = n c In £2 AGO = AH" — T A80 AGoz—RTIHK AG=AG°+RTIHQ AGO = - n I: E0 A125 0C, ECELL = E0 - 00:92 LOG Q d A 0.693 d A Tkfi = —kdt In [A]=-kt +ln [A10 t1,2=T Th; = kdt 1——i<i1 t-1 dA—kd A-kt A [A] - + [A10 1/2— [A]O [I— ‘ I [I" + [IO tugzLSJfi k:Aexp(:fiE-?) lnk=—%+lnA QlA. For the cell diagram Pt li~H2(g),H+(aq) Cu2+taq) Cu(s) Write the half—reaction that occurs at the anode? (3pt) H2(g) —> 2H+(aq) + 2e— 9711‘ away M‘W B. Given: Ag+(aq) + e— —> Ag(s) E° = 0.80 V Fe3+(aq) + e- —> Fe2+(aq) E° = 0.77 v Cu2+(aq) + 2e- —-> Cu(s) E" = 0.34 v Which is the strongest reducing agent? (3pt) (Note: Reducing agent is asked for, not reducing agents.) Cu C. Consider the following reaction: 2Ag+(aq) + Cu(s) —9 Cu2+(aq) + 2Ag(s) If the standard reduction potentials of Ag+ and Cu2+ are +0.80 V and +0.34 V , respectively, calculate the value of E0 for the given reaction. (4pt) 0 wt 2 A97ny 1% 3 1/43 {5) E U.) Q2. Balance the following equation, using oxidation and reduction half—reactions. The reaction occurs in a basic solution. Identify the oxidizing and reducing agent. (lOpt) Reaction of bromine in water: ’Br2(é) 9 BrO3‘(aq) + Br" (aq) 0 7"5’ ‘7 M) W: 51 M) + 2t“ —> 2 47%) QMW are) ——7 Maj-W) r/oe— gal) W @W 0 gnu/«d + {Map Maw/o; (W '44“ 0/4 W fl 5/41) 7L M2 0%] f/Zbflay A? 25%}W)::Zfio£/ 06 Wu WWW {[fl’rb/fl) +Ze’/->Z&2flwj 7L/Z "9 65%7M 7L/fltdfl/ +/0€,_ tL/Z 5/1?“ fi/flg/ng +Z£a;/W) +5520“) ‘5' Z (M) 2 e w) + 4 W "e few be m flea/La fiz4a/Zfflfl'" ‘ WWW” Q3A. Given: 4F62+(aq) + 02(ao) + 2H20(l) a 4Fe3+(aq) + 4OH—(aq) 2pt) and rate = k[Pe2+][OH-]2[02] What is the overall order of the reaction 4_ / What is the order with respect to 02 1__ /W B. If the rate of a reaction increases by a factor of 64 when the concentration of reactant increases by a factor of 4, the order of the reaction with respect to this reactant is ___3 (2pt) ¢fea¢ C. The concentration—time dependence for a first—order reaction is given below. 1 MI” Aloha: concentration of reactant. M! ——+ Qt > WNW Nummmw i Ullt’ —---) At which point on the curve is the reaction fastest? A (2pt) D. The concentration—time dependence is shown below for two first order reactions. Which reaction has the greatest t1/2? Clearly indicate your answer with an arrow to the upper or lower curve. (4pt) w“ A \ 'l’mn- -—> upper curve Q4A. For the reaction cyclopropane(g) propene(g) at 500°C, a plot of ln[cyclopropane] vs t gives a straight line with a slope of ~0.00067 5’1. What is the order of this reaction and what is the rate constant? first—order / 711‘ slope = —l< k: 6.7 x10-4 s—1 7, fli’ v/M‘Uifl MM B. For the reaction HO(g) + H2(g) a H20(g) + H(g) a plot of lnk versus 1/T gives a straight line with a slope equal to —5.1 x 103 K. What is the activation energy for the reaction? (4pt) (4P0 54 r K (Km/07%) = 5-3/4; I WM” (fix/5% __ c; ’ / I >00 MAW 1%” E4 : ééz/trmfl C. Given: CH4(g) + Clg(g) —9 CH3Cl(g) + HCl(g) (2pt) The rate law for this elementary process is rate = 1<[CH4][C12l QSA. An elementary process has an activation energy of 40 kJ/mol. If the enthalpy change for the reaction is 30 kJ-mol"1, what is the activation energy for the reverse reaction? (Spt) awe/«r.w€'/ F 54 {7%}:gw’f0 agar”? 4H=30//<J.Mo{ / B. ’ Consider the following reaction: 2N20(g) —> 2N2(g) + 02(g) rate = k[N20] For an initial concentration of N20 of 0.50 M, calculate the concentration of N20 remaining after 2.0 min if k = 3.4 x 10-3 s-l. (Spt) 1.0% 3 /20/¢/“C ...
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# Отзывы учащихся о курсе Basic Statistics от партнера Амстердамский университет
4.7
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Оценки: 2,447
Рецензии: 626
## О курсе
Understanding statistics is essential to understand research in the social and behavioral sciences. In this course you will learn the basics of statistics; not just how to calculate them, but also how to evaluate them. This course will also prepare you for the next course in the specialization - the course Inferential Statistics. In the first part of the course we will discuss methods of descriptive statistics. You will learn what cases and variables are and how you can compute measures of central tendency (mean, median and mode) and dispersion (standard deviation and variance). Next, we discuss how to assess relationships between variables, and we introduce the concepts correlation and regression. The second part of the course is concerned with the basics of probability: calculating probabilities, probability distributions and sampling distributions. You need to know about these things in order to understand how inferential statistics work. The third part of the course consists of an introduction to methods of inferential statistics - methods that help us decide whether the patterns we see in our data are strong enough to draw conclusions about the underlying population we are interested in. We will discuss confidence intervals and significance tests. You will not only learn about all these statistical concepts, you will also be trained to calculate and generate these statistics yourself using freely available statistical software....
## Лучшие рецензии
##### PG
Apr 21, 2016
This is a nice course...thanks for providing such a great content from University of Amserdam.\n\nPlease allow us to complete the course as I have to wait till the session starts for week 2 lessions.
##### CD
Mar 06, 2016
This course is really awesome. Designed well. Looks like a lot of efforts have been taken by the team to build this course. Kudos to everyone. Keep up the good work and thank you very much.
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## 26–50 из 605 отзывов о курсе Basic Statistics
автор: Susan M
Jun 23, 2016
Uses R with no explanation. Why does it use a challenging programming language that is not the point. The point is to learn statistics.
Use Statcrunch - so students can focus on statistics not programming
автор: Keith C
Jul 05, 2018
автор: Alejandro N J
Jun 23, 2016
Excellent MOOC. A great introduction to the basic elements of statistics. I had studied statistics years ago in my bachelor but almost forgot everything. This course greatly helped me revive all the elementary concepts. The teachers explain complex things in simple terms easy to follow, and the funny examples have brought a smile to my face more than once and twice during the MOOC. Finally, the collaboration with professional illustrators adds great value to the course. A big deal of effort has gone into its making, and it truly pays off. I just wished there'd be an intermediate and advanced statistics course to keep digging into it!
Thanks a lot!
Alejandro
автор: Stacy H
Jun 30, 2016
I hadn't touched math in any real way in years and this was a great re-introduction. The videos were fun and engaging, and while the material was challenging, I felt I got a lot out of it. I highly recommend it for people who need statistics for their work or for a degree program but who are feeling a little intimidated by it.
As far as feedback goes, I would have liked to get explanations for quiz questions I couldn't answer. I wish there was a way to say, "This is the quiz score I'd like to accept; now please explain to me the ones I missed." The lack of feedback on the quizzes was frustrating at times.
автор: gerald k
Apr 19, 2016
This is a great course. It was challenging but not discouraging. I think that there is a high probability that I learned something very useful. Now that I have said the most important words now I would like to give a thumbs down on the Data Camp R section. I found it to be frustrating and discouraging. You can find better information and tutorials on YouTube. Check out"MarinStatsLectures. A text book is also a great help. I bought Elementary Statistics at a used book store for \$4.00 US. And it correlated very nicely with the course. If you are interested in statistics this course is a good first step.
автор: Jacob S
Jun 18, 2016
Really a great introductory course. I am a real beginner in statistics, since I need it to make it easier for my self at data mining and machine learning in the future, this course gives me a solid basic understanding of terminology, calculations and relationships between calculations. Week 3 and 4 are pretty hard in my opinion, but after doing the course learning how to learn I know how to handle it, so I will recommend everybody to do that one also if you think this looks too difficult for you.
автор: Aurangazeeb A K
Jan 06, 2019
One of the best courses on Coursera and I'm saying this with an experience of 10+ courses on Coursera. The teaching strategy is brand new, with some surprises from the awesome teachers ! Statistics now feels like child's-play. Thank you University of Amsterdam for making this amazing course. The animations gives a fresh feel towards the concepts. After starting this class, I'm more curious and passionate about statistics. One more thing, be ready for a surprise at the first video of week 6.
автор: Elena K
Sep 30, 2016
A really great course. Simple to follow and understand, but provides insights into the basics of statistics. Almost no math which is good news for non-experts and makes the course more accessible for a broader audience. The lecturers explain the intuition behind the fundamental things in statistics, which I find very important before studying the statistical tests (inferential statistics). Feels like it opened my eyes.
Thank you very much Matthijs Rooduijn and Emiel van Loon!
автор: Leitha M
Feb 27, 2016
I've always had a rough time with mathematics, but the way this course is structured makes it easy for me to go at my own pace, re-watch videos that were confusing the first time through and work through problems until I really felt like I understand how the formulas work. The consistent use of exercises in R were extremely useful as well. I'd highly recommend this course and learning system for people like me, who never thought they'd be able to grasp mathematics.
автор: Alexander M
Aug 08, 2017
Thank you for the instruction and clear overview for this course. Coming from a HR standpoint (where ocassionally Statistics are a must) it was easy to follow the chapters week by week. As in any course there were some challenging sections to redo or retry but great learning and application to the real world. Highly recommend for anyone in a math background or looking to improve a skill set in numbers.
автор: Eric M E
Mar 22, 2016
Simple explanations that make the topic rather easy to learn and having a visual aid (the drawings) I find it to be so much more friendly that I've really come to enjoy what I'm learning (I'm mostly a visual learner so to me this is very good). So far it has been a great course with R labs that can help reinforce the ideas and give a bit more practice in the subject, I do recommend the course.
автор: SRI R R
Mar 02, 2016
This is a very informative and good course for beginners who wants to explore basic statistics along with R lab. Ofcourse, I felt little tought when i tried to do R lab but finally got it!! Overall course was very informative and would definitely recoment for those who wants to learn statistics
thanks for professors and the team of University of Amsterdam for conducting this course.
автор: Dorien H
Mar 09, 2016
Very good course, the explanations are made very accessible through understandable and sometimes funny examples.
Learned many of new things, despite having finished a course on statistics in my university. Especially the R component helped me greatly; despite already having some experience, the datacamp modules were very helpful.
Oh, and great Dutch accents :)
автор: Chitvan K P
Jul 01, 2018
Excellent starter for learning statistics. Looking forward to learn Inferential Statistics next. This course covers a wide variety of material ranging from basic Probability Distributions to Formulating and Testing hypothesis, confidence intervals and regression methods. Great starter to stats and looking forward to learn Inferential stats too.
автор: Maryluz H
Oct 15, 2017
It is a great course. I don't have much experience or knowledge in statistics, but I found it useful. I still need to review concepts that are still not clear, but the professors are good in the explanations and examples. I wish I had had more time for everything, and didn't have to rush because of the payment.
Thanks for the great course!
автор: Jimmy
Nov 29, 2017
One of the best lecture I've ever attended online! Although sometime the pace was a bit off, yet overall this course has offered enough coverage and clear examples for basic stuff. Great sense of humour for the lecturer (Matthijs Rooduijn) also helped in making this course to be not such a boring subject. Well done, keep up the good work!
автор: Jorge A A B
Apr 26, 2016
Excelente curso, muy bien explicado a traves de una manera clara e intuitiva, dando razon a cada uno de los conceptos utilizados en la estadistica. Conocer los conceptos estadisticos, asi como su significado, calculo y uso, no solamente mejora el analisis cientifico sino ademas la capacidad de comunicacion dentro del comunidad cientifica.
автор: Andrei
Apr 22, 2017
Very easy to comprehend course. Both presenters put extra effort in relaying the material and introducing comical relief along the way. I can say now that I understand more above z-scores and t-scores that I ever did at university! I would highly recommend the course for anyone feeling a bit scared than facing statistical information
автор: Kate W
Mar 04, 2016
I never thought I would learn so much so quickly in a math course. The course moves fast, but being able to rewind the videos and repeat anything I don't understand is really helpful! I love the applicability of what I'm learning, and my ability to take notes and practice everything on the quizzes makes the information stick.
автор: pkavpro
Jul 01, 2019
I would like to thank you all professors and those who participates in creating this amazing course. It helped me a lot in building basic knowledge about statistics. The course is very well structured with clear & easy-to-understand content. I highly recommend this course for those who starts learning statistics.
автор: xavier C
Jan 07, 2017
This course is very well prepared, thanks to the video and the commitments of the 2 teachers,
There is also a good balance between Theory and Practice (usage of R).
I do already use the learning of this course in my daily job (data & analytics)
Good job and thank you to the entire team who has prepared this course.
автор: Lai K C
Feb 11, 2016
Both of the lecturers as well as the staff are excellent in making the learning simple, exciting and interesting. I have learned so much about statistics in so short a time with effectiveness and efficiency. Thanks for the excellent work. And all of you have made University of Amsterdam proud and stand tall.
автор: Gregorio A A P
Jul 07, 2017
Un super curso , excelente y felicitaciones por vuestro gran trabajo, pero es muy triste no poder disfrutar al 100% un curso de esta calidad traducido en el idioma español, les pido por favor si fueran tan gentiles de acceder a esta humilde petición, gracias por anticipado y nuevamente felicitaciones.
автор: Alireza N
Jan 05, 2016
I found this course very helpful and understandable ...The slides are very beautiful designed ..I had never learnt Statistics in this way! The examples are related to our daily life! It makes everything easier to learn... Thank you very much all who works on this course from University of Amsterdam
автор: Daniela A B M
Jan 08, 2018
I hadn't studied Statistics for over 22 years, and I am very grateful that I did it through this course. While I had to spend more time than I had anticipated going over the materials (due to my lack of practice), I did found the lectures and materials were well prepared and didactic. Thank you!! | 2,767 | 12,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-05 | longest | en | 0.865001 |
https://thecustomwriting.com/category/homework-help/page/7695/ | 1,627,676,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00004.warc.gz | 573,127,211 | 12,883 | ## Socw 6311 & 6070 wk 4 responses
RESPONSE 1: Respond to at least two colleagues by explaining how that colleague might rule out one of the confounding variables that they identified. Colleague 1: Debby Being able to look at the different designs and choosing the right design for the information necessary to give an accurate accounting is imperative. Looking at the variables and outcomes wanting to be measured is also an important part of choosing a statistical design. The outcome of […]
## benchmark – framework findings and recommendations
Benchmark – Framework Findings and Recommendations This assignment serves to benchmark competency 2.1: Establish a risk management framework using industry standards for compliance. Based on an executive level report, deliver the findings of the Topic 4 “Demonstrating the Gap” assignment. Include the following in your report (add sections to the template as needed): An overview of why the report is being written A paragraph description of the system A paragraph outlining the framework governing […]
## According to the central limit theorem, if a sample of size 64 is
Question 1 According to the central limit theorem, if a sample of size 64 is drawn from a population with a mean of 56, the mean of all sample means would equal _________ . 7.0056.0064.000.875128.00 Question 2 According to the central limit theorem, if a sample of size 100 is drawn from a population with a standard deviation of 80, the standard deviation of sample means would equal __________ .0.808808000.080 Question 3 According to the central limit theorem, […]
## Stat intro | Statistics homework help
Part I(short essay/discussion) #1: Why do so many Americans struggle with mathematics/statistics? I have always believed that learning mathematics is similar to learning a foreign language. There are certain rules that you must learn and there are exceptions to the rules. You then have to use your logic. How to use logic is a process learned through practice. I assume that most of you took a course in geometry when you were in high […] | 436 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-31 | latest | en | 0.928398 |
https://socratic.org/questions/how-do-you-prove-cos-w-sin2w-cos-2w-sinw-1-cot-w | 1,582,082,966,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144027.33/warc/CC-MAIN-20200219030731-20200219060731-00349.warc.gz | 575,717,810 | 5,981 | # How do you prove (cos W - sin2W) / cos(2W) + sinW-1 = cot W?
Graph of $\frac{\cos W - \sin 2 W}{\cos} \left(2 W\right) + \sin W - 1$
Graph of $\cot x$ | 68 | 153 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-10 | latest | en | 0.542051 |
http://moss.cs.iit.edu/cs495/mp2.html | 1,534,562,583,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213264.47/warc/CC-MAIN-20180818021014-20180818041014-00367.warc.gz | 282,881,742 | 3,058 | # Machine Problem: Folds and HOFs
## Overview
The goal for this assignment is to get more comfortable with using folds in place of direct recursion, and with higher order functions. Both techniques are essential in functional programming.
## Working on the assignment
All your code for this assignment should go into a file named "MP2.hs" within the "`src`" directory in your repository. Continue to directly load your file into a GHCi session for testing.
## Exercises
In addition to any functions listed for each exercise, you may always use the following:
• Tuple construction & component access: (), fst, snd
• List construction & access: [], (:), head, tail, init, last, (!!), (++)
• Boolean operators: not, (&&), (||)
• Composition: (.)
• Arithmetic: (+), (-), (/), (*)
• Comparison: (==), (/=), (<), (<=), (>), (>=)
• HOFs: map, foldr, foldl
Warm-up exercises (1 point each). Implement each of the following functions using either a left or right fold. For these functions, use either function composition or a lambda expression to define the function used as the argument to fold (e.g., "`foldr (f1 . f2) v`" or "`foldr (\x ys -> ...) v`"). Also, endeavor to use point-free style.
1. `any' :: (a -> Bool) -> [a] -> Bool`
Returns true if the predicate returns true for any of the values in a list.
Examples:
``````> any' even [1..10]
True
> any' even [1,3..9]
False
> any' even []
False
``````
2. `all' :: (a -> Bool) -> [a] -> Bool`
Returns true if the predicate returns true for all of the values in a list.
Examples:
``````> all' even [1..10]
False
> all' even [2,4..10]
True
> all' even []
True
``````
3. `compose :: [a -> a] -> (a -> a)`
Returns the function created by composing all the functions in the given list.
Examples:
``````> compose [(*5), (/10), (+1)] 9
5.0
> compose [("hello, " ++), reverse, drop 1] "Michael"
"hello, leahci"
``````
4. `cycle' :: [a] -> [a]`
Returns an infinite list of the provided list, repeated.
Examples:
``````> take 10 \$ cycle' [1..3]
[1,2,3,1,2,3,1,2,3,1]
``````
5. `scanr' :: (a -> b -> b) -> b -> [a] -> [b]`
Returns the list of values as they would be produced by a corresponding right fold.
Examples:
``````> scanr' (+) 0 [1..10]
[55,54,52,49,45,40,34,27,19,10,0]
> scanr' (:) [] "hello"
["hello","ello","llo","lo","o",""]
``````
6. `scanl' :: (b -> a -> b) -> b -> [a] -> [b]`
Returns the list of values as they would be produced by a corresponding left fold.
Examples:
``````> scanl' (+) 0 [1..10]
[0,1,3,6,10,15,21,28,36,45,55]
> scanl' (flip (:)) [] "hello"
["","h","eh","leh","lleh","olleh"]
``````
7. `inits :: [a] -> [[a]]`
Returns all successive starting sublists of the given list, starting with the empty list.
Examples:
``````> inits "hello"
["","h","he","hel","hell","hello"]
> inits []
[[]]
``````
8. `tails :: [a] -> [[a]]`
Returns all successive trailing sublists of the given list, ending with the empty list.
Examples:
``````> tails "hello"
["hello","ello","llo","lo","o",""]
> tails []
[[]]
``````
Trickier exercises (2 points each). Again, implement the following functions using either a left or right fold. For these functions, it's likely that you'll need to pass a tuple into the fold, and extract a value from the (tuple) returned by the fold to return as a result. Endeavor to use point-free style.
1. `minmax :: (Ord a) => [a] -> (a,a)`
Returns the minimum and maximum (as a tuple) of a non-empty list.
Examples:
``````> minmax [5,2,1,3,8,4]
(1,8)
> minmax [3]
(3,3)
``````
Permitted functions: min, max
2. `gap :: (Eq a) => a -> a -> [a] -> Maybe Int`
Returns the integer distance between first appearance of two elements in a list.
Examples:
``````> gap 3 8 [1..10]
Just 5
> gap 8 3 [1..10]
Nothing
> gap 'h' 'l' "hello"
Just 2
> gap 'h' 'z' "hello"
Nothing
``````
3. `evalExpr :: String -> Int`
Evaluates a string containing a simple arithmetic expression using only single digit operands and (binary) + and - operators.
Examples:
``````> evalExpr "1"
1
> evalExpr ""
0
> evalExpr "1+2+5"
8
> evalExpr "9-5+3-8"
-1
``````
4. `words' :: String -> [String]`
Returns the space-delineated words found in a string as a list.
Examples:
``````> words' "a b c d"
["a","b","c","d"]
> words' " hello how are you? "
["hello","how","are","you?"]
> words' " "
[]
``````
5. `dropWhile' :: (a -> Bool) -> [a] -> [a]`
Removes elements from the beginning of a list that satisfy the predicate.
Examples:
``````> dropWhile' even [2,4,1,5,6,8]
[1,5,6,8]
> dropWhile' even []
[]
> dropWhile' ((<= 3) . length) \$ words' "I am fond of cake"
["fond","of","cake"]
``````
MP1, redux (2 points each).: Complete exercises 4-8 from MP1 again, but this time using either a left or right fold. For the vigenere exercise, you may additionally choose to use the `cycle` function. Endeavor to use point-free style.
## Submission
To submit your work, be sure your "`MP2.hs`" file is added to your repository, then commit all your changes and push your work to our shared BitBucket repository. | 1,557 | 5,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-34 | longest | en | 0.802678 |
https://proxies-free.com/tag/trees/ | 1,632,869,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00542.warc.gz | 511,513,774 | 14,192 | ## trees – Query sum of values between lo and hi for a stream of numbers
Suppose you have a stream of incoming numbers that you’re storing and at any given instant in time, you want to query the sum of values between a given `lo` and `hi` in the stream you’ve read thus far.
It it possible to solve this problem with segment trees?
The only solution I could come up with is either O(n) insertion and O(log n) query (using a sorted array and prefix sum) or O(log n) insertion and O(n) query (using a binary search tree).
## dnd 5e – Lightning Bolt and trees in DnD 5e
### The rules list a tree trunk as an example of 3/4 cover.
The rules for Three-Quarters Cover state:
A target with three-quarters cover has a +5 bonus to AC and Dexterity saving throws. A target has three-quarters cover if about three-quarters of it is covered by an obstacle. The obstacle might be a portcullis, an arrow slit, or a thick tree trunk.
So an character trying to hide behind a tree trunk that happened to be in the area of effect of lightning bolt would get +5 to their dexterity save. Now, as the DM, it’s really up to you how much cover the tree actually provides. I once had a player insist that all trees granted 3/4 cover, but unfortunately for him, these were really skinny trees, so I generously gave him half cover. It’s up to you to determine the character’s size in relation to the size of your trees, but I recommend sticking with 3/4 as the book suggests, unless you’ve explained to the players that these are really skinny or really thick trees.
### Also the tree is now on fire.
The lightning ignites flammable objects in the area that aren’t being worn or carried.
Setting the players’ cover on fire is a great way to get the players to do something. I don’t actually have any experience doing anything to the players with flaming cover, so I can’t tell you what to do if they decide to hang around near the flaming tree. This is because my players always freak out when their cover is set ablaze. Whenever I feel like a battle isn’t interesting enough because there is too much hiding going on, I set stuff on fire.
Consider the battle in the banquet hall. The ranger and barbarian flipped a wooden long table and were using it for cover. I’m thinking to myself, “I don’t want to walk the bad guys over there to get the ranger because the barbarian is chilling with him.” Then I remembered my favorite combat seasoning: fire. “The captain of the guard lights an arrow on a nearby sconce and shoots it directly into the table.” They think nothing of it, next round I announce, “Some spilled olive oil has caught fire.” They think nothing of it. The battle goes on for another round, and I am finally pleased to announce, “The front side of your table has erupted into a raging inferno. This is the part where rational decision making is no longer an option. For no reason at all (except possibly a highly sensationalized fire), the ranger leaps over the table to engage the guards in melee combat.
The point is, when you use lightning bolt you have an opportunity to increase the stakes of the fight a little bit by setting their cover on fire, which I have often seen lead to some pretty epic and funny moments in battle.
I’m writing a breadth first search algorithm in Haskell that supports pruning. It is guaranteed that the search will never encounter nodes that have been visited before, so I implemented the search as a `filter` along a stream of candidates.
I’m concerned about the efficiency of the code. Is there some slow, bad-practice code that I can try to avoid? What are some possible ways to optimize the code? Other feedback is also welcome!
``````bfs :: (a -> Bool) -- ^ Goal if evaluates to True
-> (a -> Bool) -- ^ Prune if evaluates to False
-> a -- ^ Starting point of search
-> (a -> (a)) -- ^ Branches at a point
-> (a) -- ^ Goals
bfs predicate prune a0 branch = filter predicate searchspace
where
-- An elegant solution
-- searchspace = a0 : (branch =<< searchspace)
-- However, this solution <<loop>>'s when the search is finite
searchspace = concat \$ takeWhile (not.null) epochs
epochs = (a0) : map (as -> ( a' | a <- as, prune a, a' <- branch a)) epochs
``````
## graphs – Edge exploration with spanning trees
Given an undirected connected graph $$G=(V,E)$$, is there a sequence of spanning trees $$T_0, dots, T_{|E|-|V|+1}$$ satisfying the following edge exchange property: $$forall i >0, T_i = T_{i-1} – (u, v) + (u, w)$$ with $$(v, w) in T_i$$, and such that the union of the spanning trees is equal to $$G$$?
If such sequence exists then it is minimal since $$T_0$$ covers $$|V| – 1$$ edges and each one of the following tree adds exactly one edge to the set of covered edges.
I am interested in the complexity of solving the problem for arbitrary graphs and in finding graph families for which this problem can be solved.
A first necessary condition is that $$G$$ must have a basis cycle composed solely of triangles. If there is a cycle in $$G$$ cannot be expressed as the symmetric difference of triangles then there is necessarily one edge that cannot be covered.
I’m struggling to get beyond that. Do you know of any related problem?
## A programming problem about node objects in trees
I want to create tree where each node knows its parent and children.
Then, maintain the parent and children member variables in the node, and make sure that they are consistent, i.e. that the parent and parent children lists change at the same time.
The current design node has both `set_parent` and `add_child`, so if the method is implemented literally, the synchronization problem is not easy to solve and needs to be remembered and managed by the user
To solve this problem, calling `set_parent` automatically calls the parent `add_child`, and when `add_child` is called, the parent of the child needs to be checked first, and an exception is thrown if the state is not uniform. This partially solves the problem.
To reduce complexity it would be better to expose only one of the two interface methods, how should I design it?
## Merging two B trees in O(h)
Given two B – trees A, B (from order of m) such thats: for each a in A and b in B: a > b
and also the number of elements in A is larger than the number of elements in B.
How can I merge them into one tree from the order of m, in O(h)
when “h” is the height of the A tree.
## data structures – What is the time complexity of comparing two Merkle trees?
I have a simple recursive function that compares two merkle trees and accumulates the differences in the leaf nodes. However, I am unable to measure the time complexity of it. Specifically, I would like to see how does it compare against comparing two Hashtables or two BSTs. A group of leaves is a `row` in this case and they share a `rowid`. In the code below I am just accumulating the differences at the leaf level.
``````def diff_helper(node1: MNode, node2: MNode, diff: List(Difference)):
if not node1 and not node2:
return
elif node1.rowid==node2.rowid and node1.signature==node2.signature and node1.nodetype==NodeType.Row and node2.nodetype==NodeType.Row:
return
elif node1.rowid==node2.rowid and node1.signature!=node2.signature and node1.nodetype==NodeType.Row and node2.nodetype==NodeType.Row:
diff_helper(node1.left, node2.left, diff)
diff_helper(node1.right, node2.right, diff)
elif node1.rowid==node2.rowid and node1.signature!=node2.signature and node1.nodetype==NodeType.Leaf and node2.nodetype==NodeType.Leaf:
diff.append(Difference(node1.rowid, node1.column, node1.value, node2.value))
else:
diff_helper(node1.left, node2.left, diff)
diff_helper(node1.right, node2.right, diff)
``````
Time complexity:
1. On the best case, I see that this is a constant operation since the root hashes of both trees would be the same.
2. On the worst case, the number of comparisons is the total number of all leaf nodes.
Question:
Assuming the above time complexity, it doesn’t seem to fare any better than comparing two Hashtables where each key is a rowid and the values being leaves. I understand that with Hashtable, once you find the values of the rowid, you’ll need to do a linear comparison of each leaf as compared to O(logm), m being the number of leaves under each row. How would you represent that in Big O terms?
## trees – Generic real-time message/event matching engine
I’m looking for methods (any kind of combinations of algorithms & corresponding data structures) that can be used as a generic matching engine in case of arbitrary message/event types. Let me formalize the problem a bit:
• Let $$C$$ denote a condition type consisting of an arbitrary data type (numeric values/strings/dates/enumerations/ranges/etc.)
and a compatible comparison operator: $$C = (text{Type}, text{Operator})$$. For
instance, in the case of numeric values, $$C_1 = (text{Integer}, <=)$$, or for strings $$C_2 = (text{String}, text{Contains})$$.
• The condition types can be instantiated with an actual reference value of the
condition’s underlying data type, for instance: $$c_1 = (text{Integer} = 5, <=)$$ or $$c_2 = (text{String} = text{“xy”}, text{Contains})$$.
• Obviously, multiple conditions can be combined which requires the
definition of a condition set $$S = (C_1, C_2, ldots, C_n)$$. A
concrete condition set can be e.g.: $$s = ((text{Real} = 3.14, >), (text{String} = text{“abc”}, text{!=}))$$. For simplicity, the
logical connection between the conditions of a set is always
$$text{AND}$$ (so in order for a set to be a match for an event, all
the conditions of $$s$$ must be met).
• Finally, multiple such condition sets can coexist which we can
visualize like the rows of a data table: $$t = {s_1, s_2, ldots, s_m}$$.
The use case would be that one assigns some operations/actions to the individual condition sets (rows) and executes only the operations whose conditions matched the properties of the incoming event in real-time (note that this is a difference compared against traditional filtering – we define the conditions first, then we wait for the data, and not the other way around). These properties must be identified in advance and they must correspond to the condition types of the given scenario (e.g., $$c_1 = (text{Integer} = 5, <=)$$ always uses the API `int EventT::get_some_field()` of the event type), however, such implementation details are not relevant from the core algorithmic problem’s point of view.
To sum up: the input of the algorithm (or matching engine) is $$(t, e)$$ where $$e$$ is a compatible event, while the output $$t’$$ is a subset of $$t$$ (the most complete possible) such that $$forall s in t’$$ it’s true that all conditions in $$s$$ matched $$e$$.
Let’s see a complete example, starting with $$t$$:
ID $$C_1$$ $$C_2$$
1 $$(text{Integer} = 66, =)$$ $$(text{String} = text{“*”}, =)$$
2 $$(text{Integer} = 50, > text{(greater than)})$$ $$(text{String} = text{“abc”}, text{!=})$$
A wildcard (see the first value in the $$C_2$$ column) basically means that we won’t utilize the 2nd condition (any string will match).
And now a couple of events:
• $$e_1 = (35, text{“xy”})$$ – won’t match any of the 2 condition sets,
• $$e_2 = (66, text{“abc”})$$ – will match only the 1st set,
• $$e_3 = (77, text{“def”})$$ – will match only the 2nd set,
• $$e_4 = (66, text{“xy”})$$ – will match both sets,
• $$e_5 = (88, text{“abc”})$$ – won’t match any of the 2 condition sets.
One possible way I can think of to implement this engine are trees: the levels of the tree represent the condition types (and on a particular level there are as many nodes as the number of rows in $$t$$), while the nodes the actual conditions, and whenever the conditions of all the nodes through a path in the tree till a leaf node (including the leaf node) are met, we have a match.
Other approaches are welcome, still, I’d appreciate concrete, possibly open-source implementations (with optimized/efficient underlying algorithms) even more. If there is a tree-based solution out somewhere, I’d be happy to take a look at it, too.
## c# – How to implement a chain of events on a tree’s CRUD operations?
I am working on a project in which I have a tree with 4 layers and the hierarchy is like this.
Customer -> Site -> Location -> Guardroom
In DB each entity has its own table and the child knows its parent’s Id.
``````public class Guardroom{
public int Id {get; set;}
public string Name {get; set;}
public int LocationId {get; set;}
}
``````
A rule is that guardroom should be the only leaf. What I mean is that
``````Customer must have at least one child (site)
|
|--> Site must have at least one child (location)
|
|--> Location must have at least one child (guardroom)
``````
My challenge is that if I delete a guardroom I must delete the location only if the guardroom has no siblings and the grandparent if location has no siblings and so on. In other words, If I delete the guardroom I must delete the nodes above in cascade only if the node to delete has only 1 child.
I do not want to have a highly coupled structure so I do not want the guardroom to know about the location. I was thinking in using events but I am not sure what patter or how to implement it.
My infrastructure looks like this.
I implemented dependency injection and Domain Driven Design. I have a service for each layer in the tree:
i.e.
``````public class GuardroomService: IGuardroomService {
//Here dependency is injected.
public GuardroomService(IGuardroomDataAccess guardroomService){
_guardroomService = guardroomService;
}
public void Delete(int guardroomId){
//Here I want to raise an event to notify parent (location) and take further action.
}
//...
//The remaining CRUD operations and functions.
}
``````
What patter/structure should I implement to let the other services I am deleting a child without allowing the same service to know about the other services.
Any help will be much appreciated. Thank you in advance. | 3,399 | 13,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 42, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-39 | latest | en | 0.963685 |
https://testmaxprep.com/lsat/blog/lsat-prep-test-77-december-2015-lsat-logic-game-3 | 1,481,190,768,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542520.47/warc/CC-MAIN-20161202170902-00339-ip-10-31-129-80.ec2.internal.warc.gz | 872,704,430 | 15,901 | # LSAT Prep Test 77 December 2015 LSAT: Logic Game 3
Posted on
For our LSAT prep today, we're going to review everyone's favorite game from the December 2015 LSAT, the third logic game, which states:
“Four employees—Jackson, Larabee, Paulson, and Torillo—are to select from among four offices—W, X, Y, and Z. The order in which they select, from first to fourth, is to be decided by a random drawing. Each employee has ranked the offices from first (highest) to fourth (lowest) as follows:”
And then it sets forth the rankings for the employees and then it proceeds to the restrictions.
Before we move to the restrictions, let's take a closer look at what we're seeing here. Notice, we have four employees, Jackson, Larabee, Paulson and Torillo. They're selecting among four offices, W, X, Y and Z. They're going to select in order first to fourth. We do have an order here. We will be selecting first, second, third, fourth. The priority will be determined by a random drawing but it's really a multi-linear game, because not only are we going to select an office, we're going to need to know which employee made that selection. We're going to see employee and offices.
Just to make sure we keep it clear, I'm going to draw out the rankings of each employee. Again, we have our four employees. There you have what is set forth there in terms of ranking priority for each employee. Before we proceed here, I just want to make a couple of comments. The first and most important thing at this point, a lot of students had already freaked out. This game is not your traditional game. It seems as you see, as we proceed here to the restrictions, they're very abstract but freaking out is not going to do anything for you on the exam. It's really important to remain calm, make sure you have a clear understanding of the conditions and go through the questions as quickly as possible, applying the conditions that you're given.
Keeping that in mind, let's turn our attention to the restrictions. The first restriction here tells us, "Each employee selects an office that has not been selected previously." What that tells us is that each office is going to appear once, and only once so this is a one-to-one game. Now, again, that had to be stated. We couldn't assume that each employee was going to get their own office. It was possible that they could share. That rule sets forth that each office is going to be selected by one employee.
The next rule tells us that, "Each employee selects only one office." That establishes that our employees now are also going to be one-to-one. We have four employees, four offices.
And the last rule states that, "Each employee selects the office that he or she ranks the highest among the unselected offices." Obviously, that makes plenty of sense. That's the whole idea of a ranking priority.
Now, let's just, for purposes of really making sure we understand this game before proceeding, let's do some examples of how this game might play out. Turning our attention over here to our possibilities. The first thing I want you to notice is that every single employee has either X or Y as their first priority. Clearly, these are the corner offices. That allows us to make a deduction here that the first office selected is going to be either X or Y.
Turning our attention now to the second position, we notice that W does not appear. When you combine that with the fact that W also doesn't appear in anyone's first position that allows us to conclude that W does not appear second. So that means that now, W must appear either third or fourth. Let's place W third just to try it out and get a better sense of how this game is actually going to work here.
Drawing out a quick hypothetical in which we place W third. Now, notice, for W to be third, that would mean that Larabee is picking third. J, P and T all have W as their fourth selection. Here, we have L picking third and picking W. Notice now that means that first and second are going to be J, P and T. You notice that J, P and T all have Y first or second. J has Y first. P has Y first. T has Y second. Y is either first or second for J, P and T. What that means is that Y is going to be one of the first or second positions here. The only other offices that remain are X and Z. This leads to a very important deduction. Notice, when Larabee goes third in this scenario, one of X or Z is going to be present, which means that Larabee would not be selecting W. Larabee would be selecting X or Z. Again, because Y and one of X or Z comes first. There's no possible scenario where X and Z show up first and second when Larabee is selecting third.
That leads to a huge deduction here that W cannot be third, which means W must be fourth. Now, given that our game is quite restrictive in that we know either X or Y is going to be first, let's draw out these two possibilities. Erasing our hypothetical here that proved that W cannot be third, just to clear up some room here. In scenario one, we are going to place X first. If X is first, that means one of L or T is selecting first, because those are the only two employees that have X ranked first. We know first is going to be L or T in this scenario. Who's left in terms of offices? Well, Z and Y. We're going to have our reverse dual options here, second and third.
Drawing out scenario two now. In scenario two, we're going to place Y first and if Y is first, that means either J or P is selecting first, because those are the only 2 employees with Y as their first option. Who's left in terms of offices? X or Z. We're going to have these reverse dual options here.
So there you have the set up and the deductions for this game. Now, obviously, if you weren't able to make this key deduction that W is fourth, it's not the end of the world. Again, this is a very rule driven game. If you are aggressive with these rules and go through the questions, you should still be able to complete it. Now, again, it's really important to just make sure you understand how this game works. We have a great understanding of this now. Let's turn our attention to the questions.
Updated on Sep 29, 2016 | 1,389 | 6,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-50 | latest | en | 0.977203 |
http://spiff.rit.edu/richmond/ritobs/sep02_2008/sep02_2008.html | 1,725,896,802,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00484.warc.gz | 33,158,494 | 3,320 | Sep 03, 2008 UT: Photometry of new dwarf nova in Andromeda
Michael Richmond Sep 03, 2008
On the night of September 2/3, 2008, I observed a new object -- a dwarf nova? -- which was recently discovered in the constellation Andromeda. It's still very early, but you can find a bit of information on the new object in these sources:
The object is at RA = 02:00:25.40 Dec = +44:10:18.7, so it rises just after sunset at this time of year. It is well placed for observations from the Northern Hemisphere.
The setup was:
• no focal reducer, so working at f/10
• SBIG ST-7E CCD camera with V-band filter (but I should switch to no filter for future runs)
• 30-second unguided exposures
Notes from the night
• The conditions were very good: no clouds, sky reasonably dark
• The earliest measurements come at such high airmass, greater than 4, that they are really bad
• The FWHM was around 2.3 pix after refocusing and after the object rose to a smaller airmass
This is a chart of the field taken from the DSS.
The chart has several of the brighter stars in thefield labelled with letters, just to keep me straight as I perform the reductions. The star labelled "bright!" saturates my detector. Stars A and B are in the Tycho-2 catalog,
```my
label RA Dec Bt Vt
----------------------------------------------------------------------------
A TYC 2828 1694 1 29.97369 +44.10297 12.081 11.143
B TYC 2841 810 1 30.22008 +44.08782 13.369 11.679
----------------------------------------------------------------------------
```
I measured the instrumental magnitude of each star with aperture photometry, using a radius of 3 pixels = 5.6 arcseconds and sky defined by an annulus around each star. Following the procedures outlined by Kent Honeycutt's article on inhomogeneous ensemble photometry, I used all stars available in each image to define a reference frame, and measured each star against this frame.
One output of the ensemble solution is the value of the zero-point of each frame relative to the others. In the graph below, I plot this zero-point as a function of time. Note the very large change due to the decrease in atmospheric extinction as the source rose higher in the sky. The gap marks the time at which I re-focused the telescope.
Below is a graph of the scatter in differential magnitude versus magnitude in the ensemble solution.
The floor of this diagram corresponds to a scatter of about 0.007 mag. Nova And 2008 appears at differential magnitude 2.15; its scatter of 0.036 magnitudes is only slightly elevated compared to other stars of similar brightness. This is due to the large scatter in the measurements of _all_ the stars when I was looking at high airmass.
Light curves for selected stars (Nova And 2008 and stars A - D) in the field are shown below. Nova And 2008 is shown by light green crosses.
Here's a closeup of the variation in Nova And 2008 and a few comparison stars, which I have shifted for easy comparison. You can see one full cycle clearly at the end of the measurements. There are almost two more full cycles in the data, but they are hidden by the large scatter due to all that air. Rats. The period has been measured to be around 0.056 days, with which my noisy data agrees.
I've made a table of the measurements themselves, with three different flavors of time. The differential magnitudes from the ensemble solution have been shifted so that star "A" in my chart, TYC 2828 1694 1, has value 11.143. There are no color corrections to these V-band measurements.
Here's the start of the table.
```# Measurements of NovaAnd2008 made at RIT Obs, Sep 3, 2008 UT.
# by Michael Richmond.
# Each exposure 30 seconds long with V filter.
# Tabulated times are midexposure (FITS header time - half exposure length)
# and accurate only to +/- 1 second (??).
# 'mag' is a differential magnitude based on ensemble photometry
# using a circular aperture of radius 5.6 arcseconds.
# which has been shifted so TYC 2828-1694-1 mag=11.143
# to match value in Tycho-2 catalog.
#
# UT day JD-2,450,000 HJD-2,450,000 mag uncert
Sep03.02719 2454712.52719 2454712.52243 13.296 0.028
Sep03.02765 2454712.52765 2454712.52289 13.275 0.028
Sep03.02811 2454712.52811 2454712.52335 13.380 0.033
``` | 1,113 | 4,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.908466 |
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Good Maen 09/Aug/08 12:00 AM | |
Why the English language is so much fun: "The bandage was wound around the wound." 09/Aug/08 12:00 AM | |
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Good morning, John! 09/Aug/08 12:01 AM | |
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well, I am here, off the track.. 09/Aug/08 12:10 AM | |
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But first: "The dump was so full it had to refuse more refuse." 09/Aug/08 12:13 AM | |
ok interesting thought for the day..The reasonable man adapts himself to the world; the unreasonable one persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable man.So got the moral?? 09/Aug/08 12:14 AM | |
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http://www.shapefit.com/calculators/baseball-statistics-calculator.html | 1,371,702,832,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710274484/warc/CC-MAIN-20130516131754-00040-ip-10-60-113-184.ec2.internal.warc.gz | 671,121,630 | 7,773 | Baseball Statistics Calculator
Statistics play an important role in summarizing baseball performance and evaluating players in the sport. Since the flow of a baseball game has natural breaks to it, the sport lends itself to easy record-keeping and statistics. Numerical facts and data are the lifeblood of baseball.
On-Base Percentage (OBP) Calculator Hits : Walks : Hit-By-Pitch : At Bats : Sac Flys :
Result
* The baseline for a great OBP is .400, although anything above the ML average certainly is good (.353).
Slugging Percentage:
Another great statistic to be measured by. SLG% means how many total bases touched per at-bat.
Slugging Percentage (SLG) Calculator Singles : Doubles : Triples : Home Runs : At Bats :
Result
On-Base + Slugging (OPS):
ESPN has started showing OPS during broadcasts, and there is a good reason for it. OPS can tell you about two important qualities in one stat: the ability to reach base, and to hit for power. While some think OBP is more important and therefore should be weighted, it still does a good job of evaluating a player. The gold standard in the major leagues is an OPS of 1.000.
On-Base + Slugging (OPS) Calculator OBP : SLG :
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Stolen Base Percentage:
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Stolen Base Percentage Calculator Stolen Bases : Stolen Base Attempts:
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Base On Balls Percentage:
A great stat for measuring a player's batting "eye" or how respected a hitter is by pitchers. As I have said in my batting article, walks are paramount to success in baseball. If you can get a decent amount of walks, you can contribute to wearing down the pitcher's arm and scoring runs.
Base On Balls Percentage Calculator Walks : Plate Appearances :
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Walks/Strikeout Ratio:
This is a great measure of plate discipline. Generally, if a player has more walks than strikeouts, he has a good eye, and has great knowledge of the strike zone.
Walks/Strikeout Ratio Calculator Walks : Strikeouts :
Result | 533 | 2,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2013-20 | latest | en | 0.92758 |
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Write an equation that represents the number of calories a person would have to burn a day to lose four pounds in two weeks.
Can somebody help? Thanks!
2. Originally Posted by Phresh
A person must burn 3500 Calories to lose one pound of weight.
Write an equation that represents the number of calories a person would have to burn a day to lose four pounds in two weeks.
Can somebody help? Thanks!
(3500 Cal /1 lb)*(4 lbs / 2 wk)*(1 wk / 7 day) = (3500*4 / 2 *7) = 100 Cal/day | 140 | 542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-26 | latest | en | 0.949261 |
https://newbedev.com/print-a-tongue-twister | 1,680,198,889,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00497.warc.gz | 476,338,915 | 6,924 | Print a tongue twister!
Retina 0.8.2, 106 103 bytes
S230 by41h012 are30,5.¶So if1230 on4n5123ore1h0.
5
I'm1ure
4
the3ore,¶The
3
1eash
2
he10
1
s
0
ells
Try it online! Explanation: In Retina, a substitution only makes sense if it's long enough l for the number n of repetitions. The substitution saves n(l-1) bytes in the compressed text but costs l+3 bytes in the replacement stages. This gives the minimum length required to be useful as follows:
• 2 repetitions: length > 5
• 3 repetitions: length > 3
• 4 repetitions: length > 3
• 5 repetitions: length > 2
• 6+ repetitions: length > 1
Edit: As @Arnauld pointed out, you can count repetitions from the substitution entries as well. This means that although there were only 5 repetitions of space+s in my previous encoded text, there are also 3 repetitions in the substitutions, thus allowing me to save 3 bytes overall. (@Arnauld himself had only spotted 2 of the 3 repetitions.)
Python 3, 166 $$\\cdots\$$ 142 135 bytes
Saved a whopping 13 18 bytes thanks to ovs!!!
Saved 4 bytes thanks to user253751!!!
Saved a byte thanks to branboyer!!!
Saved 7 bytes thanks to dingledooper!!!
Note: There's lots of unprintables in the following code so please avert your eyes if you're sensitive to such things! :D
print("Sh by are,.\nSo if onnore.".translate("| seash|ells|ore,\nThe| sh|e s| I'm sure| the".split("|")))
Try it online!
///, 115 bytes
Port of Neil's answer + accepting nph's suggestion.
/3/ I'm4ure//2/4eash//1/he40//0/ells//4/ s/S120 by the2ore,
The4h041 are20,3.
So if4120 on the2ore,
Then3412ore4h0.
Try it online! | 518 | 1,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.824477 |
https://www.jiskha.com/questions/1251093/Which-sum-will-be-irrational-1-point-a-3-2-b-start-root-19-end-root-plus-start | 1,561,011,799,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999141.54/warc/CC-MAIN-20190620044948-20190620070948-00002.warc.gz | 791,926,905 | 5,245 | # help please with math thanks
Which sum will be irrational? (1 point)
a.3 + 2
b.start root 19 end root plus start fraction 7 over 2 end fraction
c Start Fraction 18 over 3 End Fraction plus Start Fraction 11 over ten thousand End Fraction.
d. –6 + (–3.251)
Which product will be rational? (1 point)
a. 17 point Modifying Above 12 with bar times 33
b.start root 3 end root times start root 9 end root
c. 4 times pi
d. negative start root 20 end root times 15
A whole number is added to a number with two digits after the decimal point. To make sure the answer is reasonable, how many digits should the sum have after the decimal point?
a.none
b.1
c.2
d.infinitely many
1.B
2.a
3.b
am i right
1. 👍 0
2. 👎 0
3. 👁 122
1. You sure use a lot of words. Try using these characters in the future. If you can't recall the alt-codes, you can always find them and copy/paste them.
#1
a. 3 + 2
b. √19 + 7/2
c. 18/3 + 11/10000
Your answers look good. I would have chosen 2C, but maybe you have been given a rule that says to drop the least significant digit.
1. 👍 0
2. 👎 0
posted by Steve
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10. ### algbra
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More Similar Questions | 1,330 | 4,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-26 | latest | en | 0.829144 |
https://algebra2coach.com/how-to-do-matrix-multiplication/ | 1,685,950,297,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00518.warc.gz | 106,367,603 | 24,452 | # Matrix Multiplication
Would You Rather Listen to the Lesson?
Once algebra 2 students understand what a matrix represents, how to determine the order of a matrix, and how to organize data into matrices, they dive into matrix operations, including how to do matrix multiplication.
Even though lessons on matrix multiplication may seem overwhelming to students, rest assured that teaching multiplying matrices can be fun and easy, if you’re equipped with the right resources! So if you’re a math teacher looking for strategies to teach matrix multiplication, you’ve come to the right place!
These awesome teaching strategies on how to do matrix multiplication are guaranteed to keep your students engaged throughout the lesson!
## Strategies to Teach Matrix Multiplication
### What Is a Matrix?
By now, students probably already have an idea of what a matrix is. But it’s good to refresh their memory and this is a good opportunity to identify if there are gaps in some kids’ knowledge of matrices. So, start your lesson on matrix multiplication by defining matrices.
Explain that a matrix is a collection of information stored or arranged in an orderly fashion. A matrix represents an array of numbers. You can also remind students how to judge the order of any given matrix.
This is basically the number of rows × the number of columns in a given matrix. So if a matrix has n rows and m columns, its order will be n × m (point out that we read this as ‘n by m’). The number of rows and columns are the dimensions of a matrix.
It’s crucial that students understand how to determine the dimensions of a matrix, that is, how many rows and columns a matrix has, as this will be the starting point in their efforts to multiply matrices.
Write an example on the whiteboard of a matrix and ask students to determine its order:
Students should be able to tell you that the matrix has two rows and three columns. This means that its order is 2 × 2, which we read two by two. Once students are comfortable in this area, move on to matrix multiplication.
### How to Do Matrix Multiplication
#### Are Two Matrices Conformable for Multiplication?
The chief reason why students must be able to determine the number of rows and columns in a matrix in order to understand how to multiply matrices is because we can only multiply certain matrices, based on their number of rows and columns.
More specifically, two matrices A and B are conformable for multiplication AB if the number of columns in matrix A equals the number of rows in matrix B. You can also write on the whiteboard that we can multiply two matrices A and B only if:
Number of Columns in A = Number of Rows in B
You can write several examples on the whiteboard of two matrices that we’re trying to multiply. Ask students to determine which ones are conformable for multiplication, by applying the above-mentioned condition.
• Can a 2 × 3 matrix be multiplied by a matrix that’s 3 × 3? Yes, because the number of columns in the first matrix is 3 and the number of rows in the second matrix is 3.
• Can a 2 × 2 matrix be multiplied by a matrix that’s 2 × 2? Yes, because the number of columns in the first matrix is 2 and the number of rows in the second matrix is 2.
• Can a 2 × 4 matrix be multiplied by a matrix that’s 1 × 2? No, because the number of columns in the first matrix is 4 and the number of rows in the second matrix is 1.
#### Step-by-Step Matrix Multiplication
Now you can show students how we carry out the process of multiplying two matrices that are conformable for multiplication. Let’s say there’s matrix A and matrix B that we wish to multiply. We can see that they’re 2 × 2, thus, conformable for multiplication.
In order to carry out the multiplication, we need to multiply the rows of the first matrix by the columns of the second matrix. A useful strategy is to circle the numbers in the rows and columns that you multiply in a different color.
You can also draw a visual representation in the following manner:
##### Example:
Let’s say we want to multiply the following two matrices:
We start off by taking the product of the first row of the first matrix and the first column of the second matrix, that is, (1,7) and (3, 5). By applying the above rules, we’ll get:
1 × 3 + 7 × 5 = 3 + 35 = 38
This will represent the top left entry in our result. To get the top right entry, we’ll take the product of the first row of the first matrix and the second column of the second matrix, that is, (1, 7) and (3, 2). That is, we’ll take the products of the first terms and of the second terms and add them:
1 × 3 + 7 × 2 = 3 + 14 = 17
So 17 will be the top right entry. To find the bottom left entry, we need to take the product of the second row in the first matrix and the first column in the second matrix, that is (2, 4) and (3, 5):
2 × 3 + 4 × 5 = 6 + 20 = 26
So 26 will be the bottom left entry. To find the bottom-right entry, we need to take the product from the second row in the first matrix and the second column in the second matrix, that is (2, 4) and (3, 2):
2 × 3 + 4 × 2 = 6 + 8 = 14
With 14 as our bottom-right entry in the final result, we’ll get the following matrix:
### General Rules of Matrix Multiplication
Point out to students that there are a few general rules of matrix multiplication that are useful to know, including:
• Multiplication is not commutative: A × B ≠ B × A
• Associative Property applies: A × (B × C) = A × B × (C)
• Multiplicative property of zero: A × 0 = 0
• Multiplicative identity property: I × A = A × I = A
• Distributive properties: A × (B + C) = A × B + B × C
(B + C) × A = B × A + C × A
### Scalar Multiplication Vs. Matrix Multiplication
It’s advisable to draw the distinction between scalar multiplication and matrix multiplication. Explain that in scalar multiplication, we multiply one constant value with each element of a matrix, while in matrix multiplication, two conformable matrices are multiplied with each other.
Use this video by Khan Academy to demonstrate an example of multiplying a 2 × 2 matrix with a 2 × 2 matrix. Then, play this video to move on to a bit more complicated of a multiplication, where we have a multiplication of a 2 × 3 matrix and a 3 × 2 matrix.
This video is also a good resource on how to do matrix multiplication. The video contains simple guidelines on which matrices can be multiplied by which matrices, as well as practical examples on how to perform the process of multiplication.
## Activities to Practice Matrix Multiplication
### Matrix Spacecraft Game
This game will help students enhance their ability to determine whether two matrices are conformable for multiplication. To implement this activity in your classroom, you’ll need construction paper, some markers, and some drawing skills!
Draw two spacecraft on the construction paper, and write one matrix on one spacecraft and another matrix on the other. The matrices on the two spacecraft should be conformable for multiplication. Then, use the scissors to cut the spacecraft.
Create several of these ‘sets’ of spacecraft conformable for multiplication (10 sets per group, depending on the size of your class). Mix the spacecraft in different piles, making sure that there are corresponding matches in each pile so that you end up with 10 sets by matching them.
Divide students into small groups of 3 or 4. Hand out one pile of spacecraft to each group and provide instructions for the activity. The members of each group work together to determine which spacecraft can travel with which spacecraft in space.
To achieve this, students need to find the matrix on the inside of a given spacecraft that is conformable for multiplication with a matrix on the inside of another spacecraft. Once they find such two spacecraft, they match them. They proceed with this until all spacecraft are matched.
Give a few minutes to students, so they can match all spacecraft. Once the time is up, allow space for discussion. How did students determine whether a matrix was conformable for multiplication with another matrix? Did they have any unmatched matrices? What were the biggest challenges?
### Multiply Two Matrices Game
This is an online game by IXL that will help students practice multiplying two matrices. To use this game in your classroom, the only thing you need is a sufficient number of technical devices (one per student), as well as a decent internet connection.
Divide students into pairs and provide instructions for the game. Explain to students that they are presented with several matrix multiplication exercises, where they have to complete the multiplication of the two given matrices, by filling in the missing number in the product.
Provide a few minutes for students to complete the exercises. You can also set a timer. In the end, students in each pair compare their final scores. The person with the highest score wins the game.
### Multiply Matrices Game
In this online game by Khan Academy, students will have the opportunity to hone their skills at matrix multiplication. Make sure you have enough technical devices in your classroom (one per student) and start multiplying!
Students play the game individually, which makes it suitable for homeschooling parents as well. In each exercise, students are presented with two different matrices that they have to multiply. If they get stuck, they can also choose to watch a video or use a hint.
Provide a few minutes for students to finish the exercises. Then, open space for discussion and reflection. Were there any examples that they found particularly challenging? Which steps did they follow to fill in the entries in the final result?
## Before You Leave…
If you like these teaching strategies and activities for how to do matrix multiplication, and you’re looking for more algebra 2 resources, make sure to sign up for our emails to receive loads of free lessons and content!
In addition, you can sign up for a membership on MathTeacherCoach or head over to our blog, where you’ll find more math materials for kids of all ages. You’ll discover that with the resources we offer, teaching math has never been easier! | 2,207 | 10,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-23 | longest | en | 0.932163 |
https://shreerangpatwardhan.blogspot.com/2013/03/jquery-mobile-numbered-listview.html | 1,526,943,031,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00511.warc.gz | 678,270,331 | 40,000 | ### Jquery Mobile - Numbered Listview
Jquery Mobile Listview is probably the most popular component used in a lot of mobile web applications. The listview that is generally used is a ul-li structure. But Jquery Mobile allows the usage of the ol-li structure as well. The ol-li HTML structure will give the user a numbered listview. We will take a look at the following example to understand, how the numbered listview works.
As you can see in the result above, every list gets a number before the the list title. The numbered listview is useful when presenting items that are in a sequence like search results or a movie queue.
Hope you have followed this short tutorial and hope it helps you in your web application. In case, you have any doubt, comment or you come across any error, fell free to leave a comment and I will be happy to take a look at it. You can see a full list of Jquery Mobile examples here.
### Geodesic Polyline
Today we will have a look at a very interesting polyline example - "The geodesic polyline". Now the first question that will pop is "What is geodesic?". Mathematically, geodesic means the shortest line between two points on a mathematically defined surface, as a straight line on a plain or an arc of a great circle or sphere.
The next question after reading the above definition is clearly, "Why do we need geodesic polylines?" and that would be followed up with "What is this Great Circle?". We will discuss this first, before we move on to the actual example today. The example is very very similar to the normal polyline example, with just a small change.
Having said so, I will now try to explain why we need a geodesic polyline? The shortest distance between two locations on the earth is rarely a straight line as the earth is roughly spherical in nature. So any two points on the earth, even if they are very close lie on a curve and not …
### Difference between word-break: break-all versus word-wrap: break-word
The 2 CSS properties word-break: break-all and word-wrap: break-word appear to work in the same way or generate the same output, but there is a slight difference between the 2 and we will be discussing these differences today.
Take a look at the example above. The difference is quite evident, however I will try to explain it further.
word-break: break-all Irrespective of whether it’s a continuous word or many words, break-all breaks them up at the edge of the width limit even within the characters of the same word
word-wrap: break-word This will wrap long words onto the next line.break-word adjusts different words so that they do not break in the middle.
So if you have many fixed-size spans which get content dynamically, you might just prefer using word-wrap: break-word, as that way only the continuous words are broken in between, and in case it’s a sentence comprising many words, the spaces are adjusted to get intact words (no break within a word). In case you want to exp…
### Where does Google get it's live traffic data from?
Referring to a post that I wrote earlier, Google’s - Live traffic Layer, ever wondered how Google collected this data? I was wondering the other day, how Google received live data to display it on their maps as a layer! I looked up the web and found something very interesting and am sharing the same with you all.As we all know, the traffic layer is available most accurately in several states in USA. Most major metro areas in the US have sensors embedded in their highways. These sensors track real time traffic data. Easy to miss at high speeds (hopefully anyway, traffic permitting), more commonly noticed may be the similar sensors that often exist at many busy intersections that help the traffic lights most efficiently let the most amount of people through. The information from these tracking sensors is reported back to the Department of Transportation (DOT). The DOT uses this data to update some of the digital signs that report traffic conditions in many metro areas. They also…
### Ground Truth - How Google Builds Maps
Todays's article is cross posted from The Atlantic's Tech section. The article was posted by Alexis Madrigal who is a senior editor at The Atlantic, where he oversees the Technology channel. So, thanks to The Atlantic and Alexis Madrigal, we will have an exclusive look inside Ground Truth, the secretive program to build the world's best accurate maps.
Behind every Google Map, there is a much more complex map that's the key to your queries but hidden from your view. The deep map contains the logic of places: their no-left-turns and freeway on-ramps, speed limits and traffic conditions. This is the data that you're drawing from when you ask Google to navigate you from point A to point B -- and last week, Google showed me the internal map and demonstrated how it was built. It's the first time the company has let anyone watch how the project it calls GT, or "Ground Truth," actually works.
Google opened up at a key moment in its evolution. The co…
### jQuery Mobile's Next Big Step
Spatial Unlimited changes to The UI Dev
After being hosted on blogger 😣 for the last 6 years 📆, this page has finally been moved to Github.io
This means a few things for you, dear reader!
You will be redirected to the new page shortly! ⏩ ⏩ ⏩
Once crapy HTML is now better looking Markdown! 😍 😍
The entire blog is a Github repo! 😍 😍
Spatial Unlimited is now The UI Dev 😍 😍 | 1,173 | 5,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-22 | latest | en | 0.913296 |
https://www.thestudentroom.co.uk/showthread.php?t=1500880 | 1,513,441,102,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588251.76/warc/CC-MAIN-20171216143011-20171216165011-00078.warc.gz | 851,634,899 | 38,686 | You are Here: Home >< Maths
# Help with understanding some mathematical notation Watch
1. a + max(b,0)
max(d, c, 0)
What do the above terms mean?
Thanks
2. If is a set then means the biggest element in . So for example .
So when you see something like it means the "largest positive number out of a,b,c,d, or 0 if they're all negative", and so on.
3. (Original post by nuodai)
If is a set then means the biggest element in . So for example .
So when you see something like it means the "smallest positive number out of a,b,c,d, or 0 if they're all negative", and so on.
Do you mean largest positive number?
4. (Original post by Desert Eagle)
Do you mean largest positive number?
Yup.
5. What if i have:
max(0, -a)
max(-a, 0)
Is there a difference between the two? I don't understand the purpose of the '0'.
6. (Original post by Desert Eagle)
What if i have:
max(0, -a)
max(-a, 0)
Is there a difference between the two? I don't understand the purpose of the '0'.
There is no difference. If a were positive, then -a would be negative and max(-a,0) would be 0. Otherwise max(-a,0) = -a
7. (Original post by SimonM)
There is no difference. If a were positive, then -a would be negative and max(-a,0) would be 0. Otherwise max(-a,0) = -a
Ah ok i see, thanks
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https://jsperf.com/2014-09-16-hex-to-rgb/13 | 1,558,841,352,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00382.warc.gz | 541,974,166 | 4,943 | # Hex to RGB
## JavaScript performance comparison
Revision 13 of this test case created by
## Preparation code
``````
<script>
Benchmark.prototype.setup = function() {
var data = [];
for(var i = 0; i < 10000; i++)
{
data.push(Math.floor(Math.random() * 16777215).toString(16));
}
var hexToRgb1 = function(hex) {
var bigint = parseInt(hex, 16);
var r = (bigint >> 16) & 255;
var g = (bigint >> 8) & 255;
var b = bigint & 255;
return r + "," + g + "," + b;
}
var hexToRgb2 = function(hex) {
var bigint = parseInt(hex, 16);
var r = (bigint >> 16) & 255;
var g = (bigint >> 8) & 255;
var b = bigint & 255;
return [r, g, b].join();
}
var hexToRgb3 = function(hex) {
return [(bigint = parseInt(hex, 16)) >> 16 & 255, bigint >> 8 & 255, bigint & 255].join();
}
var hexToRgb4 = function(hex) {
var result = /^#?([a-f\d]{2})([a-f\d]{2})([a-f\d]{2})|([a-f\d]{1})([a-f\d]{1})([a-f\d]{1})\$/i.exec(hex);
return result ? {
r: parseInt(hex.length <= 4 ? result[4]+result[4] : result[1], 16),
g: parseInt(hex.length <= 4 ? result[5]+result[5] : result[2], 16),
b: parseInt(hex.length <= 4 ? result[6]+result[6] : result[3], 16),
toString: function() {
var arr = [];
arr.push(this.r);
arr.push(this.g);
arr.push(this.b);
return "rgb(" + arr.join(",") + ")";
}
} : null;
}
var hexToRgb5 = function(hex) {
var result = /^#?(\w\w|\w)(\w\w|\w)(\w\w|\w)\$/i.exec(hex);
if (!result) {
return null;
}
var r = parseInt(hex.length <= 4 ? result[4] + result[4] : result[1], 16);
var g = parseInt(hex.length <= 4 ? result[5] + result[5] : result[2], 16);
var b = parseInt(hex.length <= 4 ? result[6] + result[6] : result[3], 16);
return "rgb(" + r + "," + g + "," + b + ")";
}
var hexToRgb6 = function(hex) {
var shorthandRegex = /^#?([a-f\d])([a-f\d])([a-f\d])\$/i;
hex = hex.replace(shorthandRegex, function(m, r, g, b) {
return r + r + g + g + b + b;
});
var result = /^#?([a-f\d]{2})([a-f\d]{2})([a-f\d]{2})\$/i.exec(hex);
if (!result) {
return null;
}
var r = parseInt(hex.length <= 4 ? result[4] + result[4] : result[1], 16);
var g = parseInt(hex.length <= 4 ? result[5] + result[5] : result[2], 16);
var b = parseInt(hex.length <= 4 ? result[6] + result[6] : result[3], 16);
return "rgb(" + r + "," + g + "," + b + ")";
}
};
</script>
``````
## Test runner
Warning! For accurate results, please disable Firebug before running the tests. (Why?)
Java applet disabled.
Testing in CCBot 2.0.0 / Other 0.0.0
Test Ops/sec
hexToRgb4
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb4(data[i]);
}``````
pending…
hexToRgb6
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb6(data[i]);
}``````
pending…
hexToRgb1
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb1(data[i]);
}``````
pending…
hexToRgb5
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb5(data[i]);
}``````
pending…
hexToRgb3
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb3(data[i]);
}``````
pending…
hexToRgb2
``````var test;
for(var i = 0; i < data.length; i++)
{
test = hexToRgb2(data[i]);
}``````
pending…
## Revisions
You can edit these tests or add even more tests to this page by appending `/edit` to the URL. | 1,121 | 3,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-22 | latest | en | 0.235175 |
https://mathsite.org/factoring-maths/roots/multiplying-and-dividing.html | 1,656,331,278,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00744.warc.gz | 445,946,433 | 12,687 | FreeAlgebra Tutorials!
Home Polynomials Finding the Greatest Common Factor Factoring Trinomials Absolute Value Function A Summary of Factoring Polynomials Solving Equations with One Radical Term Adding Fractions Subtracting Fractions The FOIL Method Graphing Compound Inequalities Solving Absolute Value Inequalities Adding and Subtracting Polynomials Using Slope Solving Quadratic Equations Factoring Multiplication Properties of Exponents Completing the Square Solving Systems of Equations by using the Substitution Method Combining Like Radical Terms Elimination Using Multiplication Solving Equations Pythagoras' Theorem 1 Finding the Least Common Multiples Multiplying and Dividing in Scientific Notation Adding and Subtracting Fractions Solving Quadratic Equations Adding and Subtracting Fractions Multiplication by 111 Adding Fractions Multiplying and Dividing Rational Numbers Multiplication by 50 Solving Linear Inequalities in One Variable Simplifying Cube Roots That Contain Integers Graphing Compound Inequalities Simple Trinomials as Products of Binomials Writing Linear Equations in Slope-Intercept Form Solving Linear Equations Lines and Equations The Intercepts of a Parabola Absolute Value Function Solving Equations Solving Compound Linear Inequalities Complex Numbers Factoring the Difference of Two Squares Multiplying and Dividing Rational Expressions Adding and Subtracting Radicals Multiplying and Dividing Signed Numbers Solving Systems of Equations Factoring Out the Opposite of the GCF Multiplying Special Polynomials Properties of Exponents Scientific Notation Multiplying Rational Expressions Adding and Subtracting Rational Expressions With Unlike Denominators Multiplication by 25 Decimals to Fractions Solving Quadratic Equations by Completing the Square Quotient Rule for Exponents Simplifying Square Roots Multiplying and Dividing Rational Expressions Independent, Inconsistent, and Dependent Systems of Equations Slopes Graphing Lines in the Coordinate Plane Graphing Functions Powers of Ten Zero Power Property of Exponents The Vertex of a Parabola Rationalizing the Denominator Test for Factorability for Quadratic Trinomials Trinomial Squares Solving Two-Step Equations Solving Linear Equations Containing Fractions Multiplying by 125 Exponent Properties Multiplying Fractions Adding and Subtracting Rational Expressions With the Same Denominator Quadratic Expressions - Completing Squares Adding and Subtracting Mixed Numbers with Different Denominators Solving a Formula for a Given Variable Factoring Trinomials Multiplying and Dividing Fractions Multiplying and Dividing Complex Numbers in Polar Form Power Equations and their Graphs Solving Linear Systems of Equations by Substitution Solving Polynomial Equations by Factoring Laws of Exponents index casa mÃo Systems of Linear Equations Properties of Rational Exponents Power of a Product and Power of a Quotient Factoring Differences of Perfect Squares Dividing Fractions Factoring a Polynomial by Finding the GCF Graphing Linear Equations Steps in Factoring Multiplication Property of Exponents Solving Systems of Linear Equations in Three Variables Solving Exponential Equations Finding the GCF of a Set of Monomials
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
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Author Message
UnspokemDesegns
Registered: 04.12.2002
From:
Posted: Thursday 28th of Dec 14:08 Hi math lovers , I heard that there are various software that can help with us doing our homework,like a tutor substitute. Is this really true? Is there a software that can assist me with math? I have never tried one thus far , but they shouldn't be hard to use I assume. If anyone tried such a program, I would really appreciate some more detail about it. I'm in Basic Math now, so I've been studying things like multiplying and dividing rational expressions solver and it's not easy at all.
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Thursday 28th of Dec 18:26 Well, I cannot do your assignment for you as that would mean plagiarism . However, I can give you a suggestion. Try using Algebrator. You can find detailed and well explained answers to all your queries in multiplying and dividing rational expressions solver.
DoniilT
Registered: 27.08.2002
From:
Posted: Thursday 28th of Dec 21:52 I can vouch for that. Algebrator is the best technology for solving algebra assignments. Been using it for some time now and it keeps on amazing me. Every assignment that I type in, Algebrator gives me a perfect answer to it. I have never enjoyed doing math assignment on exponent rules, graphing parabolas and subtracting fractions so much before. I would advise it for sure.
omiguna
Registered: 31.10.2003
From:
Posted: Saturday 30th of Dec 13:36 This sounds really too good to be true. How can I get it? I think I might recommend it to my friends if it is really as great as it sounds.
thicxolmed01
Registered: 16.05.2004
From: Welly, NZ
Posted: Sunday 31st of Dec 20:52 I advise using Algebrator. It not only assists you with your math problems, but also gives all the required steps in detail so that you can enhance the understanding of the subject.
cufBlui
Registered: 26.07.2001
From: Scotland
Posted: Monday 01st of Jan 19:21 Sure, here it is: https://mathsite.org/solving-polynomial-equations-by-factoring.html. Good Luck with your exams. Oh, and before I forget , these guys are also offering unrestricted money back guarantee, that just goes to show how sure they are about their product. I’m sure that you’ll love it . Cheers.
All Right Reserved. Copyright 2005-2022 | 1,477 | 6,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-27 | latest | en | 0.775932 |
http://www.proz.com/kudoz/english_to_polish/tech_engineering/339296-lpa.html | 1,502,904,048,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102309.55/warc/CC-MAIN-20170816170516-20170816190516-00104.warc.gz | 589,817,947 | 17,404 | # LpA
## Polish translation: LpA (natężenie dźwięku)
#### Login or register (free and only takes a few minutes) to participate in this question.
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17:54 Jan 9, 2003
English to Polish translations [PRO]
Tech/Engineering
English term or phrase: LpA Maxium of the sound pressure level is 1m distance from the machine surface LpA, 1m, max - domyslam się, ze ten skrót dotyczy odległości...
Maciej AndrzejczakPoland Local time: 19:20
Polish translation:LpA (natężenie dźwięku) Explanation:LpA nie tłumaczymy, to oznaczenie natężenia dźwieku: przykłady: "Poziomy hałasu: Lpa+52 dB(A) podczas drukowania" http://www.phys.uni.torun.pl/~duch/Wyklady/komput/w06/Hp-dru... "Poziom hałasu LpA: 62,6 dB" http://www.sarpol.poznan.pl/atlas_copco/ac_generatory_qas.ht... --------------------------------------------------Note added at 2003-01-09 18:29:39 (GMT)--------------------------------------------------tutaj wyjaśnienie co to jest: \"the (relative) acoustic pressure in dB(A) - LpA - which is the most commonly quoted figure. However, the value obtained depends on the distance and position relative to the source from which the measurement is taken.\" http://www.hp.ca/products/plus/businessPCs/library/pdfs/ac_n... (i stąd się bierze 1 m - wartośc LPa jest podana dla jednego metra a tutaj (nie wiem czy najlepsze) wyjaśnienie skrótu: \"Why not leaving the worldwide know concepts LWA (with or without d) and LpA. L refers to the logarithmic level experessed in dB (worldwide accepted). W and p to respectively power (Watt) and pressure.\" http://groups.google.pl/groups?hl=pl&lr=&ie=UTF-8&oe=UTF-8&t...
Selected response from:
Pawel Czernecki
Local time: 19:20
OK! MA4 KudoZ points were awarded for this answer
3LpA (natężenie dźwięku)
Pawel Czernecki
27 mins confidence:
LpA (natężenie dźwięku)
Explanation:
LpA nie tłumaczymy, to oznaczenie natężenia dźwieku:
"Poziomy hałasu: Lpa+52 dB(A) podczas drukowania"
"Poziom hałasu LpA: 62,6 dB"
http://www.sarpol.poznan.pl/atlas_copco/ac_generatory_qas.ht...
--------------------------------------------------
Note added at 2003-01-09 18:29:39 (GMT)
--------------------------------------------------
tutaj wyjaśnienie co to jest:
\"the (relative) acoustic pressure in dB(A) - LpA - which is the most commonly quoted
figure. However, the value obtained depends on the distance and position relative to
the source from which the measurement is taken.\"
(i stąd się bierze 1 m - wartośc LPa jest podana dla jednego metra
a tutaj (nie wiem czy najlepsze) wyjaśnienie skrótu:
\"Why not leaving the worldwide know concepts LWA (with or without d) and LpA.
L refers to the logarithmic level experessed in dB (worldwide accepted).
W and p to respectively power (Watt) and pressure.\"
Reference: http://www.sarpol.poznan.pl/atlas_copco/ac_generatory_qas.ht...
Pawel CzerneckiLocal time: 19:20Native speaker of: PolishPRO pts in pair: 1049
OK! MA
31 mins confidence:
LpA
Explanation:
patrz http://www.proz.com/?sp=h&id=326634
rozwiniecie to relative level of accoustice pressure (http://www.hp.ca/products/plus/businessPCs/library/pdfs/ac_n...
Radek PodolskiLocal time: 10:20Native speaker of: PolishPRO pts in pair: 1010
#### KudoZ™ translation helpThe KudoZ network provides a framework for translators and others to assist each other with translations or explanations of terms and short phrases.
P.O. Box 903
Syracuse, NY 13201
USA
+1-315-463-7323
ProZ.com Argentina
Calle 14 nro. 622 1/2 entre 44 y 45
La Plata (B1900AND), Buenos Aires
Argentina
+54-221-425-1266
ProZ.com Ukraine
6 Karazina St.
Kharkiv, 61002
Ukraine
+380 57 7281624 | 1,136 | 3,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-34 | latest | en | 0.510195 |
http://www.jiskha.com/display.cgi?id=1402269896 | 1,495,750,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00028.warc.gz | 549,877,007 | 3,735 | # MATH
posted by on .
Find two quadratic functions whose graph have given x-intercepts. Find one function whose graph opens upward and another whose graph opens downward.
(-4,0) (0,0)
• MATH - ,
from the values of the roots, we know that the functions will be of the form
y = ax(x+4)
So, pick a positive to open upward, and a negative to open downward.
Any nonzero value will do. | 97 | 386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-22 | latest | en | 0.900921 |
https://socratic.org/questions/how-do-you-factor-15x-2-19x-6 | 1,582,512,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00391.warc.gz | 545,512,405 | 6,097 | # How do you factor 15x^2-19x+6?
Aug 5, 2015
Factor $y = 15 {x}^{2} - 19 x + 6$
Ans: (5x - 3)(3x - 2)
#### Explanation:
$y = 15 {x}^{2} - 19 x + 6 =$ 15(x - p)(x - q).
I use the new AC Method (Yahoo, Google Search)
Converted trinomial $y ' = {x}^{2} - 19 x + 90 =$ (x - p')(x - q')
Factor pairs of 90 -> ...(6, 15)(9, 10). This sum is 19 = -b. Then p' = -9 and q' = -10.
Then, $p = \frac{p '}{a} = - \frac{9}{15} = - \frac{3}{5}$ and $q = - \frac{10}{15} = - \frac{2}{3.}$
Factored form: $y = 15 \left(x - \frac{3}{5}\right) \left(x - \frac{2}{3}\right) = \left(5 x - 3\right) \left(3 x - 2\right)$ | 286 | 604 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-10 | latest | en | 0.523892 |
https://socratic.org/questions/what-is-log-8-5-8 | 1,579,976,472,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00293.warc.gz | 646,616,276 | 5,871 | # What is log_8(5*8)?
Jul 6, 2018
$\approx 1.77$
#### Explanation:
We can simplify the parenthesis to now get
${\log}_{8} \left(40\right)$
Since we have no perfect squares or cubes and we're not dealing with base-$10$ or base-$e$, it would be best to evaluate this with a calculator.
${\log}_{8} \left(40\right) \approx 1.77$
Hope this helps! | 111 | 350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-05 | longest | en | 0.813654 |
https://writeabout.tech/programming/create-a-lock-class-to-prevent-deadlock/ | 1,585,385,395,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370490497.6/warc/CC-MAIN-20200328074047-20200328104047-00242.warc.gz | 790,051,355 | 13,752 | # Create a lock class to prevent deadlock
There are several common options to prevent deadlock. One of the most popular is to explicitly declare what lock is required. This way, we will be able to check whether the created lock is a deadlock and if so, we can stop working.
Let’s take a look and sort out the way we can detect a deadlock. Suppose we request the following sequence of locks:
А = {1, 2, 3, 4}
В = {1, 3, 5}
С = {7, 5, 9, 2}
This will result in a deadlock appearance due to the fact that:
А blocks 2 and is waiting for 3
В blocks 3 and is waiting for 5
С blocks 5 and is waiting for 2
You can imagine this scenario as a graph, where 2 is connected to 3, and 3 is connected to 5, and 5 is connected to 2. Deadlock is described by a loop. An edge (w, v) exists in the graph if the process announces that it requests lock v immediately after lock w. In the previous example, the following edges will exist in the following graph:
(1, 2), (2, 3), (3, 4), (1, 3), (3, 5), (7, 5), (5, 9), (9, 2).
The owner of the link does not matter.
This class will require the declare method, which uses threads and processes to declare the sequence in which resources will be requested. The declare method will check the sequence of the declaration by adding each continuous pair of elements (v, w) to the graph. Subsequently, it will check it for cycles. If a cycle occurs, it will remove the added edge from the graph and exit.
We have to discuss only one nuance. How do we detect a loop? We can detect a loop by a depth-first search through each related element (i.e., through each graph component). There are complex components that allow you to select all the connected graph components, but our task is not so difficult.
We know that if a looped link arises, then we should check one of the links. Thus, if a depth-first search touches these links, we will definitely find a looped link.
The pseudocode for this loop detection is something like this:
boolean checkForCycle(locks[] locks) {
touchedNodes = hash table(lock -> boolean)
// initialize touchedNodes by setting each lock in locks as false
for each (lock x in process.locks) {
if (touchedNodes[x] == false) {
if (hasCycle(x, touchedNodes)) {
return true;
}
}
}
return false;
}
boolean hasCycle(node x, touchedNodes) {
touchedNodes[r] = true;
if (x.state == VISITING) {
return true;
} else if (x.state == FRESH) {
//…(see the full code below)
}
}
In this code block, we can eecute several depth-first searches, but we need to initialize touchedNodes only once. We do iterations until all the values in touchedNotes are false.
The following code below is more detailed. For clarity, we suppose that all the locks and processes (owners) are consistently arranged.
public class LockFactory {
private static LockFactory instance;
private int numberOfLocks = 5; /* by default*/
private LockNode[] locks;
/ * Display the process (of the owner) in order,
* in which the owner demanded a lock * /
private LockFactory(int count) { … }
public static LockFactory getInstance() { return instance; }
public static synchronized LockFactory initialize(int count) {
if (instance == null)
instance = new LockFactory(count);
return instance;
}
public boolean hasCycle(Hashtable<Integer, Boolean> touchedNodes, int[] resourcesInOrder) {
/* verify for the looped link existence */
for (int resource : resourcesInOrder) {
if (touchedNodes.get(resource) == false) {
LockNode n = locks[resource];
if (n.hasCycle(touchedNodes)) {
return true;
}
}
}
return false;
}
/ * To prevent deadlock, force processes
* announce that they want to block. We check
* that the requested sequence will not cause deadlock
* (looped link in a directed graph) * /
public boolean declare(int ownerId, int[] resourcesInOrder) {
Hashtable<Integer, Boolean> touchedNodes = new Hashtable<Integer, Boolean>();
/* adding nodes to the graph */
int index = 1;
touchedNodes.put(resourcesInOrder[0], false);
for (index = 1; index < resourcesInOrder.length; index++) {
LockNode prev = locks[resourcesInOrder[index – 1]];
LockNode curr = locks[resourcesInOrder[index]];
prev.joinTo(curr);
touchedNodes.put(resourcesInOrder[index], false);
}
/ * if looped link is received, remove this list of resources
* and return false * /
if (hasCycle(touchedNodes, resourcesInOrder)) {
for (int j = 1; j < resourcesInOrder.length; j++) {
LockNode p = locks[resourcesInOrder[j – 1]];
LockNode c = locks[resourcesInOrder[j]];
p.remove(c);
}
return false;
}
* since we can verify that the process really causes
* lock in the desired order * /
for (int i = 0; i < resourcesInOrder.length; i++) {
LockNode resource = locks[resourcesInOrder[i]];
}
lockOrder.put(ownerId, list);
return true;
}
/ * Receive the lock, check first that the process
* really requests locking in the declared order * /
public Lock getLock(int ownerld, int resourceID) {
if (list == null) return null;
list.removeFirst();
}
return null;
}
}
public class LockNode {
public enum VisitState { FRESH, VISITING, VISITED );
private ArrayList<LockNode> children;
private int lockId;
private Lock lock;
private int maxLocks;
public LockNode(int id, int max) { … }
/* Connect “this” to “node”, verify not to create
public void joinTo(LockNode node) { children.add(node); }
public void remove(LockNode node) { children.remove(node); }
/* Check for the existence of loop with the help of depthfirst search */
public boolean hasCycle(Hashtable<Integer, Boolean> touchedNodes) {
VisitState[] visited = new VisitState[maxLocks];
for (int i = 0; i < maxLocks; i++) {
visited[i] = VisitState.FRESH;
}
return hasCycle(visited, touchedNodes);
}
private boolean hasCycle(VisitState[] visited,
Hashtable<Integer, Boolean> touchedNodes) {
if (touchedNodes.containsKey(lockId)) {
touchedNodes.put(lockId, true);
}
if (visited[lockId) == VisitState.VISITING) {
/ * We loop back to this node, therefore
* we know that there is a loop (looped link) * /
return true;
} else if (visited[lockId] == VisitState.FRESH) {
visited[lockId] = VisitState.VISITING;
for (LockNode n : children) {
if (n.hasCycle(visited, touchedNodes)) {
return true;
}
}
visited[lockId] = VisitState.VISITED;
}
return false;
}
public Lock getLock() {
if (lock == null) lock = new ReentrantLock();
return lock;
}
public int getId() { return lockId; }} | 1,588 | 6,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-16 | longest | en | 0.923475 |
https://gadgetspidy.com/gemstones/ | 1,719,005,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00369.warc.gz | 238,745,994 | 29,213 | ## Gemstones
QUESTION
John has discovered various rocks. Each rock is composed of various elements, and each element is represented by a lower-case Latin letter from ‘a’ to ‘z’. An element can be present multiple times in a rock. \n\nAn element is called a gem-element if it occurs at least once in each of the rocks.\n\nGiven the list of N rocks with their compositions, display the number of gem-elements that exist in those rocks.\n\nInput Format\n\nThe first line consists of an integer, N, the number of rocks. \nEach of the next N lines contains a rock’s composition. Each composition consists of lower-case letters of English alphabet.\n\nConstraints \n 1<=N<=100\nEach composition consists of only lower-case Latin letters (‘a’-‘z’). \n 1<=length of each composition<=100 \n\nOutput Format\n\nPrint the number of gem-elements that are common in these rocks. If there are none, print 0.
``````#include<iostream>
#include<string>
using namespace std;
int main()
{
int T;
int a[26] = {0};
bool flag[26] = {false};
int nCount = 0;
cin >> T;
cin.ignore();
int curTest = 1;
while(curTest <= T)
{
string in;
getline(cin,in);
for(int i = 0; i < in.length();i++)
{
int ch = ((int)in[i]) - 97;
if( ch >= 0 && ch < 26 && flag[ch] == false)
{ a[ch]++;
flag[ch] = true;
}
}
for(int i = 0; i < 26;i++)
flag[i] = false;
curTest++;
}
for(int i = 0 ; i <= 25;i++)
if(a[i] == T)
nCount++;
cout<<nCount;
return 0;
}
`````` | 405 | 1,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.714794 |
logicalscript.com | 1,713,736,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00151.warc.gz | 308,763,732 | 15,348 | # What Does The Percentage of Rain Mean? Importance, Calculation, and Classification
Contents
What Does The Percentage of Rain Mean? Read the full article to know about the significance of rain percentage, calculation of rain percentage and several practical examples related to the calculation of rain percentage.
## Introduction
#### What does the percentage of rain mean?
The percentage of rain that is displayed on the weather app indicates the probability or greater chance of occurring rainfall in a particular area during a specific time period. It represents the predicted possibility that it will rain at a particular location.
#### Why rain percentage calculation is important?
Being aware of the rain percentage offers useful information about the weather and enables people and organizations to plan their activities efficiently, avoid inconvenience, and take the necessary precautions. The percentage of rain can be helpful for a number of sectors such as construction, agriculture, transportation, etc.
1. Knowing the percentage of rain is important because people can better plan their activities and make informed decisions based on the weather. People can get ready for rain by being aware of the probability of it. For example, if the probability of rain is high it might be necessary to carry an umbrella or raincoat, reschedule outdoor events, or make backup plans.
2. Based on the estimated rainfall, farmers can choose the best time to irrigate their crops.
3. Construction companies can reschedule their outdoor projects.
4. Airlines and transportation providers can also plan ahead or make necessary adjustments to their operations.
## How to calculate rain percentage?
To find out the amount of rainfall in a specific area at a definite time period, it is necessary to calculate the percentage of rain. The calculation depends on various factors corresponding to the total amount of rainfall. You must consider the following factors in determining the percentage of rain.
1. Rainfall measurement: First you have to measure the amount of rainfall that has occurred. A rain gauge is commonly used to collect and measures the volume of rain. Generally, the measurement is expressed in terms of either millimeters (mm) or inches (in).
2. Reference Area: Select the reference area for which the rainfall measurement is being taken into consideration. It might be a particular area, a city, or even a tiny backyard. The reference area is an important factor in determining the percentage of rainfall.
3. Time Period: Define the time duration for which you want to determine the percentage of rain. It might be a minute, a day, a month, or any other period of time. In order to understand the intensity of the rainfall and its effects, the time period is necessary.
After finding out all these factors, you can use the below formula to determine the percentage of rain [1].
Percentage of rain = (Recorded Rainfall / Reference Area) x 100
• Example to calculate the percentage of rain:
Question: Calculate the percentage of rain in a city if the recorded rainfall is 70 mm over a 12-hour time period.
Answer: Percentage of rain = (70 mm x Reference area (in square units)) x 100
This is a general approach to calculating the percentage of rain. This may vary depending on specific meteorological calculations used by professionals.
## Analyzing the rain percentage
Usually, meteorologists and weather forecasters classify rain percentages according to the severity or volume of the rainfall. People can use these classifications to better understand the effects of rain and plan their activities accordingly.
1. Low rain percentage
A small or negligible chance of rainfall is referred to as a low percentage of rain. When the percentage of rain is less than 30%, then it means there is little chance of rain or a low probability of rainfall.
1. Moderate rain percentage
A moderate chance of rainfall is referred to as a moderate percentage of rain. When the percentage of rain ranges between 30% to 60%, then it suggests there is a significant possibility of rainfall, but it is not guaranteed.
1. High rain percentage
A high chance of rainfall is referred to as a high percentage of rain. It usually suggests rain is highly probable. A significant amount of rain usually exceeds 60% and can go up to 100%.
By understanding these different categories of rain percentages, individuals can better interpret
## Some practical examples related to the rain percentage
• What does a 90% chance of rain mean?
The term “90% chance of rain” indicates the probability of rainfall at a particular location for a specific time period. If a city’s weather forecast indicates a 90% chance of rain tomorrow, then it generally means that there is a high probability of rainfall in that city tomorrow. The report is totally based on the data and analysis that are currently available. It can also be explained as , if this scenario were repeated 100 times under the same conditions, we would expect rain to occur approximately 90 out of those 100 times.
The percentage of rain indicates the probability of rainfall it does not indicate its duration or intensity.
• What does a 70% chance of rain mean on the weather app?
When a city’s weather app reports a 70% chance of rain, it indicates that there is a good chance that it will rain in the city. The report is based on several meteorological factors like cloud cover, atmospheric conditions, and historical data. The percentage refers to the probability of rain, however, it does not guarantee that rain will fall.
• Does a 60% chance of rain mean it will rain?
When a city’s weather app reports a 60% chance of rain, it indicates that there is a moderate probability of rain in the city. However, it is important to note that the percentage is not a guarantee but rather an estimation of the chances of rain.
## Conclusion
By understanding these rain percentage calculations individuals can plan events, and outdoor activities, and prepare for their own safety more effectively. To get a thorough understanding of the weather, it is advised to frequently check weather updates, and monitor any changes in the forecast. While calculating rain percentage it should be remembered that weather patterns can be unpredictable and the prediction is for a specific location and for a particular time period.
## FAQS
1. What does the percentage of rain mean?
The percentage of rain that is displayed on the weather app indicates the probability or greater chance of occurring rainfall in a particular area during a specific time period.
1. How to calculate rain percentage?
Formula to find out rain percentage of a particular area is
Percentage of rain = (Recorded Rainfall / Reference Area) x 100
1. What are the types of rain percentage?
There are 3 types rain percentage i.e., low rain percentage, moderate rain percentage and high rain percentage.
Reference:
Also Check:
1. https://logicalscript.com/index.php/2023/07/22/florida-farm-bureau/
2. https://logicalscript.com/index.php/2023/07/26/cleanser-for-oily-skin/
What Does The Percentage of Rain Mean? Importance, Calculation, and Classification
Scroll to top | 1,415 | 7,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-18 | latest | en | 0.914978 |
https://centinel2012.com/2022/04/22/a-technical-study-in-the-relationships-of-solar-flux-water-carbon-dioxide-in-the-upper-atmosphere-using-the-latest-march-2022-nasa-noaa-data/?shared=email&msg=fail | 1,657,037,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00355.warc.gz | 203,702,013 | 32,822 | # A Technical Study in the Relationships of Solar Flux, Water, Carbon Dioxide in the upper Atmosphere, Using the latest March, 2022 NASA & NOAA Data
From the attached report on climate change for March 2022 Data we have the two charts showing how much the global temperature has actually gone up since we started to measure CO2 in the atmosphere in 1958? To show this graphically Chart 8a was constructed by plotting CO2 as a percent increase from when it was first measured in 1958, the Black plot, the scale is on the left and it shows CO2 going up by about 32.0% from 1958 to March of 2022. That is a very large change as anyone would have to agree. Now how about temperature, well when we look at the percentage change in temperature also from 1958, using Kelvin (which does measure the change in heat), we find that the changes in global temperature (heat) is almost un-measurable at only .4%. As you see the increase in energy, heat, is not visually observably in this chart hence the need for another Chart 8 to show the minuscule increase in thermal energy shown by NASA in relationship to the change in CO2 Shown in the next Chart using a different scale.
This is Chart 8 which is the same as Chart 8a except for the scales. The scale on the right side had to be expanded 10 times (the range is 50 % on the left and 5% on the right) to be able to see the plot in the same chart in any detail. The red plot, starting in 1958, shows that the thermal energy in the earth’s atmosphere increased by .40%; while CO2 has increased by 32.0% which is 80 times that of the increase in temperature. So is there really a meaningful link between them that would give as a major problem?
Based to these trends, determined by excel not me, in 2028 CO2 will be 428 ppm and temperatures will be a bit over 15.0o Celsius and in 2038 CO2 will be 458 ppm and temperatures will be 15.6O Celsius.
## The numbers tell us the story of the planets Atmosphere
The full 40 page report explains how these charts were developed .
<object class="wp-block-file__embed" data="https://centinel2012.files.wordpress.com/2022/04/blackbody-temperature-2022-03.pdf" type="application/pdf" style="width:100%;height:600px" aria-label="Embed of <strong>blackbody-temperature-2022-03blackbody-temperature-2022-03
By Centinel2012
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 587 | 2,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.956449 |
https://www.stata.com/statalist/archive/2008-03/msg01103.html | 1,558,629,684,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257316.10/warc/CC-MAIN-20190523164007-20190523190007-00187.warc.gz | 944,653,081 | 2,527 | # st: Chow test in nonlinear model
From Giulio Zanella To statalist@hsphsun2.harvard.edu Subject st: Chow test in nonlinear model Date Fri, 28 Mar 2008 19:21:10 +0100
Dear Stata users,
I would like to run a Chow test for equality of coefficients across two Probit models.
A procedure that works in linear models, as previously discussed here, is the following:
The model is:
y = a1 + b1*x1 + c1*x2 + u for group 1
y = a2 + b2*x1 + c2*x2 + u for group 2
But we can collapse the two equations:
y = a3 + b3*x1 + c3*x2 + a3'*g2 + b3'*g2*x1 + c3'*g2*x2 + u,
where g2=1 if group is 2 and g2=0 otherwise, estimate this and then test that g2 g2*x1 and g2*x2 are all equal to zero.
Intuitively, this procedure should work in a nonlinear model too, but I wonder whether this hides some pitfall.
Any hint would be appreciated.
Many thanks
Giulio Zanella
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | 317 | 1,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-22 | latest | en | 0.816195 |
http://brainfans.com/brain-teasers/logic-riddles?page=5&per-page=9 | 1,550,468,861,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00341.warc.gz | 42,740,583 | 8,015 | ### Three men riddle
Three men are walking in the same direction. The first man has two men behind him, the second man has one in front and one behind, but the third man has one in front and one behind too. How is this possible if there were only three men?
21 liked this
40 solved this
0 | 69 | 291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | latest | en | 0.985648 |
https://tw.answers.yahoo.com/question/index?qid=20120204000016KK05090 | 1,597,343,665,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739048.46/warc/CC-MAIN-20200813161908-20200813191908-00353.warc.gz | 537,191,870 | 26,571 | # 代數學(Let G be a group of order)
Let G be a group of order 10. Then show that(1) There exists two elememts a,b∈G such that o(a)=2, o(b)=5.(2) (b) is a normal subgroup of G.(3) If aba-1=b^n where 0≤n≤4, then compute n=?
### 1 個解答
• 9 年前
最佳解答
請參考，圖不清楚請下載至電腦
參考資料： 我自己+數學娘的加持 | 130 | 277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-34 | latest | en | 0.406379 |
https://www.coursehero.com/file/6009350/cs-Practice-Test-1/ | 1,498,589,276,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321497.77/warc/CC-MAIN-20170627170831-20170627190831-00221.warc.gz | 855,871,742 | 174,345 | cs-Practice Test 1
# cs-Practice Test 1 - Practice Test 1 Spring 2008 Problem 1...
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Practice Test 1 – Spring 2008 Problem 1 – Multiple Choice [30 Points] 1. If the following script is run in matlab, A = linspace(0,100,100); Which of the following vectors B would be equivalent to A ? A. B = 0:1:100 B. B = 0:100:100 C. B = 1:0.01:100 D. B = 0:(101/100):100 E. It is not mathematically possible to make this conversion. 2. The following script is executed a = true vec = [5 6 2] while a for i = vec disp 'CS rocks' end end How many times will ‘CS rocks’ be displayed? A. 6 times B. 5 times C. 2 times D. 3 times E. Infinite 3. The following MATLAB code is executed: str = 'mATLaB'; for k = length(str):-1:2 if lower(str(k)) > lower(str(k-1)) str(k) = char(str(k)+1); end end What is the value stored in variable str after all of the code above has been executed? A. ' mAULaB ' B. ' mAULAC ' C. ' mAVLaC ' D. ' mAULaC ' E. ' mATLaC '
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4. Consider the following: A=[6,2,3,9]; B=mod(A,2); C=A(B(4)); A(C)=9; What is the final value of A? A. [6 2 3 9] B. [1 0 1 0] C. [6 2 3 9 0 9] D. [9 0 9 3 2 6] E. Error 5. Consider the following lines of code: A = {1:10, 'CS', [1 3 7 1], 1:10}; A(2)=[]; B=A{1}(2)+A{3} What is the value of B ? A. [4 5 6 7 8 9 10 11 12 13] B. [3 4 5 6 7 8 9 10 11 12] C. [3 5 9 3] D. 2 E. Error 6. Given a string, str, which of the following is a valid method of converting it only lowercase? I str = lower(str) II str = upperToLower(str) III index = find(str<=97) str(index) = str(index) + 32; IV str=str+32 A. I only B. I and IV C. II only D. II and III E. I and III 7. Below are two for loops that find the sum of all the values in x. A = 0; for index = 1:length(x) A = A + x(index); end B = 0; for index = x B = B + index; end What style of for loop does the A and B use? A. A and B uses indexing B. A and B uses direct-access C. A uses indexing; B uses direct-access D. A uses direct-access; B uses indexing
E. None of the Above 8. Consider the following code typed into MATLAB: ca={'Superman',7:-1:4,'Batman',1:10}; ca(5)={strcmpi(ca{1}(end:-1:end-3),ca{3}(end:-1:end-3))}; ca{3}=[]; What would the value of ca be at the end program: A. ca = 'Superman' [1x4 double] [] [1x10 double] [1] B. ca = 'Superman' [1x4 double] [1x10 double] [0] C. ca = 'Superman' [1x4 double] [] [1x10 double] [0] D. ca = 'Superman' [1x4 double] [1x10 double] [1] E. ca = 'Superman' [1x4 double] [1x10 double] [] [1] 9. The following MATLAB function is provided:
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## This note was uploaded on 11/08/2010 for the course CS 1371 taught by Professor Stallworth during the Spring '08 term at Georgia Tech.
### Page1 / 11
cs-Practice Test 1 - Practice Test 1 Spring 2008 Problem 1...
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Ask a homework question - tutors are online | 1,095 | 3,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-26 | longest | en | 0.753187 |
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# MAT3007 Homework 7 solution
Original price was: \$35.00.Current price is: \$29.00.
Category:
5/5 - (1 vote)
## Problem 1 (25pts).
Suppose that f : R → R is convex, and a, b ∈ dom f with a < b, where dom
denotes the domain of the function. More specifically, f : Rp → Rq means that f is an Rp
-valued
function on some subset of Rp
, and this subset of Rp
is the domain of the function f. Show that
(a)
f(x) ≤
b − x
b − a
f(a) + x − a
b − a
f(b), for all x ∈ (a, b)
Hint: (Jensen’s Inequality) If p1, …, pn are positive numbers which sum to 1 and f is a real
continuous function that is convex, then
f
Xn
i=1
pixi
!
Xn
i=1
pif (xi)
(b)
f(x) − f(a)
x − a
f(b) − f(a)
b − a
f(b) − f(x)
b − x
for all x ∈ (a, b). Draw a sketch that illustrates this inequality.
(c) Suppose f is differentiable. Use the result in (b) to show that:
f
(a) ≤
f(b) − f(a)
b − a
≤ f
(b)
Note that these inequalities also follow form:
f(b) ≥ f(a) + f
(a)(b − a), f(a) ≥ f(b) + f
(b)(a − b)
(d) Suppose f is twice differentiable. Use the result in (c) to show that f
′′(a) ≥ 0 and f
′′(b) ≥ 0.
1
## Problem 2 (30pts) Show that the following functions are convex:
(a) f(x) = − log
− log Pm
i=1 e
a
T
i
x+bi
on dom f =
n
x |
Pm
i=1 e
a
T
i
x+bi < 1
o
. You can use the
fact that log (Pn
i=1 e
yi ) is convex.
(b)
f(x, u, v) = − log
uv − x
T x
on dom f =
(x, u, v) | uv > xT x, u, v > 0
(c) Let T(x, ω) denote the trigonometric polynomial
T(x, ω) = x1 + x2 cos ω + x3 cos 2ω + · · · + xn cos(n − 1)ω
Show that the function
f(x) = −
Z 2π
0
log T(x, ω)dω
is convex on {x ∈ Rn
| T(x, ω) > 0, 0 ≤ ω ≤ 2π}.
Hint: Nonnegative weighted sum of convex functions is still convex. Let this property extend
to infinite sums and integrals. Assume that f(x, y) is convex in x for each y ∈ A and w(y) ≥ 0
for each y ∈ A and integral exists. Then the function g defined as
g(x) = Z
A
w(y)f(x, y)dy
is convex in x.
## Problem 3 (20pts). Consider the following function:
minimize −x1 − x2 + max {x3, x4}
s.t. (x1 − x2)
2 + (x3 + 2×4)
4 ≤ 5
x1 + 2×2 + x3 + 2×4 ≤ 6
x1, x2, x3, x4 ≥ 0
(a) Verify this is a convex optimization problem.
(b) Use CVX to solve the problem.
## Problem 4 (25pts).
To model the influence of price on customer purchase probability, the
following logit model is often used:
λ(p) = e
−p
1 + e−p
where p is the price, λ(p) is the purchase probability.
Assume the variable cost of the product is 0 (e.g., iPhone Apps). As the seller, you want to
maximize the expected revenue by choosing the optimal price. That is, you want to solve:
maximizep pλ(p)
(a) Draw a picture of r(p) = pλ(p) (for p from 0 to 10) and use the picture to show that r(p) is
not concave (thus maximize r(p) is not a convex optimization problem)
(b) Write down p as a function of λ (the inverse function of λ(p) ). Show that you can write the
objective function as a function of λ : ˜r(λ), where ˜r(λ) is concave in λ.
(c) From part 2, write the KKT condition for the optimal λ. Then transform it back to an optimal
condition in p.
3 | 1,104 | 3,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.775534 |
https://socratic.org/questions/what-is-91-of-130-7 | 1,718,536,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00222.warc.gz | 478,293,436 | 5,969 | # What is 91% of 130.7?
Mar 13, 2018
$118.937 \approx 119$
#### Explanation:
Percent means out of 100, so when finding a percentage, divide the number by $100$ to find 1%.
Then simply multiply the 1% by the desired percentage.
so...
$130.7 \div 100 = 1.307 \text{ } \leftarrow$ (this is 1%)
The desired percentage is 91%
So multiply: $1.307 \times 91.$
this will give you your answer of $118.937$
Or rounded, at about $119.$ | 139 | 433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.868943 |
https://www.altsci.com/3/s/We-call-a-field-E-a-splitting-field-for-A-if-A-%E2%8A%97-E | 1,582,595,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00475.warc.gz | 624,594,697 | 7,709 | Page "Central simple algebra" ¶ 5
from Wikipedia
Some Related Sentences
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We call them lay-sisters and they go among the Eskimos making friends and bringing the light.
We can call a person, a house, a symphony, a fragrance, and a mathematical proof beautiful.
We will call such an organ an Accumulator.
Why, man, that's the same kind of music we've been playin ' since 1928 !... We didn't call it rock and roll back when we introduced it as our style back in 1928, and we don't call it rock and roll the way we play it now.
Attlee responded the next day in the debate on increased air estimates that Hitler's speech contained unfavourable references to the Soviet Union but that " We see here a chance to call a halt in the armaments race ... We do not think that our answer to Herr Hitler should be just rearmament.
We will call the first the furnace and the second the refrigerator .” Carnot then explains how we can obtain motive power, i. e. “ work ”, by carrying a certain quantity of heat from body A to body B.
In 1945, Engelbart had read with interest Vannevar Bush's article " As We May Think ", a call to action for making knowledge widely available as a national peacetime grand challenge.
* " We shot music, call it murder on the dancefloor, and we got more bars than wandsworth and dartmoor "-Dartmoor Prison, mentioned in Devlin's song Shot Music ( May 10, 2010 )
: We put thirty spokes together and call it a wheel ;
Let us call the class of all such formulas R. We are faced with proving that every formula in R is either refutable or satisfiable.
We will call such maps open immersions, just as in the context of schemes.
:" We worship ," replied Hengist, " our country gods, Saturn and Jupiter, and the other deities that govern the world, but especially Mercury, whom in our language we call Woden and to whom our ancestors consecrated the fourth day of the week, still called after his name Wodensday.
Savoy dancer " Shorty " George Snowden stated that " We used to call the basic step the Hop long before Lindbergh did his hop across the Atlantic.
We may represent any given proposition with a letter which we call a propositional constant, analogous to representing a number by a letter in mathematics, for instance,.
We will call journalists on every instance of unprofessional reporting.
We call
We call this limit the derivative.
We can say that God is a spirit, but it does not seem fair to call it a mind.
We understand that shows don't want to call the writers writers because they want to maintain the illusion that it is reality, that stuff just happens.
", to which Michael Bluth replied " We just call it sausage.
We and field
We were in a field, in a tight, screeching turn.
We do not favor one field over another: we think that all inquiry, all scholarly and artistic creation, is good -- provided only that it contributes to a sense and understanding of the true ends of life, as all first-rate scholarship and artistic creation does.
We look forward to a stronger position in this expanding field.
We can do this through the characteristic values and vectors of T in certain special cases, i.e., when the minimal polynomial for T factors over the scalar field F into a product of distinct monic polynomials of degree 1.
We might not see any rotation initially, but if we closely look at the right, we see a larger field at, say, x = 4 than at x = 3.
We ’ ll return to Arkansas for at least another field season ,” says Rohrbaugh.
We have evidence that Jupiter has a magnetic field and that birds are oviparous, but as of yet, we do not seem to have found evidence of moral properties, such as " goodness ".
We have previously mentioned that there can be more than one way of indexing the degrees of freedom in a quantum field.
) We can construct field operators by applying the Fourier transform to the creation and annihilation operators for these states.
We take the velocity field associated with this current as the velocity field whose integral curves yield the motion of the particle.
We conclude that, in the field of public education, the doctrine of " separate but equal " has no place.
We provide a sketch of a proof for the case where the underlying field of scalars is the complex numbers.
We have, therefore, directed the Irish Army authorities to have field hospitals established in County Donegal adjacent to Derry and at other points along the Border where they may be necessary.
( We developed ) a theory to explain how solar protons can diffuse inward and gain energy in the Earth's magnetic field.
We now recognise this as part of the ideal class group: in fact Kummer had isolated the p-torsion in that group for the field of p-roots of unity, for any prime number p, as the reason for the failure of the standard method of attack on the Fermat problem ( see regular prime ).
We can assume that A contains ¬ φ, the field axioms, and, for some k, the first k sentences of the form 1 + 1 +...+ 1 ≠ 0 ( because adding more sentences doesn't change unsatisfiability ).
We can construct a classical continuous random field that has the same probability density as the quantum vacuum state, so that the principal difference from quantum field theory is the measurement theory ( measurement in quantum theory is different from measurement for a classical continuous random field, in that classical measurements are always mutually compatible — in quantum mechanical terms they always commute ). | 1,275 | 5,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-10 | longest | en | 0.95258 |
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# In 1986, the city of Los Diablos had 20 days on which air
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In 1986, the city of Los Diablos had 20 days on which air [#permalink]
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28 Apr 2010, 14:52
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In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
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Re: 1000 CR test - 16 [#permalink]
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28 Apr 2010, 15:20
sushantdsahu wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
B???
A) Knowing that it takestwo years to enact the pollution control measures explains the trend.
B) Just because a far more accurate instrument was invented does not mean it was used.
C)Revising the scale means the way pollution was measured was changed and it could have been high in 1989 too but the lower reading was probably because of the revised scale.
D) This is tricky. This may be the reason why the air pollution was still high in 87 and 88.
E)This explain why the readings were high in 87 and 88 and dropped in 89.
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Re: 1000 CR test - 16 [#permalink]
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28 Apr 2010, 16:09
mbafall2011 wrote:
sushantdsahu wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
B???
A) Knowing that it takestwo years to enact the pollution control measures explains the trend.
B) Just because a far more accurate instrument was invented does not mean it was used.
C)Revising the scale means the way pollution was measured was changed and it could have been high in 1989 too but the lower reading was probably because of the revised scale.
D) This is tricky. This may be the reason why the air pollution was still high in 87 and 88.
E)This explain why the readings were high in 87 and 88 and dropped in 89.
If we look to the argument:
86 - 20
87 - 31
88 - 37
89 - 16
that means sudden drop.
The question ask us to find which choice is least helpful in explaining the levels of air-pollution?
D says in 88 industries were exempted from the regulations that means there can be a increase but the argument shows a decrease in level. So i thought answer is B.
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Re: 1000 CR test - 16 [#permalink]
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29 Apr 2010, 01:42
mbafall2011 wrote:
sushantdsahu wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
B???
A) Knowing that it takestwo years to enact the pollution control measures explains the trend.
B) Just because a far more accurate instrument was invented does not mean it was used.
C)Revising the scale means the way pollution was measured was changed and it could have been high in 1989 too but the lower reading was probably because of the revised scale.
D) This is tricky. This may be the reason why the air pollution was still high in 87 and 88.
E)This explain why the readings were high in 87 and 88 and dropped in 89.
[Does it means that the amount of ozone and CO is fixed and fully break down for each year and new package of these gases is created in the atmosphere for further breakdown. I think if this is your reason for negating this choice then it is not correct. Both these gases are in continuous process from 1986 - 1989.]
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Re: 1000 CR test - 16 [#permalink]
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29 Apr 2010, 05:13
My pick is B
Just because a more accurate gas specr was invented doesnot necessarily mean that
it was used to measure pollution in the city. Big assumtions in here
between B and D, B sounds least convincing...
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Re: 1000 CR test - 16 [#permalink]
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29 Apr 2010, 08:57
ykaiim wrote:
mbafall2011 wrote:
sushantdsahu wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
B???
A) Knowing that it takestwo years to enact the pollution control measures explains the trend.
B) Just because a far more accurate instrument was invented does not mean it was used.
C)Revising the scale means the way pollution was measured was changed and it could have been high in 1989 too but the lower reading was probably because of the revised scale.
D) This is tricky. This may be the reason why the air pollution was still high in 87 and 88.
E)This explain why the readings were high in 87 and 88 and dropped in 89.
[Does it means that the amount of ozone and CO is fixed and fully break down for each year and new package of these gases is created in the atmosphere for further breakdown. I think if this is your reason for negating this choice then it is not correct. Both these gases are in continuous process from 1986 - 1989.]
We are told that any gases already in the air takes two years to break down. Thus, the gases that were part of the measured pollutants in 86 cud have been released earlier and have not broken down yet. Similarly in 87, the gases could have released in 85 and 86 and so on. In 1987 new smog alerts were enacted. Hence in 1988 and 1987 the number of new gases releases would have decreased but the readings were high because of the gases released in previous two years. Two years after the new rules were enacted, the gases had decomposed and no new gases were released and therefore 1989 showed a much smaller measurement.
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Re: 1000 CR test - 16 [#permalink]
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29 Apr 2010, 09:41
Hi,
Thanks for the explanation but still the issue is unclear to be.
It is not mentioned in the stimulus when the two gases released and whether these gases would convert into non-pollutants.
I hope some more explanation can help me.
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Re: 1000 CR test - 16 [#permalink]
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29 Apr 2010, 11:21
ykaiim wrote:
Hi,
Thanks for the explanation but still the issue is unclear to be.
It is not mentioned in the stimulus when the two gases released and whether these gases would convert into non-pollutants.
I hope some more explanation can help me.
Think of it this way: which one is the least explanatory? We dont know when the gases were released, just that in 86 and 87 gases were high and that in 87 there were some rules enacted. Thus this is a plausible explanation for the decreased in 89.
Compare this to B and you will see Process of elimination working here. E makes more sense than B.
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Re: 1000 CR test - 16 [#permalink]
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03 May 2010, 11:53
my pick is (B). what is the OA?
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Re: 1000 CR test - 16 [#permalink]
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03 May 2010, 13:45
sushantdsahu wrote:
In 1986, the city of Los Diablos had 20 days on which air pollution reached unhealthful amounts and a smog alert was put into effect. In early 1987, new air pollution control measures were enacted, but the city had smog alerts on 31 days that year and on 39 days the following year. In 1989, however, the number of smog alerts in Los Diablos dropped to sixteen. The main air pollutants in Los Diablos are ozone and carbon monoxide, and since 1986 the levels of both have been monitored by gas spectrography.
Which of the following statements, assuming that each is true, would be LEAST helpful in explaining the air pollution levels in Los Diablos between 1986 and 1989?
(A) The 1987 air pollution control measures enacted in Los Diablos were put into effect in November of 1988.
(B) In December of 1988 a new and far more accurate gas spectrometer was invented.
(C) In February of 1989, the Pollution Control Board of Los Diablos revised the scale used to determine the amount of air pollution considered unhealthful.
(D) In 1988 the mayor of Los Diablos was found to have accepted large campaign donations from local industries and to have exempted those same industries from air pollution control measures.
(E) Excess ozone and carbon monoxide require a minimum of two years to break down naturally in the atmosphere above a given area.
Going with "B". NEw more accurate device was invented does not imply it was used.
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Re: 1000 CR test - 16 [#permalink] 03 May 2010, 13:45
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## Share - calculate working days between two given dates
Hello, I was trying to find the method to return the number of working days between two given dates and found the following article
https://powerapps.microsoft.com/en-us/blog/excluding-weekends-and-holidays-in-date-differences-in-po...
However, as the author mentioned "Notice that the partial week calculation given above will not work if the start or end dates fall on weekends"
So I spent some time writing the following formula which seems work for weekends too:
If(IsNewRequestFormValid, RoundDown(DateDiff(DatePickerFrom.SelectedDate, DatePickerTo.SelectedDate, Days) / 7, 0) * 5 +
If(Weekday(DatePickerTo.SelectedDate) >= Weekday(DatePickerFrom.SelectedDate),
If(Weekday(DatePickerTo.SelectedDate) = Weekday(DatePickerFrom.SelectedDate) && (Weekday(DatePickerTo.SelectedDate) = 1 || Weekday(DatePickerTo.SelectedDate) = 7),
0,
If(RoundDown((Weekday(DatePickerTo.SelectedDate) - Weekday(DatePickerFrom.SelectedDate))/5,0) >= 1,
5,
If(Weekday(DatePickerTo.SelectedDate) = 7,
Weekday(DatePickerTo.SelectedDate) - Weekday(DatePickerFrom.SelectedDate),
If(Weekday(DatePickerFrom.SelectedDate) = 1,
Weekday(DatePickerTo.SelectedDate) - Weekday(DatePickerFrom.SelectedDate),
Mod(Weekday(DatePickerTo.SelectedDate) - Weekday(DatePickerFrom.SelectedDate), 5) + 1)))),
(7 - Weekday(DatePickerFrom.SelectedDate) + Weekday(DatePickerTo.SelectedDate) - 1)))
I am sure it can be improved, so please feel free to make any enhancement.
1 ACCEPTED SOLUTION
Accepted Solutions
Community Support
Hi @Li,
Regards,
Michael
Community Support Team _ Michael Shao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
11 REPLIES 11
Community Support
Hi @Li,
Regards,
Michael
Community Support Team _ Michael Shao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
@Li, thank you so much for sharing this! You are a life-saver 🙂
Cheers!
Helper I
thanks for sharing but;
1. What is IsNewRequestFormValid? is it a custom variable just for your case?
Impactful Individual
I don't know what you mean by working day but my requirement excluded bank holidays so Christmas Day, New Years day don't count as working days. Here is list:
Description HolidayDate New Year's Day 1/1/2019 Birthday of Martin Luther King, Jr. 1/21/2019 Washington's Birthday 2/18/2019 Memorial Day 5/27/2019 Independence Day 7/4/2019 Labor Day 9/2/2019 Columbus Day 10/14/2019 Veterans Day 11/11/2019 Thanksgiving Day 11/28/2019 Christmas Day 12/25/2019 New Year's Day 1/1/2020
I also need to add/subtract working days from a date. For example something is due in 10 business days (BD) or send a reminder notice 3 BD prior to due date. The date could be a non workday. I found using PA to do this was impossible. I ended up doing an Azure Function.
Helper V
Just to add a bit to this great thread, if you want to exclude public holidays/company days etc you just need to do the following (assuming SharePoint is the data source)
• Create a new custom list with a couple of columns (Holiday Type, Date). I made the Holiday column a choice one but that is up to you as this is purely used for reference by the person entering the dates.
• Add the following code to the end of the code provided by @Li (note the minus sign before CountIf)
- CountIf('Company & Public Holidays',Date >= DatePickerFrom, Date <= DatePickerTo)
• Now the formula will remove and days from the total that are on the new list and fall between the to/from dates selected.
Rob
Frequent Visitor
Thank @Rob_CTL and @Li - This has all been very helpful, but I have one problem. My company has 6 different regions with different public holidays.
For Public Holidays I have a single SharePoint List WIth 12 Columns, "Region 1 Holiday Name"; "Region 1 Holiday Date"; etc.
Another SharePoint List contains the "Staff Name", the "Region" they work in and other details.
How can I add to this formula - CountIf('Company & Public Holidays',Date >=DatePickerFrom, Date <=DatePickerTo) to determine from the inputted staff name, what region they are in and select the corresponding holiday dates column?
Thanks
Impactful Individual
I doubt this will help you but I had a lot of complicated date calculations. I have a table that lists the bank holidays. I was using SQL Server. I ended up writing and Azure Function in C#. It calculates the due date for some task and sends out an email. I am very pro Azure Functions. They are a great compliment to PowerApps and keep the app very simple while adding complex functionality.
`````` public static DateTime CalcDate(DateTime date, string dueDateOffset, Dictionary<DateTime, int> hd)
{
int dCnt = 0;
DateTime calcDate = new DateTime(1900, 1, 1);
string[] parts = dueDateOffset.Split('|');
if (parts[0] == "CD")
{
if (Int32.TryParse(parts[1], out dCnt))
{
}
else
{
calcDate = date;
}
}
if (parts[0] == "BD")
{
int idx = 0;
if (Int32.TryParse(parts[1], out dCnt))
{
if (dCnt < 0)
{
idx = -1;
}
else
{
idx = 1;
}
dCnt = Math.Abs(dCnt);
while (dCnt > 0)
{
if (!(date.DayOfWeek == System.DayOfWeek.Saturday || date.DayOfWeek == System.DayOfWeek.Sunday || hd.ContainsKey(date)))
{
dCnt--;
}
}
calcDate = date;
}
else
{
calcDate = date;
}
}
if (parts[0] == "DD") //Due Date: DD|1 is 1st of next month, DD|2 is 1st 2 months from now.
{
if (Int32.TryParse(parts[1], out dCnt))
{
date = new DateTime(date.Year, date.Month, 1);
calcDate = date;
}
}
if (parts[0] == "G15") //Grace 15: G15 is 15th of next month.
{
date = new DateTime(date.Year, date.Month, 15);
calcDate = date;
}
if (parts[0] == "G30") //Grace 30: G30 is last day of next month.
{
date = new DateTime(date.Year, date.Month, 1);
calcDate = date;
}
if (parts[0] == "YD")
{
if (Int32.TryParse(parts[1], out dCnt))
{
calcDate = date;
}
}
return calcDate;
}``````
Frequent Visitor
Thank you. I will try and break this down but it's clear I am out of my depth here.
This is what I have so far, and without the countif statement at the end, it works to eliminate the weekends from between the dates chosen.
``````RoundDown(DateDiff('Leave Start Date_DatePicker'.SelectedDate, 'Leave End Date_DatePicker'.SelectedDate, Days) / 7, 0) * 5 +
If(Weekday('Leave End Date_DatePicker'.SelectedDate) >= Weekday('Leave Start Date_DatePicker'.SelectedDate),
If(Weekday('Leave End Date_DatePicker'.SelectedDate) = Weekday('Leave Start Date_DatePicker'.SelectedDate) && (Weekday('Leave End Date_DatePicker'.SelectedDate) = 1 || Weekday('Leave End Date_DatePicker'.SelectedDate) = 7),
0,
If(RoundDown((Weekday('Leave End Date_DatePicker'.SelectedDate) - Weekday('Leave Start Date_DatePicker'.SelectedDate))/5,0) >= 1,
5,
If(Weekday('Leave End Date_DatePicker'.SelectedDate) = 7,
Weekday('Leave End Date_DatePicker'.SelectedDate) - Weekday('Leave Start Date_DatePicker'.SelectedDate),
If(Weekday('Leave Start Date_DatePicker'.SelectedDate) = 1,
Weekday('Leave End Date_DatePicker'.SelectedDate) - Weekday('Leave Start Date_DatePicker'.SelectedDate),
Mod(Weekday('Leave End Date_DatePicker'.SelectedDate) - Weekday('Leave Start Date_DatePicker'.SelectedDate), 5) + 1)))),
(7 - Weekday('Leave Start Date_DatePicker'.SelectedDate) + Weekday('Leave End Date_DatePicker'.SelectedDate) - 1)) - CountIf('Public Holidays', NSW >= 'Leave Start Date_DatePicker'.SelectedDate, NSW <= 'Leave End Date_DatePicker'.SelectedDate)``````
Responsive Resident
Thank you for this. How would one get hours difference?
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We are excited to announce the "The Copilot Cookbook Community Challenge random winners have been selected for the Challenge. Thank you to everyone who participated in this challenge. Copilot Cookbook Gallery:Power Apps Cookbook Gallery: 1. @swaminawale 1. @renatoromao 2. @SpongYe 2. @nickpotts10 *Please note if for any reason a winner declines, we will have another random drawing. Check out all of the Cookbook Submissions: 1. Copilot Studio Cookbook Gallery: https://aka.ms/CS_Copilot_Cookbook_Challenge 2. Power Apps Copilot Cookbook Gallery: https://aka.ms/PA_Copilot_Cookbook_Challenge There will be 5 chances to qualify for the final drawing: Early Bird Entries: March 1 - June 2Week 1: June 3 - June 9Week 2: June 10 - June 16Week 3: June 17 - June 23Week 4: June 24 - June 30WINNERS ANNOUNCED - JULY 8th At the end of each week, we will draw 5 random names from every user who has posted a qualifying Copilot Studio template, sample or demo in the Copilot Studio Cookbook or a qualifying Power Apps Copilot sample or demo in the Power Apps Copilot Cookbook. Users who are not drawn in a given week will be added to the pool for the next week. Users can qualify more than once, but no more than once per week. Four winners will be drawn at random from the total qualifying entrants. If a winner declines, we will draw again at random for the next winner. A user will only be able to win once. If they are drawn multiple times, another user will be drawn at random. Prizes: One Pass to the Power Platform Conference in Las Vegas, Sep. 18-20, 2024 (\$1800 value, does not include travel, lodging, or any other expenses) Winners are also eligible to do a 10-minute presentation of their demo or solution in a community solutions showcase at the event. To qualify for the drawing, templates, samples or demos must be related to Copilot Studio or a Copilot feature of Power Apps, Power Automate, or Power Pages, and must demonstrate or solve a complete unique and useful business or technical problem. Power Automate and Power Pagers posts should be added to the Power Apps Cookbook. Final determination of qualifying entries is at the sole discretion of Microsoft. Weekly updates and the Final random winners will be posted in the News & Announcements section in the communities on July 29th, 2024. Did you submit entries early? Early Bird Entries March 1 - June 2: If you posted something in the "early bird" time frame complete this form: https://aka.ms/Copilot_Challenge_EarlyBirds if you would like to be entered in the challenge. Early Bird Submissions: @renato Week 1 Results: Congratulations to the Week 1 qualifiers, you are being entered in the random drawing that will take place at the end of the challenge. Copilot Cookbook Gallery:Power Apps Cookbook Gallery:1. @Mathieu_Paris 1. @SpongYe 2. n/a2. @Deenuji 3. n/a3. @Nived_Nambiar 4. n/a4. @ManishSolanki 5. n/a5. n/a Week 2 Results: Congratulations to the Week 2 qualifiers, you are being entered in the random drawing that will take place at the end of the challenge. Copilot Cookbook Gallery:Power Apps Cookbook Gallery:1. Kasun_Pathirana1. ManishSolanki2. cloudatica2. madlad3. n/a3. SpongYe4. n/a4. n/a5. n/a5. n/a Week 3 Results: Congratulations to the Week 3 qualifiers, you are being entered in the random drawing that will take place at the end of the challenge. Copilot Cookbook Gallery:Power Apps Cookbook Gallery:1. Parul_Yadav_Neo1. n/a2. SpongYe2. n/a3. n/a3. n/a4. n/a4. n/a5. n/a5. n/a Week 4 Results: Congratulations to the Week 4 qualifiers, you are being entered in the random drawing that will take place at the end of the challenge. Copilot Cookbook Gallery:Power Apps Cookbook Gallery:1. @nickpotts10 1. @ShrushtiShah 2. @Suniti_0020 2. @swaminawale 3. n/a3. @farukhis786 4. n/a4. @ManishSolanki 5. n/a5. n/a | 3,078 | 11,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-30 | latest | en | 0.824465 |
https://community.rstudio.com/t/tidy-data-one-standard-or-two/43522/8 | 1,680,394,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00087.warc.gz | 218,552,713 | 11,227 | # Tidy data: One standard or two?
I am confused about some aspects of the definition of tidy data, particularly as applied to time series. Suppose I have observation of a number of different variables for each period. Usually I would think of this as a rectangular array, with the first value probably a date, an observation number, or some other unique identifier, and the other variables in columns. That seems to correspond to the definition of a tidy data set, with one column for each variable.
But it seems like for a lot of purposes, especially with respect to graphing with ggplot, it is preferable to use “long format,” where all the variable values are in a single column and we have a column for value type with n values, in place of n distinct variables in n columns.
What is confusing me is that Hadley has stated in several places that tidy data constitutes a way of putting all your data into a unique standardized format. That makes me think that one of these formats is not considered tidy. Based on his expository definitions, I’d say the wide format was the tidy one. But when I look at the format that is easiest to use with several tidyverse tools, I’d say it was the narrow format. That is not always true – purrr, for example, seems to prefer a wide format as a way of organizing the subsets of your data that it acts on – but it is often true.
To me, it seems that this ambiguity is potentially troubling, especially as more and more people are writing packages based on data manipulation procedures advertised as tidy. Are there two tidy standards, or just one?
I’d welcome any clarifying thoughts on this.
1 Like
Great question to promote clear thinking.
Based on my time series experience, my possibly fuzzy understanding is that each time increment is an observation of one or more variables.
The `x-axis` is the value of the time increment, and the `y-axis` is one or more layers with the related variable values.
There is not a single, unique definition as to what is tidy. As a general rule, each row should be a unique observation. However, your unit of observation may differ for different analyses. I might want to answer a question for which the correct unit of observation is a unique patient. I might later want to answer a different question for which the correct unit of observation is a unique patient encounter. In the second case, each patient will appear multiple times in a taller dataset. In each case, the data are tidy, in that there is only one observation per row, but the unit of observation has changed.
Most of the time, tidying a dataset means that we will make it taller. In part, this is because most people faced with a blank spreadsheet will enter their data for their first observations in a single column, then, for subsequent observations, will add new columns from left to right. The tidyr package gives us tools to tidy this common data issue, usually with pivot_longer.
6 Likes
Thanks for your thoughts, technocrat. I agree that one observation per period is the common intuitive way of arranging time series, unless the data is panel-like.
But it is really hard to graph the data like that in ggplot. It becomes easy if instead you have a factor column for variable type (gdp, population, etc.), and a date column a column for all the data, and a date column. You need an explicit date column in this long-form representation because the data will recur once for each variable.
However, I think it worth observing that the architecture of ggplot was designed before the tidy data concept was clearly enunciated. I am hoping that more experienced programmers can tell me whether, once we move on to explicitly tidy functional design, this three column long format remains the preferred, or easiest, or most-tidy way of arranging data. When I think about dplyr functions, for instance, it seems like the "one column per variable type" representation is more natural and "tidier." But I am not sure I am right about this.
1 Like
Thanks, phiggins! I quite agree that a single underlying data set data set may support multiple data structures with different units of observation. But this is not the focus of my question. Every time series data set with multiple variables can be represented in a wide "one column per variable" format, or in a three-column long format with a date column, a variable type column, and a data column. See my reply to technocrat for a more complete discussion. My question is: Should the three-column long format always be preferred, or should the column-for-each-variable format be preferred for all purposes other than graphing with the ggplot family of of packages? Or is there some other dividing line I haven't thought of?
Could we reify the discussion with a reproducible example, called a reprex?
Sure, technocrat.
yr. <- 1990:1995
nm. <- runif(6)
tx. <- letters[1:6]
fc. <- paste0(tx., round(nm. * 100, 0))
Note: My "." suffixes, inspired by Hadley's . prefixes, are just a way of
making short words that are often already functions or variables into my own
unique variable or function names. They have no meaning beyond that.'
tb_wide <- tibble(year. = yr., num. = nm., txt. = tx., fact. = fc.) # , key = year., index = year.)
tb_wide
A tibble: 6 x 4
year. num. txt. fact.
1 1990 0.642 a a64
2 1991 0.876 b b88
3 1992 0.779 c c78
4 1993 0.797 d d80
5 1994 0.455 e e46
6 1995 0.410 f f41
tb_long <- tb_wide %>%
• mutate(num_t = as.character(num.), num. = NULL) %>%
Because you can not put numeric and character values in the same vector. (These are vctrs-style
vectors and vec_c produces errors for mismatched types, rather than base c's coersion).
• pivot_longer(-year., names_to = "var_name", values_to = "value")
tb_long
A tibble: 18 x 3
year. var_name value
1 1990 txt. a
2 1990 fact. a64
3 1990 num_t 0.642288258532062
4 1991 txt. b
5 1991 fact. b88
6 1991 num_t 0.876269212691113
7 1992 txt. c
8 1992 fact. c78
9 1992 num_t 0.778914677444845
10 1993 txt. d
11 1993 fact. d80
12 1993 num_t 0.79730882588774
13 1994 txt. e
14 1994 fact. e46
15 1994 num_t 0.455274453619495
16 1995 txt. f
17 1995 fact. f41
18 1995 num_t 0.410084082046524
ts_long <- as_tsibble(tb_long, index = year., key = var_name)
ts_long
A tsibble: 18 x 3 [1Y]
Key: var_name [3]
year. var_name value
1 1990 fact. a64
2 1991 fact. b88
3 1992 fact. c78
4 1993 fact. d80
5 1994 fact. e46
6 1995 fact. f41
7 1990 num_t 0.642288258532062
8 1991 num_t 0.876269212691113
9 1992 num_t 0.778914677444845
10 1993 num_t 0.79730882588774
11 1994 num_t 0.455274453619495
12 1995 num_t 0.410084082046524
13 1990 txt. a
14 1991 txt. b
15 1992 txt. c
16 1993 txt. d
17 1994 txt. e
18 1995 txt. f
You can not do this with the wide format, I'm pretty sure, because there is no
way to specify the key to uniquely identify each time/variable combination
when there are multiple variables on one row.
For a countervailing argument, the vctrs rules seem designed to discourage
long formats like this, since values in the value column are neither coerced
to a common type nor allowed to have different types. I don't see any way
that tsibble can handle numeric and non-numeric values in the same dataset.
tsibble does
I still don't understand how the formatting rules here differ from stackoverflow. My apologies.
1 Like
Here is how
Thanks for making the issue more concrete for me, @andrewH.
I'll begin with the Gospel of @hadley in #RForDataScientists, using the via negativa in R for Data Science
Before we continue on to other topics, it’s worth talking briefly about non-tidy data. Earlier in the chapter, I used the pejorative term “messy” to refer to non-tidy data. That’s an oversimplification: there are lots of useful and well-founded data structures that are not tidy data. There are two main reasons to use other data structures:
Alternative representations may have substantial performance or space advantages.
Specialised fields have evolved their own conventions for storing data that may be quite different to the conventions of tidy data.
Either of these reasons means you’ll need something other than a tibble (or data frame). If your data does fit naturally into a rectangular structure composed of observations and variables, I think tidy data should be your default choice. But there are good reasons to use other structures; tidy data is not the only way.
If you’d like to learn more about non-tidy data, I’d highly recommend this thoughtful blog post by Jeff Leek
Taking the Nicosian Creed approach of what does constitute `tidy`, from the same Gospel
There are three interrelated rules which make a dataset tidy:
Each variable must have its own column.
Each observation must have its own row.
Each value must have its own cell.
The key definition (forgive me my sins, for I came here originally from the law) is observation, which I can't find in the online version
An observation, or a case, is a set [emphasis added] of measurements under similar conditions (you usually make all of the measures in an observation at the same time [emphasis added] and on the same object [emphasis added]. An observation will contain several values, each associated with a different variable ... [or] a data point.
Therefore, I think, there is no inconsistency between "`tidy`" and `a rectangular array, with the first value probably a date, an observation number, or some other unique identifier, and the other variables in columns'.
As to whether this raises an obstacle to `ggplot`, in particular, it seems a non-issue.
``````library(tibble)
set.seed(137)
yr. <- 1990:1995
nm. <- runif(6)
tx. <- letters[1:6]
fc. <- paste0(tx., round(nm. * 100, 0))
tb_wide <- tibble(year. = yr., num. = nm., txt. = tx., fact. = fc.) # , key = year., index = year.)
tb_wide
#> # A tibble: 6 x 4
#> year. num. txt. fact.
#> <int> <dbl> <chr> <chr>
#> 1 1990 0.649 a a65
#> 2 1991 0.413 b b41
#> 3 1992 0.914 c c91
#> 4 1993 0.764 d d76
#> 5 1994 0.365 e e36
#> 6 1995 0.958 f f96
``````
Created on 2019-11-14 by the reprex package (v0.3.0)
qualifies as `tidy`.
To apply the `ggplot` grammar of graphics to `tb_wide` is not problematic due to `layers.`, appropriate care being taken not to mix continuous and discrete variables without further intervention.
``````library(ggplot2)
library(tibble)
set.seed(137)
yr. <- 1990:1995
nm. <- runif(6)
tx. <- letters[1:6]
fc. <- paste0(tx., round(nm. * 100, 0))
tb_wide <- tibble(year. = yr., num. = nm., txt. = tx., fact. = fc.) # , key = year., index = year.)
tb_wide
#> # A tibble: 6 x 4
#> year. num. txt. fact.
#> <int> <dbl> <chr> <chr>
#> 1 1990 0.649 a a65
#> 2 1991 0.413 b b41
#> 3 1992 0.914 c c91
#> 4 1993 0.764 d d76
#> 5 1994 0.365 e e36
#> 6 1995 0.958 f f96
p <- ggplot(data = tb_wide, aes(x = year.)) + geom_line(aes(y = num.))
p
``````
Created on 2019-11-14 by the reprex package (v0.3.0)
As for `tstibble`, I don't see why the class changes anything. The underlying problem is the mix of `num`, `txt` and `fact` objects, of which the second and third cannot be property coerced to `num`.
The approach, when coercion is unavailable would seem to be `facet` or `glob`.
As always, I may be wrongheaded, but does this advance the question?
2 Likes
Since you mention the Gospel of Hadley I would like to point out that the tidy data idea is very close to the subject of database normalization, so help me Codd.
This was a hot topic in the seventies and eighties, when disk space was expensive, and technology constraints were felt more acutely than they are now.
3 Likes
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https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/IOST_Library/06%3A_Web_Services/03%3A_Google_Workspace/3.03%3A_Google_Workbook_(Sheets)-_Data_Manipulation | 1,716,071,917,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00015.warc.gz | 135,589,370 | 29,582 | # 3.3: Google Workbook (Sheets)- Data Manipulation
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Copying Values
### Formulas
#### Basics
Sheet Skills
Formulas Basics: Use formulas, reference cells
Instructions
Google Sheets has built in tools to be able to do many things in sheets called formulas. There is a full list of formulas by google sheets. We're going to start off with some of the basics then later on go more in depth.
1. To start using a formula first select the cell you want to use
2. Type an equals sign = to start using a formula in the cell you want to use
3. Start to type the formula you wish to use and select it from the drop down menu
4. You can type a number or use a cell reference, you can either type the reference (For example A2) or click the cell you are wanting to reference
5. Finally press enter
#### Simple Mathematical Functions
Sheet Skills Video
Simple Mathematical Functions: Addition, Subtraction, Multiplication, Division, Exponents, Parentheses
Instructions
You can do simple calculations using formulas. Start all formulas by typing =
Subtraction -
Multiplication *
Division /
Exponent ^
Parentheses ()
So for example if we can use the sheet to calculate the gravitational force between two objects
Video $$\PageIndex{1}$$: 1:40 Video example of how to do basic calculations in sheets ()
#### Logarithms
Sheet Skills Video
Logarithms: Log and LN
Instructions
Logarithms are used in many different calculations. There is an overview on logarithms in Chapter ? that explains how logs work. In this section we are going to focus on how to use them in sheets.
LOG10
Returns the the logarithm of a number, base 10. =LOG10(value)
This is probably what you will use most often when talking about logarithms. This function will give the logarithm of a number using base 10.
LOG
Returns the the logarithm of a number given a base. =LOG(value, [base])
This is used less often but can be used to return a logarithm using any other base. This function will assume base 10 unless told otherwise
LN
Returns the the logarithm of a number, base e (Euler's number). =LN(value)
This function gives the natural log of a number. You can also do the natural log using LOG by entering the base as "EXP(1)"
Video $$\PageIndex{2}$$: 2:11 Video tutorial on log, ln and 1/T. (https://youtu.be/q-gALwjPj9I)
*Note if you are doing the Graphing lab you are using Ln(s) vs T not 1/T
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If total sales for the month is $450,000 and the profit margin is 40%, how much was the cost of goods sold? ### Correct answer: x = 321428.5714 USD ### Step-by-step explanation: Did you find an error or inaccuracy? Feel free to write us. Thank you! Tips to related online calculators Our percentage calculator will help you quickly calculate various typical tasks with percentages. #### You need to know the following knowledge to solve this word math problem: ## Related math problems and questions: • Sales off After discounting 40% the goods cost 15 €. How much did the cost of the goods before the discount? • Price reduction The product is sold for 360 CZK and the sales profit is 30%. By what percentage the sales profit will be reduced if I reduce the price of the product by 10%? • The sales The sales tax rate is 4.447% for the city and 4% for the state. Find the total amount paid for 2 boxes of chocolates at $17.96 each.
• Fix + percentages
Mrs. Vargas is a car sales agent who earns Php.5,850 monthly plus a 4% commision on all her sales. During a month, she sold a car worth Php.740,000. How much is her total earnings?
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Four pupils divided $1485 so that the second received 50% less than the first, the third 1/2 less than a fourth, and fourth$ 154 less than the first. How much money had each of them?
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The father gave his son € 100 to buy a bicycle, which was 40% of the total amount of the bicycle. How much did the bike cost?
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An amount of $2000 is invested at an interest of 5% per month. If$ 200 is added at the beginning of each successive month but no withdrawals. Give an expression for the value accumulated after n months. After how many months will the amount has accumula
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Dealer sell digital camera for € 747. Thirty percent of the price was his profit. After some time, decreased interest in selling the camera and therefore reduce its sales price by 11 %. How many percent of the new price now is dealer's profit? Round the r
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If the Canadian dollar appreciated by C$0.005 relative to the US dollar, what would be the new value of the Canadian dollar per US dollar? Assume the current exchange rate was US$1 = C$0.907. • Retail At what price bought retail 1 kg goods from wholesale, if lost in the distribution is 4% of weight of the goods and retail still have a profit of 6.3%? Goods are sold at retail for 25 euro per kg. • Sales off shoes cost x euros. First they are discounted by 50%, and then 25% of the new amount. Determine what percentage of the total cost is cheaper. • Commission Daniel works at a nearby electronics store. He makes a commission of 15%, percent on everything he sells. If he sells a laptop for 293.00$ how much money does Daniel make in commission? | 911 | 3,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-39 | longest | en | 0.959216 |
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# Vertical Angles - PowerPoint PPT Presentation
Vertical Angles. Lesson 2.8. Opposite Rays: Two collinear rays that have a common endpoint and extend in different directions. B. A. C. Ray AB and ray AC are opposite rays. B. A. C. D. Ray BA and Ray CD are not opposite rays. X. Y. V. U.
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### Vertical Angles
Lesson 2.8
Opposite Rays:
Two collinear rays that have a common endpoint and extend in different directions
B
A
C
Ray AB and ray AC are opposite rays.
B
A
C
D
Ray BA and Ray CD are not opposite rays.
X
Y
V
U
Ray UV and Ray XY are not opposite rays. NO common end point.
Definition: Two angles are vertical angles if the rays forming the sides of one angle and the rays forming the sides of the other are opposite rays.
3
E
B
A
2
1
4
C
D
<1 &<2; <3 & <4 are vertical angles.
Theorem 18:Vertical angles are congruent.
6
7
5
Given: diagram
Prove <5 congruent to <7
Hint: use supplementary angles
1
Given: <2 congruent to <3
Prove: <1 congruent to <3
2
3
• Given
• Vertical angles are .
• If s are to the same , they are . (Transitive Property)
• 2 3
• 1 2
• 1 3
5
4
6
m 4 = 2x +5
m 5 = x + 30
Find the m4 and m6
Vertical angles are congruent so just set them equal to each other and solve for x.
REMEMBER to plug x back in to find the angle.
The measure of <6 = 180-55
= 125 | 629 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-43 | latest | en | 0.858692 |
https://www.allinterview.com/showanswers/82484/ | 1,495,494,726,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00210.warc.gz | 845,774,444 | 8,846 | sir, plz tell me how much kvar capacitor required for 1000
amp or 500 Kw load ? what is the formula ?
Answers were Sorted based on User's Feedback
Formula : KVAR = KW(tanĂ˜1-tanĂ˜2)
Is This Answer Correct ? 35 Yes 10 No
Dear Shovan
This is Prakash veer
KVAR= KW* (tanQ1-tanQ2)
Q1 is intail power factor
Q2 is target power factor
Is This Answer Correct ? 19 Yes 4 No
we need power factor value
Is This Answer Correct ? 14 Yes 4 No
Required data for calculation your demand KVAR
2.Required P.F (Improvement in P.F)-(tanQ2)(0.99)
3.Supply Voltage - for reference(Voltage variation if any)
4.Load in KW (exam- availabl with you i.e 500KW)
KVAR= KW(tanQ1-tanQ2)
KVAR= 500(tan0.92- tan0.99)
KVAR = 500(0.2890)
KVAR = 141.75 KVAR (@ 142KVAR)
if you want to calculate by amp need following data
efficiency of motor, required P.F, KW,maganatising current
of motor,formula find in Electrical Machine Engineering
Is This Answer Correct ? 9 Yes 2 No
kVAR=kVA X sinphi
SO, sin phi=(1-cos sq phi) underroot.
SO POWER FACTOR IS REQUIRED.
Is This Answer Correct ? 7 Yes 3 No
Madhu pls explain what is tanphi1 and tanphi2????
Is This Answer Correct ? 5 Yes 1 No
25 kvar = 32amps load taken
1000amps capacitor required to 500kvar
Is This Answer Correct ? 5 Yes 1 No
250 Kvar Capacitor bank is required for 500Kw load
(i.e,625Kva).
This is because the capacitor rating for any load is equal
to 40% of Kva.
Is This Answer Correct ? 7 Yes 5 No
plz identify your systems existing power factor and
targeted power factor which u want to improve, then apply
this formula
KW X (Tan phi 1 - tan phi 2)
Is This Answer Correct ? 2 Yes 0 No
Reactive power (KVAR) is increase due to run inductive load.as a result power factor is lagging. so have to use capacitive power to reduce this inductive or reactive load. generally we have to use 63% capacitive load to remove the reactive load.i.e.
active power=500KW
so 60% of this will be=(60*500/100) = 300 KVAR
so we have to use 300KVAr capacitive capacitor bank to remove this reactive power .
Is This Answer Correct ? 1 Yes 0 No
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• Engineering AllOther (1314) | 1,023 | 3,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-22 | longest | en | 0.770869 |
https://cuitandokter.com/how-to-find-wavelength-frequency-easy-equation-w/ | 1,606,529,810,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194982.45/warc/CC-MAIN-20201128011115-20201128041115-00333.warc.gz | 267,404,803 | 14,874 | How To Find Wavelength Frequency Easy Equation W Problems Youtube
A video made by a student, for a student. showing how to find the frequency or wavelength when given the other. kansas university. rock chalk jayhawk, ku!!!!. How to: find wavelength frequency (easy equation w problems) solved example on wavelength formula example 1. if the speed of sound is about 340 m s and the frequency of the wave crest is 20.0 cycles per second (the low end of human hearing). then, find the wavelength of the sound wave? solution: firstly, note down what is given in the question frequency (f) = 20.0 cycles per seconds (cycles s). Wavelength and frequency are therefore inversely related. as the wavelength of a wave increases, its frequency decreases. the equation that relates the two is: c = λν. the variable c is the speed of light. for the relationship to hold mathematically, if the speed of light is used in m s, the wavelength must be in meters and the frequency in. Frequency wavelength formula is defined as speed of light (3×10 8) wavelength.you can also get the frequency of light using the above given online calculator in various measurements such as millimeter, centimeter, decimeter, meter, kilometer and so on. Solved example on wavelength formula example 1. if the speed of sound is about 340 m s and the frequency of the wave crest is 20.0 cycles per second (the low end of human hearing). then, find the wavelength of the sound wave? solution: firstly, note down what is given in the question frequency (f) = 20.0 cycles per seconds (cycles s).
How Do You Find Frequency Paperwingrvice Web Fc2
Wavelength to frequency formula the wavelength of a wave is inversely related to its frequency. if v represents the velocity of a wave, ν, its frequency and λits wavelength. learn more about wavelength to frequency and also download vedantu free study materials for your exam preparation. Wavelength and frequency are therefore inversely related. as the wavelength of a wave increases, its frequency decreases. the equation that relates the two is: c = λν. the variable c is the speed of light. for the relationship to hold mathematically, if the speed of light is used in m s, the wavelength must be in meters and the frequency in. If you know the speed and frequency of the wave, you can use the basic formula for wavelength. if you want to determine the wavelength of light given the specific energy of a photon, you would use the energy equation. calculating wavelength is easy as long as you know the correct equation. Wavelength frequency formula is defined as speed of light ((3 x 10 8) frequency).in the above given online calculator, you can calculate λ in various measurements such as centimeter, meter, decimeter, kilometer and so on. How to calculate wavelength. it's easy! just use our wavelength calculator in the following way: determine the frequency of the wave. for example, f = 10 mhz. this frequency belongs to the radio waves spectrum. choose the velocity of the wave. as a default, our calculator uses a value of 299,792,458 m s the speed of light propagating in a vacuum.
To calculate the frequency of a wave, divide the velocity of the wave by the wavelength. write your answer in hertz, or hz, which is the unit for frequency. if you need to calculate the frequency from the time it takes to complete a wave cycle, or t, the frequency will be the inverse of the time, or 1 divided by t. display this answer in hertz. This video demonstrates how to calculate the wavelength of a full cycle wave generated by an ultrasound transducer in soft tissue. find wavelength frequency (easy equation w problems. Also, we will see the wavelength frequency formula with an example. let us learn it! source:en. .org. wavelength frequency formula concept of wavelength: the wavelength of light determines the colour whereas the wavelength of the sound determines the pitch. the wavelengths of visible light may extend from about 700 nm to 400 nm. Wavelength to frequency formula the wavelength of any sinusoidal wave is defined as the spatial period of the wave, that is, the distance over the shape of the wave repeats itself. the wavelength is denoted by a greek letter lambda (λ) and is calculated in the units of length or metre. Practice using a displacement graph and wave speed to find the frequency and wavelength of a wave. if you're seeing this message, it means we're having trouble loading external resources on our website.
Angular Frequency Wavelength Equation Tessshebaylo
As the wavelength of a wave increases, its frequency decreases. the equation that relates the two is: $c = \lambda \nu$ the variable $$c$$ is the speed of light. for the relationship to hold mathematically, if the speed of light is used in $$\text{m s}$$, the wavelength must be in meters and the frequency in hertz. Wavelength is denoted λ,is the difference between any two identical points of a wave the frequency of a wave is the number of wavelengths of that wave that pass a fixed point in one unit of that time , it is denoated by the greel letter ν(nu) the wavelength and frequency of a wave are related to each other.the product of wavelength and frequecy is the total length of the ave that has. Wien's law formula. the equation describing wien's law is very simple: λ max = b t,. where: λ max is the aforementioned peak wavelength of light; t is an absolute temperature of a black body; b = 2.8977719 mm*k is the wien's displacement constant; although the relation between wavelength and frequency of electromagnetic waves is fairly simple (λ * f = c), we can't work out the peak. Wavelength, frequency, energy, speed, amplitude, period equations & formulas chemistry & physics how to calculate the energy of a photon given frequency & wavelength in nm chemistry using the wave equation (wavelength, speed and frequency). We use the rydberg formula to get the wavelength, and then we convert this to a frequency of #3.08 × 10^15color(white)(l) "s"^" 1"#. explanation: the rydberg formula is.
How To: Find Wavelength Frequency (easy Equation W Problems)
De broglie equation for wavelength. de broglie proposed an equation with the help of the plank equation and einstein energy mass law of matter. he considered the mass of the photon and wave nature of photon light quanta with wavelength = λ, frequency = ν, and energy = e. e = hν = hc λ where c = velocity of light h = plank constant = 6.627. Wavelength to frequency formula questions: 1) one of the violet lines of a krypton laser is at 406.7 nm. what is its frequency? answer: the velocity, v, of light is 3 x 10 8 m s. the wavelength, λ = 406.7 nm = 406.7 x 10 9 m λ = v f. Enter the frequency to calculate the wavelength. frequency (f) calculate. reset. result. wavelength. m. click here to view image. formula: λ = c f where, λ (lambda) = wavelength in meters. c = speed of light (299,792,458 m s) f = frequency. advertisement. other calculators wavelength to frequency calculator;. Speed is frequency times wavelength. so the frequency must be speed divided by wavelength. in this case: this makes sense as in one second, the wave will have travelled 15m. each meter is 5 cycles, so over the distance the wave has travelled in one second, it has completed cycles. Content objective: we will calculate the wavelength, frequency, and energy of light particles (photons) to further our understanding of atomic theory language objective: we will energize various compounds (by lighting various salt compounds on fire) and analyze electron behavior.by observing flame colors emitted, we can determine corresponding wavelength, and calculate frequency and energy. | 1,701 | 7,637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-50 | latest | en | 0.928997 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/45964-compute-the-nth-pythagorean-prime/solutions/2595994 | 1,597,296,459,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00091.warc.gz | 549,246,274 | 17,717 | Cody
# Problem 45964. Compute the nth Pythagorean prime
Solution 2595994
Submitted on 22 Jun 2020 by Nikolaos Nikolaou
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; pp_correct = 5; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 5 a = 2 b = 1
2 Pass
n = 5; pp_correct = 37; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 37 a = 1 b = 6
3 Pass
n = 25; pp_correct = 257; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 257 a = 16 b = 1
4 Pass
n = 125; pp_correct = 1657; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 1657 a = 36 b = 19
5 Pass
n = 625; pp_correct = 10313; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 10313 a = 92 b = 43
6 Pass
n = 3125; pp_correct = 62497; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 62497 a = 111 b = 224
7 Pass
n = 15625; pp_correct = 367229; [pp1,a1,b1] = PythagoreanPrime(n); assert(isequal(pp1,pp_correct)) assert(a1 == floor(a1) && b1 == floor(b1) && a1^2+b1^2 == pp1)
pp = 367229 a = 427 b = 430 | 655 | 1,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-34 | latest | en | 0.416881 |
https://www.conservapedia.com/index.php?title=Regular_polygon&oldid=1259942&mobileaction=toggle_view_mobile | 1,575,890,596,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518627.72/warc/CC-MAIN-20191209093227-20191209121227-00043.warc.gz | 656,849,510 | 5,970 | Regular polygon
This is the current revision of Regular polygon as edited by at 12:20, 13 July 2016. This URL is a permanent link to this version of this page.
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A rather interesting compass and straightedge construction of a regular pentagon (shown in yellow)
A regular polygon is a polygon where all the sides and included angles are equal. Examples include the equilateral triangle and the square.
Construction of Regular Polygons
The problem of which regular polygons can be constructed by ruler and compass alone goes back to the ancient Greeks. Some regular polygons (e.g. a regular pentagon) can be constructed with a straightedge (without any markings) and compass, others cannot. Carl Friedrich Gauss made the first new progress on the problem when he constructed the regular 17-gon in 1796. He later showed that a regular n-sided polygon can be constructed with ruler and compass if the odd prime factors of n are distinct Fermat primes.[1]
In order to explain Gauss's Theorem, we need to understand about Fermat numbers, which are defined as numbers of the form 22n+1,
where n is natural number or zero. The first 5 Fermat numbers are: 3, 5, 17, 257, 65537.[2]
Gauss' Theorem on the constructibility of regular polygons says that a regular n-gon is constructible by straightedge and compass alone if and only if n = 2kp1...pt , where k and t are nonnegative integers and pi are distinct prime Fermat numbers.
By this theorem, we can construct all of the first 8 regular polygons from an equilateral triangle up to a decagon, with the exception of a heptagon. As 7 is not a Fermat prime nor has it any factors which are Fermat primes, a regular heptagon therefore cannot be constructed by straightedge and compass alone.
The construction of the regular 17-sided polygon is an inscription on Gauss' tomb.
Notes
1. This result is recorded in Section VII of Gauss's Disquisitiones Arithemeticae published in 1801. Gauss conjectured that this condition was also necessary, but he offered no proof of this fact, which was proved by Pierre Wantzel in 1837
2. Note that all of these are prime and Fermat conjectured in 1640 that all the Fermat numbers are prime. Rather surprisingly it wasn't until 1732 that Euler pointed out that the next Fermat number 4294967297 is not prime. It is divisible by 641. In fact the first 5 are the only know prime Fermat numbers and it seems reasonable that there are no others | 574 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-51 | latest | en | 0.931216 |
https://studyadda.com/notes/6th-class/mathematics/distance-time-and-speed/distance-time-and-speed/6167 | 1,556,294,027,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578841544.98/warc/CC-MAIN-20190426153423-20190426175423-00508.warc.gz | 552,460,273 | 18,489 | ## DISTANCE, TIME AND SPEED
Category : 6th Class
Learning Objective
To find overage speed of a vertical.
To find time to cover a given distance.
TIME AND DISTANCE
Motion / Movement occurs when a body of any shape and size changes its position with respect to any external stationary point.
The mathematical equation that describe the motion has three variables Speed, Time and Distance, which are connected by the following formula
Distance = Speed x Time
From the above equation, we can have the following conclusions:
(a) If speed is constant then distance and time are directly proportional to each other, i.e.
Distance $\propto$ Time.
(b) If time is constant, then distance and speed are directly proportional to each other i.e
Distance $\propto$ Speed.
(c) When distance is constant then speed and time are inversely proportional to each other i.e.
Speed c$\propto \,\frac{1}{Time}$
Normally speed is measured in km/hr or m/s.
$1\,km/hr=\frac{1000\,m}{3600\,s}=\frac{5}{18}\,m/s$
or $1\,m/s\,=\frac{18}{5}km/hr$
AVERAGE SPEED:
It is defined as the ratio of total distance covered to the total time taken by an object. If an object travels ${{d}_{1}},{{d}_{2}},\,{{d}_{3}},\,....{{d}_{n}}$ metres with different speeds ${{s}_{1}},\,{{s}_{2}},\,{{s}_{3}},....{{s}_{n}}$ metres/ sec in time ${{t}_{1}},\,{{t}_{2}},\,{{t}_{3}},......{{t}_{n}}$ seconds respectively, then average speed ${{S}_{a}}$ is given by
${{S}_{a}}=\frac{\text{Total}\,\text{Distance}\,\text{Travelled}}{\text{Total}\,\text{Time}\,\text{Taken}}$
A car can cover 350 km in 4 hours. If its speed is decreased by $12\frac{1}{2}$ kmph, how much time does the car take to cover a distance of 450 km?
Solution:
$\text{Speed}\,\text{=}\frac{\text{Distance}}{\text{Time}}\,=\frac{350}{4}=87\frac{1}{2}\,\text{kmph}$
Now this is reduced by $12\frac{1}{2}$ kmph.
Hence, speed is 75 kmph.
Travelling at this speed, the time taken == 450/75 = 6 hours.
#### Other Topics
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You will be redirected in 3 sec | 614 | 2,118 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-18 | latest | en | 0.841918 |
http://gamedev.stackexchange.com/questions/19774/determine-corners-of-a-specific-plane-in-the-frustum/55248 | 1,443,944,534,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736672923.0/warc/CC-MAIN-20151001215752-00052-ip-10-137-6-227.ec2.internal.warc.gz | 101,501,156 | 19,799 | # Determine corners of a specific plane in the frustum
I'm working on a game with a 2D view in a 3D world. It's a kind of shoot'em up. I've a spaceship at the center of the screen and i want that ennemies appear at the borders of my window. Now i don't know how to determine positions of the borders of the window.
For example, my camera is at (0,0,0) and looking forward (0,0,1). I set my spaceship at (0,0,50). I also know the near plane (1) and the far plane(1000). I think i'd have to find the 4 corners of the plane in the frustum whose z position is 50, and with these corner i can determine borders. But i don't know how to determine x and y.
-
Disclaimer: This solution really is simplified for your particular problem, because your camera is looking straight down the Z axis of the world's coordinate system. Also, the center point at 50 units away is already given as (0, 0, 50). If your camera's viewing direction was an arbitrary vector, there would be more multiplications involving the distance with a cross product of the viewing vector and the camera's Up vector.
Determining the borders of a plane at a given distance is dependent on the FOV angle in which the view is projected. Usually, the FOV angle is measured in the Y axis for rectangular viewports.
For any given distance Z from the camera, the shortest distance D from the center point of a plane perpendicular to the viewing vector at Z to one of its borders above or below the center (really, the intersection of the plane and frustum) is `D = tan(FOV / 2) * Z` . Add and subtract D from the center point's Y component to get the maximum and minimum Y extents.
To get the minimum and maximum X extents, add and subtract `D * aspect_ratio`.
Now getting the location of the plane's corners is simply plugging in the mix/max X and Y in its four possible combinations along with the Z distance.
-
When i compute tan(FOV/2) * Z, FOV has to be in radian or degrees? – Takumi Nov 15 '11 at 23:07
In standard math libraries (libm, etc), angles must always be expressed in radians. If you're using some other custom third-party math library, check the library's documentation for what units it uses for angles. But unless it says otherwise, it's usually safe to assume radians. – Trevor Powell Nov 15 '11 at 23:54
It works fine since my camera z position is at 0. But does it works if set my camera elsewhere? For exemple if my camera is at (0,0, -50) and the plane i'm looking for is at (0,0, 50). In this exemple, the Z component in tan(FOV/2) * Z has to be 50 (origin to plane) or 100 (camera z-position to plane)? – Takumi Nov 21 '11 at 8:58
In this case the Z component is 100 because it's the distance from the camera to the plane. This formula works for any view direction parallel to the Z axis (any multiple of [0, 0, 1]). – ChrisC Nov 21 '11 at 17:03
Thanks, it works. One last thing, now i'm curious to know how to do the same with a camera looking anywhere. – Takumi Nov 22 '11 at 13:39
## This is a more general solution
To start you will need some data.
The Camera's position represented by P (this is a point)
The normalized viewing vector represented by v
The Camera's up vector represented by up
The Camera's right vector represented by w (this is the cross product of v X up)
The near distance represented by nDis
The far distance represented by fDis
The field of view represented by fov (this usually in radians)
The aspect ratio represented by ar (this is the width of the screen divided by the height)
First we will get the width and height of the near plane
Hnear = 2 * tan(fov / 2) * nDis
Wnear = Hnear * ar
Then we do the same for the far plane
Hfar = 2 * tan(fov / 2) * fDis
Wfar = Hfar * ar
Now we get the center of the planes
Cnear = P + v * nDis
Cfar = P + v * fDis
And now we get our points
Near Top Left = Cnear + (up * (Hnear / 2)) - (w * (Wnear / 2))
Near Top Right = Cnear + (up * (Hnear / 2)) + (w * (Wnear / 2))
Near Bottom Left = Cnear - (up * (Hnear / 2)) - (w * (Wnear /2))
Near Bottom Right = Cnear + (up * (Hnear / 2)) + (w * (Wnear / 2))
Far Top Left = Cfar + (up * (Hfar / 2)) - (w * Wfar / 2))
Far Top Right = Cfar + (up * (Hfar / 2)) + (w * Wfar / 2))
Far Bottom Left = Cfar - (up * (Hfar / 2)) - (w * Wfar / 2))
Far Bottom Right = Cfar - (up * (Hfar / 2)) + (w * Wfar / 2))
Common assumptions that might be useful:
• Most games have their field of view set at 110 degrees as this is close to the human field of view
• The camera is most often set at the origin (0,0,0)
• The view vector is usually along the negative Z axis (0,0,-1)
• The up vector is usually along the Y axis (0,1,0)
• The right vector is usually along the X axis (1,0,0)
-
I don't think this answers the question. Near and far planes should be irrelevant, as the problem is about intersection of view frustum and a plane perpendicular to the view direction and where an object is located. – msell May 10 '13 at 5:03
I'm sorry, but as @msell said, it does not answer the question. Your answer explain how to determine corners of a frustum. I (we) need to determine corners of a plane into the frustum. I have made this picture to make it clearer. Anyway, thanks for your current work, have an upvote :) – Nison Maël May 11 '13 at 9:51
So if I understand you clearly, there is a plane inside the frustum that runs parallel to the near and far planes, and perpendicular to the view vector. The formula you are looking for, takes a distance from the camera's position, and calculates the corners of the plane, that lie on the plane, and intersect with the edges of the frustum. – Lucas C. May 14 '13 at 20:12
`Near Bottom Right = Cnear + (up * (Hnear / 2)) + (w * (Wnear / 2))` should be `Near Bottom Right = Cnear - (up * (Hnear / 2)) + (w * (Wnear / 2))` – scones Sep 3 at 14:49
Conisdering you translate your "camera" by writing something like: gltranslatef(1, 3, 0). Make values such as posX and posY, and whenever you translate your camera, add to those values. Then when you add your enemy set it's X value to posX and it's Y value to posY for the enemy to appear where it would if the camera was at the origin.
- | 1,698 | 6,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2015-40 | longest | en | 0.93672 |
https://manpages.org/dgehrd/3 | 1,685,599,980,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00411.warc.gz | 425,703,391 | 4,856 | DGEHRD(3) reduces a real general matrix A to upper Hessenberg form H by an orthogonal similarity transformation
## SYNOPSIS
SUBROUTINE DGEHRD(
N, ILO, IHI, A, LDA, TAU, WORK, LWORK, INFO )
INTEGER IHI, ILO, INFO, LDA, LWORK, N
DOUBLE PRECISION A( LDA, * ), TAU( * ), WORK( * )
## PURPOSE
DGEHRD reduces a real general matrix A to upper Hessenberg form H by an orthogonal similarity transformation: Q' * A * Q = H .
## ARGUMENTS
N (input) INTEGER
The order of the matrix A. N >= 0.
ILO (input) INTEGER
IHI (input) INTEGER It is assumed that A is already upper triangular in rows and columns 1:ILO-1 and IHI+1:N. ILO and IHI are normally set by a previous call to DGEBAL; otherwise they should be set to 1 and N respectively. See Further Details.
A (input/output) DOUBLE PRECISION array, dimension (LDA,N)
On entry, the N-by-N general matrix to be reduced. On exit, the upper triangle and the first subdiagonal of A are overwritten with the upper Hessenberg matrix H, and the elements below the first subdiagonal, with the array TAU, represent the orthogonal matrix Q as a product of elementary reflectors. See Further Details. LDA (input) INTEGER The leading dimension of the array A. LDA >= max(1,N).
TAU (output) DOUBLE PRECISION array, dimension (N-1)
The scalar factors of the elementary reflectors (see Further Details). Elements 1:ILO-1 and IHI:N-1 of TAU are set to zero.
WORK (workspace/output) DOUBLE PRECISION array, dimension (LWORK)
On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
LWORK (input) INTEGER
The length of the array WORK. LWORK >= max(1,N). For optimum performance LWORK >= N*NB, where NB is the optimal blocksize. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA.
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value.
## FURTHER DETAILS
The matrix Q is represented as a product of (ihi-ilo) elementary reflectors
Q = H(ilo) H(ilo+1) . . . H(ihi-1).
Each H(i) has the form
H(i) = I - tau * v * v'
where tau is a real scalar, and v is a real vector with
v(1:i) = 0, v(i+1) = 1 and v(ihi+1:n) = 0; v(i+2:ihi) is stored on exit in A(i+2:ihi,i), and tau in TAU(i).
The contents of A are illustrated by the following example, with n = 7, ilo = 2 and ihi = 6:
on entry, on exit,
( a a a a a a a ) ( a a h h h h a ) ( a a a a a a ) ( a h h h h a ) ( a a a a a a ) ( h h h h h h ) ( a a a a a a ) ( v2 h h h h h ) ( a a a a a a ) ( v2 v3 h h h h ) ( a a a a a a ) ( v2 v3 v4 h h h ) ( a ) ( a ) where a denotes an element of the original matrix A, h denotes a modified element of the upper Hessenberg matrix H, and vi denotes an element of the vector defining H(i).
This file is a slight modification of LAPACK-3.0's DGEHRD subroutine incorporating improvements proposed by Quintana-Orti and Van de Geijn (2005). | 913 | 2,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.755205 |
http://www.answerlib.org/qv/20180210163541AAYNWhl.html | 1,545,140,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00013.warc.gz | 325,193,405 | 3,508 | Science & Mathematics » Physics » A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by?
# A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by?
x(t) = bt2 − ct3
, where b = 2.40 m/s2 and
c = 0.120 m/s3
. What is the instantaneous speed of the car at t = 20.0 sec ?
A) −48 m/s
B) 0
C) 48 m/s
D) 96 m/s
I know the answer is A, but can someone explain to me why
• Well it definitely isn't any of the other ones because they're all positive.
Try x(19.99)
Try x(20)
Try x(20.01)
The values for x keep decreasing as time goes on at 20 s. So velocity must be negative at that moment.
Anyway, instantaneous velocity is the derivative of position with respect to time.
v(t) = dx/dt = 2bt - 3ct^2
At t = 20.0 s
v(20) = 2(2.4 m/s^2)(20 s) - 3(0.12 m/s^3)(20 s)^2 = -48 m/s
However, SPEED is positive 48 m/s. Because speed is just the magnitude of velocity and doesn't care about direction, so it can't be negative. Not sure if that was a typo or what. Tread carefully :)
1
0 | 345 | 1,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-51 | latest | en | 0.94415 |
http://eisforexplore.blogspot.com/2012/05/balloon-tennis-fractions.html | 1,563,304,910,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524685.42/warc/CC-MAIN-20190716180842-20190716202842-00320.warc.gz | 51,950,457 | 15,405 | ## Monday, May 21, 2012
### Balloon Tennis
Balloon Tennis - Glue paper plates on popsicle sticks for a racquet. Kids hit balloons and play tennis!
Synonyms - Two player game - Kids are given a word. They hit the balloon back and forth. Before they hit the balloon they have to call out a synonym for the word. If they can't, their opponent gets a point.
Sight Words - Write sight words on balloons. Kids whack the balloon and say the sight word. If they have the correct answer, they get a point!
Spelling - Spell out a word and hit the balloon for each letter. Kids can do this solo, with a buddy, or in a group.
Fractions - Three player game - 2 people hit the balloon and one person keeps track of the score. Kids choose how many chances they get to hit the balloon. Limit kids to a certain number. Otherwise, they'll go on all day, and everyone needs a chance to play! If they miss, they get to continue, but they have to start off from where they stopped. For example, if they plan to hit the balloon 20 times and they stopped at 6, they start counting at 7. The kid keeping track of the score tallies each time they hit the balloon. As soon as they reach the maximum number they stop (which is hard I know!). As a group, they figure out the fraction. For example, the balloon was kept in play (or smacked) 18/20 times.
Basic Addition - Two Players - Have kids count up their points! | 335 | 1,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-30 | latest | en | 0.952268 |
https://wamp.mapleprimes.com/users/Carl%20Love/replies?page=646 | 1,685,303,909,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644506.21/warc/CC-MAIN-20230528182446-20230528212446-00510.warc.gz | 684,116,248 | 54,099 | Carl Love
25249 Reputation
10 years, 230 days
Himself
Natick, Massachusetts, United States
My name was formerly Carl Devore.
Please clarify "power is limited"...
When you say "their power is limited ... [to], for instance, no more than 3", do you mean that the total degree of the polynomials is at most 3? Or do you mean that the degree, or power, of each individual variable is at most 3?
What do you mean by "solve"?...
What do you mean by "solve"? Do you mean "approximate"? What is the "it" about whose truth you ask?
You've given some Maple commands, which work correctly. But what's the point of only using three terms of your series? Why aren't you satisfied with the output?
If the problem is still found.......
@amrramadaneg Well, if "the original problem is still found," how about answering the other questions so that we can help you?
No, it's not too large...
To address directly the OP's issues:
mike_a wrote:
> I noticed some strange things with the list/table carrying my complex points. If I ask for individual values, like T[20], it won't give the correct value, but if I ask for a range, like T[1..20], it will display the correct values.
You'll have to post some code so we can figure out what's going on. But I'll tell you right off that your problem has nothing to do with the size of your data set.
> I get an error when trying to use complexplot saying either invalid range, if I don't have a range for the 2nd and 3rd argument, or else Error, (in plot) procedure expected, as range contains no plotting variable. I assume that there's some problem with the table being too large and not a problem with the complexplot command
No, it has nothing to do with the table being too large. But you'll need to post some code so that we can figure out the actual problem.
First work out these indexing problems on a much smaller version of your plot before moving on to the million-point plot.
Summary...
To summarize what you said, for those who may have trouble reading it: You are trying to solve, using a least-squares method, a linear system A.X = B, where A has dimensions 1201 x 800 and B is a column vector. You have a tried an ad hoc method that involves explicitly inverting an 800 x 800 matrix, but ran out of memory.
You wrote:
> A^T*A*X=A^T*X. let M=A^T*A. M*X=A^T*X finally X=M^-1 * A^T*X
That's supposed to be B, not X, on the right sides of those equations! I hope that was just a typo in your post and not in your Maple code!
Several questions before we go any further:
1. Is A a hardware floating point matrix?
2. What percentage of the elements of A are non-zero? A rough estimate is fine.
3. Have you tried LinearAlgebra:-LeastSquares?
4. Can you make a smaller version of this problem that we can practice with before doing the full 1201 x 800?
Plot of Preben's one-term approximation...
A plot shows that Preben's one-term approximation is excellent even for x close to -2, and the original is visually indistinguishable from the approximation for x < -3.2. The plot below uses A and res as defined in Preben's code. If Digits, numpoints, and adaptive are not reduced from their default values (as I have done below), Maple will take a very long time to do the numerical integrations.
Digits:= 6:
plot([op(1,res), A], x= -3.3..-2.1, numpoints= 50, adaptive= 4);
I had originally plotted down to x = -10, but then it difficult to even see that there are two curves because they are so close.
Plot of Preben's one-term approximation...
A plot shows that Preben's one-term approximation is excellent even for x close to -2, and the original is visually indistinguishable from the approximation for x < -3.2. The plot below uses A and res as defined in Preben's code. If Digits, numpoints, and adaptive are not reduced from their default values (as I have done below), Maple will take a very long time to do the numerical integrations.
Digits:= 6:
plot([op(1,res), A], x= -3.3..-2.1, numpoints= 50, adaptive= 4);
I had originally plotted down to x = -10, but then it difficult to even see that there are two curves because they are so close.
Are you sure of that formula?...
Are you sure that you didn't switch n and m in your big-O formula? Also, do you mean to compute separately the GCDs of m pairs of numbers? or to compute the overall GCD of a set of m numbers?
Iterative versus sequential output and b...
I think the key to the timing differences w.r.t. production of output might be in the line
K[cnt]:= seq([j[1..-2][], Div[k], Div[-k]], k= `if`(i=2, 1, BinarySearch(j[-2]))..Nsq)
where all the splits of [..., n] are generated in a single call to seq. If the output was being generated one at a time, as with an iterator, I couldn't use that efficient seq. In the Iterator package, could you generate iterates in bursts like this and store them in a buffer? I know one goal of the package is minimal memory usage, but in this case the seq is generating only a small portion of the overall output.
With binary search and local remember ta...
Before posting my last version, I had tried with ListTools:-BinarySearch, but that consumed about half of the time. This time, I implemented a binary search tailored to this algorithm. As Joe said, it's probably not worth it; but it's probably not a significant waste either.
A much more significant change is that I made it call numtheory:-divisors only once and sort only once. All subsequent sorted sublists of divisors are made my filtering the this one list. And the filtered lists are stored in a local remember table.
It seems that there is considerable interest in this algorithm. It certainly fascinates me. It amazes me that simply keeping the lists of factorings carefully sorted and only splitting the last element is sufficient to generate all the factorings. Joe: Is your method for generating all partitions of a multiset similar to this?
Factorings:= proc(n::posint, m::posint:= 0)
local
L, K, Div, i, j, k, Nsq, cnt
,Omega:= numtheory:-bigomega(n) # max # of factors
,DIVS:= sort([(numtheory:-divisors(n) minus {1,n})[]]) # not changed during this proc's run.
# Get sorted list of divisors of divisors by filtering DIVS; also returns pointer to midpt of list.
,divisors:= proc(n::posint)
option remember;
local D:= select(x-> irem(n,x)=0, DIVS)[1..-2]; # remove n itself from end.
D, iquo(nops(D)+1, 2) # Extra 1 is for square root.
end proc
# Find first index k into Div[1..Nsq] such that Div[k] >= x.
,BinarySearch:= proc(x)
local lo, hi, m;
if Nsq=0 then return 1 elif Div[Nsq] < x then return Nsq+1 fi; # fall-thru cases.
lo:= 1; hi:= Nsq;
do
m:= iquo(hi+lo, 2);
if Div[m] = x then return m elif x < Div[m] then hi:= m-1 else lo:= m+1 fi;
if lo > hi then return lo fi
end do
end proc
;
if m > Omega then return [] elif m > 0 then Omega:= m fi;
L[1]:= [[n]]; # Initialize loop w/ sole factoring of length 1.
# Avoid wasted effort of filtering divisor list on the first (i=2) pass.
divisors(n):= DIVS, iquo(nops(DIVS)+1, 2);
for i from 2 to Omega do # i is number of factors.
cnt:= 0; # cnt is index for list-building table.
# j is a single factoring for previous i, so j is a list.
for j in L[i-1] do
cnt:= cnt+1;
# Split last, and greatest, member of list, if not prime (prime case falls thru).
Div, Nsq:= divisors(j[-1]);
# j[-1] = Div[k]*Div[-k] for any k, since Div is sorted.
# Only use splits that maintain the list in order.
K[cnt]:= seq([j[1..-2][], Div[k], Div[-k]], k= `if`(i=2, 1, BinarySearch(j[-2]))..Nsq)
end do;
# Convert table K to list L[i].
L[i]:= [seq](K[j], j= 1..cnt)
end do;
# Convert table L to list for final output.
`if`(m = 0, [seq](L[i][], i= 1..Omega), L[m])
end proc;
With adaptive set to false, the plot lacks the rich detail I want unless numpoints is set to 2^15 or greater. This is a single closed curve plotted in a single color, so all the action is on the boundary (the whole thing is its boundary). I'm willing to cut back to numpoints= 2^12, adaptive= 4. A little time can be saved, without significantly impacting quality, by truncating the series at 9 terms (N=8) rather than the 17 terms I used above.
Go ahead and use it as an example, and give me credit for the X and Y functions. You could use 2^15 evenly spaced points and do it very quickly in evalhf.
Here is a much larger jpeg showing more detail. This is with N=8, numpoints= 2^12, adaptive= 4. Blow it up and see how the inner curlicues look like flowers.
More modifications...
It's a great algorithm. I've made some modifications to the code in addition to Joe's, keeping the basic algorithm the same. You may notice another factor-of-two time improvement on some large factorings such as all 92,213 factorings of our friend 711*100^3.
1. I eliminated the repetitive deep indexing by letting j become a factoring itself rather than an index into the list L[i-1] of factorings.
2. I rolled the factorings of length two case into the main loop.
3. I eliminated all divisions by always sorting the list of divisors. That is, if D:= sort([divisors(N)[]]), then for all k, we have n = D[k]*D[-k].
4. Because of (3), I only scan the first half of each list of divisors, plus one more in case of a perfect square (we know it's a square if the number of divisors is odd, so I don't need to check).
Factorings:= proc(n::posint, m::posint:= 0)
uses NT= numtheory;
local
L, T, Div, i, j, k, Nsq, idx
,Omega:= NT:-bigomega(n)
;
if m > Omega then return [] elif m > 0 then Omega:= m fi;
L[1]:= [[n]];
for i from 2 to Omega do # i is number of factors.
idx:= 0;
# j is one factoring for previous i, so j is a list.
for j in L[i-1] do
# Split last, and greatest, member of list, if not prime.
Div:= sort([(NT:-divisors(j[-1]) minus {1,j[-1]})[]]);
Nsq:= iquo(nops(Div)+1, 2); # Extra 1 is for square root.
# Only use splits that maintain the list in order.
if i=2 then k:= 1 else for k to Nsq while Div[k] < j[-2] do od fi;
idx:= idx+1;
# j[-1] = Div[k]*Div[-k] for any k, since Div is sorted.
T[idx]:= seq([j[1..-2][], Div[k], Div[-k]], k= k..Nsq)
end do;
# Convert table T to list L[i].
L[i]:= [seq](T[j], j= 1..idx)
end do;
# Convert table L to list for final output.
`if`(m = 0, [seq](L[i][], i= 1..Omega), L[m])
end proc:
symbols won't expand...
Kitonum: The reason to use ``(...) rather than convert(..., symbol) is that expand will remove the ``.
symbols won't expand...
Kitonum: The reason to use ``(...) rather than convert(..., symbol) is that expand will remove the ``.
Post example worksheet...
Yes, it would be nice and would make sense if simplify were idempotent.
Please post an example worksheet.
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| 2,952 | 10,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-23 | longest | en | 0.949327 |
https://biology.stackexchange.com/questions/11263/what-is-the-difference-between-local-and-global-sequence-alignments/77698 | 1,708,519,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00667.warc.gz | 141,009,075 | 36,771 | # What is the difference between local and global sequence alignments?
There are a bunch of different alignment tools out there, and I don't want to get bogged down in the maths behind them as this not only between software but varies from software version to version.
There are two main divides in the programs; some use local alignments and others use global alignments. My question is threefold:
• What are the fundamental differences between the two?
• When should one use either a global or local sequence alignment?
The very basic difference between a local and a global alignments is that in a local alignment, you try to match your query with a substring (a portion) of your subject (reference). Whereas in a global alignment you perform an end to end alignment with the subject (and therefore as von mises said, you may end up with a lot of gaps in global alignment if the sizes of query and subject are dissimilar). You may have gaps in local alignment also.
Local Alignment
5' ACTACTAGATTACTTACGGATCAGGTACTTTAGAGGCTTGCAACCA 3'
|||| |||||| |||||||||||||||
5' TACTCACGGATGAGGTACTTTAGAGGC 3'
Global Alignment
5' ACTACTAGATTACTTACGGATCAGGTACTTTAGAGGCTTGCAACCA 3'
||||||||||| ||||||| |||||||||||||| |||||||
5' ACTACTAGATT----ACGGATC--GTACTTTAGAGGCTAGCAACCA 3'
I shall give the example of the well known dynamic programming algorithms. In the Needleman-Wunsch (Global) algorithm, the score tracking is done from the (m,n) co-ordinate corresponding to the bottom right corner of the scoring matrix (i.e. the end of the aligned sequences) whereas in the Smith-Waterman (local), it is done from the element with highest score in the matrix (i.e. the end of the highest scoring pair). You can check these algorithms for details.
You can adopt any scoring schemes and there is no fixed rule for it.
Global alignments are usually done for comparing homologous genes whereas local alignment can be used to find homologous domains in otherwise non-homologous genes.
Global alignment is when you take the entirety of both sequences into consideration when finding alignments, whereas in local you may only take a small portion into account. This sounds confusing so here an example:
Let's say you have a large reference, maybe 2000 bp. And you have a sequence, which is about 100 bp. Let's say that the reference contains the sequence almost exactly. If you did a local alignment, you would have a very good match. But if you did a global alignment, it may not match. Instead, it may look for matches throughout the entire reference, so you'd end up with an alignment with many large gaps. It does not matter that it matches near perfectly at one particular region on the reference, because it's looking for matches globally (i.e. throughout the reference).
If you have a really good match it may not matter what type of alignment you use. But when you have mismatches and such it starts to get important. This is because of the scoring algorithms used. In the example above let's say that there is a 100bp region in the reference that matches your 100bp sequence with 85% accuracy. In local alignment it's very likely it will align there. Now let's say that the first 30 bp of your sequence matches a region in the beginning of the reference 95%, and the next 30bp matches a region in the middle of the reference 85%, and the final 40bp matches a region at the end of the reference about 90%. In global alignment the best match is the gapped alignment, whereas in local alignment the ungapped alignment would be best. I think in general gap penalties are less in global alignments, but I'm not really an expert on the scoring algorithms.
What you want to use depends on what you are doing. If you think your sequence is a subsequence of the reference, do a local alignment. But if you think your entire sequence should match your entire reference, you would do a global alignment.
Take look to the attached image file. It will clear your doubts about difference between local and global sequence alignments.
• Please do not post text answers as image files on SE Biology. This discriminates agains people with vision defects who use screen readers, and does not allow proper indexing. Sep 27, 2018 at 21:53 | 946 | 4,226 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.887529 |
https://techstalking.com/finding-the-index-of-an-item-in-a-list/ | 1,657,113,051,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00479.warc.gz | 579,274,610 | 27,603 | # Finding the index of an item in a list
Each Answer to this Q is separated by one/two green lines.
Given a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, how do I get its index `1`?
``````>>> ["foo", "bar", "baz"].index("bar")
1
``````
Reference: Data Structures > More on Lists
# Caveats follow
Note that while this is perhaps the cleanest way to answer the question as asked, `index` is a rather weak component of the `list` API, and I can’t remember the last time I used it in anger. It’s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about `list.index` follow. It is probably worth initially taking a look at the documentation for it:
``````list.index(x[, start[, end]])
``````
Return zero-based index in the list of the first item whose value is equal to x. Raises a `ValueError` if there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
## Linear time-complexity in list length
An `index` call checks every element of the list in order, until it finds a match. If your list is long, and you don’t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give `index` a hint. For instance, in this snippet, `l.index(999_999, 999_990, 1_000_000)` is roughly five orders of magnitude faster than straight `l.index(999_999)`, because the former only has to search 10 entries, while the latter searches a million:
``````>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
``````
## Only returns the index of the first match to its argument
A call to `index` searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
``````>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
``````
Most places where I once would have used `index`, I now use a list comprehension or generator expression because they’re more generalizable. So if you’re considering reaching for `index`, take a look at these excellent Python features.
## Throws if element not present in list
A call to `index` results in a `ValueError` if the item’s not present.
``````>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
``````
If the item might not be present in the list, you should either
1. Check for it first with `item in my_list` (clean, readable approach), or
2. Wrap the `index` call in a `try/except` block which catches `ValueError` (probably faster, at least when the list to search is long, and the item is usually present.)
One thing that is really helpful in learning Python is to use the interactive help function:
``````>>> help(["foo", "bar", "baz"])
Help on list object:
class list(object)
...
|
| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value
|
``````
which will often lead you to the method you are looking for.
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use `enumerate()`:
``````for i, j in enumerate(['foo', 'bar', 'baz']):
if j == 'bar':
print(i)
``````
The `index()` function only returns the first occurrence, while `enumerate()` returns all occurrences.
As a list comprehension:
``````[i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar']
``````
Here’s also another small solution with `itertools.count()` (which is pretty much the same approach as enumerate):
``````from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ['foo', 'bar', 'baz']) if j == 'bar']
``````
This is more efficient for larger lists than using `enumerate()`:
``````\$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 174 usec per loop
\$ python -m timeit "[i for i, j in enumerate(['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 196 usec per loop
``````
To get all indexes:
``````indexes = [i for i,x in enumerate(xs) if x == 'foo']
``````
`index()` returns the first index of value!
| index(…)
| L.index(value, [start, [stop]]) -> integer — return first index of value
``````def all_indices(value, qlist):
indices = []
idx = -1
while True:
try:
idx = qlist.index(value, idx+1)
indices.append(idx)
except ValueError:
break
return indices
all_indices("foo", ["foo","bar","baz","foo"])
``````
A problem will arise if the element is not in the list. This function handles the issue:
``````# if element is found it returns index of element else returns None
def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
``````
``````a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']
``````
You have to set a condition to check if the element you’re searching is in the list
``````if 'your_element' in mylist:
print mylist.index('your_element')
else:
print None
``````
If you want all indexes, then you can use NumPy:
``````import numpy as np
array = [1, 2, 1, 3, 4, 5, 1]
item = 1
np_array = np.array(array)
item_index = np.where(np_array==item)
print item_index
# Out: (array([0, 2, 6], dtype=int64),)
``````
All of the proposed functions here reproduce inherent language behavior but obscure what’s going on.
``````[i for i in range(len(mylist)) if mylist[i]==myterm] # get the indices
[each for each in mylist if each==myterm] # get the items
mylist.index(myterm) if myterm in mylist else None # get the first index and fail quietly
``````
Why write a function with exception handling if the language provides the methods to do what you want itself?
## Finding the index of an item given a list containing it in Python
For a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, what’s the cleanest way to get its index (1) in Python?
Well, sure, there’s the index method, which returns the index of the first occurrence:
``````>>> l = ["foo", "bar", "baz"]
>>> l.index('bar')
1
``````
There are a couple of issues with this method:
• if the value isn’t in the list, you’ll get a `ValueError`
• if more than one of the value is in the list, you only get the index for the first one
### No values
If the value could be missing, you need to catch the `ValueError`.
You can do so with a reusable definition like this:
``````def index(a_list, value):
try:
return a_list.index(value)
except ValueError:
return None
``````
And use it like this:
``````>>> print(index(l, 'quux'))
None
>>> print(index(l, 'bar'))
1
``````
And the downside of this is that you will probably have a check for if the returned value `is` or `is not` None:
``````result = index(a_list, value)
if result is not None:
do_something(result)
``````
### More than one value in the list
If you could have more occurrences, you’ll not get complete information with `list.index`:
``````>>> l.append('bar')
>>> l
['foo', 'bar', 'baz', 'bar']
>>> l.index('bar') # nothing at index 3?
1
``````
You might enumerate into a list comprehension the indexes:
``````>>> [index for index, v in enumerate(l) if v == 'bar']
[1, 3]
>>> [index for index, v in enumerate(l) if v == 'boink']
[]
``````
If you have no occurrences, you can check for that with boolean check of the result, or just do nothing if you loop over the results:
``````indexes = [index for index, v in enumerate(l) if v == 'boink']
for index in indexes:
do_something(index)
``````
### Better data munging with pandas
If you have pandas, you can easily get this information with a Series object:
``````>>> import pandas as pd
>>> series = pd.Series(l)
>>> series
0 foo
1 bar
2 baz
3 bar
dtype: object
``````
A comparison check will return a series of booleans:
``````>>> series == 'bar'
0 False
1 True
2 False
3 True
dtype: bool
``````
Pass that series of booleans to the series via subscript notation, and you get just the matching members:
``````>>> series[series == 'bar']
1 bar
3 bar
dtype: object
``````
If you want just the indexes, the index attribute returns a series of integers:
``````>>> series[series == 'bar'].index
Int64Index([1, 3], dtype="int64")
``````
And if you want them in a list or tuple, just pass them to the constructor:
``````>>> list(series[series == 'bar'].index)
[1, 3]
``````
Yes, you could use a list comprehension with enumerate too, but that’s just not as elegant, in my opinion – you’re doing tests for equality in Python, instead of letting builtin code written in C handle it:
``````>>> [i for i, value in enumerate(l) if value == 'bar']
[1, 3]
``````
## Is this an XY problem?
Why do you think you need the index given an element in a list?
If you already know the value, why do you care where it is in a list?
If the value isn’t there, catching the `ValueError` is rather verbose – and I prefer to avoid that.
I’m usually iterating over the list anyways, so I’ll usually keep a pointer to any interesting information, getting the index with enumerate.
If you’re munging data, you should probably be using pandas – which has far more elegant tools than the pure Python workarounds I’ve shown.
I do not recall needing `list.index`, myself. However, I have looked through the Python standard library, and I see some excellent uses for it.
There are many, many uses for it in `idlelib`, for GUI and text parsing.
The `keyword` module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming.
In Lib/mailbox.py it seems to be using it like an ordered mapping:
``````key_list[key_list.index(old)] = new
``````
and
``````del key_list[key_list.index(key)]
``````
In Lib/http/cookiejar.py, seems to be used to get the next month:
``````mon = MONTHS_LOWER.index(mon.lower())+1
``````
In Lib/tarfile.py similar to distutils to get a slice up to an item:
``````members = members[:members.index(tarinfo)]
``````
In Lib/pickletools.py:
``````numtopop = before.index(markobject)
``````
What these usages seem to have in common is that they seem to operate on lists of constrained sizes (important because of O(n) lookup time for `list.index`), and they’re mostly used in parsing (and UI in the case of Idle).
While there are use-cases for it, they are fairly uncommon. If you find yourself looking for this answer, ask yourself if what you’re doing is the most direct usage of the tools provided by the language for your use-case.
### Getting all the occurrences and the position of one or more (identical) items in a list
With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.
``````>>> alist = ['foo', 'spam', 'egg', 'foo']
>>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo']
>>> foo_indexes
[0, 3]
>>>
``````
### Let’s make our function findindex
This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.
``````def indexlist(item2find, list_or_string):
"Returns all indexes of an item in a list or a string"
return [n for n,item in enumerate(list_or_string) if item==item2find]
print(indexlist("1", "010101010"))
``````
Output
``````[1, 3, 5, 7]
``````
## Simple
``````for n, i in enumerate([1, 2, 3, 4, 1]):
if i == 1:
print(n)
``````
Output:
``````0
4
``````
All indexes with the `zip` function:
``````get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]
print get_indexes(2, [1, 2, 3, 4, 5, 6, 3, 2, 3, 2])
print get_indexes('f', 'xsfhhttytffsafweef')
``````
Simply you can go with
``````a = [['hand', 'head'], ['phone', 'wallet'], ['lost', 'stock']]
b = ['phone', 'lost']
res = [[x[0] for x in a].index(y) for y in b]
``````
Another option
``````>>> a = ['red', 'blue', 'green', 'red']
>>> b = 'red'
>>> offset = 0;
>>> indices = list()
>>> for i in range(a.count(b)):
... indices.append(a.index(b,offset))
... offset = indices[-1]+1
...
>>> indices
[0, 3]
>>>
``````
# And now, for something completely different…
… like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices — even if it is an empty list. It works with strings as well.
``````def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
``````
When pasted into an interactive python window:
``````Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
>>> def indices(the_list, val):
... """Always returns a list containing the indices of val in the_list"""
... retval = []
... last = 0
... while val in the_list[last:]:
... i = the_list[last:].index(val)
... retval.append(last + i)
... last += i + 1
... return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
``````
# Update
After another year of heads-down python development, I’m a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.
Something like this:
``````def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
``````
Which, when pasted into an interactive python window yields:
``````Python 2.7.14 |Anaconda, Inc.| (default, Dec 7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
>>> def indices(l, val):
... """Always returns a list containing the indices of val in the_list"""
... return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
``````
And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn’t even see that answer, because I didn’t understand it. I hope that my somewhat more verbose example will aid understanding.
If the single line of code above still doesn’t make sense to you, I highly recommend you Google ‘python list comprehension’ and take a few minutes to familiarize yourself. It’s just one of the many powerful features that make it a joy to use Python to develop code.
A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:
``````>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
``````
You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.
Finding index of item x in list L:
``````idx = L.index(x) if (x in L) else -1
``````
This solution is not as powerful as others, but if you’re a beginner and only know about `for`loops it’s still possible to find the first index of an item while avoiding the ValueError:
``````def find_element(p,t):
i = 0
for e in p:
if e == t:
return i
else:
i +=1
return -1
``````
My friend, I have made the easiest code to solve your question. While you were receiving gigantic lines of codes, I am here to cater you a two line code which is all due to the help of `index()` function in python.
``````LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
``````
Output:
``````1
``````
I Hope I have given you the best and the simplest answer which might help you greatly.
There is a chance that that value may not be present so to avoid this ValueError, we can check if that actually exists in the list .
``````list = ["foo", "bar", "baz"]
item_to_find = "foo"
if item_to_find in list:
index = list.index(item_to_find)
print("Index of the item is " + str(index))
else:
print("That word does not exist")
``````
There is a more functional answer to this.
``````list(filter(lambda x: x[1]=="bar",enumerate(["foo", "bar", "baz", "bar", "baz", "bar", "a", "b", "c"])))
``````
More generic form:
``````def get_index_of(lst, element):
return list(map(lambda x: x[0],\
(list(filter(lambda x: x[1]==element, enumerate(lst))))))
``````
Python `index()` method throws an error if the item was not found. So instead you can make it similar to the `indexOf()` function of JavaScript which returns `-1` if the item was not found:
``````try:
index = array.index('search_keyword')
except ValueError:
index = -1
``````
``````name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
new_list=[]
for item in list:
new_list.append(item[0])
print(new_list)
try:
location= new_list.index(name)
except:
location=-1
print (location)
``````
This accounts for if the string is not in the list too, if it isn’t in the list then `location = -1`
### For one comparable
``````# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2
``````
### Custom predicate
``````some_list = [item1, item2, item3]
# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
i for i, item in enumerate(some_list)
if item.matches_your_criteria())
``````
### Finding index of all items by predicate
``````index_of_staff_members = [
i for i, user in enumerate(users)
if user.is_staff()]
``````
It just uses the python function `array.index()` and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function `indexOf()`).
``````fruits = ['apple', 'banana', 'cherry']
try:
pos = fruits.index("mango")
except:
pos = -1
``````
In this case “mango” is not present in the list `fruits` so the `pos` variable is -1, if I had searched for “cherry” the `pos` variable would be 2.
Since Python lists are zero-based, we can use the zip built-in function as follows:
``````>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]
``````
where “haystack” is the list in question and “needle” is the item to look for.
(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)
If you are going to find an index once then using “index” method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect.
Using bisect module on my machine is about 20 times faster than using index method.
Here is an example of code using Python 3.8 and above syntax:
``````import bisect
from timeit import timeit
def bisect_search(container, value):
return (
index
if (index := bisect.bisect_left(container, value)) < len(container)
and container[index] == value else -1
)
data = list(range(1000))
# value to search
value = 666
# times to test
ttt = 1000
t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)
print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")
``````
Output:
``````t1=0.0400, t2=0.0020, diffs t1/t2=19.60
``````
I find this two solution is better and I tried it by myself
``````>>> expences = [2200, 2350, 2600, 2130, 2190]
>>> 2000 in expences
False
>>> expences.index(2200)
0
>>> expences.index(2350)
1
>>> index = expences.index(2350)
>>> expences[index]
2350
>>> try:
... print(expences.index(2100))
... except ValueError as e:
... print(e)
...
2100 is not in list
>>>
``````
List comprehension would be the best option to acquire a compact implementation in finding the index of an item in a list.
``````a_list = ["a", "b", "a"]
print([index for (index , item) in enumerate(a_list) if item == "a"])
``````
The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 . | 6,095 | 21,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.864151 |
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Correct answers: 3 question: Why are cells essential to our body working as it should? The Ka value for HC2H3O2 is 1.8 x 10^-5. Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, We've added a "Necessary cookies only" option to the cookie consent popup. Thus, phenolphthalein is an appropriate indicator used for this titration of weak acid and strong base. Chemistry >. Why did the Soviets not shoot down US spy satellites during the Cold War? Buffer is a solution of weak acid and its conjugate base or weak base and its conjugate acid in which the Ph cant be altered very much after adding a certain amount of acid and base. Ca(NO_3)_2, Write the net ionic equation for the reaction. It is considered an organic acid (which means it is principally made of carbon and hydrogen atoms, not anything about food being made without addition of man-made chemicals.) Acetic acid has also shown use in cleaning agents and sanitizing products. Formua w systemie Hill C2H5NaO3: Obliczanie masy molowej (masa molowa) Aby obliczy mas molow zwizku chemicznego, wpisz jego wzr i wcinij 'oblicz' We wzorze chemicznym mona uy: Plugging in the values found for the equilibrium concentration as found on the ICE table for the equation Ka = [H3O+][C2H3O2]/[HC2H3O2] allows the value of Ka to be solved in terms of x. Express your answer as a chemical equation. Thus, from the definition of buffer, it is clear that the mixture of CH3COOH and NaOH is not a buffer. (a) CaCl_2 (b) (NH_4)_2SO_4 (c) NaC_2H_3O_2. Convert between NaOH*HC2H3O2 weight and moles. Is variance swap long volatility of volatility? Label each compound (reactant or product) in the equation with a variable to represent the unknown coefficients. Write the chemical equation and the K_a expression for the acid dissociation for the aqueous solution: HSO_4^-, Write an equation that show the dissociation of the compounds in water. HC2H3O2 is the chemical formula for the organic compound acetic acid. However, there is a conjugate acid-base pair in this system. write the following electrolytes as a balanced equation for their dissociation in water. Aqueous solutions of acetic acid and barium hydroxide. Finally, calculate the molarity of acetic acid in vinegar from the moles of $$\ce{HC2H3O2}$$ and the volume of the vinegar sample used. Write an equation to show how acetic acid reacts with water to produce ions in solution. The equivalence point was reached when a total of 20.0mL . write the following electrolytes as a balanced equation for their dissociation in water. Can I use a vintage derailleur adapter claw on a modern derailleur. Read our article on how to balance chemical equations or ask for help in our chat. A burette is a device that allows the precise delivery of a specific volume of a solution. The balanced net equation of acetic acid and NaOH is-. Then allow the liquid to drain from the pipette. To transfer the solution, place the tip of the pipette against the wall of the receiving container at a slight angle. What is the percent yield of the reaction where 0.185 Mg of salicylic acid and 125 kg of acetic anhydride react to form 1.82 x 1020 fg of aspirin? Using the following reaction:OH- + HC2H3O2 > H2O + C2H3O2-a 75.0 mL sample of potassium hydroxide, KOH, of unknown concentration is titrated with42.5 mL of 0.575 M acetic acid. Use substitution, Gaussian elimination, or a calculator to solve for each variable. The net ionic equation for the neutralization reaction that occurs during the titration is represented above. Unlike water, it is miscible with oils and like water, it can dissolve most organic compounds. This acid is an organic compound that comes in a liquid form that is colorless. Initial moles of acid and base in buffer is (2.00mol/L)(0.500L) = 0.100 Example: (NaCl --> Na+ + Cl- (b) (NH4)2SO4 (c) sodium acetate (NaC2H3O2) (d) copper(II) perchlora. A weak acid CH3COOH cant be fully dissociated in aqueous solution. Schematic Diagram Of Cao Ca Oh 2 Tes 50 52 Download Scientific . There are three main steps for writing the net ionic equation for HC2H3O2 + K2CO3 = KC2H3O2 + CO2 + H2O (Acetic acid + Potassium carbonate). 1. Track your food intake, exercise, sleep and meditation for free. Why is there a memory leak in this C++ program and how to solve it, given the constraints? Much of the chemical bonding character of acetic acid is due to its carboxyl group and accompanying OH group. The Greek philosopher Theophrastus described how vinegar could be combined with various metals to produce pigments for art and the Romans boiled vinegar to make a sweet confectionery syrup called sapa. Insert the tip of the pipette into the beaker of solution so that it is about a quarter inch from the bottom. $$K_c=\frac{\ce{[H3C-COOH]}\ce{[K+]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}\ce{[H2O]}}$$, And you can then rearrange to form the base dissociation constant: Answer to Question #56388 in General Chemistry for kelly. Acetic acid is also commonly used in the kitchen, in the form of vinegar. RV coach and starter batteries connect negative to chassis; how does energy from either batteries' + terminal know which battery to flow back to? Write the equation for the dissociation of KMnO4 in water. Omit water from the equation because it is understood to be present. The polar hydrogen end of the hydroxyl group will attract the polar negative oxygen atom in the carbonyl group of a neighboring molecule of acetic acid, forming a strong electrostatic attraction. Please elaborate. Acetic acid is one of the integral components in cellular respiration. Synthesized polymer in the form of a spiral staircase. Write the balanced equation, complete ionic equation, and net ionic equation for the reaction between CH3COOH and NH3. Step 2: Step 3: Buffer Solutions | Boundless Chemistry A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an . Acetic acid can be produced a few ways, natural and synthetic. Write the net ionic equation for the reaction between sodium hydroxide and carbonic acid. To learn more, see our tips on writing great answers. ISSN: 2639-1538 (online). Due to the fact that HC2H3O2 is an acid, it is also sometimes used as a solvent. First, be sure to count all of H and O atoms on each side of the chemical equation. Write net ionic equations for the reaction, if any. Write a net ionic equation for the reaction that occurs when aqueous solutions of acetic acid and ammonia are combined. It is also an example of double displacement reaction because both the reactants exchange ions with each other to obtain the sodium acetate salt and water. The formula of calculating the pH of the neutralization reaction between a strong base and weak acid is-. Suppose you had titrated your vinegar sample with barium hydroxide instead of sodium hydroxide: What volume (in mL) of 0.586 M $$\ce{Ba(OH)2}$$ (. Do not allow the solution to be sucked into the bulb itself. Created by LABScI at Stanford 3 Materials: 1 clear jar (You will need to use a measuring cup to label volume measurements along the side of the jar.) When finished, dispose of your chemical waste as instructed. In aqueous solution, themajor species is or area. New substances are formed as a result of the rearrangement of the original atoms. Candidate of Chemical Sciences, editor-in-chief of Guide-scientific.com. In the above reaction, pkw = 14 and c is the concentration of salt (CH3COONa). A)(E) HClO4, or perchloric acid, is the strongest acid. CH_3COOH (aq) + NaOH (aq). The reaction between acetic acid and sodium hydroxide is an example of neutralization reaction. Here, the titrant is an aqueous solution of ~0.1 M sodium hydroxide ($$\ce{NaOH}$$) and the analyte is vinegar. You can use parenthesis () or brackets []. Give the balanced equation for the dissociation equilibrium of acetic acid. Acetic acid is an effective solvent and can dissolve not only polar compounds but non-polar compounds as well. $$K_b=\frac{\ce{[H3C-COOH]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}}$$. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Acetobacter is still the main method of vinegar production in the food industry. The solution is stirred and its pH is measured to be 4.73. Then perform a final rinse, but this time use vinegar. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. Write the molecular equation, complete ionic equation, and net ionic equation for the reaction that occurs between sodium hydroxide and acetic acid. (Use the lowest possible coefficients. Part A If 10.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.70 L. Part A If 10.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solution? Finally, calculate the mass percent of acetic acid in vinegar from the mass of $$\ce{HC2H3O2}$$ and the mass of vinegar. Answers >. For more information, see the site's. When Alex isn't nerdily stalking the internet for science news, he enjoys tabletop RPGs and making really obscure TV references. The Ph of the titration lies between 6.5 to 10 and pH of the phenolphthalein ranges from 8.2 to 10. substitutue 1 for any solids/liquids, and P, (assuming constant volume in a closed system and no accumulation of intermediates or side products). Write the reaction for the dissociation of KHT (s) in water. between: a. Acetic acid (0.20 M) and sodium acctale (0.20 M) b. Acetic acid (0.20 M) and hydrochloric acid (0.10 M) c- Acctic acid (0.20 M) an. Acceleration without force in rotational motion? of NaOH*HC2H3O2 is 100.0491 g/mol. To balance any chemical reaction, firstly the unbalanced equation should be written. Vinegar is roughly 3-9% acetic acid by volume, making acetic . To decide whether it is buffer solution or not first we have to clear the concept of buffer solution. Write the equation to represent the dissociation of calcium hydroxide in aqueous solution. As such, it has a higher boiling point than other compounds of an analogous structure. The name for HC2H3O2 is acetic acid. A solution is prepared by adding 100 mL of 1.0 M HC2H3O2(aq) to 100 mL of 1.0 M NaC2H3O2(aq). Write a net ionic equation for the reaction of aqueous acetic acid and aqueous potassium hydroxide. Write the chemical equation and the Ka expression for the acid dissociation of each of the following acids in aqueous solution. Detailed instructions on how to use a pipette are also found on the last page of this handout. how to critically analyse a case law; where does deadpool fit in the mcu timeline; joe montana high school stats. This high boiling point is also explained by the tendency for the hydroxyl group to make hydrogen bonds with other nearby molecules. Label Each Compound With a Variable. This is a special point in the titration called the _________________________ point. NaOH = Na + + OH -. Due to the fact that HC2H3O2 is an acid, it is also sometimes used as a solvent. Calcium perchlorate, Write a balanced chemical equation for the dissociation of the ionic compound shown below in the water. The two central carbon atoms bond, leaving the three hydrogens from the methyl at one end, and the hydroxyl group from the carboxyl on the other. Get the appropriate amount of the solution you wish to pipette in a clean, dry beaker. You may want to do this several times for practice. Never pipette directly out of the stock bottles of solution. Ammonium bromate. When the bottom of the meniscus is even with the volume mark, press your index finger firmly on the top of the pipette so no liquid leaks out. Create a System of Equations. Calculate the pH of a buffer solution containing 0.200 M acetic acid, HC2H3O2, plus 0.150 M sodium acetate. Write a balanced molecular equation, and net ionic equation for the following reaction. If any NaOH spills on you, rinse immediately under running water for up to 15 minutes and report the accident to your instructor. copyright 2003-2023 Homework.Study.com. Be Careful When Speaking About Lead Pollution: The Good, The Bad, And The Ugly! Write a balanced equation for the reaction of acetic anhydride with water. This molecule acts as the basic reactant. This is the reaction between sodium bicarbonate and acetic acid.. Through human history, the activity of Acetobacter has been the main method of producing vinegar for culinary and industrial use. Acetic acid reacts with sodium hydroxide in the following fashion: HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2(aq). Neutralization reaction is one type chemical of reaction in which acid and base reacts with each other in a quantitative amount to form salt and water. 322166814/www.reference.com/Reference_Mobile_Feed_Center3_300x250, The Best Benefits of HughesNet for the Home Internet User, How to Maximize Your HughesNet Internet Services, Get the Best AT&T Phone Plan for Your Family, Floor & Decor: How to Choose the Right Flooring for Your Budget, Choose the Perfect Floor & Decor Stone Flooring for Your Home, How to Find Athleta Clothing That Fits You, How to Dress for Maximum Comfort in Athleta Clothing, Update Your Homes Interior Design With Raymour and Flanigan, How to Find Raymour and Flanigan Home Office Furniture. a. C6H5COOH. This is the reaction between weak acid and strong base and any reaction involving acid and base will be an example of neutralization reaction. Write the balanced equation for the neutralization reaction between aqueous sodium hydroxide and acetic acid. The fermentation of glucose creates ethanol and minute amounts of acetic acid. Balance the equation HC2H3O2 + NaOH = H2O + NaC2H3O2 using the algebraic method. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. Suppose you added 40 mL of water to your vinegar sample instead of 20 mL. Cool C7H603 + C4H6O3 References . Write the chemical equation for the dissociation of CH3COOH in water. What is the anion for HC2H3O2? Write the equation for the dissociation of LiF in water. The solvent properties of acetic acid make it good for removing residue and stains on surfaces, hence why it and other acetate-derived compounds are often used in the lab to clean equipment and glassware. Write balanced equations in net-ionic form for: a. [{Blank}](reac. Write an equation that show the dissociation of the compounds in water. Most people recognize acetic acid, when it is diluted with water, as vinegar. The pH of the solution will be 8.203 - 0.069 = 7.846. C34_Group5_Expt1 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Is sulfide ion a stronger base than hydroxide ion? Polyvinyl acetate derived glues are used for carpentry, book-binding, envelopes, and as a base in gum chewing products. Two conjugate products are formed and the reaction is reversible. Because it is a mixture of weak acid and strong base not the weak acid and its conjugate base or weak base and its conjugate acid. An indicator solution is used to indicate when all the acetic acid has been consumed and that the reaction in complete. (I also corrected your charges.). As the titration is performed, the following data will be collected: Using this data, the molarity and mass percent of acetic acid in vinegar can be determined by performing a series of solution stoichiometry calculations (see Calculations Section). rev2023.3.1.43269. Is there a more recent similar source? The dissociation constant for HC2H3O2 is 1.76 x 10-5. CaCl_2 \text{ and } (NH_4)2SO_4. a. CH 3 COOH = CH 3 COO - + H +. Write a net ionic equation to show that acetic acid, C H 3 C O O H , behaves as an acid in water. The purpose of the experiment, "Flame Test Lab", is for one to determine if an element can be identified by the light they give off when inserted in the flame of a Bunsen Burner. Answer (1 of 2): The acid you asked about is acetic acid. Using the pipette bulb, draw the water into the pipette up above the 5-mL mark, then allow it to drain out through the tip. All rights reserved. The concentration of acetic acid in vinegar may be expressed as a molarity (in mol/L): Molarity = Moles of Acetic Acid Volume of Vinegar (in L) or as a mass percent. Show all physical states. In small doses and dilute concentrations, the ingestion of acetic acid is harmless and potentially beneficial to humans. Websites. Write out the dissociation reaction for sodium hydroxide. (a) the dissociation of perchloric acid in water (b) the dissociation of propanoic acid. DO NOT blow out the remaining solution. Write the balanced neutralization reaction that occurs between sodium hydroxide and acetic acid. HC2H3O2(aq)+OH(aq)C2H3O2(aq)+H2O(l) A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The above reaction is a neutralization reaction between a weak acid (CH3COOH) and strong base (NaOH). The reaction between acetic acid and sodium hydroxide is an example of neutralization reaction. The compound has a charge of -1. If 2.49 mL of vinegar requires 34.9 mL of 0.1006 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? HC2H3O2 is the chemical formula for the organic compound acetic acid. It is the second main constituent of common household vinegar, after water. Weight, g. HC2H3O2. Createyouraccount. Be specific. Molar mass of HC2H3O2 is 60.052208 g/mol Molar mass of CaSO4 is 136.14162 g/mol Molar mass of KH2PO4 is 136.0856862 g/mol Molar mass of H2O is 18.015324 g/mol Molar mass of AlOH3 is 46.00480968 g/mol Molar mass of SrF2 is 125.6178065 g/mol Molar mass of CoN5H15Cl3 is 250.4456005 g/mol Molar mass of MgSO4(H2O)7 is 246.475548 g/mol Red wine vinegar has some personality as well as acidity. Alex Guarnaschelli. Though a more efficient chemical reaction thanAcetobacter,most of these kinds of bacteria are ironically not acid tolerant. 3. Why does Jesus turn to the Father to forgive in Luke 23:34? Include physical states for all species. Write the ionic equation and balance the equation. This is called the equivalence point of the titration. The coefficients show the number of particles (atoms or molecules), and the indices show the number of atoms that make up the molecule. Type of Chemical Reaction: For this reaction we have a chemical reaction. Write the equation for the dissociation of AlCl_3 in water. Learn more about Hc2H3O2 Strong Or Weak from our Websites analysis here on IPAddress.com. Write out the reaction for the dissociation of KOH in water. Allow the distilled water to drain out through the tip in order to ensure that the tip is also rinsed. It is an extremely caustic and reactive acid that, if not handled carefully, can be hazardous. Aspirin (C9H8O4) is produced from salicylic acid (C7H6O3) and acetic anhydride (C4H6O3): C7H6O3 + C4H6O3 --> C9H8O4 + HC2H3O2. Na_2 CO_3. Write the balanced equation, complete ionic equation, and net ionic equation for the reaction between CH3COOH and NaOH. Therefore, it dissociates at least one covalent "H-X" bond, such that a proton can be donated to a water molecule. First, we balanc. Therefore, an equilibrium state can be established at a constant solution temperature. The bubbles you see when you mix backing soda with vinegar are the CO2 found on the product side of the equation. Convert between HC2H3O2 weight and moles. This is an example of acid neutralization reaction with weak acid and strong base. Write out the reaction for the dissociation of the electrolyte AlCl3 in water. Now rinse the burette with a small amount of $$\ce{NaOH}$$ (. Acetic acid is also known as ethanoic acid. Compound states [like (s) (aq) or (g)] are not required. Calculating Changes in a Buffer Solution, Example 1: Step 1: HC2H3O2(aq) H+(aq)+C2H3O 2(aq) HC 2 H 3 O 2 ( aq) H + ( aq) + C 2 H 3 O 2 ( aq) Recall that sodium acetate, NaC 2 H 3 O 2, dissociates . Use your two best sets of results along with calculated values in the previous table to determine the mass percent of acetic acid in vinegar. (First show the reaction with H+ (aq) as a product and then with the hydronium ion.) That would be $\ce{[C2H3O2]-}$ as $\ce{K+}$ has highly limited Brnsted-Lowry acid/base properties in water. Formula in Hill system is C2H5NaO3: Computing molar mass (molar weight) To calculate molar mass of a chemical compound enter its formula and click 'Compute'. It is also a natural byproduct of the fermentation of produce, such as fruit, grain, rice, and potatoes. it will form acetates and water when introduced to a basic environment, and it can be reduced by the addition of hydrogen to form ethanol (alcohol). Write an equation that show the dissociation of the compounds in water. Write an equation that show the dissociation of the compounds in water. The oxidation of metals with acidic compounds is sometimes used to create industrial amounts of hydrogen gas. francisco Record this volume of vinegar (precise to two decimal places) on your report. Substances that react are called starting materials or reactants. First, convert the moles of $$\ce{HC2H3O2}$$ in the vinegar sample (previously calculated) to a mass of $$\ce{HC2H3O2}$$, via its molar mass. {/eq} in water. Mg (s) + HC2H3O2 (aq) --> Mg(C2H3O2)2 (aq) + H2 (g) thanks. Aqueous solutions of carbonic acid and potassium hydroxide. Then remove the pipette tip from the beaker of solution. The organization of the atoms gives acetic acid a pseudo-tetrahedral structure, with three hydrogens serving as the base and the carboxyl group as the tip. Most of the problems people have are due to errors in adding up the atoms. 4.74 0.00 4.74 2.00 2.00 log log1.8 10 5 log = + = = + = + M M x acid base pH pKa 2.0 What is the new pH after 2.00 mL of 6.00M HCl is added to this buffer ? Apart from its crucial role in biology, acetic acid is an important industrial chemical that is used to produce a number of consumer goods. The acid creates an oxidizing environment that damages the membranes of bacteria and the cell walls of fungi. S. 1. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Write the ionic equation and balance the equation. Other bacteria are capable of directly converting sugars or carbon dioxide into acetic acid without ethanol as an intermediate reactant. CH_3COOH is a weak electrolyte. This problem has been solved! You are right however I should have probably expanded on my answer but when I posted it appeared that the OP had not given much thought to the question so I merely wished to nudge the OP in the right direction. Temperature Has A Significant Influence On The Production Of SMP-Based Dissolved Organic Nitrogen (DON) During Biological Processes. It is also used in the mass production of wine, chemicals, starch and lacquers and occurs naturally in both plants and animals. First, rinse the inside of the volumetric pipette with distilled water. Write the full balanced equation for the reaction between acetic acid and sodium hydroxide. Moles. Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced. Write the chemical equation and the K_a expression for the acid dissociation for the aqueous solution: CH_3NH_3^+, Write an equation that show the dissociation of the compounds in water. If a solution initially contains 0.210 M HC2H3O2, what is the equilibrium concentration of H3O . I am preparing for an exam on equilibrium and I have two questions: How would I go about calculating this? Vinegar is a dilute solution of acetic acid (HC2H3O2). Get control of 2022! Convert between NaOH*HC2H3O2 weight and moles. Write net ionic equations for the reaction of hydronium ion with SrCO3. All vinegar is typically 4% acetic acid by volume. And the acid dissociation constant (ka) of acetic acid. Type of chemical reaction, pkw = 14 and c is the chemical formula for the reaction between hydroxide! All of H and O atoms on each side of the pipette against the wall of the compound... Or ( g ) ] are not required rinse the inside of the compounds in water or [... = 14 and c is the concentration of salt ( CH3COONa ) spiral.. Download Scientific the pH of the original atoms main constituent of common household vinegar, after water read article. An exam on equilibrium and I have two questions: how would I go about calculating this or. Formed as a balanced chemical equation for the reaction in complete analysis here on IPAddress.com hydroxide. Dissolved organic Nitrogen ( DON ) during Biological Processes Pollution: the Good, the activity acetobacter. Compound shown below in the kitchen, in the mcu timeline ; joe montana high stats. And aqueous potassium hydroxide several times for practice about is acetic acid ( ). Therefore, an equilibrium state can be established at a slight angle perchlorate, write a net ionic for... Special point in the equation to represent the dissociation of propanoic acid: for titration!, it is also sometimes used to create industrial amounts of hydrogen gas than other compounds of analogous! Father to forgive in Luke 23:34 \ ) ( NH_4 ) 2SO_4 compounds of an analogous structure on... Answer ( 1 of 2 ): the acid creates an oxidizing environment that damages the membranes of bacteria capable! Is there a memory leak in this C++ program and how to critically analyse a case law ; where deadpool... Diagram of Cao ca Oh 2 Tes 50 52 Download Scientific ; where does deadpool fit in the kitchen in! Net ionic equation, complete ionic equation for the dissociation equilibrium of acetic acid more see! The original atoms substances that react are called starting materials or reactants balanced chemical equation ( to! This high boiling point is also used in the water acid without ethanol as an intermediate reactant an solution! In adding up the atoms - free Download as PDF File (.txt ) or read online for.. Out the reaction between CH3COOH and NaOH envelopes, and net ionic equation for the neutralization reaction CH3COOH! When it is an example of neutralization reaction with H+ ( aq ) as a balanced chemical equation the... Of Cao ca Oh 2 Tes 50 52 Download Scientific, the ingestion of acetic acid and strong.! And c is the reaction between a strong base, it is an example of neutralization reaction between sodium and. From the pipette caustic and reactive acid that, if not handled carefully, can hazardous... Into the beaker of solution it dissociates at least one covalent H-X '',. Ph of a specific volume of vinegar use in cleaning agents and sanitizing products the hydroxyl group to make bonds! Base ( NaOH ) gum chewing products of KHT ( s ) ( NH_4 ).. Essential to our body working as it should handled carefully, can be produced a few ways, and... Montana high school stats the titration, and net ionic equation for the dissociation AlCl_3. Of vinegar ( precise to two decimal places ) on your report to produce ions in solution ) 2SO_4 against. When aqueous solutions of acetic anhydride with water, it has a Significant Influence on last. H-X '' bond, hc2h3o2 + naoh limiting reactant as fruit, grain, rice, and net ionic equation for dissociation. Solution temperature KHT ( s ) ( E ) HClO4, or perchloric acid is! Vinegar sample instead of 20 mL is understood to be sucked into the bulb itself as it?! On equilibrium and I have two questions: how would I go about calculating this algebraic method equivalence. Between a strong base and industrial use on each side of the pipette tip from bottom! The fermentation of glucose creates ethanol and minute amounts of hydrogen gas write... Parenthesis ( ) or ( g ) ] are not required not allow the solution to be sucked the! Answers: 3 question: why are cells essential to our body working as it should H. And as a solvent, Gaussian elimination, or a calculator to solve it, given the constraints working it. Smp-Based Dissolved organic Nitrogen ( DON ) during Biological Processes are combined = 14 c. Much of the titration called the _________________________ point chemical waste as instructed an example of acid neutralization reaction,... Https: //status.libretexts.org and then with the hydronium ion with SrCO3 that allows the precise delivery of a spiral.! Phenolphthalein is an acid, is the chemical equation for the dissociation of KHT s! Chemical waste as instructed equation HC2H3O2 + NaOH = H2O + NaC2H3O2 the! ( s ) ( aq ) or read online for free the original atoms help in chat... A water molecule a dilute solution of acetic anhydride with water to drain from beaker. Wish to pipette in a clean, dry beaker net ionic equation for the neutralization reaction and potatoes both and! Go about calculating this 8.203 - 0.069 = 7.846 equation to represent the unknown coefficients salt ( )! Of common household vinegar, after water without ethanol as an intermediate reactant aq ) by tendency... Balance chemical equations or ask for help in our chat book-binding, envelopes, and acid. Pipette against the wall of the equation because it is buffer solution or not first we have a reaction... Buffer, it is about a quarter inch from the definition of buffer solution containing 0.200 acetic! A spiral staircase or ask for help in our chat Influence on the product side the! Conjugate acid-base pair in this system in aqueous solution if not handled carefully, can be established at constant! To represent the dissociation of calcium hydroxide in aqueous solution NaOH = H2O + NaC2H3O2 using the method... Below in the above reaction is a neutralization reaction between CH3COOH and NH3 H+ ( aq as! Bubbles you see when you mix backing soda with vinegar are the CO2 found the! When aqueous solutions of acetic acid and strong base and weak acid and ammonia are combined wall of neutralization. And the reaction for the dissociation of the pipette tip from the bottom like... Ka value for HC2H3O2 is 1.76 x 10-5 of vinegar still the main method of producing vinegar for culinary industrial... Does Jesus turn to the fact that HC2H3O2 is 1.76 x 10-5 indicator used for carpentry, book-binding,,! Is measured to be present H-X '' bond, such that a proton can be hazardous a base in chewing... Acid CH3COOH cant be fully dissociated in aqueous solution under running water for up to 15 minutes report. A strong base and weak acid and strong base and any reaction involving and! Fruit, grain, rice, and potatoes solution initially contains 0.210 M HC2H3O2, what is the concentration. Intermediate reactant a memory leak in this system our Websites analysis here on IPAddress.com, elimination. Is understood to be present, phenolphthalein is an example of neutralization reaction between and. Influence on the product side of the compounds in water vinegar are the CO2 on... The volumetric pipette with distilled water are capable of directly converting sugars or carbon dioxide into acetic acid been. Much of the fermentation of produce, such that a proton can be produced a ways. The pH of a buffer dispose of your chemical waste as instructed mcu timeline ; montana. Diagram of Cao ca Oh 2 Tes 50 52 Download Scientific sure to count of. Schematic Diagram of Cao ca Oh 2 Tes 50 52 Download Scientific a base in gum chewing.! Clean, dry beaker acid creates an oxidizing environment that damages the membranes of bacteria are ironically not acid.! Neutralization reaction between acetic acid by volume 3-9 % acetic acid ( HC2H3O2 ) also a byproduct. 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Are not required LiF in water still the main method of vinegar production in the kitchen, in the,. The organic compound acetic acid has been the main method of producing vinegar for culinary and industrial use of! Drain out through the tip is also rinsed how would I go about calculating this compound acetic.! | 8,318 | 33,699 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.898865 |
https://avsconsultants.co.in/valuable-techniques-for-what-is-the-most-difficult-math-that-you-can-use-starting-today/ | 1,718,966,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00225.warc.gz | 99,550,923 | 13,312 | # The Upside to What Is the Most Difficult Math
That the reader was told this does not absolutely indicate he knows what the section is all about, at all. Moreover, in the instance of Ethereum, the issue total isn’t limited. About this moment, the idea of zero was developed.
The problem isn’t to choose which is which. The mechanic wouldn’t need to even observe the carjust apply the equation. There are a couple of techniques to solve it, but Talwalkar presents a very simple shortcut.
## What Is the Most Difficult Math – the Conspiracy
In Math, problems can always be separated into a set of simpler calculations, but nevertheless, it might be difficult to understand how right away. If you locate an efficient algorithm, the issue is definitely class P. Experiment 2 Using the identical circuit as in the prior experiment, the student tried three distinct resistors.
You’re attempting to teach me fractions! Though a new differential equation solver provides theoretical mathematical understanding, its most important applications are diverse and extremely practical. https://www.gcu.edu/degree-programs/bachelor-arts-christian-studies-youth-ministry Setting up an equation for average is only going to get you up to now.
Every undertaking or job has many constraints whether it be time or price range. It is very important to maintain a check on one’s ability as it aids in organizing the thinking and utilize it accordingly to recall the data which will initially alter the success throughout the curriculum. I found it quite hard to grasp the sense supporting the limit statement and took me long hours and lots of books to have a hold of it.
## Gossip, Lies and What Is the Most Difficult Math
Your child should know the basics before he or she is able to move on to new topics. Rulers aren’t simple to use accurately should they don’t start at zero. Thus, pupils aren’t able to use the skill to new contexts.
It is believed to be that foundation that initially underlies the mathematical logic and additionally the remaining portion of the mathematics too. It goes quite in depth, and it is a superb means to start. Thus the very good practice encourages a more secure comprehension and offers pupils having a more cohesive approach to their learning.
Other folks believe financial scholarships ought to be awarded too. Only a masochist would select a college major depending on the simple fact that it’s challenging. In our experience, students discover that a few of the toughest ACT questions take several actions to fix.
Obviously, the Big Four are large, broad regions of math. The thing which makes math difficult for many students is it requires patience and persistence. In this way, you’ll have time to digest the info over a couple of weeks, plus a chance to ask me personally about the curriculum.
## What Is the Most Difficult Math – the Story
It’s been a few months since I joined this website and I repeatedly encounter questions posted with similar dilemmas and all of these have flimsy logical foundation. Neither True nor False can supply an answer to the subsequent question. It is proof-of-work.
Well, whoever solves these problems first will obtain a financial incentive in the kind of this online cash. Distributed users determine the issue quantity of rewards with no control from anyone, therefore it prevents the chance of centralized powers to intervene. The system is reportedly trustless’.
These silent drives provide you with the opportunity to find out solutions to your problems and provide you with the ability to be creative. Several issues complicate using non-standard measures. People have such very inadequate awareness of time, Barooah states, and without good time calibration, it’s much more difficult to find the outcome of your actions.
One, we apply an extremely careful collection of our writers. I like to create everything crystal clear. This very first shape is only a scaled model.
## Get the Scoop on What Is the Most Difficult Math Before You’re Too Late
To put it simply, the GRE can be a very challenging test, particularly if you’ve been out of school for some time. You might be able to bring an algebra class, but nevertheless, it will be very challenging for you. Figure out the proportion of questions in each one of the four primary GRE math types.
For example, if your child only needs an occasional answer to finish a math homework assignment, then you shouldn’t have to cover regular lessons. If you don’t have time, there are games made to help practice math concepts wherever your little one might not even realize he or she’s learning. In truth, it is sensible to employ a tutor that will help you review when you discover that you’ve sturggled in a math class!
In any event, get prepared to utilize your noggin. The total price of doing this generally is only a bit below the price of 1 ounce of gold. It is presently in your possession.
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It ensures an attacker can’t place an overwhelming number of malicious nodes within a shard. As a result of random shuffling and assignment, attackers find it impossible to opt for the shard they wish to validate. They have to answer ja in the event the answer they are devoted to give is da and this they cannot do.
Hence, mathematics is a choice vehicle for developing pupil’s logical thinking and higher order abilities and also playing a major part in lots of other subject and skilled field. Digital resources also offer assessment opportunities that may identify short-term gaps until they become long-term gaps. Thus, pupils aren’t able to use the skill to new contexts.
It is believed to be that foundation that initially underlies the mathematical logic and additionally the remaining portion of the mathematics too. At a particular point, it’s time to return and relearn the concepts, or find a tutor to assist with the fundamentals. Thus the very good practice encourages a more secure comprehension and offers pupils having a more cohesive approach to their learning.
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Therefore, this type of work generally presents an extensive method of mathematical foundations in addition to a very low amount of expectation in order to comprehend the student’s abilities because it generally limits the opportunities. Actually, the most popular college majors are often a number of the leastdifficult choices. In this way, you’ll have time to digest the info over a couple of weeks, plus a chance to ask me personally about the curriculum.
A number of months before, Barooah started to wean himself from coffee. Seriation can be challenging for young children to comprehend. It turned out to be a very long day for you, she explained.
## Top What Is the Most Difficult Math Choices
For many students, it boils to the option to drop out or get assistance with math homework. Algebra is hard for some people due to the fact that they have not learned the skills required to get started. Mastering math sometimes takes a great deal of work, but by means of the appropriate skills and continuing practice, it doesn’t need to be difficult!
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## Gossip, Deception and What Is the Most Difficult Math
What makes them difficult is that you’ve got to work through multiple actions as a way to fix the issue. Actually, there’s a great deal of power and specific machinery called ASICs that’s necessary to mine. Since budget is truly significant in lots of projects, a superior programmer will make a software using fewer resources.
Nevertheless, in a dynamical system, there’s a fixed rule that is initially described by the time dependence in a geometrical way. This ability to process data is known as scaling, and it’s critical to wide blockchain adoption. As an institutional crypto investor, it’s your right to actively take part in the system.
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That the reader was told this does not absolutely indicate he knows what the section is all about, at all. Moreover, in the instance of Ethereum, the issue total isn’t limited. You may have to remind the controller many times, always in a sort, neutral tone.
The problem isn’t to choose which is which. A superb programmer knows the way to handle the project requirements and is quite flexible. There are a couple of techniques to solve it, but Talwalkar presents a very simple shortcut.
The ticket sales supply the amount in. This is usually the scenario, and it’ll help you save you time that you may reinvest in different problems on the exam. Difficulty Inevitably, exactly like any other competition with a monetary prize, there’ll be increasing demand trying to win these precious bitcoin.
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The physician told you which you don’t have Lyme. Your brain will begin going into overdrive and you’ll produce some wonderful ideas! If you’re considering taking the ACT, you may have some anxiety about the questions that you’ll encounter.
In Math, problems can always be separated into a set of simpler calculations, but nevertheless, it might be difficult to understand how right away. To be able to be the miner to bring the next block, you’ve got to win a competition to discover a suitable hash that solves a tough math issue. Students who get variables normally have no issue with functions.
You’re attempting to teach me fractions! The write every potential book algorithm is all about as not-P as you’re able to get. The Basel Problem This equation claims that whenever you take the reciprocal of all of the square numbers, then add all of them together, you get pi squared over six.
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It’s also important to consider that just because two numbers appear different doesn’t mean they’re not the very same price. Ultracoin employs the Scrypt-Chacha algorithm, which ensures it can’t be mined by ASICs such as other cryptocurrencies. Think in terms of what number may be the smallest possible price.
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The ticket sales supply the amount in. It’s dangerous work in a location where law is thin. Utilize your time efficiently.
## The Bizarre Secret of What Is the Most Difficult Math
It ensures an attacker can’t place an overwhelming number of malicious nodes within a shard. The total price of doing this generally is only a bit below the price of 1 ounce of gold. It is presently in your possession.
It is quite a bit easier and faster to process the question by taking a look at a drawing than attempting to recreate it in your head. You always wind up serving whatever it’s you love. It could be uncomfortable initially, but after a while, you learn how to appreciate everything around you.
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## Facts, Fiction and What Is the Most Difficult Math
The problems basically begin to appear in the center and higher school and this is because students often migrate to some other grade or new subject. For students striving to excel in the area of mathematics, math competitions supply a simple avenue for highlighting your abilities and talent. Therefore, the curriculum must concentrate on important mathematics that is well worth the time and attention of pupils and that is going to prepare them for continued study and for solving problems in a number of school, house, and work settings.
This type of teaching did not seem to be difficult as I had anticipated but I did avoid using laptops for the principal lesson. The thing which makes math difficult for many students is it requires patience and persistence. In this way, you’ll have time to digest the info over a couple of weeks, plus a chance to ask me personally about the curriculum.
Centralized scarcity is the consequence of a central authority producing or issuing the merchandise. Information theory is regarded to be a branch of the applied mathematics and additionally for the electrical engineering that duly requires the quantification of any given info. These systems work in a typical pyramid scheme of incentivizing another participant to have a position as a way to increase the worth of the very first participants’ holdings.
As we have observed, large parts of the mining industry have been collected into pools which are primarily accountable for nearly all the mining rewards and security of the chain. Several issues complicate using non-standard measures. Usually, miners purchase and put money into hardware for the sole intent of generating cryptocurrencies including Bitcoin, Monero, Ethereum, and Siacoin.
This dilemma makes it simple to execute double-spending or other sorts of attacks that arrive with forking the blockchain. In the majority of situations, as soon as an individual or a society has an issue, they should solve it for themselves. Football can have a violence problem. however, it’s not one that we really need to address.
## Choosing What Is the Most Difficult Math
That the reader was told this does not absolutely indicate he knows what the section is all about, at all. Moreover, in the instance of Ethereum, the issue total isn’t limited. About this moment, the idea of zero was developed.
The problem isn’t to choose which is which. The mechanic wouldn’t need to even observe the carjust apply the equation. There are a couple of techniques to solve it, but Talwalkar presents a very simple shortcut.
SAT is used by nearly every school. Below, you will find ten of the most difficult questions in Magoosh’s Online ACT Prep. Have a look at our guide on how to acquire a perfect 800 on the SAT math section, written by means of a perfect-scorer.
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} | 3,098 | 15,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-26 | latest | en | 0.945933 |
https://www.graphclasses.org/classes/gc_318.html | 1,686,154,694,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653930.47/warc/CC-MAIN-20230607143116-20230607173116-00511.warc.gz | 857,396,611 | 16,124 | # Graphclass: hereditary dually chordal
Definition:
A graph is an hereditary dually chordal graph iff every induced subgraph is a dually chordal graph.
[+]Details
## Inclusions
The map shows the inclusions between the current class and a fixed set of landmark classes. Minimal/maximal is with respect to the contents of ISGCI. Only references for direct inclusions are given. Where no reference is given, check equivalent classes or use the Java application. To check relations other than inclusion (e.g. disjointness) use the Java application, as well.
[+]Details
[+]Details
[+]Details
## Speed
Speed [?] The speed of a class $X$ is the function $n \mapsto |X_n|$, where $X_n$ is the set of $n$-vertex labeled graphs in $X$.Depending on the rate of growths of the speed of the class, ISGCI distinguishes the following values of the parameter: Constant Polynomial Exponential Factorial Superfactorial ($2^{o(n^2)}$ ) Superfactorial ($2^{\Theta(n^2)}$ ) unknown [+]Details
## Parameters
acyclic chromatic number [?]The acyclic chromatic number of a graph $G$ is the smallest size of a vertex partition $\{V_1,\dots,V_l\}$ such that each $V_i$ is an independent set and for all $i,j$ that graph $G[V_i\cup V_j]$ does not contain a cycle. Unbounded [+]Details bandwidth [?]The bandwidth of a graph $G$ is the shortest maximum "length" of an edge over all one dimensional layouts of $G$. Formally, bandwidth is defined as $\min_{i \colon V \rightarrow \mathbb{N}\;}\{\max_{\{u,v\}\in E\;} \{|i(u)-i(v)|\}\mid i\text{ is injective}\}$. Unbounded [+]Details book thickness [?]A book embedding of a graph $G$ is an embedding of $G$ on a collection of half-planes (called pages) having the same line (called spine) as their boundary, such that the vertices all lie on the spine and there are no crossing edges. The book thickness of a graph $G$ is the smallest number of pages over all book embeddings of $G$. Unbounded [+]Details booleanwidth [?]Consider the following decomposition of a graph $G$ which is defined as a pair $(T,L)$ where $T$ is a binary tree and $L$ is a bijection from $V(G)$ to the leaves of the tree $T$. The function $\text{cut-bool} \colon 2^{V(G)} \rightarrow R$ is defined as $\text{cut-bool}(A)$ := $\log_2|\{S \subseteq V(G) \backslash A \mid \exists X \subseteq A \colon S = (V(G) \backslash A) \cap \bigcup_{x \in X} N(x)\}|$. Every edge $e$ in $T$ partitions the vertices $V(G)$ into $\{A_e,\overline{A_e}\}$ according to the leaves of the two connected components of $T - e$. The booleanwidth of the above decomposition $(T,L)$ is $\max_{e \in E(T)\;} \{ \text{cut-bool}(A_e)\}$. The booleanwidth of a graph $G$ is the minimum booleanwidth of a decomposition of $G$ as above. Unbounded [+]Details branchwidth [?]A branch decomposition of a graph $G$ is a pair $(T,\chi)$, where $T$ is a binary tree and $\chi$ is a bijection, mapping leaves of $T$ to edges of $G$. Any edge $\{u, v\}$ of the tree divides the tree into two components and divides the set of edges of $G$ into two parts $X, E \backslash X$, consisting of edges mapped to the leaves of each component. The width of the edge $\{u,v\}$ is the number of vertices of $G$ that is incident both with an edge in $X$ and with an edge in $E \backslash X$. The width of the decomposition $(T,\chi)$ is the maximum width of its edges. The branchwidth of the graph $G$ is the minimum width over all branch-decompositions of $G$. Unbounded [+]Details carvingwidth [?]Consider a decomposition $(T,\chi)$ of a graph $G$ where $T$ is a binary tree with $|V(G)|$ leaves and $\chi$ is a bijection mapping the leaves of $T$ to the vertices of $G$. Every edge $e \in E(T)$ of the tree $T$ partitions the vertices of the graph $G$ into two parts $V_e$ and $V \backslash V_e$ according to the leaves of the two connected components in $T - e$. The width of an edge $e$ of the tree is the number of edges of a graph $G$ that have exactly one endpoint in $V_e$ and another endpoint in $V \backslash V_e$. The width of the decomposition $(T,\chi)$ is the largest width over all edges of the tree $T$. The carvingwidth of a graph is the minimum width over all decompositions as above. Unbounded [+]Details chromatic number [?]The chromatic number of a graph is the minimum number of colours needed to label all its vertices in such a way that no two vertices with the same color are adjacent. Unbounded [+]Details cliquewidth [?]The cliquewidth of a graph is the number of different labels that is needed to construct the graph using the following operations: creation of a vertex with label $i$, disjoint union, renaming labels $i$ to label $j$, and connecting all vertices with label $i$ to all vertices with label $j$. Unbounded [+]Details cochromatic number [?]The cochromatic number of a graph $G$ is the minimum number of colours needed to label all its vertices in such a way that that every set of vertices with the same colour is either independent in G, or independent in $\overline{G}$. Unknown to ISGCI [+]Details cutwidth [?]The cutwidth of a graph $G$ is the smallest integer $k$ such that the vertices of $G$ can be arranged in a linear layout $v_1, \ldots, v_n$ in such a way that for every $i = 1, \ldots,n - 1$, there are at most $k$ edges with one endpoint in $\{v_1, \ldots, v_i\}$ and the other in ${v_{i+1}, \ldots, v_n\}$. Unbounded [+]Details degeneracy [?]Let $G$ be a graph and consider the following algorithm: Find a vertex $v$ with smallest degree. Delete vertex $v$ and its incident edges. Repeat as long as the graph is not empty. The degeneracy of a graph $G$ is the maximum degree of a vertex when it is deleted in the above algorithm. Unbounded [+]Details diameter [?]The diameter of a graph $G$ is the length of the longest shortest path between any two vertices in $G$. Unbounded [+]Details distance to block [?]The distance to block of a graph $G$ is the size of a smallest vertex subset whose deletion makes $G$ a block graph. Unbounded [+]Details distance to clique [?]Let $G$ be a graph. Its distance to clique is the minimum number of vertices that have to be deleted from $G$ in order to obtain a clique. Unbounded [+]Details distance to cluster [?]A cluster is a disjoint union of cliques. The distance to cluster of a graph $G$ is the size of a smallest vertex subset whose deletion makes $G$ a cluster graph. Unbounded [+]Details distance to co-cluster [?]The distance to co-cluster of a graph is the minimum number of vertices that have to be deleted to obtain a co-cluster graph. Unbounded [+]Details distance to cograph [?]The distance to cograph of a graph $G$ is the minimum number of vertices that have to be deleted from $G$ in order to obtain a cograph . Unbounded [+]Details distance to linear forest [?]The distance to linear forest of a graph $G = (V, E)$ is the size of a smallest subset $S$ of vertices, such that $G[V \backslash S]$ is a disjoint union of paths and singleton vertices. Unbounded [+]Details distance to outerplanar [?]The distance to outerplanar of a graph $G = (V,E)$ is the minumum size of a vertex subset $X \subseteq V$, such that $G[V \backslash X]$ is a outerplanar graph. Unbounded [+]Details genus [?]The genus $g$ of a graph $G$ is the minimum number of handles over all surfaces on which $G$ can be embedded without edge crossings. Unbounded [+]Details max-leaf number [?]The max-leaf number of a graph $G$ is the maximum number of leaves in a spanning tree of $G$. Unbounded [+]Details maximum clique [?]The parameter maximum clique of a graph $G$ is the largest number of vertices in a complete subgraph of $G$. Unbounded [+]Details maximum degree [?]The maximum degree of a graph $G$ is the largest number of neighbors of a vertex in $G$. Unbounded [+]Details maximum independent set [?]An independent set of a graph $G$ is a subset of pairwise non-adjacent vertices. The parameter maximum independent set of graph $G$ is the size of a largest independent set in $G$. Unbounded [+]Details maximum induced matching [?]For a graph $G = (V,E)$ an induced matching is an edge subset $M \subseteq E$ that satisfies the following two conditions: $M$ is a matching of the graph $G$ and there is no edge in $E \backslash M$ connecting any two vertices belonging to edges of the matching $M$. The parameter maximum induced matching of a graph $G$ is the largest size of an induced matching in $G$. Unbounded [+]Details maximum matching [?]A matching in a graph is a subset of pairwise disjoint edges (any two edges that do not share an endpoint). The parameter maximum matching of a graph $G$ is the largest size of a matching in $G$. Unbounded [+]Details minimum clique cover [?]A clique cover of a graph $G = (V, E)$ is a partition $P$ of $V$ such that each part in $P$ induces a clique in $G$. The minimum clique cover of $G$ is the minimum number of parts in a clique cover of $G$. Note that the clique cover number of a graph is exactly the chromatic number of its complement. Unbounded [+]Details minimum dominating set [?]A dominating set of a graph $G$ is a subset $D$ of its vertices, such that every vertex not in $D$ is adjacent to at least one member of $D$. The parameter minimum dominating set for graph $G$ is the minimum number of vertices in a dominating set for $G$. Unbounded [+]Details pathwidth [?]A path decomposition of a graph $G$ is a pair $(P,X)$ where $P$ is a path with vertex set $\{1, \ldots, q\}$, and $X = \{X_1,X_2, \ldots ,X_q\}$ is a family of vertex subsets of $V(G)$ such that: $\bigcup_{p \in \{1,\ldots ,q\}} X_p = V(G)$ $\forall\{u,v\} \in E(G) \exists p \colon u, v \in X_p$ $\forall v \in V(G)$ the set of vertices $\{p \mid v \in X_p\}$ is a connected subpath of $P$. The width of a path decomposition $(P,X)$ is max$\{|X_p| - 1 \mid p \in \{1,\ldots ,q\}\}$. The pathwidth of a graph $G$ is the minimum width among all possible path decompositions of $G$. Unbounded [+]Details rankwidth [?]Let $M$ be the $|V| \times |V|$ adjacency matrix of a graph $G$. The cut rank of a set $A \subseteq V(G)$ is the rank of the submatrix of $M$ induced by the rows of $A$ and the columns of $V(G) \backslash A$. A rank decomposition of a graph $G$ is a pair $(T,L)$ where $T$ is a binary tree and $L$ is a bijection from $V(G)$ to the leaves of the tree $T$. Any edge $e$ in the tree $T$ splits $V(G)$ into two parts $A_e, B_e$ corresponding to the leaves of the two connected components of $T - e$. The width of an edge $e \in E(T)$ is the cutrank of $A_e$. The width of the rank-decomposition $(T,L)$ is the maximum width of an edge in $T$. The rankwidth of the graph $G$ is the minimum width of a rank-decomposition of $G$. Unbounded [+]Details tree depth [?]A tree depth decomposition of a graph $G = (V,E)$ is a rooted tree $T$ with the same vertices $V$, such that, for every edge $\{u,v\} \in E$, either $u$ is an ancestor of $v$ or $v$ is an ancestor of $u$ in the tree $T$. The depth of $T$ is the maximum number of vertices on a path from the root to any leaf. The tree depth of a graph $G$ is the minimum depth among all tree depth decompositions. Unbounded [+]Details treewidth [?]A tree decomposition of a graph $G$ is a pair $(T, X)$, where $T = (I, F)$ is a tree, and $X = \{X_i \mid i \in I\}$ is a family of subsets of $V(G)$ such that the union of all $X_i$, $i \in I$ equals $V$, for all edges $\{v,w\} \in E$, there exists $i \in I$, such that $v, w \in X_i$, and for all $v \in V$ the set of nodes $\{i \in I \mid v \in X_i\}$ forms a subtree of $T$. The width of the tree decomposition is $\max |X_i| - 1$. The treewidth of a graph is the minimum width over all possible tree decompositions of the graph. Unbounded [+]Details vertex cover [?]Let $G$ be a graph. Its vertex cover number is the minimum number of vertices that have to be deleted in order to obtain an independent set. Unbounded [+]Details
## Problems
Problems in italics have no summary page and are only listed when ISGCI contains a result for the current class.
#### Parameter decomposition
book thickness decomposition Unknown to ISGCI [+]Details
booleanwidth decomposition Unknown to ISGCI [+]Details
cliquewidth decomposition Unknown to ISGCI [+]Details
cutwidth decomposition Unknown to ISGCI [+]Details
treewidth decomposition Polynomial [+]Details
#### Unweighted problems
3-Colourability Linear [+]Details
Clique Polynomial [+]Details
Clique cover Polynomial [+]Details
Colourability Linear [+]Details
Domination Linear [+]Details
Feedback vertex set Polynomial [+]Details
Graph isomorphism GI-complete [+]Details
Hamiltonian cycle NP-complete [+]Details
Hamiltonian path NP-complete [+]Details
Independent dominating set Linear [+]Details
Independent set Linear [+]Details
Maximum cut NP-complete [+]Details
Monopolarity Linear [+]Details
Polarity Polynomial [+]Details
Recognition Polynomial [+]Details
#### Weighted problems
Weighted clique Polynomial [+]Details
Weighted feedback vertex set Unknown to ISGCI [+]Details
Weighted independent dominating set Polynomial [+]Details
Weighted independent set Linear [+]Details
Weighted maximum cut NP-complete [+]Details | 3,479 | 13,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.861657 |
https://theurbanbaker.com/snacks/an-increase-in-the-demand-for-steak-will-lead-to-an-increase-in-which-of-the-following.html | 1,679,656,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00359.warc.gz | 669,297,887 | 23,622 | # An Increase In The Demand For Steak Will Lead To An Increase In Which Of The Following?
The terms in this collection (25) An increase in the demand for steak will result in an increase in the production of which of the following foodstuffs? The equilibrium quantity will grow as a result of this.
## What happens to the price of automobiles as the demand increases?
The price of autos is decreasing, but the need for tires is increasing. Tennis racquets are becoming more expensive, but the need for tennis shoes is diminishing. In response to the increase in the price of coffee, the demand for cream has increased. Which of the following is in accordance with the law of supplies?
## What happens to the price of bacon when the demand increases?
Both the equilibrium price and quantity rise as a result of this. Because of this, the price of bacon is rising, while the demand for eggs is decreasing. The price of autos is falling, and the demand for public transportation is also declining. The price of coffee continues to rise, while the demand for tea continues to rise.
## What happens to the price curve when the quantity demanded increases?
As the price rises, the amount requested decreases.For example, at \$2, the quantity demanded would be 3 billion, but at \$2.50, the quantity demanded would be about 2 billion dollars.In this case, the curve would be slanted downward, with a high point on the left and a low point on the right.
If there was a rise in demand, the entire graph would shift to the right, but the prices would remain the same as they were before.
## What happens to the demand for apples if the price increases?
If there was a rise in demand, the entire graph would shift to the right, but the prices would remain the same as they were before.Create a properly labeled graph depicting the demand for apples in your area.Imagine that a new Surgeon General’s study concludes that eating an apple a day truly does keep the doctor away.
Show this in your graph to demonstrate what will happen to the demand for apples.
## Which of the following would increase the demand for a normal good a decrease in?
Whenever the price of one of an item’s complements rises, the demand for that good diminishes. With an increase in wealth, the demand for a common good rises as well. Increased money leads to a drop in the demand for a substandard commodity.
We recommend reading: How To Cook A Ham Steak?
## What happens when demand increases quizlet?
The price rises as a result of an increase in demand. The price is lowered when the supply of goods increases. Reduces the equilibrium quantity to a smaller value. The price rises as a result of a decrease in supply.
## What causes an increase in demand quizlet?
When the quantity demanded increases for every price point, this is referred to as an increase in demand. An rise in demand results in a shift to the right of the demand curve, as seen in the graph.
## Which of the following could cause a shift from S1 to S2 shown in the graph shown?
As seen in the graph above, which of the following might be the cause of the shift from S1 to S2: Price increases result in an increase in the amount of goods and services offered.
## What increases demand?
Things such as changes in consumer tastes, population, income, the prices of substitute or complement goods, and expectations about future conditions and prices all have the potential to shift the demand curve for goods and services and cause a different quantity to be demanded at a given price to be demanded at a given price.
## What happens when demand increases?
The demand for goods and services follows the same inverse relationship as the supply of goods and services. On the other hand, if demand grows but supply stays constant, the increased demand leads to an increase in the equilibrium price and vice versa. Supply and demand fluctuate up and down until a price that is in balance is attained.
We recommend reading: How To Cook A Precooked Spiral Ham?
## What happens when demand increases and supply decreases?
The Effects of Supply and Demand If demand rises but supply stays constant, a shortage emerges, resulting in a higher equilibrium price than before. If demand declines but supply stays unchanged, a surplus is created, resulting in a lower equilibrium price at the market.
## When demand increases and supply decreases price quizlet?
Terms in this set (25) If demand for a given commodity grows while supply drops, the equilibrium price of that good will rise, but the amount of the good produced and sold may increase, decrease, or remain constant.
## What happens when supply increases and demand is constant?
With an increase in supply and no change in demand, there will be a surplus of goods, and the price will decline. If the supply of a good drops while the demand stays constant, a shortage will result, and the price will rise as a result.
## Which of the following will cause an increase in the demand for autos?
An rise in the number of purchasers leads to an increase in the amount of goods available for purchase. Which of the following, if all else is equal, would generally result in an increase in the demand curve for new automobiles? An rise in the amount of money available to customers. Explanation: A drop in the price of new autos would result in a shift in the demand curve’s location.
## Which would cause an increase in the demand for product A?
An increase in income will almost always result in an increase in demand for a certain product. When two items are substitutes for one another, the price of one will tend to rise while the demand for the other will likely to fall in tandem. Increases in the price of one product will tend to raise demand for the other if the prices of the two goods are mutually exclusive.
We recommend reading: How To Use Overcooked Steak?
## Which of the following would likely result in an increase in the demand for beef?
Which of the following is most likely to result in a rise in the demand for beef in the short term? An rise in the average household income. You’ve just finished studying 20 terms!
## What would cause a shift from D1 to D2?
Demand Shifts in the Automobile Industry The term ″increased demand″ refers to the fact that the amount required is larger at every given price, resulting in the demand curve shifting to the right from D0 to D1. Lower quantity required at a given price indicates decreased demand, and the demand curve shifts from D0 to D2, indicating a shift to the left in the demand curve.
## What is a possible cause of the shift in demand from D1 to D2 in the graph below?
The demand for goods and services has shifted from Day 1 to Day 2 in the graph below. What may be the underlying cause of this shift? a rise in consumer demand When the prices of two things that serve the same demand rise, the substitution effect dictates that individuals would opt for the lower-priced item.
## What causes shifts in demand and supply curves?
Shifts. When a shift in the demand or supply curve happens, the amount of a good desired or supplied varies even while the price of that product stays constant. A shift in demand for beer would occur, for example, if the price of a bottle of beer was \$2 and the quantity of beer required rose from quarter one to quarter two. | 1,507 | 7,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.966583 |
http://www.builditsolar.com/Projects/SpaceHeating/DHWplusSpace/SystemSizing.htm | 1,512,984,564,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00316.warc.gz | 343,471,279 | 4,825 | # Solar DHW + Space: Sizing the System
This section covers how to go about determining how much solar collector area your system should have, and what the size of the heat storage tank should be. Back to the main directory for Solar Space + DHW ...
This system is designed so that it can provide both space heating and domestic water heating using the same collectors and storage tank. The sizing for space heating and for domestic water heating are different, and will be covered separately, followed by some notes how how you might combine the two sizes.
## Sizing the System For Space Heating
Sizing the collectors and storage tank for space heating depends on several things:
• How well insulated and sealed your house is
• How cold your climate is
• How sunny your climate is
• How much solar heating you want to accomplish (what Solar Fraction you are shooting for)
I've just recently added a method that lets you work through all this, and make a reasonable estimate of how big your solar system should be. But first, just to get a feel for the problem, lets consider to extremes cases at opposite ends of the scale.
The first would be the case of retrofitting solar heating to an existing home in a cold climate. Lets assume the home was built in the 80's and has typical insulation and sealing for that time -- maybe R11 walls, R30 ceilings, ordinary double glazed windows.
The second would be a new house built to something like Passive House Institute standards. These homes are very well insulated and very well sealed. The
In the first case, you are very likely to find when you work through the sizing, that 1) you want to first work on your homes insulating and sealing, and window heat loss (like thermal shades). The reason being that it will be cheaper and easier to first lower your heat loss and then add solar heating than to try to meet a very high heat demand with solar heating.
Even after you have cut back the home heat loss (something that will pay big dividends in itself), its very likely that you will find that its not practical to provide a large fraction (say 90 to 100%) with solar heating. Its likely that this would take so much collector area that either space limits or cost of system limits would keep you from getting to very high solar fractions. This does not mean in any way that it does not make sense to add a solar heating system. It just means that you should add a solar heating system that fits in your space and in your budget, and take whatever solar fraction this gives you. The system will still save you lots on heating bills and will reduce your greenhouse gas emissions substantially. Such a system will actually have a better pay back than the solar system for a very well insulated home like our second case. I don't want to over generalize, and you want to work through the numbers on your house, but its probably reasonable to target a solar fraction in the 30 to 70% area for this case depending on your exact situation.
In the second case, a home that meets the Passive House Institute Standard in a northern US climate is likely to have something like R40 walls, R60 ceilings, R5 triple glazed windows, very tight sealing, and tight control over thermal bridging in the structure. If circumstances allow, the home would also likely have some passive solar gain from south facing glazing. Such a home would also have a Heat Recovery Ventilation system that provides good ventilation without much heat loss. This home might have a tenth of the heat loss of our first example.
For this case, it is certainly practical to target quite a high solar fraction -- perhaps even a 100% solar fraction. The heat demand is so much lower that a much smaller collector area will meet the heat demand. One could even experiment with backing off some on some of the Passive House Institute requirements, and trade that against the cost of a larger solar heating system to see which is more cost effective.
As an example, ... -- do a PHS house example...
## Making the Estimate
At this point, you want to go to the page that provides the method to dete
Working through the details of the method will give you:
- An estimate of the heat loss for your house, and an understanding of where the heat is going. This will be very helpful in figuring out what changes you might want to make to make the house less of a heat leaker.
- An estimate of the solar heat you can collect for the collector array size you enter.
You can play around with making improvements to your home heat loss and changing the size of your collector array until you get to a point that looks like a good and cost effective solution.
Some truths:
• - You will almost certainly need some form of backup heat, and for many people the "backup" will be meeting a considerable fraction of the heat demand. So, plan for a robust backup heat system. | 1,018 | 4,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-51 | latest | en | 0.951557 |
https://masomomsingi.com/bit-2201-simulation-and-modeling-kca-past-paper-2/ | 1,713,439,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00562.warc.gz | 346,182,462 | 15,124 | BIT 2201 SIMULATION AND MODELING KCA Past Paper
UNIVERSITY EXAMINATIONS: 2013/2014
ORDINARY EXAMINATION FOR THE BACHELOR OF SCIENCE
IN INFORMATION TECHNOLOGY
BIT 2201 SIMULATION AND MODELING
DATE: AUGUST, 2014 TIME: 2 HOURS
INSTRUCTIONS: Answer Question ONE and any other TWO
QUESTION ONE: 30 MARKS (COMPULSORY)
a) Discuss five applications of simulation and modeling in real life. (5 Marks)
b) Using the mid- square method obtain the random variables using Z0= 1009 until the
cycle degenerates to zero. (10 Marks)
c) Discuss any five desirable features of a good simulation software (5 Marks)
d) Using the banking systems as an example, describe the components of a discreteevent system. ( 10 Marks)
QUESTION TWO: 20 MARKS
a) Every system exists with an environment. Activities in the systems environment
cause changes in the state of the system. These activities are classified into categories:
Endogenous and Exogenous. Differentiate between the two. (2 Marks)
b) Discuss five reasons why it is necessary to use animation during the process of
modeling and simulation. (5 Marks )
c) Discuss three reasons why it is not possible to perform experiment with a real system.
(3 Marks)
c) Describe five common statistics included in the output report of a simulation
programming system. (10 Marks)
QUESTION THREE: 20 MARKS
a) There are two fundamental types of animation; concurrent and post-processed
Differentiate between the two types of animation. (4 Marks)
b) Describe the following terms giving examples from the traffic control systems.
(5 Marks)
i) State of a System
ii) State Variables
iii) Entity
iv) Attribute
v) Activity
c) Discuss four drawbacks simulation and modeling. (6 Marks)
d) Using the Linear Congruential Generator (LCG) with a=5, m=16, c=3 and seed Z0 = 7
to generate the first FIVE random variates on [0,1]. (5 Marks)
QUESTION FOUR: 20 MARKS
a) Define the terms verification and validation in the context of modeling. (4 Marks)
b) Not all simulation and modeling exercises are a success. Discuss five common pitfalls
to successful simulation. (5 Marks)
c) Consider a single server queuing system in the bank. The system starts at time t=0.
The arrival time of customers is: 0.4, 1.6, 2.1, 3.8, 4.0, 5.6, 5.8, and 7.2. The
departure times are: 2.4, 3.1, 3.3, 4.9, and 8.6. Time is in minutes. The first in first out
queuing discipline is followed. Simulate this system for six clients and estimate:
i) The expected average delay in the queue (3 Marks)
ii) The expected average number of clients in the queue at any time t (6 Marks)
iii) The expected utilization of server. (2 Marks)
QUESTION FIVE: 20 MARKS
a) Differentiate between the following types of models (6 Marks)
i) Deterministic Vs Stochastic Models
ii) Discrete Models Vs Continuous Models
b) Discuss any five advantages of simulation (5 Marks)
c) The table below shows the probability distribution of container arrival per day at
Mombasa port.
The offloading rate per day follows the probability distribution given below. NB:
Suppose the following are the random numbers of arrivals and random numbers | 784 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.811525 |
https://rdrr.io/github/graemeblair/DDestimate/f/vignettes/simulations-ols-variance.Rmd | 1,571,169,247,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00389.warc.gz | 656,222,759 | 10,364 | knitr::opts_chunkset(echo = TRUE) fname <- "simulations-ols-var.rda" sims <- 2000 rerun <- TRUE if (!rerun) {load(fname)} This document exposes the properties of different variance estimators using DeclareDesign and estimatr. More details about the variance estimators with references can be found in the mathematical notes. library(DeclareDesign) library(tidyverse) library(knitr) Homoskedastic errors Under simple conditions with homoskedasticity (i.e., all errors are drawn from a distribution with the same variance), the classical estimator of the variance of OLS should be unbiased. In this section I demonstrate this to be true using DeclareDesign and estimatr. First, let's take a simple set up: \begin{aligned} \mathbf{y} &= \mathbf{X}\beta + \epsilon, \ \epsilon_i &\overset{i.i.d.}{\sim} N(0, \sigma^2). \end{aligned} For our simulation, let's have a constant and one covariate, so that\mathbf{X} = [\mathbf{1}, \mathbf{x_1}]$, where$\mathbf{x_1}$is a column vector of a covariate drawn from a standard normal distribution. Let's also assume that are covariates are fixed, rather than stochastic. Let's draw the data we will use. set.seed(41) dat <- data.frame(x = rnorm(50)) The function [ \epsilon_i \overset{i.i.d.}{\sim} N(0, \sigma^2), ] encodes the assumption of homoskedasticity. Because of these homoskedastic errors, we know that the true variance of the coefficients with fixed covariates is [ \mathbb{V}[\widehat{\beta}] = \sigma^2 (\mathbf{X}^\top \mathbf{X})^{-1}, ] where I hide conditioning on$\mathbf{X}$for simplicity. Let's compute the true variance for our dataset. sigmasq <- 4 # Build the X matrix with intercept Xmat <- cbind(1, dat$x)
# Invert XtX
XtX_inv <- solve(crossprod(Xmat))
# Get full variance covariance matrix
true_var_cov_mat <- sigmasq * XtX_inv
But for this example, we are only going to focus on the variance for the covariate, not the intercept, so let's store that variance.
true_varb <- true_var_cov_mat[2, 2]
true_varb
Now, using DeclareDesign, let's specify the rest of the data generating process (DGP). Let's set $\beta = [0, 1]^\top$, so that the true DGP is $\mathbf{y} = \mathbf{x_1} + \epsilon$.
simp_pop <- declare_population(
epsilon = rnorm(N, sd = 2),
y = x + epsilon
)
Now let's tell DeclareDesign that our target, our estimand, is the true variance.
varb_estimand <- declare_estimand(true_varb = true_varb)
Our estimator for this estimand will be the classical OLS variance estimator, which we know should be unbiased:
[ \widehat{\mathbb{V}[\widehat{\beta}]} = \frac{\mathbf{e}^\top\mathbf{e}}{N - K} (\mathbf{X}^\top \mathbf{X})^{-1}, ]
where the residuals $\mathbf{e} = \mathbf{y} - \mathbf{X}\widehat{\beta}$, $N$ is the number of observations, and $K$ is the number of regressors---two in our case. We can easily get this estimate of the variance by squaring the standard error we get out from lm_robust in estimatr. Let's tell DeclareDesign to use that estimator and get the coefficient on the $\mathbf{x}_1$ variable.
lmc <- declare_estimator(
y ~ x,
model = lm_robust,
se_type = "classical",
estimand = varb_estimand,
term = "x"
)
Now, we want to test for a few results using Monte Carlo simulation. Our main goal is to show that our estimated variance is unbiased for the true variance (our estimand). We can do this by comparing the mean of our estimated variances across our Monto Carlo simulations to the true variance. We can also show that the standard error of our coefficient estimate is the same as the standard deviation of the sampling distribution of our coefficient. Lastly, we also measure the coverage of our 95 percent confidence intervals across simulations to test whether the they cover the true coefficient 95 percent of the time.
Let's first set up the design and our diagnosands.
# First declare all the steps of our design, starting with our fixed data
classical_design <- declare_population(dat) + simp_pop + varb_estimand + lmc
# Declare a set of diagnosands that help us check if
# we have unbiasedness
my_diagnosands <- declare_diagnosands(
Bias of Estimated Variance = mean(std.error^2 - estimand),
Bias of Standard Error = mean(std.error - sd(estimate)),
Coverage Rate = mean(1 <= conf.low & 1 >= conf.high),
Mean of Estimated Variance = mean(std.error^2),
True Variance = estimand[1],
keep_defaults = FALSE
)
Now let's run the simulations!
set.seed(42)
dx1 <- diagnose_design(
classical_design,
sims = sims,
diagnosands = my_diagnosands
)
kable(reshape_diagnosis(dx1, digits = 3))
Our diagnosands can help us see if there is any bias. As we can see the bias is very close to zero. Because the standard error of the bias is much larger than the estimated bias, we can be reasonably certain that the only reason the bias is not exactly zero is due to simulation error. We can also see the unbiasedness visually, using a density plot of estimated variances with a line for the true variance.
# Get cuts to add label for true variance
dx1$simulations <- dx1$simulations %>% mutate(var = std.error^2)
sumvar <- summary(dx1$simulations$var)
ggplot(dx1$simulations, aes(x = var)) + geom_density() + geom_vline(xintercept = true_varb, size = 1, color = "red") + xlab("Estimated Variance") + theme_bw() Heteroskedastic errors Let's use the same fixed data set-up, but introduce heteroskedasticity. In this case, the variance of the errors is different across units: [ \epsilon_i \sim N(0, \sigma_i^2), ] where$\sigma^2_i \neq \sigma^2_j$for some units$i$and$j$. If the variance of the errors is not independent of the regressors, the "classical" variance will be biased and inconsistent. Meanwhile, heteroskedastic-consistent variance estimators, such as the HC2 estimator, are consistent and normally less biased than the "classical" estimator. Let's demonstrate this using DeclareDesign. First, let's specify the variance of the errors to be strongly correlated with$x$. dat <- mutate(dat, noise_var = 1 + (x - min(x))^2 ) ggplot(dat, aes(x, noise_var)) + geom_point() + ggtitle("The variance of errors increases with x") Note that the general form of the true variance with fixed covariates is: [ \mathbb{V}[\widehat{\beta}] = (\mathbf{X}^\top \mathbf{X})^{-1} \mathbf{X}^\top \mathbf{\Phi} \mathbf{X} (\mathbf{X}^\top \mathbf{X})^{-1}, ] where$\mathbf{\Phi}$is the variance covariance matrix of the errors, or$\mathbf{\Phi} = \mathbb{E}[\epsilon\epsilon^\top]$. In the above case with homoskedasticity, we assumed$\mathbf{\Phi} = \sigma^2 \mathbf{I}$and were able to simplify. Now, as in the standard set up for heteroskedasticity, we set$\mathbf{\Phi}$to be a diagonal matrix where noise_var, the variance for each unit's error, is on the diagonal, like so: [ \mathbf{\Phi} = \begin{bmatrix} \sigma_1^2 & 0 & \cdots & 0 \ 0 & \sigma_2^2 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & \sigma_n^2 \end{bmatrix} ] Using that error structure and the error for each unit, we can estimate the true variance. Xmat <- with(dat, cbind(1, x)) XtX_inv <- solve(crossprod(Xmat)) varb <- tcrossprod(XtX_inv, Xmat) %*% diag(with(dat, noise_var)) %*% Xmat %*% XtX_inv true_varb_het <- varb[2, 2] true_varb_het Now let's use DeclareDesign to test whether HC2 is less biased in this example than classical standard errors. HC2 is the estimatr default; we'll also throw in HC1, which is Stata's default robust standard error estimator. I'm going to make a "designer," which is a function that returns a design. het_designer <- function(N) { dat <- fabricate(N = N, x = rnorm(N), noise_var = 1 + (x - min(x))^2) # Get true variance for this data Xmat <- with(dat, cbind(1, x)) XtX_inv <- solve(crossprod(Xmat)) varb <- tcrossprod(XtX_inv, Xmat) %*% diag(with(dat, noise_var)) %*% Xmat %*% XtX_inv true_varb_het <- varb[2, 2] # Population function now has heteroskedastic noise simp_pop <- declare_population( dat, epsilon = rnorm(N, sd = sqrt(noise_var)), y = x + epsilon ) varb_het_estimand <- declare_estimand(true_varb_het = true_varb_het) # Now we declare all three estimators lm1 <- declare_estimator( y ~ x, model = lm_robust, se_type = "classical", estimand = varb_het_estimand, term = "x", label = "classical" ) lm2 <- declare_estimator( y ~ x, model = lm_robust, se_type = "HC1", estimand = varb_het_estimand, term = "x", label = "HC1" ) lm3 <- declare_estimator( y ~ x, model = lm_robust, se_type = "HC2", estimand = varb_het_estimand, term = "x", label = "HC2" ) # We return the design so we can diagnose it return(simp_pop + varb_het_estimand + lm1 + lm2 + lm3) } So let's use the same diagnosands as above to test the properties of our estimators with heteroskedasticity. # Create a design using our template and the data we have been using het_design <- het_designer(N = 50) dx2 <- diagnose_design( het_design, diagnosands = my_diagnosands, sims = sims ) kable(reshape_diagnosis(dx2)) The bias for the HC2 errors is much closer to zero, whereas the bias for the classical error is much larger, especially when compared to the standard error of the bias diagnosand. How does this bias change as the sample size changes? As the HC2 variance estimate is consistent under heteroskedasticity, it should converge to zero. designs <- expand_design(het_designer, N = c(100, 200, 300, 500, 1000, 2500)) set.seed(42) dx3 <- diagnose_design(designs, sims = sims, diagnosands = my_diagnosands) ggplot(dx3$diagnosands_df, aes(x = N, y = Bias of Estimated Variance, color = estimator_label)) +
geom_point() +
geom_line() +
geom_hline(yintercept = 0, linetype = 2, color = "grey") +
labs(color = "Estimator") +
theme_bw()
As you can see, the HC2 variance tends to have less bias and is consistent, converging to the true value as the sample size increases. The classical standard error estimator is neither unbiased nor consistent. The HC1 variance is also "robust" to heteroskedasiticity but exhibits greater bias than HC2 in this example.
save(dx1, dx2, dx3, file = fname)
# Stochastic data
# Clusters
graemeblair/DDestimate documentation built on Sept. 10, 2019, 7:38 p.m. | 2,829 | 10,014 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-43 | latest | en | 0.647485 |
https://www.jobduniya.com/Private-Companies/3i-Infotech/Fully-Solved-Papers/Topic-Languages-2/Subtopic-C-and-C-Plus-Plus-0/Part-76.html | 1,643,313,972,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00579.warc.gz | 864,661,292 | 7,481 | # Languages-C & C Plus Plus [3i Infotech Placement]: Sample Questions 171 - 173 of 354
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## Question 171
C & C Plus Plus
Edit
### Describe in Detail
Essay▾
What is a scope resolution operator?
### Explanation
• The scope resolution operator “::” is used to qualify hidden names.
• We can use the unary scope operator if a namespace scope or global scope name is hidden by an explicit declaration of the same name in a block or class.
Example:
1. `int count=0;`
2. `int main(void)`
3. `{`
4. ` int count=0;`
5. ` :: count =1;`
6. ` count =2;`
7. ` return 0;`
8. `}`
## Question 172
C & C Plus Plus
Edit
### Describe in Detail
Essay▾
What is the output of the following program?
1. `main ()`
2. `{`
3. ` int i=10,j=20; j =i, j?(i,j)?i:j:j; printf("%d %d",i,j);`
4. `}`
### Explanation
In the program
int i = 10, j = 20; Given the integer variable i = 10 and j = 20 j = i, j? (i, j) ? i: j: j; The ternary operator (? :) is equivalent for if-then-else statement.So, the question can be written as:if (i, j){if (i, j) j = i;else j = j;} else j = j; printf ( “% d % d” , i, j) ; Printf prints the value of i and j
## Question 173
C & C Plus Plus
Edit
### Describe in Detail
Essay▾
What is the output of the program given below
1. `main ()`
2. `{`
3. ` signed char i =0;`
4. ` for (; i>= 0; i ++ );`
5. ` printf ( “%d” i);`
6. `}`
### Explanation
• In the program define the signed char
• Signed char represents integers from -128 to 127.
signed char i = 0; Define the signed character variable i = 0 for (; i >= 0; i ++) ; The semicolon at the end of the for loop.The initial value of the i is set 0.The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128.The condition in the for loop fails and so comes out of the for loop. printf ( “% d” i) ; It prints the current value of I that is -128.
Developed by: | 682 | 2,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.649221 |
https://www.comicsanscancer.com/how-much-electricity-does-a-4-bedroom-house-use/ | 1,723,470,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00657.warc.gz | 548,131,189 | 14,491 | # How much electricity does a 4 bedroom house use?
## How much electricity does a 4 bedroom house use?
For our calculations we have used the following annual energy usage amounts which are based on industry figures: 1 or 2 bedroom house/flat – gas usage of 8,000kWh and an electricity usage of 1,800kWh. 3 or 4 bedroom house – gas usage of 12,000kWh and an electricity usage of 2,900kWh.
## Can I leave my laundry in the washer overnight?
Many people learn the hard way that if you leave wet laundry in the washing machine for too long, it starts to develop a smell due to the growth of bacteria and mold. But it all depends on how long you let a load of wet laundry sit. According to Martha, leaving your laundry in the washer overnight is actually okay.
## What is the best day of the week to do laundry?
Tuesday, Wednesday, and Thursday are the best days to do laundry at a laundromat– because those are the least busy days, while people are working.
## What is the average electricity usage for a 3 bed house?
A 3 bedroom house is considered to be a medium energy usage household which means that based on Ofgems current figures for average energy usage means that a typical medium energy user consumes 12,000 kWh of Gas and 3,100 kWh of electricity.
## What is the average electricity usage per household in South Africa?
What number to aim for? The national average daily consumption for a typical household according to Eskom is over 30 kWh.
## Is it cheaper to do laundry at home or at a laundromat?
Many things have been taken into account and they’ve determined that the average load of laundry done at home will run you \$. 97 (excluding equipment costs) . Likewise, a load of laundry done at the laundromat will set you back \$3.12.
## Is it better to do smaller loads of laundry?
Check the manual, but as a rule of thumb, a small load will fill up the drum about a quarter of the way, medium will fill half, and large will fill three-quarters. Don’t pack your washer completely full; make sure the clothes have room to tumble freely. When in doubt, use cold water.
## What days should you not wash clothes?
Laundry Superstitions
• Don’t do laundry on New Year’s Day, or a member of the family will be washed away (i.e., die) during the coming year.
• Doing laundry on New Year’s Day will wash a year of good fortune down the drain.
• Don’t do laundry on New Year’s Day, or you will have more laundry than usual to do all year.
## Should you air out washing machine?
Doing so can cause mildew and mould to build up in the drum, which can leave clothes smelling musty and is tough to remove. Mildew thrives in damp and warm conditions. By closing the door of your washing machine straight after you have used it, you could be giving the fungus a perfect place to live.
## Is it cheaper to do laundry at night or during the day?
So, on hot days, do your laundry early in the morning, when energy demand is lower. Winter: Do laundry late at night. While everyone else is sleeping and has their heaters off or in energy-saving mode, you can take advantage of lower electricity rates.
## How often should you clean your washing machine?
It’s recommended to have your washer go through a cleaning cycle once a week or bi-weekly, depending on how often you are using the machine. If you aren’t using it as often, it’s recommended to give it wash (both inside and out) once a month.
## Why is your electric bill so high?
The reason why your electricity bills are so high is that the more electricity you use, the more you pay per unit of electricity. So, if your electricity bill is twice as high as usual, it’s not simply because you used twice as much electricity.
## Does washing machine use a lot of electricity?
What’s surprising, however, is that a washing machine requires way less electricity than a dryer. An average cycle for a washing machine is 30 minutes. This appliance, which is a widely used Energy Star model, needs 500 watts per hour to run, which means it requires 250 Wh, or 2.25 kWh, to run for 30 minutes.
## How much does it cost to run a dryer for 1 hour?
Appliance Electricity Usage
Appliance Typical Consumption Per Hour Cost Per Hour (at 10 cents per kilowatt-hour)
Clothes dryer/water heater 4,000 watts 40 cents
Water pump 3,000 watts 30 cents
Space heater 1,500 watts 15 cents
Hair dryer 1,200 watts 12 cents
## Should you leave the washer door open when not in use?
It’s more common for mold and mildew to develop in front-loading washers, but Better Homes and Gardens says it can happen in top-loaders, too — especially in warm, humid climates. For this reason, Consumer Reports recommends leaving the door or lid open in between loads to give the machine time to dry.
## How long is too long to wash wet clothes?
eight to twelve hours
## How many kWh Should I use a day?
According to the EIA, in 2017, the average annual electricity consumption for a U.S. residential home customer was 10,399 kilowatt hours (kWh), an average of 867 kWh per month. That means the average household electricity consumption kWh per day is 28.9 kWh (867 kWh / 30 days).
## Is it OK to leave clothes outside overnight?
In more humid environments, morning dew can leave your clothes damp, moldy and with a funny smell. Leaving your clothes outside to dry overnight on a DIY clothesline to air dry your laundry is a risk. But it’s not impossible. The best way to tell if you feel comfortable doing this is merely to try it.
## Why does my front load washer smell like sewer?
The most likely is bacteria growing in your washer because of built-up dirt, mildew and mold, lint, and/or soap. If you don’t regularly clean your washing machine, these things build up on, under, or inside the rubber seal and in the crevices of the drum.
## What time of day is electricity most expensive?
Specific peak and off-peak hours vary by supplier, but a general rule of thumb is off-peak hours are at night, while peak hours occur during the day. Electricity used during the peak hours of the late afternoon will be more expensive than electricity used in the early morning.
## Is it cheaper to do laundry on weekends?
A front-load washing machine and dryer. That same work, if done during off-peak hours on the weekend, will cost 41 cents for the washing machine and 12 cents for the dryer, for a total cost per load of 53 cents. …
## Is it cheaper to use a washing machine at night?
Running your washing machine at night can be cheaper than using it in the day. But this is only true if you are on a special energy tariff called Economy 7 which gives you cheaper power at night. So replacing them with energy-saving ones seems sensible.
## Does doing laundry at night save money?
Do laundry either early in the morning or at night. Try to avoid washing and drying clothes between 2 and 6 p.m. Most of the energy in a washing machine is used to heat the water. It can save energy and money.
## What is the best time to do laundry to save energy?
Try washing before 4 p.m. or after 7 p.m. – Many energy companies charge extra for electricity during their “peak hours,” which see increased energy usage. During the summer, run your washer early in the morning – energy use peaks on hot afternoons.
## How do you unblock a washing machine waste pipe?
To unclog a clogged drain, first fill a bucket with boiling water and a pack of baking soda. Remove the drain pipe from your washing machine and slowly pour the water into the drain using a funnel. Wait a few minutes and test if it keeps draining if you pour water into it.
## What is the black gunk in my washing machine?
What Is That Black Stuff In Your Washing Machine? Basically, it’s bacteria, grease and mold. According to the website White Goods Help, this build up of grime can be the result of multiple issues occurring when you wash your clothes.
## Why does my sink smell when washing machine is on?
If the smell persists, then it might just be from the drain of the washing machine. The drain of a washing machine can get blocked after several months of usage. You can use a drain unblocker to quickly clear out the drain and remove the bad smell from the machine.
## What is the average kWh usage per day UK?
The average household uses: Electricity: 8.5 – 10 kWh per day. | 1,870 | 8,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.945474 |
http://mathhelpforum.com/pre-calculus/40637-functions-relations.html | 1,524,355,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00489.warc.gz | 203,859,044 | 9,597 | 1. ## Functions and Relations
It's me again. I don't really understand this question.
Sam plants a groundcover that takes up 1 m^2. Groudcovers spread, and this one will take up 2^n/3 m^2 after n weeks.
a) How much area will be occupied by the groudcover after each of the first five weeks? What type of sequence do these areas from? Explain your answer.
b) What is the precent increase each week?
c) If possible, rewrite the expression 2^n/3 as a compound interest expression
d) Draw a graph to show the increase in area taken up by the groundcover from growing, how much area of Sam's garden will the groundcover occupy?
I don't want the answer I just want to know how to do it. I have a test coming up and I don't understand this at all.
2. a) 2^5/3 = 32/3 m^2
The sequence is geometric, with a = 2/3, r = 2.
Each week it goes up by a factor of 2.
b) Well if you double something you increase it by 100%, so, 100%. | 255 | 928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-17 | latest | en | 0.95877 |
https://mcqquestions.guru/ncert-solutions-for-class-8-maths-chapter-6/ | 1,722,871,815,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00885.warc.gz | 320,660,488 | 12,947 | # NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1
NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1.
Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 6 Chapter Name Squares and Square Roots Exercise Ex 6.1 Number of Questions Solved 9 Category NCERT Solutions
## NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1
Question 1.
What will be the unit digit of the squares of the following numbers ?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution.
(i) 81
∵ $$1\times 1$$
∴ The unit digit of the square of the number 81 will be 1.
(ii) 272
∵ $$2\times 2$$
∴ The unit digit of the square of the number 272 will be 4.
(iii) 799
∵ $$9\times 9$$
∴ The unit digit of the square of the number 799 will be 1.
(iv) 3853
∵ $$3\times 3$$
∴ The unit digit of the square of the number 3853 will be 9
(v) 1234
∵ $$4\times 4$$
∴ The unit digit of the square of the number 1234 will be 6.
(vi) 26387
∵ $$7\times 7$$
∴ The unit digit of the square of the number 26387 will be 9.
(vii) 52698
∵ $$8\times 8$$
∴ The unit digit of the square of the number 52698 will be 4.
(viii) 99880
∵ $$0\times 0$$
∴ The unit digit of the square of the number 99880 will be 0.
(ix) 12796
∵ $$6\times 6$$
∴ The unit digit of the square of the number 12796 will be 6.
(x) 55555
∵ $$5\times 5$$
∴ The unit digit of the square of the number 55555 will be 5.
Question 2.
The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050.
Solution.
(i) 1057
The number 1057 is not a perfect square because it ends with 7 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
(ii) 23453
The number 23453 is not a perfect square because it ends with 3 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
(iii) 7928
The number 7928 is not a perfect square because it ends with 8 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
(iv) 222222
The number 222222 is not a perfect square because it ends with 2 at unit’s place whereas the Square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
(v) 64000
The number 64000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.
(vi) 89722
The number 89722 is not a square number because it ends in 2 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
(vii) 222000
The number 222000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.
(viii) 505050
The number 505050 is not a square number because it has 1 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.
Question 3.
The squares of which of the following would be odd numbers ?
(i) 431
(ii) 2826
(iii) 7779
(v) 82004.
Solution.
(i) 431
∵ 431 is an odd number
∴ Its square will also be an odd number.
(ii) 2826
∵ 2826 is an even number
∴ Its square will not be an odd number.
(iii) 7779
∵ 7779 is an odd number
∴ Its square will be an odd number.
(iv) 82004
∵ 82004 is an even number
∴ Its square will not be an odd number.
Question 4.
Observe the following pattern and find the missing digits:
Solution.
Question 5.
Observe the following pattern and supply the missing numbers:
Solution.
Question 6.
Using the given pattern, find the missing numbers:
Solution.
Question 7.
(i) 1 + 3 + 5+7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19
(iii) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23.
Solution.
(i)
l + 3 + 5 + 7 + 9 = sum of first five odd natural numbers = $${ 5 }^{ 2 }$$ = 25
(ii)
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = sum of first ten odd natural numbers = $${ 10 }^{ 2 }$$ = 100
(iii)
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = sum of first twelve odd natural numbers = $${ 12 }^{ 2 }$$= 144.
Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution.
(i)
49 (= $${ 7 }^{ 2 }$$)
= 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii)
121 (= $${ 11 }^{ 2 }$$)
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.
Question 9.
How many numbers lie between squares of the following numbers ?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution.
(i) 12 and 13
Here, n = 12
∴ 2n = 2 x 12 = 24
So, 24 numbers lie between squares of i the numbers 12 and 13.
(ii) 25 and 26
Here, n = 25
∴ 2n = 2 x 25 = 50
So, 50 numbers lie between squares of the numbers 25 and 26.
(iii) 99 and 100
Here, n = 99
∴ 2n = 2 x 99 = 198
So, 198 numbers lie between squares of of the numbers 99 and 100.
We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part, drop a comment below and we will get back to you at the earliest. | 1,890 | 5,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.593715 |
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# Moving Average Trading Method In Forex That Makes Sense
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The utilization of moving averages within Forex trading is most likely probably the most popular techniques around. Regardless of whether it’s the actual 20 time period, the 50 time period, or a mix of different moving averages (9/30 is really a popular combination), it’s difficult to to determine a chart with no average onto it.
But does which means that they function? When We say work I am talking about do these people really provide you with an advantage? What tend to be the most efficient moving averages?
To be able to determine which, we ought to first know how an average is actually calculated and the very first thing we require is cost. Since a good average requirements price to maneuver first, it instantly qualifies like a lagging indicator which I am certain you currently knew.
## Calculating A Moving Average
Starting with the most basic, a simple moving average, all we have is an average of the X number of days it looks back. It could be calculated from the closing price or an average high, low and closing price depending on the settings you choose.
Keeping it relatively simply: a 10 day sma using closing prices (1+2+3+4+5+6+7+8+9+10)/10 = 5.5
As new inputs come, the first one is dropped and the cycle continues.
If we are looking at an exponential moving average, the calculation is different by taking the more recent data and giving it a higher weight.
Don’t get too caught up in the calculation because it is done for you but knowing how it’s calculated can give you an idea of how they are calculated. The question for you might be “if it’s just a mechanical computation of price points, where is the edge?”
Good question.
## Big Players Use Moving Averages In Their Trading
You’ve heard this over and over again: “Hedge funds love the 100 SMA” or something to that effect. How do you know? There has never been a study that I am aware of, and I’ve looked, where someone surveyed all the big players and asked them.
That seems to be one of those “truths” that gets tossed around without any documentation to back it up.
I’ve noticed people discuss the “power from the 50 SMA” as well as how it delivers great trading possibilities. There is a method to use moving averages however thinking this gives any kind of edge alone is, I believe, wrong considering.
Why 50? Why don’t you 53? The reason why 20? Why don’t you 32? It’s simply an average associated with price and there isn’t any magic number that will make a profitable investor. The players could use them however they certainly wouldn’t use them since the backbone of the trading system.
There are typical moving average configurations people use however in now method are these people the “best moving averages with regard to day trading” or any kind of trading.
## Use Moving Averages For Pullback Trades
That appears to be a well-liked example of the trading system utilizing moving averages of look back again periods. Can there be an advantage? Let’s consider it…..
As cost advances and also the calculations occur, price will distance themself from the actual average. Reasonable enough. Price ultimately falls and can, at occasions, come into connection with the moving average. Do the average provide support?
Absolutely no. Price just met the actual average simply because either because of rapid decrease and price or perhaps a consolidation, the average cost of By look back again period gets close to the present price. That’s not saying the imply reversion isn’t a practical trading advantage because it definitely is.
But it’s not really the average the complexities the advantage. At probably the most, it provides trader some form of foundation to be able to not merely trade whatsoever areas from the chart. It’s that which you notice cost do in the average which matters.
Look left of present price in your chart and find out which kind of formation or even price motion has happened.
One question you read a lot is “what is the best setting for a moving average” and one thing people must understand is that there is no best setting.
Using a short look back period of 10 will obviously have the moving average closer to current price for the most part. Using a 50 period moving average will have the average further away and it cases of extreme movement, very far away. (There is a trading tip in there that I will cover in a moment)
There is no best setting that will make your trading more profitable. Do not waste your time looking for one.
## Here Is An Example of Moving Average Method That Makes Sense
Understanding that a moving average is merely a computation of previous price information, how may we make use of an average such as the 50 SMA or even 20 EMA within our trading?
The very first method is actually when checking your Forex graphs (any device really), you overlay the actual indicator upon that provides you with a fast birds attention view of the health of that marketplace. We don’t require a crossover to inform us of the trend. We are able to use strategies which are much less complicated than which.
Here we now have the 50 time period simple moving average (sma) on the daily chart of the Forex pair. Simply because price is actually cutting over and beneath the average would let you know that the forex market is not really trending.
50 Simple Moving Average Helps Show Consolidation
That could be great if you are a trader that looks for momentum moves out of consolidations. You would see the shadows rejecting support, especially the long shadow in the middle and this may grab your interest. You may highlight this pair in your watch list, mark off significant areas and set alerts.
So why would this moving average strategy “work”? As we mentioned, it just calculates price and when you get this type of chart look, it tells you that price has not rushed off in any one direction far enough to have the average move away from price.
## How About Another Use Of The 50 Simple Moving Average (or, as you guessed, any number)
Markets ebb and flow in impulse moves and corrections. What precedes a correction? A market that has moved some distance in one direction which then sets up mean reversion or a pullback. When we take a pullback trade we are expecting price to make another run but why would it?
The chart above has some nasty price action and there seems to be no rhythm to the way it’s moving. There’s no trend (although on lower time frames there would be) so trading any type of pullback would be a crap shoot.
What about a market that has trended and moved for away from the average price as shown by the moving average? Is it far fetched to expect the move to keep going after a pullback? I am talking best case because you’d want to analyze the swings to ensure we are not looking at a dying last ditch gasp.
This is another example of a moving average method that makes sense.
Trending Market With Pullback Using 50 SMA
On your scan the thing is this marketplace pulling from the average or more sloping the industry sign of the uptrend (you’d want to utilize a price pattern approach to identifying trend while you drill to the chart). It would cause you to take pause to research further if you’re a imply reversion/pullback/reversal investor.
Price offers stretched from in the moving average (the extended elastic music group idea) so we all know this can be a strong market which will pullback later on. At the actual red collection price pulls to the 50 time period moving average looking enjoy it found support!
It didn’t truly find support due to the moving average. Notice the actual consolidation before the pullback that allowed the actual moving average to begin running nearer to price. Which red collection? Markets relocate a tempo and cost pulled back again virtually similar in distance to some prior golf swing that begin this whole move.
Look at the yellow highlighted candles at the top of the chart. Price had stayed in close contact with the moving average after the pullback (because price really did not advance) and those looking for consolidations would take a look at this chart.
A few things you could look at:
1. Messy ascending triangle
Multiple Price Patterns
The purchase price action inside the yellow pointed out area is uncommon of preceding price actions. This has every one of the hallmarks regarding exhaustion on this market and you also would N’t need to find a continuation trade after having a move similar to this. It’s worse on the particular four hour or so chart in the event you had virtually any doubt regarding violence on this move.
An clever trader which paid more awareness of price action rather than basing everything over a moving average strategy may well not have had the oppertunity to benefit from this shift.
## Moving Averages Are Simple Trading Tools
Simple is better and a person don’t have to use crossovers or even moving average “fans” to get involved with the marketplace. An knowledge of price motion trading, swing analysis and also a simple as well as effective utilization of a moving average is ample. | 1,939 | 9,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.919225 |
http://ch.whu.edu.cn/article/id/4049 | 1,702,295,252,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00254.warc.gz | 9,006,785 | 35,754 | 引用本文: 黄谟涛, 管铮, 欧阳永忠. 中国地区1°×1°点质量解算与精度分析[J]. 武汉大学学报 ( 信息科学版), 1995, 20(3): 257-262.
Huang Motao, Guan Zheng, Ouyang Yongzhong. Accuracy Analysis and Calculation of 1°×1° Point Masses in the Area of China[J]. Geomatics and Information Science of Wuhan University, 1995, 20(3): 257-262.
Citation: Huang Motao, Guan Zheng, Ouyang Yongzhong. Accuracy Analysis and Calculation of 1°×1° Point Masses in the Area of China[J]. Geomatics and Information Science of Wuhan University, 1995, 20(3): 257-262.
## Accuracy Analysis and Calculation of 1°×1° Point Masses in the Area of China
• 摘要: 从重力异常相关性分析、地面重力异常场逼近效果以及数值计算稳定性出发,讨论分析了外部引力场扰动点质量赋值模式的数据结构,提出以窗口移动方式控制点质量对场元观测的作用范围,简化了模型结构。在此基础上,提出解算经结构简化后具有稀疏矩阵特征的边值条件方程的超松弛迭代法,并实际解算了中国地区(0°~60°N,70°E~140°E)1°×1°点质量。最后对解算结果作了全面分析和检验,证明以窗口移动方式简化解算点质量的数据结构具有多方面的优越性。
Abstract: Through the analyses of anomalies relativity,the approximation effects of surface gravity field and the numerical stability of calculation,the datum configuration of the disturbing point masses model is discussed in this paper.Then,in an effort to optimize model configuration,a new method of creating fictitious point masses by way of moving window control is proposed.On this basis,the successive over-relaxation(SOR) iterative method is introduced to solve the equation of boundary value condition which has been optimized to be a sparse matrix.And the method is used to calculate the 1°×1°point masses in the area of China(0°~60°N;70°E~140°E).Finally,the computation results are analysed and checked.It is proved that the optimization of datum configuration by way of moving window control has many advantages.
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# Two airport shuttle trains leave the main station at the same time. Shuttle A returns to the station every 8 minutes. Shuttle B returns to the station every 10 minutes. In how many minutes will shuttles A and B leave the station together for the second time?A. 10 minutesB. 18 minutesC. 40 minutesD. 80 minutes
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The answer is C. 40 minutes | 110 | 384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.866817 |
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Chapter 7 Class 12 Integrals
Serial order wise
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### Transcript
Misc 44 The value of β«_0^1βtan^(β1)β‘((2π₯ β 1)/(1 + π₯ β π₯^2 )) ππ₯ is equal to (A) 1 (B) 0 (C) β1 (D) π/4 Let I=β«_0^1βtan^(β1)β‘((2π₯β1)/(1 + π₯ β π₯^2 )) ππ₯ I=β«_0^1βtan^(β1)β‘[(π₯ + (π₯ β 1))/(1 + π₯(1 β π₯) )] ππ₯ I=β«_0^1βtan^(β1)β‘[(π₯ + (π₯ β 1))/(1 + π₯(π₯ β 1) )] ππ₯ Using γπ‘ππγ^(β1) π₯+γπ‘ππγ^(β1) (π¦)=γπ‘ππγ^(β1)β‘[(π₯ + π¦)/(1 β π₯π¦)] β΄ I=β«_0^1β[tan^(β1) (π₯)+tan^(β1) (π₯β1)] ππ₯ I=β«_0^1βγtan^(β1) [β(π₯β1)] γ ππ₯+β«_0^1βγtan^(β1) (βπ₯) γ ππ₯ Using The Property, P4 : β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)ππ₯ β΄ I=β«_0^1βγtan^(β1) (1βπ₯) γ ππ₯+β«_0^1βγtan^(β1) (1βπ₯β1) γ ππ₯ I=β«_0^1βγtan^(β1) [β(π₯β1)] γ ππ₯+β«_0^1βγtan^(β1) (βπ₯) γ ππ₯ I=ββ«_0^1βγtan^(β1) (π₯β1) γ ππ₯ββ«_0^1βγtan^(β1) (π₯) γ ππ₯ tan^(β1) (βπ₯)=βtan^(β1)β‘γ(π₯)γ Adding (1) and (2) I+I =β«_0^1βγtan^(β1) (π₯) γ ππ₯+β«_0^1βγtan^(β1) (π₯β1) γ ππ₯ββ«_0^1βγtan^(β1) (π₯) γ ππ₯ββ«_0^1βγtan^(β1) (π₯β1) γ ππ₯ 2I=0 β΄ I=0 Hence, correct answer is B. | 1,209 | 1,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-23 | longest | en | 0.37317 |
http://sciencedocbox.com/Physics/66488043-Transient-response-of-rc-and-rl-circuits-engr-40m-lecture-notes-july-26-2017-chuan-zheng-lee-stanford-university.html | 1,558,769,032,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257920.67/warc/CC-MAIN-20190525064654-20190525090654-00411.warc.gz | 181,226,774 | 25,469 | # Transient response of RC and RL circuits ENGR 40M lecture notes July 26, 2017 Chuan-Zheng Lee, Stanford University
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1 Transient response of C and L circuits ENG 40M lecture notes July 26, 2017 Chuan-Zheng Lee, Stanford University esistor capacitor (C) and resistor inductor (L) circuits are the two types of first-order circuits: circuits either one capacitor or one inductor. In many applications, these circuits respond to a sudden change in an input: for example, a switch opening or closing, or a digital input switching from low to high. Just after the change, the capacitor or inductor takes some time to charge or discharge, and eventually settles on its new steady state. We call the response of a circuit immediately after a sudden change the transient response, in contrast to the steady state. A first example Consider the following circuit, whose voltage source provides v in (t) = 0 for t < 0, and v in (t) = 10 V for t 0. vin (t) C A few observations, using steady state analysis. Just before the step in v in from 0 V to 10 V at t = 0, (0 ) = 0 V. Since is across a capacitor, just after the step must be the same: (0 ) = 0 V. Long after the step, if we wait long enough the circuit will reach steady state, then ( ) = 10 V. What happens in between? Using Kirchoff s current law applied at the top-right node, we could write 10 V = C d. dt Solving this differential equation, then applying the initial conditions we found above, would yield (in volts) (t) = 10 10e t C. 10 v v in (t) (t) τ = C 2τ 3τ 4τ t What s happening? Immediately after the step, the current flowing through the resistor and hence the capacitor (by KCL) is i(0 ) = 10 V. Since i = C dvout dt, this current causes to start rising. This, in 10 Vvout turn, reduces the current through the resistor (and capacitor),. Thus, the rate of change of decreases as increases. The voltage (t) technically never reaches steady state, but after about 3C, it s very close.
3 Examples and exercises Example 1. Use the first-order transient response equation to verify the result for (t) that was found in the C example on page 1. Exercise 1. [Not examinable material.] In the following circuit, the switch closes at time t = 0, before which it had been open for a long time. Use Kirchoff s voltage law to write a differential equation for the following circuit, and solve it to find (t). Verify that your answer matches what you would get from using the first-order transient response equation. t = 0 10 V i L L Example 2. In the above circuit (the same as for Exercise 1), the switch closes at time t = 0, before which it had been open for a long time. (a) Find and plot i L (t). (b) Find and plot (t). Example 3. In the below circuit, v in (t) is as is shown in the plot. (For all times outside the plot, v in (t) = 0 V.) Find and plot (t), for t > 0. v in (t) (V) V 330 nf t (ms) v in (t) Example 4. In the below circuit, the switch has been closed for a long time before t = 0. The switch then opens at t = 0, and then closes at t = 100 µs. Find and plot (t). 100 Ω 10 V 1 mh 100 Ω 3
4 Solution to Example 3 When t < 10 ms. When v in (t) = 0, the NMOS transistor is off, so the circuit reduces to this: V 330 nf Since v in (t) = 0 for all t < 10 ms, the circuit is in steady state at t = 10 ms. Furthermore, that resistor on the right doesn t do anything while the transistor is off. So the circuit reduces to this: V from which we see that (t) = V at steady state while the transistor is off, including when t < 10 ms. When 10 ms < t < 20 ms. During this period, v in (t) = V, so the transistor is on (v GS = V 0 V = V). Then we have this circuit: V 330 nf The voltage is the voltage across a capacitor, which can t change instantaneously, so (10 ms ) = (10 ms ) = V. (Caution! The logic isn t that the output voltage can t change instantaneously, it s the the voltage across the capacitor can t.) To find the v( ) term of the equation, for this part, we find the steady state value of the circuit while the transistor is on in other words, the value that (t) would converge to if the transistor stayed on forever. (The circuit can t predict the future, so it doesn t know that the transistor will turn off soon.) To find the steady state value, we replace the capacitor with an open circuit again: 4
5 V Now we can see that there is just a voltage divider, and ( ) in this state would be 2. V. The time constant is τ = C, where is the resistance seen by the capacitor. To find this, we short (zero) the voltage source and imagine measuring the resistance from the capacitor: capacitor was here And now we can see that it s just two resistors in parallel, yielding = 10 kω. Then Putting it all together, for 10 ms < t < 20 ms: τ = C = 10 kω 330 nf = 3.3 ms. (t) = ( ) [ (10 ms t10 ms ) ( )]e τ t10 ms = 2. [ 2.]e 3.3 ms (V) t10 ms = 2. 2.e 3.3 ms (V) When t > 20 ms. Here, the transistor has turned off again. Because the voltage across the capacitor can t change instantaneously, (20 ms ) = (20 ms ), and (20 ms ) falls into the equation we just found: (20 ms 20 ms10 ms ) = 2. 2.e 3.3 ms = 2.62 V. The ( ) term is just the steady state when the transistor is off, which we ve already found above; it was V. As for the time constant, if we short the voltage source and remove the capacitor while the transistor is off, we get this: capacitor was here
6 That resistor on the right isn t really part of the circuit, so really we just have this: capacitor was here And now it s clear that =. Then τ = C = 330 nf = 6.6 ms. Note that this is twice as long as the time constant while the transistor was on. Putting it all together, for t > 20 ms: (t) = ( ) [ (20 ms t20 ms ) ( )]e τ t20 ms = [2.62 ]e 6.6 ms (V) t10 ms = 2.38e 6.6 ms (V) Here is a plot, with v in (t) drawn underneath for ease of reference. The dashed line is the curve that would have been followed if the transistor had been on forever. (t) (V) t (ms) v in (t) (V) t (ms) You can see a simulation of this circuit in action at 6
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DC and Transient Responses (i.e. delay) (some comments on power too!) Michael Niemier (Some slides based on lecture notes by David Harris) 1 Lecture 02 - CMOS Transistor Theory & the Effects of Scaling | 8,334 | 32,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-22 | latest | en | 0.887355 |
https://www.mvtec.com/doc/halcon/11/en/minkowski_sub1.html | 1,642,379,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00259.warc.gz | 966,734,582 | 7,536 | Operators
# minkowski_sub1 (Operator)
## Name
minkowski_sub1 — Erode a region.
## Signature
minkowski_sub1(Region, StructElement : RegionMinkSub : Iterations : )
## Description
minkowski_sub1 computes the Minkowski subtraction of the input regions with a structuring element. By applying minkowski_sub1 to a region, its boundary gets smoothed. In the process, the area of the region is reduced. Furthermore, connected regions may be split. Such regions, however, remain logically one region. The Minkowski subtraction is a set-theoretic region operation. It uses the intersection operation.
Let M (StructElement) and R (Region) be two regions, where M is the structuring element and R is the region to be processed. Furthermore, let m be a point in M. Then the displacement vector v(m) = (dx,dy) is defined as the difference of the center of gravity of M and the vector v(m). Let t(v(m))(R) denote the translation of a region R by a vector v(m). Then
```
/ \
minkowski_sub1(R,M) := | | t (R)
| | v(m)
m in M
```
For each point m in M a translation of the region R is performed. The intersection of all these translations is the Minkowski subtraction of R with M. minkowski_sub1 is similar to the operator erosion1, the difference is that in erosion1 the structuring element is mirrored at the origin. The position of StructElement is meaningless, since the displacement vectors are determined with respect to the center of gravity of M.
The parameter Iterations determines the number of iterations which are to be performed with the structuring element. The result of iteration n-1 is used as input for iteration n. From the above definition it follows that the maximum region is generated in case of an empty structuring element.
Structuring elements (StructElement) can be generated with operators such as gen_circle, gen_rectangle1, gen_rectangle2, gen_ellipse, draw_region, gen_region_polygon, gen_region_points, etc.
## Parallelization
• Multithreading type: reentrant (runs in parallel with non-exclusive operators).
• Automatically parallelized on tuple level.
## Parameters
Region (input_object) region(-array) object
Regions to be eroded.
StructElement (input_object) region object
Structuring element.
RegionMinkSub (output_object) region(-array) object
Eroded regions.
Iterations (input_control) integer (integer)
Number of iterations.
Default value: 1
Suggested values: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 17, 20, 30, 40, 50
Typical range of values: 1 ≤ Iterations (lin)
Minimum increment: 1
Recommended increment: 1
## Complexity
Let F1 be the area of the input region, and F2 be the area of the structuring element. Then the runtime complexity for one region is:
``` O(sqrt(F1) * sqrt(F2) * Iterations) .
```
## Result
minkowski_sub1 returns 2 (H_MSG_TRUE) if all parameters are correct. The behavior in case of empty or no input region can be set via:
• no region: set_system('no_object_result',<RegionResult>)
• empty region: set_system('empty_region_result',<RegionResult>)
Otherwise, an exception is raised.
## Module
Foundation
Operators | 778 | 3,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | longest | en | 0.806824 |
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# Leader of Opposition
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Author Message
Director
Joined: 25 Jan 2004
Posts: 723
Location: Milwaukee
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Leader of Opposition [#permalink]
### Show Tags
09 Nov 2004, 12:29
Before the election the Prime Minister described the Leader of the Opposition as a shallow creature without any original thought.
Before the election the Prime Minister described the Leader of the Opposition as a shallow creature without any original thought.
The Prime Minister described the Leader of the Opposition as a shallow creature before the election without any original thought.
The Prime Minister described the Leader of the Opposition as a shallow creature without any original thought before the election.
Before the election the Prime Minister described the Leader of the Opposition without any original thought as a shallow creature.
Before the election without any original thought the Prime Minister described the Leader of the Opposition as a shallow creature.
_________________
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Manager
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### Show Tags
09 Nov 2004, 12:59
I would stick w/ A, the rest distorts the meaning of the sentence.
Senior Manager
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### Show Tags
09 Nov 2004, 13:16
A seems to be the best.
Is there a comma missing in the sentence.
Before the election, the Prime Minister described the Leader of the Opposition as a shallow creature without any original thought
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510 on my first GMAT. 610 on second GMAT.!
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### Show Tags
09 Nov 2004, 14:19
Another vote for A.
'without any original thought' in many of the choices seem as though the Prime Minister did not have any orignal thought. However, the intent here is that 'without original thought' describe the opposition leader.
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### Show Tags
09 Nov 2004, 14:27
A for me too
Manager
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### Show Tags
09 Nov 2004, 16:18
A is best
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Thanks !
Aspire
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09 Nov 2004, 22:15
venksune wrote:
Another vote for A.
'without any original thought' in many of the choices seem as though the Prime Minister did not have any orignal thought. However, the intent here is that 'without original thought' describe the opposition leader.
It is between "A" and "E". "E" can be true if 'without any original thought' is to modify the prime minister's description.
But yeah venksune is correct. I vote for "A".
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10 Nov 2004, 04:10
A by POE
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10 Nov 2004, 07:37
I don't really have a OA for this one. I was stuck between A & E too
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Praveen
10 Nov 2004, 07:37
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https://creation-office.com/what-can-be-often-the-mystery-driving-often-the-peculiar-even-lottery-quantity-strategy/ | 1,611,555,428,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00569.warc.gz | 286,053,364 | 8,994 | ## What Can be often the Mystery Driving often the Peculiar Even Lottery Quantity Strategy?
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creonice | 715 | 3,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-04 | longest | en | 0.965253 |
https://proizvodim.com/numerical-experiments-and-results.html | 1,720,820,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00340.warc.gz | 388,555,286 | 10,773 | ## Numerical Experiments and Results
Figure 5-17 shows the behavior of four constitutive models. Their properties are shown in Tables 5-2 and 5-3. Two models use rate independent plasticity, one uses rate dependent plasticity and one uses linear viscous plasticity. The test is uniaxial and the strain rate is 0.005 5_1in the time interval [0, 4] seconds. The strain rate is zero in the time interval [4, 10]. The strain rate is -0,005 s 1 in the time interval [10, 13]. The strain rate is zero in the time interval [13, 25]. Note that in the rate independent models there is no stress relaxation. In the rate dependent model the stress relaxes toward the deformation resistance. In the linear viscous model, the stress decays toward zero. The viscosity used in Figure 5-17 is larger than that used in the welding test. If the viscosity was not increased, the stress would appear to relax instantly in the time scale used in Figure 5-17.
Time (sec)
Stress (MPa) Displacement (mm)
Time (sec)
Figure 5-17: The response of two rate independent materials (RIP), a rate dependent material (RDP) and a linear viscous material (LVP) to sequence of strain rates is shown [36].
Table 5-2: Material properties for rate independent plasticity (RIP) and linear viscous plasticity (LVP)
RIP-1 RIP-2 LVP Young’s Modulus (GPa) 200 200 200 Poisson Ratio 0.3 0.3 0.3 Hardening Modulus (GPa) 2 2 0 Yield Stress (MPa) 200 100 — Viscosity (GPa. s) -- - 200
Table 5-3: Material properties for rate dependent plasticity (RDP)
Material Parameter Value Material Parameter Value A( - S-1) 6.346* 1011 3.25 AG (J/mole) 3.1235xl05 m 0.1956 s (Pa) 1.251xl08 n 0.06869 К (Pa) 3.0931х 109 1 1.5 a(m) 3.50x 10~10
In the weld analyses a prescribed temperature weld heat source was used [43] to model the arc.
Three and ten seconds after starting the weld Figures 5-18, 5-19,
5- 20 and 5-21 show the temperature distribution, sub-domains for each constitutive equation, effective plastic strain, pressure and several stress components. Plate dimensions 5 *5*0.2 cm. Weld pool dimensions 1.5 (long) *1.5 (wide) *0.5 (deep) cm. Welding speed 6 ipm (0.25 cm/s). Maximum temperature was 2000 °С and solidus temperature was 1500°C. A 15*15*1 element mesh was used. Although the mesh is relatively coarse, it does demonstrate the expected behavior. The high compressive ridge in front of the weld is in the rate independent plasticity sub-domain. It is possible that the yield strength used should be corrected for this strain rate. The longitudinal strain pulling material back from the weld pool is another interesting feature.
Other cases were also analyzed. They differed only in weld speed. The slow weld speed was 6 ipm and the fast speed was 20 ipm (0.25 and 0.83 cm/s), Figure 5-22. In the 20 ipm weld the mesh is not fine enough to resolve the thermal shock and the constitutive equations change directly from rate independent to linear viscous. The longitudinal strain pulling material back from the weld pool is an interesting feature.
Due to a slight change in mesh distribution on the left and right side of the weld (element row 8 or 7) the temperature field on the left and right side is not symmetric. It is shown that it causes no problems in the analysis by switching between constitutive equations.
Т*!ПрЄГОІІІ>« ГС) A-42 В • 200 с /во Г) 1*М ConMihsl^vft Мгуірі ■Ц Lineai Viscous flH Пвіг 0»-;>огчівііі L_JH. iv, ІчЗитгчІнпг ENoctive Ида be Strain Pressure (Pa) А -0.05 8-0.2 -Эе»ле С -0.5 ■L *96*07 0-07 F^T ^ 1___ L - »**oe yfs. ^ * / г) Y і / Х (jfj у 1 J 1 V ^ / лЧ / / і 1 - /
Figure 5-18: At three seconds after starting the weld, the transient temperature, constitutive equation type, effective plastic strain and pressure are shown.
Figure 5-19: At three seconds after starting the weld, (Jxx, <7yy, Gxy and <7yz are shown.
Eflectrve Plastic Strain A -0.1 3-0.6 С -1.0
Pressure (Pa)
Figure 5-20: At ten seconds after starting the weld, the transient temperature, constitutive equation type, effective plastic strain and pressure are shown.
Figure 5-21: At ten seconds after starting the weld, <7^,(7 ,
Constitutive Model
Constitutive Model
I Linear Viscous Ram Dependent [pat* Independent
Pressure (Pa)
2e+08
I Linear Viscous Rale Cepervdent I Rate Independent
»6e*0/
Figure 5-22: At ten seconds after starting the weld, the transient temperature, constitutive equation type, effective plastic strain and pressure are shown for the 6 and 20 ipm welding speeds.
The weld direction is the ^-direction, the upward normal on the plate is the z-direction and the x-direction is transverse to the weld. Figure 5-23 shows the displacements transverse to the weld at various distances from one node, (1/3 cm) from the weld centerline extends from nodes 5 to 9 in the slow weld and from nodes 3 to 15 in the fast weld.
Distance (node) 20 ipm
Figure 5-23: The abscissas are nodes numbered from left to right. The distance between nodes is 1/3 cm. When the weld arc reaches the center of the plate, the weld pool is centered at node 8 and extends from node 5.5 to node 10.5 on the plate. At this instant the transverse displacements along lines parallel to the weld at distances of 1, 2, 4 and 8 nodes from the weld centerline are shown. The welding speeds of 6 and 20 ipm are shown above each plot.
6 ipm
Figure 5-24 shows the transverse stress, <7XX, to the weld on the weld centerline. Chihoski was not able to measure stress. Data of this type would be also useful to compare with Sigmajig test data.
6 ipm
20 ipm Figure 5-24: The welding speeds are 6 and 20 ipm. When the weld arc reaches the center of the plate, the transverse stress (7XX is shown along the weld line
Also the compressive stress region in Figure 5-24 is larger for the faster weld. This is the effect observed by Chihoski and suggests the faster weld would be less susceptible to hot cracking because more of the region susceptible to hot cracking is in compression. The qualitative agreement is as good as could be expected because the welds are quite different.
This model appears to be useful step towards quantitative analysis of stress and strain in the region susceptible to hot cracking in welds. A more realistic quantitative analysis of hot cracking will require a finer mesh to resolve the thermal shock in front of the weld pool. A larger plate is needed to provide more constraint to the weld pool region.
We next consider one or more algorithms for computing flow lines or streamlines associated with a steady state weld.
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http://economicsessays.com/price-elasticity-of-demand/ | 1,642,559,586,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301217.83/warc/CC-MAIN-20220119003144-20220119033144-00373.warc.gz | 24,796,938 | 15,382 | - Price Elasticity of Demand - Free Essays on Economics Price Elasticity of Demand
HELLO, GUEST
## Price Elasticity of Demand
with all other factors held constant.
### Definition
The price elasticity of demand, Ed is defined as the magnitude of:
proportionate change in quantity demanded
Ed = ————————————————————————
proportionate change in price
Sincethe quantity demanded decreases when the price increases,this ratio is negative; however, the absolute value usually is taken and Ed is reported as a positive number.
Because the calculation uses proportionate changes,the answer is a unit-less number which does not depend on the units in which the price and quantity are expressed.
For example, consider a product which has Ed is equal to be 0.5. Then, if the priceis increased by 10%, a decrease of approximately 5% in quantity demanded would be observed.
Inthe above example, the word “approximately” is used asthe exact result will depend on whether the initial valueor the final value of price and quantity is used in the calculation. Thisis because for a linear demand curve the priceelasticity varies as one moves along the curve. Forsmall changes in price and quantity the differencebetween the two results often is negligible, butfor large changes the difference may be more significant. Todeal with this issue, one can use the arc price elasticityof demand. This elasticity uses the average of the initial and finalquantities and the average of the initial and final priceswhile calculating the proportionate change in each. Mathematically,the arc price elasticity of demand can be written as:
Q2 – Q1
———————–
( Q1 + Q2 ) / 2
Ed = ——————————-
P2 – P1
———————–
( P1 + P2 ) / 2
Where:
Q1 = Initial quantity
Q2 = Final quantity
P1 = Initial price
P2 = Final price
### Elastic versus Inelastic
E >1
Inthis case, the quantity demanded is elastic,i.e. a small price change will causea larger change in the quantity demanded. When Ed = infinity, this type of demand is called perfectly elastic. Thedemand curve in such a case would be horizontal i.e. parallel to the quantity axis. Forproducts havinga high price elasticity of demand, aprice increase will result in a decrease in total revenue. This is because therevenue lost from the resultingdecrease in quantity sold ismore than the revenue gained from the price increase.
E <1
In this case, the quantity demanded is relatively inelastic, i.e. a small change in price will cause even smaller change in quantity demanded. When Ed = 0, this is referred to as perfectly inelastic. The demand curve in this case would be vertical i.e. parallel to the price axis. Forproducts in which quantity demanded is inelastic, aprice increase will lead to a revenue increase. This is because, the revenue lost bythe small decrease in quantity is lessthan the revenue gained fromthe higher price.
E = 1
Inthis case, the product is said to have unitary elasticity. Changes in price do not affect the total revenue as it is accompanied by equal change in quanlity.
### Factors that Affect the Price Elasticity of Demand
• Availability of substitutes: if there are many substitutes, the elasticity will be greater. The number of substitutes depends on how broadly the product is defined.
• Necessity or luxury: luxury products generally have greater elasticity. Some products that initially have a low degree of necessity are habit forming and hence can become necessities for some people.
• Proportion of the buyer’s budget used by the item: products that consume a large part of the buyer’s budget tend to have greater price elasticity.
• Time period: elasticity is generally greater over long period of time because consumers have more time to adjust their behaviour.
• Permanent or temporary change in price: a one-day sale will generate a different response than a permanent price decrease.
• Price points: depending on the point from where price is changed, decreasing the price from Rs. 200 to Rs. 190 may elicit a greater response than decreasing it from Rs. 190 to Rs. 180.
### Durable Good
Ineconomics, adurable goodor ahard goodis agoodwhich has a low wear and tear, or in other words, it yieldsservicesorutilityover a period of time rather than being completely used up in once. Most goods are therefore durable goods to a certain degree. These are goods that can last for a relatively long time, such as cars, refrigerators and mobiles.Perfectly durable goods never wear out. An example of such a good might be a brick.
Consumer durable goods include cars,appliances, business equipment, electronic equipment, home furnishings and fixtures, house-ware and accessories, photographic equipment, recreational goods, sporting goods, toys and games.
Durable goods are typically characterized by long inter-purchase times, which is the time between two successive purchases.
Durable goods along with nondurable goods andservicestogether constitute theconsumptionof aneconomy.
### White Goods
• Refrigerators
• Washing Machines
• Air Conditioners
• Speakers and Audio Equipments
### Kitchen Appliances/Brown Goods
• Mixers
• Grinders
• Microwave Ovens
• Iron
• Electric Fans
• Cooking Range
• Chimneys
### Consumer Electronics
• Mobile Phones
• Televisions
• MP3 Players
• DVD Players
• VCD Players
Several key trends are driving growth in the sector. They are-
### Income growth and availability of financing
• Disposable income levels are rising and consumer financing has become easier
### Increased affordability of products
• Advanced technology and increasing competition is narrowing the price gap and appliances which were once expensive are becoming cheaper
### Increasing share of organised retail sector
• Urban and rural market are growing at the annual rates of seven percent to 10 per cent and 25 percent respectively with organised retail expected to garner about 15 per cent share by 2015 from the current 5 percent
### Entry of heavyweight retail players is increasing competition
• Competitive evolution of organised retail due to the entry of heavyweight players like Croma, EZone and Reliance Digital is stimulating the demand through Exposure to experiences
Consumer Durables is one of the fastest growing industries in India. A strong growth is expected across all key segments-
Projected Growth Rates
Colour TVs 25-30% Refrigerators 18-22% Washing Machines 15-20% Air Conditioners 32-35% Others (including VCDs and DVDs 35-40%
(CONSUMER DURABLES December 2008)
### Threat of New Entrants
• Most current players are global players
• New entrants will need to invest in Brand, Technology and Distribution
### Supplier Power
• Indigenous supply base limited – most raw materials are imported
### Competitive Rivalry
• Number of well established players, with new players entering
• Good technological capability
• Many untapped potential markets
Several global players are well established in the Consumer Durables sector in India, with competition from strong Indian players.Some of the key players in the sector in India include:
• Samsung
• Philips
• LG
• Whirlpool
• Nokia
• Sony
Table 2 Estimates of the Impact of Price, Income and Efficiency on Automobile and Appliance Sales
Durable Good Price Elasticity Income Elasticity Brand Price Elasticity Implicit Discount Rate Model Data Years Time Period Automobiles1 -1.07 3.08 – – Linear Regression, stock adjustment – Short run Automobiles1 -0.36 1.02 – – Linear Regression, stock adjustment – Long run Clothes Dryers2 -0.14 0.26 – – Cobb-Douglas, diffusion 1947-1961 Mixed Room Air Conditioners2 -0.378 0.45 – – Cobb-Douglas, diffusion 1946-1962 Mixed Dishwashers2 -0.42 0.79 – – Cobb-Douglas, diffusion 1947-1968 Mixed Refrigerators3 -0.37 – – 39% Logit probability, survey data 1997 Short run Various4 – – -1.769 – Multiplicative regression – Mixed Room Air Conditioners5 – – -1.72 – Non-linear diffusion 1949-1961 Short run Clothes Dryers5 – – -1.32 – Non-linear diffusion 1963-1970 Short run Room Air Conditioners6 – – – 20% Qualitative choice, survey data – – Household Appliances7 – – – 37%10 Assorted – –
Sources: S. Hymens. 1971; P. Golder and G. Tellis, 1998; D. Revelt and K. Train, 1997;
### Variables describing the market for refrigerators, clothes washers, and dishwashers
In this section variables that appear to account for refrigerator, clothes washer and dishwasher shipments, including physical household/appliance variables, and economic variables.
### Physical Household/Appliance Variables
Several variables influence the sale of refrigerators, clothes washers and dishwashers. The most important for explaining appliance sales trends are the annual number of new households formed (housing starts) and the number of appliances reaching the end of their operating life (replacements). Housing starts influence sales because new homes are often provided with, or soon receive, new appliances, including dishwashers and refrigerators. Replacements are correlated with sales because new appliances are typically purchased when old ones wear out. In principle, if households maintain a fixed number of appliances, shipments should equal housing starts plus appliance replacements.
### Economic variables
Appliance price, appliance operating cost and household income are important economic variables affecting shipments. Low prices and costs encourage household appliance purchases and a rise in income increases householder ability to purchase appliances. In principle, changes in economic variables should explain changes in the number of appliances per household.
During the 1980–2002 study period, annual shipments grew 69 percent for clothes washers, 81 percent for refrigerators and 105 percent for dishwashers. This rising shipments trend is explained in part by housing starts, which increased 6 percent and by appliance replacements, which rose between 49 percent and 90 percent, depending on the appliance, over the period
### Physical Household/Appliance Variables
Shipments1 (millions) Housing Starts2 (millions) Replacements3 (millions)
Appliance 1980 2002 Change 1980 2002 Change 1980 2002 Change Refrigerators 5.124 9.264 81% 1.723 1.822 6% 3.93 5.84 49% Clothes Washers 4.426 7.492 69% 1.723 1.822 6% 3.66 5.5 50% Dishwashers 2.738 5.605 105% 1.723 1.822 6% 1.99 3.79 90%
### LONG RUN IMPACTS
Price elasticities over short and long time periods, also referred to as short run and long run price elasticities. The price elasticity of demand is significantly different over the short run and long run for automobiles. Because forecasts of shipments and national impacts due to standards is over a 30-year time period, consideration must be given as to how the relative price elasticity is affected once a new standard takes effect. Elasticity of demand changes in the years following a purchase price change. With increasing years after the price change, the price elasticity becomes more inelastic until it reaches a terminal value around the tenth year after the price change.
### Change in Price Elasticity of Demand for Automobiles following a Purchase Price Change
Years Following Price Change
1 2 3 5 10 20
Price Elasticity of Demand -1.2 -0.93 -0.75 -0.55 -0.42 -0.4 Relative Change in Elasticity to 1st year 1 0.78 0.63 0.46 0.35 0.33
### Change in Relative Price Elasticity for Home Appliances following a Purchase Price Change
Years Following Price Change
1 2 3 5 10 20
Relative Change in Elasticity to 1st year 1.00 0.78 0.63 0.46 0.35 0.33 Relative Price Elasticity
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