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https://www.studypool.com/discuss/1004993/how-do-you-solve-the-integral-of-1-cos-3t-sin-3t-dt-when-the-lower-limit-is-0?free | 1,480,772,926,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540932.10/warc/CC-MAIN-20161202170900-00507-ip-10-31-129-80.ec2.internal.warc.gz | 1,015,097,485 | 14,033 | ##### How do you solve the integral of (1-cos(3t))sin(3t)dt when the lower limit is 0
Calculus Tutor: None Selected Time limit: 1 Day
How do you solve the integral of (1-cos(3t))sin(3t)dt when the lower limit is 0 and the upper is pi/6
May 28th, 2015
Int [(1-cos(3t))sin(3t)]dt=int[sin(3t)-cos(3t)sin(3t)]dt = (-1/3) cos(3t) -(1/2) int[sin(6t)]dt
=(-1/3) cos(3t) +(1/12) cos(6t)
Now calculate in limits
int from 0 to pi/6=(-1/3) cos(pi/2) +(1/12) cos(pi) -[(-1/3)+(1/12)]=0+(1/12)(-1) +1/3 - (1/12)= 1/3 - 2/12 =1/3 -1/6
= 1/6
May 28th, 2015
...
May 28th, 2015
...
May 28th, 2015
Dec 3rd, 2016
check_circle | 293 | 617 | {"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-2016-50 | longest | en | 0.809719 |
https://kidsactivitiesblog.com/753/learning-fun-with-apples/ | 1,709,400,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00278.warc.gz | 336,180,488 | 70,562 | Thanks to Nicole I discovered “Muffin Tin Mondays”. My little gal loved yesterday’s assignment. We took two apples and named the apple parts, learned about fractions, shapes, tastes and textures. … … … … First we took the two apples and cut them in half different ways, one we cut from the stem down, the other we cut horizontally. We identified the stem, the peel, the pulp, the seed pod and the seeds. Then we discussed fractions as we cut one half in half and then in half again and counted the slices we had. My little gal ate a slice and we recounted. I love learning math with preschoolers! . I chopped the untouched half into cubes and we then looked at all the apple pieces (two halves cut horizontally, some slices and a handful of cubed apple) and the other snack “accessories” (unsweetened baking chocolate bits, a handful of pretzels, a diced cheese stick, a smear of peanut butter and some cinnamon and sugar in a jar). We then tried to identify the shapes that we saw: a circle, triangles, oval, star, squares and rectangles. … OOhhh! the fun part: eating! As we ate, we discussed what foods were salty, sweet and bitter and we felt the textures:
• smooth – cheese
• sticky – our hands after playing with our food
• hard – the seeds and the chocolate
• bumpy – the salt on the pretzel
• creamy – the peanut butter
All in all, we really enjoyed this snack time! Thank you to Muffin Tin Monday for inspiring our learning adventure of the day! … This activity is part of our learning activity theme. For more activity ideas click the link below:
Welcome to Kids Activities!
My name is Holly Homer & I am the Dallas mom of three boys… | 381 | 1,662 | {"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} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.922828 |
http://www.popflock.com/learn?s=Range_(aeronautics) | 1,611,257,546,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00368.warc.gz | 163,993,807 | 19,202 | Range (aeronautics)
Get Range Aeronautics essential facts below. View Videos or join the Range Aeronautics discussion. Add Range Aeronautics to your PopFlock.com topic list for future reference or share this resource on social media.
Range Aeronautics
The maximal total range is the maximum distance an aircraft can fly between takeoff and landing, as limited by fuel capacity in powered aircraft, or cross-country speed and environmental conditions in unpowered aircraft. The range can be seen as the cross-country ground speed multiplied by the maximum time in the air. The fuel time limit for powered aircraft is fixed by the fuel load and rate of consumption. When all fuel is consumed, the engines stop and the aircraft will lose its propulsion.
Ferry range means the maximum range the aircraft can fly. This usually means maximum fuel load, optionally with extra fuel tanks and minimum equipment. It refers to transport of aircraft without any passengers or cargo. Combat range is the maximum range the aircraft can fly when carrying ordnance. Combat radius is a related measure based on the maximum distance a warplane can travel from its base of operations, accomplish some objective, and return to its original airfield with minimal reserves.
## Derivation
For most unpowered aircraft, the maximum flight time is variable, limited by available daylight hours, aircraft design (performance), weather conditions, aircraft potential energy, and pilot endurance. Therefore, the range equation can only be calculated exactly for powered aircraft. It will be derived for both propeller and jet aircraft. If the total weight ${\displaystyle W}$ of the aircraft at a particular time ${\displaystyle t}$ is:
${\displaystyle W}$ = ${\displaystyle W_{0}+W_{f}}$,
where ${\displaystyle W_{0}}$ is the zero-fuel weight and ${\displaystyle W_{f}}$ the weight of the fuel (both in kg), the fuel consumption rate per unit time flow ${\displaystyle F}$ (in kg/s) is equal to
${\displaystyle -{\frac {dW_{f}}{dt}}=-{\frac {dW}{dt}}}$.
The rate of change of aircraft weight with distance ${\displaystyle R}$ (in meters) is
${\displaystyle {\frac {dW}{dR}}={\frac {\frac {dW}{dt}}{\frac {dR}{dt}}}=-{\frac {F}{V}}}$,
where ${\displaystyle V}$ is the speed (in m/s), so that
${\displaystyle {\frac {dR}{dt}}=-{\frac {V}{F}}{\frac {dW}{dt}}}$
It follows that the range is obtained from the definite integral below, with ${\displaystyle t_{1}}$ and ${\displaystyle t_{2}}$ the start and finish times respectively and ${\displaystyle W_{1}}$ and ${\displaystyle W_{2}}$ the initial and final aircraft weights
${\displaystyle R=\int _{t_{1}}^{t_{2}}{\frac {dR}{dt}}dt=\int _{W_{1}}^{W_{2}}-{\frac {V}{F}}dW=\int _{W_{2}}^{W_{1}}{\frac {V}{F}}dW\quad \quad (1)}$
### Specific range
The term ${\displaystyle {\frac {V}{F}}}$, where ${\displaystyle V}$ is the speed, and ${\displaystyle F}$ is the fuel consumption rate, is called the specific range (= range per unit weight of fuel; S.I. units: m/kg). The specific range can now be determined as though the airplane is in quasi steady-state flight. Here, a difference between jet and propeller driven aircraft has to be noticed.
### Propeller aircraft
With propeller driven propulsion, the level flight speed at a number of airplane weights from the equilibrium condition ${\displaystyle P_{a}=P_{r}}$ has to be noted. To each flight velocity, there corresponds a particular value of propulsive efficiency ${\displaystyle \eta _{j}}$ and specific fuel consumption ${\displaystyle c_{p}}$. The successive engine powers can be found:
${\displaystyle P_{br}={\frac {P_{a}}{\eta _{j}}}}$
The corresponding fuel weight flow rates can be computed now:
${\displaystyle F=c_{p}P_{br}}$
Thrust power, is the speed multiplied by the drag, is obtained from the lift-to-drag ratio:
${\displaystyle P_{a}=V{\frac {C_{D}}{C_{L}}}W}$ ; here W is a force in newtons
The range integral, assuming flight at constant lift to drag ratio, becomes
${\displaystyle R={\frac {\eta _{j}}{gc_{p}}}{\frac {C_{L}}{C_{D}}}\int _{W_{2}}^{W_{1}}{\frac {dW}{W}}}$ ; here W is the mass in kilograms, therefore standard gravity g is added. Its exact value depends on the distance to the centre of gravity of earth, but it averages 9.81 m/s2.
To obtain an analytic expression for range, it has to be noted that specific range and fuel weight flow rate can be related to the characteristics of the airplane and propulsion system; if these are constant:
${\displaystyle R={\frac {\eta _{j}}{gc_{p}}}{\frac {C_{L}}{C_{D}}}\ln {\frac {W_{1}}{W_{2}}}}$
### Electric aircraft
An electric aircraft with battery power only will have the same mass at takeoff and landing. The logarithmic term with weight ratios is replaced by the direct ratio between ${\displaystyle W_{battery}/W_{total}}$
${\displaystyle R=E^{*}{\frac {1}{g}}\eta _{total}{\frac {L}{D}}{\frac {W_{battery}}{W_{total}}}}$
where ${\displaystyle E^{*}}$ is the energy per mass of the battery (e.g. 150-200 Wh/kg for Li-ion batteries), ${\displaystyle \eta _{total}}$ the total efficiency (typically 0.7-0.8 for batteries, motor, gearbox and propeller), ${\displaystyle L/D}$ lift over drag (typically around 18), and the weight ratio ${\displaystyle {\frac {W_{battery}}{W_{total}}}}$ typically around 0.3.[1]
### Jet propulsion
The range of jet aircraft can be derived likewise. Now, quasi-steady level flight is assumed. The relationship ${\displaystyle D={\frac {C_{D}}{C_{L}}}W}$ is used. The thrust can now be written as:
${\displaystyle T=D={\frac {C_{D}}{C_{L}}}W}$ ; here W is a force in newtons
Jet engines are characterized by a thrust specific fuel consumption, so that rate of fuel flow is proportional to drag, rather than power.
${\displaystyle F=c_{T}T=c_{T}{\frac {C_{D}}{C_{L}}}W}$
Using the lift equation, ${\displaystyle {\frac {1}{2}}\rho V^{2}SC_{L}=W}$
where ${\displaystyle \rho }$ is the air density, and S the wing area.
the specific range is found equal to:
${\displaystyle {\frac {V}{F}}={\frac {1}{c_{T}}}{\sqrt {{\frac {C_{L}}{C_{D}^{2}}}{\frac {2}{\rho SW}}}}}$
Inserting this into (1) and assuming only ${\displaystyle W}$ is varying, the range (in meters) becomes:
${\displaystyle R={\frac {1}{c_{T}}}{\sqrt {{\frac {C_{L}}{C_{D}^{2}}}{\frac {2}{g\rho S}}}}\int _{W_{2}}^{W_{1}}{\frac {1}{\sqrt {W}}}dW}$ ; here ${\displaystyle W}$ is again mass.
When cruising at a fixed height, a fixed angle of attack and a constant specific fuel consumption, the range becomes:
${\displaystyle R={\frac {2}{c_{T}}}{\sqrt {{\frac {C_{L}}{C_{D}^{2}}}{\frac {2}{g\rho S}}}}\left({\sqrt {W_{1}}}-{\sqrt {W_{2}}}\right)}$
where the compressibility on the aerodynamic characteristics of the airplane are neglected as the flight speed reduces during the flight.
### Cruise/climb
For long range jet operating in the stratosphere (altitude approximately between 11 and 20 km), the speed of sound is constant, hence flying at fixed angle of attack and constant Mach number causes the aircraft to climb, without changing the value of the local speed of sound. In this case:
${\displaystyle V=aM}$
where ${\displaystyle M}$ is the cruise Mach number and ${\displaystyle a}$ the speed of sound. W is the weight in kilograms (kg). The range equation reduces to:
${\displaystyle R={\frac {aM}{gc_{T}}}{\frac {C_{L}}{C_{D}}}\int _{W_{2}}^{W_{1}}{\frac {dW}{W}}}$
where ${\displaystyle a={\sqrt {{\frac {7}{5}}R_{s}T}}}$ ; here ${\displaystyle R_{s}}$is the specific heat constant of air 287.16 ${\displaystyle {\frac {J}{kgK}}}$ (based on aviation standards) and ${\displaystyle \gamma =7/5=1.4}$ (derived from ${\displaystyle \gamma ={\frac {c_{p}}{c_{v}}}}$ and ${\displaystyle c_{p}=c_{v}+R_{s}}$). ${\displaystyle c_{p}}$ and ${\displaystyle c_{v}}$ are the specific heat capacities of air at a constant pressure and constant volume respectively.
Or ${\displaystyle R={\frac {aM}{gc_{T}}}{\frac {C_{L}}{C_{D}}}ln{\frac {W_{1}}{W_{2}}}}$, also known as the Breguet range equation after the French aviation pioneer, Breguet.
## References
• G. J. J. Ruijgrok. Elements of Airplane Performance. Delft University Press.[page needed]ISBN 9789065622044.
• Prof. Z. S. Spakovszky. Thermodynamics and Propulsion, Chapter 13.3 Aircraft Range: the Breguet Range Equation MIT turbines, 2002
• Martinez, Isidoro. Aircraft propulsion. Range and endurance: Breguet's equation page 25. | 2,362 | 8,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 57, "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} | 3.75 | 4 | CC-MAIN-2021-04 | latest | en | 0.920913 |
https://www.conversion-metric.org/volume/cubic_centimeter-to-drop | 1,726,741,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00645.warc.gz | 649,328,619 | 8,620 | Cubic Centimeter to Drop Conversion (cm³ to drop)
1 cm³ = 15.419629805528 drop
Swap » Drop to Cubic Centimeter
cm³: Cubic Centimeter, drop: Drop
Convert Volume Units
How Many Drop in a Cubic Centimeter?
There are 15.419629805528 drop in a cubic centimeter.
1 Cubic Centimeter is equal to 15.419629805528 Drop.
1 cm³ = 15.419629805528 drop
Cubic Centimeter to Drop Conversions
1 cm³ = 15.41963 drop
5 cm³ = 77.098149 drop
2 cm³ = 30.83926 drop
0.5 cm³ = 7.709815 drop
3 cm³ = 46.258889 drop
0.1 cm³ = 1.541963 drop
1000000 cm³ = 15419629.805528 drop
0.136 cm³ = 2.09707 drop
10 cm³ = 154.196298 drop
0.000266 cm³ = 0.004102 drop
0.25 cm³ = 3.854907 drop
25 cm³ = 385.490745 drop
50 cm³ = 770.98149 drop
8.75 cm³ = 134.921761 drop
0.0116 cm³ = 0.178868 drop
4103 cm³ = 63266.741092 drop
4.103 cm³ = 63.266741 drop
228 cm³ = 3515.675596 drop
7 cm³ = 107.937409 drop
90 cm³ = 1387.766682 drop
0.23 cm³ = 3.546515 drop
180 cm³ = 2775.533365 drop
0.025 cm³ = 0.385491 drop
0.003 cm³ = 0.046259 drop
0.06 cm³ = 0.925178 drop
210 cm³ = 3238.122259 drop
75 cm³ = 1156.472235 drop
18 cm³ = 277.553336 drop
0.4 cm³ = 6.167852 drop
250 cm³ = 3854.907451 drop
3.5 cm³ = 53.968704 drop
0.93 cm³ = 14.340256 drop
0 cm³ = 0 drop
30 cm³ = 462.588894 drop
8 cm³ = 123.357038 drop
5772 cm³ = 89002.103238 drop
Cubic Centimeter Definition
A cubic centimeter is a unit of volume accepted and widely used in the countries recognizing the SI standards. This unit is corresponding to the volume equal to the one of a cube with 1 cm each side. Therefore, one cubic centimeter is equal to 1/1 000 000th of a cubic meter. It is also roughly equal to 1 ml, or 1/1000th of cubic liter. The symbol of a cubic centimeter is cc, and this volume unit is very widely used in such fields as medicine, engineering, technology, etc.
Convert Cubic Centimeter
Drop Definition
Used very commonly in medicine and cooking, a drop is a unit of volume defined as the volume of a liquid in one drop. It has been used since the 19th century and is considered one of the most roughly specified units of volume. Undoubtedly, the volume of a drop depends on the method or device used for producing drops.
Convert Drop
This is a very easy to use cubic centimeter to drop converter. First of all just type the cubic centimeter (cm³) value in the text field of the conversion form to start converting cm³ to drop, then select the decimals value and finally hit convert button if auto calculation didn't work. Drop value will be converted automatically as you type.
The decimals value is the number of digits to be calculated or rounded of the result of cubic centimeter to drop conversion.
You can also check the cubic centimeter to drop conversion chart below, or go back to cubic centimeter to drop converter to top.
Cubic Centimeter to Drop Conversion Chart
Cubic CentimeterDrop
1 cm³15.419629805528 drop
2 cm³30.839259611055 drop
3 cm³46.258889416583 drop
4 cm³61.67851922211 drop
5 cm³77.098149027638 drop
6 cm³92.517778833166 drop
7 cm³107.93740863869 drop
8 cm³123.35703844422 drop
9 cm³138.77666824975 drop
10 cm³154.19629805528 drop
11 cm³169.6159278608 drop
12 cm³185.03555766633 drop
13 cm³200.45518747186 drop
14 cm³215.87481727739 drop
15 cm³231.29444708291 drop
16 cm³246.71407688844 drop
17 cm³262.13370669397 drop
18 cm³277.5533364995 drop
19 cm³292.97296630502 drop
20 cm³308.39259611055 drop
21 cm³323.81222591608 drop
22 cm³339.23185572161 drop
23 cm³354.65148552713 drop
24 cm³370.07111533266 drop
25 cm³385.49074513819 drop
26 cm³400.91037494372 drop
27 cm³416.33000474925 drop
28 cm³431.74963455477 drop
29 cm³447.1692643603 drop
30 cm³462.58889416583 drop
31 cm³478.00852397136 drop
32 cm³493.42815377688 drop
33 cm³508.84778358241 drop
34 cm³524.26741338794 drop
35 cm³539.68704319347 drop
36 cm³555.10667299899 drop
37 cm³570.52630280452 drop
38 cm³585.94593261005 drop
39 cm³601.36556241558 drop
40 cm³616.7851922211 drop
41 cm³632.20482202663 drop
42 cm³647.62445183216 drop
43 cm³663.04408163769 drop
44 cm³678.46371144321 drop
45 cm³693.88334124874 drop
46 cm³709.30297105427 drop
47 cm³724.7226008598 drop
48 cm³740.14223066532 drop
49 cm³755.56186047085 drop
50 cm³770.98149027638 drop
Cubic CentimeterDrop
50 cm³770.98149027638 drop
55 cm³848.07963930402 drop
60 cm³925.17778833166 drop
65 cm³1002.2759373593 drop
70 cm³1079.3740863869 drop
75 cm³1156.4722354146 drop
80 cm³1233.5703844422 drop
85 cm³1310.6685334698 drop
90 cm³1387.7666824975 drop
95 cm³1464.8648315251 drop
100 cm³1541.9629805528 drop
105 cm³1619.0611295804 drop
110 cm³1696.159278608 drop
115 cm³1773.2574276357 drop
120 cm³1850.3555766633 drop
125 cm³1927.453725691 drop
130 cm³2004.5518747186 drop
135 cm³2081.6500237462 drop
140 cm³2158.7481727739 drop
145 cm³2235.8463218015 drop
150 cm³2312.9444708291 drop
155 cm³2390.0426198568 drop
160 cm³2467.1407688844 drop
165 cm³2544.2389179121 drop
170 cm³2621.3370669397 drop
175 cm³2698.4352159673 drop
180 cm³2775.533364995 drop
185 cm³2852.6315140226 drop
190 cm³2929.7296630502 drop
195 cm³3006.8278120779 drop
200 cm³3083.9259611055 drop
205 cm³3161.0241101332 drop
210 cm³3238.1222591608 drop
215 cm³3315.2204081884 drop
220 cm³3392.3185572161 drop
225 cm³3469.4167062437 drop
230 cm³3546.5148552713 drop
235 cm³3623.613004299 drop
240 cm³3700.7111533266 drop
245 cm³3777.8093023543 drop
250 cm³3854.9074513819 drop
255 cm³3932.0056004095 drop
260 cm³4009.1037494372 drop
265 cm³4086.2018984648 drop
270 cm³4163.3000474925 drop
275 cm³4240.3981965201 drop
280 cm³4317.4963455477 drop
285 cm³4394.5944945754 drop
290 cm³4471.692643603 drop
295 cm³4548.7907926306 drop | 2,089 | 5,587 | {"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-2024-38 | latest | en | 0.670779 |
https://byjus.com/question-answer/dfrac-3-4-dfrac-5-8-dfrac-11-8-dfrac-12-8-dfrac-13-8/ | 1,643,453,993,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00109.warc.gz | 214,374,034 | 18,367 | Question
# $$\dfrac {3}{4} + \dfrac {5}{8}$$.
A
118.
B
128.
C
138.
D
238.
Solution
## The correct option is A $$\dfrac {11}{8}$$.We have,$$\,\,\,\dfrac{3}{4} + \dfrac{5}{8}$$$$= \dfrac{{6 + 5}}{8}\,\,\,\left[ {L.C.M\,of\,4\& 8 = 8} \right]$$$$= \dfrac{{11}}{8}$$Hence, option (A) is true.Mathematics
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View More | 156 | 356 | {"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} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.486674 |
https://teachics.org/sensors-notes-lessons/rvdt-rotary-variable-differential-transformer/ | 1,611,613,316,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704792131.69/warc/CC-MAIN-20210125220722-20210126010722-00275.warc.gz | 584,293,167 | 34,356 | # RVDT – Rotary Variable Differential Transformer
RVDT (Rotary Variable Differential Transformer) is a passive transducer, that works on the principle of mutual induction. It is used to measure the angular displacement. The design of RVDT is similar to LVDT, except for the design of the core.
## Construction of RVDT
RVDT consists of one primary coil and two secondary coils wounded on a cylindrical core. RVDT uses a cam-type core made up of a ferromagnetic material and it can be twisted among the two windings using the shaft.
The primary winding is connected to an AC source. The two secondary windings S1 and S2 have an equal number of turns and are set up in series opposition.
## Operation of RVDT
The working of RVDT is same as LVDT. When an alternating voltage is applied in the primary windings of an RVDT, an emf is induced in the secondary windings.
Suppose V1 is the voltage induced across coil S1 and V2 is the voltage induced across S2. The overall output voltage across the secondary winding(V0) is the difference between V1 and V2.
So the differential output is
Based on the movement of the shaft, following three conditions will be occured.
• When the core is in the centre position. (NULL Position)
• When the core rotates in a clockwise direction.
• When the core rotates in the anti-clockwise direction.
When the core is placed at the centre position, voltages induced across secondary windings are equal although reversed in phase. That is V1=V2. Then, the resultant voltage V0=0.
When the shaft is rotated in the clockwise direction, then more emf is generated in coil S1. That is V1>V2. Here V0 will have a positive value.
Similarly, anti-clockwise rotation of shaft leads to increase in voltage V1. That is V2>V1. Here V0 will have a negative value.
• Durable.
• Low cost.
• Easy to handle.
• High accuracy.
• Long life.
• Excellent linearity.
## Application of RVDT
• Robotics.
• Modern machine tools.
• The fuel control system of engines.
• Fuel valve as well as hydraulics.
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# How much is 25gb?
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### How much will gamestop buy a 25gb used ps3 slim?
Store credit about \$100 and cash for about \$50 and im guessing you mean 250gb not 25gb
### What is 87 percent of 25GB?
87% of 25GB= 87% * 25= 0.87 * 25= 21.75GB
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### How much data does blu-ray disk hold?
25GB single layer, 50GB double layer
### What is 25GB equal to?
its equal to 25000 mb
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8.5GB Standard DVD 25GB DualLayer
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The next step up in optical media is the 25GB Blu-Ray disc.
### How much data can a BD-R hold?
The format was developed to enable recording, rewriting and playback of high-definition video (HD), as well as storing large amounts of data. The format offers more than five times the storage capacity of traditional DVDs and can hold up to 25GB on a single-layer disc and 50GB on a dual-layer disc. [In short] :- Single Layered:25gb Double Layered: 50gb
### Which optical disc format supports data capacity of 25GB?
Single-sided, single-layer Blu-ray disc.
### How long does it take to burn a 25GB blu ray disc?
It depends on your burner, but it can take over an hour or more on the slower ones.
### What is the capacity of blueray?
single layer blue-ray DVD can have the storage of 25gb where as double layer blue ray DVD can have the storage of 50gb
### How much memory is in a Blu-ray DVD?
A single-layer disc can hold 25GB. A dual-layer disc can hold 50GB. To ensure that the Blu-ray Disc format is easily extendable (future-proof) it also includes support for multi-layer discs, which should allow the storage capacity to be increased to 100GB-200GB (25GB per layer) in the future simply by adding more layers to the discs. | 542 | 2,002 | {"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-2024-18 | latest | en | 0.908833 |
https://acmewriters.com/climate-and-energy-modelling/ | 1,686,076,974,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00753.warc.gz | 97,000,952 | 12,458 | # Climate and Energy Modelling
Please check the attached screen shots(I need experts who are good at those calculations) Course outline, Overview of climate and energy modeling, Basic matrix algebra2 Â?Â?Â? 2021-03-08 2021-03-14Â?Â?Â? Basic concepts of input-output table, techniques of input-output analysis Ronald Miller and Peter Blair, 2009, Input-Output Analysis: Foundations and Extensions ( Global Value Chain (GVC) analysis, Multiregional input-output (MRIO) table, GTAP DB Angel Aguiar, Maksym Chepeliev, Erwin Corong, Robert McDougall, and Dominique van der Mensbrugghe, 2019, The GTAP Data Base: Version 10, Journal of Global Economic Analysis4 Â?Â?Â? 2021-03-22 2021-03-28Â?Â?Â? Basics of micro economics (I): Basic concepts of production functions (Leontief, CES, Cobb-Douglas), profit maximization, utility functions, demand functions, elasticities Micro-economics textbook5 Â?Â?Â? 2021-03-29 2021-04-04Â?Â?Â? Basics of micro economics (II): Basic concepts of production functions (Leontief, CES, Cobb-Douglas), profit maximization, utility functions, demand functions, elasticities Micro-economics textbook6 Â?Â?Â? 2021-04-05 2021-04-11Â?Â?Â? Concepts and mathematical formulation of General Equilibrium Lofgren, Hans, et al., 2002, A Standard Computable General Equilibrium (CGE) Model in GAMS, IFPRI (4.5. – 4.7.) Course withdrawal period 7 Â?Â?Â? 2021-04-12 2021-04-18Â?Â?Â? GAMS coding of static single-region open-economy CGE (Computable General Equilibrium) model GAMS Development Corporation, 2018, GAMS Documentation.For more information on Climate and Energy Modelling read this:https://en.wikipedia.org/wiki/Energy_modeling
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Type of paper | 637 | 2,485 | {"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} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.642754 |
http://www.peterhenderson.co/NER | 1,544,518,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823614.22/warc/CC-MAIN-20181211083052-20181211104552-00568.warc.gz | 444,409,794 | 7,572 | Deep Networks for Named Entity Recognition
Following Socher’s CS224d course @ Stanford, the second part of the assignment requires developing deep networks for the task of named entity recognition (NER). For this we define a two layer network of the form:
We know, $y \in {\mathbb R}^5$ is a one hot vector for various classes of possible named entities (person, organization, location, miscellaneous, or not a named entity). We define the input
as a window of words. Each $x_i$ is a one-hot index into a word vector matrix $L$.
Now we need to compute gradients.
First, let’s compute $\frac{\delta J}{\delta \theta}$. First, the sum is really only relevant for a single one hot vector so we can kind of ignore it a little bit. If you see it missing, that’s why.
And for when $i=j$
Now we move onto other derivatives with the help of our old friend, the chain rule.
Now, we want to avoid parameters exploding or becoming highly correlated, so we need to augment our cost with a gaussian prior as follows. “This tends to push parameter weights closer to zero, without constraining their direction, and often leads to classifiers with better generalization ability.”
Maximizing log likelihood with respect to the Gaussian prior results in a formulated regularization parameter:
The the combined loss function becomes:
Our update gradients will thus include a new term for each respectively:
Now, we want to create something to have random initializations to try to avoid local minima. Apparently, the following equation has been found to work well. For a matrix $A$ of dimension $m \times n$, select values $A_{ij}$ uniformly in range $[-\epsilon, \epsilon]$. Where:
In code we can implement this as a function as seen here.
Written on August 10, 2015 | 389 | 1,762 | {"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} | 2.59375 | 3 | CC-MAIN-2018-51 | longest | en | 0.893126 |
http://learneconomicsonline.com/maxmaths.php | 1,544,857,470,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00303.warc.gz | 164,928,083 | 6,290 | LearnEconomicsOnline
Home maxmaths
Maximisation Maths Required Knowledge: Differentiation We can use differentiation to find the rate of change which has useful applications in economics. For example to find something that is marginal (one more) we can differentiate. If you are not familiar with differentiating please see this lesson here. Finding the Marginal So to find the marginal cost when total cost = 6q2 + 7q + 5 we would differentiate to find dTC/dq = 12q + 7, therefore MC = 12q + 7. If asked to find the marginal cost when quantity = 5, then we would differentiate the total costs and substitute q = 5. This is the same as if we wanted to find marginal profit or marginal revenue. Total Revenue Total Revenue = Price*Quantity For example, find the total revenue when p=200-2q and q=2. P=196, Q=2, therefore TR = 196*2 = 392. Profit Maximisation We know that profit maximisation occurs when the different between total revenue and total costs is at its greatest. This is the same as the output level when marginal cost equals marginal revenue (MR=MC). If given marginal revenue and marginal cost to find maximum profit (denoted by ∏) we would set them equal. Example 1 For example MR = 70-10q and MC = 25q, find the maximum profit. 70-10q = 25q 70 = 35q q=2 Therefore the output level at which profit is maximised is 2. We can calculate the profits at this level by substituting q into ∏=TR-TC. Example 2 p = 10 - 4q TC = 15+q2 Calculate the output required to maximise profit and find the profit level at this point. TR = (10-4q)*q = 10q-4q2 TC = 15+q2 ∏= TR-TC = (10q-4q2)-(15+q2) = 10q-15-5q2 Maximum profit occurs at MR = MC MR = dTR/dq = 10-8q MC = dTC/dq = 2q 10-8q = 2q 10 = 10q q = 1 Therefore profit is maximised at an output level of 1. To find the profit level at this point we will substitute q=1 back into (10q-15-5q2) which is TR-TC 10-15-5 = -10. Therefore the maximum profit the firm can make is -10 at an output level of 1. NB: As highlighted in this example it is important to remember that a firm doesn't always make a profit even if it is trying to maximise profits, although the profit in the example above is negative, this is the best outcome, if it had produced an output level that wasn't 1 it would have made even less money. Maximising other Functions Alternatively firms may not be trying to maximise profit and may instead be trying to maximise total revenue. To calculate the quantity they should produce in this case we would find dTR/dq and set this equal to zero (mathematically we are looking for a turning point, as this would be the maximum or the minimum, we do this by differentiating and setting the differential to 0). We can then re-arrange the equation to calculate q. We may also be asked to prove that this is a maximum and not a minimum, to do this find the second differential (d2y/dx2) and substitute q in, if the answer is positive then we have a minimum, and if the answer is negative then we have a maximum. Example For example, TR=200q-0.5q2 find the output level needed in order to maximise TR and prove that this is a maximum. dTR/dq = 200-q 200 - q = 0 q = 200 d2TR/dq2 = -1; Negative therefore our point must be a maximum Hence an output of 200 units should be produced in order to maximise total revenue and we have proven that this will maximise it, and not minimise it! Questions 1. Find an equation for the marginal cost when TC = 4q3 + 7q2 + 2 2. Find the marginal cost when TC = 12q2 + 14q + 2 when q=3 3. Find the marginal revenue when TR = 6q2 + 5 and q=1 and an equation for marginal revenue when TR=6q2 + 5 4. Find the total revenue, in terms of q, when p=6q2 + 9q + 7 5. Find the marginal revenue, in terms of q, when p=5q + 6 6. Calculate the maximum profit and the quantity produced when p = 300-3q and TC = 15+2q2. 7. Find the quantity needed to maximise total revenue when TR = 100q-5q2 and hence prove this is a maximum Answers 1. TC = 12q2 + 14q 2. MC = 86 3. MR = 12, MR = 12q 4. TR= q(6q2 + 9q + 7) => 6q3 + 9q2 + 7q 5. MR = 10q + 6 (first of all calculate the total revenue by multiplying p by q and then differentiate to find the marginal revenue) 6. Maximum profit is £2485 at an output level of q=10 7. q=10, d2TR/dq2 = -10, negative therefore maximum point Page last updated on 20/10/13 | 1,237 | 4,295 | {"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} | 4.5625 | 5 | CC-MAIN-2018-51 | longest | en | 0.882971 |
https://crypto.stackexchange.com/questions/25440/how-to-calculate-if-probability-is-negligible-or-not | 1,638,991,368,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00005.warc.gz | 260,206,218 | 36,106 | # How to calculate if probability is negligible or not
If i have probability $Q = 2C(A\times B)$ where $A$ and $B$ are unknown probabilities and $C$ is a non-negligible probability, what can i speculate about probability $Q$ and how can i calculate bounds on probabilities $A$ and $B$ in order to make this value non-negligible?
• How is this related to cryptography? Seems like a straight math question. May 5 '15 at 15:39
• I tried to post in math exchange but apparently the question wasn't 'formulated correctly' and I could not get it to post. It is for a crypto proof. May 5 '15 at 16:52
• I suggest you go back to math exchange and play with the markdown formatting until it posts. This will get closed here as Off Topic. May 5 '15 at 16:58
• ok, thanks, ill delete it, but already spend half an hour trying to get it to post! just have to struggle away on my own... May 5 '15 at 17:09
• I'm voting to close this question as off-topic because it is not about cryptography as described in our help-center. (It seems to be more fitting for Math.SE) May 6 '15 at 0:24
First of all, keep in mind that all meaningful probabilities must be between $0$ and $1$. In particular, this means that, if $X$ and $Y$ are probabilities, $0 \le XY \le \min(X, Y)$.
In cryptography, a probability is considered "negligible" if it is very small. What actually counts as "negligible" depends on context, but typically, we're talking about probabilities on the order of $1/2^{128}$ or less.
Conversely, a "non-negligible" probability may be high, perhaps even close to $1$. In fact, since we typically want probabilities (of an attack succeeding) to be small, we may safely assume that, in the worst case, all non-negligible probabilities are $\approx 1$. This leads to the following "rules":
• The product of two negligible probabilities is (always) negligible.
• The product of a negligible and a non-negligible probability is also (always) negligible.
• The product of two non-negligible probabilities is (usually) not negligible.
• Corollary: For the product of several probabilities to be negligible, at least one of them must (usually) be negligible.
• Multiplying a probability by a constant $\approx 1$ (say, between $1/2^{16}$ and $2^{16}$) does not (usually) change its negligibility.
• The sum of two negligible probabilities is (usually) negligible.
• The sum of a negligible and a non-negligible probability (or of two non-negligible probabilities) is never negligible (and might not always, strictly speaking, be a probability).
Thus, without knowing anything about $A$ and $B$, except that they are probabilities, the most we can say about $Q = 2C(AB)$ is that it is somewhere between $0$ and $2C$. Furthermore if $C$ is not negligible, then $2C$ is definitely not negligible; thus, for $Q$ to be negligible, $AB$ must (generally) be negligible.
For $AB$ to be negligible, it's sufficient for either $A$ or $B$ to be negligible. If neither $A$ nor $B$ is negligible, their product cannot be assumed to be negligible either — although it's possible for $AB$ to be negligible (e.g. $AB \approx 1/2^{128}$) even if both $A$ and $B$ are only "nearly negligible" (e.g. $A \approx B \approx 1/2^{64}$).
Conversely, for $Q = 2C(AB)$ not to be negligible, neither $A$ nor $B$ may be negligible. (Strictly speaking, for any specific threshold of negligibility, the factor of $2 > 1$ could bump a probability just below the line to just above it; but the whole point of using terms like "negligible" is to note that we're speaking loosely and usually ignoring such factors of $\approx 1$.) It's still possible for all of $A$, $B$ and $C$ to be above our negligibility threshold, but for their product to be below it (if two or more of them are "almost negligible"), but we cannot assume this without knowing their values more precisely.
• This treatment is a little strange: I think using an asymptotic definition for negligible probabilities would lead to a more precise discussion.
– Reid
May 5 '15 at 19:27
• It's weird that the sum of one nn and on n probability is nn while for this for the product does not hold. May 5 '15 at 22:15
• Negligible probability is not a fixed threshold but a function, which has to decrease faster than any polynomial. Most commonly this references to exponential decrease, but it also could be $\frac{1}{g(x)}$, where $g(x)$ is super-polynomial. If specific thresholds for probability are mentioned, most commonly $\frac{1}{2^{80}}$ is considered the upper limit of improbable.
– tylo
May 6 '15 at 8:59
To be negligible is not a property of fixed numbers, it is a property of functions of some parameter. It does not make sense to talk about a probability being negligible unless you give a parameter relative to which the probability is negligible. In cryptography we will often consider probability being negligible in the security parameter of the given scheme. So if something is stated to be negligible without specifying what it is negligible in, it is usually taken to be the security parameter.
Formally a function $f(x)$ is said to be negligible in $x$ if for all polynomials $g$ there exists a $k$ so that $|f(x')| < 1/|g(x')|$ for all $x' > k$.
Now to answer your question you first have to think of probabilities $Q$, $A$, $B$ and $C$ as functions of some parameter $x$. Now what you can say about $Q$, given that $C$ is non-negligible in $x$, is that $Q$ can only be non-negligible if both $A$ and $B$ are non-negligible.
Why? Well say $B$ is negligible in $x$. Then since $C$ and $A$ are probabilities we know they are upper-bounded by 1, so we know that $$Q(x) = 2C(x)(A(x)B(x)) \leq 2B(x)$$ Now since $B$ is negligible in $x$ for all polynomials $g$ we know there exists a $k$ so that $B(x') < 1/g(x')$ for all $x' > k$. Since $g'(x) = \frac{1}{2}g(x)$ is also a polynomial if $g$ is a polynomial we can conclude that $Q$ will also be negligible. | 1,554 | 5,908 | {"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} | 3.9375 | 4 | CC-MAIN-2021-49 | latest | en | 0.963265 |
https://www.jiskha.com/display.cgi?id=1365647170 | 1,516,413,873,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00105.warc.gz | 910,522,795 | 4,149 | Math
posted by .
An ordered triple of real numbers
(a,b,c) is called friendly, if each
number is equal to the product of
the other 2. How many friendly
triples are there?
Details and assumptions
The numbers need not be equal to
each other.
The numbers need not to be
pairwise distinct (which
means that no two of them
are the same). I got 31 but it isn't
right ...
• Math -
brilliant na??
• ma, ththth -
stuck on that one also :((
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https://au.mathworks.com/matlabcentral/cody/problems/541-use-of-regexp/solutions/1628545 | 1,596,848,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00229.warc.gz | 216,910,680 | 16,104 | Cody
Problem 541. Use of regexp
Solution 1628545
Submitted on 15 Sep 2018 by HH
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
x = 'I played piano. John played football. Anita went home. Are you safe?'; y = {'I played piano.' 'Anita went home.' 'Are you safe?'}; assert(isequal(lazy(x),y))
ans = 1×3 cell array {'I played piano.'} {'Anita went home.'} {'Are you safe?'}
2 Pass
x = 'Are you okay? Who are you? Olga will call you. Sam saw me.'; y = {'Olga will call you.'}; assert(isequal(lazy(x),y))
ans = 1×1 cell array {'Olga will call you.'}
3 Pass
x = 'One is more. Than what? No it''s not. But why? Angela said so.'; y = {'One is more.' 'Angela said so.'}; assert(isequal(lazy(x),y))
ans = 1×2 cell array {'One is more.'} {'Angela said so.'}
4 Pass
x = 'One plus two. Is four. No, that''t not right. It''s three.'; y = {'One plus two.' 'It''s three.'}; assert(isequal(lazy(x),y))
ans = 1×2 cell array {'One plus two.'} {'It's three.'}
5 Pass
x = 'I went home. After the game. It was sad. It was lame. It was great!'; y = {'I went home.' 'After the game.' 'It was lame.'}; assert(isequal(lazy(x),y))
ans = 1×3 cell array {'I went home.'} {'After the game.'} {'It was lame.'}
6 Pass
x = 'One, two, three. Climb the tree. Four, five, six. It''s not here. Eight and nine. That''s fine.'; y = {'One, two, three.' 'It''s not here.' 'Eight and nine.'}; assert(isequal(lazy(x),y))
ans = 1×3 cell array {'One, two, three.'} {'It's not here.'} {'Eight and nine.'}
7 Pass
x = 'Either one is fine. Why? Because he said so.'; y = {'Either one is fine.'}; assert(isequal(lazy(x),y))
ans = 1×1 cell array {'Either one is fine.'} | 587 | 1,762 | {"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} | 3.96875 | 4 | CC-MAIN-2020-34 | latest | en | 0.878478 |
tecnoteamwork.it | 1,620,370,042,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00082.warc.gz | 574,114,411 | 11,059 | ### Best impulse responses 2020
However, the truth of the inverse and converse of a statement are logically unrelated to the truth of the initial statement. Consider the true mathematical statement, "if a figure is a square, then it is a rectangle," which has false converse and inverse statements. The Euler Diagram below represents the statement if A, then B. best lighting for woodworking shop ... you can make. Come get inspired with a variety of types of wood planters. ... Head over to Houseful Of Handmade for the tutorial and free plans. ...
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For multiple-choice and true/false questions, simply press or click on what you think is the correct answer. For fill-in-the-blank questions press or click on the blank space provided. If you have difficulty answering the following questions, learn more about this topic by reading our Balance Sheet (Explanation). Q. Alex goes into the garden and digs up a shovel full of dirt. This is a _____ mixture.
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Handling BLANK in Boolean expressions. There is a particular behavior when a column defined as Boolean data type is involved in an expression. Former versions of DAX allowed BLANK results from a Boolean expression, whereas the more recent versions (Power BI/Excel 2016/SSAS Tabular 2016) only return TRUE or FALSE from a logical expression such as IF.
An if statement can be followed by an optional else statement, which executes when the Boolean expression is false.. Syntax. The syntax of an if...else statement in C programming language is − True False 1. True/False – Mark “T” for statements you believe are true, and “F” for statements you believe are false. _____ Reliability refers to the consistency of assessment results. _____ The best time to consider the nature of assessment activities for a unit is during the unit.
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Mar 15, 2018 · We have previously seen that accuracy can be largely contributed by a large number of True Negatives which in most business circumstances, we do not focus on much whereas False Negative and False Positive usually has business costs (tangible & intangible) thus F1 Score might be a better measure to use if we need to seek a balance between ...
True or False: Usually, people are able to cash checks for free or for a much reduced fee (less than a few dollars a month) at the bank where they have a savings or checking account. A. True B. False —False Religions — “For false Christs and false prophets shall rise, and shall shew signs and wonders, to seduce, if it were possible, even the elect.” — Mark 13:22. Religion is the WORST thing that has ever happened to this world; Countless Billions of souls have been doomed to H ell fire by false religion. Learn what the Bible has ...
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Jan 07, 2015 · So im trying to create a quiz using c++ that incorporates three different types of questions. These are the standard one answer question, true false questions and multiple choice questions. Can anyone tell me how to create the true false and multiple choice questions? One way that i came up with looks like this Question A answer1 B answer2;
following may be taken as the official definition of ‘statement’. A statement is a declarative sentence , which is to say a sentence that is capable of being true or false . The following are examples of statements. it is raining I am hungry 2+2 = 4 God exists On the other hand the following are examples of sentences that are not statements. Recall: The only way for p → q to be false is if we know that p is true but q is false. Rationale: If p is false, p → q doesn't guarantee anything. It's true, but it's not meaningful. If p is true and q is true, then the statement “if p is true, then q is also true” is itself true. If p is true and q is false, then the statement “if
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Verbal Ability questions and answers with explanation for interview, competitive examination and entrance test. Fully solved examples with detailed answer description, explanation are given and it would be easy to understand.
May 29, 2018 · False ANSWER: True 21. _____ specialize in basic processes such as the nervous system, sensation and perception, learning and memory, thought, motivation, and emotion. a. Forensic psychologists b. It is a false statement since whole numbers belong to the sets of integers and rational numbers, but not to the set of natural numbers. Example 7 : Classify the number zero, 0 . Definitely not a natural number but it is a whole , an integer , a rational , and a real number.
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Perjury, False Statements, and Obstruction of Justice. Perjury; Perjury, criminalized at 18 U.S.C. § 1621, is perhaps the most recognizable law against lying. The statute makes it a crime to “willfully and contrary to [an] oath state[] or subscribe[] any material matter which he does not believe to be true.”
false positive as variable; specificity. false positive as variable; Because false negatives (spam goes to the inbox) are more acceptable than false positives (non-spam is caught by the spam filter) Fraudulent transaction detector (positive class is "fraud"): Optimize for sensitivity. FN as a variable | 1,269 | 5,539 | {"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-2021-21 | latest | en | 0.858074 |
https://www.numbersaplenty.com/3348 | 1,660,106,424,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00696.warc.gz | 811,561,658 | 3,678 | Search a number
3348 = 223331
BaseRepresentation
bin110100010100
311121000
4310110
5101343
623300
712522
oct6424
94530
103348
112574
121b30
1316a7
141312
15ed3
hexd14
3348 has 24 divisors (see below), whose sum is σ = 8960. Its totient is φ = 1080.
The previous prime is 3347. The next prime is 3359. The reversal of 3348 is 8433.
Added to its reverse (8433) it gives a triangular number (11781 = T153).
It is a super-2 number, since 2×33482 = 22418208, which contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (18).
Its product of digits (288) is a multiple of the sum of its prime divisors (36).
It is a plaindrome in base 10.
It is a nialpdrome in base 15.
It is a self number, because there is not a number n which added to its sum of digits gives 3348.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3343) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 93 + ... + 123.
23348 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 3348, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (4480).
3348 is an abundant number, since it is smaller than the sum of its proper divisors (5612).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3348 is a wasteful number, since it uses less digits than its factorization.
3348 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 44 (or 36 counting only the distinct ones).
The product of its digits is 288, while the sum is 18.
The square root of 3348 is about 57.8619045660. The cubic root of 3348 is about 14.9598928567.
Adding to 3348 its reverse (8433), we get a triangular number (11781 = T153).
It can be divided in two parts, 33 and 48, that added together give a 4-th power (81 = 34).
The spelling of 3348 in words is "three thousand, three hundred forty-eight". | 638 | 2,231 | {"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} | 3.453125 | 3 | CC-MAIN-2022-33 | longest | en | 0.917006 |
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# Food Num. of calories per kilogram Num. of grams of protein
Author Message
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Joined: 30 Mar 2006
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Food Num. of calories per kilogram Num. of grams of protein [#permalink]
### Show Tags
09 May 2006, 16:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Food Num. of calories per kilogram Num. of grams of protein per kilo
S 2,000 150
T 1,500 90
36. The table above gives the number of calories and grams of protein per kilogram of food S and T. If a total of 7 kilograms of S and T are combined to make a certain food mixture, how many kilograms of food S are in the mixture?
(1) The mixture has a total of 12,000 calories.
(2) The mixture has a total of 810 grams of protein
I REALLY NEED A GOOD EXPLAINATION ON THIS ONE.
_________________
Andre Crompton
andre.crompton@cit.com
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### Show Tags
09 May 2006, 18:06
andrecrompton wrote:
Food Num. of calories per kilogram Num. of grams of protein per kilo
S 2,000 150
T 1,500 90
36. The table above gives the number of calories and grams of protein per kilogram of food S and T. If a total of 7 kilograms of S and T are combined to make a certain food mixture, how many kilograms of food S are in the mixture?
(1) The mixture has a total of 12,000 calories.
(2) The mixture has a total of 810 grams of protein
I REALLY NEED A GOOD EXPLAINATION ON THIS ONE.
i. using trail and error, s = 3 and t = 4
= 3x2000+4x1500 = 12,000
i. using the similar approach also s = 3 and t = 4
= 3x150 + 4x 90 = 810
no any other value for s and t satisfy the above constraoints...
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### Show Tags
09 May 2006, 19:51
Assuming s kg of food S are used, then T would be (7-s)kg.
St1:
2000s + 1500(7-s) = 12,000
We can solve for s to get the amt of food S used. Sufficient.
St2:
150s + 90(7-s) = 810.
Similarly, we can solve for s to get the amount of food S used. Sufficient.
Ans D
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### Show Tags
10 May 2006, 10:15
Using (1)
2000x + 1500y = 12,000
x + y = 7
We can get x and y, hence sufficient.
Using (2)
Same way we can get x and y. So sufficient
Hence D
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### Show Tags
11 May 2006, 00:15
Hi,
Agree with D, because:
Statement 1 is sufficient: suppose x is number of kg of S and y is number of kg of K
2000*x+1500*y=12000
x+y=7, combining both equation we can find x and y
Statement 2 is sufficient: 150*x+ 90*y=810, x+y=7. We can find x.
11 May 2006, 00:15
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Showing posts from 2014
### Multivibrator as a state machine
It's pretty hard to imagine, but a multivibrator is one of the most common terms used in the electronic world. How many times did you use a PWM signal to control the rotational speed of a DC motor? Did you ever think about a CPU clock generator? Well, a PWM signal as well as the CPU clock generator belongs to the class of a multivibrator. In this blog post, you will see that a multivibrator is a square wave generator modeled by a state machine. Multivibrator classes There are three types of multivibrators. Classification is related to a state stability. But, before I continue with the classification, I would like to say a few words about nature of a multivibrator. A multivibrator is a state machine with two discrete states and a two transitions in between. Generally speaking, with a two states and a two transitions we are able to construct a three different types of a multivibrators independent of an implementation: monostable multivibrator bistable multivibrator and a
### Monostable multivibrator as analog differential motion controller
Thinking about robots is the same as thinking about microcontrollers or some other digital computing device able to control the motion. That is correct and pretty logic way of thinking. But, the motion could be also controlled with an analog device. What we need is a square wave generator able to control an H-Bridge circuits which drive a DC motor. This simple example demonstrates a monostable multivibrator as the analog device able to control one set of the differential motion kinematics. Differential robot Before we start with a monostable multivibrator , I would like to say something about a differential robot. The differential robot belongs to a class of simple robotic system. It has two standard DC motors (with two wires), each connected to an H-Bridge device controlled by a microprocessor or analog device, what we do have in this example. Kinematics When both DC motors rotates in the same direction, the robot goes forward or backward. When one DC motor rotates in
### PWM Generator for Standard DC Motors
This example is related to simple PWM generator for standard DC motor. Complete example is based on my previous example related to Injection System Behavior . PWM modulation signal is generated by transitions through state machine states. State machine has four states: Idle, PWM, Timer and Direction. Transition matrix has 17 bytes. x2Mode is used for state encoding. Provided material for download contains two bin files related to different behavior of the system. DCSpeedController.bin - rotation in one direction with speed regulation. DCSpeedDirectionController.bin - rotation in both direction with speed regulation. Electronics #1: beside priority encoder for external commands encoding, interface between 8051 MCU and DC motor is done over device driver BA12004 IC (replacement: ULN2XXX series - motor driver kit). To create H-Bridge for bidirectional rotation, two additional resistors are used. Electronics #2: beside priority encoder for external commands encodi
### Stepper Motor Controller
This stepper motor controller is base on my previous example related to Injection System Behavior . As You can see from related YouTube video, stepper motor behavior could be easily changed by injection into the generic state machine. Injection is done over RS232 and USART terminal. Version: 1.0.0.0 Provided material for download contains several bin files related to stepper motor behavior. ContinuousRotateLeft.bin - transition matrix - stepper behavior #1 ContinuousRotateLeftAngle.bin - transition matrix - stepper behavior #2 ContinuousRotateRight.bin - transition matrix - stepper behavior #3 StepByStepRotateLeft.bin - transition matrix - stepper behavior #4 StepByStepRotateRight.bin - transition matrix - stepper behavior #5 For more details please look at Injection System Behavior example. Electronics: beside priority encoder for external commands encoding, interface between 8051 MCU and stepper motor is done over device driver BA12004 IC (re
### Injection System Behavior
This simple example present ability of the system modeled by state machine to execute different tasks by injecting behavior. It is based on 8 bit field transition matrix which stores information about system behavior. If we change transition matrix, system behavior will be changed also. Current example provides only 4 states (A, B, C and D) and 4 external commands (interrupts, events or signals). This example present generic solution for all systems modeled by state machine. Version 1.0.0.0 Usage examples: 1) Embedded Robotics: external commands are signals from sensors: IR sensors, mechanical sensors, ultrasonic sensors and so on. 2) Human - Machine games. 3) Low level driver / firmware software etc. Generally, this idea could be used for each system which includes component able to store information about states (like micro-controller unit). This example is written in ANSI C programming language and supports all hardware platforms powered by MikroEle | 1,021 | 5,125 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.939169 |
https://examsfunda.in/view-questions.php?id=UXVhbnRpdGF0aXZlIEFwdGl0dWRlIC0gSUk=&cid=OA== | 1,553,102,028,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.64/warc/CC-MAIN-20190320170159-20190320192159-00374.warc.gz | 480,730,809 | 11,180 | Directions - Solve the following questions -
1. A rectangular garden is 100 m x 80m. There is a path along the garden and just outside it. Width of the path is 10 m. The area of the path is
a) 1900 sq m
b) 2400 sq m
c) 3660 sq m
d) 4000 sq m
2. A dealer offered a machine for sale for Rs. 27500 but even if he had charged 10% less, he would have made a profit of 10%. The actual cost of the machine is
a) Rs. 22000
b) Rs. 24250
c) Rs. 22500
d) Rs. 22275
3. An employer reduces the number of employees in the ratio 8 : 5 and increases their wages in the ratio 7 : 9. As a result, the overall wages bill is
a) increased-in the ratio 56 : 69
b) decreased in the ratio 56 : 45
c) increased in the ratio 13 : 17
d) decreased in the ratio 17 : 13
4. The average age of a jury of 5 is 40. If a member aged 35 resigns and a man aged 25 becomes a member, then the average age of the new jury is
a) 30 yr
b) 38 yr
c) 40 yr
d) 42 yr
5. With average speed of 40 km/h, a train reaches its destination in time. If it goes with an average speed of 35 km/h, it is late by 15 min. The total journey is
a) 30 km
b) 40 km
c) 70 km
d) 80 km | 381 | 1,130 | {"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-2019-13 | latest | en | 0.92194 |
https://meshthefont.xyz/?post=9229 | 1,624,064,004,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00127.warc.gz | 363,864,126 | 4,984 | # What is exothermic energy
## Exothermic reaction
As exothermic is a chemical reaction in which energy, e.g. B. in the form of heat, is released into the environment. The opposite is the endothermic reaction. In the case of an exothermic reaction, the so-called enthalpy of reaction \$ \ Delta_ \ mathrm {R} H \$ is negative.[1] The enthalpy \$ H \$ is the sum of the internal energy of a system and the product of pressure and volume. It is the heat content of a system at constant pressure.[2][3]
The term exothermic is not with the term exergon to be confused (see: Demarcation).
Typical exothermic reactions are:
• Fire (combustion) as well
• Setting (= hardening) of concrete.
The mixing of substances (heat of the mixture) or the adsorption and absorption of substances on activated carbon or zeolites, for example, are often exothermic, albeit to a much lesser extent.
\$ \ Delta H \$ denotes the difference between the enthalpies of the final (\$ H_2 \$) and starting materials (\$ H_1 \$), i.e. the recorded Energy, applies to exothermic reactions \$ \ Delta H = H_ {2} - H_ {1} <0 \$.
In physics, too, a nuclear reaction that releases energy is called exothermic. An exothermic nuclear fusion is the burning of hydrogen, for example, as it happens in the sun.
### procedure
The starting materials are initially in a metastable state. By briefly supplying a certain amount of energy, the activation energy (activation enthalpy), the system is lifted into the unstable state. Activation sets the reaction in motion and runs independently without any additional energy input. In the overall balance, the chemical system gives off energy to the environment; it is called the enthalpy of reaction. The products are now in a stable condition. For the stability of systems see also system properties.
An example of adding activation energy is igniting a fire by rubbing a match on the friction surface or, in the case of gases, with an electric spark.
Legend:\$ \! \ H: = \ text {enthalpy} \$\$ \! \ \ Delta ^ {\ ddagger} H: = \ text {Activation enthalpy} \$\$ \! \ \ Delta_ \ mathrm {R} H: = \ text {enthalpy of reaction} \$left: initial state of the starting materials: metastablemiddle: transition state of the activated complex: unstableright: final state of the products: stable
Example: Carbon burns with the oxygen in the air, generating heat to form carbon dioxide: \$ \ mathrm {C + O_2 \ \ xrightarrow {\ bigtriangledown} \ CO_2 + W \ ddot arme} \$.
The enthalpy of reaction (enthalpy difference) \$ \ Delta H \$ of this reaction is negative. It can be calculated from the standard enthalpies of formation.
If the activation energy is very low, the reaction can be set in motion without additional external energy input. The necessary activation energy is withdrawn from the environment. The reaction takes place exergonically (spontaneously).
### Demarcation
If one speaks of an exothermic process, no statement is made about whether a reaction takes place voluntarily. A distinction is made here between exergonic and endergonic reactions.
### Individual evidence
1. ↑ PAC, 1996, 68, 149 (A glossary of terms used in chemical kinetics, including reaction dynamics [IUPAC Recommendations, 1996]), page 165.
2. ^ PAC, 1996, 68, 957 (Glossary of terms in quantities and units in Clinical Chemistry [IUPAC-IFCC Recommendations, 1996]), page 972.
3. ^ PAC, 1990, 62, 2167 (Glossary of atmospheric chemistry terms [Recommendations, 1990]), page 2187. | 853 | 3,479 | {"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-2021-25 | latest | en | 0.913616 |
https://educationwithfun.com/course/view.php?id=10§ion=8 | 1,718,658,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00304.warc.gz | 206,514,304 | 16,759 | ## Topic outline
• ### Comparison of Numbers (Greater Than – Less Than – Equal To)
• Comparison of Numbers
Greater Than – Less Than – Equal To
Greater Than
Let’s observe this.
Here are 6 monkeys and 5 trees.
If each monkey sits on a tree, then one monkey is left without a tree.
So, 6 is more than 5 or we can say, 6 is greater than 5.
We write, 6>5
Here, ‘>’ is the ‘Greater than’ sign
The number on the left is greater than the number on the right; 6 > 5
Less Than
Let’s observe this.
Here are 5 kids and 7 Choco pies.
If each eats up 1 Choco pie, 2 Choco pies remain.
So, 5 is less than 7.
We write, 5<7
Here, ‘<’ is the ‘Less than’ sign
The number on the right is greater than the number on the left; 5>7
Equal To
Let’s observe this.
There are 8 girls and 8 candies.
If each gets 1 candy, then we are left with no extra candy. This means that the number of girls is equal to the number of candies.
We write, 8=8.
Here, ‘=’ is the ‘Equal to’ sign
The number on the right is equal to the number on the left; 8=8
Things to Remember
1. To remember which way around the "<" and ">" signs go, just remember:
a. BIGGER number > smaller number
b. Smaller number < BIGGER number
c. The “small” end always points to the smaller number and “open” end always points to the bigger number.
2. When one value is bigger than another we use a "greater than" sign i.e. ‘>’.
Example: 8>3
3. When one value is smaller than another we use a "less than" sign i.e. ‘<’.
Example: 5<7
4. When two values are equal we use the "equals" sign i.e. ‘=’.
Example: 4=4 | 463 | 1,583 | {"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} | 4.59375 | 5 | CC-MAIN-2024-26 | latest | en | 0.847268 |
https://forums.adobe.com/thread/439582 | 1,526,931,468,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00193.warc.gz | 574,184,985 | 30,058 | 9 Replies Latest reply on May 29, 2009 1:43 PM by kentgbaker
# Odd or Even Page selection
Is there a typical way to define all odd pages in a document, as well as all even.
I have scripts (with help from these forums) that do page rotations on the current page
example:
var crtpage = this.pageNum
var crtpagerotate = this.getPageRotation({nPage:crtpage});
try {
this.setPageRotations({nStart:crtpage, nEnd:crtpage, nRotate:(90+crtpagerotate)%360});
}
catch(e)
{
}
But I would like to be able to rotate all odd pages 90, and all even pages 270
Any help is appreciated
• ###### 1. Re: Odd or Even Page selection
Here's one way you can determine if the page number is odd or even:
((myPageNumber/2) == parseInt(myPageNumber/2))
This will return true for even pages and false for odd pages. Rotate 270 if true, or 90 if false.
• ###### 2. Re: Odd or Even Page selection
You could also use the "%" modulos operator to obtain the remainder of the division by 2 (modulos) of 1 (odd page) or zero (even page).
if(this.pageNum % 2) {
// result true not 1 so even page
app.alert("even - this.pageNum % 2: " + (this.pageNum % 2) );
} else {
// result false 1 so odd page
app.alert("odd - this.pageNum % 2: " + (this.pageNum % 2) );
}
So one could even use this result to either add 180 or 0 degree rotation adjusment.
if(this.pageNum % 2) {
// result true not 1 so even page
app.alert("even - this.pageNum % 2: " + (this.pageNum % 2) );
} else {
// result false 1 so odd page
app.alert("odd - this.pageNum % 2: " + (this.pageNum % 2) );
}
var pRotate = 90;
pRotate += ((this.pageNum % 2) % 2) * 180;
app.alert("rotation: " + pRotate);
1 person found this helpful
• ###### 3. Re: Odd or Even Page selection
great,
but I need to have the rotations happen on all pages of the document. I can use setPageRotations and define only the nEnd to get to all pages.......but how to get one rotation if odd, and a different rotation if even.
Is there someway to loop through pages? then do the check for odd or even, then move on to the next page
• ###### 4. Re: Odd or Even Page selection
Yes, you can use the JavaScript "for() {" loop control and use the "numPages" of the doucment object for the control value for the number of pages in the PDF.
var pRotate; // degrees for page rotation
for(i = 0; i < this.numPages, i++) {
pRotate = 90 + ((this.pageNum % 2) * 180);
this.setPageRotations({nStart: i, nEnd: i, nRotate: pRotate });
} // end page loop
1 person found this helpful
• ###### 5. Re: Odd or Even Page selection
Sorry,
console reports missing ; after for-loop condition, and I can't figure out what it needs.
• ###### 6. Re: Odd or Even Page selection
sorry, it was only the this.numPages;
• ###### 7. Re: Odd or Even Page selection
here is where I am, stop me if this is the long way:
for(i = 0; i < this.numPages; i++)
{
if(this.pageNum % 2) {
var crtpagerotate = this.getPageRotation({nPage:?});
this.setPageRotations({nStart: ?, nEnd: ?, nRotate:(270+crtpagerotate)%360});
}
else {
var crtpagerotate = this.getPageRotation({nPage:?});
this.setPageRotations({nStart: ?, nEnd: ?, nRotate:(90+crtpagerotate)%360});
}
}
notice the nStart: ?, nEnd: ?, nPage:? this needs to be the page the loop is currently executing. Is this possible?
• ###### 8. Re: Odd or Even Page selection
the variable 'i' will be your page number, so this should be what you want (untested)
for(i = 0; i < this.numPages; i++)
{
if(i % 2) {
var crtpagerotate = this.getPageRotation({nPage:i});
this.setPageRotations({nStart: i, nEnd: i, nRotate:(270+crtpagerotate)%360});
}
else {
var crtpagerotate = this.getPageRotation({nPage:i});
this.setPageRotations({nStart: i, nEnd: i, nRotate:(90+crtpagerotate)%360});
}
}
• ###### 9. Re: Odd or Even Page selection
Thanks everybody. I had used i as the variable at one point and could not get it to work. I will blame it on Friday. Thanks very much Mark. | 1,151 | 3,914 | {"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-2018-22 | latest | en | 0.608476 |
https://eatradingacademy.com/what-is-pip-cryptocurrency-trading/ | 1,723,643,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00100.warc.gz | 168,722,688 | 50,254 | # What is a Pip in Cryptocurrency Trading
## What is Pip? The difference between Bitcoin pips and points
What is a Pip in crypto trading? This is one of the questions I get from the students very often.
Pip, short for percentage in point, is a unit of measure used in the trading of currencies, cryptocurrencies, and other financial instruments. In the context of Bitcoin trading, a pip is a unit of measurement that represents the smallest change in value that a Bitcoin can experience.
Hello dear traders, my name is Petko Aleksandrov from EA Trading Academy, and in this lecture, I will make it clear what is a pip and what is a point. But I will talk about Bitcoin pips and points.
I had many students asking me is what is pip, is this in points, how many points is the spread, the swap, and I want to make it clear for everybody.
Therefore probably I will update some of the courses that I have ready, but the difficulty here comes with the cryptocurrencies mostly because these are the ones with a huge spread, and some brokers are having after the second or the third digit.
So most people do not understand what is a pip in cryptocurrency trading because they do not look at the price in detail. And Pip Bitcoin is harder because Bitcoin is expensive.
### How to calculate Bitcoin pips?
One of the tools that traders use to calculate the value of a pip is a pip calculator. A Bitcoin pip calculator is a tool that allows traders to determine the value of a pip in terms of the currency that they are trading. For example, a trader who is trading Bitcoin against the US dollar would use a Bitcoin pip calculator to determine the value of a pip in terms of USD.
Calculating the value of a pip can be particularly important for traders who are using leverage to trade Bitcoin. Leverage allows traders to trade larger amounts of Bitcoin than they would be able to trade without it, but it also increases the risk of losses. By calculating the value of a pip, traders can determine the amount of profit or loss that they can expect from a trade and make informed decisions about whether or not to enter a trade.
In addition to calculating the value of a pip, traders may also need to calculate the value of a point. A point is a unit of measure that is similar to a pip, but it is used in the context of trading instruments that have a higher degree of price precision. For example, some Bitcoin exchanges quote the price of Bitcoin to five decimal places, in which case a point would be equal to 0.00001 BTC.
But I will explain that in detail below in the post.
• 2 digits after the crypto decimal comma that is a Bitcoin pip
• 3 digits, that is a point
Also, for the currencies, most of the brokers have five digits, some with four still, but anyway, I want to make it clear what is a crypto pip for everybody.
### To make it clear what is cryptopip, we need to know what digit is?
If you have a price of 12345, this means you have 0 digits. And we are talking about the digits after the decimal comma. So, let’s say this is stock. And the price is 12345, and there is nothing after that, there is no .05 or 0.5 nothing. It is a whole number, meaning digits are 0, pip is 1, the point is one as well.
When there is nothing after a point, so if there is no point at all, there are no digits. Said what it is after the point, after the comma or whatever you call it is called digit and in this case, the Bitcoin pips equal to the point.
The next example is when you have “.3”, for example, after the number. Usually, nowadays, you will not see such prices with currencies, with cryptocurrencies, or with stocks and commodities.
This is because, on the online trading platforms, we have a minimal difference between the bid and the ask prices (the buy and the sell prices). Not like in the bank, when we exchange currencies for a vocation, you pay a lot for that huge difference.
Try asking someone in the bank, “What is Pip?”. They do not know. They do not know that there could be such a small difference between the bid and the ask prices.
Now, if you have already trading experience, you know what I am talking about. But if you do not have, let’s have some more examples so you will have a better idea about it.
And in this case, you have pip 0.1, and you have point 0.1. So, when you have only one digit after the point this means that the pip and the point are equal.
The next case is when you have two digits after the point, so 120.12,
### Now the pip equals the point.
This is the most common case for Bitcoin pips:
As you can see the crypto Pip equals the points because there are just 2 digits after the decimal comma.
And the next case is when you have three digits after the point. This comes typically for the USDJPY at the moment or any of the JPY pairs. We have three digits, and here comes the difference between the pip and the point.
So, 12 pips and 3 points-this is how we say, we pronounce the price as 120.12, we usually do not mean the point, but this is the smallest change you can see on the price. Here already we have three digits, we have pip of 0.1, so 12 pips and the point is simply the last one.
### Here already, what is pip? It is ten times bigger than the point.
The next example what I will show you is when we have four digits, let’s say this is the price of EURUSD, it doesn’t matter what example I would take, but it is a currency. You will see four digits after the point and here basically, you do not have a point because the point is equal to the pip.
### Here at 1.2345, you have your pip as 45, and point is as well 45.
The point is here, but it does not make any difference from what is a pip.
And the next line is now what you see with most of the brokers where we have pips, and the last one is the point.
In the old days, guys, all the brokers were till the fourth one, till the fourth digit, and the pip was equal to the point. Of course, at that time the crypto Pip was not there to confuse the traders.
And now because there are so many brokers on the market and because of the huge competition over there most of them started providing spread lower than one pip.
### And because there was nothing before lower than one pip, they created the point.
Now often, you will see brokers providing you spread 12345 points. Which of course is good for everybody that is on the trading side, because you are paying less spread. Right here, when you have five digits, the price is 1.2345 pips and 4 points, but we usually do not say the 4 points when you pronounce the price. But the points are there and
### The pip is ten times bigger than the points.
An easy way to remember it is when you have three digits and five digits are where you have the pip ten times bigger than the point and when you have two, and you have 4, you are having pip same as a point.
You can say there are no points, or you can say that pip is the same as the point. I hope it is clear.
And for the Bitcoin pips, if there are 3 digits behind the crypto price only then we will have a point. But if there are just 2, that is a Pip Bitcoin.
I do not think you will find a broker nowadays that will show you a one-digit price or without digit price, but I just made it this way, so it will be easier for you to understand me.
### And the next question is about the spread, the swap, the Stop Loss, and the Take Profit.
If you have spread and swap, usually the spread and the swap are in points or USD or any other currency.
For example, you can see some of the brokers for the cryptocurrency trading provide swap negative of \$25, and there are no points to calculate there, or you can see it as well in points, but the spread nowadays is mostly in points.
Of course, there are brokers with a considerable spread. Simply you can stay away from these brokers.
#### So what is the pip in cryptocurrency trading broker? It is the measurement of spread and swap
And the Stop Loss and the Take Profit it’s generally in pips, could be in points as well but especially the software that we are using, for example, EA Studio works with pips and FSB Pro works with points.
This is how you should work with such pieces of software if you are, and this is how the prices of the stocks, of the cryptocurrencies, of the regular currencies are basically structured.
I hope it is clear what is Pip in Cryptocurrency trading and you already know how to calculate Bitcoin pips, guys, thank you very much for reading.
If you are interested in the Cryptocurrency algorithmic trading course and you want to know more about cryptocurrencies, you can find it on our website. All our cryptocurrency courses you can find in the section cryptocurrency trading.
In summary, a pip is a unit of measure used in the trading of financial instruments, including Bitcoin, that represents the smallest change in value that the instrument can experience. A Bitcoin pip calculator is a tool that traders can use to calculate the value of a pip in terms of the currency that they are trading, and a point is a similar unit of measure that is used in the context of instruments with a higher degree of price precision.
## What is Pip?
Pip stands for a point in percentage. This is how traders measure the change of any asset or currency pair. A smaller movement than the pip is the point.
## What is the difference between a point and a pip?
The currency prices normally are with 5 digits after the decimal comma. The 5th digit is the point and this is the smallest movement measured in the price. The Pips are the 3rd and the 4rt digits.
## How the pip is different in cryptocurrency trading?
This depends on the broker or the exchange that offers the cryptocurrency. Most providers offer 3 digits after the decimal comma. The first 2 are the pips and the last one is the point.
## How to calculate pips in Cryptocurrency trading?
If we divide the exchange rate by 0.0001 we will see what is the value of 1 pip in cryptocurrency trading. In other words, this is the smallest movement possible.
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written by Kyle Forinash and Wolfgang Christian
This is a collection of interactive tutorials on the fundamentals of waves. The tutorials emphasize concepts that are not usually easy to illustrate in textbooks. Java applets are used to illustrate the physics. The lessons begin with very simple wave properties and end with an examination of nonlinear wave behavior.
Although this material deals strictly with classical wave properties, it can be used as a remedial tutorial or to introduce concepts such as group velocity and interference in a quantum mechanics class.
Subjects Levels Resource Types
Education Practices
- Active Learning
= Modeling
Optics
- Geometrical Optics
- Polarization
Oscillations & Waves
- General
- Wave Motion
= Doppler Effect
= Impedance and Dispersion
= Interference and Diffraction
= Longitudinal Pulses and Waves
= Phase and Group Velocity
= Reflection and Refraction (Sound)
= Standing Waves
= Transverse Pulses and Waves
- High School
- Collection
- Instructional Material
= Interactive Simulation
= Tutorial
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- Learners
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Formats:
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Mirror:
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Access Rights:
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Keywords:
fundamentals of waves, group velocity, wave behavior, wave dispersion, wave interference
Record Creator:
Metadata instance created August 17, 2005 by kyle forinash
Record Updated:
April 27, 2024 by Wolfgang Christian
Last Update
when Cataloged:
August 9, 2005
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### AAAS Benchmark Alignments (2008 Version)
#### 4. The Physical Setting
4F. Motion
• 6-8: 4F/M4. Vibrations in materials set up wavelike disturbances that spread away from the source. Sound and earthquake waves are examples. These and other waves move at different speeds in different materials.
• 6-8: 4F/M6. Light acts like a wave in many ways. And waves can explain how light behaves.
• 6-8: 4F/M7. Wave behavior can be described in terms of how fast the disturbance spreads, and in terms of the distance between successive peaks of the disturbance (the wavelength).
• 9-12: 4F/H5ab. The observed wavelength of a wave depends upon the relative motion of the source and the observer. If either is moving toward the other, the observed wavelength is shorter; if either is moving away, the wavelength is longer.
• 9-12: 4F/H6ab. Waves can superpose on one another, bend around corners, reflect off surfaces, be absorbed by materials they enter, and change direction when entering a new material. All these effects vary with wavelength.
• 9-12: 4F/H6c. The energy of waves (like any form of energy) can be changed into other forms of energy.
#### 11. Common Themes
11B. Models
• 6-8: 11B/M4. Simulations are often useful in modeling events and processes.
### Common Core State Standards for Mathematics Alignments
#### High School — Algebra (9-12)
Creating Equations? (9-12)
• A-CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
• A-CED.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.
#### High School — Functions (9-12)
Interpreting Functions (9-12)
• F-IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.?
• F-IF.5 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes.?
• F-IF.9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions).
Trigonometric Functions (9-12)
• F-TF.5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline.?
### Common Core State Reading Standards for Literacy in Science and Technical Subjects 6—12
Craft and Structure (6-12)
• RST.11-12.4 Determine the meaning of symbols, key terms, and other domain-specific words and phrases as they are used in a specific scientific or technical context relevant to grades 11—12 texts and topics.
Range of Reading and Level of Text Complexity (6-12)
• RST.11-12.10 By the end of grade 12, read and comprehend science/technical texts in the grades 11—CCR text complexity band independently and proficiently.
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### Waves: An Interactive Tutorial:
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# In the last twenty years, despite the chauvinism of European
Author Message
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Manager
Joined: 21 Jun 2004
Posts: 238
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In the last twenty years, despite the chauvinism of European [#permalink]
### Show Tags
14 May 2005, 08:05
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In the last twenty years, despite the chauvinism of European connoisseurs, California wines are respected throughout the world.
are respected
are becoming better respected
which have gained respect
have gained respect
have since become respected
[/u]
Director
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### Show Tags
14 May 2005, 08:08
D
Present perfect fits in the best to describe a continuous event (gaining respect)
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### Show Tags
14 May 2005, 10:09
agree with Vithal.
Shunned A and B for passive construction, C is wrong because it introduces a dependent clause and we need a present perfect as in D.
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### Show Tags
14 May 2005, 10:09
A) Wrong. In the last 20 years California wines are respected.
B) Awkward and wordy. In the last 20 years California wines are becoming better respected.
C) WRONG. CA wines which have gained respect.
D) CORRECT. California wines have gained respect.
E) Cleary wrong.
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### Show Tags
17 Jun 2007, 02:41
Vithal wrote:
D
Present perfect fits in the best to describe a continuous event (gaining respect)
Agree. In the last twenty years implies action from the past to the present point in time, thus, present perfect. (has/have + past participle)
have/has gained.
wines have gained respect.
Therefore D.
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Display posts from previous: Sort by | 897 | 3,171 | {"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} | 3.03125 | 3 | CC-MAIN-2016-36 | longest | en | 0.88573 |
https://www.daveramsey.com/blog/building-unstoppable-momentum/ | 1,501,164,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428257.45/warc/CC-MAIN-20170727122407-20170727142407-00221.warc.gz | 657,057,922 | 27,594 | # Building Unstoppable Momentum
In 1971, the final episode of The Beverly Hillbillies was broadcast on television sets across the country. After 247 episodes, Jed, Granny, Elly May and Jethro performed for the final time.
Nearly 40 years later, though, The Beverly Hillbillies theme song lives on. You’re probably humming it right now, aren’t you?
Isn’t it interesting that a television show that filmed its last episode in 1971 made such an impact—not only on people who watched the show during its run, but also on people born well after it ended?
Local experts you can trust.
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Think about this: If a television show can make that type of impact, imagine what the body of Christ could do! So what are you doing today that will affect your church and community 50 years from now?
Winning at anything starts with momentum. But momentum doesn’t just happen. It’s not some sort of random occurrence that miraculously appears. We’ve got to create momentum. We can’t just sit back and wait for God to “bless us” with success without acknowledging that we have a role to play.
St. Augustine said, “You’ve got to pray like it all depends on God and work like it all depends on you.” James 2:26 says, “For as the body without the spirit is dead, so faith without works is dead also.” There is a balance between works and faith, so you have a role to play in this deal.
Momentum isn’t just some abstract concept. In fact, Dave discovered that the power of momentum can be summed up in the Momentum Theorem: Fi/T(G) = M Focused intensity, over time, multiplied by God, equals unstoppable momentum.
Let’s dig into this formula a little deeper:
## Focus
Without focus, you’re going to be like the rest of our A.D.D. culture. When you try to do everything at once, you don’t get anything done. Like a wide receiver who has to focus on the football instead of the defender, you’ve got to have tunnel vision to win. And when you actually start to focus, you’ll begin to stand out in a culture that doesn’t have any.
## Intensity
The poster boy for intensity is the lunatic fan who’s painted in red, sitting on the front row, and living and dying with every score. Now put that kind of intensity into your marriage, your job, your calling—things that matter. What would that look like? Intensity moves things, makes things happen. Combine intensity with focus, and you are ready to start building momentum.
## Time
Yes, intensity moves things. But if you’re only focused and intense for a few days, you aren’t going to move anything. This is the hardest part of the Momentum Theorem. Paul said to “run the race in such a way as to win the prize.” You’re not in a sprint; you’re in a marathon. The tortoise wins the race every single time. He stays the course, keeps his eye on the finish line, and never, ever quits.
## God
This is the most important part of the theorem. Sure, you can do all these things by yourself. Staying focused and intense over time will lead to good results. But it will wear you out. That’s when you just have to let things go—take that focused intensity over time and give it to God. When you do that, God steps into your effort and gives you energy and the ability to win.
## Unstoppable Momentum
If you’ll stay focused and shed all the distractions, and if you’ll consistently, over a long period of time, stay intense and passionate about your calling, then you’ll begin to see unstoppable momentum in your life and in your church.
Decades from now, your kids and grandkids—and future generations in your church—will look back and know that you were doing something that mattered, something that made a difference. That’s the kind of impact unstoppable momentum can make.
Want to create unstoppable momentum in your congregation? Learn about Momentum, Dave's church-wide program.
## More from the Blog
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Take a peek into some of the new money problems we face today, and how we can overcome those issues. | 991 | 4,447 | {"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} | 2.5625 | 3 | CC-MAIN-2017-30 | latest | en | 0.953882 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/980/2/c/b/ | 1,632,482,389,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00249.warc.gz | 885,589,559 | 57,526 | # Properties
Label 980.2.c.b Level $980$ Weight $2$ Character orbit 980.c Analytic conductor $7.825$ Analytic rank $0$ Dimension $8$ CM discriminant -20 Inner twists $8$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$980 = 2^{2} \cdot 5 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 980.c (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$7.82533939809$$ Analytic rank: $$0$$ Dimension: $$8$$ Coefficient field: 8.0.3317760000.3 Defining polynomial: $$x^{8} - 4 x^{6} + 7 x^{4} - 36 x^{2} + 81$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$2^{4}$$ Twist minimal: no (minimal twist has level 140) Sato-Tate group: $\mathrm{U}(1)[D_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\beta_{1} q^{2} + \beta_{3} q^{3} + 2 q^{4} + \beta_{6} q^{5} + ( -\beta_{5} + \beta_{6} ) q^{6} -2 \beta_{1} q^{8} + ( -1 + \beta_{2} ) q^{9} +O(q^{10})$$ $$q -\beta_{1} q^{2} + \beta_{3} q^{3} + 2 q^{4} + \beta_{6} q^{5} + ( -\beta_{5} + \beta_{6} ) q^{6} -2 \beta_{1} q^{8} + ( -1 + \beta_{2} ) q^{9} + ( \beta_{3} + \beta_{7} ) q^{10} + 2 \beta_{3} q^{12} + ( 3 \beta_{1} - \beta_{4} ) q^{15} + 4 q^{16} + ( 2 \beta_{1} - 2 \beta_{4} ) q^{18} + 2 \beta_{6} q^{20} + ( -2 \beta_{1} + 3 \beta_{4} ) q^{23} + ( -2 \beta_{5} + 2 \beta_{6} ) q^{24} -5 q^{25} + ( -2 \beta_{3} + \beta_{7} ) q^{27} + ( -3 + 2 \beta_{2} ) q^{29} + ( -5 + \beta_{2} ) q^{30} -4 \beta_{1} q^{32} + ( -2 + 2 \beta_{2} ) q^{36} + ( 2 \beta_{3} + 2 \beta_{7} ) q^{40} + ( 6 \beta_{5} + \beta_{6} ) q^{41} + ( -4 \beta_{1} - \beta_{4} ) q^{43} + ( 5 \beta_{5} - \beta_{6} ) q^{45} + ( 1 - 3 \beta_{2} ) q^{46} + ( -3 \beta_{3} - 3 \beta_{7} ) q^{47} + 4 \beta_{3} q^{48} + 5 \beta_{1} q^{50} + ( 3 \beta_{5} - \beta_{6} ) q^{54} + ( 5 \beta_{1} - 4 \beta_{4} ) q^{58} + ( 6 \beta_{1} - 2 \beta_{4} ) q^{60} + ( -4 \beta_{5} - 3 \beta_{6} ) q^{61} + 8 q^{64} + ( -\beta_{1} + 5 \beta_{4} ) q^{67} + ( -8 \beta_{5} + 5 \beta_{6} ) q^{69} + ( 4 \beta_{1} - 4 \beta_{4} ) q^{72} -5 \beta_{3} q^{75} + 4 \beta_{6} q^{80} + ( 4 + \beta_{2} ) q^{81} + ( -5 \beta_{3} + 7 \beta_{7} ) q^{82} + ( 4 \beta_{3} - 7 \beta_{7} ) q^{83} + ( 9 + \beta_{2} ) q^{86} + ( -11 \beta_{3} + 2 \beta_{7} ) q^{87} + ( 3 \beta_{5} + 4 \beta_{6} ) q^{89} + ( -6 \beta_{3} + 4 \beta_{7} ) q^{90} + ( -4 \beta_{1} + 6 \beta_{4} ) q^{92} -6 \beta_{6} q^{94} + ( -4 \beta_{5} + 4 \beta_{6} ) q^{96} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8q + 16q^{4} - 8q^{9} + O(q^{10})$$ $$8q + 16q^{4} - 8q^{9} + 32q^{16} - 40q^{25} - 24q^{29} - 40q^{30} - 16q^{36} + 8q^{46} + 64q^{64} + 32q^{81} + 72q^{86} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} - 4 x^{6} + 7 x^{4} - 36 x^{2} + 81$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$-4 \nu^{7} + 7 \nu^{5} + 35 \nu^{3} + 81 \nu$$$$)/189$$ $$\beta_{2}$$ $$=$$ $$($$$$-\nu^{6} + 4 \nu^{4} + 2 \nu^{2} + 18$$$$)/9$$ $$\beta_{3}$$ $$=$$ $$($$$$\nu^{7} - 4 \nu^{5} + 7 \nu^{3} - 9 \nu$$$$)/27$$ $$\beta_{4}$$ $$=$$ $$($$$$-\nu^{7} + 4 \nu^{5} - 7 \nu^{3} + 63 \nu$$$$)/27$$ $$\beta_{5}$$ $$=$$ $$($$$$-8 \nu^{6} + 14 \nu^{4} - 56 \nu^{2} + 225$$$$)/63$$ $$\beta_{6}$$ $$=$$ $$($$$$-\nu^{6} + 22$$$$)/7$$ $$\beta_{7}$$ $$=$$ $$($$$$\nu^{7} - \nu^{5} + 4 \nu^{3} - 24 \nu$$$$)/9$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{4} + \beta_{3}$$$$)/2$$ $$\nu^{2}$$ $$=$$ $$($$$$\beta_{6} - 2 \beta_{5} + \beta_{2} + 2$$$$)/2$$ $$\nu^{3}$$ $$=$$ $$($$$$\beta_{7} + \beta_{3} + 7 \beta_{1}$$$$)/2$$ $$\nu^{4}$$ $$=$$ $$($$$$-4 \beta_{6} + \beta_{5} + 4 \beta_{2} + 1$$$$)/2$$ $$\nu^{5}$$ $$=$$ $$($$$$7 \beta_{7} + 5 \beta_{4} - 12 \beta_{3} + 7 \beta_{1}$$$$)/2$$ $$\nu^{6}$$ $$=$$ $$-7 \beta_{6} + 22$$ $$\nu^{7}$$ $$=$$ $$($$$$21 \beta_{7} + 29 \beta_{4} + 8 \beta_{3} - 21 \beta_{1}$$$$)/2$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/980\mathbb{Z}\right)^\times$$.
$$n$$ $$101$$ $$197$$ $$491$$ $$\chi(n)$$ $$-1$$ $$-1$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
979.1
−1.01575 − 1.40294i 1.72286 − 0.178197i 1.72286 + 0.178197i −1.01575 + 1.40294i 1.01575 − 1.40294i −1.72286 − 0.178197i −1.72286 + 0.178197i 1.01575 + 1.40294i
−1.41421 2.80588i 2.00000 2.23607i 3.96812i 0 −2.82843 −4.87298 3.16228i
979.2 −1.41421 0.356394i 2.00000 2.23607i 0.504017i 0 −2.82843 2.87298 3.16228i
979.3 −1.41421 0.356394i 2.00000 2.23607i 0.504017i 0 −2.82843 2.87298 3.16228i
979.4 −1.41421 2.80588i 2.00000 2.23607i 3.96812i 0 −2.82843 −4.87298 3.16228i
979.5 1.41421 2.80588i 2.00000 2.23607i 3.96812i 0 2.82843 −4.87298 3.16228i
979.6 1.41421 0.356394i 2.00000 2.23607i 0.504017i 0 2.82843 2.87298 3.16228i
979.7 1.41421 0.356394i 2.00000 2.23607i 0.504017i 0 2.82843 2.87298 3.16228i
979.8 1.41421 2.80588i 2.00000 2.23607i 3.96812i 0 2.82843 −4.87298 3.16228i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 979.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
20.d odd 2 1 CM by $$\Q(\sqrt{-5})$$
4.b odd 2 1 inner
5.b even 2 1 inner
7.b odd 2 1 inner
28.d even 2 1 inner
35.c odd 2 1 inner
140.c even 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 980.2.c.b 8
4.b odd 2 1 inner 980.2.c.b 8
5.b even 2 1 inner 980.2.c.b 8
7.b odd 2 1 inner 980.2.c.b 8
7.c even 3 1 140.2.s.a 8
7.c even 3 1 980.2.s.a 8
7.d odd 6 1 140.2.s.a 8
7.d odd 6 1 980.2.s.a 8
20.d odd 2 1 CM 980.2.c.b 8
28.d even 2 1 inner 980.2.c.b 8
28.f even 6 1 140.2.s.a 8
28.f even 6 1 980.2.s.a 8
28.g odd 6 1 140.2.s.a 8
28.g odd 6 1 980.2.s.a 8
35.c odd 2 1 inner 980.2.c.b 8
35.i odd 6 1 140.2.s.a 8
35.i odd 6 1 980.2.s.a 8
35.j even 6 1 140.2.s.a 8
35.j even 6 1 980.2.s.a 8
35.k even 12 2 700.2.p.b 8
35.l odd 12 2 700.2.p.b 8
140.c even 2 1 inner 980.2.c.b 8
140.p odd 6 1 140.2.s.a 8
140.p odd 6 1 980.2.s.a 8
140.s even 6 1 140.2.s.a 8
140.s even 6 1 980.2.s.a 8
140.w even 12 2 700.2.p.b 8
140.x odd 12 2 700.2.p.b 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
140.2.s.a 8 7.c even 3 1
140.2.s.a 8 7.d odd 6 1
140.2.s.a 8 28.f even 6 1
140.2.s.a 8 28.g odd 6 1
140.2.s.a 8 35.i odd 6 1
140.2.s.a 8 35.j even 6 1
140.2.s.a 8 140.p odd 6 1
140.2.s.a 8 140.s even 6 1
700.2.p.b 8 35.k even 12 2
700.2.p.b 8 35.l odd 12 2
700.2.p.b 8 140.w even 12 2
700.2.p.b 8 140.x odd 12 2
980.2.c.b 8 1.a even 1 1 trivial
980.2.c.b 8 4.b odd 2 1 inner
980.2.c.b 8 5.b even 2 1 inner
980.2.c.b 8 7.b odd 2 1 inner
980.2.c.b 8 20.d odd 2 1 CM
980.2.c.b 8 28.d even 2 1 inner
980.2.c.b 8 35.c odd 2 1 inner
980.2.c.b 8 140.c even 2 1 inner
980.2.s.a 8 7.c even 3 1
980.2.s.a 8 7.d odd 6 1
980.2.s.a 8 28.f even 6 1
980.2.s.a 8 28.g odd 6 1
980.2.s.a 8 35.i odd 6 1
980.2.s.a 8 35.j even 6 1
980.2.s.a 8 140.p odd 6 1
980.2.s.a 8 140.s even 6 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{4} + 8 T_{3}^{2} + 1$$ acting on $$S_{2}^{\mathrm{new}}(980, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( -2 + T^{2} )^{4}$$
$3$ $$( 1 + 8 T^{2} + T^{4} )^{2}$$
$5$ $$( 5 + T^{2} )^{4}$$
$7$ $$T^{8}$$
$11$ $$T^{8}$$
$13$ $$T^{8}$$
$17$ $$T^{8}$$
$19$ $$T^{8}$$
$23$ $$( 4489 - 136 T^{2} + T^{4} )^{2}$$
$29$ $$( -51 + 6 T + T^{2} )^{4}$$
$31$ $$T^{8}$$
$37$ $$T^{8}$$
$41$ $$( 10609 + 226 T^{2} + T^{4} )^{2}$$
$43$ $$( 1089 - 96 T^{2} + T^{4} )^{2}$$
$47$ $$( 90 + T^{2} )^{4}$$
$53$ $$T^{8}$$
$59$ $$T^{8}$$
$61$ $$( 9 + 186 T^{2} + T^{4} )^{2}$$
$67$ $$( 33489 - 384 T^{2} + T^{4} )^{2}$$
$71$ $$T^{8}$$
$73$ $$T^{8}$$
$79$ $$T^{8}$$
$83$ $$( 25281 + 408 T^{2} + T^{4} )^{2}$$
$89$ $$( 2809 + 214 T^{2} + T^{4} )^{2}$$
$97$ $$T^{8}$$ | 4,266 | 8,247 | {"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} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.382125 |
https://www.mtgprofessor.com/a%20-%20arms/choosing_between_arm_and_frm.htm | 1,563,601,298,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526446.61/warc/CC-MAIN-20190720045157-20190720071157-00241.warc.gz | 762,456,602 | 8,599 | Choosing Between Fixed & Adjustable Rate Mortgages
February 2, 2004, Revised November 14, 2008, Reviewed February 12, 2011
"I am trying to choose between a 3/1 ARM at 4.625% and a FRM at 5.875%, both 30 years. I don’t expect to be out of my house within 3 years. What is the best way to make this decision?"
Whether the adjustable rate mortgage (ARM) or fixed rate mortgage (FRM) turns out better depends on what happens to interest rates in the future, which no one knows. Shoppers faced with this decision should ask themselves "Is this a risk worth taking," and "can I afford to take it?"
## Scenario Analysis of ARM Versus FRM
The best way I know to deal with these questions is by determining what will happen to the rate and payment on the ARM if market interest rates change in ways that you specify. This "scenario analysis" provides a measure of the risk if rates increase, and the benefit if they don’t. It also allows you to determine the extent to which you can reduce the risk on the ARM by making the larger payment that you would have made had you selected the FRM.
A side benefit is that you can’t do scenario analysis without knowing all the features of the ARM that affect future rates and payments. The information you are forced to compile for this purpose you should have anyway. Otherwise, you don’t know whether you have found the best deal on your ARM.
## Example of a Scenario Analysis
For example, you told me that your 3/1 ARM had a rate of 4.625%, but that rate holds for only 3 years, after which the rate adjusts every year. You did not tell me what I needed to know to calculate the rate and payment after the 3 years. I found out that your ARM rate was tied to the one-year Treasury index, which had a recent value of 1.28%, and had a margin of 2.75%. After 3 years, your rate would equal the index at that time plus 2.75%, subject to an adjustment cap of 2% (no rate change can exceed 2%) and a maximum rate of 10.625%.
You need all that to do scenario analysis, but you also want it for shopping. If you could find the same 3/1 ARM with a 2.5% margin, you should grab it.
The numbers cited below all assume loan amounts of \$100,000, and came from calculator 7b Monthly Payment Calculator: Adjustable-Rate Mortgages Without Negative Amortization.
## Stable Rate Scenario
A stable-rate scenario provides the best measure of the potential benefit of the ARM. The payment would be \$514.14 for the first 36 months, and \$481.76 thereafter, as compared to \$591.54 on the FRM. If you made the \$591.54 payment on the ARM, you would pay it off in 257 months.
## Rising Rate Scenarios
I used 4 rising rate scenarios of gradually increasing severity: 1. Small rate increase: after 2 years, the index increases by .5 % /year for 3 years. 2. Moderate rate increase: after 1 year, the rate index increases by .75%/year for 4 years. 3. Larger rate increase: starting immediately, the index increases by 1%/year for 5 years. 4. Worst case: the index rises to 100% in month 2.
With the small rate increase scenario, the payment remains lower on the ARM than on the FRM over the entire 30 years. If the borrower makes the FRM payment, he will pay off in 304 months. The borrower thus benefits if rates are stable or decline, or have a delayed rise of 1.5% over 3 years.
With the larger rate-increase scenarios, the benefits of the ARM over the first 3 or 4 years are followed by losses. Skipping to the worst case, the payment would rise from \$514.14 to \$630.64 in month 37, to \$754.44 in month 49, and to \$883.74 in month 61 where it would remain until payoff. It is useful to know whether you could deal with these increases, even though the likelihood of their occurring is very low.
These payment increases could be reduced by making the larger FRM payment in the first 3 years. If you paid \$591.54 rather than \$514.14 for 36 months, you would reduce the worst case payment in months 61-360 from \$883.74 to \$856.01. The complete results for all the scenarios are shown below.
Scenario analysis doesn’t provide definitive answers to the questions posed at the beginning of this article. However, it does allow you to make an informed judgment based on all available information. In the face of an uncertain future, that’s the best anyone can do.
ARM Features
Initial Interest Rate on ARM 4.625% Initial Rate Period 3 Years Subsequent Adjustment Period 1 Year Most Recent Index Value 1.28% Margin 2.75% First Rate Adjustment Cap 2.000% Later Adjustment Caps 2.000% Maximum Interest Rate 10.625% Loan Term (in years) 30 Rate on FRM Loan Used as Comparison 5.875% FRM Payment \$591.54
Interest Rates and Monthly Payments Under 5 Interest Rate Scenarios
SCENARIO Months No Change Small Increase Moderate Increase Large Increase Worst Case Rate % Pmt \$ Rate % Pmt \$ Rate % Pmt \$ Rate % Pmt \$ Rate % Pmt \$ 1-36 4.625 514.14 4.625 514.14 4.625 514.14 4.625 514.14 4.625 514.14 37-48 4.03 481.76 4.53 508.90 5.53 565.46 6.625 630.64 6.625 630.64 49-60 4.03 481.76 5.03 536.01 6.28 608.58 8.03 716.66 8.625 754.44 61-360 4.03 481.76 5.53 563.01 7.03 651.97 9.03 779.13 10.625 883.74 Borrower Makes the FRM Payment So Long As It Is Larger Than the ARM Payment 1-36 4.625 591.54 4.625 591.54 4.625 591.54 4.625 591.54 4.625 591.54 37-48 4.03 591.54 4.53 591.54 5.53 591.54 6.625 610.85 6.625 610.85 49-60 4.03 591.54 5.03 591.54 6.28 591.54 8.03 694.17 8.625 730.76 61-360 4.03 591.541 5.53 591.542 7.03 627.26 9.03 754.67 10.625 856.01
1Pays off in 257 months
2Pays off in 304 months
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https://economistwritingeveryday.com/2022/10/28/cheers-to-sumproduct/ | 1,670,211,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711003.56/warc/CC-MAIN-20221205032447-20221205062447-00084.warc.gz | 253,731,284 | 29,055 | ## Cheers to Sumproduct!
I teach macroeconomics, finance, and other things.
Often, I use Excel to complete repetitive calculations for my students. The version that I show them is different from the version that I use. They see a lot more mathematical steps displayed in different cells, usually with a label describing what it is. But when I create an answer calculator or work on my own, I usually try to be as concise as possible, squeezing what I can into a single cell or many fewer cells. That’s what brings me to to the sumproduct excel function that I recently learned. It’s super useful I’ll illustrate it with two examples.
Example 1) NGDP
One way to calculate NGDP is to sum all of the expenditures on the different products during a time period. The expenditures on a good is simply the price of the good times the quantity that was purchased during the time period. The below image illustrates an example with the values on the left, and the equations that I used on the right. That’s the student version. There is an equation for each good which calculates the total expenditure on the individual goods. Then, there is a final equation which sums the spending to get total expenditures, or NGDP.
Granted, autofill makes creating the third column relatively fast and easy. And, it’s good for explaining content to students. But what would *I* do? I’d forgo that third column entirely, thanks to sumproduct. Below is the same calculation more concisely. I dispense with the expenditures on individual goods and just skip to the part that I care about: the sum of the products. Of course, I can imagine a case in which I might want to know the spending on individual goods. But, it’s nice to skip it when I don’t care.
Example 2) Expected Return
What if you have a portfolio of assets? Or a portfolio of debt? Or a menu of possible outcomes, each with a different probability? Below is the expected performance of an asset and the probability that each outcome occurs. If we want to find the expected return, then we’d calculate the sum of weighted returns. Again, we can save ourselves the third column of work by using the sumproduct function.
Example 3) Division
‘Big whoop’ you might say. The function appears to save only one step and to have limited applications. Nay, I say! the sumproduct function has flexibility. Namely, you can transform the values in an array prior to the multiplication. Below is the student-version of calculating the expected return variance. The variance is the sum of the weighted squared deviations from the mean. What a mouthful! Not only that, what a spreadsheet-ful! The simple operator functions use four columns of calculation, plus a cell for the expected value and a cell for the variance. Of course, one could combine the smaller steps in order to reduce the calculated columns from four to two. But there’d still be ten cells of formulas.
Below is the truncated method of calculating the expected variance that uses the sumproduct function. First, it uses only one cell of of formula. That’s a 90% reduction from the simplified version of the example above. Specifically, I want you to notice how the values in the arrays can be manipulated. First, I nested the sumproducts so that I could calculate the average return like I did in example 2). I subtract that average from each possible return in column C, then square that difference. Only after these manipulations does the sumproduct function multiply the values of the arrays and then sum them. How about them apples!
At first blush, the sumproduct function in excel saves a negligible amount of time and appears to be a uni-tasker. But, thanks to some developer foresight and convenient syntax, we can complete a great number of operations all in one cell. It might take some time to get used to it. After all, an advantage of spreadsheets is that we can break problems down into their components across different cells. That makes it easier for us to understand the various parts and to debug. But, once you become very familiar with a process and you don’t need to decompose it every time, putting everything in a single cell improves efficiency and let’s you spend your time on your other priorities. And for that, cheers to the sumproduct function!
### 2 thoughts on “Cheers to Sumproduct!”
1. Jackson Jules October 28, 2022 / 2:43 pm
Hmm I think I like this.
One thing that bothers me is when derived quantities (ones that are calculated from other quantities) are not properly differentiated from fundamental quantities (data that had to be collected). My reducing columns that contain derived quantities, it helps with this confusion.
Also, there is just the principle of parsimony. No one likes cluttered spreadsheets. The fewer the columns the better.
Like | 999 | 4,793 | {"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-2022-49 | longest | en | 0.945519 |
https://archive.gamedev.net/archive/reference/articles/article222.html | 1,623,975,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00021.warc.gz | 121,430,309 | 4,947 | Simple Fire Effect
In this article I'll try to explain how to do realtime fire by simulating the structure of flames. I'll explain the method for "8-Bit-Indexed-Color-Mode" or better: VGA mode 13h (320x200x256). But once you understood this, it shouldn't be hard to port this to other graphic-modes. I learned it from a book 1 a few weeks ago, but I think this effect was originaly invented by JARE of Iguana in 1993. This effect is so old and well known that I think everybody should know how it works ;)
But before we're going to talk about the algorithm we have to define two 2d-arrays [x,y] with equal sizes. Let's call them Buffer1 and Buffer2. The size of those arrays depends on the size you want to have the fire at (e.g. 320x200 = 64000 bytes).
The next thing we need to define is a palette with gradient colors like black-orange-white (0-255) (but every other palette is also possible). This is IMHO the hardest part ... to create a good palette ;-)
Let's take a closer look at the structure of a flame:
• Fire is constantly moving up
• It's hot at the bottom (white) and cools down at the top (black)
• Fire doesn't look pixelated but smoothed
• Sometimes we see some hot sparks/spots
Okay that should be enough ;-) To create a nice fire-clone on the screen now, we need to do 3 steps in a loop:
1. Update the fire
2. Scroll fire up and smooth it
3. Visualize fire (copy to VGA) and swap arrays
I'm going to explain each step:
# Step 1
Put some coal on the ground which should be burned: Fill the bottom of Buffer1 with a value between, let's say, 100-150 (white/yellow) ... play arround with this!
```for x <- 0 to 319
{
Buffer1 [x , 198] = random_value (100-150)
Buffer1 [x , 199] = random_value (100-150)
}
```
In this case above I painted the last two lines (198 & 199) but this could also have been 1 or ? lines. If you want to have some additional big sparks/hotspots then put some (10-50) big blocks of 9 pixels at the bottom with color 255 (bright white). It'll take more time till they fade out and so they go higher than the normal flames/sparks/spots created above:
```for x <- 0 to random_value (10-50)
{
Buffer1 [x-1,197] = 255
Buffer1 [x ,197] = 255
Buffer1 [x+1,197] = 255
Buffer1 [x-1,198] = 255
Buffer1 [x ,198] = 255
Buffer1 [x+1,198] = 255
Buffer1 [x-1,199] = 255
Buffer1 [x ,199] = 255
Buffer1 [x+1,199] = 255
}
```
This will create a few 3x3 squares of bright white pixels. Okay, this was step one and now we know how to do the white ground of the fire and hotspots. The rest will be done in ...
# Step 2
Now ... to move the fire up we will have to copy each pixel to the line above. To cool the fire down at the top we just decrement the value of each pixel before we move it up. But if we did only this two steps the fire would look very pixelated ... one couldn't even say "fire" to it ;) What we are doing to let it look more realistic is smoothing: Instead of copying the decremented value of each pixel to the line above, we copy the average value of the surrounding pixels to the line above (decremented by one of course - to cool down). Sounds confusing ?!? Okay let me explain it. If we only scrolled the fire up and cooled it down it would look like this: Attention: the value of the pixel will be read from Buffer1 but written to Buffer2! Otherwise it could look a little bit crazy (or at least it won't be the effect we want to see) %-)
```for x <- 0 to 319
{
for y <- 1 to 199
{
Buffer2 [x,y-1] = Buffer1 [x,y] - 1
}
}
```
To smooth the flames, we take the average value of the 4 sourrounding pixels (left, right, above, under) and decrement this value...we don't even need/want the value of the center-pixel:
Notice: you can also take the 8 surrounding pixels (upper left, upper, upper, right, left,...)
```for x <- 0 to 319 {
for y <- 1 to 199 {
Buffer2 [x,y-1] = (Buffer1 [x-1,y] + Buffer1 [x+1,y]
+ Buffer1 [x,y-1] + Buffer1 [x,y+1]) / 4 - 1
}
}
```
One bug remains if the average value is zero: if you decrement it know, you'll get the value 255! So before you decrement it, you should check for zero first ! If you don't do it, it'll look ugly !
# Step 3
Know you have to copy Buffer2 to the VGA memory and then you should copy Buffer2 into Buffer1 so that this will be the destination array in the next loop ! Notice: to copy Buffer2 into Buffer1 it's sufficient to swap the memory-adresses of the 2 arrays (you should have used two virtual screens instead of normal arrays and those buffers should be accessed through direct memory access (SEGMENT : INDEX), now just swap the segment adresses... if you don't understand it or have problems with this, go and read Denthor's Tutorials in the graphics section (Virtual screens)).
## Literature
[1] PC UNDERGROUND, Data Becker, ISBN: 3-8158-1185-6 (german) | 1,324 | 4,781 | {"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} | 3.203125 | 3 | CC-MAIN-2021-25 | latest | en | 0.926128 |
http://mathhelpforum.com/geometry/8989-help-centroid-circumcenter-orthocenter-print.html | 1,529,550,996,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864019.29/warc/CC-MAIN-20180621020632-20180621040632-00145.warc.gz | 204,769,322 | 4,119 | # HELP ON CENTROID, CIRCUMCENTER, and ORTHOCENTER
• Dec 17th 2006, 09:46 AM
hm_498
HELP ON CENTROID, CIRCUMCENTER, and ORTHOCENTER
Can someone help me find the centroid, circumcenter, and the orthocenter of a traingle for the following points:
(-3,6)
(-9,-3)
(4,2)
• Dec 18th 2006, 06:43 AM
Soroban
Hello, hm_498!
I'll give you a small start . . .
Quote:
Find the centroid, circumcenter, and the orthocenter of the triangle with vertices:
. . $\displaystyle A(\text{-}3,6),\;B((\text{-}9,\text{-}3),\;C(4,2)$
I must assume you are familiar with slopes and equations of lines . . .
The centroid is the intersection of the medians.
A median joins a vertex to the midpoint of the opposite side.
First, find some of the midpoints of the sides.
Let $\displaystyle D$ be the midpoint of side $\displaystyle BC$. .Then: $\displaystyle D\left(-\frac{5}{2},\,-\frac{1}{2}\right)$
. . Write the equation of $\displaystyle M_1$, the line through $\displaystyle A$ and $\displaystyle D.$
Let $\displaystyle E$ be the midpoint of side $\displaystyle AC$. .Then: $\displaystyle E\left(\frac{1}{2},\,4\right)$
. . Write the equation of $\displaystyle M_2$, the line through $\displaystyle B$ and $\displaystyle E.$
The centroid is the intersection of $\displaystyle M_1$ and $\displaystyle M_2$.
Hints for the rest of the problem:
The circumcenter is the intersection of the perpendicular bisectors of the sides.
The orthocenter is the intersection of the altitudes to the sides.
• Dec 18th 2006, 01:30 PM
Quick
Here is a diagram of ortho, circum, and centroid.
Press the buttons to make things easier to see.
You can move the red points.
"Mp" stands for "Midpoint"
Red lines are perpendicular.
Magenta Lines are Altitudes.
This is just to give you a visual.
Does it Help?
EDIT: Added Euler's Line (I was bored, ok?)
• Dec 18th 2006, 01:37 PM
ThePerfectHacker
I would like to mention that they always line (if you have 3 points) on a single line!
It is called "Euler's Line".
He discovered it.
When I was about your age, I proved it using some analytic geometry. The algebra was intense (as you can imagine) but it worked out nicely in the end.
However, I think Euler proved it classically. I wonder how he did that :rolleyes: . I have some idea but never been able to. Anybody know? | 655 | 2,297 | {"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} | 4.03125 | 4 | CC-MAIN-2018-26 | latest | en | 0.847606 |
https://community.fabric.microsoft.com/t5/Desktop/Calculated-2-years-previous/m-p/130313/highlight/true | 1,718,609,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00420.warc.gz | 152,293,257 | 114,826 | cancel
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Post Prodigy
## Calculated 2 years previous
Hello,
Does anyone know how I would write a correct formula?
Year 2 = CALCULATE([YTD Budget],ALLSELECTED('Calendar'[Year]) -2,ALL('Calendar'[Month]))
In the above formula I want to be able to select any year in the calendar table and have the formula go back 2 years in time and grab that entire years budget.
Adding the -2 breaks th eformula but you can see what I was trying to do. What should I be using?
4 REPLIES 4
Community Champion
@lcasey
```YTD From 2 Ago =
CALCULATE ( [YTD Budget], DATEADD ( 'Calendar'[Date], - 2, YEAR ) )```
If [YTD Budget] uses TOTALYTD - then the above will give you the YTD Total but from 2 years ago!
Post Prodigy
Thanks,
My YTD budget is useing:
YTD Budget = IF(SUM('00-GLSummary'[Revenue]) > 0, Abs(SUM('00-Budget'[BUDGETAMT])),sum('00-Budget'[BUDGETAMT]))
So this formula is blank for now. I am going to try and convert my YTD Budget to use TotalYTD
That sounds like a more sdtandard way to calculate amounts anyway.
Post Prodigy
Weird,
I changed my formula to:
IF(SUM('00-GLSummary'[Revenue]) > 0, TOTALYTD(Sum('00-Budget'[BUDGETAMT]),'Calendar'[Date]),sum('00-Budget'[BUDGETAMT]))
and the YTD From 2 Ago = CALCULATE ( [YTD Budget], DATEADD ( 'Calendar'[Date], - 2, YEAR ) )
is still blank.
Super User
Hi @lcasey
Just to confirm that you do not have any filters in the Visual, Page or Report levels?
What I would suggest to troubleshoot, is to create a table and put the indivudual measures into the table with your dates. In doing so this can help visually see where the possible error is occuring.
I have done that in the past, which has helped me identify where I was going wrong.
Proud to be a Super User!
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2_Math141_Sample_C_Solutions
# 2_Math141_Sample_C_Solutions - WW5 I R;IATH 141 2[EXAMII l...
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Unformatted text preview: WW5 I R;’IATH 141 2 [EXAMII l SAMPLEC‘ 1 F. dl I. . I. l—cosx +oo ' m he "mt \$11.11;; I2 ' 5. Find the sum of the series 2 {tan—101+ 1}—tan_1(n)]. Hz! a) w a) 1r b) 1 b) E J 1 4 c _ 2 c) E l 2 d) —- . ‘2 dfr d _._ ) 4 e) r1"he series diverges. 2. Find the limit _1i:nn(1 — mi “’0 a +3 6. If the series 2 an = 10, then find Iim ( u n-~+oo 2 a.) 0 “:1 b} 1 a) 0 c} -2 3 b _ "g ) 2 d) r: 13 e) 1n‘2 c) 2 d) ‘2 .1 l 3. Evaluate the integral / —5 dz. 3) 00 . ._1 :12 a) U '3: 7. Determine whether the series is convergent or divergent Z W. b) l H'.:I C) ‘1 a} convergent. by the root test. d) ‘2 b) convergent by the comparison test 3) The integral diVerges- c) divergent by the divergence test d) divergent by the root test +03 n—l 4. Find the sum of the series 2 2 . 6) divergent by the rat-i0 tth n:1 a) Z 8. For each ofthe series (I) and (II) given below choose the right. answer. 8 +90 2 +oc . — 1 1 + 3m :1. 8 (I) Z n (M) Z e—H— b — 4 ‘ 'n I } .1, “:1 Em. +1 112“ 6 c) 8 2 a) Only (I) converges. d} :f h) Only (II) converges. e) 4 (3} Both diverge. .d} Bot-h converge. e} None of the above. +m . . s . _1 n+1 _ ‘_ t 9. What is the minimum number of terms of the series Z --——( T33 D If It 1" Emerge“ "=1 we need to add to find the sum with 1 error I g 0.001? +30 (-1)“ 13. Z a) 3 “:1 n + 1 b) 5 +00 2 sin n (_l)n c.) 9 E R) d) 1] +00 n ' 15. —1 n e) 1.5 "2:; ) 2 + 1 +00 10. The terms of a series 2 an are defined recursively by equations 15. a] : 1 and ‘l + in n. _ . . a.“ H = M can. determine whether the series converges or dl- verges. n 17. a) The series converges by the root test. 1)) The series converges by the ratio test. c] The series diverges by the test for divergence. d) The series diverges by the comparison test. e) The series diverges by the ratio test. +30 22" II. The series 2 — is l “:1 n' a) convergent by the ratio test. h} cmlvergent by the root test. c) divergent by the integral test. d} a divergent geometric series. e) divergent by the ratio test. 1 nlnn +"XJ 12. The series 2 {1:2 is a) convergent by the ratio test. b} convergent by the integral test. e) divergent by the integral test. d) divergent by the divergence test. e) divergent by the ratio test. For Problems 13- 17, (each worth 3 points) determine whether each series is absolutely convergent. conditionally convergent. Cir diver— gent. Code on your acantron sheet: A - if the series is Absolutely convergent, C — if it is Conditionally Convergent, SAMPLE C.‘ MATH 141 EXAM II SAMPLE C 18. {15 points) Determine whether the given sequence converges or 19. [10 pts.) Determine whether the following integral is convergent or diverges‘ If it converges, calculate the limit. If the series diverges circle the answer. 5“ a) a“ = 3n+4 diverges l b) an = n sin — Tl. diverges c} an = (—1)""' diverges 4 A Lil an: diverges RCOSTI. n2 + 1 dive rgee' 6‘.) an 3 i n+1 converges to converges to converges to co 1] verges to (To nve r gas to divergent. If convergent, find its value: 1 lnz —,dx. f” a Justify each step carefully. a) EXAM Il- FORM A b)1.C‘2.D3.E4.B5.BS.B?.C8.D9.CIU.HII.A 12. 013. 014.1% 15. D16. (31?. A c) 18. a) Diverges; b) Converges to 1: c} converges 1-0 0'. d} converges to 1‘I e) converges to 0. cl) 19. Converge-5 to —4 {l ...
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https://www.convertunits.com/from/bucket+%5BUS%5D/to/femtolitre | 1,601,353,782,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00191.warc.gz | 694,736,274 | 8,488 | ## ››Convert bucket [US] to femtoliter
bucket [US] femtolitre
Did you mean to convert bucket [US] bucket [UK] to femtolitre
How many bucket [US] in 1 femtolitre? The answer is 5.2834410248312E-17.
We assume you are converting between bucket [US] and femtoliter.
You can view more details on each measurement unit:
bucket [US] or femtolitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 52.834410248312 bucket [US], or 1.0E+18 femtolitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between buckets and femtoliters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of bucket [US] to femtolitre
1 bucket [US] to femtolitre = 1.8927059E+16 femtolitre
2 bucket [US] to femtolitre = 3.7854118E+16 femtolitre
3 bucket [US] to femtolitre = 5.6781177E+16 femtolitre
4 bucket [US] to femtolitre = 7.5708236E+16 femtolitre
5 bucket [US] to femtolitre = 9.4635295E+16 femtolitre
6 bucket [US] to femtolitre = 1.13562354E+17 femtolitre
7 bucket [US] to femtolitre = 1.32489413E+17 femtolitre
8 bucket [US] to femtolitre = 1.51416472E+17 femtolitre
9 bucket [US] to femtolitre = 1.70343531E+17 femtolitre
10 bucket [US] to femtolitre = 1.8927059E+17 femtolitre
## ››Want other units?
You can do the reverse unit conversion from femtolitre to bucket [US], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Femtoliter
The SI prefix "femto" represents a factor of 10-15, or in exponential notation, 1E-15.
So 1 femtoliter = 10-15 liter.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 636 | 2,095 | {"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} | 3.21875 | 3 | CC-MAIN-2020-40 | latest | en | 0.715086 |
https://dreme.stanford.edu/news/halloween-pumpkin-design-ideas-that-support-math-learning/ | 1,726,796,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00762.warc.gz | 192,900,315 | 28,152 | # Halloween Pumpkin Design Ideas That Support Math Learning
Whether creating Halloween decorations for your home or classroom, incorporating pumpkin design into your activities offers many opportunities to develop early math skills.
Key Points
• Decorating jack-o’-lanterns offers opportunities for children to explore early math concepts like shape, size, and measurement.
• You can do this activity with things commonly found at home and in classrooms—no pumpkin required.
• Explore more ideas for incorporating math into pumpkin designs in this handout.
Whether creating Halloween decorations for your home or classroom, incorporating pumpkin design into your activities offers many opportunities to develop early math skills. Children can practice shape, measurement, and spatial skills while designing the face of a jack-o’-lantern. You don’t even need a real pumpkin for this fun math activity. Use things that you already have on hand, such as scrap paper and repurposed containers, to design a jack-o’-lantern and engage in math learning together.
## Pumpkin Design Activity Instructions
To be sure, carving pumpkins can be messy and difficult, and traditional pumpkin carving tools are not appropriate for young hands. Instead, you and your children can design a jack-o’-lantern with washable markers applied directly to the pumpkin. Another idea is to paste or trace shapes onto a pumpkin, piece of paper, or things that otherwise might end up in the recycling bin. For example, you can use plastic lids from empty oatmeal containers as the pumpkin and cut out the shapes from leftover envelopes that came in the mail. Here are the instructions for designing a jack-o’-lantern with a focus on math exploration.
### Step 1: Gather the materials.
Choose what will serve as your pumpkin base—an actual pumpkin, a piece of paper, or some other object. You’ll need paper, scissors, and a measuring tool, such as a ruler or measuring tape. The ruler will help you cut equal sides for the squares and decide how big to make your shapes. Have tape or glue ready to affix the shapes to your pumpkin. Washable markers are optional for tracing shapes and adding embellishments.
### Step 2: Create an assortment of shapes.
Cut or draw the shapes that will be used to build your jack-o’-lantern. For tips on what shapes to include, download and print our handout.
Preparing the shapes opens opportunities to engage in conversations around key early math concepts, including:
• Size. What will fit on your pumpkin?
• Number. How many triangles will we need to make the eyes and nose?
• Estimation. About how many triangles do you think we will need to make a mouth out of triangles?
### Step 3:Design the jack-o’-lantern face.
Planning around and arranging the shapes invites conversations about:
• Shape. What shape should we use for the nose?
• Shape attributes. These rectangles are different. How come they are both rectangles?
• Distance. How far apart should we place the eyes?
• Spatial relations. The nose goes in the middle of the pumpkin. The eyes are above the nose. If we rotate the shape, the nose looks different.
• Size. Is the nose triangle bigger or smaller than the triangles we used for the eyes?
• Number. How many squares will fit where we want to make the mouth?
### Adaptation Ideas to Support Learning
To extend children’s math learning and offer different challenges, try these ideas:
Create several jack-o’-lanterns and use just one different shape for each. In the end, you could have a jack-o’-lantern made entirely of triangles, another one made of squares, another one made of octagons, and another one made of trapezoids.
If there is more than one child, consider giving each child the same combination of shape pieces but keep the design open-ended. See all the different ways they create jack-o’-lantern faces!
A jack-o’-lantern usually has two eyes, one nose, and one mouth. There’s nothing wrong with this classic design, but you can engage children in deciding whether to add additional flair: A third eye? Drawn-on ears? Hair, and if so, will the hair be made with straight, curvy, curly, or zigzag lines?
Help children build new, complex shapes out of the shapes they already have. Overlapping triangles and squares can create different looking eyes and mouths. For example, see how different the two jack-o’-lanterns pictured on the right look just by changing the kind of triangles we used to make the noses.
Pumpkin decoration opens opportunities for children to learn about shapes and broaden their math vocabulary. You can point out that no matter the direction that a triangle faces, it is a closed shape with three sides and three angles. For children who are interested in learning about the different types of triangles, explain that some triangles have special features:
### Step 4: Enjoy and talk about the designs.
Like any craft, each result is unique. To keep the math conversations going, explore why and how each jack-o’-lantern looks different. Have children look for other jack-o’-lanterns in their school and neighborhood. Instead of pointing out the shapes yourself, first listen for whether your children now spontaneously notice shape, size, and spatial features in the pumpkin designs. Then, ask questions about what they see and enjoy exploring math together! | 1,112 | 5,340 | {"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} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.913011 |
https://www.peterswar.net/how-to-calculate-interest-on-hard-money-loans-1654740108/ | 1,657,069,437,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00653.warc.gz | 929,894,468 | 13,179 | ‹ Articles
Tips to Calculate Interest Only Loan – Hard Money Loan Calculator
Hard Money Loans, Real Estate & Total Interest Cost
As a money lender, I ’ thousand much asked about our sake only loans, how they work, and the cost of taking out the hard money loan. here ’ s some information about how to calculate an matter to only loan and how you can calculate the payments. It is what we call our hard money loanword calculator .
Let ’ s start with the basics. Hard money lenders necessitate pastime payments on the hard money loans principal libra merely like any other fiscal institution .
sake is the monetary cathexis for using borrowed money and is normally expressed as an annual share rate ( APR ). Our loans are considered to be “ interest-only ” loans.
This means the only requital due monthly is the interest on the loan for the former month. At the goal of the loan term, the entire chief come will be due and collectible .
How to Calculate Cost of a Hard Money Interest Only Loan
If a hard money lender uses the bare interest method, here ’ s how you can calculate the sum of pastime you will have to pay on an interest-only lend and what your monthly cost will be. This is the hard money calculator .
You’ll need:
1. the total principal of the loan amount
2. the hard money loan interest rate you’re being charged
3. and the length of the loan in years.
EXAMPLE:
Let ’ s say you take out a five-year loan for \$100,000, and your lender gives you an annual percentage rate (APR) of 10% .
Calculate Interest On A Loan – Method #1
The first lend calculator formula is as follows :
• Principal Loan Amount x Interest Rate x Time (# of years) = Total Interest Dollars You’ll Pay For Loan
• \$100,000 x .10 x 5 = \$50,000 total interest will be paid for the loan.
To determine how much you will have to pay each calendar month, take the interest dollars from the formula above and divide it by the number of months you ’ ll have the lend. eminence, to get the length of your hard money loan in months, multiply the number of years you ’ ll have the loan x 12 months .
Calculate Interest On A Loan – Method #2
The irregular rule is :
• Total Interest \$’s / # of months you’ll have the loan = Your Monthly Payment
• \$50,000 / 60 months = \$833.33
so, in this exercise, you would pay \$ 833.33 every calendar month for five years until the principal balance is paid in fully .
Real Estate | Hard Money | Interest Rate
The Benefit of Hard Money Interest Only Loans
Keeping in mind the saying a dollar today is more than a dollar tomorrow, here are some facts about interest merely mortgage loans that are interesting to consider .
Interest on virtually all mortgages is paid in arrears. This means that the payment you make each month pays the interest due for the previous calendar month. The benefit is that you use the funds for a calendar month before you have a payment due .
Most hard money lenders will prorate sake at close up, so the payment is due on the first of the month, the following month. If you close on January 30th, the lender will prorate two days of interest, and your first sake payment will not be until March 1st. rent is paid in improvement, and concern is paid in arrears. indeed, with pastime, you hold onto your money long, giving you more time to invest it elsewhere and earn off it .
Know All of Your Costs – Loan Figures, Penalty Charges & Loan Origination Fees
many lenders charge a prepayment punishment if a loan is paid up before the term of the loanword ends or at another pin down time. With DFW Hard Money, we do not charge prepayment penalties no matter how promptly a lend is paid in full. This is a significant advantage to the person taking out the lend .
We require the monthly concern payment to be drafted from your explanation on the 1st of every calendar month, and our interest rates can vary up to 13 %. Check out our 5 Step Money Process that outlines the steps from application to loan wages .
Get Expert Help & Use A Hard Money Loan Calculator
once you know the actual monetary value of your lend, you can make smart decisions – a hard money lend calculator will help with this. We are here if you end up needing help oneself with this or any view of a intemperate money loans , and it is what we do!
Who Is Trusted For Hard Money Loans?
DFW Hardmoney is a true private money lender providing honest and bare loans for veridical estate of the realm investing. We specialize in 24-48-hour fund with competitive terms, no appraisal, and no predetermined polish requital required. We are the hard money lenders you can trust.
Interest Rate Whether you want to fix and flip houses, need funds for commercial, transactional funding, very estate developments, or you ’ re a first-time or experience investor, we ’ rhenium here to help in all matters related to hard money & hard money commercial loans .
Contact us with any hard money loanword calculator questions, queries about how how to calculate an interest merely lend, or equitable a absolve consultation call (817)200-7575 .
source : https://www.peterswar.net
Category : Finance
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Select ’ s editorial team works independently to review fiscal products and write articles we think our readers will find useful. We earn a perpetration from affiliate… | 1,375 | 6,309 | {"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} | 3.640625 | 4 | CC-MAIN-2022-27 | latest | en | 0.9306 |
https://www.hackmath.net/en/example/1036 | 1,563,315,626,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00525.warc.gz | 698,345,945 | 2,488 | Rectangle - sides
#0 xxx() called at [/LinSys.php:671]
#1 LinSys::tryIntegerEquations(Array ([0] => D,[1] => F,[2] => a,[3] => c,[4] => d,[5] => e,[6] => f,[7] => m,[8] => o,[9] => p,[10] => q,[11] => r,[12] => t,[13] => x), Array ([0] => a^2 +5a -266 =0,[1] => p=1,[2] => q=5,[3] => r=-266,[4] => D = q^2 - 4pr = 5^2 - 4** 1 ** (-266) = 1089,[5] => ax{1.2} = frac{ -q pm sqrt{ D } }{ 2p } = frac{ -5 pm sqrt{ 1089 } }{ 2 },[6] => ax{1.2} = frac{ -5 pm 33 }{ 2 },[7] => ax{1.2} = -2.5 pm 16.5,[8] => ax{1} = 14,[9] => ax{2} = -19,[10] => (a -14) (a +19) = 0), 1) called at [/LinSys.php:344]
#2 LinSys::SolveInner(a^2 +5a -266 =0
p=1; q=5; r=-266
D = q^2 - 4pr = 5^2 - 4** 1 ** (-266) = 1089
D>0
ax{1.2} = frac{ -q pm sqrt{ D } }{ 2p } = frac{ -5 pm sqrt{ 1089 } }{ 2 }
ax{1.2} = frac{ -5 pm 33 }{ 2 }
ax{1.2} = -2.5 pm 16.5
ax{1} = 14
ax{2} = -19
text{ Factored form: }
(a -14) (a +19) = 0, , 1, linsys, 1, , 1, 1) called at [/LinSys.php:220]
#3 LinSys::Solve(a^2 +5a -266 =0
p=1; q=5; r=-266
D = q^2 - 4pr = 5^2 - 4** 1 ** (-266) = 1089
D>0
a_{1,2} = \frac{ -q \pm \sqrt{ D } }{ 2p } = \frac{ -5 \pm \sqrt{ 1089 } }{ 2 }
a_{1,2} = \frac{ -5 \pm 33 }{ 2 }
a_{1,2} = -2.5 \pm 16.5
a_{1} = 14
a_{2} = -19
\text{ Factored form: }
(a -14) (a +19) = 0, , 1, linsys, 1, , 1) called at [/Example_Generic.php:87]
#4 Example_Generic->GenerateSolveVector(stdClass Object ([example_id] => 1036,[title_sk] => Obĺžnik - strany,[title_en] => Rectangle - sides,[title_cz] => Obdélník - strany,[add_date] => 2014-05-11 15:06:44,[img] => colored_squares.jpg,[visible] => 1,[text_sk] => Aký je obvod obdĺžnika, ktorého obsah je $S cm^2 a dĺžka kratšej strany je o$dx cm kratšia ako dĺžka dlhšej strany?,[text_en] => What is the perimeter of a rectangle with area $S cm^2 if length of the shorter side is$dx cm shorter than the length of the longer side?,[text_cz] => Jaký je obvod obdélníku, jehož obsah je $S cm^2 a délka kratší strany je o$dx cm kratší než délka delší strany?,[input_vector] => do
{
$a = rand(10,40);$b = rand(10,40);
}while($a==$b);
$dx = abs($a-$b);$S = $a*$b;,[output_vector] => $o = new StdClass();$o->in="x=#cm";
$o->val = "2*($a+$b)";$eq = QTex(1,$dx, -$S, 'a', true);
$o->tex = "ab =$S
b = a + $dx a(a+$dx) = $S a^2 +$dx a - $S =0$eq
a = $a b=$b
o = 2(a+b) = RES";,[user_id] => 12,[approved] => 1,[cnt_views] => 12006,[cnt_solved] => 413,[cnt_solved_ok] => 20,[focus] => 1,[preview_sk] => Aký je obvod obdĺžnika, ktorého obsah je 266 cm^2 a dĺžka kratšej strany je o 5 cm kratšia ako dĺžka dlhšej strany?,[preview_en] => What is the perimeter of a rectangle with area 266 cm^2 if length of the shorter side is 5 cm shorter than the length of the longer side?,[preview_cz] => Jaký je obvod obdélníku, jehož obsah je 266 cm^2 a délka kratší strany je o 5 cm kratší než délka delší strany?,[preview_vector_sk] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[preview_vector_en] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[preview_vector_cz] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[last_regenerate] => 2019-07-09 16:16:26,[external_url] => http://forum.matematika.cz/viewtopic.php?id=73651,[suggestion_id] => 0,[fulltext_sk] => ~ oblznik strany aky je obvod obdlznika ktoreho obsah 266 cm^2 a dlzka kratsej o 5 cm kratsia ako dlhsej kvadraticka rovnica algebra planimetria obdlznik 9 rocnik stredna skola 1036 ~,[fulltext_en] => ~ rectangle sides what is the perimeter of a rectangle with area 266 cm^2 if length shorter side 5 cm than longer quadratic equation algebra planimetrics shape 9t 9 th grade 14y 4 y high school 1036 ~,[fulltext_cz] => ~ obdelnik strany jaky je obvod obdelniku jehoz obsah 266 cm^2 a delka kratsi o 5 cm nez delsi kvadraticka rovnice algebra planimetrie obdelnik 9 rocnik stredni skola 1036 ~,[english_last_modified] => 0000-00-00 00:00:00,[title] => Rectangle - sides,[text] => What is the perimeter of a rectangle with area $S cm^2 if length of the shorter side is$dx cm shorter than the length of the longer side?), stdClass Object ([a] => 14,[b] => 19,[dx] => 5,[S] => 266)) called at [/Example_Generic.php:869]
#5 Example_Generic->Run(stdClass Object ([example_id] => 1036,[title_sk] => Obĺžnik - strany,[title_en] => Rectangle - sides,[title_cz] => Obdélník - strany,[add_date] => 2014-05-11 15:06:44,[img] => colored_squares.jpg,[visible] => 1,[text_sk] => Aký je obvod obdĺžnika, ktorého obsah je $S cm^2 a dĺžka kratšej strany je o$dx cm kratšia ako dĺžka dlhšej strany?,[text_en] => What is the perimeter of a rectangle with area $S cm^2 if length of the shorter side is$dx cm shorter than the length of the longer side?,[text_cz] => Jaký je obvod obdélníku, jehož obsah je $S cm^2 a délka kratší strany je o$dx cm kratší než délka delší strany?,[input_vector] => do
{
$a = rand(10,40);$b = rand(10,40);
}while($a==$b);
$dx = abs($a-$b);$S = $a*$b;,[output_vector] => $o = new StdClass();$o->in="x=#cm";
$o->val = "2*($a+$b)";$eq = QTex(1,$dx, -$S, 'a', true);
$o->tex = "ab =$S
b = a + $dx a(a+$dx) = $S a^2 +$dx a - $S =0$eq
a = $a b=$b
o = 2(a+b) = RES";,[user_id] => 12,[approved] => 1,[cnt_views] => 12006,[cnt_solved] => 413,[cnt_solved_ok] => 20,[focus] => 1,[preview_sk] => Aký je obvod obdĺžnika, ktorého obsah je 266 cm^2 a dĺžka kratšej strany je o 5 cm kratšia ako dĺžka dlhšej strany?,[preview_en] => What is the perimeter of a rectangle with area 266 cm^2 if length of the shorter side is 5 cm shorter than the length of the longer side?,[preview_cz] => Jaký je obvod obdélníku, jehož obsah je 266 cm^2 a délka kratší strany je o 5 cm kratší než délka delší strany?,[preview_vector_sk] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[preview_vector_en] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[preview_vector_cz] => O:8:"stdClass":4:{s:1:"a";i:14;s:1:"b";i:19;s:2:"dx";i:5;s:1:"S";i:266;},[last_regenerate] => 2019-07-09 16:16:26,[external_url] => http://forum.matematika.cz/viewtopic.php?id=73651,[suggestion_id] => 0,[fulltext_sk] => ~ oblznik strany aky je obvod obdlznika ktoreho obsah 266 cm^2 a dlzka kratsej o 5 cm kratsia ako dlhsej kvadraticka rovnica algebra planimetria obdlznik 9 rocnik stredna skola 1036 ~,[fulltext_en] => ~ rectangle sides what is the perimeter of a rectangle with area 266 cm^2 if length shorter side 5 cm than longer quadratic equation algebra planimetrics shape 9t 9 th grade 14y 4 y high school 1036 ~,[fulltext_cz] => ~ obdelnik strany jaky je obvod obdelniku jehoz obsah 266 cm^2 a delka kratsi o 5 cm nez delsi kvadraticka rovnice algebra planimetrie obdelnik 9 rocnik stredni skola 1036 ~,[english_last_modified] => 0000-00-00 00:00:00,[title] => Rectangle - sides,[text] => What is the perimeter of a rectangle with area $S cm^2 if length of the shorter side is$dx cm shorter than the length of the longer side?)) called at [/index_real.php:185]
#6 HackMath->ExampleDetail() called at [/index_real.php:316]
#7 HackMath->ExampleAction() called at [/index_real.php:461]
#8 HackMath->Run() called at [/index_real.php:815]
#9 include(/index_real.php) called at [/index.php:41] | 2,916 | 7,165 | {"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} | 3.59375 | 4 | CC-MAIN-2019-30 | latest | en | 0.283645 |
https://www.airmilescalculator.com/distance/sdu-to-sjk/ | 1,627,421,028,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00217.warc.gz | 639,477,213 | 17,229 | # Distance between Rio De Janeiro (SDU) and São José Dos Campos (SJK)
Flight distance from Rio De Janeiro to São José Dos Campos (Santos Dumont Airport – São José dos Campos Airport) is 173 miles / 279 kilometers / 151 nautical miles. Estimated flight time is 49 minutes.
Driving distance from Rio De Janeiro (SDU) to São José Dos Campos (SJK) is 215 miles / 346 kilometers and travel time by car is about 4 hours 17 minutes.
## Map of flight path and driving directions from Rio De Janeiro to São José Dos Campos.
Shortest flight path between Santos Dumont Airport (SDU) and São José dos Campos Airport (SJK).
## How far is São José Dos Campos from Rio De Janeiro?
There are several ways to calculate distances between Rio De Janeiro and São José Dos Campos. Here are two common methods:
Vincenty's formula (applied above)
• 173.203 miles
• 278.743 kilometers
• 150.509 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 172.936 miles
• 278.314 kilometers
• 150.278 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Santos Dumont Airport
City: Rio De Janeiro
Country: Brazil
IATA Code: SDU
ICAO Code: SBRJ
Coordinates: 22°54′37″S, 43°9′47″W
B São José dos Campos Airport
City: São José Dos Campos
Country: Brazil
IATA Code: SJK
ICAO Code: SBSJ
Coordinates: 23°13′45″S, 45°51′41″W
## Time difference and current local times
There is no time difference between Rio De Janeiro and São José Dos Campos.
-03
-03
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 51 kg (111 pounds).
## Frequent Flyer Miles Calculator
Rio De Janeiro (SDU) → São José Dos Campos (SJK).
Distance:
173
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
173
Round trip? | 515 | 2,012 | {"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-2021-31 | latest | en | 0.764456 |
https://zhangruochi.com/Machine-Learning-Andraw-Ng/2019/02/15/ | 1,628,002,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00044.warc.gz | 1,126,754,872 | 47,119 | Reference From Coursera Course Machine Learning. I am also really sorry that I did not write down some sources when I cite from the web. I was too young and did not get any academic training when I wrote these notes.
Let me thank Ng first. This course Changed me. I still remember that afternoon when I was just a sophomore. I discovered the Machine Learning course lectured by Andrew Ng on Coursera and was fascinated by the underlying algorithms. It was amazing to see that a simple yet elegant mathematical model could make predictions on new data after being trained with large amounts of training sets for analysis and fitting. I spent several days on the assignments and developed a classifier to filter spam emails and an Optical Character Recognition program. Never expecting that I could program the machine to gain the cognitive ability, I was so thrilled and resolved to advance my knowledge in this area. About two years later, I took this course again and reviewed some important concepts to prapare the interviews. I wrote down notes and paid for the course this time
This is my course certificate. I really want all of guys who want to dive into the area of machine learning to take this course on Coursera.
## What is Machine Learning?
Two definitions of Machine Learning are offered.
• Arthur Samuel described it as: “the field of study that gives computers the ability to learn without being explicitly programmed.“ This is an older, informal definition.
• Tom Mitchell provides a more modern definition: “A computer program is said to learn from experience E with respect to some class of tasks T and performance measure P, if its performance at tasks in T, as measured by P, improves with experience E.”
Example: playing checkers.
• E = the experience of playing many games of checkers
• T = the task of playing checkers.
• P = the probability that the program will win the next game.
In general, any machine learning problem can be assigned to one of two broad classifications:
Supervised learning and Unsupervised learning.
### Supervised Learning
In supervised learning, we are given a data set and already know what our correct output should look like, having the idea that there is a relationship between the input and the output.
Supervised learning problems are categorized into “regression“ and “classification“ problems. In a regression problem, we are trying to predict results within a continuous output, meaning that we are trying to map input variables to some continuous function. In a classification problem, we are instead trying to predict results in a discrete output. In other words, we are trying to map input variables into discrete categories.
• Example 1:
• Given data about the size of houses on the real estate market, try to predict their price. Price as a function of size is a continuous output, so this is a regression problem.
• We could turn this example into a classification problem by instead making our output about whether the house “sells for more or less than the asking price.” Here we are classifying the houses based on price into two discrete categories.
• Example 2:
• Regression - Given a picture of a person, we have to predict their age on the basis of the given picture
• Classification - Given a patient with a tumor, we have to predict whether the tumor is malignant or benign.
### Unsupervised Learning
Unsupervised learning allows us to approach problems with little or no idea what our results should look like. We can derive structure from data where we don’t necessarily know the effect of the variables.
We can derive this structure by clustering the data based on relationships among the variables in the data.
With unsupervised learning there is no feedback based on the prediction results.
Example:
Clustering: Take a collection of 1,000,000 different genes, and find a way to automatically group these genes into groups that are somehow similar or related by different variables, such as lifespan, location, roles, and so on.
### Model Representation
To establish notation for future use, we’ll use x(i) to denote the “input” variables (living area in this example), also called input features, and y(i) to denote the “output” or target variable that we are trying to predict (price). A pair (x(i),y(i)) is called a training example, and the dataset that we’ll be using to learn—a list of m training examples (x(i),y(i));i=1,…,m—is called a training set. Note that the superscript “(i)” in the notation is simply an index into the training set, and has nothing to do with exponentiation. We will also use X to denote the space of input values, and Y to denote the space of output values. In this example, X = Y = ℝ.
To describe the supervised learning problem slightly more formally, our goal is, given a training set, to learn a function h : X → Y so that h(x) is a “good” predictor for the corresponding value of y. For historical reasons, this function h is called a hypothesis. Seen pictorially, the process is therefore like this:
When the target variable that we’re trying to predict is continuous, such as in our housing example, we call the learning problem a regression problem. When y can take on only a small number of discrete values (such as if, given the living area, we wanted to predict if a dwelling is a house or an apartment, say), we call it a classification problem.
## Linear Regression with Multiple Variables
### Cost Function
We can measure the accuracy of our hypothesis function by using a cost function. This takes an average difference (actually a fancier version of an average) of all the results of the hypothesis with inputs from x’s and the actual output y’s.
To break it apart, it is $\frac{1}{2}\bar{x}$ where $\bar{x}$ is the mean of the squares of $h_θ(x_i)−y_i$ , or the difference between the predicted value and the actual value.
This function is otherwise called the “Squared error function”, or “Mean squared error”. The mean is halved ($\frac{1}{2}$) as a convenience for the computation of the gradient descent, as the derivative term of the square function will cancel out the $\frac{1}{2}$ term. The following image summarizes what the cost function does:
#### Cost Function - Intuition I
If we try to think of it in visual terms, our training data set is scattered on the x-y plane. We are trying to make a straight line (defined by $h_θ(x)$) which passes through these scattered data points.
Our objective is to get the best possible line. The best possible line will be such so that the average squared vertical distances of the scattered points from the line will be the least. Ideally, the line should pass through all the points of our training data set. In such a case, the value of $J(θ_0,θ_1)$ will be 0. The following example shows the ideal situation where we have a cost function of 0.
When $θ_1=1$, we get a slope of 1 which goes through every single data point in our model. Conversely, when θ1=0.5, we see the vertical distance from our fit to the data points increase.
This increases our cost function to 0.58. Plotting several other points yields to the following graph:
Thus as a goal, we should try to minimize the cost function. In this case, $θ_1=1$ is our global minimum.
#### Cost Function - Intuition II
A contour plot is a graph that contains many contour lines. A contour line of a two variable function has a constant value at all points of the same line. An example of such a graph is the one to the right below.
Taking any color and going along the ‘circle’, one would expect to get the same value of the cost function. For example, the three green points found on the green line above have the same value for $J(θ_0,θ_1)$ and as a result, they are found along the same line. The circled x displays the value of the cost function for the graph on the left when $θ_0 = 800$ and $θ_1= -0.15$. Taking another h(x) and plotting its contour plot, one gets the following graphs:
When $θ_0 = 360$ and $θ_1 = 0$, the value of $J(θ_0,θ_1)$ in the contour plot gets closer to the center thus reducing the cost function error. Now giving our hypothesis function a slightly positive slope results in a better fit of the data.
The graph above minimizes the cost function as much as possible and consequently, the result of $θ_1$ and $θ_0$ tend to be around 0.12 and 250 respectively. Plotting those values on our graph to the right seems to put our point in the center of the inner most ‘circle’.
So we have our hypothesis function and we have a way of measuring how well it fits into the data. Now we need to estimate the parameters in the hypothesis function. That’s where gradient descent comes in.
Imagine that we graph our hypothesis function based on its fields $θ_0$ and $θ_1$ (actually we are graphing the cost function as a function of the parameter estimates). We are not graphing x and y itself, but the parameter range of our hypothesis function and the cost resulting from selecting a particular set of parameters.
We put $θ_0$ on the x axis and $θ_1$ on the y axis, with the cost function on the vertical z axis. The points on our graph will be the result of the cost function using our hypothesis with those specific theta parameters. The graph below depicts such a setup.
We will know that we have succeeded when our cost function is at the very bottom of the pits in our graph, i.e. when its value is the minimum. The red arrows show the minimum points in the graph.
The way we do this is by taking the derivative (the tangential line to a function) of our cost function. The slope of the tangent is the derivative at that point and it will give us a direction to move towards. We make steps down the cost function in the direction with the steepest descent. The size of each step is determined by the parameter α, which is called the learning rate.
For example, the distance between each ‘star’ in the graph above represents a step determined by our parameter α. A smaller α would result in a smaller step and a larger α results in a larger step. The direction in which the step is taken is determined by the partial derivative of $J(θ_0,θ_1)$. Depending on where one starts on the graph, one could end up at different points. The image above shows us two different starting points that end up in two different places.
repeat until convergence:
where
j=0,1 represents the feature index number.
At each iteration j, one should simultaneously update the parameters $θ_1$,$θ_2$,…,$θ_n$. Updating a specific parameter prior to calculating another one on the $j^{(th)}$ iteration would yield to a wrong implementation.
In this video we explored the scenario where we used one parameter $θ_s1$ and plotted its cost function to implement a gradient descent. Our sformula for a single parameter was:
Repeat until convergence:
Regardless of the slope’s sign for \frac{d}{dθ_1}J(θ_1), $θ_1$ eventually converges to its minimum value. The following graph shows that when the slope is negative, the value of $θ_1$ increases and when it is positive, the value of θ1 decreases.
On a side note, we should adjust our parameter α to ensure that the gradient descent algorithm converges in a reasonable time. Failure to converge or too much time to obtain the minimum value imply that our step size is wrong.
How does gradient descent converge with a fixed step size α?
The intuition behind the convergence is that $\frac{d}{dθ_1}J(θ_1)$ approaches 0 as we approach the bottom of our convex function. At the minimum, the derivative will always be 0 and thus we get:
#### Gradient Descent For Linear Regression
When specifically applied to the case of linear regression, a new form of the gradient descent equation can be derived. We can substitute our actual cost function and our actual hypothesis function and modify the equation to :
repeat until convergence:{
}
where m is the size of the training set, $θ_0$ a constant that will be changing simultaneously with θ1 and xi,yiare values of the given training set (data).
Note that we have separated out the two cases for $θ_j$ into separate equations for $θ_0$ and $θ_1$; and that for $θ_1$ we are multiplying $x_i$ at the end due to the derivative. The following is a derivation of $\frac{∂}{∂θ_j}J(θ)$ for a single example :
The point of all this is that if we start with a guess for our hypothesis and then repeatedly apply these gradient descent equations, our hypothesis will become more and more accurate.
So, this is simply gradient descent on the original cost function J. This method looks at every example in the entire training set on every step, and is called batch gradient descent. Note that, while gradient descent can be susceptible to local minima in general, the optimization problem we have posed here for linear regression has only one global, and no other local, optima; thus gradient descent always converges (assuming the learning rate α is not too large) to the global minimum. Indeed, J is a convex quadratic function. Here is an example of gradient descent as it is run to minimize a quadratic function.
The ellipses shown above are the contours of a quadratic function. Also shown is the trajectory taken by gradient descent, which was initialized at (48,30). The x’s in the figure (joined by straight lines) mark the successive values of θ that gradient descent went through as it converged to its minimum.
### Matrices and Vectors
Matrices are 2-dimensional arrays:
The above matrix has four rows and three columns, so it is a 4 x 3 matrix.
A vector is a matrix with one column and many rows:
So vectors are a subset of matrices. The above vector is a 4 x 1 matrix.
Notation and terms:
• $A_{ij}$ refers to the element in the ith row and jth column of matrix A.
• A vector with ‘n’ rows is referred to as an ‘n’-dimensional vector.
• $v_i$ refers to the element in the ith row of the vector.
In general, all our vectors and matrices will be 1-indexed. Note that for some programming languages, the arrays are 0-indexed.
• Matrices are usually denoted by uppercase names while vectors are lowercase.
• “Scalar” means that an object is a single value, not a vector or matrix.
• ℝ refers to the set of scalar real numbers.
• ℝ𝕟 refers to the set of n-dimensional vectors of real numbers.
Addition and subtraction are element-wise, so you simply add or subtract each corresponding element:
Subtracting Matrices:
To add or subtract two matrices, their dimensions must be the same.
In scalar multiplication, we simply multiply every element by the scalar value:
In scalar division, we simply divide every element by the scalar value:
Experiment below with the Octave/Matlab commands for matrix addition and scalar multiplication. Feel free to try out different commands. Try to write out your answers for each command before running the cell below.
#### Matrix-Vector Multiplication
Below is an example of a matrix-vector multiplication. Make sure you understand how the multiplication works. Feel free to try different matrix-vector multiplications.
#### Matrix-Matrix Multiplication
We multiply two matrices by breaking it into several vector multiplications and concatenating the result.
An m x n matrix multiplied by an n x o matrix results in an m x o matrix. In the above example, a 3 x 2 matrix times a 2 x 2 matrix resulted in a 3 x 2 matrix.
To multiply two matrices, the number of columns of the first matrix must equal the number of rows of the second matrix.
For example:
#### Matrix Multiplication Properties
• Matrices are not commutative: A∗B≠B∗A
• Matrices are associative: (A∗B)∗C=A∗(B∗C)
The identity matrix, when multiplied by any matrix of the same dimensions, results in the original matrix. It’s just like multiplying numbers by 1. The identity matrix simply has 1’s on the diagonal (upper left to lower right diagonal) and 0’s elsewhere.
When multiplying the identity matrix after some matrix (A∗I), the square identity matrix’s dimension should match the other matrix’s columns. When multiplying the identity matrix before some other matrix (I∗A), the square identity matrix’s dimension should match the other matrix’s rows.
In other words:
### multivariate Linear Analytically
#### Multiple Features
Linear regression with multiple variables is also known as “multivariate linear regression”.
We now introduce notation for equations where we can have any number of input variables.
• $x^{(i)}_j$ = value of feature j in the $i^{th}$ training example
• $x^{(i)}$ =the input (features) of the $i^{th}$ training example
• m = the number of training examples
• n = the number of features
The multivariable form of the hypothesis function accommodating these multiple features is as follows:
In order to develop intuition about this function, we can think about θ0 as the basic price of a house, θ1 as the price per square meter, θ2 as the price per floor, etc. x1 will be the number of square meters in the house, x2 the number of floors, etc.
Using the definition of matrix multiplication, our multivariable hypothesis function can be concisely represented as:
This is a vectorization of our hypothesis function for one training example; see the lessons on vectorization to learn more.
Remark: Note that for convenience reasons in this course we assume $x^{(i)}_0=1$ for (i∈1,…,m). This allows us to do matrix operations with theta and x. Hence making the two vectors ‘θ’ and $x^{(i)}$ match each other element-wise (that is, have the same number of elements: n+1).]
### Gradient Descent For Multiple Variables
The gradient descent equation itself is generally the same form; we just have to repeat it for our ‘n’ features:
repeat until convergence:{
}
In other words:
repeat until convergence: {
}
The following image compares gradient descent with one variable to gradient descent with multiple variables:
#### Gradient Descent in Practice I - Feature Scaling
We can speed up gradient descent by having each of our input values in roughly the same range. This is because θ will descend quickly on small ranges and slowly on large ranges, and so will oscillate inefficiently down to the optimum when the variables are very uneven.
The way to prevent this is to modify the ranges of our input variables so that they are all roughly the same. Ideally:
or
These aren’t exact requirements; we are only trying to speed things up. The goal is to get all input variables into roughly one of these ranges, give or take a few.
Two techniques to help with this are feature scaling and mean normalization. Feature scaling involves dividing the input values by the range (i.e. the maximum value minus the minimum value) of the input variable, resulting in a new range of just 1. Mean normalization involves subtracting the average value for an input variable from the values for that input variable resulting in a new average value for the input variable of just zero. To implement both of these techniques, adjust your input values as shown in this formula:
Where μi is the average of all the values for feature (i) and si is the range of values (max - min), or si is the standard deviation.
For example, if xi represents housing prices with a range of 100 to 2000 and a mean value of 1000, then, $x_i:=\frac{price−1000}{1900}$.
#### Gradient Descent in Practice II - Learning Rate
Debugging gradient descent. Make a plot with number of iterations on the x-axis. Now plot the cost function, $J_{(θ)}$ over the number of iterations of gradient descent. If $J_{(θ)}$ ever increases, then you probably need to decrease α.
Automatic convergence test. Declare convergence if $J_{(θ)}$ decreases by less than E in one iteration, where E is some small value such as 10−3. However in practice it’s difficult to choose this threshold value.
It has been proven that if learning rate α is sufficiently small, then $J_{(θ)}$ will decrease on every iteration.
To summarize:
• If α is too small: slow convergence.
• If α is too large: may not decrease on every iteration and thus may not converge.
### Features and Polynomial Regression
We can improve our features and the form of our hypothesis function in a couple different ways.
We can combine multiple features into one. For example, we can combine $x_1$ and $x_2$ into a new feature x3 by taking $x_1⋅x_2$.
Polynomial Regression
Our hypothesis function need not be linear (a straight line) if that does not fit the data well.
We can change the behavior or curve of our hypothesis function by making it a quadratic, cubic or square root function (or any other form).
For example, if our hypothesis function is $h_θ(x)=θ_0+θ_1x_1$ then we can create additional features based on x1, to get the quadratic function $h_θ(x)=θ_0+θ_1x_1+θ_2x^2_1$ or the cubic function $hθ(x)=θ_0+θ_1x_1+θ_2x^2_1+θ_3x^3_1$
In the cubic version, we have created new features $x_2$ and $x_3$ where $x_2=x^2_1$ and $x_3=x^3_1$.
To make it a square root function, we could do: $h_θ(x)=θ_0+θ_1x_1+θ_2\sqrt{x_1}$
One important thing to keep in mind is, if you choose your features this way then feature scaling becomes very important.
eg. if $x_1$ has range 1 - 1000 then range of $x^2_1$ becomes 1 - 1000000 and that of $x^3_1$ becomes 1 - 1000000000
### Normal Equation
Gradient descent gives one way of minimizing J. Let’s discuss a second way of doing so, this time performing the minimization explicitly and without resorting to an iterative algorithm. In the “Normal Equation” method, we will minimize J by explicitly taking its derivatives with respect to the θj ’s, and setting them to zero. This allows us to find the optimum theta without iteration. The normal equation formula is given below:
There is no need to do feature scaling with the normal equation.
The following is a comparison of gradient descent and the normal equation:
|:—-|:—-|
|Need to choose alpha| No need to choose alpha|
|Needs many iterations| No need to iterate|
|$O(kn^2)$|$O(n^3)$ | need to calculate inverse of $X^TX$|
|Works well when n is large|Slow if n is very large|
With the normal equation, computing the inversion has complexity $O(n^3)$. So if we have a very large number of features, the normal equation will be slow. In practice, when n exceeds 10,000 it might be a good time to go from a normal solution to an iterative process.
#### Normal Equation Noninvertibility
When implementing the normal equation in octave we want to use the ‘pinv’ function rather than ‘inv.’ The ‘pinv’ function will give you a value of θ even if $X^TX$ is not invertible.
If $X^TX$ is noninvertible, the common causes might be having :
• Redundant features, where two features are very closely related (i.e. they are linearly dependent)
• Too many features (e.g. m ≤ n). In this case, delete some features or use “regularization” (to be explained in a later lesson).
Solutions to the above problems include deleting a feature that is linearly dependent with another or deleting one or more features when there are too many features.
## Logistic Regression
### Classification
To attempt classification, one method is to use linear regression and map all predictions greater than 0.5 as a 1 and all less than 0.5 as a 0. However, this method doesn’t work well because classification is not actually a linear function.
The classification problem is just like the regression problem, except that the values we now want to predict take on only a small number of discrete values. For now, we will focus on the binary classification problem in which y can take on only two values, 0 and 1. (Most of what we say here will also generalize to the multiple-class case.) For instance, if we are trying to build a spam classifier for email, then $x^{(i)}$ may be some features of a piece of email, and y may be 1 if it is a piece of spam mail, and 0 otherwise. Hence, y∈{0,1}. 0 is also called the negative class, and 1 the positive class, and they are sometimes also denoted by the symbols “-” and “+.” Given $x^{(i)}$, the corresponding $y^{(i)}$ is also called the label for the training example.
### Hypothesis Representation
We could approach the classification problem ignoring the fact that y is discrete-valued, and use our old linear regression algorithm to try to predict y given x. However, it is easy to construct examples where this method performs very poorly. Intuitively, it also doesn’t make sense for hθ(x) to take values larger than 1 or smaller than 0 when we know that y ∈ {0, 1}. To fix this, let’s change the form for our hypotheses $h_θ(x)$ to satisfy $0≤h_θ(x)≤1$. This is accomplished by plugging $θ^Tx$ into the Logistic Function.
Our new form uses the “Sigmoid Function,” also called the “Logistic Function”:
The following image shows us what the sigmoid function looks like:
The function g(z), shown here, maps any real number to the (0, 1) interval, making it useful for transforming an arbitrary-valued function into a function better suited for classification.
hθ(x) will give us the probability that our output is 1. For example, $h_θ(x)=0.7$ gives us a probability of 70% that our output is 1. Our probability that our prediction is 0 is just the complement of our probability that it is 1 (e.g. if probability that it is 1 is 70%, then the probability that it is 0 is 30%).
### Decision Boundary
In order to get our discrete 0 or 1 classification, we can translate the output of the hypothesis function as follows:
The way our logistic function g behaves is that when its input is greater than or equal to zero, its output is greater than or equal to 0.5:
Remember.
Again, the input to the sigmoid function g(z) (e.g. θTX) doesn’t need to be linear, and could be a function that describes a circle (e.g. $z=θ_0+θ_1x_2^1+θ_2x_2^2$) or any shape to fit our data.
### Cost Function
We cannot use the same cost function that we use for linear regression because the Logistic Function will cause the output to be wavy, causing many local optima. In other words, it will not be a convex function.
Instead, our cost function for logistic regression looks like:
When y = 1, we get the following plot for $J(θ)$ vs $h_θ(x)$:
Similarly, when y = 0, we get the following plot for $J(θ)$ vs $h_θ(x)$:
If our correct answer ‘y’ is 0, then the cost function will be 0 if our hypothesis function also outputs 0. If our hypothesis approaches 1, then the cost function will approach infinity.
If our correct answer ‘y’ is 1, then the cost function will be 0 if our hypothesis function outputs 1. If our hypothesis approaches 0, then the cost function will approach infinity.
Note that writing the cost function in this way guarantees that J(θ) is convex for logistic regression.
### Simplified Cost Function and Gradient Descent
We can compress our cost function’s two conditional cases into one case:
Notice that when y is equal to 1, then the second term $(1−y)log(1−h_θ(x))$ will be zero and will not affect the result. If y is equal to 0, then the first term $−ylog(h_θ(x))$ will be zero and will not affect the result.
We can fully write out our entire cost function as follows:
Remember that the general form of gradient descent is:
We can work out the derivative part using calculus to get:
### Multiclass Classification: One-vs-all
Now we will approach the classification of data when we have more than two categories. Instead of y = {0,1} we will expand our definition so that y = {0,1…n}.
Since y = {0,1…n}, we divide our problem into n+1 (+1 because the index starts at 0) binary classification problems; in each one, we predict the probability that ‘y’ is a member of one of our classes.
We are basically choosing one class and then lumping all the others into a single second class. We do this repeatedly, applying binary logistic regression to each case, and then use the hypothesis that returned the highest value as our prediction.
The following image shows how one could classify 3 classes:
To summarize:
Train a logistic regression classifier $h_θ(x)$ for each class to predict the probability that y = i, To make a prediction on a new x, pick the class that maximizes $h_θ(x)$.
### Solving the problem of Overfitting
#### The Problem of Overfitting
Consider the problem of predicting y from $x \in R$. The leftmost figure below shows the result of fitting a $y = θ_0+θ_1x$ to a dataset. We see that the data doesn’t really lie on straight line, and so the fit is not very good.
Instead, if we had added an extra feature $x_2$ , and fit $y=θ_0+θ_1x+θ_2x_2$ , then we obtain a slightly better fit to the data (See middle figure). Naively, it might seem that the more features we add, the better. However, there is also a danger in adding too many features: The rightmost figure is the result of fitting a 5th order polynomial $y=\sum^5_{j=0}θ_jx_j$. We see that even though the fitted curve passes through the data perfectly, we would not expect this to be a very good predictor of, say, housing prices (y) for different living areas (x). Without formally defining what these terms mean, we’ll say the figure on the left shows an instance of underfitting—in which the data clearly shows structure not captured by the model—and the figure on the right is an example of overfitting.
• Underfitting, or high bias, is when the form of our hypothesis function h maps poorly to the trend of the data. It is usually caused by a function that is too simple or uses too few features.
• Overfitting, or high variance, is caused by a hypothesis function that fits the available data but does not generalize well to predict new data. It is usually caused by a complicated function that creates a lot of unnecessary curves and angles unrelated to the data.
This terminology is applied to both linear and logistic regression. There are two main options to address the issue of overfitting:
• Reduce the number of features:
• Manually select which features to keep.
• Use a model selection algorithm (studied later in the course).
• Regularization
• Keep all the features, but reduce the magnitude of parameters $θ_j$.
• Regularization works well when we have a lot of slightly useful features.
#### Cost Function
If we have overfitting from our hypothesis function, we can reduce the weight that some of the terms in our function carry by increasing their cost.
Say we wanted to make the following function more quadratic:
We’ll want to eliminate the influence of $θ_3x_3$ and $θ_4x_4$ . Without actually getting rid of these features or changing the form of our hypothesis, we can instead modify our cost function:
We’ve added two extra terms at the end to inflate the cost of $θ_3$ and $θ_4$. Now, in order for the cost function to get close to zero, we will have to reduce the values of $θ_3$ and $θ_4$ to near zero. This will in turn greatly reduce the values of $θ_3x^3$ and $θ_4x^4$ in our hypothesis function. As a result, we see that the new hypothesis (depicted by the pink curve) looks like a quadratic function but fits the data better due to the extra small terms $θ_3x^3$ and $θ_4x^4$.
We could also regularize all of our theta parameters in a single summation as:
The λ, or lambda, is the regularization parameter. It determines how much the costs of our theta parameters are inflated.
Using the above cost function with the extra summation, we can smooth the output of our hypothesis function to reduce overfitting. If lambda is chosen to be too large, it may smooth out the function too much and cause underfitting. Hence, what would happen if λ=0 or is too small ?
#### Regularized Linear Regression
We can apply regularization to both linear regression and logistic regression. We will approach linear regression first.
We will modify our gradient descent function to separate out $θ_0$ from the rest of the parameters because we do not want to penalize $θ_0$.
repeat until convergence:{
}
The term $\frac{λ}{m}θ_j$ performs our regularization. With some manipulation our update rule can also be represented as:
The first term in the above equation,$1−α\frac{λ}{m}$ will always be less than 1. Intuitively you can see it as reducing the value of $θ_j$ by some amount on every update. Notice that the second term is now exactly the same as it was before.
#### Normal Equation
Now let’s approach regularization using the alternate method of the non-iterative normal equation.
To add in regularization, the equation is the same as our original, except that we add another term inside the parentheses:
L is a matrix with 0 at the top left and 1’s down the diagonal, with 0’s everywhere else. It should have dimension (n+1)×(n+1). Intuitively, this is the identity matrix (though we are not including $x_0$), multiplied with a single real number \lambda.
Recall that if m < n, then $X^TX$ is non-invertible. However, when we add the term $\lambda \cdot l$, then $X^TX + \lambda⋅L$ becomes invertible.
#### Regularized Logistic Regression
We can regularize logistic regression in a similar way that we regularize linear regression. As a result, we can avoid overfitting. The following image shows how the regularized function, displayed by the pink line, is less likely to overfit than the non-regularized function represented by the blue line:
Cost Function
Recall that our cost function for logistic regression was:
We can regularize this equation by adding a term to the end:
he second sum,$\sum^n_{j=1}θ^2_j$ means to explicitly exclude the bias term, $θ_0$. I.e. the θ vector is indexed from 0 to n (holding n+1 values, $θ_0$ through $θ_n$), and this sum explicitly skips $θ_0$, by running from 1 to n, skipping 0. Thus, when computing the equation, we should continuously update the two following equations:
## Neural Networks
### Model Representation
#### Model Representation I
Let’s examine how we will represent a hypothesis function using neural networks. At a very simple level, neurons are basically computational units that take inputs (dendrites) as electrical inputs (called “spikes”) that are channeled to outputs (axons). In our model, our dendrites are like the input features $x_1 \dots x_n$, and the output is the result of our hypothesis function. In this model our $x_0$ input node is sometimes called the “bias unit.” It is always equal to 1. In neural networks, we use the same logistic function as in classification, $\frac{1}{1+e^{−θ^Tx}}$, yet we sometimes call it a sigmoid (logistic) activation function. In this situation, our “theta” parameters are sometimes called “weights“.
Visually, a simplistic representation looks like:
The values for each of the “activation” nodes is obtained as follows:
This is saying that we compute our activation nodes by using a 3×4 matrix of parameters. We apply each row of the parameters to our inputs to obtain the value for one activation node. Our hypothesis output is the logistic function applied to the sum of the values of our activation nodes, which have been multiplied by yet another parameter matrix $Θ^{(2)}$ containing the weights for our second layer of nodes.
Each layer gets its own matrix of weights, $Θ^{(j)}$.
The dimensions of these matrices of weights is determined as follows:
If network has $s_j$ units in layer j and $s_{j+1}$ units in layer j+1, then $Θ^{(j)}$ will be of dimension $s_{j+1}×(s_j+1)$.
The +1 comes from the addition in $Θ^{(j)}$ of the “bias nodes,” $x_0$ and $Θ^{(j)}_0$. In other words the output nodes will not include the bias nodes while the inputs will. The following image summarizes our model representation:
Example: If layer 1 has 2 input nodes and layer 2 has 4 activation nodes. Dimension of $Θ^{(1)}$ is going to be 4×3 where $s_j=2$ and $s_{j+1}=4$, so $s_{j+1}×(s_j+1)=4×3$.
#### Examples and Intuitions I
A simple example of applying neural networks is by predicting x1 AND x2, which is the logical ‘and’ operator and is only true if both x1 and x2 are 1.
The graph of our functions will look like:
Remember that x0 is our bias variable and is always 1.
Let’s set our first theta matrix as:
This will cause the output of our hypothesis to only be positive if both $x_1$ and $x_2$ are 1. In other words:
So we have constructed one of the fundamental operations in computers by using a small neural network rather than using an actual AND gate. Neural networks can also be used to simulate all the other logical gates. The following is an example of the logical operator ‘OR’, meaning either $x_1$ is true or $x_2$ is true, or both:
#### Examples and Intuitions II
The $Θ^{(1)}$ matrices for AND, NOR, and OR are:
We can combine these to get the XNOR logical operator (which gives 1 if x1 and x2 are both 0 or both 1).
For the transition between the first and second layer, we’ll use a $Θ^(1)$ matrix that combines the values for AND and NOR:
For the transition between the second and third layer, we’ll use a $Θ^(2)$ matrix that uses the value for OR:
Let’s write out the values for all our nodes:
And there we have the XNOR operator using a hidden layer with two nodes! The following summarizes the above algorithm:
### Multiclass Classification
To classify data into multiple classes, we let our hypothesis function return a vector of values. Say we wanted to classify our data into one of four categories. We will use the following example to see how this classification is done. This algorithm takes as input an image and classifies it accordingly:
We can define our set of resulting classes as y:
Each $y^{(i)}$ represents a different image corresponding to either a car, pedestrian, truck, or motorcycle. The inner layers, each provide us with some new information which leads to our final hypothesis function. The setup looks like:
Our resulting hypothesis for one set of inputs may look like:
In which case our resulting class is the third one down, or $h_Θ(x)_3$, which represents the motorcycle.
### Cost Function
Let’s first define a few variables that we will need to use:
• L = total number of layers in the network
• $s_l$ = number of units (not counting bias unit) in layer l
• K = number of output units/classes
Recall that in neural networks, we may have many output nodes. We denote $h_Θ(x)_k$ as being a hypothesis that results in the kth output. Our cost function for neural networks is going to be a generalization of the one we used for logistic regression. Recall that the cost function for regularized logistic regression was:
For neural networks, it is going to be slightly more complicated:
We have added a few nested summations to account for our multiple output nodes. In the first part of the equation, before the square brackets, we have an additional nested summation that loops through the number of output nodes.
In the regularization part, after the square brackets, we must account for multiple theta matrices. The number of columns in our current theta matrix is equal to the number of nodes in our current layer (including the bias unit). The number of rows in our current theta matrix is equal to the number of nodes in the next layer (excluding the bias unit). As before with logistic regression, we square every term.
Note:
• the double sum simply adds up the logistic regression costs calculated for each cell in the output layer
• the triple sum simply adds up the squares of all the individual Θs in the entire network.
• the i in the triple sum does not refer to training example i
## Advice for Applying Machine Learning
### Evaluating a Hypothesis
Once we have done some trouble shooting for errors in our predictions by:
• Getting more training examples
• Trying smaller sets of features
• Trying polynomial features
• Increasing or decreasing λ
We can move on to evaluate our new hypothesis.
A hypothesis may have a low error for the training examples but still be inaccurate (because of overfitting). Thus, to evaluate a hypothesis, given a dataset of training examples, we can split up the data into two sets: a training set and a test set. Typically, the training set consists of 70% of your data and the test set is the remaining 30%.
The new procedure using these two sets is then:
Learn Θ and minimize $J_{train}(Θ)$ using the training set
Compute the test set error $J_{test}(Θ)$
### The test set error
1. For linear regression: $J_{test}(Θ)=\frac{1}{2m_{test}}\sum^{m_{test}}_{i=1}(h_Θ(x^{(i)}_{test})−y^{(i)}_{test})^2$
For classification ~ Misclassification error (aka 0/1 misclassification error):
This gives us a binary 0 or 1 error result based on a misclassification. The average test error for the test set is:
This gives us the proportion of the test data that was misclassified.
### Model Selection and Train/Validation/Test Sets
Just because a learning algorithm fits a training set well, that does not mean it is a good hypothesis. It could over fit and as a result your predictions on the test set would be poor. The error of your hypothesis as measured on the data set with which you trained the parameters will be lower than the error on any other data set.
Given many models with different polynomial degrees, we can use a systematic approach to identify the ‘best’ function. In order to choose the model of your hypothesis, you can test each degree of polynomial and look at the error result.
One way to break down our dataset into the three sets is:
• Training set: 60%
• Cross validation set: 20%
• Test set: 20%
We can now calculate three separate error values for the three different sets using the following method:
1. Optimize the parameters in Θ using the training set for each polynomial degree.
2. Find the polynomial degree d with the least error using the cross validation set.
3. Estimate the generalization error using the test set with $J_{test}(Θ^{(d)})$, (d = theta from polynomial with lower error);
This way, the degree of the polynomial d has not been trained using the test set.
### Diagnosing Bias vs. Variance
In this section we examine the relationship between the degree of the polynomial d and the underfitting or overfitting of our hypothesis.
• We need to distinguish whether bias or variance is the problem contributing to bad predictions.
• High bias is underfitting and high variance is overfitting. Ideally, we need to find a golden mean between these two.
The training error will tend to decrease as we increase the degree d of the polynomial.
At the same time, the cross validation error will tend to decrease as we increase d up to a point, and then it will increase as d is increased, forming a convex curve.
• High bias (underfitting): both $J_{train}(Θ)$ and $J_{CV}(Θ)$ will be high. Also, $J_{CV}(Θ) \approx J_{train}(Θ)$.
• High variance (overfitting): $J_{train}(Θ)$ will be low and $J_{CV}(Θ)$ will be much greater than $J_{train}(Θ)$.
The is summarized in the figure below:
### Regularization and Bias/Variance
In the figure above, we see that as λ increases, our fit becomes more rigid. On the other hand, as λ approaches 0, we tend to over overfit the data. So how do we choose our parameter λ to get it ‘just right’ ? In order to choose the model and the regularization term λ, we need to:
1. Create a list of lambdas (i.e. $λ \in \left\{0,0.01,0.02,0.04,0.08,0.16,0.32,0.64,1.28,2.56,5.12,10.24\right\}$);
2. Create a set of models with different degrees or any other variants.
3. Iterate through the λs and for each λ go through all the models to learn some Θ.
4. Compute the cross validation error using the learned Θ (computed with λ) on the $J_{CV}(Θ)$ without regularization or λ = 0.
5. Select the best combo that produces the lowest error on the cross validation set.
6. Using the best combo Θ and λ, apply it on $J_{test}(Θ)$ to see if it has a good generalization of the problem.
### Learning Curves
Training an algorithm on a very few number of data points (such as 1, 2 or 3) will easily have 0 errors because we can always find a quadratic curve that touches exactly those number of points. Hence:
• As the training set gets larger, the error for a quadratic function increases.
• The error value will plateau out after a certain m, or training set size.
#### Experiencing high bias:
• Low training set size: causes $J_{train}(Θ)$ to be low and $J_{CV}(Θ)$ to be high.
• Large training set size: causes both $J_{train}(Θ)$ and $J_CV(Θ)$ to be high with $J_{train}(Θ) \approx J_{CV}(Θ)$.
If a learning algorithm is suffering from high bias, getting more training data will not (by itself) help much.
#### Experiencing high variance:
• Low training set size: causes $J_{train}(Θ)$ to be low and $J_{CV}(Θ)$ to be high.
• Large training set size: $J_{train}(Θ)$ increases with training set size and $J_{CV}(Θ)$ continues to decrease without leveling off. Also, $J_{train}(Θ) < J_{CV}(Θ)$ but the difference between them remains significant.
If a learning algorithm is suffering from high variance, getting more training data is likely to help.
### Deciding What to Do Next Revisited
Our decision process can be broken down as follows:
• Getting more training examples: Fixes high variance
• Trying smaller sets of features: Fixes high variance
• Adding features: Fixes high bias
• Adding polynomial features: Fixes high bias
• Decreasing λ: Fixes high bias
• Increasing λ: Fixes high variance.
#### Diagnosing Neural Networks
• A neural network with fewer parameters is prone to underfitting. It is also computationally cheaper.
• A large neural network with more parameters is prone to overfitting. It is also computationally expensive. In this case you can use regularization (increase λ) to address the overfitting.
Using a single hidden layer is a good starting default. You can train your neural network on a number of hidden layers using your cross validation set. You can then select the one that performs best.
## Support Vector Machines
### Cost Function
recall the cost function of logistic Regression:
We replace some terms with new terms and delete some terms which will not influence the results:
### kernel
#### 低维线性不可分到高维线性可分的简单例子
(前后轴为x轴，左右轴为y轴，上下轴为z轴）注意到绿色的平面可以完美地分割红色和紫色，也就是说，两类数据在三维空间中变成线性可分的了。而三维中的这个判决边界，再映射回二维空间中是这样的：
### SVM or Logistic Regression
• 如果特征维数很高，往往线性可分（SVM解决非线性分类问题的思路就是将样本映射到更高维的特征空间中），可以采用LR或者线性核的SVM；
• 如果样本数量很多，由于求解最优化问题的时候，目标函数涉及两两样本计算内积，使用高斯核明显计算量会大于线性核，所以手动添加一些特征，使得线性可分，然后可以用LR或者线性核的SVM；
• 如果不满足上述两点，即特征维数少，样本数量正常，可以使用高斯核的SVM。
## Clustering
### K-means algorithm
K-means算是一个很简单的聚类算法，而聚类与决策树、SVM等不同，是一种无监督的学习，所谓无监督学习（Unsupervised learning）是和监督学习相对应的，不同于监督学习，无监督学习所给的训练集是不包含标签的，所有数据集都只包括特征xi
#### K-Means的描述如下：
• $x^{(i)}$为第i个数据点；
• $c^{(i)}$ 为x^{(i)}的簇；
• $u_j$为第j个簇的质心点；
1. 首先需要初始化质心点，在K-Means中，通常采用随机的方法对质心点进行初始化。更好的办法是：随机选择m(m>k)个数据，再从中选择k个数据点作为质心点；
2. 第一个for循环主要用于给数据点$x^{(i)}$赋值$c^{(i)}$，称为 cluster assignment steps，对每一个数据点，都会计算她与所有质心点的距离，而后将数据点分配到与它距离最近的簇；
3. 第二个for循环主要用于更新质心点的位置，称为move centroid steps，而$u_j$这里的计算公式所代表的意思就是，分母：统计所有$c_i=j$的点的个数；分子是所有$c_i=j$的点的坐标和。那整体的意思就很明确了，就是求这些点的平均值，作为新的质心点的位置。
#### 优化目标
J(c,u)实际上是一个单调递减的函数，且是一个非凸函数，只要我们能找到拐点，那我们就已经达到了收敛，又称这种方法为elbow function。偶尔也有可能陷入局部最优情况，或出现震荡情况，这样一定是有问题了。
#### 总结
K-Means虽然简单，容易实现，但是也会收敛到局部最小值，这种情况下可以采用K-Means的改进算法：二分K-均值算法。算法的思想就是：首先将所有点做为一个簇，然后将该簇一分为二。之后选择其中一个簇进行继续划分，选择哪一个簇进行划分则取决于对其划分是否可以最大程度降低SSE的值，不断划分直到达到用户所指定的K值。
## Dimensionality Reducion
why we need to do Dimensionality Reducion?
• data compression
• visually
### PCA实例
1. 分别求x和y的平均值，然后对于所有的样例，都减去对应的均值。这里x的均值是1.81，y的均值是1.91，那么一个样例减去均值后即为（0.69,0.49），得到
2. 求特征协方差矩阵，如果数据是3维，那么协方差矩阵是：
这里只有x和y，求解得
对角线上分别是x和y的方差，非对角线上是协方差。协方差是衡量两个变量同时变化的变化程度。协方差大于0表示x和y若一个增，另一个也增；小于0表示一个增，一个减。如果ｘ和ｙ是统计独立的，那么二者之间的协方差就是０；但是协方差是０，并不能说明ｘ和ｙ是独立的。协方差绝对值越大，两者对彼此的影响越大，反之越小。协方差是没有单位的量，因此，如果同样的两个变量所采用的量纲发生变化，它们的协方差也会产生树枝上的变化。
3. 求协方差的特征值和特征向量，得到：
上面是两个特征值，下面是对应的特征向量，特征值0.0490833989对应特征向量为，这里的特征向量都归一化为单位向量。
4. 将特征值按照从大到小的顺序排序，选择其中最大的k个，然后将其对应的k个特征向量分别作为列向量组成特征向量矩阵。这里特征值只有两个，我们选择其中最大的那个，这里是1.28402771，对应的特征向量是(-0.677873399, -0.735178656)T。
FinalData(101) = DataAdjust(102矩阵) x 特征向量(-0.677873399, -0.735178656)T
得到的结果是： | 12,037 | 47,881 | {"found_math": true, "script_math_tex": 1, "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": 15, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-31 | latest | en | 0.975901 |
http://hackage.haskell.org/packages/archive/SG/1.0/doc/html/Data-SG-Vector-Basic.html | 1,369,253,003,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702444272/warc/CC-MAIN-20130516110724-00036-ip-10-60-113-184.ec2.internal.warc.gz | 122,437,992 | 2,866 | SG-1.0: Small geometry library for dealing with vectors and collision detection
Data.SG.Vector.Basic
Description
Some types that are very basic vectors. Most of the use that can be made of the vectors is in their type-class instances, which support a powerful set of operations. For example:
``` fmap (*3) v -- Scales vector v by 3
pure 0 -- Creates a vector filled with zeroes
v + w -- Adds two vectors (there is a 'Num' instance, basically)
```
Plus all the instances for the classes in Data.SG.Vector, which allows you to use `getX` and so on.
You will probably want to create more friendly type synonyms, such as:
``` type Vector2 = Pair Double
type Vector3 = Triple Double
type Line2 = LinePair Double
type Line3 = LineTriple Double
```
Synopsis
# Documentation
newtype Pair a Source
A pair, which acts as a 2D vector.
Constructors
Pair (a, a)
Instances
Functor Pair Applicative Pair Foldable Pair Traversable Pair VectorNum Pair Coord2 Pair Coord Pair IsomorphicVectors Pair Rel2' IsomorphicVectors Pair Point2' IsomorphicVectors Rel2' Pair IsomorphicVectors Point2' Pair Geometry Pair Pair LinePair Eq a => Eq (Pair a) (Show a, Eq a, Num a) => Num (Pair a) Ord a => Ord (Pair a) Read a => Read (Pair a) Show a => Show (Pair a)
newtype Triple a Source
A triple, which acts as a 3D vector.
Constructors
Triple (a, a, a)
Instances
Functor Triple Applicative Triple Foldable Triple Traversable Triple VectorNum Triple Coord3 Triple Coord2 Triple Coord Triple IsomorphicVectors Triple Rel3' IsomorphicVectors Triple Point3' IsomorphicVectors Rel3' Triple IsomorphicVectors Point3' Triple Geometry Triple Triple LineTriple Eq a => Eq (Triple a) (Show a, Eq a, Num a) => Num (Triple a) Ord a => Ord (Triple a) Read a => Read (Triple a) Show a => Show (Triple a)
newtype Quad a Source
A quad, which acts as a 4D vector.
Constructors
Quad (a, a, a, a)
Instances
newtype LinePair a Source
A pair of (position vector, direction vector) to be used as a 2D line.
Constructors
LinePair (Pair a, Pair a)
Instances
Geometry Pair Pair LinePair Eq a => Eq (LinePair a) Ord a => Ord (LinePair a) Read a => Read (LinePair a) Show a => Show (LinePair a)
newtype LineTriple a Source
A pair of (position vector, direction vector) to be used as a 3D line.
Constructors
LineTriple (Triple a, Triple a)
Instances
Geometry Triple Triple LineTriple Eq a => Eq (LineTriple a) Ord a => Ord (LineTriple a) Read a => Read (LineTriple a) Show a => Show (LineTriple a) | 646 | 2,488 | {"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} | 2.734375 | 3 | CC-MAIN-2013-20 | latest | en | 0.792936 |
https://www.pryor.com/blog/the-dichotomies-of-data-understanding-data-opposites/ | 1,726,699,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00213.warc.gz | 858,146,054 | 46,647 | # The Dichotomies of Data: Understanding Data Opposites
Understanding data involves understanding differences in data types. This article describes dichotomies, opposites or dualities that help us broaden our view about the types of data that may be important for our organization and in our jobs. These dichotomies highlight different aspects or perspectives that are important in understanding and analyzing data.
• What question or problem are you trying to solve? What need are you trying to fill? Starting with the end in mind helps you evaluate data against a baseline need or goal. Without this focus, all data can become seductive – and lead to “analysis paralysis” without an end in mind.
• What type of data am I looking for to answer that question or need – what’s the data about? You might want data about a product, about different types of events over time (like growth, sales or employment patterns), or about perceptions people have about something.
Apply these questions to think about a case study you have in mind. Then, walk through the following common data dichotomies to see what might fit.
## Quantitative and Qualitative Data
Data can be quantitative, consisting of numerical values that can be measured and analyzed using mathematical and statistical methods, or qualitative, consisting of non-numerical information such as text, images, observations that require interpretation and analysis. These data types differ in their characteristics, collection methods and analysis techniques. Here’s an overview of their differences.
## SAVE \$10 AND TRAIN ON THIS TOPIC TODAY
Quantitative Data
• Quantitative data generally include objective numerical data that can be counted or measured. These data generally capture quantities, measurement and statistical relationships.
• Quantitative data may come from organization records, like employee data or sales and inventory records; or industry reports, like market size and buyer demographics. Data can also be collected using structured methods such as surveys with closed-ended questions, or by gathering concrete measurements.
• Quantitative data are often represented in the form of numbers, in tables, charts or graphs.
• Quantitative data analysis involves statistical techniques such as data summarization, descriptive statistics, and inferential statistics. It aims to identify patterns, trends, correlations or cause-and-effect relationships. For example, these data may provide useful information for business forecasts based on previous results.
Quantitative data is often well-organized and formatted according to a predefined structure and scheme. In highly structured data, relationships between data elements are clearly defined. These data sources generally follow a consistent and rigid structure, and can be more easily queried, sorted and analyzed using structured query languages (SQL) or other database tools.
Qualitative Data
• Qualitative data include descriptive and subjective data that capture qualities, opinions, experiences and behaviors.
• Qualitative data are typically collected through methods such as interviews, observations, focus groups or surveys with open-ended questions. Asking a customer to respond in text boxes is a common way to gather these types of data.
• Qualitative data are often presented in the form of narratives, transcripts, quotes or themes; however, they can also be analyzed and then presented in quantitative form, like a count of satisfied or unsatisfied employees, based on interviews or open-ended survey questions.
• Qualitative data analysis involves interpreting and making sense of the data by identifying patterns, themes, or meanings. It often involves coding, categorizing and interpreting textual or visual data. It can be used to conduct root cause analyses and to support strategic planning.
Qualitative data is often unstructured or semi-structured and may not adhere to a predefined schema or format. It could be generated in real-time from various sources such as social media feeds, text documents or multimedia content. This provides greater context for use, but may be challenging to process and analyze using traditional methods. New advanced analysis techniques include natural language processing, machine learning, or data mining for analysis. The analysis may involve identifying patterns, sentiment analysis, topic modeling or extracting insights from the data.
## Internal and External
Internal data sources refer to data that are generated, collected and stored within an organization. These sources are typically proprietary and specific to the organization. Here are some examples of internal data sources that can be used to answer a variety of business questions:
• Databases: Organizations maintain databases and data warehouses to store structured data, such as customer information, sales data and inventory records.
• Enterprise Resource Planning (ERP) Systems: ERP systems integrate business processes and provide a centralized platform for managing data related to finance, human resources and supply chains. They may also include knowledge bases that help employees do their jobs better.
• Customer Relationship Management (CRM) Systems: CRM systems store data about customers, their interactions, preferences, purchase history and other relevant information.
• Transactional Systems: These systems record and store data related to daily business transactions, such as point-of-sale systems, online payment gateways and order management systems.
• Internal Analytics and Reporting Systems: Organizations often have specialized systems and tools for collecting and analyzing data, generating reports and tracking key performance indicators (KPIs). These may be complex systems customized to meet the unique business setting, or as simple as Excel spreadsheets held by one team.
• Records and File Management Systems: A lot of qualitative data may be held in different types of documents generated using different applications; as such, the file management system itself is a source of data.
External data sources refer to data that is obtained from outside the organization. These sources provide additional insights, context or information that complements internal data. Some examples of external data sources include:
• Publicly Available Datasets: These include government data, census data, public surveys, weather data, economic indicators and other publicly accessible information.
• Third-Party Data Providers: There are various companies and associations that specialize in collecting and aggregating data on specific topics, such as market research, industry data, consumer behavior data or social media data. Organizations can purchase or license this data for their analytical needs.
• Data Exchanges and Marketplaces: Online platforms exist where organizations can exchange or purchase data from other organizations. These platforms facilitate data sharing and collaboration between different entities.
• Partner and Vendor Data: Data obtained from business partners, suppliers, vendors or collaborators can provide valuable insights when integrated with internal data.
It is important to stay compliant with data privacy, security and legal regulations when using both internal and external data sources. Know where the data come from, and how they can be used.
## Other Data Opposites
Here are other dichotomies to help you assess the types of data you both have and need to answer your organizational question or problem:
• Primary vs. Secondary: Primary data is collected firsthand for a specific purpose, often through surveys, interviews, experiments or observations. Primary data is often internal data. Secondary data, on the other hand, is pre-existing data collected by someone else for a different purpose and made available for analysis, such as government reports, or published studies.
• Historical (Retrospective) vs. Real-time (Prospective): Historical data refers to past records or events, allowing for retrospective analysis and trend identification. Real-time data, on the other hand, is generated and captured in the present moment, providing immediate insights and enabling real-time decision-making. So, historical data helps you understand what services or products may be growing or losing popularity over time, where real-time data tells you what is needed right now (e.g., through an inventory or ticketing system).
• Privacy vs. Utility: The dichotomy between privacy and utility concerns the trade-off between protecting individuals’ personal information and extracting value or insights from data. Balancing privacy concerns with the utility of data is crucial in ethical and responsible data practices.
These dichotomies provide different dimensions to consider when working with data, highlighting the diversity and complexity of data sources, types and characteristics. Understanding these aspects can help inform data collection, analysis and decision-making processes. | 1,600 | 9,007 | {"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} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.913632 |
https://ww2.mathworks.cn/matlabcentral/cody/problems/44445-tax-calculator/solutions/1810081 | 1,600,938,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00691.warc.gz | 612,067,732 | 18,300 | Cody
# Problem 44445. Tax Calculator
Solution 1810081
Submitted on 8 May 2019 by E. Cheynet
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
income = 0; tax_correct = 0; assert(isequal(taxFor(income),tax_correct))
2 Pass
income = 100; tax_correct = 10; assert(isequal(taxFor(income),tax_correct))
3 Pass
income = 1000; tax_correct = 100; assert(isequal(taxFor(income),tax_correct))
4 Pass
income = 2000; tax_correct = 200; assert(isequal(taxFor(income),tax_correct))
5 Pass
income = 2500; tax_correct = 300; assert(isequal(taxFor(income),tax_correct))
6 Pass
income = 3000; tax_correct = 400; assert(isequal(taxFor(income),tax_correct))
7 Pass
income = 5000; tax_correct = 1000; assert(isequal(taxFor(income),tax_correct))
8 Pass
filetext = fileread('taxFor.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden') | 301 | 979 | {"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} | 2.875 | 3 | CC-MAIN-2020-40 | latest | en | 0.644689 |
https://math.stackexchange.com/questions/2312553/determine-the-convergence-of-the-series-sum-k-1-infty-frac2-1k5k/2312557 | 1,643,290,515,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00691.warc.gz | 429,369,987 | 29,239 | # Determine the Convergence of the series $\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k}$ [closed]
So I've been searching for a way to determine whether or not the following series:
$$\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k}$$
converges or not.
I've used the integral test, the comparison test, the limit comparison test, the ratio test and the root test and I didn't manage to accomplish anything.
• Can you see that $2\sum_{k\geq 1}\left(\frac{1}{5}\right)^k + \sum_{k\geq 1}\left(-\frac{1}{5}\right)^k$ is obviously convergent? Jun 6 '17 at 21:08
Hint: $-1 \leq (-1)^k \leq 1$.
Recall that the geometric series, $$\sum_{n=1}^\infty ax^n=\frac{ax}{1-x}$$ converges if and only if $|x|<1$, where $a\in\Bbb R$. So, compare the series you have with the geometric series where $x=\frac{1}{5}$.
Addendum: in fact, using the above formula for the geometric series, one can compute the explicit value of this series by splitting the given series into two geometric series. Note that $$\sum_{n=1}^\infty (a_n+b_n)=\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty b_n$$ provided the individual series converge (i.e. the limits of partial sums exist individually for $\{a_n\}$ and $\{b_n\}$).
I see two geometric series: $$\sum_{k=1}^\infty \frac{2+(-1)^k}{5^k} = \frac{2}{5} \sum_{k=0}^\infty (\frac{1}{5})^k-\frac{1}{5} \sum_{k=0}^\infty (\frac{-1}{5})^k$$ Both geometric series are convergent and their combined value is: $$=\frac{2}{5} \frac{1}{1-\frac{1}{5}}-\frac{1}{5} \frac{1}{1+\frac{1}{5}} =\frac{1}{2} - \frac{1}{6} =\frac{1}{3}$$ | 551 | 1,521 | {"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} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.76265 |
https://www.daniweb.com/software-development/computer-science/threads/419019/ieee-short-real-floating-point-format | 1,427,724,774,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131299360.90/warc/CC-MAIN-20150323172139-00007-ip-10-168-14-71.ec2.internal.warc.gz | 989,216,506 | 11,263 | Hi!
I have a decimal number (e.g 6433) and would like to see the process of converting this to IEEE short real floating point format!
Any help is appreciated ;)
See the wikipedia page on the topic: https://en.wikipedia.org/wiki/IEEE_754-2008
Basically you have a sign bit, an exponent part, and a mantissa part, and they represent the number
`(1-2*sign_bit) * (1.0 + mantissa) * 2^(exponent - K)` where K is a constant, exponent is interpreted as an unsigned integer, and mantissa is interpreted as the bits (of a base 2 number) after the decimal point, forming the number 1.abcdefghijklm... where abcdefghijklm... are the bits of the mantissa. And sign_bit is 0 or 1 of course.
You
This article has been dead for over six months: Start a new discussion instead
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# Chapter 3 Demand Theory
3
INTRODUCTION manager of a firm.
DEMAND THEORY
You would have done demand theory in your diploma course. Your lecturer would have discussed the theoretical aspect of demand. Now you will be applying this theory to make decisions in a firm. In managerial economics you have to assume that you are a
For your firm to survive or to be established it needs sufficient demand. It cannot survive if there is no demand even though it has the most efficient production method. As a manager you have to understand and analyse the forces that determine the demand for your product. For example how will customers react to an increase in the price of haircut during festive season and off-festive seasons? How does the recession affect the demand for your product? Therefore it can be seen that the demand theory is important for the creation, survival and profitability of your firm. This chapter will begin by explaining the basic differences between demand function and demand curve function, followed by market demand curve. Then attention will be given to different revenue functions. Finally the concept of elasticity and its importance in decision making will be discuss .The mechanics of demand will also be analyses using algebraic equation and graph.
83
## Chapter 3 Demand Theory
Key terms for review: Determinants of demand Multiplicative demand function Total revenue Average revenue price elasticity of demand Cross elasticity Complements Luxury goods Inferior goods Demand function Linear demand function Demand curve Marginal revenue Elasticity Price elasticity Income elasticity Substitutes Basic necessities Normal goods
84
CHAPTER OVERVIEW
85
## Chapter 3 Demand Theory
Learning Objectives After reading this chapter, the students should be able to: 1. 2. 3. 4. 5. Explain a demand function. Derive a demand curve function from the demand function Derive the total, marginal and average revenue function and describe the relationship between them. Calculate and interpret price, income and cross elasticity. Analyse the importance of elasticity in decision making.
86
3.1
## THE DEMAND FUNCTION
A textbook definition of effective demand is - demand that is backed by the purchasing power. This definition can further be analysed as the desire, willingness and ability to buy a product at a specific price at given point of time. Demand function refers to the relationship that exists between the quantity demanded and the determinants of demand. The determinants of demand are the factors that influence demand. For example, the demand for shoes (Qs) is influenced by price of shoes (Ps), advertising expenditure (A), income (I) and (T). The demand function can be written in a functional form as: Qs = f (Ps, A, I, T ) Eqn 3a
FORMS OF DEMAND FUNCTIONS There are various forms of demand functions. The specific form of the demand function is determined empirically. At this stage we shall analyse two forms of demand function. They are:
a)
Linear Demand Function The determinants of demand for product X in a linear form can be written as : Qx = - 0 Px + 1 Py + 2 I + 3 T Eqn 3b
where Px is price of product X, Py is the price of related product y. I is income and T is taste, 0 to 3 = constant and includes all the other variables not mentioned in the demand function. = coefficients of the demand function.
87
## Chapter 3 Demand Theory
The coefficients indicate the marginal impact of each independent variable on the quantity demanded. For example if the price of X change by one unit the quantity demanded will change by 0 units. The sign (+/-) before the coefficient shows the (positive or negative) relationship between the independent and dependent variable.
b)
Multiplicative Demand Function Q = Px 0 Py 1 I 2 T 3 Where 0 to 3 = = constant measures the percentage change in the Eqn 3c
dependent variable relative to the percentage change in the respective independent variable. The coefficients are replaced by their numerical equivalent after they are estimated using regression analysis.
Example 1 Suppose the demand function for product X has been estimated and is as follow: Qx = 1.0 - 2.0Px + 1.5I + 0.8Py + 2.0T and coefficients are replaced by their numerical equivalents, the positive or
negative signs before the respective coefficient indicates the direction of the relationship that variable has on Qx. For example, coefficient -2.0 Px , explains that when the price of X increase by one unit the quantity demanded of X will decrease by 2.0 units. If we assume Px = 2, I = 4, Py = 2.5 , and T = 2 , The quantity demanded will be: (Substitute these values into the equation) Qx = 1.0 - 2.0(2) + 1.5(4) + 0.8(2.5) + 2.0(2) = 9 Units
88
## Chapter 3 Demand Theory
Based on the equation we can predict that, at the current values of the independent variables, quantity demanded of product X will be 9 units, ceteris paribus.
3.2
## The Demand Curve
The Law of demand seeks to explain the relationship between quantity demanded and price, that is quantity and price are inversely related (price increases, quantity demanded decrease and vice versa). The Law of demand can be depicted in a functional form by a demand curve function. A demand curve function shows the relationship between quantity demanded and the price of the product assuming all the other factors influencing its demand to remain constant (Ceteris Paribus). The demand curve function is a special sub-case of the demand function. In other word it is derived from the demand function. In functional form a demand curve function is written as: Qx = f (Px) ceteris paribus. In a linear form it will be: Qx = A - 0Px A is derived Eqn 3e Eqn 3d
by compressing (adding) all the factors that determine demand except for
price of the product, based on equation Eqn 3b, A is: A = + 1Py + 2I + 3T Note: Since traditionally price is being placed on the y-axis and quantity on the x-axis, the above equation can be rewritten to make Px the subject. But one has to bear in mind that quantity (Qx) is the dependent variable and price (Px) is the independent variable. Rewritten in the conventional way (making P the subject):
89
Px=
A 0
## _ 1Q 0 is a and 1/ is b. Therefore the
For easy reference we shall assume that A/ demand curve function will be as follows: Px = a b Qx
Eqn 3f
## Diagram 3.1 Demand Curve
Example 2 This example will show you how to derive a demand curve function from demand function. Based on the estimated demand function from Example 1 where: Qx = 1.0 - 2.0Px + 1.5I + 0.8Py + 2.0T
90
## = 1.0 + 1.5 (4) + 0.8 (2.5) + 2.0 (2) = 13
and then by substituting it into the demand curve function: Qx = In the conventional form: Px = 6.5 0.5Qx 13 - 2.0Px
## Diagram 3.2 Demand curve
NOTES The price is expressed as a linear function of quantity demanded, 6.5 is the intercept on the vertical axis (a) and 0.5 is the slope term (b). The negative sign before the slope term or coefficient shows an inverse relationship between price and quantity. The intercept on the horizontal axis is 13 (the value of A)
91
3.3
## THE MARKET DEMAND CURVE
Market demand of a product or service is the sum of individual demand. It shows the total quantity demanded in the market at various prices. To get the market demand function we add the individual demand functions horizontally, that is:
Eqn 3g
## = (10 - 4p) + (6 - 3p) + (15 - 0.9p) = 31 - 7.9p or PM = 3.92 - 0.126 QP
The market demand curve can also be determined graphically by adding the quantity demanded at a given price. For simplicity we assume that there are two individuals with the following demand curves.
92
## Chapter 3 Demand Theory
Diagram 3.3 The market demand curve Referring to the diagram 3.3 when the price is RM10, Individual 1 buys 3 units and individual II buys 2 units. The market demand at this price will be 5 units (2+3). At price RM5 the market demand is (5+6) 11 units.
3.4
## TOTAL, MARGINAL AND AVERAGE REVENUE FUNCTIONS
TOTAL REVENUE FUNCTION The total revenue shows the total dollar sales of a firm at a certain period of time. Managers are interested in the relationship between price and quantity since it influences total revenue. When the price changes the total revenue will change. To get the total revenue function, total quantity demanded (sold) is multiplied by the price of the product (P x Q). That is : Total revenue is: Px = a b Qx TR = P x Q
93
## TR x = (a b Qx) Qx Therefore the total revenue function is
TRx = a Qx bQx2
Eqn 3h
Example 4 Based on Example 2 : Px = 6.5 - 0.5Qx the TR function is equal to TR = Q (6.5 - 0.5Qx) = 6.5Q - 0.5Q2 The total revenue function can be represented graphically by plotting quantity against the total revenue. For example, if the quantity is 2 the total revenue will be: 6.5(2) - 0.5 (2)2 = 9 and at quantity 6.5 and 7 the total revenue is 21.125 and 14 respectively. Using these values the total revenue curve can be plotted as show in diagram 3.4.
94
## Chapter 3 Demand Theory
To find the total revenue maximising quantity, differentiate the equation, set it equal to zero and solve for Q The process is shown below. TR = 6.5Q - 0.5Q2 TR = 6.5 - Q = 0 Q Q = 6.5 We studied optimization in chapter 2. Make a point to revise the chapter if you find difficulties
At quantity 6.5 units the total revenue is 6.5 (6.5) - 0.5 (6.5) 2 which is 21.125.
THE MARGINAL REVENUE FUNCTION The marginal revenue is defined as the change in the total revenue that results from one unit change in the quantity demanded. It can be express as a first derivative of total revenue with respect to Qx That is: TRx MRx = a Qx bQx2 = TR = a 2bQx Qx Diagram 3.5 shows a graphical representation of marginal revenue function. Eqn 3i
Example 5 Based on Example 4: TR = 6.5Q - 0.5Q2 The marginal revenue function is MR = TR = 6.5 - Q Q
95
## Chapter 3 Demand Theory
The marginal revenue slope downwards and the intercept is at 6.5. marginal curve is shown in Diagram 3.5 MR/p 6.5 P = 6.5 -0.5Q
The
MR = 6.5 Q or a 2bQx
6.5
Qtty
Diagram 3.5 The Marginal Revenue Curve The marginal revenue is zero at quantity 6.5. MR = 6.5 - Q = 0 Q = 6.5
THE AVERAGE REVENUE FUNCTION Average revenue is the revenue earned per unit of output sold. It is calculated by dividing total revenue by the quantity. TRx = a Qx bQx2
AR x
= TR Qx
= a bQx
Eqn 3j
96
## Chapter 3 Demand Theory
NOTES The average revenue function is also the demand curve function.
Example 6 The average revenue function can be computed from the above total revenue function. Given: TR AR = 6.5Q - 0.5Q2 = TR = 6.5 - 0.5Q Qx The average revenue function is similar to the demand curve function P = 6.5 - 0.5Q.
The relationship between demand curve, marginal revenue and the total revenue curves. The relationship between demand curve, marginal revenue and the total revenue curve functions can be analyse diagrammatically.
97
## Again, section 2.2 previously explained this relationship
Diagram 3.5 The relationship between Total Revenue and Marginal Revenue Curve. Based on diagram 3.5 first we begin by comparing the similarities and differences between the average /demand curve and marginal revenue functions. a. b. AR and MR has the same intercept, this indicates that both curves must begin from the same point on Y axis. The slope of marginal revenue curve is twice (2 b Qx) of the demand curve (b Qx ) Next we shall look at the relationship between marginal revenue and total revenue. a. b. c. When marginal revenue is zero the total revenue is at its maximum. (refer to quantity Q*) When marginal revenue is positive the total revenue is increasing (refer to quantity O- Q*) When marginal revenue is negative the total revenue is decreasing (refer to quantity after Q*)
98
## Chapter 3 Demand Theory
QUESTIONS
1.
Given the demand curve function for caps as: Q = 3200 - 12P Q - Quantity demand for caps P - Price of caps. a) b) c) Rewrite the demand curve function in the conventional way. How many caps will be sold at RM10 each? At what price would sales equal zero?
2.
There are three individuals with different demand curve function wanting to buy badges (the quantity is in hundreds). individual is as follows. Individual I Individual II Individual III a) b) PI PII PIII = 5 - 5 Q1 = 4 - 3.5Q = 8 - 1.5Q Their demand curve function for each
What is the market demand curve function? How many badges will be bought in the market at price RM2 and how will this be distributed among the individuals?
3.
Suppose the demand function for special Sukom files is Qd = 2500 - 6P a) b) c) Derive the total revenue function? At what output is the total revenue maximum? Derive the marginal revenue function?
99
## Chapter 3 Demand Theory
d) e) 4.
At what output is marginal revenue zero? What can you conclude from the answers in (b) and (d)
A firm selling hot-dogs estimates the demand function for hot-dogs as Qh = 150.3 + 10.2 A + 9.6 Py - 15.4 Ph. Where Qh is the quantity of hot dogs, A is advertising expenditure, Py is the price of hot dog competitor and Ph is the price of hot dog Given A is RM100; Py is RM2; and Ph is RM18.00. a) b) Determine the demand curve function and plot the demand curve using any three price-quantity combinations. Derive the marginal revenue function.
SUGGESTED SOLUTIONS
1.
a)
Q P P
b)
When P Q
c)
When Q P P
2.
a)
100
## Chapter 3 Demand Theory
(make Q the subject) PI Q1 Q1 PII Q11 QII PIII QIII QIII QM QM PM b) When P QM = 5 - 5 Q1 = 5 = 1.0 - 0.20P = 4 - 3.5Q = 4P 3.5 = 1.14 - 0.29P = 8 - 1.5Q = 8-P 1.5 = 5.33 - 0.67P = QI + QII + QIII = 7.47 - 1.16P = 6.44 - 0.86Q =2 = 7.47 - 1.16 (2) = 5.15 Each Individual will buy: QI QII QIII 3. a) P TR TR b) = = = = 1-0.2(2) = 1.14 - 0.29(2) = 5.33 - 0.67(2) 416.67 - 0.17Q PQ 416.67Q - 0.17Q2 = 0.6 = 0.56 = 3.99 5- P
TR maximising quantity.
101
TR Q
= 416.67 - 0.34Q
## Q = 1225.5 c) MR = TR = 416.67 - 0.334Q Q d) MR 416.67 - 0.34 Q Q = 0 =0 = 1225.5
e) 4. a)
When marginal revenue is zero the total revenue is maximum. Qh A = A - 15.4 Ph = 150.3 + 9.6 (2) + 10.2 (100) = 150.3 + 19.2 + 1020 = 1189.5
The demand curve function is: Qh = 1189.5 - 15.4 Ph Written in the convention form: Ph = 77.24 - 0.065 Qh To plot the demand curve we can take any three values of Q.
Points A B C
102
## 61.24 100 b) TR MR = 77.24 Q - 0.065 Q2 = TR Q = 77.24 - 0.13 Q 250
C D Quantity
3.5
ELASTICITY
Elasticity is a measure of responsiveness of the quantity demanded when there is a change in its determinants You would have learnt about concept of elasticity and how to calculate it in the economics course at diploma level. You would have used a standard formula to calculate elasticity. For example, the price elasticity is calculated by using following formula: (basic formula) EP = percentage change in quantity demanded percentage change in price or EP = OD - ND X OP OP-NP OQ
103
## Chapter 3 Demand Theory
Where OD is original demand, ND is new demand, OP is original price and NP is new price. This method of measuring elasticity is quite sensitive to which value is chosen as old and which is chosen as the new. However, in this course you are introduced to two other approaches to compute elasticity. They are arc price elasticity (where the basic formula is the same) and point price elasticity. Arc elasticity measures the elasticity over a range of price and takes into consideration averages. The point elasticity is used to measure a very small or infinitesimally small change in the price.
PRICE ELASTICITY OF DEMAND (Ex) Price elasticity of demand measures the responsiveness of quantity demanded when there is a change in its price. Basically elasticity can be divided into three broad categories and two extreme cases. Price elasticity of demand is seen a positive integer (Note: When elasticity is interpreted the negative sign can be ignored for normal goods) a) b) c) Elastic (E > 1). Consumers are responsive to the change in price. Inelastic (E < 1). Consumers are not very responsive to the change in price. Unitary elasticity (E = 1) Consumer are proportionately responsive to the change in price. There are two extreme cases. a) Perfectly elastic (E = ) consumers are very responsive to the change in price. When there is a slight increase in price, the quantity demanded will fall to zero. The demand curve is a horizontal straight line. When there is a slight decrease in the price, the consumers will buy up all that available in the market i.e. price elasticity is infinity.
104
## Chapter 3 Demand Theory
b) Perfectly inelastic (E = 0) consumers are not responsive to change in price. The demand curve is a vertical straight line. Quantity demanded remain unchanged when the price changes.
MEASUREMENT OF PRICE ELASTICITY The price elasticity is measured by dividing the percentage change in quantity demanded by the percentage change in price. Arc Price Elasticity: As mention earlier the basic formula is the same, however, the method of calculating the percentage differs. It takes into consideration the average of the old quantity and the new quantity on the denominator and the average of old price and new price on the numerator The formula is:
Q2 - Q1 P2 - P1
P2 + P1 Q2 + Q1
Where: Q1 Q2 P1 P2
is original quantity demanded, is new quantity demanded, is original price and is new price
Point Elasticity: Point elasticity calculates elasticity for an infinitesimally small change in the independent variable (price). The formula is:
Ep =
Qx P x
Px Qx
eqn 3k
The first term is the partial derivative of the demand function in terms of Px. Px and Qx is the price and the quantity demanded of the product X respectively. Based on Eqn 3.1 the demand function is
105
## Chapter 3 Demand Theory
Qx = - 0 Px + 1 Py + 2 I + 3 T The point price elasticity based on the above formula will be;
Ex = - 0
Px /Qx
Example 7 Based on Example 1 where : Qx = 1.0 - 2.0Px + 1.5I + 0.8Py + 2.0T Given Px = RM2 Py = RM2.5 I = 4 and T = 2
Qx = 9 units The point elasticity can be calculated by differentiating the equation in terms of Px and then multiplying by the ratio of Px to Qx. Qx = - 2.0 Px The point price elasticity is Ex = - 2.0 x 2/9 = - 0.44 (inelastic) When the price changes by one percent the quantity demanded will change by 0.44 percent. Qx = 9 Px = 2
106
## Chapter 3 Demand Theory
ELASTICITY ALONG A DEMAND CURVE In a linear demand curve function the elasticity along the demand curve varies from zero to infinity. This is because QX / PX is constant but Px / Qx will increase as price increase. (You may want to check it out yourselves. The elasticity at different points along the demand curve can be determined by calculating the quantity demanded at different price level and then substituting it into the elasticity formula). Diagram 3.6 shows elasticity along a demand curve at different points. Price E= E>1 E = 1 (Mid - point) E<1 E = 0 Quantity Diagram 3.6 Elasticity along the Demand Curve
THE IMPORTANCE OF PRICE ELASTICITY The concept of elasticity is useful in decision-making. It helps managers to understand the relationship between elasticity, total revenue and marginal revenue. It assist the manager in terms of pricing decision that is, whether to change or maintain the price of the product when the objective is to maximise revenue. The relationship between price, marginal revenue and total revenue can be summarized in a single number known as price elasticity of demand. That is: TR MR = PQ = dTR dQ = P + Q P Q
107
## Chapter 3 Demand Theory
P P
1 + Q Q P
Given the elasticity formula as E = Q - P P and next: MR = P 11 E For normal good the price elasticity is negative, the positive sign in the above equation becomes negative when multiplied by a negative. Q
1 = Q - Q E P P
MR
1+1 E eqn 3l
Bases on the above equation the following relationship can be develop: a. b. When price elasticity is greater than one the marginal revenue is positive (Ex > 1, MR > 0) When price elasticity is less than one the marginal revenue is negative (E x < 1, MR < 0) c. When price elasticity is equal to one the marginal revenue is zero (E x = 1, MR = 0)
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## Chapter 3 Demand Theory
The relationship can be diagrammatically represented as in diagram 3.7 Price E>1 TR DD MR = 0 Quantity MR Diagram 3.7. Elasticity, Total Revenue and Marginal Revenue A producer will not sell his output in the region where elasticity is less than one. This is because the total revenue is decreasing and the marginal revenue is negative as more units are sold. By reducing the output the producer will be able to increase his profits: as total revenue will increases and total cost decreases. E=1 E<1
## DETERMINANTS OF DEMAND ELASTICITY
a)
Availability of substitutes Substitutes play an important role in determining the elasticity of the product. Products with substitute have a higher price elasticity compared to products with poor or no substitutes. For example if the price of coca-cola increases, the Thus, the demand for coca-cola is demand for its substitutes will increase. The demand for coca-cola will decrease rapidly with a slight increase in price. relatively elastic.
b)
Proportion of income spent on the product. The demand for the product is said to be elastic if the expenditure on the product forms a large proportion of the total expenditure For example, students who are
109
## Chapter 3 Demand Theory
staying outside campus and are renting rooms, if the rent increases, these students will find alternative accommodation as a large proportion of their income is spent on rental. The demand for rented rooms can be considered elastic. On the other hand, demand is inelastic for products that account for a small proportion of total expenditure e.g. salt.
c)
Length of the period In the long run demand is more elastic than in the short runs as consumers are able to adjust to new prices. For example, demand for electricity. In the short run, when the price of electricity increases, consumers will not reduce their consumption of electricity. Consumers may still use their electrical appliances. But given some time (long run), consumers are able to find other energy sources, i.e. replace electrical appliances with non electrical appliances or to use them less frequently, reducing the consumption of electricity in the long run. Consumers are more responsive to the changes in price in the long run.
INCOME ELASTICITY (EI) Income elasticity measures the responsiveness of demand to the change in income.
MEASUREMENT OF INCOME ELASTICITY Income elasticity is measured by dividing the percentage change in quantity demanded by the percentage change in income. Arc Income Elasticity: The formula is very much like the arc price elasticity. For arc
income elasticity the variable price is changed to income. The formula is: EI = Where: Q1 Q2 I1 I2 Q2- Q1 I2 - I1 X I2 + I1 Q2+ Q1
is old quantity demanded, is new quantity demanded, is original income and is new income
110
## Chapter 3 Demand Theory
Point Income Elasticity: Where the change in income is infinitesimally small. The formula is:
Ei
Qx x I
I Qx eqn 3m
## Based on demand function in equation 3.1: Qx = - 0 Px + 1 Py + 2 I + 3 T The income elasticity is: EI = 2 X I / Qx
Example 8 Based on Example the income elasticity for product X is: Qx Given Px Qx = 1.0 - 2.0Px + 1.5I + 0.8Py + 2.0T = RM2 Py = RM2.5 = 9 Qx = 1.5; Qx = 9, I = 4 I EI EiI = = 1.5 x 4/9 0.67
111
I = 4 and T = 2
## Chapter 3 Demand Theory
When income increase by one percent the quantity demanded will increase by 0.67 percent. This product can be classified as a basic necessity. level (economic growth). It is also recession proof product since the impact on quantity demand is smaller compare to the fall in the income
IMPORTANCE OF INCOME ELASTICITY As a decision maker income elasticity is important. It shows how the consumers react towards the quantity demand when there is a change in the income. Income elasticity is useful to:
a)
## Income elasticity can either be positive or negative.
associated with normal goods, i.e. when income increases the quantity demanded for these goods increase. Normal goods can further be divided into basic necessities and luxury goods. For basic necessities the income elasticity is less than one or close to zero, whereas for luxury good the income elasticity is greater than one. On the other hand, negative income elasticity is associated with inferior goods. That is, when income increases, the spending power increases the demand for an inferior good will decrease. Consumers will buy better quality goods and reduce their purchase of inferior quality goods. .
b)
## Forecasting the future demand
A producer will need to forecast the future demand and be prepared to supply the amount of quantity demanded. One-way to forecast the demand the product is to determine the phase of business cycle in the economy, that is in which phase (recession, expanding, peak and contraction) is the economy. If the economy is expanding or during economic prosperity, consumers are willing to spent more on luxury goods and the demand for luxury goods will increase. The rate of increase in demand for these good is greater than the rate of income (economic growth). The reverse is true during recession;
112
## Chapter 3 Demand Theory
the demand for luxury goods is weak. During this time producers should concentrate on basic necessities. Basic necessities are said to be recession proof as the percentage decrease in demand is much smaller compare to the percentage decrease in income The consumers spending on these goods is consistent or the impact of decreasing income is very small.
c)
## Income elasticity is also useful in promoting or marketing a product.
that are consumed by higher income-group are advertised in media that reaches them. Their promotional strategies would differ in terms of service and are given personal touch. These products are also be displayed in an exclusive manner Whereas normal products can be promoted in daily media and be displayed to reach the general public .
CROSS ELASTICITY (Exy) Cross elasticity measures the responsiveness of the demand of one product (X) when the price of another changes (Y).
MEASUREMENT OF CROSS ELASTICITY The cross elasticity is measured by dividing the percentage change in quantity demanded of one product (X) by the percentage change in the price of another product (Y). Arc Cross Elasticity:: The formula is : Exy = Qx, 2 - Qx,1 Py, 2 - Py,1 X Py,2 + Py,1 Qx,2 + Qx,1
## Where: Qx,1 Qx, 2 Py, 1 Py, 2
is old quantity demanded, is new quantity demanded, is original price of product y and is new price of product y
113
## Chapter 3 Demand Theory
Point Cross price Elasticity: Measure the change in the quantity demand of a product when there is an infinitesimally small change in price of another product. The formula is :
Exy
Qx Py
Py Qx eqn 3 n
## Based on equation 3a demand function: Qx = - 0 Px + 1 Py + 2 I + 3 T The elasticity is: Ex = 1 X Py /Qx
IMPORTANCE OF CROSS ELASTICITY The cross elasticity is used to it determine the relationship between the two goods. That is If the cross elasticity is positive the two goods are substitutes, for example Pepsi and Coke. The higher the value of the cross elasticity the better the substitutes. If the cross elasticity is negative the two goods are complementary, for example pen and ink. If the cross elasticity is zero, it shows that the two products are not related.
114
## Chapter 3 Demand Theory
Example 9 From Example 1 above, the cross elasticity for product X is Qx = 1.0 - 2.0Px + 1.5I + 0.8Py + 2.0T Given Px = RM2 Py = RM2.5 Qx = 9 The cross elasticity is Qx = 0.8, Py Exy = = 0.8 x 2.5 / 9 0.22 Qx = 9, Py = 2.5 I = 4 and T = 2
When the price of X changes by one percent the quantity demanded for X will change by 0.22 percent. The two products are said to be poor substitutes
115
## Chapter 3 Demand Theory
QUESTIONS
1.
The demand function for Rainbow Lemonade has been estimated as QL = 26 - 20PL + 28 Pc - 17Ac + 20AL + 5I Where QL is the quantity of Rainbow Lemonade (in cases) PL is the price of Rainbow Lemonade in RM PC is the price of competitor in RM AL is the advertising expenditure of Rainbow lemonade. AC is the advertising expenditure of competitor. I is the income per capita in RM The current values of the independent value are AL = RM10, a) b) c) d) PL = RM 1.50, PC = RM 1.8, I = RM12, AC = RM 15
Derive the demand curve function and the total revenue function. At what output is the total revenue maximum? Calculate the price at this output? If the income decrease by 10% what is the impact on Rainbow Lemonade. (Calculate income elasticity)? What is the price and cross elasticity? Interpret the elasticity
2.
Given the following demand function Qx = 7 - 2PX + 5Py + 8I Where Qx and Px is the quantity and price of X respectively PY is the price of another product; and I is per capita income. Given Px = RM5, Py = 5.50 and I = 4.
116
## Chapter 3 Demand Theory
a) b) c)
Calculate the price elasticity. Interpret the elasticity. Calculate the income elasticity. What can you say about the product? Calculate the cross elasticity. What is the relationship between this two product?
3.
A firm XYZ has written childrens comic books. The estimated total revenue function for the sales is TR = 150Q - 20Q2 a) b) c) Over what range is demand elastic? Derive the demand curve function. Given the current price is RM50, should the firm maximise its total revenue? change the price to
4.
A consultant estimates the price-quantity relationship for Alice Pizza to be P = 75 - 25 Q a) b) c) At what output is demand unitary elastic? At what output is marginal revenue zero? At price RM7, what is the price elasticity?
5.
The price elasticity for fish is estimated to be -0.6 and income elasticity is 0.7. At a price of RM8.00 per kg and a per capita income of RM30, 000, the demand for fish is 25,000 kg. a) b) c) Is fish a basic or Luxury good? Explain. If the per capita increase to RM30, 500, what will the quantity demanded? If the price of fish increases to RM9.00 when per capita is RM30, 000, how much should the per capita income change for the quantity demanded to remain at 25,000 kg?
117
## Chapter 3 Demand Theory
SUGGESTED SOLUTIONS
1.a)
QL QL PL TR
= 26 - 20PL + 28 (1.8) - 17 (15) + 20 (10) +5(12) = 81.4 - 20PL = 4.07 - 0.05 QL = 4.07Q - 0.05QL2 Total Revenue function Demand curve function
b)
TR maximising output is TR = 4.07 - 0.1Q Q Q P = 40.7 = 4.07 - 0.05 (40.7) = 2.035 = Q Marginal Revenue function
Total revenue maximising quantity is 40.7 cases and the priceRM2.035. c) Income elasticity: QL QL EI = 26 20(1.5) + 28 (1.8) - 17 (15) + 20 (10) +5(12) = 51.4 = QL I EI = 5 x 51.4
118
I QL
12
= 1.17
## Chapter 3 Demand Theory
When income falls by 10% quantity demanded will decrease by 1.17 x 10 = 11.7 %. d) Price elasticity Ep = QL PL QL QL Ep x PL QL
= 26 20(1.5) + 28 (1.8) - 17 (15) + 20 (10) +5(12) = 51.4 = -20 X 1.5 51.4 = - 0.58
When price increase by 1% the quantity demanded decreases by 0.58%. Cross elasticity ELC = QL x PC ELC = 28 x PC QL 1.8 = 0.98 51.4 When the price of competitor increases by 1% the quantity demanded for Rainbow Lemonade will increase by 0.98%. 2.a) The quantity demanded is Qx = 7 - 2 (5) + 5 (5.50) + 8 (4) Qx = 56.5
119
## Chapter 3 Demand Theory
Price elasticity Ep = Qx P x Ep = -2 x 5 56.5 When the price increases by 1% the quantity demanded decrease by 0.18%. b) Income elasticity EI = Qx I EI = 8x 4 56.5 When income increases by 1% the quantity demanded increases by 0.57%. The product is a basic necessity. c) Cross elasticity Exy = Qx Py Exy = 5 x 56.5 = 0.49 x Py Qx 5.5 x I Qx = 0.57 x Px Qx = -0.18
When the price of Y increases by 1% the quantity demanded for X will increase by 0.49%. The two products are substitutes.
120
3.a)
## TR maximizing quantity is: TR TR Q Q = 3.75 = 150Q - 20Q2 = 150 - 40Q =0
When the total revenue is at the maximum, elasticity is unitary. Therefore the demand curve is elastic just before the total revenue maximizing quantity is achieved. In this case it will be before 3.75 units. The elastic range will be 0 to3.75 units b) Demand curve function is: P = TR = 150 - 20 Q Q c) To find the total revenue maximising price substitute the total revenue quantity calculated in (a) into the demand curve function: TR P P = 150 Q - 20 Q2 = 150 - 20 (3.75) = RM 75 Demand curve function
The firm should increase the price to maximise to RM75 to increase total revenue. 4.a) Unitary elasticity (calculate the quantity that maximise total revenue). TR = 75 Q - 25 Q2
121
## Chapter 3 Demand Theory
TR Q
= 75 - 50 Q Q
= 0 = 1.5
When total revenue is maximum when elasticity is unitary at Q = 1.5. c) MR MR =0 = 75 - 25 Q Q c) TR P = 75 Q - 25 Q2 = TR Q Given P = 7 , substitute into the demand curve function 7 Q = 75 - 25Q = 2.72 =75 -25Q Demand curve function = 0 = 1.5
122
## Chapter 3 Demand Theory
5. a)
Fish is an Basic good because the income elasticity is less than one (EI.= 0.7)
b)
The percentage change is income is 30500 - 30000 X 100 = 1.67% 30000 When the income changes by 1% quantity demanded will change by 0.7%.If income changes by 1.67 quantity demanded will change by : 1.67 x 0.7 = 1.17%, that is The change in quantity is: 1.17 x 25000 = 292.5 units 100 The quantity demanded will be (25000 + 292.5) = 25292.5.5
c)
At price RM9.00 the quantity demanded will be. Percentage change in price of fish 9 - 8 x 100 = 12.5% 8 The percentage change in quantity demanded is: 12.5 x 0.6 = 7.5% The total quantity is 7.5% x 25000 = 1875. The percentage change income to offset the decrease in quantity demanded will be.
123
## Chapter 3 Demand Theory
7.5% = 10.7 % 0.7% The income will increases by 10.7% x 30,000 = 3214.28
SUMMARY
This chapter explained how to derive the demand curve, total marginal and average revenue curve function from a demand function.
It showed how the above three functions are interrelated. You will also notice how elasticity is calculated. Here emphasis was given on point elasticity and it looked at price, income and cross elasticity.
## As a manager you will be made aware of the importance of elasticity in decisionmaking.
PRACTISE QUESTIONS
Q1.
The demand function for TV set is given as follow Qx = 350 3.7Px + 0.2Y + 4.2 Pz Where Qx is quantity of TV set demanded per month Px is the price of TV set Y is the per capita income Pz is the price of a competitive brand a. What is the demand curve function if Px = RM1200 Y = RM5500 Pz = RM1500.
124
## Chapter 3 Demand Theory
b. If the firm wishes to increase its total revenue, should the price be increased or decreased? c. At what price and output would total revenue maximize? Q2. The market demand for good X is Q = 70 3.5P 0.6M + 4Pz Where Q is the number of units of good X demanded P is the price of good X M is income of buyers Pz is the price of related good. a) Assuming P = 10, M= 20, and Pz = 6, calculate Q. hence, compute the price, income and cross elasticity of demand of good X. b) Is X a normal or inferior good? Explain. c) Are X and Z substitute or complements? Explain. d) Is demand for X elastic or inelastic? Why do you say so? e) Suppose the producer of X wishes to increase total revenue by changing price, should he increase or decrease price? State your reason. Q3. Zaids Frozen Pizza has enjoyed rapid growth in West Malaysia. From the analysis it was found that the demand curve follows this pattern. Q = 1000 3000P + 10A Where Q = quantity demanded P = product price (in RM) A = advertising expenditure (in RM) a) Assume that P = 3 and A = 2000 Suppose the firm reduces price. Would this increases total revenue? Explain. b) Assume that P =4 and A = 2100 Suppose the firm reduces price. Would this increase total revenue? Explain. Q4. Hardwood cutters offers seasoned, split fireplace logs to consumers in Kampung Parit, Perak. The company is the low-cost provider of firewood in this market with
125
## Chapter 3 Demand Theory
fixed costs of RM10 000 per year, plus variable costs of RM25 for each unit of firewood. Annual demand for the company is: P= 225 0.125 Q Where p is the price of firewood per unit and Q is the number of units of firewood. a) Determine the marginal revenue and the marginal cost function. b) Calculate the profit maximizing quantity, price and profit. c) Calculate the price elasticity at the profit maximizing price. Q5. Syarikat Alpha-Beta has estimated the demand curve for its product as follows: Q = 8000 - 5P
where Q is quantity sold per week and P is the price per unit. a) on the estimated demand curve, write the equation for: i. ii. iii. Based total revenue Average revenue Marginal revenue
b) What is the maximum total revenue per week that Syarikat Alpha-Beta can obtain from sales of its product? c) Calculate the arc price elasticity of demand for Syarikat Alpha-Betas product between Q = 3,000 and Q = 3,200. Q6. The marketing department for a firm that manufactures vehicles has determined the following demand function for their vehicles: QV = 5000 0.6PV + 0.2PC + 0.04Y + 0.02A Where: QV PV PC Y A = the number of the firms vehicles sold weekly = the price of the firms vehicle = the price of a close competitors vehicle = average household income, and = weekly advertising dollars spent
126
## Chapter 3 Demand Theory
a) If PV = RM10, 000, PC = RM8000, Y = RM12,000 and A = RM4000, find the price elasticity of demand. b) Is demand elastic, unitary elastic, or inelastic? Why? If the firm decreases price, what will happen to total revenue? c) Assuming values of the variable is as given in part (a) above, determine the income elasticity. Interpret your answer. d) Calculate the cross price elasticity of demand and interpret your answer. Q7. The following functions describe the demand of 3 small firms: A, B and C that sell hand phone accessories to the customers. Firm A : P = 500 0.5QA Firm B : P = 300 0.25QB Firm C : P = 200 0.125QC Where Q is the quantity demanded and P is the price. a) Calculate the market demand function. b) Using the market demand function, determine the total revenue maximizing price and quantity. c) The industry supply equation is given by QS = -1000 + 10P. Determine the market equilibrium price and quantity Q8. The research department of White Pigeon wished to estimate the demand function for its new product, Cage XP. A demand function had been estimated on 120 respondents using regression analysis. The demand function is as follows: Q = 12,000 8P Where Q = Quantity demanded for Cage XP cages P = Price of Cage XP cages (RM70) A = Advertising expense, in thousands (RM54)
127
+ 1300A + 5Pc + 2I
## Chapter 3 Demand Theory
Pc = Price of competitor's product (RM80) I = Average monthly income (RM400) a) Derive an expression for the firms conventional demand curve for the new product, Cage XP cages If the firms objective is to maximize total revenue from the sales of Cage XP cages, at what price should the firm charge? b) Should White Pigeon Company consider reducing its price in order to increase its total revenue? Explain. c) Calculate income elasticity of demand. Is the Cage XP cages a luxury, inferior or necessity good? d) Calculate the cross elasticity of demand. Are Cage XP cages and interpret it.
STUDY NOTES
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1. ## What is the maximum number of intersects on a coplaner parabola and a hyperbola? var addthis_config = {"data_track_clickback":false};
What is the maximum number of intersects on a coplaner parabola and a hyperbola?
I know the relevant answer is 4, however, someone asked if it was possible that the parabola took the same curve path of one of the hyperbolas. Is it possible that in that case, the maximum number could be infinite?
2. The correct answer is 4 indeed.
The Cartesian coordinates of common points of two 2nd degree curves are the solutions of the system of two 2nd degree equations with 2 unknowns. The elimination of one of them leads to a single algebraic equation with 1 unknown of at most 4th degree that may have 0, 1, 2, 3, 4 or infinitely many (the latter case possible if it is identity) real roots.
The equations of the parabola and the hyperbola are structurally different (they have different discriminants - follow the link in Sources below for details), what leads to different geometrical properties of both curves, for example:
1) Parabola has a single branch, hyperbola has two;
2) Parabola has no asymptotes, hyperbola has two;
3) Parabola's eccentricity is 1, hyperbola's - greater than 1 (see an animation in the article, 'Parameters' Section), etc.
That's why it is impossible for a parabola to be identical with a branch of hyperbola, so both curves can not have more than 4 common points.
Here is an example:
parabola y² = x + 11 /"c"-shaped/,
hyperbola x² - y² = 1 /") ("-shaped/,
Common points (-3, -?8), (-3, ?8), (4, -?15) and (4, ?15)
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## The inability to find any record of Harrison J. Bounel means he must be a real person!
The American Thinker published an article, “Barack Hussein Obama and Harrison J. Bounel,” by criminologist Jason Kissner today that argues that the association between the name Harrison J. Bounel and Barack Obama in some unspecified transaction in a public database is much more than coincidence or an attempt at identity fraud. While the argument is couched in the language of probability, no actual math was harmed in the writing of the article.1
Detractors of the birther Social Security fraud theory point out that no one has been able to locate this Harrison J. Bounel who birthers claim is a real person to whom the social security number used by Barack Obama was originally assigned. If such a person existed, debunkers say, then one ought to be able to find some independent record of him.
Kissner turns the tables and says that the inability to find other records of Bounel is proof that he is a real person (are you confused yet?). Kissner’s argument is that the database record could not be fraudulent because it would have been too difficult to actually find someone with no other record in order to perpetrate the fraud.
Kissner’s claim is inserted in a straw man argument refuting the idea that some anti-Obama person created the fake record for Bounel. I suppose someone might have speculated on the possibility that the public record for Bounel with Obama’s SSN was created for the purpose of creating an anomaly in Obama’s record (I might have even done it), but I think it is more likely to be an error or an attempt at identity theft. The straw man context is not important because Kissner’s argument fits the real argument of random error or fraud as well as it does the straw man.
Kissner’s fallacy, however, is the ad hoc assignment of probability to something that’s already happened. It’s like looking at the winning lottery number and saying “what are the chances this number would come up?” and then arguing that the lottery must be rigged. In order to make the probability argument, one has to set the criteria in advance, or they have to be necessary. The fact that the name “Harrison J Bounel” doesn’t belong to anybody is not necessary to the hypothesis and so it’s not significant, no matter how improbable it is.
But the probability of coming up with a name belonging to nobody isn’t that low; in fact, it’s easy. I took the names “Kissner” and “Bounel” and used them with the first and middle names of the members of my immediate family (6 total). I got only one hit on Google for the 12 names I tried.
And finally Kissner’s argument works equally well against Obama fraudulently using an SSN belonging to Bounel: How could Obama have found someone so totally devoid of any record?
What Kissner won’t address is: assuming Harrison J Bounel is a real person (birther hypothesis), what are the chances that there is no record of him anywhere, not birth announcement, obituary, immigration record, census, city directory, grave, genealogy nor criminal record? | 675 | 3,118 | {"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-2015-22 | latest | en | 0.952838 |
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` the ellipse with its focus and centre and passing through point given how to find equation of ellipse`
5 months ago
1808 Points
``` hello jadhav. this is an excellent ques tbh. lets say one of the given focus is (r, s) and centre is (c, d). then we can find the coordinates of the second focus by using the fact that the centre of an ellipse lies in the centre (or mid point) of both the foci. so assume that coordinates of the second focus are (p, q). then by section formula, we can write(r+p)/2= c and (s+q)/2= d..........from these we obtain p and q.now, by the definition of ellipse, for every point (x, y) the sum of its distances from two other points (the foci) is constant.so that √[(x – r)^2+(y – s)^2] + √[(x – p)^2+(y – q)^2]= K (a const)now, we are also given a point (say m, n) on the ellipse.. so substitute that value in above eqn.we get √[(m – r)^2+(n – s)^2] + √[(m– p)^2+(n – q)^2]= K so that we obtain the value of K. so the eqn becomes√[(x – r)^2+(y – s)^2] + √[(x – p)^2+(y – q)^2]= √[(m – r)^2+(n – s)^2] + √[(m– p)^2+(n – q)^2]now if you wish you can be simplify this eqn by getting rid of the square roots.kindly approve ;)
```
5 months ago
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• Previous Year Exam Questions | 576 | 1,907 | {"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-2020-05 | longest | en | 0.758763 |
https://a4accounting.com.au/running-total-in-a-formatted-table/ | 1,723,552,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00567.warc.gz | 63,037,545 | 17,842 | # Running Total in a Formatted Table
### Using the SUM function
Formatted Tables allow you to create formulas that automatically copy down as the table expands. To create a running total in a column you have a couple of options.
In the example below we want to create a running total in column C.
There are two solutions.
Structured references only
Formatted tables include structured references that are similar to range names. We can use those structured references to create the running total formula.
We can use the SUM function but there is trick to creating the formula. We need to use the colon : symbol to separate the starting cell and the last cell for the SUM range. The problem with this technique is that sometimes Excel won’t let you use the colon. It will let you use the comma , so that is the workaround. Try creating it using the colon and if it doesn’t work use the comma and then change that to a colon – see images below.
In cell C2 type
`=SUM(`
And click on cell B1
This enters the headers reference as above, type a colon if it doesn’t allow you the type the comma and click on cell B2 and add the closing bracket to the formula.
Then change the comma into a colon if necessary.
Press Enter.
The final formula is
`=SUM(tblData[[#Headers],[Amount]]:[@Amount])`
This formula is identical in all the formula cells in column C.
This formula works because the SUM function treats text entries as zero – e.g. the heading in cell B1 is treated a zero.
Structured and a cell reference
The second technique combines a structured reference with a relative cell reference.
We will add another column to demonstrate this technique. In cell D2 type
`=SUM(`
Then click on cell B2 and type a comma then type D1 and press Enter.
The final formula is
`=SUM([@Amount],D1)`
This formula works because the SUM function treats text entries (cell D1) as zero and doesn’t display a #VALUE! error.
`=[@Amount]+D1`
It would display have displayed a #VALUE! error.
Referring to other rows
In general you don’t refer to another row in a table, but in this case the formula works even when the table is sorted.
Please note: I reserve the right to delete comments that are offensive or off-topic.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 502 | 2,302 | {"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-2024-33 | latest | en | 0.835558 |
https://kidsworksheetfun.com/arithmetic-sequence-worksheet-with-answers/ | 1,702,025,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00683.warc.gz | 389,311,907 | 25,643 | # Arithmetic Sequence Worksheet With Answers
Apart from the stuff given in this section if you need any other stuff in math please use our google custom search here. Special line segments in.
50 Arithmetic Sequence Worksheet With Answers In 2020 Arithmetic Sequences Arithmetic Persuasive Writing Prompts
### Given the first term and the common difference of an arithmetic sequence find the explicit formula and the three terms in the sequence after the last one given.
Arithmetic sequence worksheet with answers. 23 a 21 1 4 d 0 6 24 a 22 44 d 2 25 a 18 27 4 d 1 1 26 a 12 28 6 d 1 8 given two terms in an arithmetic sequence find the recursive formula. If asked to extend either of these patterns the next 3 terms you need to know the last term and then you can go on from there. 4 3 arithmetic and geometric sequences worksheet determine if the sequence is arithmetic.
A sequence is a function. Prior to dealing with arithmetic sequence worksheet with answers you need to know that training is our answer to an even better tomorrow in addition to mastering doesn t only stop the moment the education bell rings that remaining claimed many of us offer you a a number of uncomplicated but helpful content in addition to web themes manufactured well suited for any helpful purpose. Area and perimeter worksheets.
Complementary and supplementary worksheet. The following diagrams give the formulas for arithmetic sequence and arithmetic series. Given a term in an arithmetic sequence and the common difference find the recursive formula and the three terms in the sequence after the last one given.
Apart from the stuff given above if you want to know more about arithmetic sequence worksheet please click here. Given a term in an arithmetic sequence and the common difference find the 52nd term and the explicit formula. 32 26 20 14 8 this is a decreasing arithmetic sequence with a common difference of 6.
After having gone through the stuff we hope that the students would have understood arithmetic sequence worksheet. The first term is 1 and the last term is 1000 and the common difference is equal to 1. An arithmetic sequence is a sequence that has the pattern of adding a constant to determine consecutive terms.
17 a 16 105 and a 30 203 18 a 17 95 and a 38. The above sequence has 1000 terms. These are examples of.
Complementary and supplementary word problems worksheet. This is an increasing arithmetic sequence with a common difference of 3. S 1000 1000 1 1000 2 500500 problem 6.
Examples solutions videos activities and worksheets that are suitable for a level maths to help students answer questions on arithmetic sequence and arithmetic series. Scroll down the page for more examples and solutions. Types of angles worksheet.
Properties of parallelogram worksheet. We say arithmetic sequences have a common difference. 27 a 18.
13 a 32 622 d 20 14 a 18 166 d 8 15 a 9 74 d 6 16 a 28 231 d 10 given two terms in an arithmetic sequence find the explicit formula. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms see formula above. Proving triangle congruence worksheet.
What is the domain and range of the following sequence. Given the formula for the arithmetic sequence determine the first 3. Sum of the angles in a triangle is 180 degree worksheet.
If it is find the common difference.
Arithmetic Sequence Practice At 3 Levels Arithmetic Sequences Arithmetic Complex Sentences Worksheets
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Geometric Sequences Worksheet Answers Best Of Arithmetic And Geometric Sequences Worksh In 2020 Arithmetic Sequences Persuasive Writing Prompts Word Problem Worksheets | 1,155 | 5,752 | {"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-2023-50 | latest | en | 0.91959 |
https://kubicle.com/lessons/diagram-the-problem | 1,527,369,061,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867904.94/warc/CC-MAIN-20180526210057-20180526230057-00372.warc.gz | 581,710,886 | 14,811 | 3. Diagram the Problem
Subtitles Enabled
Overview
The first stage in building the model is drawing an influence diagram - which we will learn how to do in this lesson.
Lesson Notes
Influence diagrams
- Before we jump into building the model, we need to first structure the model
- Influence diagrams are a graphical way of structuring the model
- They are allow you to experiment with different structures without getting mired in detail
Transcript
Let's start by adding Total Cost as an outcome. I know that total cost is equal to total variable cost plus fixed cost. Fixed costs like rent, light, and heating cannot be changed in the near term and so I've included as parameters. Variable cost, on the other hand, is a function of unit cost and the quantity sold, so it's actually connected to the total revenue model. To finish up our Influence diagram I'll create a profit output and link both total revenue and total cost to this outcomes. Total revenue and total cost are now intermediate variables. So in the final version of the influence diagram I've changed their shape to take account of this fact. Armed with this influence diagram, we can now quickly create an excel profit model which would have been more difficult if we had skipped this step.
Even for a simple profit model such as this one, you can see how the influence diagram can serve to quickly structure your thinking. In the next lesson, we'll see how influence diagrams are even more necessary for complex problems such as Zippy Airways.
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2 mins | 360 | 1,665 | {"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-2018-22 | latest | en | 0.939182 |
https://www.iqenergy.org.ua/answers/646566-the-school-s-guidance-department-compares-the-grade-point-averages-and-standardized-state-test-scores-for-10-students-in-each | 1,709,398,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00345.warc.gz | 813,577,743 | 7,674 | This is because both have a negative correlation coefficient, meaning they are showing regression, or decrease.
Step-by-step explanation:
## Related Questions
The probability of event A is x, and the probaility of event B is y. If the two events are independent, which condition must be true?
P(A|B) = x
The condition P(A|B) = x is true.
Step-by-step explanation:
Given : The probability of event A is x, and the probability of event B is y.
If the two events are independent.
To find : Which condition must be true?
Solution :
P(A)=x and P(B)=y
If two events are independent
Then
The formulas for conditional probabilities are
If we substitute the independent condition in conditional probability formulas we get,
Applying the given condition,
Therefore, The condition P(A|B) = x is true.
6-2x=2(15-x)
A. No solutions
B. -5
C. 5
D. All real numbers
6-2x=2(15-x)
distribute
6-2x=30-2x
6=30
fasle
no solution
Determine which system below will produce infinitely many solutions. A) −6x + 3y = 18
4x − 3y = 6
B)2x + 4y = 24
6x + 12y = 36
C)3x − y = 14
−9x + 3y = −42
D) 5x + 2y = 13
−x + 4y = −6
Infinitely many solution means the lines have to be the same
3x - y = 14....multiply everything by -3
-9x + 3y = - 42...notice that this is the same as the other equation...therefore, it is the same line with infinite solutions...so ur answer is C
Is 9.99 rational or irrational | 421 | 1,397 | {"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} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.893608 |
https://larnakaband.com/when-was-cool-math-games-created-latest-2023/ | 1,675,139,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00082.warc.gz | 368,872,677 | 15,624 | # When Was Cool Math Games Created latest 2023
You are searching about When Was Cool Math Games Created, today we will share with you article about When Was Cool Math Games Created was compiled and edited by our team from many sources on the internet. Hope this article on the topic When Was Cool Math Games Created is useful to you.
Page Contents
## The Art of Reading Backwards: The Magic of Palindromes
Have you ever tried to read words backwards? I tried once when I was a kid. My name was Ilek. My cat’s backwards name was Niffum, and my favorite food was etalocohc. I walked around saying words no one else understood. It was like speaking my own language.
But then I came across some very surprising words that stayed the same when I read them backwards. “Dad” remained dad and “mum” remained mum, and even my friend Hannah remained the same! I didn’t know it then, but I had discovered the very cool world of palindromes.
What is a palindrome?
The palindromes remain the same when you read them forward or backward. These can be words:
eye
popular
It can be phrases or sentences:
never odd or even
don’t agree
Was it a cat that I saw?
Do geese see God?
These can be numbers:
505
62326
12:21
And some palindromes are even the same upside down, upside down, upside down, and upside down:
108801
S.O.S.
NOON
Palindrome Magic Trick
Here’s a trick you can do to turn almost any number into a palindrome. First, think of a number and write it down:
3241
Then, write that number upside down and add the two:
3241
+ 1423
4664
In some cases, it doesn’t work the first time. When this happens, keep reversing and adding numbers until you get your palindrome:
4972
+ 2794
7766
+6677
14443
+34441
48884
Are you a Palindrome?
Warning: palindromes can be addictive! Some people start looking for them everywhere and can get a little crazy in their search for palindromes.
You may be at risk of being a palindromanic if:
• You have friends named Ana, Eve, Izzi, Otto or Bob.
• Your favorite times of day are 10:01 a.m., 11:11 a.m., 5:55 a.m., etc.
• You love kayaking and have always wanted to drive a race car.
• Your favorite car is a Toyota.
Stock up on palindromane with these books and resources:
For young children: Mom and dad are palindromesby Mark Shulman and Adam McCauley, or If you were a palindrome, by Michael Dahl. These two fun picture books are a great introduction for children to the world of palindromes.
For adults and adults: Jon Agee is the king of palindrome books like Go hang a salami, I’m a lasagna pigand Palindromania. Cartoony line art is paired with wacky captions for fun.
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You are searching about What Is The E In Math, today we will share with you article about What Is The E In Math was compiled and… | 1,068 | 4,464 | {"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-2023-06 | longest | en | 0.924736 |
https://stats.stackexchange.com/questions/68351/robust-residual-standard-error-in-r | 1,716,633,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00751.warc.gz | 465,698,072 | 39,404 | # Robust Residual standard error (in R)
I have a question regarding to the concept of robust standard errors. What I found about that topic is, that one can estimate the robust standard error for regression coefficients to eliminate problems with heteroscedasticity (when one wants to interpret a model). I want to know if there is a way not only to determine robust standard errors of coefficients but also of the standard error of the overall regression (residual standard error). When its possible, how can I calculate such a value in general?
Because I'm using R its also interesting for me if there is a R-function for this problem (I only know the sandwich-package for the normal robust SE of the coefficients).
Thanks.
• Is it still the policy of Cross Validated to refer problems relating to statistical packages to StackOverflow? Aug 26, 2013 at 15:29
• It depends, @CesareCamestre. If the question is about how to use R, such that an explanation of the ideas w/o reference to R, or demonstrated w/ different software, eg Stata, would not answer the question, then it belongs on SO. But if the issue is w/ understanding the ideas, then it can stay here, even if they as about R as well. (Which of the above applies to this Q is not clear to me yet.) Aug 26, 2013 at 15:33
• Welcome to the site, @Meiner. It is not clear to me if you are wondering about the nature of sandwich estimates, or if you are only wondering about another way to implement them in R. If the latter, this question would be off-topic for CV (see above), but on-topic on Stack Overflow. If your question is about the underlying ideas, please edit to clarify; if it's about implementation in R, flag your Q & we'll migrate it for you (please don't cross-post, though). Aug 26, 2013 at 15:37
• Do you mean forming confidence bands for predicted values or creating something like a global test for all regression parameters against a null intercept-only model? Feb 13, 2018 at 21:26
If you are interested in the conditional mean $$\mathop{\mathbb{E}} \bigl[ y_j|X_j \bigr] = X_j' \beta$$, where $$X_j$$ may be in or out of sample, then of course you can get the standard error for that as the square root of $$X_j' \, \hat v[\hat \beta] \, X_j$$ where $$\hat v[\hat \beta]$$ is the heteroskedasticity-corrected/sandwich variance estimator. But that conditional mean is rarely of huge interest; I believe you are interested in characterizing what the whole distribution of $$y_j = X_j + \varepsilon_j$$ may have looked like. Without knowing more about the distribution of $$\varepsilon_j$$ you, of course, won't be able to say much about what the variance of $$y_j$$ will be. | 647 | 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": 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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.906361 |
https://web2.0calc.com/questions/solution_10 | 1,720,838,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00196.warc.gz | 495,029,233 | 5,648 | +0
# solution
0
59
1
How many solutions are there to the equation
u + v + w + x + y + z = 10
where u, v, w, x, y, and z are nonnegative integers, and x is at most 2 or y is at least 3?
Jun 27, 2023
#1
+23
0
There are 21 solutions to the equation u + v + w + x + y + z = 10 where u, v, w, x, y, and z are nonnegative integers, and x is at most 2 or y is at least 3.
To solve this problem, we can use the following cases:
Case 1: x=0 and y>=3. In this case, the only solution is u+v+w+z=10. There is 1 solution in this case.
Case 2: x=1 and y>=3. In this case, the possible solutions are u+v+w+z=9, u+v+w+z=8, and u+v+w+z=7. There are 3 solutions in this case.
Case 3: x=2 and y>=3. In this case, the possible solutions are u+v+w+z=8, u+v+w+z=7, and u+v+w+z=6. There are 3 solutions in this case.
Case 4: x=0 and y<3. In this case, the possible solutions are u+v+w+x+y+z=10, u+v+w+x+y+z=9, u+v+w+x+y+z=8, u+v+w+x+y+z=7, u+v+w+x+y+z=6, u+v+w+x+y+z=5, u+v+w+x+y+z=4, and u+v+w+x+y+z=3. There are 8 solutions in this case.
The total number of solutions is 1+3+3+8=21
Jun 27, 2023
edited by ollysuper Jun 27, 2023 | 434 | 1,123 | {"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} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.848087 |
https://www.studypool.com/discuss/505279/statistics-question-a-3-21?free | 1,480,904,584,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541518.17/warc/CC-MAIN-20161202170901-00481-ip-10-31-129-80.ec2.internal.warc.gz | 1,034,314,300 | 13,819 | ##### Statistics question a.3
Statistics Tutor: None Selected Time limit: 1 Day
Apr 30th, 2015
n1=10 and n2=20,therefore df=30-2=28
critical value is t=1.701
Apr 30th, 2015
...
Apr 30th, 2015
...
Apr 30th, 2015
Dec 5th, 2016
check_circle | 98 | 244 | {"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-2016-50 | longest | en | 0.790978 |
https://ro.scribd.com/document/402556616/Lesson-2-2-Frequent-Pattern-Analysis | 1,575,828,705,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514475.44/warc/CC-MAIN-20191208174645-20191208202645-00339.warc.gz | 521,912,497 | 78,108 | Sunteți pe pagina 1din 54
# Frequent Pattern Analysis
## CS 822 Data Mining
Anis ur Rahman
Department of Computing
NUST-SEECS
## September 24, 2018
1 / 54
Mining Frequent Patterns, Associations and Correlations
Basic Concepts
Apriori Algorithm
Optimization Techniques
FP-Growth
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Mining Frequent Patterns, Associations and Correlations
## Frequent Pattern. a pattern (a set of items, subsequences,
substructures, etc.) that occurs frequently in a dataset
## Goal. Finding inherent regularities in data
What products were often purchased together? Juice and diapers?
What are the subsequent purchases after buying a PC?
What kinds of DNA are sensitive to this new drug?
Can we automatically classify Web documents
Applications
Basket data analysis, cross-marketing, catalog design, sale campaign
analysis, Web log (click stream) analysis, and DNA sequence analysis.
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Mining Frequent Patterns, Associations and Correlations
Frequent Patterns
## Itemset. A set of one or more items
K-itemset X = {x1 , · · · , xk }
## (absolute) support, or, support count of X: Frequency or
occurrence of an itemset X
(relative) support, s, is the fraction of transactions that contains
X (i.e., the probability that a transaction contains X)
## An itemset X is frequent if X’s support is no less than a minsup
threshold.
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Mining Frequent Patterns, Associations and Correlations
Association Rules
## Q. Find all the rules X → Y with minimum support and confidence
threshold
Support, s, probability that a transaction contains X ∪ Y
Confidence, c, conditional probability that a transaction having
X also contains Y
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Mining Frequent Patterns, Associations and Correlations
Association Rules
## Q. Find all the rules X → Y with minimum support and confidence
threshold
Tid Items bought
10 Juice, Nuts, Diaper
20 Juice, Coffee, Diaper
30 Juice, Diaper, Eggs
40 Nuts, Eggs, Milk
50 Nuts, Coffee, Diaper, Eggs, Milk
## Let minsup = 50%, minconf = 50%
Freq. Pat. Juice:3, Nuts:3, Diaper:4, Eggs:3, {Juice, Diaper}:3
Association rules. (many more!)
Juice → Diaper (60%, 100%)
Diaper → Juice (60%, 75%)
Rules that satisfy both minsup and minconf are called strong rules.
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Mining Frequent Patterns, Associations and Correlations
## A long pattern contains a combinatorial number of sub-patterns,
e.g., {a1 , · · · , a100 } contains = 2100 − 1 = 1.27 ∗ 1030 sub-patterns!
Solution. Mine closed patterns and max-patterns instead
## An itemset X is closed if X is frequent and there exists no
super-pattern Y ⊃ X , with the same support as X
An itemset X is a max-pattern if X is frequent and there exists no
frequent super-pattern Y ⊃ X
## Closed pattern. is a lossless compression of freq. patterns
Reducing the number of patterns and rules
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Mining Frequent Patterns, Associations and Correlations
Min_sup = 1
## 1 What is the set of closed itemset?
< a1 , · · · , a100 >: 1
< a1 , · · · , a50 >: 2
2 What is the set of max-pattern?
< a1 , · · · , a100 >: 1
3 What is the set of all patterns?
!!
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Mining Frequent Patterns, Associations and Correlations
Computational Complexity
## Q. How many itemsets are potentially to be generated in the worst
case?
The number of frequent itemsets to be generated is sensitive to
the minsup threshold
When minsup is low, there exist potentially an exponential number
of frequent itemsets
## The worst case: MN where M: # distinct items, and N : max length of
transactions
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Mining Frequent Patterns, Associations and Correlations
## The downward closure property of frequent patterns
Any subset of a frequent itemset must be frequent
If {Juice, Diaper, Nuts} is frequent, so is {Juice, Diaper}
i.e., every transaction having {Juice, Diaper, Nuts} also contains
{Juice, Diaper}
## Apriori pruning principle
If there is any itemset which is infrequent, its superset should not be
generated/tested.
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Mining Frequent Patterns, Associations and Correlations
1 Basic Concepts
2 Apriori Algorithm
3 Optimization Techniques
4 FP-Growth
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Mining Frequent Patterns, Associations and Correlations
Apriori: Method
## 1 Initially, scan DB once to get frequent 1-itemset
2 Generate length (k + 1) candidate itemsets from length k frequent
itemsets
3 Test the candidates against DB
4 Terminate when no frequent or candidate set can be generated
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Mining Frequent Patterns, Associations and Correlations
Apriori Example
Database
Tid Items
10 A, C, D
20 B, C, E
30 A, B, C, E
40 B, E
Supmin = 2
1st scan
Itemset
Itemset sup
Itemset sup
{A, B}
{A} 2
{A} 2 {A, C}
{B} 3 ⇒ {B} 3 ⇒ {A, E}
{C} 3
{C} 3 {B, C}
{D} 1
{E} 3 {B, E}
{E} 3
{C, E}
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Mining Frequent Patterns, Associations and Correlations
Apriori Example
Database
Tid Items
10 A, C, D
20 B, C, E
30 A, B, C, E
40 B, E
Supmin = 2
2nd scan
Itemset sup
Itemset sup
{A, B} 1
{A, C} 2 {A, C} 2 Itemset
{A, E} 1 ⇒ {B, C} 2 ⇒
{B, C, E}
{B, C} 2 {B, E} 3
{B, E} 3 {C, E} 2
{C, E} 2
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Mining Frequent Patterns, Associations and Correlations
Apriori Example
Database
Tid Items
10 A, C, D
20 B, C, E
30 A, B, C, E
40 B, E
Supmin = 2
3rd scan
Itemset sup
{B, C, E} 2
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Mining Frequent Patterns, Associations and Correlations
Apriori Algorithm
## Ck : Candidate itemset of size k
Lk : frequent itemset of size k
L1 = frequent items;
for (k = 1; Lk ! = φ; k + +) do begin
Ck +1 = candidates generated from Lk ;
for each transaction t in database do
increment the count of all candidates in Ck +1 that are contained in t
Lk +1 = candidates in Ck +1 with min_support
end
S
return k Lk ;
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Mining Frequent Patterns, Associations and Correlations
Candidate Generation
## How to generate candidates?
Step 1. self-joining Lk
Step 2. pruning
Join Lk p with Lk q, as follows:
insert into Ck +1
select {p.itemi }{1,··· ,k −1} , p.itemk , q.itemk
from Lk p, Lk q
where {p.itemi }{1,··· ,k −1} = {q.itemi }{1,··· ,k −1} and p.itemk < q.itemk
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Mining Frequent Patterns, Associations and Correlations
## Suppose we have the following frequent 3-itemsets and we would like to
generate the 4-itemsets candidates
L3 = {{I1 , I2 , I3 }, {I1 , I2 , I4 }, {I1 , I3 , I4 }, {I1 , I3 , I5 }, {I2 , I3 , I4 }}
Self-joining: L3 ∗ L3 gives:
{I1 , I2 , I3 , I4 } from {I1 , I2 , I3 } , {I1 , I2 , I4 }, and {I2 , I3 , I4 }
{I1 , I3 , I4 , I5 } from {I1 , I3 , I4 } and {I1 , I3 , I5 }
L4 = {I1 , I2 , I3 , I5 }
Pruning: {I1 , I3 , I4 , I5 } is removed because {I1 , I4 , I5 } is not in L3
L4 = {I1 , I2 , I3 , I5 }
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Mining Frequent Patterns, Associations and Correlations
## Once the frequent itemsets have been found, it is straightforward
to generate strong association rules that satisfy:
minimum support
minimum confidence
Relation between Support and Confidence
support_count(A ∪ B )
Confidence(A ⇒ B ) = P (B |A ) =
support_count(A )
## Support_count(A ∪ B ) is the number of transactions containing the
itemsets A ∪ B
Support_count(A) is the number of transactions containing the
itemset A.
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Mining Frequent Patterns, Associations and Correlations
## For each frequent itemset L , generate all non empty subsets of L
For every non empty subset S of L , output the rule:
S ⇒ (L − S )
If (support_count(L)/support_count(S)) ≥ min_conf
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Mining Frequent Patterns, Associations and Correlations
Example
## Suppose the frequent Itemset
L = {I1 , I2 , I5 } Transactional Database
Subsets of L are:
TID List of item IDS
{I1 , I2 }, {I1 , I5 }, {I2 , I5 }, {I1 }, {I2 }, {I5 }
Association rules: T100 I1 , I2 , I5
T200 I2 , I4
I1 ∧ I2 ⇒ I5 confidence = 2/4 = 50% T300 I2 , I3
T400 I1 , I2 , I4
I1 ∧ I5 ⇒ I2 confidence = 2/2 = 100% T500 I1 , I3
I2 ∧ I5 ⇒ I1 confidence = 2/2 = 100% T600 I2 , I3
T700 I1 , I3
I1 ⇒ I2 ∧ I5 confidence = 2/6 = 33% T800 I1 , I2 , I3 , I5
I2 ⇒ I1 ∧ I5 confidence = 2/7 = 29% T900 I1 , I2 , I3
## I5 ⇒ I1 ∧ I2 confidence = 2/2 = 100%
Question: If the minimum confidence = 70%, what are the strong rules?
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Mining Frequent Patterns, Associations and Correlations
## Strong Rules are Not Necessarily Interesting
confidence=66%]
This rule is strong but it is misleading
The probability of purshasing videos is 75% which is even larger
than 66%
In fact computer games and videos are negatively associated
because the purchase of one of these items actually decreases the
The confidence of a rule A ⇒ B can be deceiving
It is only an estimate of the conditional probability of itemset B
given itemset A
It does not measure the real strength of the correlation implication
between A and B
Need to use Correlation Analysis
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Mining Frequent Patterns, Associations and Correlations
## Use Lift, a simple correlation measure
The occurrence of itemset A is independent of the occurrence of
itemset B if P (A ∪ B ) = P (A )P (B ), otherwise itemsets A and B are
dependent and correlated as events
The lift between theoccurences of A and B is given by
Lift(A , B ) = P (A ∪ B )/P (A )P (B )
If > 1, then A and B are positively correlated (the occurrence of one
implies the occurrence of the other)
If <1, then A and B are negatively correlated
If =1, then A and B are independent
Example: P ({game, video}) = 0.4/(0.60 × 0.75) = 0.89
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Mining Frequent Patterns, Associations and Correlations
## Major computational challenges
Huge number of candidates
Multiple scans of transaction database
Tedious workload of support counting for candidates
Improving Apriori: general ideas
Shrink number of candidates
Reduce passes of transaction database scans
Facilitate support counting of candidates
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Mining Frequent Patterns, Associations and Correlations
Basic Concepts
Apriori Algorithm
Optimization Techniques
FP-Growth
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Mining Frequent Patterns, Associations and Correlations
## DHP has two major features:
Efficient generation for large itemsets
Effective reduction on transaction database size
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Mining Frequent Patterns, Associations and Correlations
## DHP: Direct Hashing with efficient Pruning
Itemset sup
Tid Items Itemset
{A} 2
10 A, C, D {A}
⇒ {B} 3 ⇒
20 B, C, E {B}
{C} 3
30 A, B, C, E {C}
{D} 1
40 B, E {E}
{E} 3
## Tid Items 2-itemsets
10 A, C, D {A,C}, {A, D}, {C,D}
20 B, C, E {B,C}, {B, E}, {C,E}
30 A, B, C, E {A,B}, {A, C}, {A,E}, {B,C}, {B, E}, {C,E}
40 B, E {B, E}
Note. min-support=2
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Mining Frequent Patterns, Associations and Correlations
## Tid Items 2-itemsets
10 A, C, D {A,C}
20 B, C, E {B,C}, {B, E}, {C,E}
30 A, B, C, E {A,B}, {A, C}, {A,E}, {B,C}, {B, E}, {C,E}
40 B, E {B, E}
## h ({x, y}) = ((order of x) ∗ 10 + (order of y)) mod 7
Hash codes 0 1 2 3 4 5 6
Buckets {A,C} {B,C}
{A,B} {B,E}
{A,C} {C,E}
{A,E} {B,C}
{B,E}
{C,E}
{B,E}
Buckets counters 0 0 4 0 0 7 0
binary vector 0 0 1 0 0 1 0
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Mining Frequent Patterns, Associations and Correlations
Itemset sup
Itemset
{A,B} 1
{A,C} 2 {A,C}
{A,E} 1 ⇒ {B,C}
{B,C} 2 {B,E}
{B,E} 3 {C,E}
{C,E} 2
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Mining Frequent Patterns, Associations and Correlations
## Tid Items 2-itemsets
10 A, C, D φ
20 B, C, E {B,C,E}
30 A, B, C, E {A,B,C}, {A,B,E}, {A,C,E}, {B,C,E}
40 B, E φ
## h ({x, y, z}) = ((order of x) ∗ 100 + (order of y) ∗ 10 + (order of z)) mod 7
Hash codes 0 1 2 3 4 5 6
Buckets {A,B,C} {B,C,E}
{A,B,E} {B,C,E}
{A,C,E}
Buckets counters 0 0 3 0 0 2 0
binary vector 0 0 1 0 0 1 0
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Mining Frequent Patterns, Associations and Correlations
## DHP: Direct Hashing with efficient Pruning
Itemset sup
{A,B,C} 1 Itemset
{A,B,E} 1 ⇒
{B,C,E}
{A,C,E} 1
{B,C,E} 2
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Mining Frequent Patterns, Associations and Correlations
## DHP: Direct Hashing with efficient Pruning
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Mining Frequent Patterns, Associations and Correlations
## Subdivide the transactions of D into k non overlapping partitions
Each partition can fit into main memory, thus it is read only once
D1 + D2 + · · · + Dk = D
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Mining Frequent Patterns, Associations and Correlations
## Step1. Partition database and find local frequent patterns (1
scan)
Step2. Consolidate global frequent patterns (1 scan)
The database is scanned only twice
Correctness. The key to correctness is that any potential large
itemset appears as a large itemset in at least one of the partitions
D1 + D2 + · · · + Dk = D
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Mining Frequent Patterns, Associations and Correlations
## The counts for the candidate itemsets are generated using a
structure called tidlist
item tidlist
Tid Items
A 10,30
10 A, C, D
20 B, C, E ⇒ B 20,30,40 ⇒ {B,C} — 20,30
C 10,20,30
30 A, B, C, E
D 10
40 B, E
E 20,30,40
Correctness:
It is easy to see that the intersection of tidlists gives the correct
support for an itemset
Since the partitions are non overlapping, a cumulative count over
all partitions gives the global support for an itemset
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Mining Frequent Patterns, Associations and Correlations
## Partition Large Databases
#Transactions= 200,000
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Mining Frequent Patterns, Associations and Correlations
Basic Concepts
Apriori Algorithm
Optimization Techniques
FP-Growth
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Mining Frequent Patterns, Associations and Correlations
## Adopts a divide and conquer strategy
Compress the database representing frequent items into a
frequent–pattern tree or FP-tree
Retains the itemset association information
Divide the compressed database into a set of conditional
databases, each associated with one frequent item
Mine each such databases separately
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Mining Frequent Patterns, Associations and Correlations
Example FP-Growth
## The first scan of data is the same as Apriori
Derive the set of frequent 1-itemsets
Let min-sup=2
Generate a set of ordered items
Transactional Database
## TID List of item IDS
T100 I1 , I2 , I5 Item ID Support count
T200 I2 , I4 I2 7
T300 I2 , I3 I1 6
T400 I1 , I2 , I4 I3 6
T500 I1 , I3 I4 2
T600 I2 , I3 I5 2
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
T100 I1 , I2 , I5
T200 I2 , I4 Create a branch for each transaction
T300 I2 , I3 Items in each transaction are processed in order
T400 I1 , I2 , I4
T500 I1 , I3
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T100 : {I2 , I1 , I5 }
T500 I1 , I3 2 Construct the first branch: < I2 : 1 >, < I1 : 1 >, < I5 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T200 : {I2 , I4 }
T500 I1 , I3 2 Construct the second branch: < I2 : 1 >, < I4 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T300 : {I2 , I3 }
T500 I1 , I3 2 Construct the third branch: < I2 : 2 >, < I3 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T400 : {I2 , I1 , I4 }
T500 I1 , I3 2 Construct the fourth branch: < I2 : 3 >, < I1 : 1 >, < I4 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T400 : {I2 , I1 , I4 }
T500 I1 , I3 2 Construct the fourth branch: < I2 : 3 >, < I1 : 1 >, < I4 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## TID List of item IDS
Create a branch for each transaction
T100 I1 , I2 , I5 Items in each transaction are processed in order
T200 I2 , I4
T300 I2 , I3
T400 I1 , I2 , I4 1 Order the items T400 : {I1 , I3 }
T500 I1 , I3 2 Construct the fifth branch: < I1 : 1 >, < I3 : 1 >
T600 I2 , I3
T700 I1 , I3
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## Item ID Support count
I2 7
I1 6
I3 6
I4 2
I5 2
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Mining Frequent Patterns, Associations and Correlations
## The problem of mining frequent patterns in databases is transformed to
that of mining the FP-tree
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Mining Frequent Patterns, Associations and Correlations
I2 7
I1 6
I3 6
I4 2
I5 2
## Occurrences of I5 :< I2 , I1 , I5 > and < I2 , I1 , I3 , I5 >
Two prefix Paths < I2 , I1 : 1 > and < I2 , I1 , I3 : 1 >
Conditional FP tree contains only < I2 : 2, I1 : 2 >, I3 is not
considered because its support count of 1 is less than the
minimum support count.
Frequent patterns {I2 , I5 : 2}, {I1 , I5 : 2}, {I2 , I1 , I5 : 2}
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Mining Frequent Patterns, Associations and Correlations
I2 7
I1 6
I3 6
I4 2
I5 2
## TID Conditional Pattern Base Conditional FP-tree
I5 {{I2 , I1 : 1}, {I2 , I1 , I3 : 1}} < I2 : 2, I1 : 2 >
I4 {{I2 , I1 : 1}, {I2 , 1}} < I2 : 2 >
I3 {{I2 , I1 : 2}, {I2 : 2}, {I1 : 2}} < I2 : 4, I1 : 2 >, < I1 : 2 >
I1 {I2 , 4} < I2 : 4 >
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Mining Frequent Patterns, Associations and Correlations
I2 7
I1 6
I3 6
I4 2
I5 2
## TID Conditional FP-tree Frequent Patterns Generated
I5 < I2 : 2, I1 : 2 > {I2 , I5 : 2}, {I1 , I5 : 2}, {I2 , I1 , I5 : 2}
I4 < I2 : 2 > {I2 , I4 : 2}
I3 < I2 : 4, I1 : 2 >, < I1 : 2 > {I2 , I3 : 4}, {I1 , I3 : 4}, {I2 , I1 , I3 : 2}
I1 < I2 : 4 > {I2 , I1 : 4}
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Mining Frequent Patterns, Associations and Correlations
FP-growth properties
## FP-growth transforms the problem of finding long frequent
patterns to searching for shorter ones recursively and
concatenating the suffix
It uses the least frequent suffix offering a good selectivity
It reduces the search cost
If the tree does not fit into main memory, partition the database
Efficient and scalable for mining both long and short frequent
patterns
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Mining Frequent Patterns, Associations and Correlations
## Both Apriori and FP-growth use horizontal data format
Alternatively data can also be represented in vertical format
Transform by scanning the database once
## TID List of item IDS
T100 I1 , I2 , I5 itemset TID_set
T200 I2 , I4
T300 I2 , I3 I1 {T100 , T400 , T500 , T700 , T800 , T900 }
T400 I1 , I2 , I4 ⇒ I2 {T100 , T200 , T300 , T400 , T600 , T800 , T900 }
T500 I1 , I3 I3 {T300 , T500 , T600 , T700 , T800 , T900 }
T600 I2 , I3 I4 {T200 , T400 }
T700 I1 , I3 I5 {T100 , T800 }
T800 I1 , I2 , I3 , I5
T900 I1 , I2 , I3
## The support count of an itemset is simply the length of the TID_set
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Mining Frequent Patterns, Associations and Correlations
## Frequent 1-itemsets in vertical format
itemset TID_set
I1 {T100 , T400 , T500 , T700 , T800 , T900 }
I2 {T100 , T200 , T300 , T400 , T600 , T800 , T900 }
I3 {T300 , T500 , T600 , T700 , T800 , T900 }
I4 {T200 , T400 }
I5 {T100 , T800 }
The frequent k-itemsets are used to construct candidate (k+1)-itemsets based on the Apriori
property
## Frequent 2-itemsets in vertical format (consider min_sup = 2)
itemset TID_set
{I1 , I2 } {T100 , T400 , T800 , T900 }
{I1 , I3 } {T500 , T700 , T800 , T900 }
{I1 , I4 } {T400 }
{I1 , I5 } {T100 , T800 }
{I2 , I3 } {T300 , T600 , T800 , T900 }
{I2 , I4 } {T200 , T400 }
{I2 , I5 } {T100 , T800 }
{I3 , I5 } {T800 }
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Mining Frequent Patterns, Associations and Correlations
## Frequent 3-itemsets in vertical format
itemset TID_set
{I1 , I2 , I3 } {T800 , T900 }
{I1 , I2 , I5 } {T100 , T800 }
## This process repeats, with k incremented by 1 each time, until no
frequent items or no candidate itemsets can be found
Properties of mining with vertical data format
Take the advantage of the Apriori property in the generation of
candidate (k + 1)-itemset from k -itemsets
No need to scan the database to find the support of (k + 1) itemsets,
for k ≥ 1
The TID_set of each k -itemset carries the complete information
required for counting such support
The TID-sets can be quite long, hence expensive to manipulate
Use diffset technique to optimize the support count computation
54 / 54 | 7,164 | 20,911 | {"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-2019-51 | latest | en | 0.837662 |
http://www.nag.com/numeric/CL/nagdoc_cl23/examples/source/f07jnce.c | 1,438,199,898,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-32/segments/1438042986625.58/warc/CC-MAIN-20150728002306-00246-ip-10-236-191-2.ec2.internal.warc.gz | 604,582,508 | 1,619 | /* nag_zptsv (f07jnc) Example Program. * * Copyright 2004 Numerical Algorithms Group. * * Mark 23, 2011. */ #include #include #include #include int main(void) { /* Scalars */ Integer exit_status = 0, i, j, n, nrhs, pdb; /* Arrays */ Complex *b = 0, *e = 0; double *d = 0; /* Nag Types */ NagError fail; Nag_OrderType order; #ifdef NAG_COLUMN_MAJOR #define B(I, J) b[(J-1)*pdb + I - 1] order = Nag_ColMajor; #else #define B(I, J) b[(I-1)*pdb + J - 1] order = Nag_RowMajor; #endif INIT_FAIL(fail); printf("nag_zptsv (f07jnc) Example Program Results\n\n"); /* Skip heading in data file */ scanf("%*[^\n]"); scanf("%ld%ld%*[^\n]", &n, &nrhs); if (n < 0 || nrhs < 0) { printf("Invalid n or nrhs\n"); exit_status = 1; goto END; } /* Allocate memory */ if (!(b = NAG_ALLOC(n*nrhs, Complex)) || !(e = NAG_ALLOC(n-1, Complex)) || !(d = NAG_ALLOC(n, double))) { printf("Allocation failure\n"); exit_status = -1; goto END; } #ifdef NAG_COLUMN_MAJOR pdb = n; #else pdb = nrhs; #endif /* Read the lower bidiagonal part of the tridiagonal matrix A and */ /* the right hand side b from data file */ for (i = 0; i < n; ++i) scanf("%lf", &d[i]); scanf("%*[^\n]"); for (i = 0; i < n - 1; ++i) scanf(" ( %lf , %lf )", &e[i].re, &e[i].im); scanf("%*[^\n]"); for (i = 1; i <= n; ++i) for (j = 1; j <= nrhs; ++j) scanf(" ( %lf , %lf )", &B(i, j).re, &B(i, j).im); scanf("%*[^\n]"); /* Solve the equations Ax = b for x using nag_zptsv (f07jnc). */ nag_zptsv(order, n, nrhs, d, e, b, pdb, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_zptsv (f07jnc).\n%s\n", fail.message); exit_status = 1; goto END; } /* Print solution */ printf("Solution\n"); for (i = 1; i <= n; ++i) { for (j = 1; j <= nrhs; ++j) printf("(%8.4f, %8.4f)%s", B(i, j).re, B(i, j).im, j%4 == 0?"\n":" "); printf("\n"); } /* Print details of factorization */ printf("\nDiagonal elements of the diagonal matrix D\n"); for (i = 0; i < n; ++i) printf("%7.4f%s", d[i], i%8 == 7?"\n":" "); printf("\n\nSub-diagonal elements of the Cholesky factor L\n"); for (i = 0; i < n-1; ++i) printf("(%8.4f, %8.4f)%s", e[i].re, e[i].im, i%8 == 7?"\n":" "); END: if (b) NAG_FREE(b); if (e) NAG_FREE(e); if (d) NAG_FREE(d); return exit_status; } | 826 | 2,184 | {"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-2015-32 | latest | en | 0.409536 |
https://mix.office.com/watch/uga4q92gv40x | 1,524,588,651,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946807.67/warc/CC-MAIN-20180424154911-20180424174911-00454.warc.gz | 669,319,710 | 17,594 | # Solve Systems By Substitution (Harder Version)
Systems of Equations
# Solve Systems By Substitution (Harder Version)
Created 3 years ago
Duration 0:08:08
52
Slide Content
1. ### Slide 1 - 2) Solve the system using substitution
• 3y + x = 7
• 4x – 2y = 0
• Step 1: Solve an equation for one variable.
• Step 2: Substitute
• It is easiest to solve the
• first equation for x.
• 3y + x = 7
• -3y -3y
• x = -3y + 7
• 4x – 2y = 0
• 4(-3y + 7) – 2y = 0
2. ### Slide 2 - 2) Solve the system using substitution
• 3y + x = 7
• 4x – 2y = 0
• Step 4: Plug back in to find the other variable.
• 4x – 2y = 0
• 4x – 2(2) = 0
• 4x – 4 = 0
• 4x = 4
• x = 1
• Step 3: Solve the equation.
• -12y + 28 – 2y = 0
• -14y + 28 = 0
• -14y = -28
• y = 2
3. ### Slide 3 - 2) Solve the system using substitution
• 3y + x = 7
• 4x – 2y = 0
• Step 5: Check your solution.
• (1, 2)
• 3(2) + (1) = 7
• 4(1) – 2(2) = 0
• When is solving systems by substitution easier to do than graphing?
• When only one of the equations has a variable already isolated (like in example #1).
4. ### Slide 4 - If you solved the first equation for x, what would be substituted into the bottom equation.
• 2x + 4y = 4
• 3x + 2y = 22
• -4y + 4
• -2y + 2
• -2x + 4
• -2y+ 22
5. ### Slide 6 - 3) Solve the system using substitution
• 2x + y = 4
• 4x + 2y = 8
• Step 1: Solve an equation for one variable.
• Step 2: Substitute
• The first equation is
• easiest to solved for y!
• y = -2x + 4
• 4x + 2y = 8
• 4x + 2(-2x + 4) = 8
• Step 3: Solve the equation.
• 4x – 4x + 8 = 8
• 8 = 8
• This is also a special case.
• Does 8 = 8? TRUE!
• When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS. | 703 | 1,658 | {"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} | 4.59375 | 5 | CC-MAIN-2018-17 | latest | en | 0.706461 |
http://oeis.org/A014949/internal | 1,558,689,472,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00056.warc.gz | 144,724,062 | 2,792 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A014949 Numbers n such that n divides 8^n - 1. 12
%I
%S 1,7,49,343,889,2359,2401,6223,16513,16807,43561,112903,115591,117649,
%T 299593,304927,790321,794983,809137,823543,2033143,2097151,2134489,
%U 5532247,5564881,5663959,5764801,13549249,14232001,14338681,14680057,14941423,34591207,38048311,38725729
%N Numbers n such that n divides 8^n - 1.
%C For all k, 7^k is in the sequence. - _Benoit Cloitre_, Mar 05 2002
%C From _Alexander Adamchuk_, May 16 2010: (Start)
%C 7 divides a(n) for n>1.
%C Prime divisors of a (n) in the order of their first appearance are {7, 127, 337, 2287, 15241, 14407, 18199, 42463, ...}. (End)
%H M. F. Hasler, <a href="/A014949/b014949.txt">Table of n, a(n) for n = 1..64</a>
%o (PARI) is(n)=Mod(8,n)^n==1 \\ _Charles R Greathouse IV_, Nov 04 2016
%K nonn
%O 1,2 | 378 | 938 | {"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} | 3.171875 | 3 | CC-MAIN-2019-22 | latest | en | 0.638593 |
http://www1.maths.leeds.ac.uk/~ppmartin/GRAD/blob.html | 1,600,468,495,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00791.warc.gz | 248,254,365 | 2,541 | BACK or Mathematics Department Home
Paul Martin: Reference Section : Projects...
Consider the Brauer algebra B_n as a subalgebra of the partition algebra. (See here for the partition algebra.) Consider in particular the diagram basis (certain pictures of partitions of a set of n+n elements drawn in a rectangular frame), and the algebra multiplication formulated in terms of a corresponding diagram juxtaposition. From this one sees that there is a subalgebra with basis the subset of non-crossing diagrams. This is the TL algebra - an algebra of considerable significance in many areas of mathematics and physics. The blob algebra is a generalisation of the TL algebra as we now indicate.
The non-crossing idea involves drawing diagrams on a rectangle. It is natural (particularly from a computational physics perspective) to consider a generalisation where one draws diagrams on a cylinder, and distinguishes closed loops (formed in diagram composition) that are non-contractible. Rather than address this algebra directly, it is convenient to study an algebra in which the cohomological data is encoded in a blob on the `seam' (a line drawn on the cylinder to cut it open, and hence return to the rectangle). This is the idea of the blob algebra. (Although it can also be characterised in a number of other interesting ways...)
One of the most interesting aspects of the study of the blob algebra concerns its representation theory. Its `reductive' representation theory over algebraically closed fields is quite completely understood (see [Cox et al] and references therein). But a number of other interesting questions about it and its generalisations remain open.
Some nice looking recent papers on the blob algebra, indicating a lot of interesting open problems (!):
Some original refs:
• Saleur, The blob algebra and the periodic TL algebra, LMP
• Saleur, On an algebraic approach to higher dimensional statistical mechanics, CMP
• Woodcock, On the structure of the blob algebra, J Alg
• Woodcock, Generalised blob algebras and alcove geometry, LMS JCM
• Cox, Graham and Martin The blob algebra in positive characteristic
• R Green et al, Symplectic blob algebra...
• J de Gier et al
• T tom Dieck
• A Doikou
• S Ryom-Hansen
• ...
Some other refs:
• These papers study a very mild generalisation of the blob:
• R Green, Generalised TL algebras and decorated tangles
• R Green, Decorated tangles and canonical bases | 515 | 2,429 | {"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-2020-40 | latest | en | 0.878951 |
https://studylib.net/doc/25324634/abc | 1,582,678,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00339.warc.gz | 573,879,258 | 14,682 | Uploaded by Ibrahim Mohamed
# abc
```Activity-Based Costing
Cost Allocation
The basic approach in product costing is to allocate costs in the cost pools to the individual cost
objects, which are the products or services of interest.
We assign, or allocate, these costs to the individual cost objects by using appropriate cost
allocation bases or cost drivers.
Plantwide Allocation method
LO 1
Compare and contrast plantwide and
department allocation methods.
Plantwide allocation method
All overhead costs are recorded in one
cost pool and applied to products using
one overhead allocation rate.
One cost pool for the entire plant
The cost pool is the entire plant.
This method uses one overhead allocation rate, or one set of rates, to allocate overhead to
products for all departments in a particular plant.
Companies using a single plantwide rate generally use an allocation base related to the volume
of output, such as direct labor-hours, machine-hours, units of output, or materials costs.
Department allocation method
Department allocation method
Overhead costs are traced to separate
departments and applied to products
using a department allocation rate.
One cost pool for each department
A company has a separate cost pool for each department. The company establishes a separate
overhead allocation rate for each department.
Each production department is a separate cost pool.
Example: Plantwide Vs. Department allocation
Main Street Ice Cream Company uses a plantwide allocation method to allocate overhead based on
direct labor-hours at a rate of \$3 per labor-hour. Strawberry and vanilla flavors are produced in
Department SV. Chocolate is produced in Department C. Sven manages Department SV and Charlene
manages Department C. The product costs (per thousand gallons) follow:
Required:
a. If the number of hours of labor per 1,000 gallons is 50 for strawberry, 55 for vanilla, and 75 for
chocolate, compute the total cost of 1,000 gallons of each flavor using plantwide allocation.
b.Charlene’s department uses older, outdated machines. She believes that her department is being
allocated some of the overhead of Department SV, which recently bought state-of-the-art machines.
After she requested that overhead costs be broken down by department, the following information was
discovered:
Using machine-hours as the department allocation base for Department SV and labor-hours as the
department allocation base for Department C, compute the allocation rate for each.
c. Compute the cost of 1,000 gallons of each flavor of ice cream using the department allocation rates
computed in requirement ( b ) if the number of machine-hours for 1,000 gallons of each of the three
flavors of ice cream are as follows: strawberry, 50; vanilla, 55; and chocolate, 150. Direct labor hours by
product remain the same as in requirement a.
Solution: Plantwide Vs. Department allocation
b. Department SV has an overhead allocation rate of \$4.20 per machine-hour (\$105,840 ÷ 25,200
machine hours). Department C has an overhead allocation rate of \$1.32 per labor-hour (\$23,760
÷ 18,000 labor-hours).
c.
a\$210 = 50 machine-hours x \$4.20 per machine-hour.
b\$231 = 55 machine-hours x \$4.20 per machine-hour.
c\$99 = 75 labor-hours x \$1.32 per labor-hour.
Choice of Allocation Methods
Which method is appropriate?
The choice of whether to use a plantwide rate or departmental rates depends on the products
and the production process.
•
If the company manufactures products that are quite similar and that use the same set of
resources, the plantwide rate is probably sufficient.
If multiple products use the manufacturing facilities in many different ways, departmental rates
provide a better picture of the use of manufacturing resources by the different products.
Managers need to make a decision about plantwide versus departmental rates based on the
costs and benefits of the information inherent in each system.
•
Selecting more complex allocation methods requires more time and skill to collect and process
accounting information. Such incremental costs of additional information must be justified by an
increase in benefits from improved decisions.
Two-Stage Cost Allocation
First stage:
Allocate overhead costs to departments.
Second stage:
Allocate department overhead costs
to the products or services.
Activity-Based Costing (ABC)
Assume that you are thinking about going into business offering music to be downloaded
(legally) over the Internet.
One of the first steps is to develop a business plan that includes a financial analysis.
One aspect of the financial analysis is estimating the cost of downloads to help you assess the
profitability of your venture.
Because you are not now in the business, you have no accounting records to use to help you.
Instead, you need to use the engineering approach to estimate cost. You would probably
proceed by identifying the activities that you would need to perform.
The activities would include:
obtaining permission from various artists and studios to include their songs in your catalog.
Negotiating a royalty payment for each download would be necessary.
Buying and maintaining a Web site,
processing orders, collecting payments,
keeping records, and so on.
Once you had identified each activity that you would have to accomplish, you would estimate
the cost of completing each one.
In this description of the process you would follow, the italicized words are actions that
represent tasks that you would complete to make the product (service) available for sale. You
do not attempt to determine what department or what overhead account would be used. You
use activities.
Applying this approach to the two-stage cost allocation system, you assign costs to activities, not
departments or buildings, in the first stage. In the second stage, you “allocate” costs to your
single product, downloads, using the appropriate cost drivers for each activity.
LO 2
Explain how activity-based costing and
a two-stage product system are related.
ABC is a costing method that first assigns costs
to activities and then assigns them to products
based on the products’ consumption of activities.
Stage 1:
Assign costs to activities.
Stage 2:
Assign costs to products based
on the use of each activity
Why ABC?
Step 1: Identify the activities that consume resources
and assign costs to them.
Step 2: Identify the cost driver(s) associated with
each activity.
Step 3: Compute a cost rate per cost driver unit
or transaction.
Step 4: Assign costs to products by multiplying the
cost driver rate by the volume of cost driver
units consumed by the product.
Identifying Activities That Use Resources
Often the most interesting and challenging part of the exercise is identifying activities that use
resources because doing so requires understanding all the activities required to make a product.
Example, the activities involved in making a bottle of water:
Ordering;
receiving and inspecting materials;
bottling the water;
packing the cases;
shipping the cases.
Choosing Cost Drivers
Cost drivers are factors that cause or “drive”
an activity’s costs.
The best cost driver is one that is
causally related to the cost being allocated
Computing a Cost Rate per Cost Driver
In general, predetermined rates for allocating indirect costs to products are computed as
follows:
Costs are allocated to a product by multiplying each activity’s predetermined rate by the volume
of activity used in making it.
In the ABC two-stage cost system, the first stage consists of activities, not departments. Instead
of a department rate, activity-based costing computes a cost driver rate for each activity center.
Assigning Costs to Products
The final step in the activity-based costing system is to assign the activity costs to products.
•
We multiply the cost driver rates by the number of units of the cost driver in each product.
Cost Hierarchies
This is the classification of cost drivers
into general
levels of activity; volume, batch, product, and so on.
LO 3
Compute product costs using activity-based costing.
Janis, she’s the cost accountant at Joplin—interviews the production managers to determine the
major activities used to produce cameras in the Port Arthur facility. She learns that the Assembly
building has three major activities—setting up, handling material, and assembling—and that the
Packaging building has two major activities—inspecting and packing, and shipping.
Step 1: Identify the activities that consume resources and assign costs to them.
setting up
handling material
assembling
inspecting and packing
shipping.
•
Step 2: Identify the cost driver(s) associated with each activity.
•
Step 3: Compute a cost rate per cost driver unit or transaction.
•
Step 4: Assign costs to products by multiplying the cost driver rate by the volume of cost driver
units consumed by the product.
•
Cost Flow Diagram – ABC System
•
Match each of the following activities with the appropriate category
Solution\
طالب جميع الفرق برجاء قراءه البوست جيدا (مواعيد االسبوع القادم المراجعه)
الفرقه االولي المجموعه االساسيه
الحصه االولي السبت الساعه 1تعاد االحد الساعه 1او 2:30
الحصه الثانيه الثالثاء الساعه 11:30
الحصه الثالثه االربعاء الساعه 11:30
الحصه الرابعه الخميس الساعه 11:30
الحصه الخامسه الجمعه الساعه 9تعاد الساعه 1:30
الفرقه االولي المجموعه الجديده (المراجعه)
الحصه االولي االحد الساعه 11:30تعاد االثنين الساعه 2:30
الحصه الثانيه الثالثاء الساعه 11:30
الحصه الثاثه االربعاء الساعه 1
الحصه الرابعه الخميس الساعه 1
الحصه الخامسه الجمعه الساعه 2:30
الفرقه االولي BISمراجعه report
البدايه يوم الخميس الساعه 1
الفرقه الثانيه
الحصه االولي السبت الساعه 9
الحصه الثانيه الثالثاء الساعه 9
الفرقه الثالثه
الحصه االولي االحد الساعه 9
الحصه الثانيه االثنين الساعه 11:30
الحصه الثالثه الثالثاء الساعه 1
الفرقه الرابعه
الحصه االولي االثنين الساعه 9
الحصه الثانيه االربعاء الساعه 9
الفرقه الثانيه Bis
يوم السبت الساعه 11:30تعاد الساعه 2:30وباقي الموعيد في الحصه
``` | 2,472 | 10,114 | {"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} | 3.171875 | 3 | CC-MAIN-2020-10 | longest | en | 0.906427 |
https://www.jiskha.com/questions/1655191/A-certain-sample-of-a-pure-quartz-crystal-has-a-mass-of-276-grams-and-a-volume-95-8cm-3 | 1,534,833,910,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217970.87/warc/CC-MAIN-20180821053629-20180821073629-00419.warc.gz | 899,204,952 | 5,295 | # physics
A certain sample of a pure quartz crystal has a mass of 276 grams and a volume 95.8cm^3.
-what is the mass density of the sample?
-what would be the volume of a larger quartz crystal of the same density if its mass is 1.138 kg?
Help please!! Have been struggling with this problem for an hour now and still can't get anywhere. Thank you!!
1. density = mass / volume
volume = mass/density
I would do it all in cm^3 and grams
1 kg = 1,000 grams
posted by Damon
2. thank you!!
posted by Anonymous
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An unknown compound contains only the three elements C,H, and O. A pure sample of the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass. a.)Determine the empirical formula of the compound. b.)A
10. ### physics
Show that for 1 kg of pure gold the volume of water displaced is 51.8 cm^3 I know that the density of pure gold is 19.32 gm/cm^3 so 1 kg/19.32 gm/cm^3= 0.05175 ? I have come along way with this problem and feel like I am close,
More Similar Questions | 756 | 2,835 | {"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} | 3.359375 | 3 | CC-MAIN-2018-34 | latest | en | 0.910332 |
http://surveying2012.blogspot.com/2014/08/ | 1,529,465,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00185.warc.gz | 313,387,407 | 17,430 | Monday, August 4, 2014
Surveying- II (N.S.) B.Tech. 4th Sem Examination (CE-224) - HPTU- 2014
Please find here the question paper set by HPTU for B.Tech. 4th Semester Examination June 2014, for Surveying-II (CE-224).
------------------------------------------------------------------------------------------------------------
B.Tech. 4th Semester Examination
Surveying -II (N.S.)
CE-224
Time: 3 Hours Max. Marks: 100
The candidates shall limit their answer precisely within the answer book (40 pages) issued to them and no supplementary/continuation sheet will be issued.
Note: Attempt five questions in all selecting one question from each section A, B, C and D and all subparts of Section E are compulsory.
Section - A
1. (a) Two parallel railway lines are to be connected by a reverse curve. If the lines are 10 m apart, the maximum distance between tangent points measured parallel to the straight is 50 m, find:
(b) radius R if R1= 50m. Also caculate the length of both curves. (14)
(b) Discuss the characteristics of a transition curve by the method of tangential method. (6)
2. Describe the weight of quantities. How weight of different quantities are allocated? Discuss various laws of weights. (20)
Section - B
3. (a) A base line was measured with steel tape which was exactly 30 m at 20 degree Celsius at a pull of 100 N. The measured length was 1500.00 m. If the temperature during measurement was 28 degrees Celsius and pull applied was 150 N, determine correct length of line if cross section area of tape = 2.5 mm^2, coefficient of expansion 3.5*10^(-6) per degree Celsius. Modulus of elasticity = 2.1 * 10^3 N/mm^2. (15)
(b) Differentiate between triangulation & trilateration. (5)
4. Two triangulation stations A & B 60 km above having elevations of 265 m and 385 m respectively, the intervening ground may be assumed to have a uniform elevation of 220 m. Find the minimum height of signal at B so that the line of sight may not pass neat the ground less than 3 m. (20)
Section - C
5. Draw the expression to determine the height of the object when the two instruments stations are not in the same vertical plane. (20)
6. Following observations were made in trignometric levelling:
Observed altitude = 3 Degree 10 minute 49 seconds.
Height of instrument = 1.24 m
Height of signal = 5.32 m
Horizontal distance = 4935 m
Coefficient of refraction = 0.07
Rsinj" = 30.88m
Correct the observed altitude for the height of signal, refraction and curvature. (20).
Section - D
7. Explain in detail the basic geometric characteristics of aerial photographs. (20)
8. (a) A rectangular agricultural field measures 8.65 cm long & 5.13 cm wide on a vertical photograph having scale of 1:20,000. Find are of field. (6)
(b) What is the significance of equation of time? How do we calculate the local time at a location.? (14)
Section - E
9. Attempt all parts:
(a) If the first chord gradient is 0.16, calculate the gradient of fourth chord.
(b) What is the basic criteria for the design of transition curves?
(c) Differentiate between triangulation and traversing.
(d) Define "extension of base line".
(e) How do you account correction for curvature of earth in trigonometric levelling?
(f) What is axis-signal correction?
(g) Define "Tilt displacement" in photogrammetry.
(i) Explain most probable value. (2*10 = 20)
----------------------------------------------------------------------------------------------------------- | 875 | 4,043 | {"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} | 3.734375 | 4 | CC-MAIN-2018-26 | latest | en | 0.819353 |
https://www.iaswww.com/apr/Science/Math/Topology/ | 1,652,981,581,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00777.warc.gz | 898,054,397 | 4,683 | Topology Math Science
The primary functions of this list are: providing abstracts of papers posted to the Hopf archive, providing information about topology conferences, and serving as a forum for topics related to algebraic topology. The site also serves as an archive of link
Top: Science: Math: Topology
See Also:
• British Topology Home Page - A source of pointers to Topology-related sites, including archives topology and conference announcements.
• Topology Course Lecture Notes - By Aisling McCluskey and Brian McMaster. HTML with symbol fonts, science DVI and PostScript.
• Links to Low-dimensional Topology - Topics: General, Conferences, Pages of Links, Knot Theory, science 3-manifolds, Journals.
• MAA Basic Library List in Topology - MAA recommended books in General, Geometric, Algebraic and topology Differential Topology.
• Topology - Descriptions and illustrations of several topological and differential geometry related notions.
• Topology Glossary - Definitions of over 100 terms in topology.
• The Cantor Set - Article in the Platonic Realms, describing the Cantor topology discontinuum, a science favorite example of topology. Includes topology examples and illustrations.
• Algebraic Topology Discussion List - The primary functions of this list are: providing topology abstracts of papers posted to the Hopf archive, topology providing information about topology conferences, and serving as topology a forum for topics related to algebraic topology. topology The site also serves as an archive of topology link
• TTT on WWW - The Transpennine Topology Triangle is a topology seminar topology partially supported by the London Mathematical Society topology with vertices at Leicester, Manchester and Sheffield.
• Topics in Mathematics - Topology - In the Mathematics Archives at University of Tennessee, math Knoxville.
• Differentiable manifolds - Lecture notes by Mariusz Wodzicki in postscript or pdf.
• Mazes and Mathematics - History and mathematical analysis of labyrinths.
• Differential Topology - Course notes by Matthew G. Brin in PostScript including "Introduction to Differential Topology", "Introduction to Seifert fibered 3-manifolds", "Groups acting on 1-dimensional spaces", and "Presentations, conjugacy, roo
• Planar Machines - an Invitation to Topology. - Java applets exploring configuration spaces.
• Topology Atlas - Preprints, abstracts, calendar, links, other resources.
MySQL - Cache Direct | 503 | 2,443 | {"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} | 2.671875 | 3 | CC-MAIN-2022-21 | longest | en | 0.840417 |
https://convertoctopus.com/255-9-cubic-centimeters-to-gallons | 1,680,359,351,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00092.warc.gz | 227,922,795 | 7,460 | ## Conversion formula
The conversion factor from cubic centimeters to gallons is 0.00026417205124156, which means that 1 cubic centimeter is equal to 0.00026417205124156 gallons:
1 cm3 = 0.00026417205124156 gal
To convert 255.9 cubic centimeters into gallons we have to multiply 255.9 by the conversion factor in order to get the volume amount from cubic centimeters to gallons. We can also form a simple proportion to calculate the result:
1 cm3 → 0.00026417205124156 gal
255.9 cm3 → V(gal)
Solve the above proportion to obtain the volume V in gallons:
V(gal) = 255.9 cm3 × 0.00026417205124156 gal
V(gal) = 0.067601627912715 gal
The final result is:
255.9 cm3 → 0.067601627912715 gal
We conclude that 255.9 cubic centimeters is equivalent to 0.067601627912715 gallons:
255.9 cubic centimeters = 0.067601627912715 gallons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 gallon is equal to 14.79254318093 × 255.9 cubic centimeters.
Another way is saying that 255.9 cubic centimeters is equal to 1 ÷ 14.79254318093 gallons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred fifty-five point nine cubic centimeters is approximately zero point zero six eight gallons:
255.9 cm3 ≅ 0.068 gal
An alternative is also that one gallon is approximately fourteen point seven nine three times two hundred fifty-five point nine cubic centimeters.
## Conversion table
### cubic centimeters to gallons chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to gallons
cubic centimeters (cm3) gallons (gal)
256.9 cubic centimeters 0.068 gallons
257.9 cubic centimeters 0.068 gallons
258.9 cubic centimeters 0.068 gallons
259.9 cubic centimeters 0.069 gallons
260.9 cubic centimeters 0.069 gallons
261.9 cubic centimeters 0.069 gallons
262.9 cubic centimeters 0.069 gallons
263.9 cubic centimeters 0.07 gallons
264.9 cubic centimeters 0.07 gallons
265.9 cubic centimeters 0.07 gallons | 564 | 2,104 | {"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 | 4 | CC-MAIN-2023-14 | latest | en | 0.674773 |
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# Cost Based Pricing
Imagine you are a lawyer and planning how to price your services. Running your small law firm costs you $100,000 a year. You want to make at least$80,000 a year. How much should you charge for your services? Should you take into account the costs you incur? Look no further because we've got you covered with the ultimate guide to cost-based pricing! From the definition and strategy to the formula and real-life examples, we'll explain everything you need to know about this pricing method. Plus, we'll explore the advantages of cost-based pricing and highlight the key differences between cost-based and value-based pricing.
## Cost-Based Pricing Definition
Cost-based pricing is a pricing method where the cost of manufacturing and distributing a product is taken into account. With this method, a percentage markup is added to the cost of production and distribution to determine the selling price of the product. This means that the price of a product is determined by adding up all the expenses involved in making and delivering it, and then adding a profit markup.
Cost-based pricing is a pricing method based on the cost of production and distribution.
Let's say a company produces and sells a product for $50. The cost of production and distribution for each unit is$30. To determine the selling price, the company adds a 20% profit margin to the cost of production and distribution, which is $6. Therefore, the company would set the selling price at$36 to ensure they cover their costs and make a profit.
## Cost-Based Pricing Strategy
A cost-based pricing strategy aims for businesses to achieve a specified profit margin over and above the entire cost of production and manufacturing. A cost-based pricing strategy enables companies to cover production expenses and make a profit.
There are two cost-based pricing strategies—namely, cost-plus pricing strategy and break-even pricing strategy.
### Cost-Plus Pricing Strategy
The cost-plus or markup pricing strategy is one of the most common types of cost-based pricing strategies.
The cost-plus pricing strategy works by adding a set markup to the total cost of production.
Take, for example, construction companies. Before submitting bids for a project they might be undertaking, they estimate the total cost of production. After they estimate and come up with the costs, they add a markup for profit.
Lawyers, accountants, and other professionals typically price by adding a standard markup to their costs.
Applying a standard markup is common for various reasons. To begin with, retailers are often more concerned with the cost of their products than their demand level. When sellers simplify pricing by connecting it to the cost of production, they eliminate the need to make frequent modifications in response to shifts in demand.
Second, since prices are more likely comparable when all companies in an industry utilise the same pricing mechanism, price rivalry is reduced significantly.
Third, pricing based on costs plus a markup is usually more equitable for both consumers and sellers. When there is high customer demand, sellers profit from their investments but do not take advantage of customers.
### Break-Even Pricing Strategy
Break-even pricing, also known as target-return pricing, is the second pricing approach based on costs.
Without adding a markup, the price of a product is determined by adding up the costs of its creation, production, and distribution.
Instead of marking up each unit to earn income, this method calculates how many units a firm needs to sell to cover the manufacturing expenses.
The formula companies use to determine the break-even volume is
$$\hbox{Break-even volume}=\frac{\hbox{Fixed costs}}{\hbox{Price - Variable cost}}$$
This formula helps companies learn about the number of units they need to sell at a particular price to become profitable.
Let's say a company has invested $3,000 in manufacturing pens and the variable cost per pen is$1. If the firm sells the pen for $2, it needs to sell: $$\hbox{Break-even volume}=\frac{3,000}{2-1}= 3,000$$ The company needs to produce 3,000 pens to break even. Fixed costs refer to costs that do not change as the level of sales or production changes, while variable costs change directly with the level of production. ## Cost-Based Pricing Formula ### Cost-Plus Pricing Formula $$\hbox{Selling Price}=\hbox{Cost of production and distribution per unit}+\hbox{Markup}$$ In this formula, the "Cost of Production and Distribution" is the total cost of producing and delivering the product or service, including all direct and indirect costs. The "Markup" is a fixed percentage or amount that's added to the cost of production and distribution to determine the selling price. The markup is usually determined by factors such as industry standards, competition, and desired profit margin. ### Break-Even Pricing Formula $$\hbox{Selling price}=\frac{\hbox{Total fixed costs}}{\hbox{Number of units sold}}+ \hbox{Variable cost per unit}$$ In this formula, the "Total Fixed Costs" are the expenses that remain constant regardless of the number of units sold, such as rent, salaries, and insurance. The "Variable Cost per Unit" is the cost of producing one unit of the product or service, including materials, labor, and other variable costs. By using this formula, a company can determine the minimum selling price that they need to charge to cover their total costs and break even. ## Cost-Based Pricing Example Let's now take a look at some cost-based pricing examples. ### Cost-Plus Pricing Example Let's take a look at an example of a cost-plus pricing strategy. Let's say a company manufactures a product that costs$50 to produce and distribute, including all materials, labour, and overhead expenses. The company wants to earn a profit margin of 20% on each unit sold.
To determine the selling price using cost-plus pricing, the company would add the desired profit margin to the cost of production:
Cost of production = $50 Desired profit margin = 20% $$\hbox{Selling price} = 50 + (20\%\times50=50+10=60$$ So the company would set the selling price of the product at$60 to ensure that it covers all costs and generates a profit margin of 20%. If the company's costs of production or desired profit margin change, it can adjust the selling price accordingly using the same cost-plus pricing formula.
### Break-Even Pricing Example
A bakery is considering launching a new type of cake and needs to determine the minimum selling price required to break even. The bakery estimates that the fixed costs of producing the cake, including ingredients, equipment, and labor, are $1,500. The variable cost per cake is$5, which includes the cost of ingredients and packaging. The bakery expects to sell 500 cakes in the first month.
To calculate the break-even point using the break-even pricing method, the bakery would add up the fixed costs and variable costs and divide the total by the expected number of cakes sold:
$$\hbox{Total cost = Fixed cost + (Variable cost per cake x Expected number of cakes sold) Total cost = 1,500 + (5 x 500)}$$
$$\hbox{Total cost = Fixed cost + (Variable cost }\times\hbox{Expected numbers of cakes sold}$$
$$\hbox{Total cost}= 1,500+(5\times 500)$$
$$\hbox{Total cost}= 4,000$$
$$\hbox{Break-even price per cake}= \frac{\hbox{Total cost}}{\hbox{Expected number of cakes sold}}$$
$$\hbox{Break-even price per cake}=\frac{4,000}{500}= 8$$
## Test your knowledge with multiple choice flashcards
While a business might have a price ceiling determined by ___________, the price floor is determined by _________.
__________ refers to the cost that does not change as the level of sales or production changes.
___________changes directly with the level of production.
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A voltage $1000 \sin\omega t\:\: \text{Volts}$ is applied across $YZ.$ Assuming ideal diodes, the voltage measured across $WX$ in $\text{Volts},$ is
1. $\sin \omega t$
2. $(\sin \omega t \:+ \mid \sin \omega t \mid)/2$
3. $(\sin \omega t \: - \mid \sin \omega t \mid)/2$
4. $0$ for all $t$ | 114 | 300 | {"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} | 2.578125 | 3 | CC-MAIN-2021-49 | longest | en | 0.611758 |
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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:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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math scale factor
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tiddgnmm
Registered: 14.11.2005
From:
Posted: Saturday 30th of Dec 07:22 Hello math wizards, I need some urgent help. I have a set of math questions that I need to solve and I am hopelessly lost. I don’t know where to begin or how to go about and this paper is due next week. Kindly let me know if you are good in graphing parabolas or if there is a good site which can assist me.
Vofj Timidrov
Registered: 06.07.2001
From: Bulgaria
Posted: Saturday 30th of Dec 16:10 Well, I cannot do your assignment for you as that would mean cheating. However, I can give you a suggestion. Try using Algebrator. You can find detailed and well explained solutions to all your queries in math scale factor.
Jrahan
Registered: 19.03.2002
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Mibxrus
Registered: 19.10.2002 | 576 | 2,136 | {"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-2018-09 | latest | en | 0.915027 |
https://learn.careers360.com/engineering/question-pleaseplease-help-me-kinematics-jee-main/ | 1,716,739,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00013.warc.gz | 303,132,433 | 35,651 | #### A vector is rotated by a small angle radians to get a new vector .In that case is : Option 1) Option 2) Option 3) Option 4)
As we discussed in
Triangle law of vector Addition -
If two vector are represented by both magnitude and direction by two sides of triangle taken in same order then their resultant is represented by 3rd side of triangle.
- wherein
Represents triangle law of vector Addition
By Triangle Rule
Again
So
Option 1)
Incorrect
Option 2)
Incorrect
Option 3)
Correct
Option 4)
Incorrect | 125 | 528 | {"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} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.888378 |
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posted by .
Determine where the given function is concave up or where it is concave down:
f(x)=9x/x^2+49
• calculus -
f = 9x/(x^2 + 49)
f is concave up where f'' > 0
f' = -9(x^2 - 49) / (x^2 + 49)^2
f'' = 18x(x^2 - 147)/(x^2 + 49)^3
f'' < 0 where 0 < x < √147 or x < -√147
f'' > 0 where -√147 < x < 0 or x > √147
## Respond to this Question
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Determine the intervals on which the function is concave up and concave down f(x) = 17x^2 = x^4 by differentiating I get f'(x) = 34x+4x^3 and then I get lost. Can someone finish this up?
8. ### Calculus
THANK YOU TUTORS SO MUCH FOR YOUR HELP Determine the intervals on which the function is concave up or concave down. I finally understand that when f'' is greater than 0 it means concave up. f(X) = 18x^2 + x^4 so I differentiate that …
9. ### Calculus
For R"(x) = -15[(x-1)(e^-x)-(e^-x)] with 0 < x < 7 What interval is the graph concave and and concave down?
10. ### Calculus
Determine over what interval(s) is the function y=4x-6arctan(x) is concave up/concave down I took the second derivative and ended up with x=0. I do not believe I did this right. IF I happen to be right, wouldn't our answer just be …
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https://www.instasolv.com/question/9-300-175-25-30-10x-4-1-110-2-120-3-162-4-135-ze9nr1 | 1,611,299,573,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00420.warc.gz | 823,888,380 | 10,696 | 9. (300-175+25) + 30+10X 4 = ? (1) ...
Question
# 9. (300-175+25) + 30+10X 4 = ? (1) 110 (2) 120 (3) 162 (4) 135
11th - 12th Class
Maths
Solution
94
4.0 (1 ratings)
9. (3) ? ( =(overline{300-175}+25)+30 div 10 times 4 ) ( =(125+25)+30 div 10 times 4 ) ( =150+30 div 10 times 4 ) ( =150+frac{30}{10} times 4 ) ( =150+3 times 4 ) ( =150+12 ) ( =162 )
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http://www.kilicaslan.nom.tr/en/calculations/equation/equation.aspx | 1,624,607,285,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00288.warc.gz | 60,724,272 | 4,882 | Calculations
# Nonlinear Equation System Roots
Variable Number :
Variable symbols
Equations$$\mathcal{F}\left(\mathcal{X}\right)=0$$
$f_{1}\left ( x,y\right)=$
$f_{2}\left ( x,y\right)=$
Iteration Initial Vector
$x_{0}=$
$y_{0}=$
Max. Iter. Number Max. Error
Functions to be used in the equation:
$\begin{array}{lll|lll} x^a & : & \mathrm{pow(x,a)} \\\sin\, x & : & \mathrm{sin(x)} &\cos\,x & : & \mathrm{cos(x)} \\\tan\,x & : &\mathrm{tan(x)} &\ln\,x & : & \mathrm{log(x)} \\e^x & : & \mathrm{exp(x)} &\left|x\right| & : & \mathrm{abs(x)} \\\arcsin\,x & : & \mathrm{asin(x)} &\arccos\,x & : & \mathrm{acos(x)} \\\arctan\,x & : & \mathrm{atan(x)} &\sqrt{x} & : & \mathrm{sqrt(x)} \\ \\\pi & : & \mathrm{pi} &e \mathrm{ sayısı} & : & \mathrm{esay} \\\ln\,2 & : &\mathrm{LN2} & \ln\,10 & : & \mathrm{LN10} \\\log_{2}\,e & : & \mathrm{Log2e} & \log_{10}\,e & : & \mathrm{Log10e} \end{array}$
Equation Solution Nonlinear Equation System Roots Linear Equation System Solution Cubic Equation Solution Quartic Equation Solution Quintic Equation Solution Sextic Equation Solution Differential Equations Differential Equation Solution Higher Order Differential Equation
Documents Products Calculator Unit Conversion Reference Contact Pipe Calculations Air Ducts Equation Solver Kenan KILIÇASLAN 2012© Copyright. Designed by Nuit | 457 | 1,357 | {"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": 2, "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} | 3.40625 | 3 | CC-MAIN-2021-25 | longest | en | 0.280013 |
https://community.deeplearning.ai/t/c1w2-gaussian-elimination/573343 | 1,713,964,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00832.warc.gz | 163,857,549 | 5,426 | # C1W2 Gaussian Elimination
I don’t understand why my first exercise is not correct, because I get all test passed. Maybe it is not correct?
w2_unittest.test_reduced_row_echelon_form(reduced_row_echelon_form)
[0. 0. 0.]
[0. 0.]
[0.]
[0.]
All tests passed
1 Like
You posted this under AI Questions, but it would be better to post it under the course specific Q&A. This looks like M4ML C1 W2, so I will move it for you.
Please give us a little more info here: is it a different test in the notebook that fails? Or are you saying that the test in the notebook passes, but the grader fails that function? If so, please show us the actual grader output that you getting.
2 Likes
I’m talking about unit test inside of notebook, which checks your function.
Sure, I can show you result I get after my func.
``````A = np.array([[1,2,3],[0,0,0], [0,0,5]])
B = np.array([[1], [2], [4]])
reduced_row_echelon_form(A,B)
array([[1. , 2. , 3. , 1. ],
[0. , 0. , 1. , 0.8],
[0. , 0. , 0. , 2. ]])
``````
1 Like
I reformatted your output using the “{}” formatting tool to make it easier to read.
Here’s my output for that test cell:
``````A = np.array([[1,2,3],[0,0,0], [0,0,5]])
B = np.array([[1], [2], [4]])
reduced_row_echelon_form(A,B)
array([[1., 2., 3., 1.],
[0., 0., 0., 1.],
[0., 0., 1., 0.]])
``````
As you can see, it’s not the same as your result. So you need to look a bit more closely at what is supposed to happen there and how your logic works.
1 Like | 481 | 1,463 | {"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-2024-18 | latest | en | 0.832611 |
https://www.doorsteptutor.com/Exams/KVPY/Stream-SB-SX/Physics/Questions/Part-48.html | 1,500,854,843,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00597.warc.gz | 763,959,485 | 20,892 | # KVPY Stream SB-SX (Class 12 & 1st Year B.Sc.) Physics: Questions 151 - 154 of 268
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 268 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 151
MCQ▾
### Question
If an oscillator completes 200 oscillations, its amplitude decreases to of its initial value. When it completes 400 oscillations, the amplitude will be –
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 152
» Thermodynamics » Laws of Thermodynamics » First Law of Thermodynamics
MCQ▾
### Question
An ideal gas expands from to at a constant pressure of . The heat energy supplied to the gas in this process is –
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 153
Appeared in Year: 2015
MCQ▾
### Question
The intensity of sound during the festival season increased by 100 times. This could imply a decibel level rise from,
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 154
MCQ▾
### Question
A cart is moving along x direction with a velocity of 3 ms-1. A person on the cart throws a stone with a velocity of 8 ms-1 relative to himself. In the frame of reference of the cart the stone is thrown in plane making an angle of 600 with vertical y axis. The speed of stone at highest point of the trajectory the stone is –
### Choices
Choice (4) Response
a.
2 ms-1
b.
5 ms-1
c.
4 ms-1
d.
2.5 ms-1
f Page | 417 | 1,566 | {"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-2017-30 | longest | en | 0.754412 |
http://nosuch.com/music/expresso.cgi?generations=8&seed=99336069&ntracks=1&patches=no&randomseed=off&x= | 1,508,587,795,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00356.warc.gz | 242,676,484 | 2,842 | Home : Tune Toys : Expresso
Expresso takes a simple musical expression (literally "X") and mutates it. In each generation, transformations are applied to components of the expression. This is a fractal technique known as an L-system. The more generations there are, the larger and more complex the expression gets (and the longer it can take to compute). The output varies wildly, from boring to fascinating. Press "Mutate" a few times till you get something that looks interesting, then click on the image to play it. See Composer's Quarry for examples of output.
Transformations:
Seed: Randomize seed: Generations: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Phrase for X: Basic 1 - c,e,gBasic 2 - c,g,e,b-Basic 3 - c,gBasic 4 - cBasic 5 - co3,co4,co3,co2Chord 1 - c e gChord 2 - c e- gChord 3 - c e- g b-Chord 4 - c f gFile 1 - bachinv1.midFile 2 - bachinv2.midFile 3 - bachinv3.midFile 4 - bachinv4.midFile 5 - bachinv5.midFile 6 - bachinv6.midFile 7 - bachinv7.midFile 8 - bachinv8.mid # of Tracks: 1234 Randomize patches: noyes Results may not appear for a few seconds.Be patient!
Do you like this result? Do you want to let other people listen to it? Save this one in theTune Trove! See what's alreadyin the Tune Trove!
After 8 generations the expression "X" became this: step((transpose((transpose(shuffle((X|X) ) ,-7) +arpeggio(X) ) ,4) +transpose(transpose(step(arpeggio(X) ,12) ,-5) ,12) ) ,12) random seed used was 99336069
The algorithm for Expresso is written in KeyKit, and here's the source code. | 472 | 1,527 | {"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-2017-43 | longest | en | 0.779148 |
https://blogdredd.blogspot.com/2015/06/the-question-is-how-much-acceleration_25.html | 1,558,364,393,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00519.warc.gz | 409,206,922 | 26,774 | ## Thursday, June 25, 2015
### The Question Is: How Much Acceleration Is Involved In SLR - 8?
Fig. 1 Arctic sea ice extent / area 6/27/15
In this series I have pointed out that sea level rise (SLR) projections must consider acceleration of the volume of ice sheets that are entering the oceans per year.
The old way of linear projections, by older software models, has been proven to be wrong.
Those models have consistently underestimated the rate of SLR caused by loss of ice volume on Greenland and Antarctic ice sheets.
So, Greenland and Antarctic ice sheet volume loss, in terms of contribution to SLR, is now determined by rate of acceleration of ice volume lost per year.
If the old models continue to be used, their linear formulas will faithfully underestimate the rate of loss of ice volume at the ice sheets of both Greenland and Antarctica (The Question Is: How Much Acceleration Is Involved In SLR?, 2, 3, 4, 5, 6, 7).
The exercise becomes, then, an issue of how much acceleration of ice volume loss is taking place (ibid, cf. The Evolution of Models, 2, 3, 4, 5, 6, 7, 8, 9, 10).
II. "Doubling" Is Another Way of Saying "Acceleration"
In general, acceleration is difficult to determine using only a year to year basis.
An alternate method of projection has been expressed by top scientists in terms of "doubling" of ice volume loss over some span of time:
The effect that 2, 3, 5, 7, and 10 year doubling has is that it determines when SLR reaches the 1 m / 3 ft. level:
10 yr = 2067
7 yr = 2055
5 yr = 2045
3 yr = 2035
2 yr = 2031
(The Question Is: How Much Acceleration Is Involved In SLR - 5?). We focus most on 1 m / 3ft. because it is a point where many locations that are tidal will experience serious problems, especially with coastal infrastructure and ports (The 1% May Face The Wrath of Sea Level Rise First).
(The Question Is: How Much Acceleration Is Involved In SLR - 7?). That "1 m / 3ft." level of SLR will be reached at a time that is dependent on the rate of acceleration of ice volume loss.
If acceleration of the amount of land ice which reaches the ocean, in the form of melt water or icebergs, doubles each two years from now on, then that "1 m / 3 ft." level of SLR will take place circa 2031, but if it doubles each 10 years from now on, then that "1 m / 3 ft." level of SLR will take place circa 2067.
III. Catastrophe Is What We Want To Avoid
Since that "1 m / 3ft." level of SLR is the catastrophic level, that is what we must keep our eyes on.
What will happen at that level and above is "unthinkable" (Greenland & Antarctica Invade The United States, The 1% May Face The Wrath of Sea Level Rise First, Why The Military Can't Defend Against The Invasion, Why Sea Level Rise May Be The Greatest Threat To Civilization, 2, 3, 4).
It would change civilization as we have known it (Weekend Rebel Science Excursion - 44, Will This Float Your Boat?, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11).
IV. The Wildcard Is "The Surge"
I found scientific papers which examine the evidence for historical surges in SLR (The Surge: A Forgotten Aspect of Sea Level Rise).
Surges or "pulses" are abrupt increases in acceleration of SLR.
The evidence indicates several surges in SLR (e.g. "Pulse 1A ... 1B and 1C") in the paleoclimatology records of ice cores, soil strata, etc. (ibid).
The larger the increase the slower they tend to be, taking more time to happen.
The only surge I am interested in is "pulse 1C", a "1 m / 3 ft." SLR which took place about 7-8,000 years ago, which left marks and other evidence in North America.
It is the most interesting surge, because "1 m / 3 ft." of SLR was all said and done in "a few years or less" (ibid).
Two vast lakes of melt water had been held back from entering the ocean by an ice dam which eventually collapsed as temperatures warmed, releasing all that water into the ocean.
There are no vast lakes now, but there are hundreds of "moulins" where melt water flows under Greenland's ice sheet and then disappears (The Question Is: How Much Acceleration Is Involved In SLR? - 4).
It is not known whether or not the water is being captured in sub-glacial canyons, thereby creating lakes, or is flowing out into the ocean.
Similar events now take place in Antarctica too (ibid).
V. We Know Less About Critical Issues Here Than There
As late as 2005, it was written:
"In spite of its importance, the mass balance (the net volumetric gain or loss) of the Antarctic ice sheet is poorly known; it is not known whether the ice sheet is growing or shrinking."
(The Agnotology of Sea Level Rise Via Ice Melt). We know more about the ice caps on Mars it would seem.
Yet, Greenland and Antarctica have enough ice sheet volume that if only 1.14% of their ice melts or otherwise makes it to the ocean, we will have the catastrophe that a "1 m / 3 ft." SLR can bring about (Why Sea Level Rise May Be The Greatest Threat To Civilization - 4).
We seem to know more about other planets, moons, comets, and the like, than we know about the planet where we are (You Are Here).
VI. There Is Some Uncertainty About "When"
The non-linearity mentioned in the Introduction means that we are made somewhat uncertain as to the timing of events by the nature of those events.
Notice Fig. 1, which shows that this year's Arctic sea ice extent or area was the lowest on record for several months, but has, as of a few days ago, fallen into second place (but "it ain't over 'til its over").
The year of record for lowest extent, 2012, was also a year of extreme melt events on the surface of the Greenland ice sheet.
During the years since then, we have not had those same record breaking events.
The point is that there are both surges and then non-linear years (not the same ice volume loss as the year before) following surges, where ups and downs take place.
Having said that, it is not the same as saying the ice volume loss stops.
Loss continues in a non-linear fashion, but overall there is an acceleration in the loss of ice volume in both Greenland and Antarctica (the ice volume loss graph line trend is upward).
VII. Conclusion
Even though there have been popular events such as the Papal encyclical, the view from astute observers is still quite pessimistic realistic (Greenland & Antarctica Invade The United States - 2).
Realistic in an extinction kind of way.
The previous post in this series is here.
David Puttnam on climate change: | 1,599 | 6,424 | {"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} | 3.015625 | 3 | CC-MAIN-2019-22 | latest | en | 0.918938 |
https://brainmass.com/business/auditing/sunbelt-company-bank-charges-120324 | 1,624,227,841,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00388.warc.gz | 149,197,114 | 75,003 | Explore BrainMass
# Sunbelt Company Bank Charges
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In June, the Sunbelt Company utilizes direct sends for which its bank charges fees of \$100. The bank's earned credit ratio and reserve requirement are 5% and 12%, respectively. What is the minimum balance required to compensate the bank for the June fees?
© BrainMass Inc. brainmass.com March 4, 2021, 7:44 pm ad1c9bdddf
#### Solution Preview
To offset some or all of the service charge, an earnings credit is given on the ...
#### Solution Summary
Simple calculation showing a minimum balance needed to offset bank fees with regards to the bank's earned credit ratio and the reserve requirement.
\$2.49 | 191 | 856 | {"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} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.930391 |
https://www.weegy.com/?ConversationId=VFIXGEMQ | 1,601,426,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00731.warc.gz | 1,098,064,133 | 10,077 | The median of a sample will always equal the Options A-mode B-mean C-50th percentile D- all of the above answers are correct
The median of a sample will always equal to the 50th percentile.
s
Question
Updated 4/12/2014 9:24:02 AM
Confirmed by debnjerry [4/12/2014 9:24:02 AM]
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Current Assets minus current liabilities is equal to (a) Gross working capital (b) Capital employed (c) Net worth (d) Net working capital.
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https://www.valiotti.com/leftjoin/all/beautiful-bar-charts-with-python-and-matplotlib/ | 1,627,495,838,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00434.warc.gz | 1,099,233,943 | 10,892 | # Beautiful Bar Charts with Python and Matplotlib
Estimated read time – 5 min
The Matplotlib library provides a wide range of tools for Data Visualisation, allowing us to create compelling, expressive visualizations. But why then so many plots look so bland and boring? Back in 2011 we built a simple yet decent diagram for a telecommunication company report and named it ‘Thermometer’. Later this type of bars was exposed to a wide audience on Chandoo, which was a popular blog on Excel. By the way, here’s what it looks like:
Times change, and today we’ll recall the way to plot this type of diagrams with the help of Matplotlib
When should one use this type of diagram?
The best way to plot this type of diagrams is when comparing target values with actual values because it reflects underfulfilment and overfulfilment of planned targets. A diagram may reflect data in percentages as well as in real figures. Let’s view an example using the latter.
We’ll use data stored in an excel file and already familiar python libraries for data analysis:
``````import numpy as np
import pandas as pd
import matplotlib.pyplot as plt``````
Read our file as a DataFrame:
``df = pd.read_excel('data.xlsx')``
That’s what it looks like:
Now, we need to extract columns from the table. The first column called «Sales» will be displayed under each bar. Some values may be of a string type if there is a comma between two values. We need to convert these type of values by replacing a comma with a dot and converting them to floats.
``````xticks = df.iloc[:,0]
try:
bars2 = df.iloc[:,1].str.replace(',','.').astype('float')
except AttributeError:
bars2 = df.iloc[:,1].astype('float')
try:
bars1 = df.iloc[:,2].str.replace(',','.').astype('float')
except AttributeError:
bars1 = df.iloc[:,2].astype('float')``````
As we don’t know for sure if the table includes such values, our actions may cause an AttributeError . Fortunatelly for us, Python has a built-in try – except
method for handling such errors.
Let’s plot a simple side-by-side bar graph, setting a distance between two related values using a NumPy array:
``````barWidth = 0.2
r1 = np.arange(len(bars1))
r2 = [x + barWidth for x in r1]
plt.bar(r1, bars1, width=barWidth)
plt.bar(r2, bars2, width=barWidth)``````
And see what happens:
Obviously, this is not what we expected. Let’s try to set a different bar width to make bars overlapping each other.
``````barWidth1 = 0.065
barWidth2 = 0.032
x_range = np.arange(len(bars1) / 8, step=0.125)``````
We can plot the bars and set its coordinates, color, width, legend and signatures in advance:
``````plt.bar(x_range, bars1, color='#dce6f2', width=barWidth1/2, edgecolor='#c3d5e8', label='Target')
plt.bar(x_range, bars2, color='#ffc001', width=barWidth2/2, edgecolor='#c3d5e8', label='Actual Value')
for i, bar in enumerate(bars2):
plt.text(i / 8 - 0.015, bar + 1, bar, fontsize=14)``````
Add some final touches – remove the frames, ticks, add a grey line under the bars, adjust font size and layout, make a plot a bit wider and save it as a .png file.
``````plt.xticks(x_range, xticks)
plt.tick_params(
bottom=False,
left=False,
labelsize=15
)
plt.rcParams['figure.figsize'] = [25, 7]
plt.axhline(y=0, color='gray')
plt.legend(frameon=False, loc='lower center', bbox_to_anchor=(0.25, -0.3, 0.5, 0.5), prop={'size':20})
plt.box(False)
plt.savefig('plt', bbox_inches = "tight")
plt.show()``````
And here’s the final result:
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https://csedoubts.gateoverflow.in/tag/ricestheorem | 1,627,963,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00466.warc.gz | 204,323,437 | 11,942 | # Recent questions tagged ricestheorem
If a language satisfies monotonic property then is it REL or REC or NOT REL ?? And What about CO- REL?? #RICE THEOREM
What are the languages, RE or not RE? Please explain using Rice’s theorem or otherwise. 1) L={⟨M⟩|TM halts on empty string} 2) L={⟨M⟩|TM halts on empty string} 3) L={⟨M⟩|TM halts on no input} Not sure how rice’s theorem is working here.
Here in place of or if and was given then is it correct: $T_{yes}=\{a,b\} \ \ T_{no}=\{\ \Sigma^* \}$ Here for $T_{yes}$ x = a or b is in the language, but let y = ab is not in the langugage, in $T_{no}$ there is no x or y which ... since the language of $T_{no}$ is universal language also $T_{yes}$ is a proper subset of $T_{no}$ so the language is not RE. Please provide your valuable inputs. | 245 | 790 | {"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} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.794486 |
https://www.jiskha.com/questions/594309/how-do-u-circumscribe-and-inscribe-a-hexagoneabout-a-2inch-circle | 1,606,791,915,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00243.warc.gz | 721,757,457 | 5,095 | geometry
how do u circumscribe and inscribe a hexagoneabout a 2inch circle?
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3. 👁 112
1. Draw a 2 inch circle.
Divide the circle into six 60º sectors.
Connect the points where the sector radii intersect the circumference giving you an inscribed hexagon.
Construct six lines tangent to the circle and parallel to the sides of the inscribed hexagon giving you a circumscribed hexagon.
1. 👍 0
2. 👎 0
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https://www.physicsforums.com/threads/a-rigorous-definition-of-a-limit-and-advanced-calculus.947729/ | 1,713,696,879,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00361.warc.gz | 858,293,042 | 23,349 | # A rigorous definition of a limit and advanced calculus
• I
• gibberingmouther
In summary, the individual is reviewing calculus for fun and to prepare for school. They struggle with the rigorous definition of the limit but found a helpful web page and plan to practice more. They ask if there are other difficult concepts in advanced calculus and express interest in learning more about it. They also mention the length of Andrew Wile's proof of Fermat's Last Theorem and ask for tips on understanding the epsilon delta definition. The expert suggests practicing with the definition and seeking help from others.
gibberingmouther
i'm trying to review calculus and look a little deeper into proofs/derivations/etc. I'm doing this both for fun and to review before i go back to school.
am i the only one who has difficulty understanding the "rigorous" definition of the limit? i found this web page: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx
i think i can get this if i spend some more time with the explanations and exercises on this page. my question is, are there lots of hard to understand pieces of logic like this in advanced calculus or analysis? i don't have to take analysis for my degree but i want to learn a little bit of it just to expand my understanding of how the universe works (which is fun).
when i took calculus i pretty much worked through the textbook and i remember having difficulty with the definition of the limit in that book as well. for those who have taken analysis, i guess, what is it like? what can you say about it to someone with beginner/intermediate level experience like myself? i remember reading that Andrew Wile's proof of Fermat's Last Theorem was over 100 pages long. is there anything in advanced calculus like that, or are the main theorems relatively easy to demonstrate?
also, any tips on how to understand this epsilon delta thing would be appreciated.
edit: maybe the super scientists of the future will be AIs? apparently there are some super long proofs that were done by computers - not really AIs, but sophisticated programs. this got me thinking so i did some google searches.
Last edited:
Aufbauwerk 2045
The way I like to think of the epsilon delta stuff is as a kind of two player game. We are given a function f() and an x coordinate.
a. I pick a number y, propose it as the limit of f() at x and play begins.
b. The other player picks an epsilon -- an non-zero error tolerance in the y direction.
c. My job is then to pick a delta -- a non-zero distance in the x direction.
d. Finally, the other player picks an x' within delta of x (but not equal to x).
If f(x') is more than epsilon different from y then the other player wins.
If f(x') is less than epsilon different from y then I win.
The limit is defined and equal to y if I can come up with a strategy that always wins.
Highwire, Mark44 and gibberingmouther
gibberingmouther said:
i'm trying to review calculus and look a little deeper into proofs/derivations/etc. I'm doing this both for fun and to review before i go back to school.
am i the only one who has difficulty understanding the "rigorous" definition of the limit? i found this web page: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx
This is the definition of what it means for a function to be continuous at a point ##x=a##. It is a bit tricky, because it is very formal. It might help to imagine, which functions are not continuous at ##a##.
The definition has to rule out these situations.
i think i can get this if i spend some more time with the explanations and exercises on this page. my question is, are there lots of hard to understand pieces of logic like this in advanced calculus or analysis?
This depends on what you want to learn and what you consider advanced. There is a long way to the top.
i don't have to take analysis for my degree but i want to learn a little bit of it just to expand my understanding of how the universe works (which is fun).
As so much in life: all a question of practice.
when i took calculus i pretty much worked through the textbook and i remember having difficulty with the definition of the limit in that book as well. for those who have taken analysis, i guess, what is it like? what can you say about it to someone with beginner/intermediate level experience like myself?
You can always come to PF and ask. Usually, in I think in your case always, there will be someone who can help you. I've had similar problems as I first saw this definition. And still today I prefer to look up the order of the "for alls" and "exists" if I need them.
i remember reading that Andrew Wile's proof of Fermat's Last Theorem was over 100 pages long. is there anything in advanced calculus like that, or are the main theorems relatively easy to demonstrate?
As far as I know is his proof mainly algebra and I think the theory of modular forms. Trying to understand it is in itself a lifetime goal. Most mathematicians don't.
also, any tips on how to understand this epsilon delta thing would be appreciated.
Play a little with the definition and try to understand, why the functions in my image fail to be continuous at ##x=a## and why they are continuous elsewhere: step, gap, asymptote.
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gibberingmouther
thanks guys, both of those were helpful. if today weren't game night for me, with me on my laptop as my friends and i were watching the Dark Tower and a movie about bigfoot, i wouldn't have had the patience and persistence to sit and study this. i was good at solving calculus problems, but this type of thing is another layer of theory and difficulty. i don't quite "grok" it yet, but i think as i do some examples and think about it i will become comfortable enough with it. attention span is important for this kind of thing!
i thought i could just understand theory and not do problems, but i guess you've got to do a bit of both sometimes if you want to wrap your mind around a concept.
gibberingmouther said:
what can you say about it to someone with beginner/intermediate level experience like myself?
Before you worry about the particulars of the definition of limit, study a little logic, especially the logic of the quantifiers "for each" and "there exists". After you understand how to deal with logic and quantifiers, then return to the particular example of the definition of a limit. There is much material on the web about elementary logic and the logic of quantifiers - e.g. https://www.whitman.edu/mathematics/higher_math_online/section01.02.html
Understanding the logic of quantified statements is essential for studying calculus from a rigorous point of view.
gibberingmouther
Theres a nice little cheap book that has a great explanation of logic quantifiers. Levin: "Discrete Mathematics." It can be found for free, as it is an open stacks book, or purchased for less than 20 dollars. I suggest the second edition, since the logic is put in the beginning sections and not towards the middle. Also, Hammock: Book of Proof. It has great explanations for the fundamentals of mathematics. i.e. relations, functions, etc.
The "epsilon-delta" definition of a limit is used everywhere in analytical mathematics. If you want to be good at analytical mathematics, you will want to become very comfortable with this type of definition and with proofs using it. The definition may seem obscure at first, but learning it will pay off and you will use it so much that it will become second nature.
In english:
No matter how close (##\forall \epsilon > 0##) you want to keep a function to it's limit, ##L##, (##|f(x)-L| < \epsilon ##) for x near ##a##,
You can find a ##\delta## (## \exists \delta \text{ such that} ##), where keeping x that close to ##a## (## |x-a| < \delta ##) will keep f(x) close enough to ##L##. (##|f(x)-L| < \epsilon ##)
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gibberingmouther
I will add my two cents worth. I had this math teacher at U. once who boasted about how he didn't care about "delta-epsilonics" since he was too busy working on practical stuff like matrix calculations for jet fighter aerodynamics or whatever. Anyway, please don't listen to guys like that particular math teacher of mine, who build up the definition of a limit using delta and epsilon into some kind of big obstacle, or something for the pedantic types which you really don't need to worry about. I really hate it when teachers shortchange their students like that guy did.
Actually if you want to really learn calculus, you should spend a lot of time on limits and learn how to prove a limit using delta-epsilon. It takes some practice to get the idea.
I recommend Bob Miller's math books in general because he usually explains things very clearly. You will find a very good introduction to limits in chapter 1 of his Calculus for the Clueless Part 1. I'm not familiar with any better introduction.
The "rigorous definition" is the one you need to know. Who wants to learn non-rigorous calculus? It makes no sense. Miller gently leads you from the basic idea to the hard core rigorous version. After Miller, I would look for more problems in limits from other books and work them all until you are very skilled in the topic. You will be very glad you did.
P.S. since I'm on the topic, I recommend to everyone who wants to learn math to sit down with a book, paper and pencil, and a calculator, read very slowly, pay attention to every detail, and work tons of problems. Old fashioned advice I know, but it works wonders. You don't really understand it until you have worked lots of problems and burned the topic into your long term memory.
Miller has a good story on one topic he wanted to make sure his students really learned. He said if he woke them up in the middle of the night and asked them a certain math question, they would be able to answer it without hesitation, and then tell him to go away so they could get back to sleep.
Last edited by a moderator:
FactChecker
gibberingmouther said:
i'm trying to review calculus and look a little deeper into proofs/derivations/etc. I'm doing this both for fun and to review before i go back to school.
am i the only one who has difficulty understanding the "rigorous" definition of the limit? i found this web page: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx
i think i can get this if i spend some more time with the explanations and exercises on this page. my question is, are there lots of hard to understand pieces of logic like this in advanced calculus or analysis? i don't have to take analysis for my degree but i want to learn a little bit of it just to expand my understanding of how the universe works (which is fun).
when i took calculus i pretty much worked through the textbook and i remember having difficulty with the definition of the limit in that book as well. for those who have taken analysis, i guess, what is it like? what can you say about it to someone with beginner/intermediate level experience like myself? i remember reading that Andrew Wile's proof of Fermat's Last Theorem was over 100 pages long. is there anything in advanced calculus like that, or are the main theorems relatively easy to demonstrate?
also, any tips on how to understand this epsilon delta thing would be appreciated.
edit: maybe the super scientists of the future will be AIs? apparently there are some super long proofs that were done by computers - not really AIs, but sophisticated programs. this got me thinking so i did some google searches.
A few more thoughts on your question.
No, you are not the only one who has difficulty understanding the rigorous definition of the limit. It took even great mathematicians a while to figure it out. We are lucky, because it's already been figured out for us. As I stated earlier, it's not so hard if you follow Bob Miller's explanation step by step. I would write out all of this examples and go over them until you start dreaming about it.
I think the hardest part of learning calculus is understanding the basis which is limits and continuity. One you understand limits, you can understand the definition of a derivative quite easily. But you still need to apply limits of all sorts to make progress. Not everything is as simple as differentiating a polynomial. You can also understand the definition of the integral. Then you can understand the relationship between the derivative and the integral.
Once you get that down in one variable, I think it's relatively easy to extend to two and three variables. Then it's mainly a matter of doing lots of problems and seeing how calculus is applied. Remember, it was developed because the mathematics was needed to solve real world problems, not just as some kind of interesting puzzle. When you see what kind of problems can be solved with calculus, then you see how awesome it is.
It gets even much greater when you learn about advanced topics. I will single out Green's Theorem. Speaking of genius -- an overused word to be sure -- he was one of the top geniuses. Especially when you consider he only had one year of formal schooling. There is genius, then there is genius. Green was one of the latter.
Even the symbols used in calculus are beautiful. Continue to the stage where you learn about partial differential equations and you have the symbols of the universe at your fingertips. It's the greatest. Beauty, power, understanding -- what more motivation does one need?
To sum up, I would spend two or three weeks doing nothing but limits and also to review anything from pre-calculus you are not sure about, such as logs and exponentials, trig functions, and series. Series are super important in so many ways, not just in calculus.
To make calculus seem more approachable, just remember that it is based on two problems. Problem #1 : find the slope of a line tangent to a curve at a given point. Problem #2: find the area under a curve between two given points. Differential calculus answers problem #1. Integral calculus answers problem #2. The two problems are in fact related, as you learn in first year calculus. These two problems needed to be solved in order for physics to make progress during the time of Newton and Leibniz.
Finally, for now, here is a mystery question for science fiction movie fans. In which movie does a scientist say to some of the other characters, "I have something you don't have -- Calculus!"
Last edited by a moderator:
gibberingmouther
## 1. What is a limit in calculus?
A limit in calculus is a fundamental concept that describes the behavior of a function as the input approaches a certain value. It is used to define the continuity of a function and to determine the behavior of a function at points where it is not defined.
## 2. How is a limit calculated?
A limit is calculated by evaluating the function at points approaching the desired input value. This can be done algebraically or graphically, depending on the function and the type of limit being evaluated.
## 3. What is the difference between a one-sided and two-sided limit?
A one-sided limit only considers the behavior of a function as the input approaches the desired value from one side, either the left or the right. A two-sided limit takes into account the behavior from both sides of the input value.
## 4. How is a rigorous definition of a limit different from an intuitive understanding?
A rigorous definition of a limit is based on the mathematical concept of epsilon-delta, where an epsilon value represents a small distance from the limit and a delta value represents the corresponding distance from the input. This definition eliminates any ambiguity and provides a precise understanding of the limit, whereas an intuitive understanding may be based on general observations and may not hold for all cases.
## 5. What is the importance of limits in advanced calculus?
Limits are essential in advanced calculus as they are used to define important concepts such as derivatives and integrals. They also play a crucial role in understanding the behavior of functions and solving complex problems in mathematics and other fields such as physics and engineering.
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ACI Mix Design. The American Concrete Institute (ACI) mix design method is but one of many basic concrete mix design methods available today. This section summarizes the ACI absolute volume method because it is widely accepted in the and continually updated by the ACI. Keep in mind that this summary and most methods designated as "mix ...
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• Eurocode 2: Design of concrete structures EN199211
Design criteria Aggressivity of environment Specified service life Design measures Sufficient cover thickness Sufficiently low permeability of concrete (in combination with cover thickness) Avoiding harmfull cracks parallel to reinforcing bars Other measures like: .
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• Types of Concrete Mix Ratio Design and their Strengths ...
IS has designated the concrete mixes into a number of grades as M10, M15, M20, M25, M30, M35 and M40. In this designation the letter M refers to the mix and the number to the specified 28 day cube strength of mix in N/mm 2 . The mixes of grades M10, M15, M20 and M25 correspond approximately to the mix proportions (1:3:6), (1:2:4), (1:1 ...
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• What is the mix ratio of M30 concrete?
Design of M30 Grade Concrete mix According to IS 10262:2009 Assume : Type of Cement: 53 Grade Specific Gravity of Cement : Coarse Aggregate : 20 mm Crushed Granite Specific Gravity of Coarse Aggregate : Specific Gravity of Fine Aggregate...
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• American Concrete Institute
Total volume from mix design calculations should equal cf. Measured Batch Weight is the weights of all materials used to mix your concrete (excluding admixtures). Batch Dosage (oz) Official Mix Design and Cost Form (Page 1 of 2) Official Mix Design and Cost Form (Page 2 of 2) More than 20% = 1; more than 35% = 2; more than 50% = 3
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• Eurocode 2: Design of concrete structures EN199211
22 February 2008 1 Eurocode 2: Design of concrete structures EN199211 Symposium Eurocodes: Backgrounds and Appliions, Brussels 1820 February 2008
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• Pdf File Of Concrete Ratio Of Mix Design Code From M15 To M80
Pdf File Of Concrete Ratio Of Mix Design Code From M15 To M80. Get Latest Price. Abstract The aim of this study is to investigate the characteristics exhibited by three different grades of concrete using mix design approach From the result of the sieve analysis it shows that the sands used for the experiment is a well graded sand of zone 1 of BS882
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• pdf file of concrete ratio of mix design code from m15 to m80
File Of Concrete Ratio Of Mix Design Code From M15 To M80. File Of Concrete Ratio Of Mix Design Code From M15 To M80. Concrete mix design as per indian standard code determine the concrete mix proportions for the first trial mix 10 prepare the concrete using the calculated proportions and cast three cubes of 150 mm size and test them wet after 28days moist curing and check for the .
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• Manual for the design of reinforced concrete building ...
IStructE EC2 (Concrete) Design Manual 9 Foreword The Eurocode for the Design of Concrete Structures(EC2) is likely to be published as a Euronorm (EN) in the next few years. The prestandard (ENV) for EC2 has now been available since 1992. To facilitate its familiarisation the Institution of Structural Engineers and
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• E*
the new 2002 Design Guide for New Rehabilitated Pavements. The guide is based on mechanistic principles and requires a modulus, analogous to E for steel, to compute stress and strains in hotmix asphalt (HMA) pavements. In 1999, the NCHRP Panel for Project 137A selected E* for this purpose. The selection was based on a paper authored by M.
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• How to Calculate Quantities of Cement, Sand and Aggregate ...
Incase you want to convert the requirement of Sand and Aggregate in Cum; 1 Cum = Cft Eyeopener:Many popular blogs claim M20 nominal mix as 1::3,however we strongly differ by this blog,we are also trying to address the same myth which is being carried forward since last 4 decades.. The reason being: With constant research and development in the field of cement .
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• September 1, 2003 CONCRETE MANUAL MIX DESIGN .
September 1, 2003 CONCRETE MANUAL MIX DESIGN NOTE: FOR PROJECTS REQUIRING CONTRACTOR MIX DESIGN, THE DESIGN ... Specific gravity of a material is the ratio of the mass (weight) of a given volume to the mass (weight) of an equal volume of water. Water is us ed as a standard because of its uniformity.
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• IS 10262 (2009): Guidelines for concrete mix design ...
earlier title 'Recommendedguidelines for concrete mix design'. b) The applicability ofthe standard has been specified for ordinary and standard concrete grades only. c) Various requirements have been modified in line with the requirements of IS 456 : 2000 'Plain and reinforced concrete Code ofpractice (fourth revision)'.
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• pdf file of concrete ratio of mix design code from m15 to m80
Primary mobile crushing plant. Independent operating combined mobile crushing station. Mobile secondary crushing plant. Fine crushing and screening mobile station
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• Reinforced Concrete Design
concrete design P u = factored column load calculated from load factors in concrete design R = shorthand for rain or ice load R n = concrete beam design ratio = M u /bd 2 s = spacing of stirrups in reinforced concrete beams S = shorthand for snow load t = name for thickness T = name for a tension force = =shorthand for thermal load
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• Mix Design Of Concrete British Doe Method B
(PDF) Concrete Mix Design (DOE) ... Testing Concrete Mix Ratio The design strength of concrete is the strength after 30 days of curing. ... as M80, M100 can also be prepared. The workability requirements of each mix can also meet using this method from zero slump to the 150 mm
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• Reinforced Concrete Design
= reinforcement ratio in concrete beam design = A s /bd = balanced reinforcement ratio in concrete beam design = shear strength in concrete design Reinforced Concrete Design Structural design standards for reinforced concrete are established by the Building Code and Commentary (ACI 31811) published by the American Concrete
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# 0.995″ to mm – 0.995 inch to mm
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Welcome to 0.995″ to mm, our page dedicated to converting 0.995 inch to mm.
Here we are going to show you how many millimeters is 0.995 inches.
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## 0.995 inch to mm
To convert 0.995 inch to mm you have to multiply 0.995 by 25.4 as one inch equals 25.4 mm.
The inch to mm formula is [mm] = [in] * 25.4.
0.995″ to mm = 25.273 millimeters
0.995 inch to mm = 25.273 millimeters
Now you know how to convert 0.995″ to mm and that 0.995 in mm = 25.273 millimeters.
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## 0.995″ in mm
Getting 0.995″ in mm is really easy as we have shown above.
For converting 0.995 to mm you only have to do a quick multiplication or insert the value into our calculator.
Apart from 0.995 inch in mm you may also be interested in the other units of length in the International System of Units:
0.995″ to cm = 2.5273 centimeter
0.995″ to dm = 0.25273 decimeter
0.995″ to m = 0.02527 meter
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# AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] watch
1. (Original post by mattyt307)
I got x<-6 x>3 and 1<x<2 lol
SAME. I GOT THIS EXACTLY (but the less than/more than and EQUAL to sign)
2. (Original post by KB_97)
No way!! That's too high. Loads of people found it so hard. There's people (including myself) that have straight up lost out chunks of marks as well as stupid mistakes. I think we should do a poll for what people think an A grade boundary will be.
See the updated post. Most people had complaints on the last question, but you could work out part b and c without part a. (Adding the two equations gives you 2secx=-5.2
, secx = -13/5
cosx=-5/13
On the integration by parts of ln(3x)/x multiplied by something or whatever it was, i got the integration by parts..... but then with the rotating around an axis... how did people make the numerator ln(3x) because it was squared???
you bring the 2 down because its ln, so it will be 2ln3x/x anf then bring the 2 out next to pie and do and it will ln3x/x then continue
4. I got f(x) </= 8ln8 -8
5. Anyone got x<-6 and x>3 for the third question?
Also, what did you get for substitution? I got something like 18/3 or 18/2 (can't remember really)
The second part of integration by parts question: anyone had to use integration by parts twice?
I was able to do every single past paper in less than an hour and get an A*, but not this one.. so, I believe grade boundaries should be much lower
6. I'm hoping for about 50-53 for an A because it looks like people either ran out of time/missed out some questions/silly mistakes and people found it generally more difficult than their other papers. Someone please make a poll thing so we can vote
7. Pretty sure x^2>>mod so it was my answer but after this exam I have no clue. Was getting 100% in my c3 but that test was a disgrace.
8. (Original post by tanyapotter)
65 for an A?! 64 was an A* last year, and that paper was really easy.
Are you trolling?
deffo a troll
9. (Original post by Mowerharvey)
What did everyone get for tbe range?
Think I got f(x) </= 8ln8 -8 or something
10. (Original post by hish11)
Got straight A*/A students saying they dropped 15 marks, at least.
Reckon
a*-61
a-57
b-54
c-51
d-47
It will never been 60 for an A*. Maths is the most subscribed subject for AQA , inevitably with students skewing the grade boundaries.
Add 3 to each, then it'll a be more sensible prediction. Saying that, it was not a hard paper.
11. (Original post by KB_97)
SAME. I GOT THIS EXACTLY (but the less than/more than and EQUAL to sign)
remember its the modulus of 5x-something or whatever, so if you sketch the graphs/have a graphical calculator you'll see they dont interest at -6
12. I should get an A* (assuming an A* will be 30 marks).
That was an unacceptably hard paper and it's unfair for AQA to make this year's paper so hard in comparison to all the papers. I can say a goodbye to my firm uni that's for sure, and now I genuine think I'll be going through clearing.
13. (Original post by khuman97)
you bring the 2 down because its ln, so it will be 2ln3x/x anf then bring the 2 out next to pie and do and it will ln3x/x then continue
I was thinking of doing that, but I thought that ln(3x)^2 was different to ln(3x^2)? Like can't you only bring down the 2 if it's in the ln?
14. (Original post by Pablo Picasso)
I did the paper. Excpet from one, no questions were challenging. It's all overexaggeration
Joes like you are the reason ISIS exists
Posted from TSR Mobile
15. Got straight A*/A students saying they dropped 15 marks, at least.
Reckon
a*-61
a-55
b-50
c-45
d-40
16. Range question
17. (Original post by Marelitza)
I was able to do every single past paper in less than an hour and get an A*, but not this one.. so, I believe grade boundaries should be much lower
Same. I used to finish the paper in less than an hour but I was working till the last second for this one. They should be much lower
I was thinking of doing that, but I thought that ln(3x)^2 was different to ln(3x^2)? Like can't you only bring down the 2 if it's in the ln?
Correct, you had to change the integral to
(lnx) * (lnx)/(x^2)
set u=lnx, v'=lnx/x^2
and go from there
I'm sticking by
100 UMS - 70 (maybe 69)
A* - 62
A - 55
B - 51
C - 46
D - 41
E - 36
19. (Original post by the1pedro)
I got f(x) </= 8ln8 -8
Me too! I was getting scared. Hopefully its right.
I was thinking of doing that, but I thought that ln(3x)^2 was different to ln(3x^2)? Like can't you only bring down the 2 if it's in the ln?
You're right it was (ln3x)^2 which can't be written as anything else.
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https://physics.stackexchange.com/questions/770072/calculate-change-in-tank-psi-per-breath-at-depth | 1,722,992,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00808.warc.gz | 366,824,377 | 40,994 | Calculate Change in Tank PSI per breath at depth
I am a scuba diver, trying to understand some physics that occurs throughout a dive. Primarily I am trying to understand the surface air consumption (SAC) derivation.
For those unfamiliar, SAC is the change in pressure (PSI or bar) per unit time at the surface (1 atm). SAC is calculated by measuring the change in PSI at the dive depth and dividing it by the pressure at depth (expressed in atm).
$$SAC = \frac{\Delta PSI* P_{sea level}}{P_{depth}}$$
Since the pressure at sea-level is 1 atmosphere the equation can be simplified by expressing the pressure at depth in atmospheres as $$P_{depth} = N_{atm}*1atm$$. $$N_{atm}$$ is the number of atmospheres the pressure at depth is equivalent to.
In this case, SAC is expressed as
$$SAC = \frac{\Delta PSI * 1atm}{N_{ATM}*1atm} = \frac{\Delta PSI}{N_{ATM}}$$
I am trying to determine the assumptions/derivation to arrive at this equation.
So far, what I have done is assumed that
• The volume of air inhaled per breath is constant regardless of depth
• Breathing rate is constant regardless of depth
• Temperature of the gas is independent of depth
Using these assumptions and Boyle's law, it is easy to show that one breath at depth would be equivalent to $$N_{ATM}$$ breaths at sea level.
I am not sure about what steps to take next. I need to figure out how much my tank's PSI will change for a breath, given the volume of 1 breath, the ambient pressure, and the tank's volume. I also need to show the PSI change per breath is independent of the PSI of the cylinder. I am rusty on my thermodynamics and looking for advice on how to derive/prove these two quantities.
• You are mentioning too many units for pressure in ways that they contradict your equations. Commented Jun 29, 2023 at 21:13
• @Themis can you clarify a bit how the different units are contradicting the equations? I realized I am using PSI and atmosphere which I agree is probably a bit confusing. I am going to edit to more clearly show that I assume the pressure at sea-level is 1 atmosphere Commented Jul 7, 2023 at 20:30
• I meant that you are referring to PSI, atm and bar at the same time. Commented Jul 9, 2023 at 19:07
I need to figure out how much my tank's PSI will change for a breath, given the volume of 1 breath, the ambient pressure, and the tank's volume.
Consider applying the ideal gas law both in the tank and in the lungs, with the connection being the amount of gas $$\Delta n$$ transferred from the former to the latter:
$$\text{Tank, constant volume and temperature:}\;\Delta P_\text{tank}V_\text{tank}=\Delta nRT;$$
$$\text{Lungs, constant pressure and temperature:}\;P_\text{ambient}\Delta V_\text{breath}=\Delta nRT;$$
$$\Delta P_\text{tank}=\frac{P_\text{ambient}\Delta V_\text{breath}}{V_\text{tank}}.$$
• Hi Thank you for the answer it is very helpful. My only question is : is the ideal gas law still applicable at pressures in the thousands of PSI? Typically a scuba tank is used from 3000psi to around 1000 psi Commented Jul 12, 2023 at 23:06
• The ideal gas law can be safety applied if the compressibility factor is suitably close to 1 at the conditions of interest. Commented Jul 12, 2023 at 23:20
• Awesome, thank you so much for the answer you really clarified these concepts for me Commented Jul 15, 2023 at 11:02 | 862 | 3,338 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.931755 |
http://physics.stackexchange.com/questions/45922/action-for-a-point-particle-in-a-curved-spacetime?answertab=oldest | 1,462,266,374,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121090.75/warc/CC-MAIN-20160428161521-00052-ip-10-239-7-51.ec2.internal.warc.gz | 228,284,474 | 16,470 | # Action for a point particle in a curved spacetime
Is this action for a point particle in a curved spacetime correct? $$\mathcal S =-Mc \int ds = -Mc \int_{\xi_0}^{\xi_1}\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$$
-
Yes, but if you're trying to derive the geodesic equation, you lose nothing but headaches by substituting this action with $S = \int ds (g_{ab}{\dot x}^{a}{\dot x}^{b})$, where ${\dot x}^{a} = \frac{\partial x^{a}}{\partial s}$, since a minimum of $\int f(x)$ is also going to be a minimum of $\int (f(x))^{2}$ – Jerry Schirmer Dec 4 '12 at 21:32
The end points of your action must be events, therefore must be an d+1 dimensional object. Action must be extremized over all paths that start and end at the given space time points.
A path can be be parametrized by 4 functions of space and time, $x^\mu(\xi)$ of a one parameter object $\xi$. So it would be wrong to label the end points in terms of $\xi$. Instead $\xi$ must be treated as intermediate label to describe paths. Otherwise the functional form the integral is correct.
-
It depends what you mean; that is the action of a test particle in a background gravitational field given by a metric $g_{\mu\nu}$. If you minimize it, you will get the geodesic equation. That is NOT the dynamical action for the gravitational field; your test particle does not change the curvature of the background spacetime. The action for the gravitational field the Einstein-Hilbert one,
$S=\frac{1}{\kappa}\int RdV$
where $R$ is the scalar curvature, $\kappa$ is the coupling constant.
-
It is misleading to write $\mbox d^4 V$. That makes it seem like it is integrating over a $4(4)=16$ dimensional manifold. – centralcharge Jul 17 '13 at 14:07
You are 100% correct. I will edit that. – levitopher Jul 17 '13 at 20:29
Is this action for a point particle in a curved spacetime correct? $$\mathcal S =-Mc \int ds = -Mc \int_{\xi_0}^{\xi_1}\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$$
Yes, it is.
The second expression/term/whatever comes from trivially integrating the energy $\mathrm{d}t$.
You can also try deriving the geodesic equation from this, as a fun homework exercise.
-
## protected by Qmechanic♦Jan 4 '13 at 11:07
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https://support.minitab.com/en-us/minitab/19/help-and-how-to/modeling-statistics/anova/how-to/analysis-of-means/perform-the-analysis/enter-your-data/ | 1,582,888,754,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00441.warc.gz | 565,599,868 | 5,057 | Enter your data for Analysis of Means
Stat > ANOVA > Analysis of Means
Select the option that best describes your data.
Enter normally distributed data
1. In Response, enter the numeric column of data that you want to analyze. Normally distributed data are typically measurement data, such as weight. With normally distributed data, Minitab compares the mean of each group to the overall mean.
2. Under Distribution of Data, select Normal.
3. In Factor 1, enter the column that contains the levels for the first factor. If you enter a single factor, Minitab produces a single plot showing the means for each level of the factor.
4. (Optional) In Factor 2, enter a column that contains the levels for the second factor. If you enter two factors, Minitab produces an interaction plot and a main effects plot for each factor.
In this worksheet, Density is the response and contains density measurements. Minutes and Strength are factors 1 and 2 and may explain differences in the density measurements.
C1 C2 C3
Density Minutes Strength
0 10 1
5 15 1
2 18 2
4 10 2
Enter binomial data
1. In Response, enter the column that contains the counts of events in each sample, such as the number of defectives. With binomial data, Minitab compares the proportion of each sample to the overall proportion.
2. Under Distribution of Data, select Binomial.
3. In Sample size, enter the number of observations contained in each sample. Each sample must have the same number of observations. The sample size must be large enough to ensure that the normal distribution adequately approximates the binomial distribution because the decision limits are based on the normal distribution. The normal distribution is adequate when np > 5 and n(1 − p) > 5, where n is the sample size and p is the proportion of events.
In this worksheet, Pipes is the response. Each row represents the number of defective pipes counted in samples of 100 pipes. For example, inspectors record 1 defective pipe in the first sample and 6 defective pipes in the second sample
C1
Pipes
1
6
3
9 | 459 | 2,052 | {"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-2020-10 | latest | en | 0.887632 |
https://www.jiskha.com/display.cgi?id=1355960119 | 1,503,162,927,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105700.94/warc/CC-MAIN-20170819162833-20170819182833-00319.warc.gz | 919,301,422 | 4,223 | # math//algebra
posted by .
Eye color is determined by genetic combination. Let R represent the gene for brown eyes and l represent the gene for blue eyes. Any gene combination including R results in brown eyes. Consider the offspring of a parent with a homozygous brown-eyed, RR, and a parent with a heterozygous brown-eyed, Rl, gene combination.
Show how you could use a polynomial to model the possible genetic combinations of the offspring.
What percent of the possible genetic combinations result in brown eyed offspring?******Based on a Punnett square, I think this is 75%, but not sure
What percent of the possible genetic combinations are carriers for the blue eyed gene?****** Also based on the square i think this is 50%, but still not sure
• math//algebra -
All offspring will have the dominant R gene and brown phenotype, since that is all the RR parent can give. Did you actually use a Punnett square?
But you are right with half as carriers.
• math//algebra -
She needs the polynomial... it isn't hard to figure out the 50%
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Why do I have brown eyes? The genes we inherit from our parents determine things like our height, looks, hair color and eye color. This passing of characteristics from parent to child is called heredity. If your mother has brown eyes,
2. ### Statistics
For the gene pair that determines eye color, each gene trait may be one of two types: dominant brown (B) or recessive blue (b). A person possessing the gene pair BB or Bb has brown eyes, whereas the pair bb produces blue eyes. a) Suppose …
3. ### Genetics
A recessive eye color mutation (se) in fruit flies leading to a color called "sepia" is determined by an autosomal gene. Another recessive mutation called "brown" (bw) causes brown eye color, and is determined by an autosomal gene …
4. ### math
Eye color is determined by genetic combination. Let R represent the gene for brown eyes and l represent the gene for blue eyes. Any gene combination including R results in brown eyes. Consider the offspring of a parent with a homozygous …
5. ### algebra
Eye color is determined by genetic combination. Let R represent the gene for brown eyes and l represent the gene for blue eyes. Any gene combination including R results in brown eyes. Consider the offspring of a parent with a homozygous …
6. ### Algebra
Eye color is determined by genetic combination. Let R represent the gene for brown eyes and l represent the gene for blue eyes. Any gene combination including R represents browns eyes. Consider the offspring of a parents with homozygous …
7. ### Genetics
Percentage phenotype in eye color in humans In humans, brown eyes are dominant to blue eyes. If a blue-eyed man marries a brown-eyed woman whose mother has blue eyes, what percentage of the brown-eyed offspring do you expect will be …
8. ### Biology
In humans, the allele for brown eyes is dominant to the allele for blue eyes. If a man with blue eyes and a woman with brown eyes who is homozygous for eye color have children, what is the probability of having a child with blue eyes?
9. ### algebra
Eye color is determined by genetic combination. Let R represent the gene for brown eyes and L represent the gene for blue eyes. Any gene combination including R results in brown eyes. Consider the offspring of a parent with homozygous …
10. ### algebra
Will someone please check my answer: If R is the gene for brown eyes,and L is the gene for blue eyes, consider the children of a parent with a homozygous brown-eyed,RR,gene combination and a parent with a homozygous brown-eyed,RL,gene …
More Similar Questions | 797 | 3,626 | {"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} | 3.390625 | 3 | CC-MAIN-2017-34 | latest | en | 0.905833 |
https://www.numbersaplenty.com/3504 | 1,713,757,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00190.warc.gz | 828,254,277 | 3,325 | Search a number
3504 = 24373
BaseRepresentation
bin110110110000
311210210
4312300
5103004
624120
713134
oct6660
94723
103504
1126a6
122040
131797
1413c4
151089
hexdb0
3504 has 20 divisors (see below), whose sum is σ = 9176. Its totient is φ = 1152.
The previous prime is 3499. The next prime is 3511. The reversal of 3504 is 4053.
3504 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (12).
It is a nialpdrome in base 8 and base 16.
It is a self number, because there is not a number n which added to its sum of digits gives 3504.
It is a congruent number.
It is an unprimeable number.
3504 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 12 + ... + 84.
23504 is an apocalyptic number.
3504 is the 32-nd nonagonal number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 3504, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (4588).
3504 is an abundant number, since it is smaller than the sum of its proper divisors (5672).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3504 is a wasteful number, since it uses less digits than its factorization.
3504 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 84 (or 78 counting only the distinct ones).
The product of its (nonzero) digits is 60, while the sum is 12.
The square root of 3504 is about 59.1945943478. The cubic root of 3504 is about 15.1887266363.
3504 divided by its sum of digits (12) gives a palindrome (292).
Adding to 3504 its reverse (4053), we get a palindrome (7557).
The spelling of 3504 in words is "three thousand, five hundred four". | 572 | 2,019 | {"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} | 3.515625 | 4 | CC-MAIN-2024-18 | latest | en | 0.90536 |
https://www.arduino.cc/reference/en/language/structure/control-structure/if/ | 1,718,983,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00011.warc.gz | 575,801,426 | 5,842 | Change language
# if
[Control Structure]
### Description
The `if` statement checks for a condition and executes the following statement or set of statements if the condition is 'true'.
### Syntax
``````if (condition) {
//statement(s)
}``````
### Parameters
`condition`: a boolean expression (i.e., can be `true` or `false`).
### Example Code
The brackets may be omitted after an if statement. If this is done, the next line (defined by the semicolon) becomes the only conditional statement.
``````if (x > 120) digitalWrite(LEDpin, HIGH);
if (x > 120)
digitalWrite(LEDpin, HIGH);
if (x > 120) {digitalWrite(LEDpin, HIGH);}
if (x > 120) {
digitalWrite(LEDpin1, HIGH);
digitalWrite(LEDpin2, HIGH);
}
// all are correct``````
### Notes and Warnings
The statements being evaluated inside the parentheses require the use of one or more operators shown below.
Comparison Operators:
```x == y (x is equal to y)
x != y (x is not equal to y)
x < y (x is less than y)
x > y (x is greater than y)
x <= y (x is less than or equal to y)
x >= y (x is greater than or equal to y)```
Beware of accidentally using the single equal sign (e.g. `if (x = 10)` ). The single equal sign is the assignment operator, and sets `x` to 10 (puts the value 10 into the variable `x`). Instead use the double equal sign (e.g. `if (x == 10)` ), which is the comparison operator, and tests whether `x` is equal to 10 or not. The latter statement is only true if `x` equals 10, but the former statement will always be true.
This is because C++ evaluates the statement `if (x=10)` as follows: 10 is assigned to `x` (remember that the single equal sign is the (assignment operator)), so `x` now contains 10. Then the 'if' conditional evaluates 10, which always evaluates to `TRUE`, since any non-zero number evaluates to TRUE. Consequently, `if (x = 10)` will always evaluate to `TRUE`, which is not the desired result when using an 'if' statement. Additionally, the variable `x` will be set to 10, which is also not a desired action. | 533 | 2,018 | {"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-2024-26 | latest | en | 0.825088 |
https://www.hiveworkshop.com/threads/bounce-angle-formula.94686/ | 1,685,860,625,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649518.12/warc/CC-MAIN-20230604061300-20230604091300-00761.warc.gz | 884,348,121 | 20,903 | # [Trigger]Bounce Angle Formula
#### lnfernal
Level 8
Okey. I tried with this formula but didn't work. I'm trying to make a bounce angle.
• If (All Conditions are True) then do (Then Actions) else do (Else Actions)
• If - Conditions
• (Facing of Unit1[(Integer A)]) Greater than or equal to 180.00
• Then - Actions
• Set RM_Angle[(Integer A)] = (540.00 - (Facing of Unit1[(Integer A)]))
• Else - Actions
• Set RM_Angle[(Integer A)] = (360.00 - (Facing of Unit1[(Integer A)]))
This trigger works great. Except in some cases, it can't check if the wall is either south or north, west or right. Example with this code, let's say the red line is 45 degrees, which means it's reflection should be 315 degrees. And in this case it works great.
But when I use this formula and try to bounce to the south direction, it went like this.
The red line is still 45 degrees, but this time it also reflect 315 degrees, right in the wall, the black line. But it should have bounced 135 degrees, the white line. I can solve this by doing 180-Facing of unit, then the south wall will work, but then the right wall will be messed up. So I need to find a way to somehow check where I hit.
Level 12
1stly, you shouldn't be doing this with angles but rather X,Y that way all you have to do is set one of them to: Itself x -1
#### lnfernal
Level 8
Don't know how to use x/y. Is that the only way and will it work? Can I post the code and ask for someone to do that? That's JASS right?
#### PurplePoot
Level 40
~Moved to Triggers & Scripts.
#### WildField
Level 3
360 - facing(for east/west) , 180 - facing (for north/south)
#### lnfernal
Level 8
That's what I'm doing right now.. Read the problem.
If you shoot a ball at 45 degrees, it can either bounce 315 degrees OR 135 degrees depending which side you throw it of the wall.. My formula only support 315 degrees. And I can't seem to find a way to make it support both 135 and 315 degrees depending where the wall is.
Level 17
#### lnfernal
Level 8
I've read the whole thread. And their solution seem to be 180-a. Isn't that what my trigger is?
#### PurplePoot
Level 40
If you take that approach, it's 180-a relative to the surface, and the relative to the surface bit is exactly the problem you're having right now.
I've never really done any stuff with physics, but I'll see if I can come up with something.
#### lnfernal
Level 8
Don't worry. Someone found a way. But he said it may bug so he didn't realease it as a tutorial (Didn't say what bug).
#### GhostWolf
Level 29
You simply put the negative value of the angle, or you do what Poot said.
Both of them don't work for 2 angles:
-a - 0 and 180
180-a - 90 and 270
So, whichever you use, remember to check first if it is not one of the angles that don't work.
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439 | 804 | 2,894 | {"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-2023-23 | latest | en | 0.91977 |
https://www.physicsforums.com/threads/sum-of-all-moments-problem.151804/ | 1,576,408,442,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00007.warc.gz | 815,549,690 | 15,625 | # Sum of all moments problem
If the sum of all moments about a point A of a body is 0, will it then be 0 anywhere else on the body? (I'm working in 2-D)
Related Advanced Physics Homework Help News on Phys.org
russ_watters
Mentor
Maybe. It depends entirely on the particulars of the problem.
Hootenanny
Staff Emeritus
Gold Member
Not necessarily.
So I don't necesarrily have equilibrium if I find that F(res) = 0 and that M(A) = 0 ? What more do I need to `show eq.?
I assume that it works the other way around: if I know that there's equilibrium, then the sum of all moments around any point is 0.
Hootenanny
Staff Emeritus | 170 | 629 | {"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-2019-51 | latest | en | 0.87149 |
https://spectre-code.org/namespacestrahlkorpertags | 1,550,808,510,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513222.88/warc/CC-MAIN-20190222033812-20190222055812-00156.warc.gz | 689,247,883 | 4,847 | StrahlkorperTags Namespace Reference
Holds tags and ComputeItems associated with a Strahlkorper. More...
## Namespaces
aliases
Defines type aliases used in Strahlkorper-related Tags.
## Classes
struct CartesianCoords
CartesianCoords(i) is $x_{\rm surf}^i$, the vector of $(x,y,z)$ coordinates of each point on the surface. More...
D2xRadius(i,j) is $\partial^2 r_{\rm surf}/\partial x^i\partial x^j$. Here $r_{\rm surf}=r_{\rm surf}(\theta,\phi)$ is the function describing the surface, which is considered a function of Cartesian coordinates $r_{\rm surf}=r_{\rm surf}(\theta(x,y,z),\phi(x,y,z))$ for this operation. More...
DxRadius(i) is $\partial r_{\rm surf}/\partial x^i$. Here $r_{\rm surf}=r_{\rm surf}(\theta,\phi)$ is the function describing the surface, which is considered a function of Cartesian coordinates $r_{\rm surf}=r_{\rm surf}(\theta(x,y,z),\phi(x,y,z))$ for this operation. More...
struct InvHessian
InvHessian(k,i,j) is $\partial (J^{-1}){}^k_j/\partial x^i$, where $(J^{-1}){}^k_j$ is the inverse Jacobian. InvHessian is not symmetric because the Jacobians are Pfaffian. InvHessian doesn't depend on the shape of the surface. More...
struct InvJacobian
InvJacobian(0,i) is $r\partial\theta/\partial x^i$, and InvJacobian(1,i) is $r\sin\theta\partial\phi/\partial x^i$. Here $r$ means $\sqrt{x^2+y^2+z^2}$. InvJacobian doesn't depend on the shape of the surface. More...
struct Jacobian
Jacobian(i,0) is $\frac{1}{r}\partial x^i/\partial\theta$, and Jacobian(i,1) is $\frac{1}{r\sin\theta}\partial x^i/\partial\phi$. Here $r$ means $\sqrt{x^2+y^2+z^2}$. Jacobian doesn't depend on the shape of the surface. More...
$\nabla^2 r_{\rm surf}$, the flat Laplacian of the surface. This is $\eta^{ij}\partial^2 r_{\rm surf}/\partial x^i\partial x^j$, where $r_{\rm surf}=r_{\rm surf}(\theta(x,y,z),\phi(x,y,z))$. More...
struct NormalOneForm
NormalOneForm(i) is $s_i$, the (unnormalized) normal one-form to the surface, expressed in Cartesian components. This is computed by $x_i/r-\partial r_{\rm surf}/\partial x^i$, where $x_i/r$ is Rhat and $\partial r_{\rm surf}/\partial x^i$ is DxRadius. See Eq. (8) of [2]. Note on the word "normal": $s_i$ points in the correct direction (it is "normal" to the surface), but it does not have unit length (it is not "normalized"; normalization requires a metric). More...
(Euclidean) distance $r_{\rm surf}(\theta,\phi)$ from the center to each point of the surface. More...
struct Rhat
Rhat(i) is $\hat{r}^i = x_i/\sqrt{x^2+y^2+z^2}$ on the grid. Doesn't depend on the shape of the surface. More...
struct Strahlkorper
Tag referring to a Strahlkorper More...
struct Tangents
Tangents(i,j) is $\partial x_{\rm surf}^i/\partial q^j$, where $x_{\rm surf}^i$ are the Cartesian coordinates of the surface (i.e. CartesianCoords) and are considered functions of $(\theta,\phi)$. More...
struct ThetaPhi
$(\theta,\phi)$ on the grid. Doesn't depend on the shape of the surface. More...
## Typedefs
template<typename Frame >
using items_tags = tmpl::list< Strahlkorper< Frame > >
template<typename Frame >
using compute_items_tags = tmpl::list< ThetaPhi< Frame >, Rhat< Frame >, Jacobian< Frame >, InvJacobian< Frame >, InvHessian< Frame >, Radius< Frame >, CartesianCoords< Frame >, DxRadius< Frame >, D2xRadius< Frame >, LaplacianRadius< Frame >, NormalOneForm< Frame >, Tangents< Frame > >
## Detailed Description
Holds tags and ComputeItems associated with a Strahlkorper. | 1,067 | 3,455 | {"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} | 2.5625 | 3 | CC-MAIN-2019-09 | latest | en | 0.66397 |
https://www.staging.electrical4u.com/wien-bridge-oscillator/ | 1,638,164,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00324.warc.gz | 1,003,886,413 | 15,590 | # Wien Bridge Oscillator: Circuit And Frequency Calculation
## What is a Wien Bridge Oscillator?
A Wien-Bridge Oscillator is a type of phase-shift oscillator which is based upon a Wien-Bridge network (Figure 1a) comprising of four arms connected in a bridge fashion. Here two arms are purely resistive while the other two arms are a combination of resistors and capacitors.
In particular, one arm has resistor and capacitor connected in series (R1 and C1) while the other has them in parallel (R2 and C2).
This indicates that these two arms of the network behave identical to that of high pass filter or low pass filter, mimicking the behavior of the circuit shown by Figure 1b.
In this circuit, at high frequencies, the reactance of the capacitors C1 and C2 will be much less due to which the voltage V0 will become zero as R2 will be shorted.
Next, at low frequencies, the reactance of the capacitors C1 and C2 will become very high.
However even in this case, the output voltage V0 will remain at zero only, as the capacitor C1 would be acting as an open circuit.
This kind of behavior exhibited by the Wien-Bridge network makes it a lead-lag circuit in the case of low and high frequencies, respectively.
## Wien Bridge Oscillator Frequency Calculation
Nevertheless, amidst these two high and low frequencies, there exists a particular frequency at which the values of the resistance and the capacitive reactance will become equal to each other, producing the maximum output voltage.
This frequency is referred to as resonant frequency. The resonant frequency for a Wein Bridge Oscillator is calculated using the following formula:
Further, at this frequency, the phase-shift between the input and the output will become zero and the magnitude of the output voltage will become equal to one-third of the input value. In addition, it is seen that the Wien-Bridge will be balanced only at this particular frequency.
In the case of Wien-Bridge oscillator, the Wien-Bridge network of Figure 1 will be used in the feedback path as shown in Figure 2. The circuit diagram for a Wein Oscillator using a BJT (Bipolar Junction Transistor) is shown below:
In these oscillators, the amplifier section will comprise of two-stage amplifier formed by the transistors, Q1 and Q2, wherein the output of Q2 is back-fed as an input to Q1 via Wien-Bridge network (shown within the blue enclosure in the figure).
Here, the noise inherent in the circuit will cause a change in the base current of Q1 which will appear at its collector point after being amplified with a phase-shift of 180o.
This is fed as an input to Q2 via C4 and gets further amplified and appears with an additional phase-shift of 180o.
This makes the net phase-difference of the signal fed back to the Wien-Bridge network to be 360o, satisfying phase-shift criterion to obtain sustained oscillations.
However, this condition will be satisfied only in the case of resonant frequency, due to which the Wien-Bridge oscillators will be highly selective in terms of frequency, leading to a frequency-stabilized design.
Wien-bridge oscillators can even be designed using Op-Amps as a part of their amplifier section, as shown by Figure 3.
However it is to be noted that, here, the Op-Amp is required to act as a non-inverting amplifier as the Wien-Bridge network offers zero phase-shift.
Further, from the circuit, it is evident that the output voltage is fed back to both inverting and non-inverting input terminals.
At resonant frequency, the voltages applied to the inverting and non-inverting terminals will be equal and in-phase with each other.
However, even here, the voltage gain of the amplifier needs to be greater than 3 to start oscillations and equal to 3 to sustain them. In general, these kind of Op-Amp-based Wien Bridge Oscillators cannot operate above 1 MHz due to the limitations imposed on them by their open-loop gain.
Wien-Bridge networks are low frequency oscillators which are used to generate audio and sub-audio frequencies ranging between 20 Hz to 20 KHz.
Further, they provide stabilized, low distorted sinusoidal output over a wide range of frequency which can be selected using decade resistance boxes.
In addition, the oscillation frequency in this kind of circuit can be varied quite easily as it just needs variation of the capacitors C1 and C2.
However these oscillators require large number of circuit components and can be operated upto a certain maximum frequency only.
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1. hi! | 1,012 | 4,733 | {"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} | 3.484375 | 3 | CC-MAIN-2021-49 | latest | en | 0.932827 |
http://forums.wolfram.com/mathgroup/archive/1999/Oct/msg00083.html | 1,713,287,200,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00769.warc.gz | 12,060,253 | 7,890 | The 196-Algorithm
• To: mathgroup at smc.vnet.net
• Subject: [mg20210] The 196-Algorithm
• From: Hans Havermann <haver at total.net>
• Date: Wed, 6 Oct 1999 21:06:29 -0400
• Sender: owner-wri-mathgroup at wolfram.com
```http://www.treasure-troves.com/math/196-Algorithm.html
Some 10 years ago, John Walker looped this algorithm (for n=196) 2415836
times to obtain a number containing one million digits. Five years later,
Tim Irvin extended the computation to obtain a number containing two
million digits.
I was wondering how well today's Mathematica fared with yesterday's
mainframes. Specifically, can we loop the algorithm 2415836 times in
reasonable time?
q[n_] := (i++; r = FromDigits[Reverse[IntegerDigits[n]]];
If[r != n, r + n, n])
i = 0; s = NestWhile[q, 89, Unequal, 2]; {i, s}
{25, 8813200023188}
NestWhile is a Mathematica 4 addition:
?NestWhile
NestWhile[f, expr, test] starts with expr, then repeatedly applies f until
applying test to the result no longer yields True. NestWhile[f, expr, test,
m] supplies the most recent m results as arguments for test at each step...
i = 0; s = NestWhile[q, 196, Unequal, 2]; {i, s}
Interrupting the evaluation and entering the inspector dialog...
i
Log[r] // N
10553
10059.9
Ignoring the matter of even comparing successive terms to see if, in fact,
we have reached a palindrome, we have, more simply...
p[n_] := n + FromDigits[Reverse[IntegerDigits[n]]]
s = Nest[p, 196, 10000]; // Timing
{253.6 Second, Null}
s = Nest[p, 196, 100000]; // Timing
{55416.7 Second, Null}
So it doesn't appear I can reach several million iterations in good time.
Any possibility of speeding this up?
```
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https://snyk.io/advisor/python/thewalrus/example | 1,686,094,065,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00782.warc.gz | 586,508,962 | 32,951 | How to use thewalrus - 10 common examples
To help you get started, we’ve selected a few thewalrus examples, based on popular ways it is used in public projects.
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XanaduAI / thewalrus / thewalrus / quantum.py View on Github
``````if not is_pure_cov(cov, hbar=hbar, rtol=1e-05, atol=1e-08):
raise ValueError("The covariance matrix does not correspond to a pure state")
rpt = i
beta = Beta(mu, hbar=hbar)
Q = Qmat(cov, hbar=hbar)
A = Amat(cov, hbar=hbar)
(n, _) = cov.shape
N = n // 2
B = A[0:N, 0:N].conj()
alpha = beta[0:N]
if np.linalg.norm(alpha) < tol:
# no displacement
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
haf = hafnian(B_rpt)
else:
haf = hafnian_repeated(B, rpt)
else:
gamma = alpha - B @ np.conj(alpha)
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
np.fill_diagonal(B_rpt, reduction(gamma, rpt))
haf = hafnian(B_rpt, loop=True)
else:
haf = hafnian_repeated(B, rpt, mu=gamma, loop=True)
if include_prefactor:
pref = np.exp(-0.5 * (np.linalg.norm(alpha) ** 2 - alpha @ B @ alpha))
haf *= pref``````
XanaduAI / thewalrus / thewalrus / quantum.py View on Github
``````N = n // 2
B = A[0:N, 0:N].conj()
alpha = beta[0:N]
if np.linalg.norm(alpha) < tol:
# no displacement
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
haf = hafnian(B_rpt)
else:
haf = hafnian_repeated(B, rpt)
else:
gamma = alpha - B @ np.conj(alpha)
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
np.fill_diagonal(B_rpt, reduction(gamma, rpt))
haf = hafnian(B_rpt, loop=True)
else:
haf = hafnian_repeated(B, rpt, mu=gamma, loop=True)
if include_prefactor:
pref = np.exp(-0.5 * (np.linalg.norm(alpha) ** 2 - alpha @ B @ alpha))
haf *= pref
return haf / np.sqrt(np.prod(fac(rpt)) * np.sqrt(np.linalg.det(Q)))``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````n1, n2 = cov.shape
if n1 != n2:
raise ValueError("Covariance matrix must be square.")
nmodes = n1 // 2
prev_prob = 1.0
mu = np.zeros(n1)
for k in range(nmodes):
probs1 = np.zeros([2], dtype=np.float64)
kk = np.arange(k + 1)
_, V_red = reduced_gaussian(mu, cov, kk)
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
O = Xmat(k + 1) @ A
indices = result + [0]
ind2 = indices + indices
probs1[0] = tor(np.complex128(reduction(O, ind2))).real
indices = result + [1]
ind2 = indices + indices
pref = np.sqrt(np.linalg.det(Q).real)
probs1a = probs1 / pref
probs2 = probs1a / prev_prob
probs2[1] = 1.0 - probs2[0]
probs1a[1] = probs2[1] * prev_prob
probs3 = np.maximum(``````
XanaduAI / thewalrus / thewalrus / operations.py View on Github
``````(array): Tensor containing the Fock representation of the Gaussian unitary
"""
# Check the matrix is symplectic
if check_symplectic:
if not is_symplectic(S, rtol=rtol, atol=atol):
raise ValueError("The matrix S is not symplectic")
# And that S and alpha have compatible dimensions
l, _ = S.shape
if l // 2 != len(alpha):
raise ValueError("The matrix S and the vector alpha do not have compatible dimensions")
# Construct its Choi expansion and then the covariance matrix and A matrix of such pure state
S_exp = choi_expand(S, r)
cov = S_exp @ S_exp.T
A = Amat(cov)
# Because the state is pure then A = B \oplus B^*. We now extract B^* and follow the procedure
# described in the paper cited above.
n, _ = A.shape
N = n // 2
B = A[0:N, 0:N].conj()
# Now we need to figure out the loops (cf. Eq. 111 of the reference above)
l = len(alpha)
alphat = np.array(list(alpha) + ([0] * l))
zeta = alphat - B @ alphat.conj()
# Finally, there are the prefactors (cf. Eq. 113 of the reference above).
# Note that the factorials that are not included here from Eq. 113 are calculated
# internally by hafnian_batched when the argument renorm is set to True
pref_exp = -0.5 * alphat.conj() @ zeta``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````n1, n2 = cov.shape
if n1 != n2:
raise ValueError("Covariance matrix must be square.")
nmodes = n1 // 2
prev_prob = 1.0
mu = np.zeros(n1)
for k in range(nmodes):
probs1 = np.zeros([2], dtype=np.float64)
kk = np.arange(k + 1)
_, V_red = reduced_gaussian(mu, cov, kk)
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
O = Xmat(k + 1) @ A
indices = result + [0]
ind2 = indices + indices
probs1[0] = tor(np.complex128(reduction(O, ind2))).real
indices = result + [1]
ind2 = indices + indices
pref = np.sqrt(np.linalg.det(Q).real)
probs1a = probs1 / pref
probs2 = probs1a / prev_prob
probs2[1] = 1.0 - probs2[0]
probs1a[1] = probs2[1] * prev_prob
probs3 = np.maximum(``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````approx (bool): if ``True``, the approximate hafnian algorithm is used.
Note that this can only be used for real, non-negative matrices.
approx_samples: the number of samples used to approximate the hafnian if ``approx=True``.
Returns:
np.array[int]: a photon number sample from the Gaussian states.
"""
N = len(cov) // 2
result = []
prev_prob = 1.0
nmodes = N
if mean is None:
local_mu = np.zeros(2 * N)
else:
local_mu = mean
A = Amat(Qmat(cov), hbar=hbar)
for k in range(nmodes):
probs1 = np.zeros([cutoff + 1], dtype=np.float64)
kk = np.arange(k + 1)
mu_red, V_red = reduced_gaussian(local_mu, cov, kk)
if approx:
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
for i in range(cutoff):
indices = result + [i]
ind2 = indices + indices
if approx:
factpref = np.prod(fac(indices))
mat = reduction(A, ind2)``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````approx (bool): if ``True``, the approximate hafnian algorithm is used.
Note that this can only be used for real, non-negative matrices.
approx_samples: the number of samples used to approximate the hafnian if ``approx=True``.
Returns:
np.array[int]: a photon number sample from the Gaussian states.
"""
N = len(cov) // 2
result = []
prev_prob = 1.0
nmodes = N
if mean is None:
local_mu = np.zeros(2 * N)
else:
local_mu = mean
A = Amat(Qmat(cov), hbar=hbar)
for k in range(nmodes):
probs1 = np.zeros([cutoff + 1], dtype=np.float64)
kk = np.arange(k + 1)
mu_red, V_red = reduced_gaussian(local_mu, cov, kk)
if approx:
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
for i in range(cutoff):
indices = result + [i]
ind2 = indices + indices
if approx:
factpref = np.prod(fac(indices))
mat = reduction(A, ind2)``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````result = []
n1, n2 = cov.shape
if n1 != n2:
raise ValueError("Covariance matrix must be square.")
nmodes = n1 // 2
prev_prob = 1.0
mu = np.zeros(n1)
for k in range(nmodes):
probs1 = np.zeros([2], dtype=np.float64)
kk = np.arange(k + 1)
_, V_red = reduced_gaussian(mu, cov, kk)
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
O = Xmat(k + 1) @ A
indices = result + [0]
ind2 = indices + indices
probs1[0] = tor(np.complex128(reduction(O, ind2))).real
indices = result + [1]
ind2 = indices + indices
pref = np.sqrt(np.linalg.det(Q).real)
probs1a = probs1 / pref
probs2 = probs1a / prev_prob
probs2[1] = 1.0 - probs2[0]
probs1a[1] = probs2[1] * prev_prob``````
XanaduAI / thewalrus / thewalrus / samples.py View on Github
``````result = []
n1, n2 = cov.shape
if n1 != n2:
raise ValueError("Covariance matrix must be square.")
nmodes = n1 // 2
prev_prob = 1.0
mu = np.zeros(n1)
for k in range(nmodes):
probs1 = np.zeros([2], dtype=np.float64)
kk = np.arange(k + 1)
_, V_red = reduced_gaussian(mu, cov, kk)
Q = Qmat(V_red, hbar=hbar)
A = Amat(Q, hbar=hbar, cov_is_qmat=True)
O = Xmat(k + 1) @ A
indices = result + [0]
ind2 = indices + indices
probs1[0] = tor(np.complex128(reduction(O, ind2))).real
indices = result + [1]
ind2 = indices + indices
pref = np.sqrt(np.linalg.det(Q).real)
probs1a = probs1 / pref
probs2 = probs1a / prev_prob
probs2[1] = 1.0 - probs2[0]
probs1a[1] = probs2[1] * prev_prob``````
XanaduAI / thewalrus / thewalrus / quantum.py View on Github
``````raise ValueError("The covariance matrix does not correspond to a pure state")
rpt = i
beta = Beta(mu, hbar=hbar)
Q = Qmat(cov, hbar=hbar)
A = Amat(cov, hbar=hbar)
(n, _) = cov.shape
N = n // 2
B = A[0:N, 0:N].conj()
alpha = beta[0:N]
if np.linalg.norm(alpha) < tol:
# no displacement
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
haf = hafnian(B_rpt)
else:
haf = hafnian_repeated(B, rpt)
else:
gamma = alpha - B @ np.conj(alpha)
if np.prod([k + 1 for k in rpt]) ** (1 / len(rpt)) < 3:
B_rpt = reduction(B, rpt)
np.fill_diagonal(B_rpt, reduction(gamma, rpt))
haf = hafnian(B_rpt, loop=True)
else:
haf = hafnian_repeated(B, rpt, mu=gamma, loop=True)
if include_prefactor:
pref = np.exp(-0.5 * (np.linalg.norm(alpha) ** 2 - alpha @ B @ alpha))
haf *= pref
return haf / np.sqrt(np.prod(fac(rpt)) * np.sqrt(np.linalg.det(Q)))``````
thewalrus
Open source library for hafnian calculation
Apache-2.0 | 3,072 | 9,087 | {"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} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.540938 |
https://www.exemplars.com/education-materials/math-k-12/units-ps21 | 1,563,822,706,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528208.76/warc/CC-MAIN-20190722180254-20190722202254-00535.warc.gz | 681,359,541 | 7,270 | Standards-based assessment and Instruction
# Units of Study
To assist educators in teaching a focused mathematics curriculum, Exemplars has grouped 21st century math concepts and skills into the following rich Units of Study for grades K–5:
### Kindergarten Units
• Counting and Cardinality
• Comparing Numbers
• Composing and Decomposing Numbers
• Geometry
• Data Analysis
• Number Concepts and Place Value
• Number Patterns
• Comparing Numbers
• Composing and Decomposing Numbers
• Strategies for Addition and Subtraction
• Money
• Equivalence
• Algebraic Reasoning
• Geometry
• Fractions
• Data Analysis
• Place Value
• Comparing Numbers
• Fractions
• Problem Solving With Money
• Multiplication and Division
• Algebraic Reasoning
• Geometry
• Composing and Decompsing Shapes
• Measurement
• Data Analysis
• Place Value
• Understanding Fractions
• Comparing Fractions
• Multiplication
• Division
• Relationship Between Multiplication and Division
• Algebraic Reasoning
• Geometry
• Measurement
• Data Analysis
• Whole Number and Decimal Place Value
• Multiplying and Dividing Whole Numbers
• Comparing Fractions
• Multiplying With Fractions
• Algebraic Reasoning
• Geometry
• Angle Measurement
• Measurement
• Data Analysis
• Decimal Place Value
• Operations With Whole Numbers
• Products and Quotients With Decimals
• Dividing With Fractions
• Algebraic Reasoning
• Geometry
• Measurement
• Coordinate Plane
• Data Analysis
Our teacher-friendly tasks are designed to support both the Common Core and Citywide instructional expectations. GO Math! alignments are also available.
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Explore our latest K-5 math material and begin using it in your classroom.
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## Here's What People Are Saying
This has been a great resource with the introduction of Common Core Math and ELA.
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Teacher
Exemplars | 271 Poker Hill Road | Underhill, Vermont 05489 | ph: 800-450-4050 | fax: 802-899-4825 | infoREMOVETHISBEFORESENDING@exemplars.com | 477 | 2,022 | {"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-2019-30 | latest | en | 0.721808 |
https://michaj.w.staszic.waw.pl/ulamki.pas | 1,675,617,858,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00623.warc.gz | 422,373,065 | 1,660 | //implementation of the simplest arithmetic on rational numbers //represented as: a b/c with sign. //author: Michal Pilipczuk //XIV Secondary School, Warsaw, Poland unit ulamki; interface type ulam=record znak:longint; //sign a:qword; b:qword; c:qword; end; //type of fraction procedure wypisz(o:ulam); //writes fraction function app(xu:ulam):ulam; //makes a fraction appropriate so by.a) then mna:=(x.ay.znak then mn:=(x.znak0) and (y.znak>0)) then add:=adda(x,y) else begin if x.znakpom1.b*x.c then begin dec(pom2.a); pom2.b:=(pom1.b*x.c+pom2.c); pom2.b:=(pom2.b-x.b*pom1.c); add:=app(pom2); end else begin pom2.b:=pom1.b*x.c-x.b*pom1.c; add:=app(pom2); end; end; end; end; end; function multi(e,f:ulam):ulam; var pomy,pomq2,pomq3:ulam; begin pomq2.znak:=1; pomq3.znak:=1; pomq2.a:=e.a*f.a; pomq2.c:=f.c; pomq2.b:=e.a*f.b; pomq2:=app(pomq2); pomq3.a:=0; pomq3.c:=e.c; pomq3.b:=f.a*e.b; pomq3:=app(pomq3); pomy:=adda(pomq2,pomq3); pomy.b:=pomy.b+e.b*f.b; pomy.znak:=e.znak*f.znak; multi:=pomy; end; function kwa(e:ulam):ulam; var pomq2:ulam; begin pomq2.znak:=1; pomq2.a:=e.a*e.a; pomq2.c:=e.c; pomq2.b:=2*(e.b*e.a); pomq2:=app(pomq2); pomq2.b:=pomq2.b*pomq2.c+e.b*e.b; pomq2.c:=pomq2.c*pomq2.c; kwa:=app(pomq2); end; end. | 515 | 1,223 | {"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} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.159635 |
https://jp.mathworks.com/matlabcentral/cody/problems/84-longest-divisor-run/solutions/734584 | 1,581,993,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00103.warc.gz | 451,738,426 | 15,869 | Cody
# Problem 84. Longest Divisor Run
Solution 734584
Submitted on 13 Sep 2015 by Jean-Yves Tinevez
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
%% a = [93 1147 473 259 629 1591 851 533 2021 86 817 2279 1763 961 1073 205] len_correct = 5; d_correct = 43; [len,d] = divisor_run(a); assert(isequal(len,len_correct) && isequal(d,d_correct))
a = Columns 1 through 14 93 1147 473 259 629 1591 851 533 2021 86 817 2279 1763 961 Columns 15 through 16 1073 205
2 Pass
%% a = [166 553 1241 869 237 395 1079 1577]; len_correct = 3; d_correct = 79; [len,d] = divisor_run(a); assert(isequal(len,len_correct) && isequal(d,d_correct))
3 Pass
%% a = [533 205 123 287 451 697 82]; len_correct = 7; d_correct = 41; [len,d] = divisor_run(a); assert(isequal(len,len_correct) && isequal(d,d_correct))
4 Pass
%% a = [3937 7493 1397 5633 6439 3151 5617 685 7261 5069 ... 959 3973 411 2603 2329 1781 262]; len_correct = 12; d_correct = 137; [len,d] = divisor_run(a); assert(isequal(len,len_correct) && isequal(d,d_correct)) | 418 | 1,137 | {"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-2020-10 | latest | en | 0.341556 |
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