url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://gmatclub.com/forum/a-lesson-in-economics-81263.html
| 1,505,880,331,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818686169.5/warc/CC-MAIN-20170920033426-20170920053426-00044.warc.gz
| 682,193,853
| 37,847
|
It is currently 19 Sep 2017, 21:05
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# A lesson in economics
Author Message
Senior Manager
Joined: 01 Mar 2009
Posts: 367
Kudos [?]: 96 [1], given: 24
Location: PDX
### Show Tags
22 Jul 2009, 22:45
1
KUDOS
Found this on craigslist - a good read
http://portland.craigslist.org/forums/? ... =131086963
_________________
In the land of the night, the chariot of the sun is drawn by the grateful dead
Kudos [?]: 96 [1], given: 24
Manager
Joined: 08 Jul 2009
Posts: 170
Kudos [?]: 28 [0], given: 26
Re: A lesson in economics [#permalink]
### Show Tags
05 Jan 2010, 15:55
This is funny.
Kudos [?]: 28 [0], given: 26
Intern
Joined: 01 Oct 2009
Posts: 9
Kudos [?]: [0], given: 0
Re: A lesson in economics [#permalink]
### Show Tags
07 Jan 2010, 09:02
What is this , exceptional ;
Kudos [?]: [0], given: 0
Re: A lesson in economics [#permalink] 07 Jan 2010, 09:02
Similar topics Replies Last post
Similar
Topics:
Amid Economic Carnage, Business Schools Mull Fixes 0 11 Mar 2009, 07:10
31 Dec 1969, 17:00
Economic situation in the US ?? Please answer.. 2 20 Mar 2013, 06:03
An Analysis of the Current Economic Situation 1 12 Sep 2011, 10:20
1 Economics Rap Battle 1 17 Jun 2011, 22:55
Display posts from previous: Sort by
| 548
| 1,799
|
{"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}
| 2.828125
| 3
|
CC-MAIN-2017-39
|
latest
|
en
| 0.777729
|
http://www.jiskha.com/display.cgi?id=1289578604
| 1,498,592,623,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00350.warc.gz
| 535,714,321
| 3,807
|
# Physics
posted by .
A tennis ball of mass= 0.058 kg and speed= 19 m/s strikes a wall at a 45 degree angle and rebounds with the same speed at 45 degrees.
1. What is the magnitude of the impulse given to the ball?
2. What is the direction of the impulse given to the ball?
• Physics -
Its quite simple... first you take the computer...and SMASH IT AGAINST YOUR FACE!!!
Then go to your mother and ask her WHY YOU ARE BORN!!!
| 109
| 430
|
{"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-2017-26
|
latest
|
en
| 0.867784
|
http://clay6.com/qa/14668/an-infinitely-long-straight-conductor-is-bent-into-the-shape-as-shown-below
| 1,516,325,945,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084887692.13/warc/CC-MAIN-20180119010338-20180119030338-00281.warc.gz
| 71,055,396
| 27,759
|
# An infinitely long straight conductor is bent into the shape as shown below. It carries a current of i Amps, and the circular loop is R meters. Then the magnitude of magnetic induction at the center of the circular loop is :
$(1)\; \frac{\mu_o i}{2 \pi R} \\(2)\;\frac{\mu_o n i}{2 R} \\ (3)\; \frac{\mu_o i}{2 \pi R}(\pi +1) \\(4)\; \frac{\mu_o i}{2 \pi R}(\pi-1)$
$(3)\; \frac{\mu_o i}{2 \pi R}(\pi +1)$
| 145
| 409
|
{"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-2018-05
|
longest
|
en
| 0.64516
|
https://hextobinary.com/unit/acceleration/from/inhmin/to/fts2
| 1,713,145,875,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00569.warc.gz
| 267,175,531
| 15,711
|
# Inch/Hour/Minute to Feet/Second Squared Converter
Acceleration
Inch/Hour/Minute
Feet/Second Squared
1 Inch/Hour/Minute = 3.858024691358e-7 Feet/Second Squared
## How many Feet/Second Squared are in a Inch/Hour/Minute?
The answer is one Inch/Hour/Minute is equal to 3.858024691358e-7 Feet/Second Squared and that means we can also write it as 1 Inch/Hour/Minute = 3.858024691358e-7 Feet/Second Squared. Feel free to use our online unit conversion calculator to convert the unit from Inch/Hour/Minute to Feet/Second Squared. Just simply enter value 1 in Inch/Hour/Minute and see the result in Feet/Second Squared.
## How to Convert Inch/Hour/Minute to Feet/Second Squared (in/h/min to ft/s2)
By using our Inch/Hour/Minute to Feet/Second Squared conversion tool, you know that one Inch/Hour/Minute is equivalent to 3.858024691358e-7 Feet/Second Squared. Hence, to convert Inch/Hour/Minute to Feet/Second Squared, we just need to multiply the number by 3.858024691358e-7. We are going to use very simple Inch/Hour/Minute to Feet/Second Squared conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Inch/Hour/Minute} = 1 \times 3.858024691358e-7 = \text{3.858024691358e-7 Feet/Second Squared}$$
## What is Inch/Hour/Minute Unit of Measure?
Inch/Hour/Minute or Inch per Hour per Minute is a unit of measurement for acceleration. If an object accelerates at the rate of 1 inch/hour/minute, that means its speed is increased by 1 inch per hour every minute.
## What is the symbol of Inch/Hour/Minute?
The symbol of Inch/Hour/Minute is in/h/min. This means you can also write one Inch/Hour/Minute as 1 in/h/min.
## What is Feet/Second Squared Unit of Measure?
Feet/Second Squared or Feet per Second Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 feet/second squared, that means its speed is increased by 1 feet per second every second.
## What is the symbol of Feet/Second Squared?
The symbol of Feet/Second Squared is ft/s2. This means you can also write one Feet/Second Squared as 1 ft/s2.
## Inch/Hour/Minute to Feet/Second Squared Conversion Table
Inch/Hour/Minute [in/h/min]Feet/Second Squared [ft/s2]
13.858024691358e-7
27.716049382716e-7
30.0000011574074074074
40.0000015432098765432
50.000001929012345679
60.0000023148148148148
70.0000027006172839506
80.0000030864197530864
90.0000034722222222222
100.000003858024691358
1000.00003858024691358
10000.0003858024691358
## Inch/Hour/Minute to Other Units Conversion Table
Inch/Hour/Minute [in/h/min]Output
1 inch/hour/minute in meter/second squared is equal to1.1759259259259e-7
1 inch/hour/minute in attometer/second squared is equal to117592592592.59
1 inch/hour/minute in centimeter/second squared is equal to0.000011759259259259
1 inch/hour/minute in decimeter/second squared is equal to0.0000011759259259259
1 inch/hour/minute in dekameter/second squared is equal to1.1759259259259e-8
1 inch/hour/minute in femtometer/second squared is equal to117592592.59
1 inch/hour/minute in hectometer/second squared is equal to1.1759259259259e-9
1 inch/hour/minute in kilometer/second squared is equal to1.1759259259259e-10
1 inch/hour/minute in micrometer/second squared is equal to0.11759259259259
1 inch/hour/minute in millimeter/second squared is equal to0.00011759259259259
1 inch/hour/minute in nanometer/second squared is equal to117.59
1 inch/hour/minute in picometer/second squared is equal to117592.59
1 inch/hour/minute in meter/hour squared is equal to1.52
1 inch/hour/minute in millimeter/hour squared is equal to1524
1 inch/hour/minute in centimeter/hour squared is equal to152.4
1 inch/hour/minute in kilometer/hour squared is equal to0.001524
1 inch/hour/minute in meter/minute squared is equal to0.00042333333333333
1 inch/hour/minute in millimeter/minute squared is equal to0.42333333333333
1 inch/hour/minute in centimeter/minute squared is equal to0.042333333333333
1 inch/hour/minute in kilometer/minute squared is equal to4.2333333333333e-7
1 inch/hour/minute in kilometer/hour/second is equal to4.2333333333333e-7
1 inch/hour/minute in inch/hour/second is equal to0.016666666666667
1 inch/hour/minute in inch/minute/second is equal to0.00027777777777778
1 inch/hour/minute in inch/hour squared is equal to60
1 inch/hour/minute in inch/minute squared is equal to0.016666666666667
1 inch/hour/minute in inch/second squared is equal to0.0000046296296296296
1 inch/hour/minute in feet/hour/minute is equal to0.083333333333333
1 inch/hour/minute in feet/hour/second is equal to0.0013888888888889
1 inch/hour/minute in feet/minute/second is equal to0.000023148148148148
1 inch/hour/minute in feet/hour squared is equal to5
1 inch/hour/minute in feet/minute squared is equal to0.0013888888888889
1 inch/hour/minute in feet/second squared is equal to3.858024691358e-7
1 inch/hour/minute in knot/hour is equal to0.00082289417166667
1 inch/hour/minute in knot/minute is equal to0.000013714902861111
1 inch/hour/minute in knot/second is equal to2.2858171435185e-7
1 inch/hour/minute in knot/millisecond is equal to2.2858171435185e-10
1 inch/hour/minute in mile/hour/minute is equal to0.000015782828282828
1 inch/hour/minute in mile/hour/second is equal to2.6304713804714e-7
1 inch/hour/minute in mile/hour squared is equal to0.0009469696969697
1 inch/hour/minute in mile/minute squared is equal to2.6304713804714e-7
1 inch/hour/minute in mile/second squared is equal to7.3068649457538e-11
1 inch/hour/minute in yard/second squared is equal to1.2860082304527e-7
1 inch/hour/minute in gal is equal to0.000011759259259259
1 inch/hour/minute in galileo is equal to0.000011759259259259
1 inch/hour/minute in centigal is equal to0.0011759259259259
1 inch/hour/minute in decigal is equal to0.00011759259259259
1 inch/hour/minute in g-unit is equal to1.1991107319277e-8
1 inch/hour/minute in gn is equal to1.1991107319277e-8
1 inch/hour/minute in gravity is equal to1.1991107319277e-8
1 inch/hour/minute in milligal is equal to0.011759259259259
1 inch/hour/minute in kilogal is equal to1.1759259259259e-8
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority.
| 1,961
| 6,272
|
{"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}
| 3.15625
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.907674
|
http://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-015-0542-3
| 1,495,933,019,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00491.warc.gz
| 11,373,717
| 20,169
|
Impact Factor 0.297
Open Access
# On some classes of difference equations of infinite order
DOI: 10.1186/s13662-015-0542-3
Accepted: 17 June 2015
Published: 10 July 2015
## Abstract
We consider a certain class of difference equations on an axis and a half-axis, and we establish a correspondence between such equations and simpler kinds of operator equations. The last operator equations can be solved by a special method like the Wiener-Hopf method.
### Keywords
difference equation symbol solvability
39B32 42A38
## 1 Introduction
Difference equations of finite order arise very often in various problems in mathematics and applied sciences, for example in mathematical physics and biology. The theory for solving such equations is very full for equations with constant coefficients [1, 2], but fully incomplete for the case of variable coefficients. Some kinds of such equations were obtained by the second author by studying general boundary value problems for mode elliptic pseudo differential equations in canonical non-smooth domains, but there is no solution algorithm for all situations [35]. There is a certain intermediate case between the two mentioned above, namely it is a difference equation with constant coefficients of infinite order. Here we will briefly describe these situations.
The general form of the linear difference equation of order n is the following [1, 2]:
$$\sum_{k=0}^{n}a_{k}(x)u(x+k)=v(x), \quad x\in M\subset{\mathbb{R}},$$
(1)
where the functions $$a_{k}(x)$$, $$k=1,\ldots,n$$, $$v(x)$$ are defined on M and given, and $$u(x)$$ is an unknown function. Since $$n\in{\mathbf{N}}$$ is an arbitrary number and all points $$x, x+1,\ldots,x+n$$, $$\forall x\in M$$, should be in the set M, this set M may be a ray from a certain point or the whole R.
A more general type of difference equation of finite order is the equation
$$\sum_{k=0}^{n}a_{k}(x)u(x+ \beta_{k})=v(x),\quad x\in M\subset{\mathbb{R}},$$
(2)
where $$\{\beta_{k}\}_{k=0}^{n}\subset{\mathbb{R}}$$.
Further, such equations can be equations with a continuous variable or a discrete one, and this property separates such an equation on a class of properly difference equations and discrete equations. In this paper we will consider the case of a continuous variable x, and a solution on the right-hand side will be considered in the space $$L_{2}({\mathbb {R}})$$ for all equations.
### 1.1 Difference equation of a finite order with constant coefficients
This is an equation of the type
$$\sum_{k=0}^{n}a_{k}u(x+ \beta_{k})=v(x), \quad x\in{\mathbb{R}},$$
(3)
and it easily can be solved by the Fourier transform:
$$(Fu) (\xi)\equiv\tilde{u}(\xi)=\int_{-\infty}^{+\infty }e^{-ix\cdot\xi}u(x) \, dx.$$
Indeed, applying the Fourier transform to (3) we obtain
$$\tilde{u}(\xi)\sum_{k=0}^{n}a_{k}e^{ i\beta_{k}\xi}= \tilde {v}(\xi),$$
or renaming
$$\tilde{u}(\xi)p_{n}(\xi)=\tilde{v}(\xi).$$
The function $$p_{n}(\xi)$$ is called a symbol of a difference operator on the left-hand side (3) (cf. [6]). If $$p_{n}(\xi )\neq0$$, $$\forall\xi\in{\mathbb{R}}$$, then (3) can easily be solved,
$$u(x)=F^{-1}_{\xi\to x} \bigl(p_{n}^{-1}(\xi) \tilde{v}(\xi) \bigr).$$
### 1.2 Difference equation of infinite order with constant coefficients
The same arguments are applicable for the case of an unbounded sequence $$\{\beta_{k}\}_{-\infty}^{+\infty}$$. Then the difference operator with complex coefficients
$$\mathcal{D}: u(x)\longmapsto\sum_{-\infty}^{+\infty }a_{k}u(x+ \beta_{k}), \quad x\in{\mathbb{R}}, a_{k}\in{\mathbb{C}},$$
(4)
has the following symbol:
$$\sigma(\xi)=\sum_{-\infty}^{+\infty}a_{k}e^{i\beta_{k}\xi}.$$
### Lemma 1
The operator $$\mathcal{D}$$ is a linear bounded operator $$L_{2}({\mathbb {R}})\to L_{2}({\mathbb{R}})$$ if $$\{a_{k}\}_{-\infty}^{+\infty}\in {l}^{1}$$.
### Proof
The proof of this assertion can be obtained immediately. □
If we consider the operator (4) for $$x\in{\mathbb{Z}}$$ only
$$\mathcal{D}: u(x_{d})\longmapsto\sum _{-\infty}^{+\infty }a_{k}u(x_{d}+ \beta_{k}), \quad x_{d}\in{\mathbb{Z}},$$
(5)
then its symbol can be defined by the discrete Fourier transform [7, 8]
$$\sigma_{d}(\xi)=\sum_{-\infty}^{+\infty}a_{k}e^{i\beta_{k}\xi }, \quad \xi\in[-\pi,\pi].$$
### 1.3 Difference and discrete equations
Obviously there are some relations between difference and discrete equations. Particularly, if $$\{\beta_{k}\}_{-\infty}^{+\infty }={\mathbb{Z}}$$, then the operator (5) is a discrete convolution operator. For studying discrete operators in a half-space the authors have developed a certain analytic technique [911]. Below we will try to enlarge this technique for more general situations.
## 2 General difference equations
We consider the equation
$$(\mathcal{D}u) (x)=v(x), \quad x\in{\mathbb{R}}_{+},$$
(6)
where $${\mathbb{R}}_{+}=\{x\in{\mathbb{R}}, x>0\}$$.
For studying this equation we will use methods of the theory of multi-dimensional singular integral and pseudo differential equations [3, 6, 12] which are non-usual in the theory of difference equations. Our next goal is to study multi-dimensional difference equations, and this one-dimensional variant is a model for considering other complicated situations. This approach is based on the classical Riemann boundary value problem and the theory of one-dimensional singular integral equations [1315].
### 2.1 Background
The first step is the following. We will use the theory of so-called paired equations [15] of the type
$$(aP_{+}+bP_{-})U=V$$
(7)
in the space $$L_{2}(\mathbb{R})$$, where a, b are convolution operators with corresponding functions $$a(x)$$, $$b(x)$$, $$x\in{\mathbb{R}}$$, $$P_{\pm }$$ are projectors on the half-axis $${\mathbb{R}}_{\pm}$$. More precisely,
$$(aP_{+}U) (x)=\int_{0}^{+\infty}a(x-y)U(y)\, dy, \qquad (bP_{-}U) (x)=\int_{-\infty}^{0}b(x-y)U(y)\, dy.$$
Applying the Fourier transform to (7) we obtain [12] the following one-dimensional singular integral equation [1315]:
$$\tilde{a}(\xi) (P\widetilde{U}) (\xi)+\tilde{b}(\xi) (Q\widetilde {U}) (\xi)=\widetilde{V}(\xi),$$
(8)
where P, Q are two projectors related to the Hilbert transform
\begin{aligned}& (Hu) (x)=\mathrm{v.p.}\, \frac{1}{\pi i}\int_{-\infty}^{+\infty} \frac {u(y)}{x-y}\, dy, \\& P=\frac{1}{2}(I+H), \qquad Q=\frac{1}{2}(I-H). \end{aligned}
Equation (8) is closely related to the Riemann boundary value problem [13, 14] for upper and lower half-planes. We now recall the statement of the problem: finding a pair of functions $$\Phi^{\pm}(\xi )$$ which admit an analytic continuation on upper ($${\mathbb{C}}_{+}$$) and lower ($${\mathbb{C}}_{-}$$) half-planes in the complex plane $$\mathbb {C}$$ and of which their boundary values on $$\mathbb{R}$$ satisfy the following linear relation:
$$\Phi^{+}(\xi)=G(\xi)\Phi^{-}(\xi)+g(\xi),$$
(9)
where $$G(\xi)$$, $$g(\xi)$$ are given functions on $$\mathbb{R}$$.
There is a one-to-one correspondence between the Riemann boundary value problem (9) and the singular integral equation (8), and
$$G(\xi)=\tilde{a}^{-1}(\xi)\tilde{b}(\xi), \qquad \widetilde{V}(\xi )= \tilde{a}^{-1}(\xi)g(\xi).$$
### 2.2 Topological barrier
We suppose that the symbol $$G(\xi)$$ is a continuous non-vanishing function on the compactification $$\dot{\mathbb{R}}$$ ($$G(\xi)\neq0$$, $$\forall\xi\in\dot{\mathbb{R}}$$) and
$$\operatorname{Ind} G\equiv\frac{1}{2\pi}\int _{-\infty}^{+\infty }d\arg G(t)=0.$$
(10)
The last condition (10), is necessary and sufficient for the unique solvability of the problem (9) in the space $$L_{2}({\mathbb{R}})$$ [13, 14]. Moreover, the unique solution of the problem (9) can be constructed with a help of the Cauchy type integral
$$\Phi^{+}(t)=G_{+}(t)P \bigl(G_{+}^{-1}(t)g(t) \bigr),\qquad \Phi ^{-}(t)=-G_{-}^{-1}(t)Q \bigl(G_{+}^{-1}(t)g(t) \bigr),$$
where $$G_{\pm}$$ are factors of a factorization for the $$G(t)$$ (see below),
$$G_{+}(t)=\exp \bigl(P \bigl(\ln G(t) \bigr) \bigr),\qquad G_{-}(t)=\exp \bigl(Q \bigl(\ln G(t) \bigr) \bigr).$$
### 2.3 Difference equations on a half-axis
Equation (6) can easily be transformed into (7) in the following way. Since the right-hand side in (6) is defined on $$\mathbb{R}_{+}$$ only we will continue $$v(x)$$ on the whole $${\mathbb {R}}$$ so that this continuation $$\mathit{lf}\in L_{2}({\mathbb{R}})$$. Further we will rename the unknown function $$u_{+}(x)$$ and define the function
$$u_{-}(x)=(\mathit{lf}) (x)-(\mathcal{D}u_{+}) (x).$$
Thus, we have the following equation:
$$(\mathcal{D}u_{+}) (x)+u_{-}(x)=(\mathit{lf}) (x),\quad x\in \mathbb{R},$$
(11)
which holds for the whole space $${\mathbb{R}}$$.
After the Fourier transform we have
$$\sigma(\xi)\tilde{u}_{+}(\xi)+\tilde{u}_{-}(\xi)=\widetilde {\mathit{lv}}(\xi),$$
where $$\sigma(\xi)$$ is called a symbol of the operator $$\mathcal{D}$$.
To describe a solving technique for (11) we recall the following (cf. [13, 14]).
### Definition
A factorization for an elliptic symbol is called its representation if it is in the form
$$\sigma(\xi)=\sigma_{+}(\xi)\cdot\sigma_{-}(\xi),$$
where the factors $$\sigma_{+}$$, $$\sigma_{-}$$ admit an analytic continuation into the upper and lower complex half-planes $${\mathbb{C}}_{\pm}$$, and $$\sigma^{\pm1}_{\pm}\in L_{\infty}({\mathbb{R}})$$.
### Example 1
Let us consider the Cauchy type integral
$$\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac {u(t)}{z-t}\, dt\equiv\Phi(z), \quad z=x\pm iy.$$
It is well known this construction plays a crucial role for a decomposition $$L_{2}({\mathbb{R}})$$ on two orthogonal subspaces, namely
$$L_{2}({\mathbb{R}})=A_{+}({\mathbb{R}})\oplus A_{-}({\mathbb{R}}),$$
where $$A_{\pm}({\mathbb{R}})$$ consists of functions admitting an analytic continuation onto $${\mathbb{C}}_{\pm}$$.
The boundary values of the integral $$\Phi(z)$$ satisfy the Plemelj-Sokhotskii formulas [13, 14], and thus the projectors P and Q are corresponding projectors on the spaces of analytic functions [15].
The simple example we need is
$$\exp(u)=\exp(Pu)\cdot\exp(Qu).$$
### Theorem 2
Let $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$, $$\operatorname{Ind}\sigma =0$$. Then (6) has unique solution in the space $$L_{2}({\mathbb {R}}_{+})$$ for arbitrary right-hand side $$v\in L_{2}({\mathbb{R}}_{+})$$, and its Fourier transform is given by the formula
$$\tilde{u}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta.$$
### Proof
We have
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi)= \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi).$$
Further, since $$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)\in L_{2}({\mathbb{R}})$$ we decompose it into two summands
$$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)=P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr)+Q \bigl( \sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi) \bigr)$$
and write
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde { \mathit{lv}}(\xi) \bigr)-\sigma^{-1}_{-}(\xi)\tilde{u}_{-}(\xi).$$
The left-hand side of the last quality belongs to the space $$A_{+}({\mathbb{R}})$$, but the right-hand side belongs to $$A_{-}({\mathbb {R}})$$, consequently these are zeros. Thus,
$$\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-P \bigl(\sigma^{-1}_{-}(\xi )\widetilde{ \mathit{lv}}(\xi) \bigr)=0$$
and
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr),$$
(12)
or in the complete form
$$\tilde{u}_{+}(\xi)=\frac{1}{2}\sigma^{-1}(\xi)\widetilde{ \mathit{lv}}(\xi )+\frac{\sigma^{-1}_{+}(\xi)}{2\pi i}\, \mathrm{v.p.}\int_{-\infty }^{+\infty} \frac{\sigma^{-1}_{-}(\eta)\widetilde{\mathit{lv}}(\eta)}{\xi -\eta}\, d\eta.$$
□
### Remark 1
This result does not depend on the continuation lv. Let us denote by $$M_{\pm}(x)$$ the inverse Fourier images of the functions $$\sigma ^{-1}_{\pm}(\xi)$$. Indeed, (12) leads to the following construction:
\begin{aligned} u_{+}(x)&=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t) (\mathit{lv}) (t)\, dt \biggr)\, dy \\ &=\int_{-\infty}^{+\infty}M_{+}(x-y) \biggl(\int _{0}^{+\infty}M_{-}(y-t)v(t)\, dt \biggr)\, dy. \end{aligned}
### Remark 2
The condition $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$ is not a strong restriction. Such symbols exist for example in the case that $$\sigma (\xi)$$ is represented by a finite sum, and $$\beta_{k}\in{\mathbb {Q}}$$. Then $$\sigma(\xi)$$ is a continuous periodic function.
## 3 General solution
Since $$\sigma(\xi)\in C(\dot{\mathbb{R}})$$, and Indσ is an integer, we consider the case $$\ae\equiv\operatorname{Ind} \sigma\in{\mathbb{N}}$$ in this section.
### Theorem 3
Let $$\operatorname{Ind} \sigma\in{\mathbb{N}}$$. Then a general solution of (6) in the Fourier image can be written in the form
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)\omega^{-\ae}(\xi)P \bigl( \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr)+( \xi-i)^{-\ae }\sigma^{-1}_{+}(\xi)P_{\ae-1}(\xi),$$
and it depends on æ arbitrary constants.
### Proof
The function
$$\omega(\xi)=\frac{\xi-i}{\xi+i}$$
has the index 1 [1315], thus the function
$$\omega^{-\ae}(\xi)\sigma(\xi)= \biggl(\frac{\xi-i}{\xi+i} \biggr)^{-\ae}\cdot\sigma(\xi)$$
has the index 0, and we can factorize this function
$$\omega^{-\ae}(\xi)\sigma(\xi)=\sigma_{+}(\xi)\sigma_{-}(\xi).$$
Further, we write after (11)
$$\omega^{\ae}(\xi)\omega^{-\ae}(\xi)\sigma(\xi)\tilde{u}_{+}(\xi)+ \tilde{u}_{-}(\xi)=\widetilde{\mathit{lv}}(\xi),$$
factorize $$\omega^{-\ae}(\xi)\sigma(\xi)$$, and rewrite
$$\omega^{\ae}(\xi)\sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\sigma ^{-1}_{-}( \xi)\tilde{u}_{-}(\xi)=\sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi).$$
Taking into account our notations we have
\begin{aligned}& \sigma_{+}(\xi)\tilde{u}_{+}(\xi)+\omega^{-\ae}(\xi)\sigma ^{-1}_{-}(\xi)\tilde{u}_{-}(\xi) \\& \quad =\omega^{-\ae}(\xi)P \bigl(\sigma ^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi) \bigr) +\omega^{-\ae}(\xi)Q \bigl(\sigma^{-1}_{-}( \xi)\widetilde{\mathit{lv}}(\xi ) \bigr), \end{aligned}
(13)
because
$$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)=P \bigl( \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi) \bigr)+Q \bigl( \sigma^{-1}_{-}(\xi)\widetilde {\mathit{lv}}(\xi) \bigr)$$
and
\begin{aligned}& \sigma_{+}(\xi)\tilde{u}_{+}(\xi)-\omega^{-\ae}(\xi)P \bigl( \sigma ^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr) \\& \quad =\omega^{-\ae}(\xi)Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi ) \bigr)-\omega^{-\ae}(\xi)\sigma^{-1}_{-}(\xi) \tilde{u}_{-}(\xi) \end{aligned}
(14)
and we conclude from the last that the left-hand side and the right-hand side also are a polynomial $$P_{\ae-1}(\xi)$$ of order $$\ae-1$$. It follows from the generalized Liouville theorem [13, 14] because the left-hand side has one pole of order æ in $${\mathbb{C}}$$ in the point $$z=i$$. So, we have
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi)\omega^{-\ae}(\xi)P \bigl( \sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr)+( \xi-i)^{-\ae }\sigma^{-1}_{+}(\xi)P_{\ae-1}(\xi).$$
□
### Remark 3
This result does not depend on the choice of the continuation l.
### Corollary 4
Let $$v(x)\equiv0$$, $$\ae\in{\mathbb{N}}$$. Then a general solution of the homogeneous equation (6) is given by the formula
$$\tilde{u}_{+}(\xi)=(\xi-i)^{-\ae}\sigma^{-1}_{+}( \xi)P_{\ae-1}(\xi).$$
## 4 Solvability conditions
### Theorem 5
Let $$-\operatorname{Ind} \sigma\in{\mathbb{N}}$$. Then (6) has a solution from $$L_{2}({\mathbb{R}}_{+})$$ iff the following conditions hold:
$${\int_{-\infty}^{+\infty}(\xi-i)^{k-1} \sigma^{-1}_{-}(\xi )\widetilde{\mathit{lv}}(\xi)\, d\xi}=0, \quad k=1,2, \ldots,|\ae|.$$
### Proof
We argue as above and use the equality (14); we write it as
\begin{aligned}& (\xi-i)^{\ae}\sigma_{+}(\xi)\tilde{u}_{+}(\xi)-( \xi+i)^{\ae }P \bigl(\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}( \xi) \bigr) \\& \quad =(\xi+i)^{\ae}Q \bigl(\sigma^{-1}_{-}(\xi)\widetilde{ \mathit{lv}}(\xi) \bigr)-(\xi+i)^{\ae}\sigma^{-1}_{-}(\xi) \tilde{u}_{-}(\xi). \end{aligned}
(15)
Since we work with $$L_{2}({\mathbb{R}})$$ both the left-hand side and the right-hand side are equal to zero at infinity, hence these are zeros, and
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi) \biggl(\frac{\xi-i}{\xi +i} \biggr)^{-\ae}P \bigl(\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}( \xi ) \bigr).$$
But there is some inaccuracy. Indeed, this solution belongs to the space $$A_{+}({\mathbb{R}})$$, but more exactly it belongs to its subspace $$A^{k}_{+}({\mathbb{R}})$$. This subspace consists of functions analytic in $${\mathbb{C}}_{+}$$ with zeros of the order −æ in the point $$z=i$$. To obtain a solution from $$L_{2}({\mathbb{R}}_{+})$$ we need some corrections in the last formula. Since the operator P is related to the Cauchy type integral we will use certain decomposition formulas for this integral (see also [1214]).
Let us denote $$\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi)\equiv g(\xi)$$ and consider the following integral:
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta }, \quad z=\xi+i\tau \in{\mathbb{C}}.$$
Using a simple formula for a kernel
$$\frac{1}{z-\eta}=\sum_{k=1}^{|\ae|} \frac{(\eta -i)^{k-1}}{(z-i)^{k}}+\frac{(\eta-i)^{|\ae|}}{(z-i)^{|\ae|}(z-\eta)}$$
we obtain the following decomposition:
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta}=\sum _{k=1}^{|\ae|}{(z-i)^{-k}} {\int _{-\infty}^{+\infty }(\eta-i)^{k-1}g(\eta)\, d\eta}+ \int_{-\infty}^{+\infty }\frac{(\eta-i)^{|\ae|}g(\eta)\, d\eta}{(z-i)^{|\ae|}(z-\eta)}.$$
So, we have a following property. If the conditions
$${\int_{-\infty}^{+\infty}(\eta-i)^{k-1}g(\eta)\, d \eta }=0, \quad k=1,2,\ldots,|\ae|,$$
hold, then we obtain
$$\int_{-\infty}^{+\infty}\frac{g(\eta)\, d\eta}{z-\eta }=(z-i)^{\ae} \int_{-\infty}^{+\infty}\frac{(\eta-i)^{|\ae |}g(\eta)\, d\eta}{z-\eta}.$$
Hence the boundary values on $${\mathbb{R}}$$ for the left-hand side and the right-hand one are equal, and thus
$$P \bigl(g(\xi) \bigr)=(\xi-i)^{\ae}P \bigl((\xi-i)^{|\ae|}g(\xi ) \bigr).$$
Substituting the last formula into the solution formula we write
$$\tilde{u}_{+}(\xi)=\sigma^{-1}_{+}(\xi) (\xi+i)^{\ae}P \bigl((\xi -i)^{|\ae|}\sigma^{-1}_{-}(\xi)\widetilde{\mathit{lv}}(\xi) \bigr).$$
□
## 5 Conclusion
It seems this approach to difference equations may be useful for studying the case that the variable x is a discrete one. We have some experience in the theory of discrete equations [911], and we hope that we can be successful in this situation also. Moreover, in our opinion the developed methods might be applicable for multi-dimensional difference equations.
## Declarations
### Acknowledgements
The authors are very grateful to the anonymous referees for their valuable suggestions. This work is supported by Russian Fund of Basic Research and government of Lipetsk region of Russia, project No. 14-41-03595-a.
## Authors’ Affiliations
(1)
Department of Mathematical Analysis, National Research Belgorod State University
(2)
Chair of Pure Mathematics, Lipetsk State Technical University
## References
1. Milne-Thomson, LM: The Calculus of Finite Differences. Chelsea, New York (1981)
2. Jordan, C: Calculus of Finite Differences. Chelsea, New York (1950) Google Scholar
3. Vasil’ev, VB: Wave Factorization of Elliptic Symbols: Theory and Applications. Introduction to the Theory of Boundary Value Problems in Non-Smooth Domains. Kluwer Academic, Dordrecht (2000)
4. Vasilyev, VB: General boundary value problems for pseudo differential equations and related difference equations. Adv. Differ. Equ. 2013, 289 (2013)
5. Vasilyev, VB: On some difference equations of first order. Tatra Mt. Math. Publ. 54, 165-181 (2013)
6. Mikhlin, SG, Prößdorf, S: Singular Integral Operators. Akademie Verlag, Berlin (1986)
7. Sobolev, SL: Cubature Formulas and Modern Analysis: An Introduction. Gordon & Breach, Montreux (1992)
8. Dudgeon, DE, Mersereau, RM: Multidimensional Digital Signal Processing. Prentice Hall, Englewood Cliffs (1984) Google Scholar
9. Vasilyev, AV, Vasilyev, VB: Discrete singular operators and equations in a half-space. Azerb. J. Math. 3(1), 84-93 (2013)
10. Vasilyev, AV, Vasilyev, VB: Discrete singular integrals in a half-space. In: Current Trends in Analysis and Its Applications, Proc. 9th Congress, Krakow, Poland, August 2013, pp. 663-670. Birkhäuser, Basel (2015)
11. Vasilyev, AV, Vasilyev, VB: Periodic Riemann problem and discrete convolution equations. Differ. Equ. 51(5), 652-660 (2015)
12. Eskin, G: Boundary Value Problems for Elliptic Pseudodifferential Equations. Am. Math. Soc., Providence (1981) Google Scholar
13. Gakhov, FD: Boundary Value Problems. Dover, New York (1981) Google Scholar
14. Muskhelishvili, NI: Singular Integral Equations. North-Holland, Amsterdam (1976) Google Scholar
15. Gokhberg, I, Krupnik, N: Introduction to the Theory of One-Dimensional Singular Integral Equations. Birkhäuser, Basel (2010)
| 7,408
| 20,965
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2017-22
|
longest
|
en
| 0.866245
|
http://forum.arduino.cc/index.php?topic=133407.msg1004419
| 1,469,922,722,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469258943369.84/warc/CC-MAIN-20160723072903-00268-ip-10-185-27-174.ec2.internal.warc.gz
| 96,384,392
| 13,019
|
Go Down
### Topic: LED Gurus? (Read 966 times)previous topic - next topic
#### brhyaa729
##### Nov 20, 2012, 11:40 pm
Hey guys - how is everyone doing? Was hoping I could get some direction in setting up an array of 940nm 8mm LEDs. I watched a few YouTube Videos and found a guy that used 850nms and gave the specs to wire them in series but I want to be sure I understand the specs on the LEDs I am purchasing. Here is are the specs of the units.
Wavelength : 940nm
Vf: 1.4~1.5V
Vr: 5V
If: 140mA
Ipusle: 700mA
Power Dissipation:200mW
Tjun: 125'C
Tsol: 5second @ 260'C
MCD=50@ 350mA, MCD=70@400mA
Angle: 30'
I also found an LED Calculator online here: http://led.linear1.org/led.wiz
Am I correct in assuming that diode forward voltage is 1.4 (or 1.5 which should I use?)V and diode forward current is 140ma? I would like to use 5 of these in series with a 9 volt battery and I am trying to determine which size resistor I would use to prevent destroying the LEDs. If anyone could tell me what the abbreviations on the specs mean I would appreciate it. Thanks guys!
#### Tom Carpenter
#1
##### Nov 20, 2012, 11:51 pm
Yup, Forward Current (aka If) = 140 [mA] = 0.14 [A]. Forward Voltage (aka Vf) is somewhere in the region of 1.4 [V] to 1.5 [V], so I would work on the theory that it is 1.45 [V] (the actual voltage will depend on the LED, temperature, current), but this approximate figure should be good enough.
If we do the math, 5 diodes in series x 1.45 [V] each = 7.25 [V] dropped across the diodes. For a nominal 9 [V] supply, this leaves 9-7.25=1.75 [V] across your resistor. V=IR, so R = V/I = 1.75/0.14 = 12.5 [Ohm]. You will need an minimum P=IV = 0.14*1.75 = 0.245 [W] rated resistor.
However... A 9V battery supplying 140mA will run flat very very very quickly. You would be lucky if you got half an hour out of it. The other thing is that you would be wasting a lot of energy in that resistor.
~Tom~
#### betomax
#2
##### Nov 20, 2012, 11:56 pm
i'm not a LED guru (in fact anything else at all !!), but the spec as in most datasheet are Maximum absolute, so you must use more conservative vals, in your link i choose a 10mA (as usual) to get started, and the result are:
Solution 0: 5 x 1 array uses 5 LEDs exactly
+9V R = 150 ohms
The wizard says: In solution 0:
each 150 ohm resistor dissipates 15 mW
the wizard says the color code for 150 is brown green brown
the wizard thinks 1/4W resistors are fine for your application
together, all resistors dissipate 15 mW
together, the diodes dissipate 75 mW
total power dissipated by the array is 90 mW
the array draws current of 10 mA from the source.
if (a.k.a. foward current) is directly linked with 'brightness' but keep it as low as you can to give long life to your LEDS !
#### brhyaa729
#3
##### Nov 21, 2012, 12:01 am
If I would only get 30 mins or so out of the 9 volt battery are there any longer life battery alternatives that would still be feasible to carry around attached to a small 3 inch x 2 inch project box?
#### brhyaa729
#4
##### Nov 21, 2012, 12:52 am
I think I have the answer to my own question, I could use two 9 volt batteries in parallel to double the life and remain at 9 volts without adding much bulk to the project.
#### retrolefty
#5
##### Nov 21, 2012, 01:12 am
I think I have the answer to my own question, I could use two 9 volt batteries in parallel to double the life and remain at 9 volts without adding much bulk to the project.
Have you priced what 9 volt batteries retail for? They are a very poor choice both from a capacity and cost point of view. Either convert to series alkaline AA batteries for use a DC power module. 9 volt batteries are for smoke alarms.
Lefty
#### Tom Carpenter
#6
##### Nov 21, 2012, 01:23 am
For maximum power density, you would need to start looking at rechargeable Li-Po batteries like those used for RC planes and helicopters.
As a gauge for how long they last, a battery rated for 1100mAH would supply 1100mA for an hour, or in your case 140mA for 1100mAH/140mA = 7.9Hours.
~Tom~
#### brhyaa729
#7
##### Nov 21, 2012, 01:52 am
I have several 9 volt rechargeables and a charger.
#### Simpson_Jr
#8
##### Nov 21, 2012, 02:36 am
Question, what are you going to use those high power IR-leds for ?
Continuous illumination will indeed require quite some power, you can use 'm, but you'll have to recharge those batteries a lot. In a "TV-remote/TVBgonekit on steroids"-project though, the leds will hardly be on and a 9v battery could be a perfect choice.
#### brhyaa729
#9
##### Nov 21, 2012, 03:33 am
An IR Illuminator to connect to a Sony Handycam with nightvision. Probably won't ever be turned on for more than a few mins at a time.
#### brhyaa729
#10
##### Nov 21, 2012, 04:06 am
I also have a ton of AA Eneloops - plus chargers for those, was thinking a caddy with 8 of those would give me 9.6 volts. This would probably be better than the 9 volts correct?
#### Chagrin
#11
##### Nov 21, 2012, 05:19 am
Read the battery; it should state a "mah" (milli amp hour) rating.
For alkaline batteries, a 9V battery has around 550mah meaning it can supply 500 milliamps (.5 amps) of current for an hour (or 250 milliamps for 2 hours, etc.). An AA battery is around 2000mah.
#### brhyaa729
#12
##### Nov 21, 2012, 12:02 pm
So would this battery give me 28 hrs of run time?
http://www.amazon.com/Venom-Group-International-4000mAh-Universal/dp/B000W7WWFW/ref=sr_1_4?ie=UTF8&qid=1353495538&sr=8-4&keywords=li-po+battery+11.1v+and+charge
Go Up
Please enter a valid email to subscribe
| 1,686
| 5,546
|
{"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}
| 3.6875
| 4
|
CC-MAIN-2016-30
|
longest
|
en
| 0.929212
|
http://nrich.maths.org/4808/clue
| 1,477,691,483,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988725475.41/warc/CC-MAIN-20161020183845-00382-ip-10-171-6-4.ec2.internal.warc.gz
| 181,618,426
| 5,282
|
### Walk and Ride
How far have these students walked by the time the teacher's car reaches them after their bus broke down?
### Fence It
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Speeding Up, Slowing Down
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
# How Far Does it Move?
##### Stage: 3 Challenge Level:
Try to approach the problem systematically, keeping one variable fixed and just altering the other one.
As you go along try to understand why the graph takes the shape that it does:
• by relating it to the rolling polygon and the journey of the red dot
• by trying to predict what will happen before you set the polygon rolling
Could the dot have been on the centre of a polygon?
Try for each of the polygons.
Could the dot have been on the centre of the base of a polygon?
Try for each of the polygons.
Could the dot have been on the centre of one of the sloping sides of a polygon?
Try for each of the polygons.
Could the dot have been on the centre of a side opposite the base of a polygon?
Try for each of the polygons.
Could the dot have been on a vertex opposite the base of a polygon?
Try for each of the polygons.
Could the dot have been on a vertex on the base of a polygon?
Try for each of the polygons...
Alternatively...
• try all possible positions of the dot in a triangle,
• and then in a square,
• and then in a pentagon,
• and then in a hexagon...
| 346
| 1,556
|
{"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}
| 3.75
| 4
|
CC-MAIN-2016-44
|
longest
|
en
| 0.940169
|
https://www.theknowledgeroundtable.com/tutorials/probability-and-statistics/
| 1,611,833,817,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00060.warc.gz
| 993,255,753
| 15,908
|
## GRE Math Tutorial
#### Intro
This is the problem in the probability and statistics asked frequenlty in the GRE-Quant section .In this problem ,we will have a look at the typical problem in the subject area which lays foundation concept for the further problems related to the probability which carries important portion in the GRE examination.
#### Sample Problem
1.There are 5 good and 6 damaged apples in a bag which look identical otherwise.You are told to take out the two apples simultaneously(not one by one ,if one by one the answer will be different). So what is the probability that you draw one good apple and one damaged apple.
5C2*6C2/11C2
5C11*6C1/11C1
5!*6!/11!
the data is not sufficient
5C1*6C1/11C2
#### Solution
1.First identify whether it is a combination problem or a permutation .This is the plcae where many of us get confused and commit the mistake .
In this case , does ‘the order in which the objects/or the subjects are arranged’ makes the outcomes different ?I mean does the order of two drawn apples like good-damaged or damaged=good are different or same.they are same ,so its a combination.
Lets take another example ,you are to make a password of 3 different digits out of 5 digits i.e. 0,1,2,3,4.Now ask yourself, does the order of the outcome matters . Lets say one outcome is 012 and another is 120 .Though they have the same objects /or subjects , the outcome are different password and the ‘ order’ matters .So this will be permutation.
2.After deciding the problem type,we confirm it is a combination problem.
If two apples are drawn from a bag of total 11( =5++6) ,they can be drawn in 11C2 ways .(C carries the meaning for mathematical operation.) 11C2= 11!/(2!*9!).=Total possible outcome
The desired outcome is the one good and one bad apple. so number of getting the result= 5C1*6C!.(As from 6 bad apples any one can be drawn in 6c1 ways , first..second..or sixth. so is the case for good apples .Since the the results are to happen at once .Multiply them .
Probability= 5C1*6C1/11C2
This is the basic but very crucial steps for learning the probability.I will present more in he upcoming tutorials .
| 534
| 2,157
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.71875
| 5
|
CC-MAIN-2021-04
|
latest
|
en
| 0.910227
|
https://jarviscodinghub.com/product/ling-473-project-5-solution/
| 1,726,012,742,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00096.warc.gz
| 301,913,747
| 26,165
|
# Ling 473 Project 5 solution
\$25.00
Original Work ?
## Description
5/5 - (1 vote)
For this project you will build a naïve Bayesian classifier which is able to classify text fragments according
to their language. Determining the language of a document may seem trivial, since the most common few
words in each language—or the character set it uses—could be thought of as an identifying signature.
But
we would like to use a more principled approach that quantitatively scores the probability of some fragment
of text being written in one language, relative to others.
Naturally, Bayes’ theorem comes to mind:
𝑃(lang|text) =
𝑃(text|lang)𝑃(lang)
𝑃(text)
As we did in the POS-tagging lecture, we can ignore the denominator of the right side. This is because we
will be interested in comparing the result of this calculation for different languages—but always for the
same text. Since we will only compare results for the same text fragment, 𝑃(text) will always be 1.0, and
it can be excluded from the calculation.
𝑃(lang|text) = 𝑃(text|lang)𝑃(lang)
Next, we will assume that each of the 15 languages are equally likely. Again, we could always multiply the
right side by 1
15
, but since we will only use the results for comparison, this term can also be dropped.
𝑃(lang|text) = 𝑃(text|lang)
Finally, we apply the naïve Bayesian assumption: all words in the text are conditionally independent of
each other. Therefore, the probability that a text fragment is written in a language will be product of the
probabilities of all of its words, according to a unigram language model of that language.
𝑃(lang|text) = ∏𝑃(word𝑖
|lang)
𝑖
Because this product may approach zero, the calculation will actually be done by adding log-probabilities.
logprob(lang|text) = ∑log10 𝑃(word𝑖
|lang)
𝑖
This calculation is repeated for each of the 15 languages, and the language with the highest probability is
predicted. If we just wanted to know the best predicted language 𝐿̂, we could use argmax to drop the left
side of the equation. As you recall, this notation discards the actual result, returning only the parameter that
yielded it.
𝐿̂ = argmax𝑗 ∑log10 𝑃(word𝑖
|lang𝑗)
𝑖
However, for this project, in addition to identifying the best prediction, you will calculate and report the
actual log-prob scores for each line of text, for each of the 15 languages.
Input
In the following location you will find the corpus: 15 unigram language model files similar to what you
built for Project 2. The models are truncated, containing only the most common 1,500 words.
/opt/dropbox/18-19/473/project5/language-models
These files use the Latin-1 encoding, an 8-bit file format. Because some languages use lexical capitalization
(for example, nouns are always capitalized in German), and because some programming languages may
lack support for correctly changing the case of letters with diacritics, this project will use case-sensitive
comparison. Do not ignore case.
In order to match the way the language model was built, you should strip punctuation from the input
sentences. A suggested set of punctuation to strip is given here: .,!¡¥\$£¿;:()”‘—–-/[]¹²³«» but you are
welcome to develop your own set, based on an examination of the language models.
In accordance with typical classification practice, you will be provided with separate training and testing
input files. These files, test.txt and train.txt are located in the /project5 dropbox. Each example is
on a single line, consisting of an identifier, a tab character, and the example text. The training examples
are labeled, so the identifier will be the correct three-letter language code for the example. The test
examples are unlabeled, so the identifier will be a numeric value. Since the training examples are taken
from the training corpus, they should all be readily identified by your classifier, and you can use them to
verify that it is working properly. You only need to submit results for the testing examples.
Smoothing
As is usual, you will encounter a lot of words in the test data which are not in the language model that you
are currently evaluating. How you deal with these is called smoothing, and you should implement a
principled approach to the problem. One possibility is to assign unseen words the same probability as a
singleton (a word that occurred once in the model) for the language under consideration.
Discuss your smoothing technique in your write-up and comment on the effect your choice has on your
classifier’s probability space.
Extra Credit (15 points)
For extra credit, also process the file extra-test.txt, which includes samples from languages that are not
included in your fifteen models. Using a threshold value or other parameters, you may determine that none
of the languages is clearly more likely than any of the others. In this case, your classifier can conclude that
it is an ‘unknown’ language (use the string “unk” for this). Use the extra-train.txt file, in which the
languages are labeled, to tune your algorithm and/or parameter(s).
Output Format
Output your results to the console (stdout). For each example you process, write the following lines:
The identifier, a tab character, and the example text
One line per language model, consisting of the language code, a tab character and a numeric value
The word “result”, a tab character, and the language code for the “best” result from your classifier.
Example output follows. Your output file will contain one section like this for each of the input examples.
The numeric values can represent whatever you wish, since they are only relative for the single example.
Depending on how you set up your math, the “result” should be the language that has either the highest—
or the lowest—value.
Submission
Include the following files in your submission:
compile.sh Contains command(s) that compile your program. If you are using python or
any other interpreted language that does not require compiling, then this file
will be empty, or contain just the single line:
#!/bin/sh
run.sh The command(s) that run your program. Be sure to include compiled
binaries in your submission so that this script will execute without first
running compile.sh
output This is the output of your program, captured from the console by executing
the following command:
\$ ./run.sh >output
compile-extra.sh Compiles the extra credit, exactly as compile.sh (optional)
run-extra.sh Runs the extra credit, exactly as run.sh (optional)
output-extra Extra credit output (optional)
special features, or anything else you’d like me to review. If you could not
complete some or all of the project’s goals, please explain what you were
able to complete.
(source code and
binary files)
All source code and binary files (jar, a.out, etc., if any) required to run and
Gather together all the required files, making sure that, for example, any PDF or other binary files are
transferred from your local machine using a binary transmission format. Then, from within the directory
2 Cé agus, is féidir cur síos a dhéanamh mar sin ar an gcóras.
dan -17.434282
deu -17.735051
dut -17.394591
eng -16.219135
fin -17.451900
fra -17.711787
gla -13.688687
ita -17.291179
nob -17.722804
pol -17.778136
por -16.278133
spa -17.402331
swe -17.856767
swh -17.201888
tgl -17.359011
result gla
tar -czf hw.tar.gz *
Notice that this command packages all files in the current directory; do not include any top-level directories.
| 1,770
| 7,435
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.1875
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.914673
|
http://www.realclearscience.com/blog/2011/12/nothing-can-go-faster-than.html
| 1,371,701,923,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368710274484/warc/CC-MAIN-20130516131754-00049-ip-10-60-113-184.ec2.internal.warc.gz
| 653,621,424
| 7,783
|
# Why Are Faster-than-Light Neutrinos Such a Big Deal?
Nothing can go faster than the speed of light! This has been a mantra of physics for the past century. The central pillar of Einstein's famous theory of relativity, this core experimental truth in physics is now facing a slightly greater challenge.
Neutrinos are created and fly out of the enormous particle collider (LHC) at CERN on the Swiss-French border. Physicists sitting nearly 500 miles away in Italy have been clocking them. Several times now, their speed has appeared to be about three thousandths of a percent higher than the speed of light.
Why would confirmation of this result be such enormous news?
The principle driving relativity theory is a simple one. All objects move through spacetime at a constant speed. That is, they move about in space and forward in time in such a way that the speed in space and the speed in time add up to be a certain total (actually the speed of light) that never changes.
A football sitting still on the ground is traveling at the speed of light through time. Say you pick the football up and throw it. If you add together the speed it is going up into the air, the speed it is going to the left or right, and the speed it is going away from you, you have its speed through space. The speed that the ball travels through time is then the constant total speed minus this speed through space:
Speed in time = Constant total spacetime speed - space speed.
This has the fascinating consequence that the faster you move, the slower you age (move through time) relative to someone sitting still! Imagine that you can climb into a spaceship and pilot it at 95% of the speed of light for one year on your watch before coming back to Earth. When you get home, your family will all be more than three years older!
There is another enormous consequence. If something were to go faster than the speed of light, it would travel backwards (negative speed) through time. Faster than light particles would be time-traveling!
Before this startling result is accepted, several possible mistakes need to be checked for. Chief among these is the synchronization of several clocks: at the source of neutrinos, at the place where they are measured and a GPS satellite in space will all have to be checked. Then, other independent experiments will have to confirm the same finding. It could be that the rate it is moving (since faster moving things keep slower time) is throwing one clock out of time. Finally, there is the test of convincing scientists that a fundamental idea that they have cherished for a century is wrong!
| 541
| 2,615
|
{"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}
| 3.078125
| 3
|
CC-MAIN-2013-20
|
longest
|
en
| 0.95878
|
www.100schooldays.com
| 1,386,465,684,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386163057146/warc/CC-MAIN-20131204131737-00016-ip-10-33-133-15.ec2.internal.warc.gz
| 201,960,804
| 6,485
|
# Dot-to-Dot Puzzles
## Carousel – 5 to 100
Ride this carousel round and round! Don’t forget to count by 5′s and connect the dots all the way to 100 along the way. Click on the thumbnail to open the coloring page in a new window.
## Puzzle Cube – 5 to 100
Do you think you are up to the challenge of solving this puzzle cube? I bet you are able to if you know how to count by 5′s! Just connect the dots from 5 to 100 and see how easy this one really is to solve! Click on the thumbnail to open the coloring page […]
## Pencil – 5 to 100
This pencil can’t write unless you will! You’ll need to count by 5′s all the way to 100 to get this one done. Are you up for the challenge? Click on the thumbnail to open the coloring page in a new window.
## Leap Frog – 5 to 100
Do you like to play leap frog? Can you tell what this handsome prince is jumping over? You’ll find out when you count by 5′s to 100 and connect the dots as you go! Click on the thumbnail to open the coloring page in a new window.
## Polaroid Camera – 5 to 100
Have you ever seen one of these classic cameras? Make one of your very own by counting by 5′s to 100 and connecting the dots along the way. Click on the thumbnail to open the coloring page in a new window.
## Witch on Broom – 1 to 100
This witch is waiting for you to help her take off! Connect the dots from 1 to 100 and see if she doesn’t fly away! Click on the thumbnail to open the coloring page in a new window.
## Teddy Bear – 10 to 100
Help this teddy bear stay out of the rain! Count by tens to one hundred and finish up his umbrella! Click on the thumbnail to open the coloring page in a new window.
## Scooter – 10 to 100
Zoom! Zoom! Make this scooter go faster connecting the dots, counting by tens all the way to 100! Click on the thumbnail to open the coloring page in a new window.
## Princess – 1 to 100
This princess is patiently waiting for you to connect her dots from 1 to 100. When you finish, how about making her a castle too? Click on the thumbnail to open the coloring page in a new window.
## Frog on Lilly Pad – 10 to 100
Ribbit, ribbit! Learn to count by tens and help this frog catch more flies by connecting his dots from 10 to 100. Click on the thumbnail to open the coloring page in a new window.
| 584
| 2,274
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2013-48
|
latest
|
en
| 0.919742
|
https://www.w3resource.com/javascript-exercises/javascript-math-exercise-112.php
| 1,674,808,236,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00277.warc.gz
| 1,081,194,131
| 28,478
|
# JavaScript: Maximum value swapping two digits in an integer
## JavaScript Math: Exercise-112 with Solution
Write a JavaScript program to get the maximum value swapping two digits in a given integer.
Test Data:
(100) -> 100
(120) -> 210
(129) -> 921
Sample Solution:
HTML Code:
``````<!DOCTYPE html>
<html>
<meta charset="utf-8">
<title>JavaScript program to Maximum value swapping two digits in an integer</title>
<body>
</body>
</html>
```
```
JavaScript Code:
``````function test(n) {
var n_str = n.toString();
for (var i = 0; i < n_str.length; i++) {
var t = n_str[i];
var index = i;
for (var x = n_str.length - 1; x > i; x--) {
if (n_str[x] > t) {
t = n_str[x];
index = x;
}
}
if (t != n_str[i]) {
var nums = n_str.split('');
var tmp = nums[i];
nums[i] = nums[index];
nums[index] = tmp;
return parseInt(nums.join(''));
}
}
return n;
}
n = 100
console.log("n = "+n)
console.log("Maximum value swapping two digits in the said integer: "+test(n));
n = 120
console.log("n = "+n)
console.log("Maximum value swapping two digits in the said integer: "+test(n));
n = 129
console.log("Maximum value swapping two digits in the said integer: "+test(n));
console.log("e: "+test(n));
```
```
Sample Output:
```n = 100
Maximum value swapping two digits in the said integer: 100
n = 120
Maximum value swapping two digits in the said integer: 210
Maximum value swapping two digits in the said integer: 921
e: 921
```
Flowchart:
Live Demo:
See the Pen javascript-math-exercise-112 by w3resource (@w3resource) on CodePen.
Improve this sample solution and post your code through Disqus
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
## JavaScript: Tips of the Day
Shorten an array using its length property
A great way of shortening an array is by redefining its length property.
```let array = [0, 1, 2, 3, 4, 5, 6, 6, 8, 9]
array.length = 4
// Result: [0, 1, 2, 3]
```
Important to know though is that this is a destructive way of changing the array. This means you lose all the other values that used to be in the array.
Ref: https://bit.ly/2LBj213
| 600
| 2,124
|
{"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}
| 3.5
| 4
|
CC-MAIN-2023-06
|
longest
|
en
| 0.559959
|
https://www.soultiply.com/post/what-is-credit-var/
| 1,725,882,707,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00361.warc.gz
| 941,008,302
| 8,954
|
# What Is Credit Var?
Author: Artie
Published: 13 Aug 2022
### VaR Assessment
The VaR assessment allows for the determination of cumulative risks from aggregated positions held by different trading desks and departments within the institution. Financial institutions can use VaR modeling to determine if they have enough capital in place to cover losses or if they need to reduce holdings. There is no standard protocol for determining asset, portfolio, or firm-wide risk.
### Risk in Financial Contracts
The other party to a financial contract is at risk of not meeting its obligations. Every trade needs a party to take the opposite side. VaR is a measure of risk.
VaR gives the probability of a certain amount of loss for a certain portfolio. A portfolio of assets with a one-month VaR of \$1 million has a 5% chance of losing more than \$1 million. The VaR can give a measure of the risk of default on a credit default swap.
3. The swap is valued using future interest rates. The swap value at that point in time is what each interest rate path implies.
### Monte Carlo Method for Valuation and Risk Measurement
It is necessary to calculate the risk and return of each asset and the correlations between them in order to calculate Value at Risk. The more assets in a portfolio, the harder it is to calculate VaR. The simplest method for calculating Value at Risk is the historical method.
The percentage change for each risk factor is calculated using market data for the last 250 days. The percentage change is calculated with the current market values to present 250 scenarios for the future value. The Monte Carlo method uses non- linear pricing models to estimate the change in value for each scenario and then calculate the VaR according to the worst losses.
The Monte Carlo method is suitable for a wide range of risk measurement problems. It assumes that there is a distribution for risk factors. If an investment is negatively correlated with the portfolio, it may contribute less risk to the portfolio than if it is positive.
### VaR: A Realistic Model of Volatility Clustering
VaR has found favor because it is easy to understand, but it is also simplification of the real world. The main behavior of the real world is captured but some features are discounted to avoid making the model too complex and historically they never played a crucial role. The volatility clustering is a market behavior that does not capture historical simulation. Every event in the past is given equal probability, but there are times when there is a market surge and there are times when it is not.
### Online Statistics Study Guide
A survey is the most common way that nominal scale data is collected. A researcher might ask 100 people what type of place they live in. Companies collect outlying scale data through surveys to get feedback on their product or service.
### Value at Risk: A Measure of the Perception and Performance
The value at risk is a statistical risk management technique that can be used to monitor and quantifies the risk level associated with an investment portfolio. The value at risk is a measure of the maximum amount of loss over a specified time horizon. The value at risk calculation is measured by backtesting.
Backtesting is the process of determining how well a strategy will perform. The value at risk is compared with actual losses at the end of the time horizon. The value at risk calculates the potential maximum losses over a specified time horizon.
The one-year value at risk of an investment portfolio is \$10 million with a confidence level of 95%. There is a 5% chance of having losses that exceed \$10 million at the end of the year. The worst expected portfolio loss will not be more than \$10 million.
If the value at risk is calculated and the actual portfolio losses have not exceeded the expected value at risk losses, then it is an appropriate measure. If the actual portfolio losses exceed the calculated value at risk losses, the expected value at risk calculation may not be accurate. When the actual portfolio losses are greater than the calculated value at risk estimated loss, it is known as a breach of value at risk.
If the portfolio loss is above the estimated value at risk only a few times, it doesn't mean that the value at risk has failed. The number of breaches needs to be determined. The daily value at risk of an investment portfolio is \$500,000, with a 95% confidence level for 250 days.
### Maverick BankCard: A Leading Electronic Payment Processor
Los Angeles based Maverick BankCard is a leading electronic payment processor. Maverick handles all processes in-house, while providing industry-leading technology, as a full-service payment provider.
### Credit Risk Assessment
Credit scoring is a method used to estimate the risk associated with a loan. The customer score is calculated using data provided in the loan application or obtained from other sources. The higher the rating it will receive, the more similar the profile of the borrowers is to those who repay their loans on time.
The scoring models enable an objective assessment of credit risk, which is a key element of the credit granting process. Banks are using ready-made systems that allow for performing credit assessment models in a point system in order to make the credit calculation as accurate, transparent and low-risk as possible. The use of such tools reduces the chance of granting doubtful loans and allows for faster credit process while reducing the risk of human error.
The risk of insolvency and bankruptcy is looked at by assigning points. Capital, debt and development strategy are taken into account. The credit calculation is made on the basis of the company's previous financial results and industry characteristics.
### Consumer Credit
The two major categories for consumer credit are open-end and closed-end. Revolving credit can be used for purchases that will be paid back monthly. Paying the full amount is not required, but interest will be added to any balance that is not paid.
Consumers can get a loan for anything they want to purchase, which tells you how many loan types are available. If you want to borrow money to make a purchase, there is someone who will lend it to you. College students and their families can get student loans to help pay for college.
The interest rates were reduced to zero when the COVID-19 pandemic hit. Federal student loans and private student loans are the two types of student loans. Federally funded loans have lower interest rates and are more borrower friendly.
The federal loans have a 0% interest rate through Sept. 30, 2021. Consumers can buy a home with a mortgage, which is a loan that is distributed by banks, credit unions and online lenders. If you fall behind on your mortgage payments, you risk being foreclosed on.
Mortgages have the lowest interest rates because they are secured. A cash advance is a loan against your credit card. You can use cash instead of a credit card to make purchases or pay for services, and you can use the cash at any bank or ATM.
### Credit Risk Analysis
The financer's cash flow is impacted when the interest is not paid. The cost of collections increases. The process of intelligent credit analysis can help mitigate the severity of complete loss of the borrowings and its recovery, even though there is a grey area in guessing who and when will default on borrowings.
Banks, financial institutions and NBFCs offer a lot of credit products and need to be careful with their credit risk analysis. Companies that offer credit, bond issuers, insurance companies, and even investors need to know the techniques of effective risk analysis. India is fast becoming digital and so is the need for a credit analysis course.
The risk assessment can be done in many ways, like the points-based system, personal appraisals by trained risk-assessors, or by departments for credit-risk assessment of loan-customers. The credit rating of bonds is looked into by investors. Bonds with a B or C low-rating are more likely to default on payments.
Credit analysis the method used to assess the creditworthiness of a business organization. It means the ability and evaluation of the person or company to honour their financial obligations. The financial audited statements of larger companies are used to rate their credit-worthiness.
Click Horse
X Cancel
No comment yet.
| 1,659
| 8,430
|
{"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-2024-38
|
latest
|
en
| 0.912404
|
https://www.difference.wiki/simple-harmonic-motion-vs-periodic-motion/
| 1,726,049,056,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00698.warc.gz
| 696,685,938
| 23,486
|
# Simple Harmonic Motion vs. Periodic Motion: What's the Difference?
Edited by Aimie Carlson || By Janet White || Published on December 23, 2023
Simple harmonic motion is a specific type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction.
## Key Differences
Simple harmonic motion (SHM) is a model for oscillatory motion where the restoring force is directly proportional to the displacement from an equilibrium position and acts in the direction opposite to the displacement. This results in a sinusoidal movement over time. In contrast, periodic motion refers to any motion that repeats at regular intervals, regardless of the nature of the forces involved. Periodic motion encompasses a broader range of motions, which include but are not limited to SHM.
The defining characteristic of simple harmonic motion is its restoring force, which is always directed towards a fixed point (equilibrium) and varies linearly with distance from it. This creates a predictable, sine or cosine wave pattern of movement. Periodic motion, on the other hand, is defined merely by the repetition of the motion in a regular cycle, such as the orbiting of planets or the swinging of a pendulum. The forces in periodic motion may or may not vary linearly with displacement.
In simple harmonic motion, the period and frequency of the oscillations are determined by the properties of the system, like mass and spring constant in a mass-spring system. This motion is symmetrical and time-independent. Whereas, in periodic motion, the period and frequency can be influenced by a wider range of factors and the motion may not necessarily exhibit symmetry or a sinusoidal pattern as in SHM.
An example of simple harmonic motion is a mass attached to a spring, oscillating back and forth. This motion is predictable and mathematically simple. In contrast, periodic motion examples include the Earth’s rotation around the sun or a ferris wheel turning, where the motion repeats but the forces involved are not necessarily proportional to displacement, nor is the motion sinusoidal.
## Comparison Chart
### Nature of Restoring Force
Directly proportional to displacement
Varies, not always proportional
### Type of Motion
Sinusoidal and symmetrical
Can be non-sinusoidal, not always symmetrical
### Determinants of Period
Mass and spring constant (in a mass-spring system)
Various factors, depending on the system
### Mathematical Complexity
Simple, described by sine/cosine functions
Can be complex, not limited to simple equations
### Dependence on Equilibrium
Motion centered around an equilibrium position
May or may not involve an equilibrium position
## Simple Harmonic Motion and Periodic Motion Definitions
#### Simple Harmonic Motion
SHM is a type of periodic motion where the system's acceleration is directly proportional to its displacement and is directed towards a fixed point.
The vibrating string of a guitar follows simple harmonic motion.
#### Periodic Motion
Periodic motion includes both linear and circular movements that are repetitive.
The vibration of a tuning fork exhibits periodic motion.
#### Simple Harmonic Motion
SHM describes a motion that follows a sinusoidal pattern over time.
The movement of a metronome exemplifies simple harmonic motion.
#### Periodic Motion
This motion is characterized by constant frequency and period, regardless of the pattern.
The rotation of a Ferris wheel is a form of periodic motion.
#### Simple Harmonic Motion
Simple harmonic motion is oscillatory motion under a linear restoring force.
The oscillation of a pendulum in a clock exhibits simple harmonic motion.
#### Periodic Motion
It refers to a motion that returns to its initial position after a fixed period.
The hands of a clock undergo periodic motion.
#### Simple Harmonic Motion
SHM occurs when the force on an object is proportional and opposite to displacement.
A mass on a spring demonstrates simple harmonic motion when pulled and released.
#### Periodic Motion
Periodic motion is any motion that repeats at regular time intervals.
The Earth’s orbit around the Sun is an example of periodic motion.
#### Simple Harmonic Motion
Simple harmonic motion is characterized by energy oscillation between kinetic and potential forms.
The bob of a simple pendulum shows simple harmonic motion at small angles.
#### Periodic Motion
Periodic motion encompasses oscillatory movements that recur in a cycle.
The back and forth swinging of a playground swing is periodic motion.
## FAQs
#### What are the characteristics of simple harmonic motion?
Characteristics include sinusoidal movement, symmetry, and a constant period and frequency determined by the system's properties.
#### What is an example of simple harmonic motion?
A mass oscillating on a spring is a classic example of simple harmonic motion.
#### What defines simple harmonic motion?
Simple harmonic motion is defined by a restoring force proportional to displacement in the opposite direction.
#### How does simple harmonic motion differ from general periodic motion?
SHM is a specific type of periodic motion with a linear restoring force, while periodic motion can involve various types of forces and patterns.
#### Is the mathematical description of simple harmonic motion complex?
No, SHM is typically described using simple sine and cosine functions.
#### Does periodic motion always involve equilibrium?
No, unlike SHM, periodic motion may or may not involve an equilibrium position.
#### What distinguishes periodic motion from simple harmonic motion in terms of forces?
In periodic motion, the forces involved can be more complex and varied compared to the linear restoring force in SHM.
#### Can all periodic motions be considered simple harmonic?
No, only those with a linear restoring force proportional to displacement are SHM.
#### Are all simple harmonic motions periodic?
Yes, all simple harmonic motions are periodic by nature.
#### What are examples of periodic motion?
Examples include the Earth’s orbit, the swinging of a pendulum, and the rotation of a Ferris wheel.
#### What is periodic motion?
Periodic motion refers to any motion that repeats at regular intervals.
#### Can friction affect periodic motion?
Yes, friction can influence the period and amplitude of periodic motion.
#### Can periodic motion have varying periods?
Yes, depending on the system, periodic motion can have varying periods.
#### Can periodic motion be non-sinusoidal?
Yes, periodic motion can be non-sinusoidal, unlike SHM which always follows a sinusoidal pattern.
#### How is the period of simple harmonic motion determined?
It's determined by the system’s properties, like mass and spring constant in a mass-spring system.
#### What role does damping play in simple harmonic motion?
Damping reduces the amplitude of SHM over time, eventually stopping the motion.
#### Can periodic motion include linear motion?
Yes, periodic motion includes both linear and circular repetitive movements.
#### Is amplitude a defining feature of simple harmonic motion?
Amplitude is a characteristic but not a defining feature; the defining feature is the linear restoring force.
#### Is the energy in simple harmonic motion always conserved?
In an ideal SHM, energy is conserved, oscillating between kinetic and potential forms.
#### What factors can change the period of a periodic motion?
Factors include the system's mass, elasticity, and external forces like gravity and friction.
| 1,433
| 7,573
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.46875
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.901037
|
https://www.clutchprep.com/chemistry/practice-problems/98946/use-the-molar-volume-of-a-gas-at-stp-to-determine-the-volume-in-l-occupied-by-30-1
| 1,618,879,955,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038921860.72/warc/CC-MAIN-20210419235235-20210420025235-00231.warc.gz
| 806,407,880
| 30,930
|
Standard Temperature and Pressure Video Lessons
Concept: Standard Temperature and Pressure
# Problem: Use the molar volume of a gas at STP to determine the volume (in L) occupied by 30.1 g of neon at STP.
###### FREE Expert Solution
From the Ideal Gas Law:
$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{\mathbf{m}}{\mathbf{M}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{PV}=\frac{m}{\overline{)M}}\mathrm{RT}\right)\overline{)\mathbf{M}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\overline{)\mathbf{P}}\mathbf{V}\overline{)\mathbf{M}}}{\overline{)\mathbf{PM}}}\mathbf{=}\frac{\mathbf{mRT}}{\mathbf{PM}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{mRT}}{\mathbf{PM}}}$
100% (293 ratings)
###### Problem Details
Use the molar volume of a gas at STP to determine the volume (in L) occupied by 30.1 g of neon at STP.
| 372
| 988
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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-2021-17
|
latest
|
en
| 0.44885
|
http://www.softmath.com/parabola-in-math/slope/solving-two-step-inequality.html
| 1,575,772,862,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00087.warc.gz
| 234,971,623
| 13,298
|
English | Español
# Try our Free Online Math Solver!
Online Math Solver
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:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
solving two step inequality pre-algebra test
Related topics:
algebra for primary grades | free math equation worksheets grade 7 | how to solve polynomials | course information for linear algebra | trinominal factoring | free solving equations worksheets | learning standards for mathematics | combinations and permutations powerpoints | maths general test papers - yr 11 | exponent lesson plan | fractions and mix number example
Author Message Author Message
GJOH
Reg.: 23.07.2004
Posted: Thursday 04th of Jan 20:00 MichMoxon
Reg.: 21.08.2001
Posted: Sunday 07th of Jan 08:42
1. Hello Guys Can someone out there help me? My Hi Friend , Algebrator helped me with my learning
algebra teacher gave us solving two step inequality sessions last month . I got the Algebrator from
pre-algebra test homework today. Normally I am good
at 3x3 system of equations but somehow I am just >. Go ahead, check that and let us know your opinion. I
stuck on this one homework. I have to turn it in by this have even suggested Algebrator to a list of of my
weekend but it looks like I will not be able to complete it friends at college.
in time. So I thought of coming online to find assistance. I
will really be thankful if a math master can help me
work this (topicKwds) out in time. Vnode
Reg.: 27.09.2001
Posted: Monday 08th of Jan 10:07
oc_rana
Reg.: 08.03.2007
Posted: Friday 05th of Jan 21:55 I remember having difficulties with exponential equations,
equation properties and absolute values. Algebrator is a
Algebrator is one of the best resources that can offer really great piece of algebra software. I have used it
a helping hand to a person like you. When I was a through several algebra classes - Pre Algebra,
novice , I took support from Algebrator. Algebrator Remedial Algebra and Intermediate algebra. I would
offers all the principles of Algebra 1. Rather than utilizing simply type in the problem and by clicking on Solve, step
the Algebrator as a line-by-line tutor to solve all your by step solution would appear. The program is highly
math assignments, you can use it as a coach that can recommended.
offer the basics of graphing parabolas, simplifying
fractions and function composition. Once you assimilate
the principles, you can go ahead and solve any tough
problem on Intermediate algebra within minutes.
| 761
| 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}
| 2.5625
| 3
|
CC-MAIN-2019-51
|
longest
|
en
| 0.865779
|
https://brainly.com/question/229247
| 1,485,205,088,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00546-ip-10-171-10-70.ec2.internal.warc.gz
| 804,812,626
| 8,967
|
2014-12-17T02:14:10-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Cm=centimetres
Mm=millimetres
1 meter=100 centimeters
1 meter=1,000 millimeters
So to solve your equation, you divide 140 by 100cm (since he is measured in centimetres)
140cm÷100cm=1.4 meters-is Ben's height
To solve Mike's height, you divide 1090 by 1,000mm (since he is measured in Millimetres)
1,090mm÷1,000mm=1.09 meters-is Mike's height
1.4 meters is greater than 1.09 meters
To see by how much, you subtract 1.4m by 1.09
Mike is taller than Mike by .31 meters or 31 in centimetres or 310 in millimetres
| 247
| 853
|
{"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-2017-04
|
latest
|
en
| 0.922073
|
https://us.metamath.org/mpeuni/qliftrel.html
| 1,695,777,211,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00635.warc.gz
| 658,772,558
| 4,497
|
Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > qliftrel Structured version Visualization version GIF version
Theorem qliftrel 8179
Description: 𝐹, a function lift, is a subset of 𝑅 × 𝑆. (Contributed by Mario Carneiro, 23-Dec-2016.)
Hypotheses
Ref Expression
qlift.1 𝐹 = ran (𝑥𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)
qlift.2 ((𝜑𝑥𝑋) → 𝐴𝑌)
qlift.3 (𝜑𝑅 Er 𝑋)
qlift.4 (𝜑𝑋 ∈ V)
Assertion
Ref Expression
qliftrel (𝜑𝐹 ⊆ ((𝑋 / 𝑅) × 𝑌))
Distinct variable groups: 𝜑,𝑥 𝑥,𝑅 𝑥,𝑋 𝑥,𝑌
Allowed substitution hints: 𝐴(𝑥) 𝐹(𝑥)
Proof of Theorem qliftrel
StepHypRef Expression
1 qlift.1 . 2 𝐹 = ran (𝑥𝑋 ↦ ⟨[𝑥]𝑅, 𝐴⟩)
2 qlift.2 . . 3 ((𝜑𝑥𝑋) → 𝐴𝑌)
3 qlift.3 . . 3 (𝜑𝑅 Er 𝑋)
4 qlift.4 . . 3 (𝜑𝑋 ∈ V)
51, 2, 3, 4qliftlem 8178 . 2 ((𝜑𝑥𝑋) → [𝑥]𝑅 ∈ (𝑋 / 𝑅))
61, 5, 2fliftrel 6884 1 (𝜑𝐹 ⊆ ((𝑋 / 𝑅) × 𝑌))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 387 = wceq 1507 ∈ wcel 2050 Vcvv 3416 ⊆ wss 3830 ⟨cop 4447 ↦ cmpt 5008 × cxp 5405 ran crn 5408 Er wer 8086 [cec 8087 / cqs 8088 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1758 ax-4 1772 ax-5 1869 ax-6 1928 ax-7 1965 ax-8 2052 ax-9 2059 ax-10 2079 ax-11 2093 ax-12 2106 ax-13 2301 ax-ext 2751 ax-sep 5060 ax-nul 5067 ax-pow 5119 ax-pr 5186 ax-un 7279 This theorem depends on definitions: df-bi 199 df-an 388 df-or 834 df-3an 1070 df-tru 1510 df-ex 1743 df-nf 1747 df-sb 2016 df-mo 2547 df-eu 2584 df-clab 2760 df-cleq 2772 df-clel 2847 df-nfc 2919 df-ne 2969 df-ral 3094 df-rex 3095 df-rab 3098 df-v 3418 df-sbc 3683 df-dif 3833 df-un 3835 df-in 3837 df-ss 3844 df-nul 4180 df-if 4351 df-pw 4424 df-sn 4442 df-pr 4444 df-op 4448 df-uni 4713 df-br 4930 df-opab 4992 df-mpt 5009 df-id 5312 df-xp 5413 df-rel 5414 df-cnv 5415 df-co 5416 df-dm 5417 df-rn 5418 df-res 5419 df-ima 5420 df-iota 6152 df-fun 6190 df-fn 6191 df-f 6192 df-fv 6196 df-er 8089 df-ec 8091 df-qs 8095 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
| 1,178
| 2,068
|
{"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-2023-40
|
latest
|
en
| 0.138891
|
https://wiki-helper.com/find-the-area-of-a-circular-ring-formed-by-circumference-of-two-cocentric-circles-whose-radii-ar-37505592-51/
| 1,716,086,999,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00185.warc.gz
| 542,294,358
| 30,556
|
# find the area of a circular ring formed by circumference of two cocentric circles whose radii are 11 and 4
find the area of a circular ring formed by circumference of two cocentric circles whose radii are 11 and 4
About the author
### 1 thought on “find the area of a circular ring formed by circumference of two cocentric circles whose radii are 11 and 4”
1. Answer:
43.98 sq. cm
Step-by-step explanation:
Let the radii of the outer and inner circles be r
2 and r
1
respectively; we have
Area=πr
2
2
−πr
1
2
=π(r
2
2
−r
1
2
)
=π(r
2
−r
1
)(r
2
+r
1
)
=π(5.7−4.3)(5.7+4.3)=π×1.4×10 sq. cm
=3.1416×14sq.cm.=43.98 sq. cm
Reply
| 231
| 664
|
{"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}
| 4.25
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.85229
|
https://independentseminarblog.com/2019/04/01/rsa-algorithm-baiting/
| 1,560,977,173,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999041.59/warc/CC-MAIN-20190619204313-20190619230313-00123.warc.gz
| 467,004,764
| 23,233
|
RSA Algorithm – Baiting
In this blog, I would like to provide updates on the M3 Challenge and my research project in Abstract Algebra for Chester County Science Fair. Then, I will introduce the RSA Algorithm, the most used information encryption algorithm in the world and how it relates to my project.
M3 Challenge was certainly a successful experience, despite the fact that we messed up several places in our math model. If you still remember, M3 Challenge this year is about drug overdoses. In Question 2, we were asked to predict the drug overdose of nicotine, alcohol, marijuana, and non-prescribed opioids. In the math model, I was trying to find the percentage of people who misuse drugs in a specific population group whereas some of my teammates were finding among all “misusers” of the drug, how many belong to each population group. We didn’t have enough effort or communication to figure out this mistake during the competition, which leads us to the incorrect results.
During the start of Spring Break, three Westtown students including me participated in the Chester County Science Fair in different subjects. This year, I was the only participant in 12th Grade Mathematics so I automatically won first place and advanced into the Delaware Valley Science Fair. During the competition, I introduced my project to two judges, and I am quite sure that neither of them had a strong understanding of Abstract Algebra. However, our conversation was still meaningful. They asked me about the application of my project, which I wasn’t ready to explain at the time. Many times, one or two math theorems don’t mean anything in real life, so I just explained maybe scientists could use these theorems to solve equations. It turned out to be a dumb response. During Spring Break, however, I discovered the RSA Algorithm, which closely relates to Fermat’s Little Theorem, which I am going to explain right here.
Open the URL above, you will find a RSA simulator so that you can play with it and see what happens!
Let’s look at an example.
1. Normally, we would pick two very large prime numbers that are hundreds or thousands digits long. However, in this example, say we pick p = 13 and q = 17.
2. Then n = 21, and phi(n) = 192.
3. Say we pick u=7, and we confirm that gcd (7, 192) = 1
4. Then we want to find v such that 7v==1 (mod 192). This implies 7v – 1 = 192m for an integer m. We then find m = 2 and v = 55.
5. n = 21 and u = 7 are the public keys. v = 55 is the private key.
6. For instance, the message we want to encrypt is x = 12. Use the public key u = 7, then the encrypted message becomes y = 12^7 = 35,831,808.
7. Send 4,782,969 to your friend. To decrypt it, they need to use the private key v = 55.
How does the RSA Algorithm relate to Fermat’s Little Theorem? Here, I am providing a quick proof.
This part proves why our encryption and decryption method works. However, it is not a strict proof. To strictly prove the RSA Algorithm, we need to prove each detail in the process. Say in Step 4 above, we need to prove that such a v always exist and is always unique. If you are interested in learning more, please visit here. Finally, I would like to share a fun fact to conclude the RSA section. You may have seen companies advertising their to be 256-bit, 512-bit, etc. and didn’t understand what that number means. In fact, if the chip uses RSA Algorithm, it is highly likely that those numbers represent the length of u, v, or n. Wow, mind blooming, right? Obviously, to find the original text without knowing the private key would take hundreds, if not millions of years. This is why RSA is widely used and so safe!
In the very end of my blog, I would like to share my plan for Independent Seminar after spring break. I will keep working and preparing for the Delaware Valley Science Fair. After that, I will shift focus toward Intro Computer Science. Even though I don’t have any background in this field yet, I believe the journey will be fun.
Works Cited
AES and RSA Encryption. http://www.boxcryptor.com/en/encryption/. Accessed 1 Apr. 2019.
Blinder, S. M. RSA Encryption and Decryption. Mar. 2011, demonstrations.wolfram.com/RSAEncryptionAndDecryption/. Accessed 1 Apr. 2019.
Milanov, Evgeny. “The RSA Algorithm.” University of Washington, 3 June 2009, sites.math.washington.edu/~morrow/336_09/papers/Yevgeny.pdf. Accessed 1 Apr. 2019.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 1,057
| 4,459
|
{"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-2019-26
|
latest
|
en
| 0.967962
|
https://www.physicsforums.com/threads/drag-force-with-differential-equations-finding-max-speed.911314/
| 1,720,908,816,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00274.warc.gz
| 833,288,958
| 18,916
|
# Drag force with differential equations, finding max speed
• gelfand
In summary, the submarine engine provides maximum constant force ##F## to propel it through the water. When the engine is switched on to full power, the submarine starts from rest and moves horizontally in a straight line. If the mass of the submarine is ##M## the maximum possible speed the submarine can reach is ##v_{max} = Ae^{-Kt}##.
gelfand
## Homework Statement
A submarine engine provides maximum constant force ##F## to propel it through the
water.
Assume that the magnitude of the resistive drag force of the water experienced
by the submarine is ##kv##, where ##k## is the drag coefficient and ##v## is the
instantaneous speed of the boat.
When the engine is switched on to full power, the submarine starts from rest and
moves horizontally in a straight line.
If the mass of the submarine is ##M##
1) Write the differential equation of motion of the submarine in terms of its
speed ##v##
2) Solve the equation of motion from part (1) to find the speed ##v(t)## as a
function of time ##t##
3) What is the maximum possible speed ##v_{max}## the submarine can reach with
this engine?
## Homework Equations
Newtons second states ##F = ma##. Then given force ##F## I have ##F = ma##.
## The Attempt at a Solution
For the drag force ##-kv## I can write this as ##-Kmv## where ##K## is some constant
such that ##-Kmv = -kv##.
This gives me
$$ma = -Kmv$$
Dividing by ##m## and noting that ##a = \frac{dv}{dt}## gives
$$\frac{dv}{dt} = -Kv$$
Which is a differential equation of motion in terms of the speed ##v##.
Solving this differential equation as
$$\frac{1}{v} dv = -K dt$$
Then integrate both sides for
$$\ln(v) = -K t + C_1$$
Here I can take the exponential of each side and note that ##e^{x + b} = Ae^x##
where ##A,b## are constants.
So I have
$$v = Ae^{-Kt}$$For the maximum speed I need to consider where the force of motion by the engine
is balanced by the force of the drag.
At this point I will have
$$kv_{max} = ma$$
Using the solution from part (2) we can sub for ##v## as
$$k\left( A e^{-K t} \right) = ma$$I'm not sure how to obtain the maximum speed here though?
The expression can be rewritten as
$$\frac{kA}{ma} = e^{Kt}$$
Given ##K## is some constant this is just (where ##D## is some constant)
$$\frac{kA}{ma} = e^{t} e^{K} = De^{t}$$
So
$$\frac{kA}{Dma} = e^{t}$$
I don't think that this is correct, or even know if it makes sense.
It's suggesting that the constant ##k## multiplied by the constant ##A## divided by
the product of mass, acceleration and ##D## is equal to ##e^t##.
So I'm confused
Thanks
Last edited by a moderator:
The submarine engine is propelling the submarine with force F and this force is being opposed by the drag -Kmv. Therefore the net force on the submarine will be F-Kmv.
gelfand
cheers -
so I can use that at the equilibrium I will have
$$F - kv = 0$$
And this is the net force, so these are also equal to ##ma##, and can be expressed as
$$F - kv = ma$$
Note that ##a = \frac{dv}{dt}## which gives
$$F - kv = m \frac{dv}{dt}$$
From here I can divide through by mass, letting ##p = \frac{1}{m}## gives
$$pF - pkv = \frac{dv}{dt}$$
Then arranging the differential for integration gives
$$\frac{1}{pF - pdk} dv = dt$$
Integrate both sides
$$\int \frac{1}{pF - pdk} dv = \int dt$$
Here I can make the sub ##u = pF - pkv## such that ##du = -(pk) dv## and
##\frac{1}{-pk} du = dv##
Then subbing these in gives
$$-\frac{1}{pk}\int \frac{1}{u} du = \int dt$$
Solving this gives
$$-\frac{1}{pk} \ln(u) = t + C_1$$
Subbing back for ##u## gives
$$- \frac{1}{pk} \ln(pF - pkv) = t + C_1$$
Then if I wanted to solve for ##v## I can multiply both sides by ##- \frac{1}{pk}##
for (using a new constant ##C_2##)
$$\ln(pF - pkv) = -pkt + C_2$$
Then take exponentials as
$$pF - pkv = e^{-pkt + C_2}$$Then note that ##e^{-pkt + C_2}## is equal to ##e^{-pkt} \times C_3##. Rearranging
from this gives
$$pkv = pF - C_3 e^{-pkt}$$
Dividing through by ##pk## gives$$v = \frac{pF - C_3 e^{-pkt}}{pk}$$
As the solution to the differential in terms of ##v##Is this approach right?
gelfand said:
Dividing through by pkpk gives
v=pF−C3e−pktpk
v = \frac{pF - C_3 e^{-pkt}}{pk}
As the solution to the differential in terms of v
A good way to check the result, would be to plug it back into the original differential equation to see that m dv/dt and F - kv are indeed equal. But, the answer does look correct to me (provided you use the boundary conditions to find the value of C3 and use m instead of the placeholder variable p).
## 1. What is drag force and how is it related to differential equations?
Drag force is a resistive force that acts on an object moving through a fluid, such as air or water. It is proportional to the velocity of the object and can be calculated using differential equations, specifically the Navier-Stokes equations.
## 2. How can differential equations be used to find the maximum speed of an object?
Differential equations can be used to model the motion of an object and determine its velocity as a function of time. By setting the derivative of velocity with respect to time equal to zero, the maximum speed can be found when there is no acceleration.
## 3. What factors affect the drag force on an object?
The drag force on an object is affected by the density of the fluid, the object's shape and size, and the object's velocity. Additionally, the viscosity of the fluid and the presence of obstacles can also impact the drag force.
## 4. Can drag force be reduced to increase maximum speed?
Yes, drag force can be reduced by changing the shape or size of an object, or by changing the properties of the fluid it is moving through. This can allow for an increase in maximum speed as the object experiences less resistance.
## 5. Are there other methods besides differential equations to calculate drag force and maximum speed?
Yes, there are other methods such as using empirical formulas or conducting experiments. However, differential equations provide a more precise and comprehensive understanding of the relationship between drag force and maximum speed.
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
41
Views
3K
• Introductory Physics Homework Help
Replies
16
Views
538
• Introductory Physics Homework Help
Replies
19
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
693
• Introductory Physics Homework Help
Replies
2
Views
948
• Introductory Physics Homework Help
Replies
12
Views
416
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
218
| 1,854
| 6,743
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.862393
|
https://www.recruitmentinboxx.com/how-to-calculate-mat-score/106071/
| 1,638,250,515,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00267.warc.gz
| 1,060,653,693
| 15,021
|
Tuesday , November 30 2021
Home / Uncategorized / How To Calculate MAT Score | Percentile 2019 Conversion – Online Check
# How To Calculate MAT Score
MAT Scores will be considers for admission to MBA and allied programmes in B-Schools, so students who are wondering to know How to Calculate MAT Score, may go through entire page. MAT Scores is calculated on the basis of candidate’s performance in each section/subject.
How To Calculate MAT Score?
MAT Score will be considered to measure and evaluating the academic skills of many individuals. Check MAT Percentile 2019 Conversion by the following the below mentioned instructions.
Check MAT Score Calculation Online
Management Aptitude Test is standardized test held by All India Management Association (AIMA) to provide admission to students into MBA and allied programmes in more than 600 B-School located across India or international
Through MAT Examination, the exam authorities tests candidate’s familiarity with Data Analysis & Sufficiency, Language Comprehension, Intelligence & Critical Reasoning, Mathematical Skills and Indian & Global Environment.
We the team member of www.recruitmentinboxx.com has well presented this page with details related to How To Calculate MAT Score, to know MAT Score calculation procedure must go through structured page.
## How To Calculate MAT Score
MAT Score provides a common measure, administered under standard conditions, with known reliability and validity for evaluating the academic skills of many individuals. Check out procedure of How To Calculate MAT Score online
MAT Score And Its Calculation
• There are 06 MAT Scores on the MAT score card including Language Comprehension, Mathematical Skills, Data Analysis & Sufficiency, Intelligence & Critical Reasoning, Composite score, Indian & Global Environment
• Language Comprehension, Mathematical Skills, Data Analysis & Sufficiency and Intelligence & Critical Reasoning are reported on a scale of 0 to 100. In this section, scores less than 20 and more than 80 are rare.
• Indian & Global Environment score is calculated separately on a scale of 0 to 100. In this section too, scores less than 20 and more than 80 are rare.
• The composite score of MAT is reported on a scale of 199 to 801. Less than 200 or more than 800 is a rarity and all scores below 200 are reported as 199 and all scores above 800 are reported 801. In means that even if you have scored less than 199, your composite score will be reported as 199. Similarly, 801 or more will be reported as 801.
• Composite score is calculated on the basis of candidate’s performance in the first four sections of the examination. The first four sections are related to the skills required over a longer period of time in management field. Further equal weightage has been assigned to the first four sections.
• The most separate feature of MAT is that the score scales of the first four sections and the composite scores are based on the performance of candidates who wrote the exam in 1996. This is so because their scores were defined in a way that a score of about 50 in a section and about 500 on a composite scale represent the average performance of that group.
• Further, for each of the six scores, a percentile below figure is given. This figure indicates the percentage of examinees who scored below the candidate based on the entire MAT testing population. The percentile below figure can change with each administration for the same scores.
Press Here For More Details About MAT Score Interpretation
MAT Marking Scheme
MAT exam consists of 200 questions and each question carries one mark. Candidates will be awarded 1 mark for each correct answer and 0.25 marks will be deducted for every wrong answer, no marks will be deducted or given for unattempted questions
MAT Result 2019
All India Management Association (AIMA) will issue the MAT Result 2019, after successfully completion of Management Aptitude Test. To download the MAT Scorecard candidates need to visit the official website and enter Registered Form number and Roll number.
Students have to download the result through online mode as authority will not send hard copy of scores to any participants. After declaring the MAT February Result, MAT Merit List will be prepared. Those who are shortlisted will be called for MAT Counseling Procedure.
How To Check MAT 2019 Score?
Students who are wondering to check Management Aptitude Test scores may follow the below mentioned step by steps guidelines –
• Firstly visit to the Official Website
• After that hot the suitable link of MAT 2019 Scores
• Enter your Registered Form number and Roll number.
• Press on submit button to check MAT Exam Scores
• At last take the printouts of Results
MAT Exam Dates 2019:
For PBT February 2019:
Events Paper Based Test (PBT) Last Date for Online Registration (phase-II PBT) 08-02-2019 Availability of MAT Admit Card 09-02-2019 MAT Exam 17-02-2019 Timings 10.00 to 12.30 Hrs MAT Results Updating Soon
Check Out List Of: MBA Entrance Exam
For CBT February 2019:
Events Computer Based Test (CBT) Last Date for Online Registration 12-02-2019 Availability of MAT 2019 Admit Card 19-02-2019 MAT Exam(CBT) 23-02-2019 Timings In single/different time slots at specific test venues; subject candidates registration MAT Results Updating Soon
Check Out List Of: Top MBA Colleges In India
Final Note:
We hope with the help of this page of www.recruitmentinboxx.com you will get the answer of question that How To Calculate Mat Score? Still of you have any query regarding the same then you may mention your comments in below given comment box.
You May Also Like To Check This Section
MAT Exam Pattern MAT Exam Dates Management Aptitude Test Quiz MAT Previous Year Question Papers MAT Syllabus How To Prepare For MAT MAT Mock Test Series MAT Sample Papers MAT Application Form What To Do After Graduation?
| 1,241
| 5,900
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.87636
|
https://hwmadeeasy.com/find-the-period-of-repetition-of-the-following-sinusoidal-waveforms-and-express-them-in-phasor-representation-v_1-t-12-cos-314-t-10-degree-v-v_2-t-4-sin-2765t-75-degree-v-i_1-t-0/
| 1,627,206,183,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00576.warc.gz
| 323,349,514
| 41,024
|
# Find the period of repetition of the following sinusoidal waveforms and express them in phasor representation: v_1 (t) = 12 cos (314 t + 10 degree) V v_2 (t) = 4 sin (2765t – 75 degree) V i_1 (t) = -0.2 sin (600pi t – 100 degree) A i_2 (t) = -2.5 cos (377pi t + 120 degree) A
This content is for Premium members only.
| 115
| 320
|
{"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
| 0.773922
|
https://www.jiskha.com/display.cgi?id=1330197733
| 1,529,918,778,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867644.88/warc/CC-MAIN-20180625092128-20180625112128-00190.warc.gz
| 842,287,188
| 4,064
|
# physics
posted by John
A ball rolls with a speed of 4.4 m/s toward the edge of a table that is 1 m above the floor. The ball rolls off the table.
a. How long is it in the air? s
b. How far out from the edge of the table does the ball hit the floor? m
1. Elena
The ball after leaving the table
takes part in two motions: horizontal uniform motion
L = v(hor) •t,
and vertical free fall
H = gt^2/2,
t = sqroot(2H/g) = sqroot(2•1/9.8) = 0.2 s.
L = v(hor) •t = 4.4•0.2=0.88 m.
2. angel
a ball rolls 5cm in 5 seconds what is the speed of the ball?
## Similar Questions
1. ### Physics - Projectile Motion
A ball rolls horizontally off the edge of a tabletop that is 1.60 m high. It strikes the floor at a point 1.58 m horizontally away from the table edge. (Neglect air resistance.) (a) How long was the ball in the air?
2. ### Math/Physics
A ball rolls horizontally off the edge of a tabletop that is 1.60 m high. It strikes the floor at a point 1.58 m horizontally away from the table edge. (Neglect air resistance.)How long was the ball in the air?
3. ### Physics
A small ball rolls horizontally off the edge of a table that is 1.20 m high. It strikes the floor at a point 1.52 m horizontally from the table edge. How long is the ball in the air?
4. ### Phyics
A rubber ball with a radius of 4.8 cm rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.83 m to the floor below. If the ball completes 0.68 revolutions …
5. ### physics
A ball of radius 0.2m rolls along a horizontal table top with a constant linear speed of 3.04 m/s. The ball rolls off the edge and falls a vertical distance of 2.02m before hitting the floor. What is the angular displacement of the …
6. ### physics
A ball rolls with a speed of 6.4 m/s toward the edge of a table that is 0.7 m above the floor. The ball rolls off the table. a. Find the magnitude of the balls velocity when it hits the floor. m/s b. At what angle, measured from the …
7. ### physics
A ball of radius 0.211 m rolls along a horizontal table top with a constant linear speed of 3.41 m/s. The ball rolls off the edge and falls a vertical distance of 2.27 m before hitting the floor. What is the angular displacement of …
8. ### Physics
A ball rolls off a tabletop 0.900 m above the floor and lands on the floor 2.60 m away from a point that is directly under the edge of the table. At what speed did the ball roll off of the table?
9. ### Physics
A small ball rolls horizontally off the edge of a tabletop of height h. It strikes the floor a distance x horizontally away from the edge of the table. (Use any variable or symbol stated above along with the following as necessary: …
10. ### Physics
A ball is rolled off a table that is 0.481 m above the floor. The ball is rolling with a velocity if 1.8 m/s as it goes off the edge of the table. At the exact instant the first ball rolls off the table a second ball is dropped from …
More Similar Questions
| 806
| 3,001
|
{"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-2018-26
|
latest
|
en
| 0.799833
|
http://otrnet.com.au/index.php/integrated-maths-modules-135/product/90-h06s-probability/category_pathway-19
| 1,542,745,755,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039746639.67/warc/CC-MAIN-20181120191321-20181120213321-00470.warc.gz
| 239,721,649
| 12,074
|
# H06S: Probability
\$4.00 each
+ –
Stage/No: H06 Length: 3 weeks CF Outcomes: 12, 14 Levels: 6, 7, 8 Due out: available now
{slider=Publishing}
H06S: Probability 3 week module Old ISBN: 1876800658 New APN: 9781876800659
{slider=CF Outcomes}
This information is found in the wrap around pages of the teachers module. It details the Curriculum Framework Outcomes addressed in the module.
Outcomes 1 & 2 should be an inherent part of all mathematics lessons.
1. Show a disposition to use mathematics to assist with understanding new situations, solving problems and making decisions, showing initiative, flexibility and persistence when working mathematically and a positive attitude to their own continued involvement in learning and doing mathematics.
2. Appreciate that mathematics has its origins in many cultures, and its forms reflect specific social and historical contexts, and understand its significance in explaining and influencing aspects of our lives.
Outcomes 3, 4 & 5 "Working Mathematically" are an integral part in the design of every module. These outcomes determined the three stage learning process developed in each module.
3. Call on a repertoire of general problem-solving techniques, appropriate technology and personal and collaborative management strategies when working mathematically.
4. Choose mathematical ideas and tools to fit the constraints in a practical situation, interpret and make sense of the results within the context and evaluate the appropriateness of the methods used.
5. Investigate, generalise and reason about patterns in number, space and data, explaining and justifying conclusions reached.
Outcomes 12 and 14 from “Chance & Data†are specific to this module.
12. Understand and use the everyday language of chance and make statements about how likely it is that an event will occur based on experience, experiments and analysis.
14. Locate, interpret, analyse and draw conclusions from data, taking into account data collection techniques and chance processes involved.
{slider=Student Outcomes}
### Major Student Outcomes
This information is found in the wrap around pages of the teachers module. It details the major student outcomes that should have been developed during the module. (Student Outcome Statement numbering is first, e.g. SOS 5.1a, Progress Map numbering is bracketed e.g. [PM 15.5])
### Working Mathematically
Working mathematically outcomes at levels 6, 7 and 8 are interwoven with the structure of these modules.
### Chance and Data
C&D6.1 [C&D12.6] Estimates probabilities and proportions based on primary or secondary data collection and assigns probabilities for one- and two-stage events by reasoning about equally likely outcomes.
C&D7.1 [C&D12.7] Estimates probabilities, proportions, means and medians based on primary and secondary data collection and assigns probabilities using complementarity and independence.
C&D6.4 [C&D14.6] Interprets, makes comparisons and describes relationships in collected and published data from tables, diagrams, plots, graphs, prose, summary statistics and databases, distinguishing sample and population data.
C&D7.4 [C&D14.7] Selects and interprets information from a wide range of collected and published data including to construct arguments to support or refute a position taken.
C&D8 [C&D12.8, 14.8] Comments on data in terms of what conclusions might reasonably be drawn from them, distinguishing between conclusions reached on the basis of evidence and those reached on other grounds, such as personal opinion or habit.
### Other Strands
To a lesser degree some outcomes from other strands are also integrated into this module.
{slider=Details}
This information is found in the wrap around pages of the teachers module. It gives details about the module.
### Module Length:
Approximately 3 weeks (12 hours). This may vary upward according to the ability and prior understandings of the students.
### Outcomes Levels:
Includes outcomes from levels 6, 7 & 8.
### Stage/Number:
Each Integrated Maths Module has been assigned a stage and number. The stage is designed to help teachers in their sequencing of modules. The number is for identification of each module.
This module is H06, that is stage H module 6.
The content of this module is:
Sample spaces
Two way tables
Complementary events
Compound events
Counting techniques
### Language development:
The following terms (or derivations of them) are an essential part of this module:
probability
relative frequency
simulation
random
random number
complementary events
independent events
sample space
odds
event
tree diagram
These terms are in bold whenever they appear in this text.
{slider=Samples}
The structure of all modules is to present materials in three stages:
### Exploration - Formalisation - Application
The files available here for downloading are a Sample Activity from the Exploration Stage of this module and a Sample Application from the Application stage of this module.
These files are available as pdf files. To view and print these files you will need a program like Adobe Acrobat Reader which is available free here.
These links are to the sample files:
Sample Activity
Sample Application
This information is found in the wrap around pages of the teachers module produced for Western Australian schools. It details the links to the old Unit Curriculum objectives for schools trying to adapt programmes etc.
Unit Curriculum Objectives covered are:
From Maths Development 6.3
• I 6.1 Construct sample spaces to develop the concepts of complementary and compound events.
• I 6.5 Read and interpret two-way tables.
• I 6.6 Use sample spaces as a method of counting in determining probabilities, including those of compound and complementary events.
• I 6.7 Make inferences from two-way tables, using notions of probability.
{/sliders}
OTRNet © 2016 | OTRNet: Online Teachers' Resource Network
| 1,249
| 5,948
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2018-47
|
longest
|
en
| 0.886567
|
https://thekidsworksheet.com/addition-and-subtraction-worksheets-for-grade-2/
| 1,718,745,423,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00199.warc.gz
| 486,854,171
| 22,454
|
Plunge into practice with our addition and subtraction worksheets featuring oodles of exercises to practice performing the two basic arithmetic operations of addition and subtraction. Using the sheets in this section will help your child to.
Kumon Publishing Kumon Publishing Grade 2 Subtraction Math Worksheets Kumon Math 1st Grade Worksheets
### These grade 2 addition worksheets span topics from adding single digit numbers to addition in columns with regrouping all worksheets are printable pdf documents with answer pages.
Addition and subtraction worksheets for grade 2. They teach how the commutative property of addition produces the same result for two different math facts and also show how subtraction is the. These word problem worksheets place 2nd grade math concepts in a context that grade 2 students can relate to. In some questions students are asked to write the addition or subtraction equation which describes.
They cover 2nd grade topics ranging from basic subtraction facts to subtracting in columns with regrouping. Worksheets math grade 2 word problems mixed addition subtraction 1 2 digits. Printable 2nd grade math worksheets addition and subtraction ccss 2 oa 1 worksheets.
Mixed addition subtraction with 1 2 digits. These 100 problem fact family worksheets are a great workout for addition and subtraction facts. Grade 2 subtraction worksheets.
Presenting a mixed review of addition and subtraction of single digit 2 digit 3 digit 4 digit and 5 digit numbers each pdf practice set is designed to suit. Use them in a no stress way to refresh fact fluency or they are perfect for a two minute timed test. These worksheets are generated automatically each time you click on a link.
Questions may have 2 or 3 terms. Math word problem worksheets for grade 2. Often when we focus on only a single type of math fact at a time progressing our way through addition subtraction multiplication and division facts we can find that students become modal when looking at a worksheet.
Showing top 8 worksheets in the category addition and subtraction. They learn the basic subtraction facts by heart subtract mentally in various ways and learn regrouping borrowing in subtraction with two and three digit numbers. Grade 2 subtraction worksheets.
Worksheets math grade 2 addition. Subtraction math worksheets pdf printable subtraction math worksheets for different grades 1st 2nd 3rd 4th 5th 6th 7th grades subtract with or without regrouping subtraction word problems subtract with pictures subtract large numbers subtract decimals. It is important when learning the basic math operations to develop the skill of looking at the operation itself on each problem.
All numbers are less than 100. Worksheets math grade 2 subtraction. Some of the worksheets displayed are year 2 maths addition and subtraction workbook addition and subtraction ks2 sats standard work grade 4 addition and subtraction word problems addition and subtraction of matrices 1 mixed addition subtraction word problems addition and subtraction of decimals one step word problems.
We provide math word problems for addition subtraction multiplication time money and fractions. Our grade 2 subtraction worksheets provide the practice needed to master basic subtraction skills. Below are six grade 2 word problem worksheets with a mix of addition and subtraction problems.
Twodigitsubtractionwithoutregrouping Jpg 1 275 1 650 Pixels Math Subtraction 2nd Grade Worksheets Subtraction Worksheets
Addition Subtraction 60 Printable Worksheets With Single Etsy First Grade Math Worksheets Kids Math Worksheets Math Workbook
5 Free Math Worksheets Second Grade 2 Subtraction Subtract 3 Digit Numbers In 2020 Addition And Subtraction Worksheets Subtraction Worksheets Math Addition Worksheets
2 Digit Subtraction Worksheet 3worksheets 2nd Grade Math Worksheets Subtraction With Regrouping Worksheets 2nd Grade Math
3 Free Math Worksheets Second Grade 2 Subtraction Subtracting 1 Digit From Addition And Subtraction Worksheets Subtraction Worksheets 2nd Grade Math Worksheets
Flamingo Fun Practice Two Digit Subtraction Worksheet Education Com Math Worksheets Subtraction Worksheets 2nd Grade Math Worksheets
Subtraction Worksheets Grade 2 Subtraction Worksheets Math Worksheets Printable Math Worksheets
4 Free Math Worksheets Second Grade 2 Addition Adding Whole Tens 3 Digits Missing Number In 2020 Fun Math Worksheets Subtraction Practice Math Subtraction
2 3 Or 4 Digits Mixed Operator Worksheets Subtraction Worksheets Addition And Subtraction Worksheets Math Subtraction
4 Free Math Worksheets Second Grade 2 Addition Adding 2 Digit Plus 1 Digit No Regroup 005 Add In 2020 2nd Grade Math Worksheets Math Worksheets Free Math Worksheets
| 913
| 4,736
|
{"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}
| 3.078125
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.896209
|
http://www.r-bloggers.com/soccer-is-all-about-money-part-2-simple-analyses/
| 1,469,612,330,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00212-ip-10-185-27-174.ec2.internal.warc.gz
| 653,655,056
| 16,459
|
# Soccer is all about money (?) – Part 2: Simple analyses
October 18, 2012
By
(This article was first published on Rcrastinate, and kindly contributed to R-bloggers)
Alright, now we have all the data we need in one dataframe. To make this code work, I assume you ran the code from Part 1. We need the dataframe big.tab.
All the data presented here is based on the data from 18/10/2012. You can run an analysis with the actual data or I can do it at some point later in the season.
Let’s plot some stuff. How about the old german saying about soccer “Geld schießt keine Tore” (Money doesn’t score goals)? Let’s look into this.
plot(big.tab\$Value, big.tab\$Goals.for, type = “n”, axes = F, xlab = “Value”, ylab = “Goals”)
text(x = big.tab\$Value, y = big.tab\$Goals.for, labels = big.tab\$Team, cex = 0.7, col = “#65656599”)
axis(side = 1)
axis(side = 2)
We get this… (clickable)
Sorry for the overlapping team names. But you get the gist: It looks like the value of a team covaries with the number of goals for that team.
Now we add a regression line. This means, we predict the number of goals for one team by the value of that team. We also add a Pearson correlation coefficient (r) and its associated p value in the subtitle of the plot. We get this…
How do we interpret this? There are several conclusions that could be drawn.
(1) The value of a team in the british Premier League is reliably correlated with the number of goals that team scored in the championship so far (after 7 games). Beware: Correlation does not imply causation.
(2) The “best guess” of predicting the number of goals by the value of a team is visualized by the dashed red line in the second plot. This means that there are teams who “over-perform” and “under-perform” in relation to their value. FC Fulham, for example, shot way “too many” goals given its value. FC Liverpool, on the other hand, should have shot more goals, because they are under the red line.
(3) One could infer from this plot that it is quite difficult for very valuable teams (e.g., ManU, ManCity and the FC Chelsea) to over-perform since the regression line is so steadily rising. So, they have to score many many goals to outperform their level on the regression line.
By the way: This also works quite good for the value of a team and the points they achieved in the championship (win = 3 points, draw = 1 point).
In the next post, I will do some more analyses and plots with this dataset. And I will try to compare different european championships.
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...
| 703
| 2,826
|
{"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-2016-30
|
latest
|
en
| 0.938377
|
https://docs.juliahub.com/General/Ai4EMetaPSE/stable/Examples/example/
| 1,721,086,573,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00542.warc.gz
| 193,730,559
| 5,459
|
# Differential Equations Json Examples
## CommonJson
using Ai4EMetaPSE
str = """{
"name": "Name",
"pkgs": [
"ModelingToolkit",
"DifferentialEquations"
],
"variables": [
"x(t) = 1.0",
"y(t) = 1.0",
"z(t) = 2.0"
],
"parameters": [
"σ = 1.0",
"ρ = 3.0",
"β = 5.0"
],
"equations": [
"der(x) = σ*(y - x)",
"der(y) = x*(ρ - z) - y",
"der(z) = x*y - β*z"
],
"u0": [
"x => 1.0",
"y => 2.0",
"z => 3.0"
],
"timespan": [0,1,0.1],
"solver": "Rosenbrock23"
}"""
solution = generatecode(str, CommonJson())
The generated codes are:
quote
using Pkg
pkgNeeds = ["ModelingToolkit", "DifferentialEquations"]
nothing
else
end
using ModelingToolkit, DifferentialEquations
#= none:1 =# @variables begin
t
x(t) = begin
#= none:1 =#
1.0
end
y(t) = begin
#= none:1 =#
1.0
end
z(t) = begin
#= none:1 =#
2.0
end
end
#= none:1 =# @parameters begin
σ = 1.0
ρ = 3.0
β = 5.0
end
begin
der = Differential(t)
eqs = []
append!(eqs, [der(x) ~ σ * (y - x)])
append!(eqs, [der(y) ~ x * (ρ - z) - y])
append!(eqs, [der(z) ~ x * y - β * z])
end
init = Dict(x => 1.0, y => 2.0, z => 3.0)
timespan = (0.0, 1.0)
Name = solve(ODEProblem(structural_simplify(ODESystem(eqs, t; name = :Model)), init, timespan), Rosenbrock23())
end
## ModelJson
using Ai4EMetaPSE
str = """{
"name": "Project Name",
"pkgs": [
"ModelingToolkit",
"DifferentialEquations",
"Ai4EComponentLib.IncompressiblePipe"
],
"components": [
{
"name": "Pump",
"type": "CentrifugalPump",
"args": {
"ω": 5000
}
},
{
"name": "A",
"type": "Sink_P",
"args": {}
},
{
"name": "B",
"type": "Sink_P",
"args": {}
},
{
"name": "Pipe1",
"type": "SimplePipe",
"args": {
"L": 2.0
}
},
{
"name": "Pipe2",
"type": "SimplePipe",
"args": {
"L": 7.0
}
},
{
"name": "Pipe3",
"type": "SimplePipe",
"args": {
"L": 7.0
}
},
{
"name": "Pipe4",
"type": "SimplePipe",
"args": {
"L": 9.0
}
},
{
"name": "Pipe5",
"type": "SimplePipe",
"args": {
"L": 5.0
}
},
{
"name": "Pipe6",
"type": "SimplePipe",
"args": {
"L": 4.0
}
},
{
"name": "Pipe7",
"type": "SimplePipe",
"args": {
"L": 5.0
}
},
{
"name": "Pipe8",
"type": "SimplePipe",
"args": {
"L": 1.0
}
},
{
"name": "Pipe9",
"type": "SimplePipe",
"args": {
"L": 10.0
}
},
{
"name": "Pipe10",
"type": "SimplePipe",
"args": {
"L": 2.0
}
},
{
"name": "Pipe11",
"type": "SimplePipe",
"args": {
"L": 2.0
}
},
{
"name": "Pipe12",
"type": "SimplePipe",
"args": {
"L": 3.0
}
},
{
"name": "Pipe13",
"type": "SimplePipe",
"args": {
"L": 2.0
}
},
{
"name": "Pipe14",
"type": "SimplePipe",
"args": {
"L": 1.0
}
},
{
"name": "Pipe15",
"type": "SimplePipe",
"args": {
"L": 2.0
}
},
{
"name": "Pipe16",
"type": "SimplePipe",
"args": {
"L": 3.0
}
},
{
"name": "Pipe17",
"type": "SimplePipe",
"args": {
"L": 6.0
}
},
{
"name": "Pipe18",
"type": "SimplePipe",
"args": {
"L": 6.0
}
},
{
"name": "Pipe19",
"type": "SimplePipe",
"args": {
"L": 6.0
}
},
{
"name": "Pipe20",
"type": "SimplePipe",
"args": {
"L": 1.0
}
},
{
"name": "Pipe21",
"type": "SimplePipe",
"args": {
"L": 1.0
}
},
{
"name": "Pipe22",
"type": "SimplePipe",
"args": {
"L": 7.0
}
},
{
"name": "Pipe23",
"type": "SimplePipe",
"args": {
"L": 3.0
}
},
{
"name": "Pipe24",
"type": "SimplePipe",
"args": {
"L": 3.0
}
},
{
"name": "Pipe25",
"type": "SimplePipe",
"args": {
"L": 2.0
}
}
],
"connections": [
[
"A.port",
"Pump.in"
],
[
"Pump.out",
"Pipe1.in"
],
[
"Pipe1.out",
"Pipe2.in",
"Pipe5.in"
],
[
"Pipe2.out",
"Pipe3.in",
"Pipe6.in"
],
[
"Pipe3.out",
"Pipe4.in",
"Pipe7.in"
],
[
"Pipe4.out",
"Pipe10.out",
"Pipe14.in"
],
[
"Pipe5.out",
"Pipe11.in",
"Pipe12.in"
],
[
"Pipe6.out",
"Pipe8.in",
"Pipe9.in"
],
[
"Pipe7.out",
"Pipe9.out",
"Pipe10.in"
],
[
"Pipe12.out",
"Pipe8.out",
"Pipe13.in"
],
[
"Pipe13.out",
"Pipe14.out",
"Pipe15.in"
],
[
"Pipe11.out",
"Pipe19.in",
"Pipe16.in"
],
[
"Pipe16.out",
"Pipe17.in",
"Pipe20.in"
],
[
"Pipe17.out",
"Pipe18.in",
"Pipe21.in"
],
[
"Pipe18.out",
"Pipe15.out",
"Pipe22.in"
],
[
"Pipe19.out",
"Pipe20.out",
"Pipe23.in"
],
[
"Pipe21.out",
"Pipe22.out",
"Pipe24.in"
],
[
"Pipe23.out",
"Pipe24.out",
"Pipe25.in"
],
[
"B.port",
"Pipe25.out"
]
],
"u0": [],
"timespan": [
0,
0,
0
],
"solver": "Rosenbrock23"
}"""
solution = generatecode(str, ModelJson())
The generated codes are:
quote
using Pkg
pkgNeeds = ["ModelingToolkit", "DifferentialEquations", "Ai4EComponentLib"]
nothing
else
end
using ModelingToolkit, DifferentialEquations, Ai4EComponentLib.IncompressiblePipe
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pump = CentrifugalPump(ω = 5000)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named A = Sink_P()
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named B = Sink_P()
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe1 = SimplePipe(L = 2)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe2 = SimplePipe(L = 7)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe3 = SimplePipe(L = 7)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe4 = SimplePipe(L = 9)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe5 = SimplePipe(L = 5)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe6 = SimplePipe(L = 4)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe7 = SimplePipe(L = 5)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe8 = SimplePipe(L = 1)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe9 = SimplePipe(L = 10)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe10 = SimplePipe(L = 2)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe11 = SimplePipe(L = 2)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe12 = SimplePipe(L = 3)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe13 = SimplePipe(L = 2)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe14 = SimplePipe(L = 1)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe15 = SimplePipe(L = 2)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe16 = SimplePipe(L = 3)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe17 = SimplePipe(L = 6)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe18 = SimplePipe(L = 6)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe19 = SimplePipe(L = 6)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe20 = SimplePipe(L = 1)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe21 = SimplePipe(L = 1)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe22 = SimplePipe(L = 7)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe23 = SimplePipe(L = 3)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe24 = SimplePipe(L = 3)
#= d:\postgraduate\project\develop\Ai4EMetaPSE.jl\src\solution.jl:120 =# @named Pipe25 = SimplePipe(L = 2)
components = [Pump, A, B, Pipe1, Pipe2, Pipe3, Pipe4, Pipe5, Pipe6, Pipe7, Pipe8, Pipe9, Pipe10, Pipe11, Pipe12, Pipe13, Pipe14, Pipe15, Pipe16, Pipe17, Pipe18, Pipe19, Pipe20, Pipe21, Pipe22, Pipe23, Pipe24, Pipe25]
eqs = [connect(A.port, Pump.in), connect(Pump.out, Pipe1.in), connect(Pipe1.out, Pipe2.in, Pipe5.in), connect(Pipe2.out, Pipe3.in, Pipe6.in), connect(Pipe3.out, Pipe4.in, Pipe7.in), connect(Pipe4.out, Pipe10.out, Pipe14.in), connect(Pipe5.out, Pipe11.in, Pipe12.in), connect(Pipe6.out, Pipe8.in, Pipe9.in), connect(Pipe7.out, Pipe9.out, Pipe10.in), connect(Pipe12.out, Pipe8.out, Pipe13.in), connect(Pipe13.out, Pipe14.out, Pipe15.in), connect(Pipe11.out, Pipe19.in, Pipe16.in), connect(Pipe16.out, Pipe17.in, Pipe20.in), connect(Pipe17.out, Pipe18.in, Pipe21.in), connect(Pipe18.out, Pipe15.out, Pipe22.in), connect(Pipe19.out, Pipe20.out, Pipe23.in), connect(Pipe21.out, Pipe22.out, Pipe24.in), connect(Pipe23.out, Pipe24.out, Pipe25.in), connect(B.port, Pipe25.out)]
init = Dict()
timespan = (0.0, 0.0)
Project_Name = solve(ODEProblem(structural_simplify(compose(ODESystem(eqs, t; name = :Model), components; name = :system)), init, timespan), Rosenbrock23())
end
## ComponentsJson
using Ai4EMetaPSE
str = """{
"name":"TransitionPipe",
"args":[
"λ1=1.0",
"λ2=1.0",
"λ3=1.0",
"n=10",
"f=0.016",
"D=0.2",
"L=100",
"T=300",
"pins=0.56e6",
"pouts=0.56e6"
],
"custom_Code":[
"RT = 287.11 * T",
"A0 = pi / 4 * D^2",
"c10 = RT / A0",
"c20 = c10 * f / 2 / D",
"dx=L/n",
"qms = sqrt(abs(pins^2 - pouts^2) / (f * L * RT / D / A0 / A0))",
"pms = map(i->sqrt(pins^2 * (1 - (i-1) / n) + pouts^2 * (i-1) / n),1:n+1)"
],
"components":[
{
"name": "inlet",
"type": "FlowPort",
"args": {
"T": 300
}
},
{
"name": "outlet",
"type": "FlowPort",
"args": {
"T": 300
}
}
],
"variablesInclude":[],
"variables": [
"qm[1:n](t)=qms",
"p[1:n+1](t)=pms"
],
"parameters": [
"A = A0*λ2",
"c1 = c10*λ1",
"c2 = c20*λ3",
"dx = L / n",
"f = f"
],
"equations": [
"[der(p[i]) = c1 * (qm[i-1] - qm[i]) / dx for i = 2:n]",
"[der(qm[i]) = (c1 * qm[i]^2 / (0.5 * (p[i+1] + p[i]))^2 - A) * (p[i+1] - p[i]) / dx + c1 * qm[i] / (0.5 * (p[i+1] + p[i])) * (qm[i-1] - qm[i+1]) / dx - c2 * qm[i] * abs(qm[i]) / (0.5 * (p[i+1] + p[i])) for i = 2:n-1]",
"p[1] = inlet.p",
"p[n+1] = outlet.p",
"qm[n] = -outlet.qm",
"qm[1] = inlet.qm",
"der(qm[1]) = (c1 * qm[1]^2 / (0.5 * (p[2] + p[1]))^2 - A) * (p[2] - p[1]) / dx + c1 * qm[1] / (0.5 * (p[2] + p[1])) * (3 * qm[1] - 4 * qm[2] + qm[3]) / dx - c2 * qm[1] * abs(qm[1]) / (0.5 * (p[2] + p[1]))",
"der(qm[n]) = (c1 * qm[n]^2 / (0.5 * (p[n+1] + p[n]))^2 - A) * (p[n+1] - p[n]) / dx + c1 * qm[n] / (0.5 * (p[n+1] + p[n])) * (-3 * qm[n] + 4 * qm[n-1] - qm[n-2]) / dx - c2 * qm[n] * abs(qm[n]) / (0.5 * (p[n+1] + p[n]))"
]
}"""
solution = generatecode(str, ComponentsJson())
The generated codes are:
function TransitionPipe(; name, λ1 = 1.0, λ2 = 1.0, λ3 = 1.0, n = 10, f = 0.016, D = 0.2, L = 100, T = 300, pins = 560000.0, pouts = 560000.0)
begin
RT = 287.11T
A0 = (pi / 4) * D ^ 2
c10 = RT / A0
c20 = ((c10 * f) / 2) / D
dx = L / n
qms = sqrt(abs(pins ^ 2 - pouts ^ 2) / ((((f * L * RT) / D) / A0) / A0))
pms = map((i->begin
#= none:1 =#
sqrt(pins ^ 2 * (1 - (i - 1) / n) + (pouts ^ 2 * (i - 1)) / n)
end), 1:n + 1)
end
#= c:\Users\A\Desktop\git\Ai4EMetaPSE.jl\src\solution.jl:127 =# @named inlet = FlowPort(T = 300)
#= c:\Users\A\Desktop\git\Ai4EMetaPSE.jl\src\solution.jl:127 =# @named outlet = FlowPort(T = 300)
components = [inlet, outlet]
#= none:1 =# @variables begin
t
(qm[1:n])(t) = begin
#= none:1 =#
qms
end
(p[1:n + 1])(t) = begin
#= none:1 =#
pms
end
end
#= none:1 =# @parameters begin
A = A0 * λ2
c1 = c10 * λ1
c2 = c20 * λ3
dx = L / n
f = f
end
begin
der = Differential(t)
eqs = []
append!(eqs, [[der(p[i]) ~ (c1 * (qm[i - 1] - qm[i])) / dx for i = 2:n]])
append!(eqs, [[der(qm[i]) ~ ((((c1 * qm[i] ^ 2) / (0.5 * (p[i + 1] + p[i])) ^ 2 - A) * (p[i + 1] - p[i])) / dx + (((c1 * qm[i]) / (0.5 * (p[i + 1] + p[i]))) * (qm[i - 1] - qm[i + 1])) / dx) - (c2 * qm[i] * abs(qm[i])) / (0.5 * (p[i + 1] + p[i])) for i = 2:n - 1]])
append!(eqs, [p[1] ~ inlet.p])
append!(eqs, [p[n + 1] ~ outlet.p])
append!(eqs, [qm[n] ~ -(outlet.qm)])
append!(eqs, [qm[1] ~ inlet.qm])
append!(eqs, [der(qm[1]) ~ ((((c1 * qm[1] ^ 2) / (0.5 * (p[2] + p[1])) ^ 2 - A) * (p[2] - p[1])) / dx + (((c1 * qm[1]) / (0.5 * (p[2] + p[1]))) * ((3 * qm[1] - 4 * qm[2]) + qm[3])) / dx) - (c2 * qm[1] * abs(qm[1])) / (0.5 * (p[2] + p[1]))])
append!(eqs, [der(qm[n]) ~ ((((c1 * qm[n] ^ 2) / (0.5 * (p[n + 1] + p[n])) ^ 2 - A) * (p[n + 1] - p[n])) / dx + (((c1 * qm[n]) / (0.5 * (p[n + 1] + p[n]))) * ((-3 * qm[n] + 4 * qm[n - 1]) - qm[n - 2])) / dx) - (c2 * qm[n] * abs(qm[n])) / (0.5 * (p[n + 1] + p[n]))])
compose(ODESystem(eqs, t, sts, ps; name = name), components)
end
end
| 5,015
| 12,029
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.178206
|
http://circuitcalculator.com/wordpress/2007/09/20/wire-parameter-calculator/
| 1,524,679,307,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00062.warc.gz
| 60,098,970
| 10,286
|
Wire Parameter CalculatorSeptember 20, 2007
This Javascript web calculator will calculate the resistance and ampacity for copper wire based on the gauge. Both metric (mm) and American Wire Gauge (AWG) are supported. Note: Ampacity is based on a curve fit to MIL-STD-975. To see the wire table that this calculator is based on as well as important information about wire insulation temperature ratings, click here.
Features:
• Results update as you type
• Several choices of units
• Units and other settings are saved between sessions
• Blog format allows user comments
Inputs:
Wire Size AWG mm
Optional Inputs:
Wire Temperature Deg. C F Wire Length m cm ft inches Number of Wires in Bundle
Results (per each wire):
Resistance Ohms Single Wire Ampacity Amps Wire Bundle Ampacity (per wire) Amps Copper Diameter mils mm AWG Copper Area mils^2 mm^2 Copper Weight kg g lbs oz
1. Gene - October 11, 2007
Having an option to select aluminum wires instead of copper would also be handy.
2. Brad - October 11, 2007
Hi Gene,
That is a good idea. The spec this calculator is based on does not cover aluminum wires so I will have to think about how to scale and adapt it. For sure the resistance is easy to calculate, but the ampacity is the tricky one.
3. Paul - October 23, 2007
Who uses Aluminum wire these days? I thought they banned that stuff in the 70’s? Aluminum melts at a relatively low temperature and doesn’t conduct as well. I wouldn’t waste time calculating that.
4. Paul - October 23, 2007
Thanks for the Calculator BTW. It’s the best one I’ve used so far.
5. al - October 25, 2007
Can you determine the different speeds of a motor by checking the windings or wires of a 3 speed single phase motor using an ohm meter?
[Al, I don’t think so - or at least I don’t know how. Brad]
6. Steve - October 26, 2007
Paul, aluminum wire is not “banned” per se. Nor was its use discontinued due to the melting temperature, but the galvanic effect when bonded with other metals causing corrosion. It is still used in controlled environments.
7. A. James Flynn - November 9, 2007
All of those drop feeders to your homes and bussiness are “ALUMINUM” so Paul you are one of those who uses “ALUMINUM” wire these days!
8. Peter Manins - November 17, 2007
Hmm,
If I put in a multistrand copper cable: 0.19 mm wire diameter, 252 strands, I get nonsense.
For a start, the calc is internally limited to 15 strands in a bundle, and the bundle amperage is lower than the single wire amperage at 0.777 Amp. Yet this is an 8 AWG power cable for ~60 Amp!
Also, I would like to see the voltage drop for the length in volts/Amp. This is important for 12 volt systems.
9. Brad - November 17, 2007
Hi Peter,
I should clarify that the “Number of wires in Bundle” is not for multi-stranded wire, but for multiple insulated wires (e.g. in a harness). When there is a bundle of multiple wires, each wire is derated to carry less current. Above 15 wires in a bundle, there is no additional derating. This is just the MIL-STD-975 way of doing things.
Calculating the voltage drop is a good idea. I will plan to add that feature.
10. Peter Manins - November 18, 2007
Right, sorry. I have a totally different use in mind with my comments. And not being familiar with the MIL standard does not help!
Nevertheless, a calculator for multistrand wire would be very useful - to me.
Voltage drop in V/A is just resistance! Took me a bit to realise that!
Peter
11. Gio - November 27, 2007
Thanks for the calculator! It is a really handy and by far the best one I have found to-date. I hate to ask but is there any chance of a downloadable version that can run on windows as out here in the boonies there is no way to log on and use it when it is most needed…
12. Brad - November 27, 2007
Hi Gio,
I’m glad you like it. To use the calculator without an internet connection, save the web page to your hard drive. For example, with IE, make sure the menu bar is displayed, and then choose file - save – save as web page complete.
13. Kelly - December 12, 2007
What about insulation temperature ratings? For example 200C Kapton wire can handle more current than 60C PVC. Some places say 12 AWG 200C wire can handle 55 Amps.
14. Brad - December 12, 2007
Hi Kelly,
You are right. Insulation temperature ratings are an important factor. They are addressed in the attachment linked in the main write-up above. I also updated the write-up to make this clearer. As for 55 Amps for a 12 AWG 200C wire, you are going to get different results depending on which spec you go by. The reason is the temperature the wire will reach due to self heating is greatly affected by the environment in which it is used and the specs are linked to the intended environment.
15. PeterJan - December 22, 2007
Hi Guys
Is it possible to make the standard in this calculator selectable so you can calculate against NEC or NFPA79 or (I’m Dutch) IEC60204?
16. Brad - December 24, 2007
PeterJan,
If you can get me a copy of the applicable part of those specs, I will look into it.
17. SHAFIQUE - December 24, 2007
Please send me the information about Formulas for using circular wire in place of flat wire. I shall be so much thankful to you.
SHAFIQUE
18. Brad - December 24, 2007
SHAFIQUE,
We have the above calculator for round wire, and for flat wire please see:
http://circuitcalculator.com/wordpress/2006/01/24/trace-resistance-calculator/
If these do not calculate what you need, please let me know.
19. Randy - December 28, 2007
I have a rectangular conductor that I would like to know the ampacity of. It is not a PCB trace, it is a turn in an electric machine.
I have the following information for an operating point (all values are per turn).
1. Operating current 245A,rms (line to neutral)
2. Operating temperature 165 C
3. Copper losses per turn 36.79W
4. Cross sectional area 4.3mm width (X) 2.84mm height =12.212mm^2
What I dont know
5. Length of the turn (I have an idea what is fesible, but I dont know exactly)
Lets call it right now a single conductor, in reality there are 4 turns per slot of this conductor. However that answers your “wires per bundle question” which I believe is more properly termed strands per turn at least in machine design lingo.
I know that the machine can do this, what I am curious about is how much current I can pump before this coil melts essentially. I know that its actually more complicated than that and that the stacking of coils, skin effect and induced fields will actually make the heating different, but it would give me a good start just where the copper itself melts.
20. Brad - December 28, 2007
Hi Randy,
That is a good question and indeed it is not a simple question. Even if we knew the power loss in the wire, we do not know the thermal resistance from the wire to the ambient temperature and I suspect this thermal resistance is very different in a machine than in a cable bundle since the heat path is very different. Of course copper/insulation melt/burn at specific temperatures not specific currents. Finding the ampacity amounts to solving a thermal problem where the heat source is the joule heating of the wire. I am afraid the general ampacity curves for wire would not be applicable to your machine problem, and I unfortunately don’t have curves or equations for machines. I think you would have to do a detailed finite element thermal analysis or determine the thermal impedances experimentally.
21. S. P. Bhardwaj - January 4, 2008
I have a generator of 250KVA and 230V 3-Phase. The length of wire is 50 meters. Of what cross-section and of how many strands cable should use?
22. Brad - January 6, 2008
S. P. Bhardwaj,
You will first need to calculate the current. Then, based on the allowable voltage drop and ohms law, choose a wire that has low enough resistance to work. In addition, the wire you choose will have to be rated to handle the current and meet all applicable local codes. The number of strands inside the cable varies by design and just the rated current is important. I added the following calculator to my site [1]. You may also find [2] helpful.
23. Larry Steenstry - January 17, 2008
Could you please settle a question we have? Please define ‘Wire Bundle Ampacity’. Is the number shown, amps per wire in the bundle, or is it the total amps in the bundle?
Thanks,
larry
24. Denny - January 18, 2008
Does your calculator apply to single wires coiled around a generator core? I’m trying to calculate the number of turns required and wire diameter required for some driver coils to saturate the cores with 12VDC @ 2A.
Thanks,
Denny
25. Brad - January 29, 2008
Denny,
See comment 20 above. I don’t think this calculator would apply for your situation either.
26. Brad - January 29, 2008
Hi Larry,
Good question. Wire Bundle Ampacity is per wire. I added a note to the calculator where results the results are displayed.
27. paul - February 7, 2008
Hi friends,
Does any one know of a free software that could help me to do load calculations of motors, required power cables, selection of breakers, short circuit calculations etc. for a medium type project? At least to save the time I thought using software is better rather than manual computing. I have gone through browsing, found all those are very expensive.
28. Rob - February 25, 2008
Hey Brad, thank you so much for this calculator. I use it probably once/week on average!
One thing I’m looking for right now and missing is the ability to calculate the mass of copper given the inputs. It should be difficult since you’ve got diameter and length. I could look up density (which is what I will do). Of course, it would just be an estimate (issues such as insulation mass and stranded vs. un-stranded) but would be useful.
Thanks again!
29. Brad - February 28, 2008
Hi Rob,
Adding the mass calculation is a good idea. I’ll put it on my “to do” list. For reference, the density I found for copper is 8.96 g/cm^3.
| 2,459
| 9,999
|
{"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-2018-17
|
latest
|
en
| 0.914554
|
http://www.talkstats.com/tags/correlation-matrix/
| 1,660,380,725,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00766.warc.gz
| 99,901,047
| 10,664
|
# correlation matrix
1. ### How to intrepret this Correlation Matrix result from R ?
This is the correlation matrix of Variable 1 against 16 other variables. I have applied the Distance correlation matrix from the R package "Correlation" which I read is suitable for non-linear data. But I am having time interpreting this result as I could not find enough resources on the...
2. ### Linear regression analysis
Hi all, I am having problems with analysis. I want to perform a multiple linear regression. 1). The dependent variable and all but one of the independent variables is normally distributed. I was made aware that the independent variables don't have to be normally distributed to use it for a...
3. ### Can i use the results from a PCA if the matrix is 'not positive definite'
Hi, I have a 'not positive definite' correlation matrix having done a principal component analysis (PCA) on SPSS. The data i have used is from a questionnaire i did using a 7 point likert type scale. There were 36 questions (36 variables) i got 16 responses (n=16). The questionnaire was very...
4. ### Input data (data view)
Hi I have a question about SPSS. This might sound really stupid though. I have to run a regression, but prior to that, I have to check for correlation in SPSS. The correlation is between my independent variables AND my independent variables on my dependent variables. I know where to find...
5. ### Euclidean distance comparison weighted by correlation
Hello, I want to calculate the Euclidean distance between two objects across a set of n variables. For those n variables I have generated a correlation matrix. From that correlation matrix I would like to produce a weight vector that I can apply before calculating the distance, so that the...
6. ### How can I get R to generate p-value when doing a correlation matrix?
How can I get R to generate p-value for each correlation when doing a correlation matrix? Here are my current steps: > M <- cor(data1) # get correlations into Matrix > library('corrplot') # import corrplot library > corrplot(M, method = "circle")...
7. ### Using a Correlation Matrix with no data in SPSS
I have a correlation matrix with the correlations between each set of five independent variables and the correlation between each IV and the DV. I would like to use this information to determine how R Squared changes for the model when I use the pair of IVs with the greatest correlation to the...
8. ### Interpreting Binary Logistic Regression
Hi, This is my first time on this forum so I don't really know how this works. For this reason I will try to explain my problem as extensive as possible. I am researching the innovativeness of different firm sizes. Beforehand I expected that their would be a positve relation between firm...
9. ### covariance matrix vs correlation matrix
Dear all, I am now studying PCA/FA and stumbled upon following problem: if variables are measured on different scales it's reasonably obvious that we should use correlation matrix. But, when one should use covariance matrix if variables have the same scale? Jolliffe wrote that there might be...
10. ### correlating many time series to find functional pairs
Hi there, I'm a neuroscientist in deep water, trying to do things that aren't standard in my field. I have a strong feeling there are established methods for this, so please give some input if you can! Data (very simplified but illustrates the issue): - From 20 humans, I have...
11. ### Dummies in a correlation matrix
Hi everyone, I am working on a multiple moderate regression. And just wanted to add the correlation matrix as well. Now I got the responds that I cannot state the mean and standard deviation of dummy variables. How does it work? Do I just leave the mean out and the standard deviation...
12. ### Evaluating clusters of variables produced by varclus & hclustvar
Background - I want to cluster analyze a mixed dataset, clustering the variables on the basis of correlational similarity. SPSS gives me this option, but doesn't allow me to evaluate the clustering solutions by providing statistical measures of heterogeneity change (e.g. pseudo F statistic) or...
13. ### testing different models using the same group of longitudinal data
Description of my research Longitudinal growth data are fitted using parametric models giving growthcurves. Differentiating these growthcurves gives velocity curves. From every curve biological parameters can be determinated. Every parameteric model has function parameters. The same...
14. ### Correlation matrix pinted into table. How?
I need to create a "decent" correlation matrix; the output of the R is really ugly and not easy to follow What are my choices? How can I export the output into a simple table? Also, I found on the net this beautiful correlation matrix at...
| 1,009
| 4,829
|
{"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-2022-33
|
latest
|
en
| 0.912034
|
https://www.thestudentroom.co.uk/showthread.php?t=2218816
| 1,527,114,564,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00344.warc.gz
| 854,738,939
| 43,160
|
You are Here: Home >< Maths
# Differentiating trigonometric functions watch
1. How would you differentiate:
y=cosx^2
2. (Original post by monstercable)
How would you differentiate:
y=cosx^2
Is that (cosx)^2 or cos(x^2)
either way ... chain rule as you have a function of a function
3. (Original post by monstercable)
How would you differentiate:
y=cosx^2
Chain rule
4. It's cos(x^2). Okay thank you, i'll give it a go
5. Okay, I still can't do it :s
6. You can use the chain rule with u = x^2
Setup up the chain rule
Spoiler:
Show
Or you could notice learn a rule that I have used.
Much the same as other trig functions this works for
- I suggest you do the chain rule method first, as this is probably the method they want you to use, but for future reference the formulas are there and are simple to remember.
7. (Original post by insparato)
You can use the chain rule with u = x^2
Setup up the chain rule
Spoiler:
Show
Or you could notice learn a rule that I have used.
Much the same as other trig functions this works for
- I suggest you do the chain rule method first, as this is probably the method they want you to use, but for future reference the formulas are there and are simple to remember.
Okay so is this the right answer:
y= cos(x^2)
Let U = (x^2), therefore DU/DX = 2x
Y= CosU, therefore DY/DU= -SinU
Therefore, DY/DX = -2XSin(X^2)
Thank you by the way!
8. (Original post by monstercable)
Okay so is this the right answer
Yes.
### Related university courses
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: January 6, 2013
Today on TSR
### Edexcel C2 Core Unofficial Markscheme!
Find out how you've done here
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams
| 547
| 2,084
|
{"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-2018-22
|
longest
|
en
| 0.932052
|
https://legacy.voteview.com/POLI272_2009_4.htm
| 1,632,095,627,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00117.warc.gz
| 406,028,954
| 2,174
|
This site is an archived version of Voteview.com archived from University of Georgia on May 23, 2017. This point-in-time capture includes all files publicly linked on Voteview.com at that time. We provide access to this content as a service to ensure that past users of Voteview.com have access to historical files. This content will remain online until at least January 1st, 2018. UCLA provides no warranty or guarantee of access to these files.
POLI 272 BAYESIAN METHODS
Fourth Assignment
Due 29 October 2009
All homeworks must be neatly typed with Microsoft Word or its equivalent with all R and WINBUGS code shown in an Appendix.
1. Follow the instructions in Running Regressions in STATA and WINBUGS and run regressions using 3 of the Excel spreadsheets posted on Excel Spreadsheet Download Page. In particular, pick 3 that are at least 20 years apart but none before H50.XLS (1887-88 U.S. House of Representatives)! For example, H60.XLS, H70.XLS, and H80.XLS are each 20 years apart.
1. Show the results from STATA and the corresponding results from WINBUGS in the fashion explained in Running Regressions in STATA and WINBUGS and shown in the handout Reagan Presidential Support Example for WINBUGS (PDF).
2. Show the statistics for the Betas and Sigma with updates at 1000, 10,000, and 100,000. How do the results change?
3. Show the Density plots for the Betas and Sigma. Do these make sense?
4. Show the Auto Correlation plots for the Betas and Sigma. Do these make sense?
5. Show the History plots for the Betas and Sigma. According to the WINBUGS Manual, "Checking Convergence" (near the end), do the chains look like they have converged?
2. For one House File (Not all three!), run the following Probit in both STATA and WINBUGS. Treat partydum as the dependent variable and run:
probit partydum x2 southdum
Report the STATA results and repeat 1a) - 1e) for this problem as well.
h105_BUGS_probit_model.txt -- WINBUGS model illustrating how to use Probit
| 485
| 1,979
|
{"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}
| 3.078125
| 3
|
CC-MAIN-2021-39
|
longest
|
en
| 0.895693
|
https://openwaterfestival.org/clothing/how-fast-did-sailing-ships-go.html
| 1,638,004,447,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00384.warc.gz
| 509,308,062
| 18,386
|
# How fast did sailing ships go?
Contents
## How fast did ancient ships sail?
Vessels could not reach their maximum speed until they met the waters south of Rhodes. When we combine all the above evidence we find that under favorable wind conditions, ancient vessels averaged between 4 and 6 knots over open water, and 3 to 4 knots while working through islands or along coasts.
## How fast did old ships move?
Ancient mariners used to gauge how fast their ship was moving by throwing a piece of wood or other floatable object over the vessel’s bow then counting the amount of time that elapsed before its stern passed the object. This method was known as a Dutchman’s log.
## How long did it take a sailing ship to cross the Atlantic?
Tell students that Henry Hudson was a European explorer traveling across the Atlantic during the colonial period. It took Hudson more than two months to sail from Amsterdam to New York City on his sailing ship, the Half Moon. A modern ocean liner, such as the Queen Mary 2, makes the trip from Europe in seven days.
## How long did a ship take to cross the Atlantic?
In the early 19th century sailing ships took about six weeks to cross the Atlantic. With adverse winds or bad weather the journey could take as long as fourteen weeks.
IT IS IMPORTANT: Can you Wakesurf behind a jet boat?
## How fast did pirate ships go mph?
With an average distance of approximately 3,000 miles, this equates to a range of about 100 to 140 miles per day, or an average speed over the ground of about 4 to 6 knots.
## What is the fastest warship ever built?
In 1968, during a shakedown cruise, the Iowa-class USS New Jersey achieved a top speed of 35.2 knots (65.2 km/h) which it sustained for six hours. As part of a brutal test of the ship’s engines, the captain then ordered the ship to go instantly from “all ahead flank” to “all back emergency”.
## How fast could the Cutty Sark sail?
How fast was Cutty Sark? Cutty Sark’s top speed was over 17 knots.
## How did old ships sail without wind?
Without having the winds in your sails, the boat will not move forward. Instead, you’ll only drift along and get stuck in the neutral. … When there are forces of the wind on the sails, it’s referred to as aerodynamics and can propel the sailboat by lifting it in the same way the winds lift an airplane wing.
## How fast were Roman ships?
Ships would usually ply the waters of the Mediterranean at average speeds of 4 or 5 knots. The fastest trips would reach average speeds of 6 knots. A trip from Ostia to Alexandria in Egypt would take about 6 to 8 days depending on the winds.
| 591
| 2,616
|
{"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}
| 2.71875
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.967714
|
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=123&t=39805&view=print
| 1,600,824,583,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00593.warc.gz
| 478,213,370
| 2,079
|
Page 1 of 1
### When is pv=nrt used?
Posted: Sun Jan 13, 2019 4:03 pm
Can someone please explain to me when the ideal gas law is used?
### Re: When is pv=nrt used? [ENDORSED]
Posted: Sun Jan 13, 2019 4:08 pm
You would use the ideal gas law if the question gave you a both concentration and pressure values which you would use to convert all the values to either concentration or pressure to solve for the equilibrium constant.
### Re: When is pv=nrt used?
Posted: Sun Jan 13, 2019 5:40 pm
Also, the constants sheet should have the gas constant, R, on it. For this, use the most appropriate form so you will have common units. As for temperature, T, it should be given in the problem. Like the previous answer stated, use the Ideal Gas Law to convert between concentration and partial pressure.
### Re: When is pv=nrt used?
Posted: Sun Jan 13, 2019 5:42 pm
and the concentration is the number of moles over the volume
### Re: When is pv=nrt used?
Posted: Mon Jan 14, 2019 3:37 pm
You can use PV=NRT in any situation when given any three out of the four: pressure, volume, number of moles, or temperature. R=.08206. In this instance, we use it to convert Kc to Kp or vice versa
### Re: When is pv=nrt used?
Posted: Mon Jan 14, 2019 7:16 pm
You can use this equation to find an unknown (pressure, volume, molarity or temp)
| 370
| 1,332
|
{"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-2020-40
|
latest
|
en
| 0.844175
|
https://tardigrade.app/question/7aecfozq
| 1,582,656,203,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00066.warc.gz
| 571,746,708
| 20,681
|
# Q. For $r = 0, 1, ... , 10,$ if $A_r,B_r$ and $C_r$ denote respectively the coefficient of $x^r$ in the expansions of $(1 + x)^{10}, (1 + x)^{20}$ and $(1 + x)^{30}$. Then, $\displaystyle \sum A_r(B_{10}B_r-C_{10}A_r)$ is equal to
IIT JEEIIT JEE 2010Binomial Theorem
Solution:
## $A_r$ = Coefficient of $x^r$ in $(1 + x)^{10} = ^{10}C_r$ $B_r$= Coefficient of $x^r$ in $(1+x)^{20} = ^{20}C_r$ $C_r$= Coefficient of $x^r$ in $(1+x)^{30} =^{30}C_r$ $\therefore \, \, \displaystyle \sum_{r=1}^{10} A_r (B_{10}B_r-C_{10}A_r) =\displaystyle \sum_{r=1}^{10} A_r B_{10} B_r- \displaystyle \sum_{r=1}^{10} A_r C_{10}A_r$ =$\displaystyle \sum_{r=1}^{10} \, ^{10}C_r \, ^{20}C_{10} \, ^{20}C_r - \displaystyle \sum_{r=1}^{10} \, ^{10}C_r \, ^{30}C_{10} \, ^{10}C_r$ =$\displaystyle \sum_{r=1}^{10} \, ^{10}C{10_r} \, ^{20}C_{10} \, ^{20}C_r - \displaystyle \sum_{r=1}^{10} \, ^{10}C{10_r} \, ^{30}C_{10} \, ^{10}C_r$ =$^{20}C_{10} \displaystyle \sum_{r=1}^{10} \, ^{10} C_{10-r} . ^{20}C_r -\, ^{30}C_{10} \displaystyle \sum_{r=1}^{10} \, ^{10} C_{10-r} \, ^{10}C_r$ $=^{20}C_{10} (^{30}C_{10}-1)- \, ^{30}C_{10}(^{20}C_{10}-1)$ $= ^{30}C_{10} - ^{20}C_{10} = C_{10}-B_{10}$
You must select option to get answer and solution
Determinants
AIEEE 2002 Sets
## 2. Let $\omega \ne 1$ be a cube root of unity and S be the set of all non-singular matrices of the form $\begin {bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end {bmatrix}$ where each of a , b and c is either $\omega \ or \ omega^2$ Then, the number of distinct matrices in the set S is
IIT JEE 2011 Determinants
## 3. Perpendicular are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}= \frac{z}{3}$ to the plane $x+ y + z = 3$. The feet of perpendiculars lie on the line
JEE Advanced 2013 Introduction to Three Dimensional Geometry
## 4. A ray of light along $x+\sqrt 3 \, y=\sqrt 3$ gets reflected upon reaching X-axis, the equation of the reflected ray is
JEE Main 2013 Straight Lines
## 5. A ray of light along $x+\sqrt 3 \, y=\sqrt 3$ gets reflected upon reaching X-axis, the equation of the reflected ray is
JEE Main 2013 Straight Lines
## 6. A ray of light along $x+\sqrt 3 \, y=\sqrt 3$ gets reflected upon reaching X-axis, the equation of the reflected ray is
JEE Main 2013 Straight Lines
## 7. The centre of the circle passing through the point (0, 1)and touching the curve $y = x^2 at (2,4)$ is
IIT JEE 1983 Conic Sections
## 8. Let AB be a chord of the circle $x^2 + y^2 = r^2$ subtending a right angle at the centre. Then, the locus of the centroid of the $\Delta PAB$ as P moves on the circle, is
IIT JEE 2001 Conic Sections
| 1,041
| 2,636
|
{"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}
| 4.21875
| 4
|
CC-MAIN-2020-10
|
latest
|
en
| 0.494887
|
https://electronics.stackexchange.com/questions/441005/constant-current-source-bjt
| 1,642,549,288,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00054.warc.gz
| 280,741,240
| 35,787
|
# Constant Current Source BJT
I am stuck trying to determine the applicable formula for the load current. I have used $$I_b = \frac{V_{\text{in}}-V_{\text{be}}}{(\beta+1)R_{\text{e}}}$$ and then $$I_{\text{L}}= \beta I_\text{b}$$
The problem is I can't find a proper relationship for why the current drops or increases as $$\R_{\text{L}}\$$ changes. Can someone please just point me in the right direction? I want to solve it myself, just need to know what I'm missing.
• Do you know what saturation is? And for what Rc value your BJT's will start to enter into saturation region?
– G36
May 29 '19 at 19:33
• When they write "assume $\beta$ is a high value" do they want you to assume it is $\beta=\infty$?
– jonk
May 29 '19 at 19:53
• I know the lab book says 100mV but doesn't say the resistor value. And this is where our text does a horrible job. It asks what the highest value of R_L can be before the basic equation is no longer valid, and I assume this is where the curve in the plot is non-linear. I just don't have any clue how to get there mathematically May 29 '19 at 19:54
• The \Beta value given is 200 so, this lab book is seriously vague on some topics May 29 '19 at 19:55
• @WARmachin3 Basically, this is just an emitter follower circuit. The emitter voltage will follow the base voltage ($V_\text{IN}$), less a $V_\text{BE}$ drop. That voltage, combined with $R_\text{E}$, determines the emitter current. This emitter current, less any base current required for recombination, is the collector current. That collector current causes a voltage drop on $R_\text{C}$. You have to subtract that voltage drop from the supply voltage rail. It stays in active mode until the collector voltage reaches the base voltage, $V_\text{IN}$. (Or less, if your text allows.)
– jonk
May 29 '19 at 19:58
In active mode (not saturated):
\\begin{align*} I_\text{L}&=I_\text{E}-\frac{I_\text{E}}{\beta+1}\\\\ I_\text{E}&=\frac{V_\text{IN}-V_\text{BE}}{R_\text{E}}\\\\ V_\text{C}&=V_\text{CC}-I_\text{L}\cdot R_\text{L} \end{align*}\
Saturation just barely begins to happen right at the point where $$\V_\text{C}=V_\text{IN}\$$, which is right at the point where the BC junction of the BJT just begins to become forward-biased. But it's not really very noticeable at that point.
If you were designing an amplifier stage where you wanted the BJT to always be in active mode and never saturated at all, then you'd definitely want to maintain $$\V_\text{C}\ge \left[V_\text{B}=V_\text{IN}\right]\$$. But if you wanted to actually readily see some significant change in the "linear" declining collector voltage behavior, this doesn't really start to happen enough to be very visible until $$\V_\text{C}\$$ nears about $$\500\:\text{mV}\$$ below the base voltage (in your case, $$\V_\text{IN}\$$.) So you use what's appropriate to the situation at hand.
Let's call the variable defining saturation as $$\V_{\text{CE}_\text{SAT}}\$$. If you set $$\V_{\text{CE}_\text{SAT}}=V_\text{BE}\$$, then you are using a definition where saturation is defined to occur when it is just barely starting to happen. On the other hand, if you set $$\V_{\text{CE}_\text{SAT}}=V_\text{BE}-500\:\text{mV}=200\:\text{mV}\$$ (in your case, let's say), then you are using a definition where saturation is defined to occur when it moving into deep saturation. This latter definition is where you really see the cornering voltage at the collector.
Either way, the general equation for when saturation begins is:
$$V_\text{E}=V_\text{IN}-V_\text{BE}=\left(V_\text{CC}-V_{\text{CE}_\text{SAT}}\right)\cdot\frac{R_\text{E}}{R_\text{E}+\frac{\beta}{\beta+1}\,R_\text{L}}$$
(Solve for $$\V_\text{IN}\$$, of course, if you want that value.)
It may not be immediately obvious to you. But this is the same equation you'd get from the following schematic:
simulate this circuit – Schematic created using CircuitLab
In effect, this as a voltage divider between $$\R_\text{E}\$$ and $$\R_\text{L}\$$. The very slight adjustments involve reducing $$\V_\text{CC}\$$ by whatever value you pick for $$\V_{\text{CE}_\text{SAT}}\$$ and by a very minor adjustment in the value of $$\R_\text{L}\$$ to account for the fact that only $$\\frac{\beta}{\beta+1}\$$ of the emitter current becomes a collector current. Otherwise, it's really simple.
You have the variation in the large signal model with Vbe (and beta) as the transistor heats, but there is more to it than that.
I think you should use the hybrid-pi model for the transistor. With a "stiff" voltage source on the base, the output resistance of your current sink is ro, which is related to the Early Voltage Va that you will find in a transistor SPICE model.
| 1,325
| 4,678
|
{"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875
| 4
|
CC-MAIN-2022-05
|
latest
|
en
| 0.849254
|
https://programmersheaven.com/discussion/368469/pascal-help
| 1,529,563,807,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00328.warc.gz
| 698,771,326
| 11,618
|
# Pascal help
Hi there, I'm taking Pascal in school and have run into troubles. I need to write a program to print a box that takes in the Length and Width of the box using symbols to create the box.
A sample output would be Length of 5, Width of 3 and Symbol of \$ would produce
\$\$\$\$\$
\$\$\$\$\$
\$\$\$\$\$
I've tried adapting another program to try and figure this problem out, but it isn't working to well. My current program is
[code]program Assignment2;
{Cole Gauthier 01/07/08}
uses wincrt;
var
Symbol:char;
Length,Width:integer;
procedure Draw(l:integer;w:integer;s:char);
var
Number_Length,Number_Width,Number_Stars:integer;
begin
for Number_Length:=1 to l do
begin
for Number_Stars:=1 to Number_Length do
write(s);
writeln
end;
end;
begin
write('Enter the length of the box. ');
write('Enter the width of the box. ');
write('Enter the symbol to be used in the design. ');
Draw(Length,Width,Symbol)
end.[/code]
Any pointers or suggestions or anything that will help me?
• : [code]: program Assignment2;
: {Cole Gauthier 01/07/08}
:
: uses wincrt;
:
: var
: Symbol:char;
: Length,Width:integer;
:
: procedure Draw(l:integer;w:integer;s:char);
: var
: Number_Length,Number_Width,Number_Stars:integer;
: begin
: for Number_Length:=1 to l do
: begin
[red]: for Number_Stars:=1 to Number_Length do[/red]
: write(s);
: writeln
: end;
: end;
:
: begin
: write('Enter the length of the box. ');
: write('Enter the width of the box. ');
: write('Enter the symbol to be used in the design. ');
: Draw(Length,Width,Symbol)
: end.[/code]:
:
: Any pointers or suggestions or anything that will help me?
:
Look at the two lines above. The inner loop looks good (red line), but you are using Number_Length which will create more of a triangle type of shape.
For the inner loop, just use [b]for Number_Stars:=1 to w do[/b].
Hope this helps, had to rewrite a bunch of stuff here as the font makes 1's and l's look alot the same!
Phat Nat
• : : [code]: : program Assignment2;
: : {Cole Gauthier 01/07/08}
: :
: : uses wincrt;
: :
: : var
: : Symbol:char;
: : Length,Width:integer;
: :
: : procedure Draw(l:integer;w:integer;s:char);
: : var
: : Number_Length,Number_Width,Number_Stars:integer;
: : begin
: : for Number_Length:=1 to l do
: : begin
: [red]: for Number_Stars:=1 to Number_Length do[/red]
: : write(s);
: : writeln
: : end;
: : end;
: :
: : begin
: : write('Enter the length of the box. ');
: : write('Enter the width of the box. ');
: : write('Enter the symbol to be used in the design. ');
: : Draw(Length,Width,Symbol)
: : end.[/code]: :
: :
: : Any pointers or suggestions or anything that will help me?
: :
:
: Look at the two lines above. The inner loop looks good (red line),
: but you are using Number_Length which will create more of a triangle
: type of shape.
: For the inner loop, just use [b]for Number_Stars:=1 to w do[/b].
:
: Hope this helps, had to rewrite a bunch of stuff here as the font
: makes 1's and l's look alot the same!
: Phat Nat
Yes, thanks so much. It worked perfectly. Without your help, I would have never figured that out
| 883
| 3,075
|
{"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-2018-26
|
latest
|
en
| 0.794407
|
https://in.mathworks.com/matlabcentral/profile/authors/10692611?page=3
| 1,653,226,632,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662545548.56/warc/CC-MAIN-20220522125835-20220522155835-00463.warc.gz
| 390,221,486
| 4,105
|
Solved
The Goldbach Conjecture
The <http://en.wikipedia.org/wiki/Goldbach's_conjecture Goldbach conjecture> asserts that every even integer greater than 2 can ...
2 years ago
Solved
Which values occur exactly three times?
Return a list of all values (sorted smallest to largest) that appear exactly three times in the input vector x. So if x = [1 2...
2 years ago
Solved
Back and Forth Rows
Given a number n, create an n-by-n matrix in which the integers from 1 to n^2 wind back and forth along the rows as shown in the...
2 years ago
Solved
Remove any row in which a NaN appears
Given the matrix A, return B in which all the rows that have one or more <http://www.mathworks.com/help/techdoc/ref/nan.html NaN...
2 years ago
Solved
Quote Doubler
Given a string s1, find all occurrences of the single quote character and replace them with two occurrences of the single quote ...
2 years ago
Solved
Summing digits
Given n, find the sum of the digits that make up 2^n. Example: Input n = 7 Output b = 11 since 2^7 = 128, and 1 + ...
2 years ago
Solved
Pizza!
Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
2 years ago
Solved
Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
2 years ago
Solved
Return the 3n+1 sequence for n
A Collatz sequence is the sequence where, for a given number n, the next number in the sequence is either n/2 if the number is e...
2 years ago
Solved
Fibonacci-Sum of Squares
Given the Fibonacci sequence defined by the following recursive relation, * F(n) = F(n-1) + F(n-2) * where F(1) = 1 and F(1)...
2 years ago
Solved
Fibonacci sequence
Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu...
2 years ago
Solved
Extra safe primes
Did you know that the number 5 is the first safe prime? A safe prime is a prime number that can be expressed as 2p+1, where p is...
2 years ago
Solved
We love vectorized solutions. Problem 1 : remove the row average.
Given a 2-d matrix, remove the row average from each row. Your solution MUST be vectorized. The solution will be tested for ac...
2 years ago
Problem
Rescale Scores
Each column (except last) of matrix |X| contains students' scores in a course assignment or a test. The last column has a weight...
2 years ago | 32 | 3841 solvers
Problem
Calculate BMI
Given a matrix |hw| (height and weight) with two columns, calculate BMI using these formulas: * 1 kilogram = 2.2 pounds * 1 ...
2 years ago | 49 | 5550 solvers
Problem
Find the Best Hotels
Given three input variables: * |hotels| - a list of hotel names * |ratings| - their ratings in a city * |cutoff| - the rat...
2 years ago | 41 | 4793 solvers
Problem
Crop an Image
A grayscale image is represented as a matrix in MATLAB. Each matrix element represents a pixel in the image. The element values ...
2 years ago | 74 | 4378 solvers
Problem
Verify Law of Large Numbers
If a large number of fair N-sided dice are rolled, the average of the rolls is likely to be close to the expected value of a sin...
2 years ago | 102 | 6172 solvers
Problem
Calculate a Damped Sinusoid
The equation of a damped sinusoid can be written as |y = A.ⅇ^(-λt).cos(2πft)| Given inputs λ ( |lambda...
2 years ago | 41 | 5720 solvers
Problem
Solve a System of Linear Equations
Given a constant input angle θ(theta) in radians, create the coefficient matrix(A) and constant vector(b) to solve the giv...
2 years ago | 52 | 7109 solvers
Problem
Plot Damped Sinusoid
Given two vectors |t| and |y|, make a plot containing a blue dashed line of |y| versus |t|. Mark the minimum value |m| of the...
2 years ago | 49 | 5263 solvers
Solved
Longest run of consecutive numbers
Given a vector a, find the number(s) that is/are repeated consecutively most often. For example, if you have a = [1 2 2 2 1 ...
3 years ago
Solved
Remove the vowels
Remove all the vowels in the given phrase. Example: Input s1 = 'Jack and Jill went up the hill' Output s2 is 'Jck nd Jll wn...
3 years ago
Solved
Make a checkerboard matrix
Given an integer n, make an n-by-n matrix made up of alternating ones and zeros as shown below. The a(1,1) should be 1. Examp...
3 years ago
Solved
Set the array elements whose value is 13 to 0
Input A either an array or a vector (which can be empty) Output B will be the same size as A . All elements of A equal to 13...
3 years ago
Solved
Breaking Out of the Matrix
Do you want to take the Red Pill, or the Blue Pill? If you take the Blue Pill, you will simply pass along to the next problem...
3 years ago
Solved
Find all elements less than 0 or greater than 10 and replace them with NaN
Given an input vector x, find all elements of x less than 0 or greater than 10 and replace them with NaN. Example: Input ...
3 years ago
Solved
Most nonzero elements in row
Given the matrix a, return the index r of the row with the most nonzero elements. Assume there will always be exactly one row th...
3 years ago
Solved
Plot Line Specifications
Create a line plot for function cos(x) where x is a vector of linearly spaced values going from 0 to input N with an increment o...
3 years ago
Problem
Plot Line Specifications
Create a line plot for function cos(x) where x is a vector of linearly spaced values going from 0 to input N with an increment o...
3 years ago | 1 | 40 solvers
| 1,484
| 5,490
|
{"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}
| 3.265625
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.807912
|
https://www.askiitians.com/forums/Magical-Mathematics%5BInteresting-Approach%5D/using-the-definition-prove-that-the-function-f_267944.htm
| 1,723,098,840,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00058.warc.gz
| 503,844,478
| 43,136
|
# Using the definition, prove that the function f : A→ B is invertible if and only if f is both oneone and onto.
Grade:12
## 1 Answers
Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student
Let us assume
f: A → B be many-one function.
Let’s assume
f(a) = p and f(b) = p
So, for inverse function we will have f^-1(p) = a and f^-1(p) = b
Thus, in this case inverse function is not defined as we have two images ‘a and b’ for one pre-image ‘p’.
But for f to be invertible it must be one-one.
Now, let f: A → B is not onto function.
Let B = {p, q, r} and range of f be {p, q}.
Here image ‘r’ has not any pre-image, which will have no image in set A.
And for f to be invertible it must be onto.
Hence, ‘f’ is invertible if and only if ‘f’ is both one-one and onto.
A function f = X → Yis invertible iff f is a bijective function.
Thanks
## ASK QUESTION
Get your questions answered by the expert for free
| 277
| 921
|
{"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-2024-33
|
latest
|
en
| 0.90986
|
http://stackoverflow.com/questions/4092348/how-to-generate-an-array-with-random-values-without-using-a-loop/4093775
| 1,469,365,813,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257824037.46/warc/CC-MAIN-20160723071024-00144-ip-10-185-27-174.ec2.internal.warc.gz
| 232,272,223
| 40,422
|
Dismiss
Announcing Stack Overflow Documentation
We started with Q&A. Technical documentation is next, and we need your help.
Whether you're a beginner or an experienced developer, you can contribute.
# How to generate an array with random values, without using a loop?
How can I generate an array in Perl with 100 random values, without using a loop?
I have to avoid all kind of loops, like "for", foreach", while. This is my exercise, from my lab. I can't find a way to do solve this, because I am new in Perl.
In C, generating this array would by very easy, but I don't know how to do it in Perl.
-
Why would you want to avoid a loop? – Oliver Charlesworth Nov 3 '10 at 22:49
Please mark problems related to school assignments with the `homework` tag. – mob Nov 3 '10 at 22:59
How would you do it in C? – Martin Broadhurst Nov 3 '10 at 23:00
You can do it with recursion in C, and Perl. – Martin Broadhurst Nov 3 '10 at 23:16
@ysth: it's educative insofar as if the student ever manages to understand all of our helpful solutions, they'll learn quite a bit more than their instructor had counted on them learning from it. – tchrist Nov 4 '10 at 6:36
``````my @rand = map { rand } ( 1..100 );
``````
But a map is just a loop with fancy window-dressing.
If you need to do something 100 times, you're going to need to use some kind of iterative structure.
-
Well... a "map is just a loop" seems to be a VERY debatable statement... literally. stackoverflow.com/questions/3019925/is-map-a-loop .I happen to agree with labelling it a lop, FWIW – DVK Nov 3 '10 at 23:17
I wouldn't call map "just a loop". It's main feature it that it's an expression, so it can be part of a statement. The other looping constructs require their own statements (even in the modifier form). – brian d foy Nov 4 '10 at 16:22
Or a recursive structure. – ripper234 Nov 9 '10 at 6:39
Recursion is the same as iteration. – friedo Nov 9 '10 at 18:35
For amusement value:
### A method that works on systems where EOL is a single character:
``````#!/usr/bin/perl
use strict;
use warnings;
\$/ = \1;
open 0;
my @r = map rand,<0>;
print "@r\n";
``````
### A possibly nondeterministic method that does not use `for`, `while`, `until`:
``````#!/usr/bin/perl
use strict; use warnings;
my @rand;
NOTLOOP:
push @rand, rand;
sleep 1;
goto NOTLOOP if 100 > time - \$^T;
print 0 + @rand, "\n";
``````
### Using regular expressions:
``````#!/usr/bin/perl
use strict; use warnings;
my \$s = '-' x 100;
\$s =~ s/(-)/rand() . \$1/eg;
my @rand = \$s=~ m/([^-]+)/g;
``````
### Copying and pasting 100 `rand` invocations by hand is really passé:
``````#!/usr/bin/perl
use strict; use warnings;
my \$s = '(' . 'rand,' x 100 . ')';
my @rand = eval \$s;
``````
### A file I/O based solution that does not require `/dev/random`:
``````#!/usr/bin/perl
use strict; use warnings;
\$/ = \1;
my @rand;
seek \*DATA, 0, 0;
NOTLOOP:
scalar <DATA>;
push @rand, rand;
goto NOTLOOP if \$. < 100;
__DATA__
``````
### No reason to use recursion with Perl's `goto`
``````#!/usr/bin/perl
use strict; use warnings;
use autodie;
\$/ = \1;
open my \$F, '<', \( 1 x 100 . 0 );
my @rand or ¬loop;
sub notloop {
my \$t = <\$F>;
\$t or return;
push @rand, rand;
goto \¬loop;
}
``````
### Here is a recursive string `eval` version:
``````#!/usr/bin/perl
use strict; use warnings; use autodie;
local \$/ = \1;
open my \$F, '<', \( 1 x 100 . 0 );
my @rand;
eval <<'NOLOOP'
my \$self = (caller(0))[6];
<\$F> or die;
push @rand, rand;
eval \$self;
NOLOOP
;
``````
Of course, all of these actually do contain loops, but they do not use the keywords you were barred from using.
NB: This question has brought out the wacko in me, but I must admit it is amusing.
-
FOR THE WIN!!!! – DVK Nov 3 '10 at 23:20
I dunno. The tail recursion would have been cooler if it had been an anon `sub{}`. :) – tchrist Nov 4 '10 at 0:24
I'm waiting for something spooky to happen to *CORE::GLOBAL::rand. – tchrist Nov 4 '10 at 0:24
@tchrist I believe I need `Sub::Current` for the anon sub solution. However, I added one based on recursive string `eval`. – Sinan Ünür Nov 4 '10 at 1:02
Oh, I dunno about that: see my anon sub solution. :) – tchrist Nov 4 '10 at 1:04
Nothing could be simpler!
``````my @rands = (rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand, rand, rand, rand, rand, rand, rand, rand, rand, rand,
rand, rand);
``````
-
+1. A valid solution to an inane problem – friedo Nov 3 '10 at 22:56
Wow. That is so noisy. Clearly `my @rands = eval sprintf q[(%s)], join q[,] => ('rand') x 100;` is much more maintainable. – rafl Nov 3 '10 at 23:02
+1 But didn't you have a loop in your brain while typing? ;) – Peter G. Nov 3 '10 at 23:10
Obligatory: search.dilbert.com/comic/Random%20Nine – Ether Nov 3 '10 at 23:23
Copying and pasting 100 `rand` s by hand is so not Perl! ;-) – Sinan Ünür Nov 3 '10 at 23:26
## Pure Regex Solution
``````perl -E'say for&{sub{"\U\x{fb01}\x{fb03}"=~/.{0,2}.{0,3}.{0,3}.{0,4}+(?{\$_[++\$#_]=rand})(*FAIL)/||pop;@_}}'
``````
## Double‐`/e` Regex Solution
This:
``````(\$_=(120.44.32)x(2+2*2**2)**2)=~s/\170/114.97.110.100/gee;
s/(.*)/64.95.61.40.\$1.41.35.89.65.78.69.84.85.84/ee;
print "@_\n";
``````
looplessly produces this:
``````0.636939813223766 0.349175195300148 0.692949079946754 0.230945990743699 0.61873698433654 0.940179094890468 0.435165707624346 0.721205126535175 0.0322560847184015 0.91310500801842 0.31596325316222 0.788125484008084 0.802964232426337 0.417745170032291 0.155032810595454 0.146835982654117 0.181850358582611 0.932543988687968 0.143043972615896 0.415793094159206 0.576503681784647 0.996621492832261 0.382576007897708 0.090130958455255 0.39637315568709 0.928066985272665 0.190092542303415 0.518855656633185 0.797714758118492 0.130660731025571 0.534763929837762 0.136503767441518 0.346381958112605 0.391661401050982 0.498108766062398 0.478789295276393 0.882380841033143 0.852353540653993 0.90519922056134 0.197466335156797 0.820753004050889 0.732284103461893 0.738124358455405 0.250301496672911 0.88874926709342 0.0647566487704268 0.733226696403218 0.186469206795884 0.837423290530243 0.578047704593843 0.776140208497122 0.375268613243982 0.0128391627800006 0.872438613450569 0.636808174464274 0.676851978312946 0.192308731231467 0.401619465269903 0.977516959116411 0.358315250197542 0.726835710856381 0.688046044314845 0.870742340556202 0.58832098735666 0.552752229159754 0.170767637182252 0.683588677743852 0.0603160539059857 0.892022266162105 0.10206962926371 0.728253338154527 0.800910562860132 0.628613236438159 0.742591620029089 0.602839705915397 0.00926448179027517 0.182584549347883 0.53561587562946 0.416667072500555 0.479173194613729 0.78711818598828 0.017823873107119 0.824805088282755 0.302367196288522 0.0677539595682397 0.509467036447674 0.906839536492864 0.804383046648944 0.716848992363769 0.450693083312729 0.786925293921154 0.078886787987166 0.417139859647296 0.9574382550514 0.581196777508975 0.75882630076142 0.391754631502298 0.370189654004974 0.80290625532508 0.38016959549288
``````
## Recursive Numeric Function Solution
As in fact, does this, if you print the array:
``````@_=(*100=sub{\$_[0]?(rand,(*{--\$_[0]}=*{\$_[0]})->(@_)):()})->(\$==100);
``````
The second solution now allows for getting different numbers of random numbers easily enough, since following the assignment above, you can do such niceties as:
`````` print for @42=42->(\$==42);
``````
And yes, that is indeed a function named `42()`. The previous assignment to `@_` created it along with a hundred other numerically named functions.
## Explanation
The first regex solution relies on Unicode’s tricky casing of the two characters matched against. It may (or may not) be more easily understood with whitespace and comments added:
``````use 5.010;
say for &{
sub { "\U\x{fb01}\x{fb03}" =~ m((?mix-poop)
#include <stdlib.h>
#include <unistd.h>
#include <regex.h>
#include "perl.h"
#include "utf8.h"
#ifndef BROKEN_UNICODE_CHARCLASS_MAPPINGS
.{0,2}
.{0,3} .{0,3}
.{0,4}
+(?{ \$_ [++\$#_] = rand() || rand() || UTF8_TWO_BYTE_LO (*PERL_UNICODE)
#else (*PRUNE)
#define FAIL (*ACCEPT)
}) (*FAIL)
#endif (*COMMIT)
)poop || pop @{ (*_{ARRAY}) }
;#; @{ (*SKIP:REGEX) }
@{ (*_{ARRAY}) }
}
}
``````
The way to understand how the second regex solution works is:
• First, reduce the compile-time constant expressions into their more customary forms so you can more easily read the literals. For example, `\170` is "x".
• Second, trim down the double `e` to a single `e` in each substitution, then print out what that leaves in the string both times.
I’m sure you’ll appreciate the comment. :)
For the recursive solution, adding whitespace may help a little:
``````(*100 = sub { \$_[0]
? ( rand, ( *{ --\$_[0] } = *{ \$_[0] } )->(@_) )
: ( )
}
)->( \$= = 100 );
``````
The need to pass a variable as an argument is due to the auto-decrement requiring an lvalue. I did that because I didn’t want to have to say `\$_[0] - 1` twice, or have any named temporary variables. It means you can do this:
``````\$N = 100;
print for \$N->(\$N);
``````
And when you’re done, `\$N == 0`, because of pass‐by‐implicit‐reference semantics.
For the cost of a bit of repetition, you can relax the need of an lvalue argument.
``````(*100 = sub { \$_[0]
? ( rand, ( *{ \$_[0] - 1 } = *{ \$_[0] } )->( \$_[0] - 1 ) )
: ( )
}
)->( 100 );
``````
Now you no longer need an lvalue argument, so you can write:
``````print for 100->(100);
``````
to get all 100 random numbers. Of course, you also have a hundred other numeric functions, too, so you can get lists of random numbers in any of these ways:
``````@vi = 6->(6);
@vi = &6( 6);
\$dozen = 12;
@dozen = \$dozen->(\$dozen);
@baker's_dozen = &\$dozen(++\$dozen);
@_ = 100;
print &0; # still a hundred of 'em
``````
(Sorry ’bout the silly colors. Must be an SO bug.)
I trust that clears everything up. ☺
-
Am I really the only one to include an informative comment in my solution to help out our hapless student? :) – tchrist Nov 4 '10 at 0:33
You win the line noise award of the month/year :) Too bad I can't upvote twice :) – DVK Nov 4 '10 at 2:00
Why are there two `e` modifiers? What does it achieve? – Zaid Nov 4 '10 at 6:04
@Zaid: That one's the easier to understand of my two solutions. But to answer your question, the second `e` does the same thing as the first `e`. Kinda. – tchrist Nov 4 '10 at 6:41
``````# I rolled a die, honestly!
my @random = (5, 2, 1, 3, 4, 3, 3, 4, 1, 6,
3, 2, 4, 2, 1, 1, 1, 1, 4, 1,
3, 6, 4, 6, 2, 6, 6, 1, 4, 5,
1, 1, 5, 6, 6, 5, 1, 4, 1, 2,
3, 1, 2, 2, 6, 6, 6, 5, 3, 3,
6, 3, 4, 3, 1, 2, 1, 2, 3, 3,
3, 4, 4, 1, 5, 5, 5, 1, 1, 5,
6, 3, 2, 2, 1, 1, 5, 2, 5, 3,
3, 3, 5, 5, 1, 6, 5, 6, 3, 2,
6, 3, 5, 6, 1, 4, 3, 5, 1, 2);
``````
-
Too bad most of the solutions focused on the non-looping part and neglected the random numbers part:
``````use LWP::Simple;
my @numbers = split /\s+/, #/ Stackoverflow syntax highlighting bug
get( 'http://www.random.org/integers/?num=100&min=1&max=100&col=1&base=10&format=plain&rnd=new' );
``````
Here's the insanity that Tom was waiting to see, but I make it slightly more insane. This solution reduces the problem to:
``````my @numbers = rand( undef, 100 ); # fetch 100 numbers
``````
Normal rand normally takes 0 or 1 arguments. I've given it a new prototype that allows a second argument to note how many numbers to return.
Notice some differences to the real `rand` though. This isn't continuous, so this has far fewer available numbers, and it's inclusive on the upper bound. Also, since this one takes two arguments, it's not compatible with programs expecting the real one since a statement like this would parse differently in each:
`````` my @array = rand 5, 5;
``````
However, there's nothing particularly special about `CORE::GLOBAL::rand()` here. You don't have to replace the built-in. It's just a bit sick that you can.
I've left some `print` statements in there so you can watch it work:
``````BEGIN {
my @buffer;
my \$add_to_buffer = sub {
my \$fetch = shift;
\$fetch ||= 100;
\$fetch = 100 if \$fetch < 100;
require LWP::Simple;
push @buffer, split /\s+/, #/ Stackoverflow syntax highlighting bug
LWP::Simple::get(
"http://www.random.org/integers/?num=\$fetch&min=1&max=100&col=1&base=10&format=plain&rnd=new"
);
};
my \$many = sub (\$) {
print "Fetching \$_[0] numbers\n";
\$add_to_buffer->(\$_[0]) if @buffer < \$_[0];
my @fetched = splice @buffer, 0, \$_[0], ();
my \$count = @fetched;
print "Fetched [\$count] @fetched\n";
@fetched
};
*CORE::GLOBAL::rand = sub (;\$\$) {
my \$max = \$_[0] || 1; # even 0 is 1, just like in the real one
my \$fetch = \$_[1] || ( wantarray ? 10 : 1 );
my @fetched = map { \$max * \$_ / 100 } \$many->( \$fetch );
wantarray ? @fetched : \$fetched[-1];
};
}
my @rand = rand(undef, 5);
print "Numbers are @rand\n\n";
@rand = rand(87);
print "Numbers are @rand\n\n";
\$rand = rand(undef, 4);
print "Numbers are \$rand\n\n";
\$rand = rand();
print "Numbers are \$rand\n\n";
\$rand = rand(undef, 200);
print "Numbers are \$rand\n\n";
``````
My source of random numbers isn't important for this technique though. You could read from /dev/random or /dev/urandom to fill the buffer if you like.
-
That is a good solution. – dawg Nov 4 '10 at 4:50
Now I know the spooky thing that's going to happen to `*CORE::GLOBAL::rand`. – tchrist Nov 4 '10 at 6:46
Thanks. Can’t upvote you again, though. :( At first I thought you could have used `&\$add_to_buffer if @buffer < \$_[0]` but I see now that that doesn’t always work quite right. – tchrist Nov 4 '10 at 17:20
Nice! Will this implementation make it into Perl 5.14 or 5.16? – Christopher Bottoms Nov 9 '10 at 0:21
Why would any implementation be in Perl? It should be a module anyway. No one will agree on their method to get random numbers. – brian d foy Nov 12 '10 at 18:51
``````sub f {
my \$n = shift;
if( \$n == 0 ) {
return @_;
}
else {
return f( \$n-1, rand, @_ );
}
}
my @random_array = f(100);
``````
-
Dammit. You had to post a recursive solution after my answer :-) – user166390 Nov 3 '10 at 23:12
I saw it blinking but read it only afterwards, a pity to delete it then. I tested it before posting otherwise I would have been earlier. – Peter G. Nov 3 '10 at 23:17
`map` is used here as a topicalizer over a single value, exempting it from loop status:
``````my @rand = map&\$_(\$_),sub{@_<=100&&goto&{push@_,rand;\$_[0]};shift;@_};
``````
or with two subs:
``````my @rand = sub{&{\$_[0]}}->(sub{@_<=100&&goto&{(@_=(rand,@_))[-1]};pop;@_});
``````
both of these are Y-combinator style self-passed subs that build up the list via iteration but one is clearly faster than the other.
you can fix the inefficiency with `s'g[^}]+'goto&{unshift@_,rand;\$_[-1]'` but then its getting a bit long.
or to sidestep the call stack:
``````my @rand = do{local*_=sub{(push@_,rand)<100?goto&_:@_};&_};
``````
or with eval, no variable assignment, no external state, one anon sub:
``````my @rand = eval'sub{@_<100?eval((caller 1)[6]):@_}->(@_,rand)';
``````
but most concise of all is:
``````my @rand = map&\$_,sub{(100^push@_,rand)?goto&\$_:@_};
``````
-
Eric, I think you win the code-golf award. – tchrist Nov 6 '10 at 1:29
While the copy'n'paste examples are novel and the map/foreach alternatives have also been mentioned, I think one approach that has not been discussed is recursion. Using recursion (an implicit loop) would need method calls and a simple 'if' statement: no for/grep/map/etc. It could be side-effecting or side-effect free.
Since this is homework, I will leave the implementation to the poster.
Happy coding.
BTW: Nobody has posted a regular expression solution yet ;-)
It's nice to see some even more innovative solutions! :-)
-
Plus one for NOT doing the posters homework for him. – philosodad Nov 5 '10 at 15:03
Using an anonymous sub and recursion:
``````use strict;
use warnings;
my @foo;
my \$random;
(\$random = sub {
push @{\$_[0]}, rand;
return if @{\$_[0]} == \$_[1];
goto \&\$random;
})->(\@foo, 100);
print "@foo\n";
print scalar @foo, "\n";
``````
Using computed goto, for all the fortran fans:
``````use strict;
use warnings;
sub randoms {
my \$num = shift;
my \$foo;
CARRYON:
push @\$foo, rand;
# goto \$#{\$foo} < 99 ? 'CARRYON' : 'STOP';
goto ( ('CARRYON') x 99, 'STOP' )[\$#\$foo];
STOP:
return @\$foo;
}
my @foo = randoms(100);
print "@foo\n";
print scalar(@foo)."\n";
``````
2 anonymous subs and an arrayref:
``````use strict;
use warnings;
my @bar = sub { return &{\$_[0]} }->
(
sub {
push @{\$_[1]}, rand;
goto \&{\$_[0]}
unless scalar(@{\$_[1]}) == \$_[2];
return @{\$_[1]};
},
[],
100
);
print "@bar\n";
print scalar(@bar), "\n";
``````
-
It's trivial if you keep a reference to it in a variable. What I wanted to do was to find a way to do everything in a solitary `sub { }` without storing the ref in a variable. – Sinan Ünür Nov 4 '10 at 1:28
Devel::Caller / callee seemed a bit much for the OP's question – MkV Nov 4 '10 at 13:14
@Sinan: bless the closure and invoke it against itself as a method. – tchrist Nov 4 '10 at 14:23
That `x99` token doesn’t look right: it needs to be two tokens. Also, `\$#{\$foo}` can always be rewritten as `\$#\$foo` due to the rule that dereferencing braces can be omitted whenever they contain nothing but a bareword symbol with zero or more leading dollar signs. I do like the double-closure solution. – tchrist Nov 4 '10 at 15:22
The world would be a better place if more Perl hackers knew about the Y combinator. – Porculus Nov 6 '10 at 17:11
``````!#/usr/bin/perl
use strict;
use warnings;
my \$x = 99;
my @rands = (rand,(*x=sub{rand,(map{*x->(\$x,sub{*x})}(\$x)x!!--\$x)})->(\$x,*x));
use feature 'say';
say for @rands;
``````
-
Nice work! It’s always nice to see people play with their symbol tables. Been a while since I’ve had to explain perl4’s pass-by-name semantics using `*x` parameters to anybody. Ah, but I see you’ve used a lexical `\$x`, which sidesteps that issue. – tchrist Nov 5 '10 at 3:39
Thanks Tom, I originally used positional parameters before realising I could keep the recursion count in a lexical and add another meaning for 'x' to the script. Wish I could have removed the map (even if it is just a glorified 'if')... – Daniel Holz Nov 5 '10 at 11:07
``` push @array, rand;
push @array, rand;
# ... repeat 98 more times
```
-
You have taken the wimp's way out with your `# ... repeat 98 more times` ! – Andy Lester Nov 3 '10 at 23:06
No, he's just being managerial and delegated the boring work to the readers :) – DVK Nov 3 '10 at 23:18
No perl loops:
``````#!/usr/bin/perl
use strict;
use warnings;
@ARGV=q!echo 'int rand(void); int printf(const char *format, ...); int main(void) { int i; for(i=0;i<100;++i)printf("%d\\\\n",rand()); return 0; }' | gcc -x c - && ./a.out |!;
chomp(my @array=<>);
``````
-
I double dog dare you to implement a Lisp interpreter in Perl without loops and then call the Lisp code :) – DVK Nov 4 '10 at 18:08
err..........no – ysth Nov 5 '10 at 1:39
<my best Biff voice>CHICKEN, McFly?!?!?! – DVK Nov 5 '10 at 16:38
@DVK -1 for no end tag!!! don't worry I got your back: </my best Biff voice> – Joel Berger Nov 14 '10 at 14:46
Someone asked for a pure regex solution. How about
``````#!/usr/bin/perl
open my \$slf, \$0;
undef \$/;
(my \$s = <\$slf>) =~ s/./rand()." "/eggs;
\$s .= rand();
``````
-
@JadeNB: Why open `\$0` when it’s already open with `__END__` and `seek(DATA,0,0)`? :) – tchrist Nov 4 '10 at 23:12
Oh for goodness sake, self-modifying code! It will only produce 99 of them unless you insert a `#` at byte 0. That or another `/g` in your omelette. :) – tchrist Nov 4 '10 at 23:17
IMHO it's NOT pure regex in a sense of you're still hard-coding a 100 char string, only as your own code (and need to fudge the code to get to 100 at that). Outside of "cool" value, I don't see it as any different than taking my original regex solution and replacing "Dx100" with a hardcoded string of 100 "D" characters. – DVK Nov 5 '10 at 16:27
My idea of a pure regex solution was to have the counting-to-100 done via some regex magic WITHOUT having 100-char string to do the implicit looping for you. – DVK Nov 5 '10 at 16:28
@DVK: Mine has no 100-char string, but good luck at catching it counting. :) – tchrist Nov 5 '10 at 20:19
Since the task seems to be either to get you to use recursion or to learn how to write an easy loop in a not-so-easy form, humbly I submit the following FULLY EXECUTABLE Perl programs:
# Camel:
``````#!/usr/bin/perl
''=~('('.'?'
.'{'.( '`'|'%').("\["^
'-').('`'| '!').('`'|',').'"'
.'\\'.'\$'. ("\`"| ',').('`'|')').('`'|
'-').'='.('^'^("\`"| '/')).('^'^('`'|'.')).
('^'^('`'|'.')).';'.( '!'^'+').('`'|'&').('`'
|'/').('['^')').'(' .'\\'.'\$'.'='.'='.(('^')^(
'`'|'/')).';'. '\\'.'\$'.'='.'<'.'='.'\\'.'\$'
.('`'|(',')).( '`'|')').('`'|'-').';'.'+'."\+".
'\\'.'\$'.('='). ')'.'\\'.'{'.('['^'+').('['^"\.").(
'['^'(').("\`"| '(').('{'^'[').'\\'.'@'.'='.','.("\{"^
'[').('['^')'). ('`'|'!').('`'|'.').('`'|'\$').'\\'.'}'.(
'!'^'+').'\\'. '\$'.'='.'='.('^'^('`'|'/')).';'.('!'^'+')
.('`'|('&')).( '`'|'/').('['^')').('{'^'[').'('.'\\'.'@'.
'='.')'.('{'^'[').'\\'.'{'.('!'^'+').('*'^'#').('['^'+').(
'['^')').('`'|')').('`'|'.').('['^'/').('{'^'[').'\\'.'"'.(
'['^')').('`'|'!').('`'|'.').('`'|'\$').('{'^'[').'\\'.'\$'.
'='.('{'^'[').('`'|'/').('`'|'&').('{'^'[').'\\'.'\$'.("\`"|
',').('`'|')').('`'|'-').'='.'\\'.'\$'.'_'.'\\'.'\\'.(('`')|
'.').'\\'.'"'.';'.('!'^'+').('*'^'#').'\\'.'\$'.'='.'+'.'+'
.';'.('!'^'+').('*'^'#').'\\'.'}'.'"'.'}'.')');\$:='.' ^((
'~'));\$~='@'|'(';\$^=')'^'[';\$/='`'|'.';\$,='('^"\}"; \$\=
'`'|'!';\$:=')'^'}';\$~='*'|'`';\$^='+'^'_'; \$/="\&"| '@'
;\$,='['&'~';\$\=','^'|';\$:='.'^"\~";\$~= '@'|'(' ;\$^
=')'^ '[';\$/='`'|'.';\$,='('^"\}";\$\= '`'|'!' ;\$:
=')'^'}';\$~='*'|'`';\$^=('+')^ '_';\$/= '&'
|'@';\$,= '['&'~';\$\ ="\,"^ '|';\$: =(
('.'))^ "\~";\$~= ('@')| '(';\$^ =(
(')'))^ "\[";\$/= "\`"| "\."; (
(\$,))= '('^'}'; (\$\) ='`'
|"\!"; \$:=(')')^ '}'; (\$~)
='*'| "\`";\$^= '+'^ '_';
(\$/)= '&'|'@' ;\$,= '['&
'~'; \$\=',' ^'|' ;\$:=
'.'^ '~' ;\$~= '@'|
'('; \$^= ')' ^((
'[' )); \$/= '`'
|(( '.' )); \$,=
'(' ^(( '}' ))
;( (\$\))= ((
(( '`')) ))
|+ "\!";\$:= ((
')' ))^+ "\}"; \$~
=(( '*'))| '`'; \$^=
'+'^ "\_";\$/= '&' |'@'
;(\$,)= ('[')&
"\~";\$\= ','^'|'
``````
# Martini:
``````#!/usr/bin/perl
''=~('(?{'.
('`'|'%').(
'['^"\-").(
'`'|"\!").(
'`'|(',')).
'"\\\$'.('`'
|',').(
'`'|')'
).('`'|
('-')).
('=').(
'^'^('`'|
('/'))).(
'^'^('`'|
('.'))).(
'^'^(('`')|
'.')).';'.(
'!'^'+').('`'
|'&').(('`')|
'/').('['^')').
'(\\\$=='.('^'^(
'`'|'/')).';\\\$='
.'<=\\\$'.('`'|',').
('`'|')').('`'|"\-").
';++\\\$=)\\{'.('['^'+')
.('['^'.').('['^'(').('`'
|'(').('{'^'[').'\\@'.('`'|
'!').('['^')').('['^"\)").(
'`'|'!').('['^'"').','.('{'
^'[').('['^')').('`'|'!').(
'`'|'.').('`'|"\\$").'\\}'.(
'!'^'+').'\\\$'.('`'|')').'='.("\^"^( '`'|'/')).';'.('!'^('+')).(
'`'|'&').('`'|'/').('['^')').('{'^ '[').'(\\@'.('`'|'!').('['^
')').('['^')').('`'|'!').('['^ '"').')'.('{'^"\[").'\\{'.(
'!'^'+').('*'^'#').('['^'+') .('['^')').('`'|')').("\`"|
'.').('['^'/').('{'^'[') .'\\"'.('['^')').('`'|'!').
('`'|'.').('`'|"\\$").( '{'^'[').'\\\$'.('`'|"\)").(
'{'^'[').('`'|'/') .('`'|'&').('{'^'[').'\\\$'.
('`'|',').("\`"| ')').('`'|'-').'=\\\$_\\\\'.
('`'|('.')). '\\";'.('!'^'+').('*'^'#').
'\\\$'.('`' |')').'++;'.('!'^'+').('*'^
"\#"). '\\}"})');\$:='.'^'~';\$~='@'
|'(' ;\$^=')'^'[';\$/='`'|"\.";\$,=
'('^ '}';\$\='`'|'!';\$:=')'^"\}";
(\$~) ='*'|'`';\$^='+'^'_';\$/='&'|
'@'; \$,='['&'~';\$\=','^('|');\$:=
'.'^ '~';\$~='@'|'(';\$^=')'^"\[";
(\$/) ='`'|'.';\$,='('^'}';\$\='`'|
'!'; \$:=')'^'}';\$~='*'|('`');\$^=
'+'^ '_';\$/='&'|'@';\$,='['&"\~";
(\$\) =','^'|';\$:='.'^'~';\$~='@'|
'('; \$^=')'^'[';\$/='`'|('.');\$,=
'('^ '}';\$\='`'|'!';\$:=')'^"\}";
(\$~) ='*'|'`';\$^='+'^'_';\$/='&'|
'@';\$,='['&'~';\$\=','^'|';\$:='.'^'~' ;\$~='@'|'(';\$^=')'^"\[";\$/=
'`'|'.';\$,='('^'}';\$\='`'|'!';\$:=')' ^'}';\$~='*'|'`';\$^='+'^'_';
``````
# For the Holidays, snowflakes with recursion rather than iteration inside:
``````#!/usr/bin/perl
'?' =~(
'('.'?' ."\{".(
'`' |'%' ).('['^"\-").( '`'| '!'
).('`'|','). '"'. '\\'.('\$').(
'`'|(',')).( '`'| ')').(('`')|
(( '-') )). +( '`' |')' ).
((( '[' ))^+ (( '/') )). '='
.('^'^ ('`'|'/')) .( '^'^("\`"| '.')).
+( '^'^('`'|'.')).';'.('!'^"\+"). ((
'\\')).'\$'.('`'|'#').('`'|'/').('['^'.').('`'|'.').(
'['^ '/').'='. (('^')^( '`'|'/') ).(';').( '!'^
'+' ).('['^ '(').( ('[')^ "\.").( '`'
|'"' ).(('{')^ ('[')).( '['^'+') .('['^'.' ).+(
'['^'(').('`'|'(').'_'.('['^')').('`'|'!').('`'|'.')
.( '`'|'\$').('{'^'[').'\\'."\{".( ((
'!'))^ '+').('{'^ (( ('[')))).( ('{')^
'[' ).( '{'^ (( '[') )). (((
(( '{') ))) ^+ '[' ).+( ((
'['))^')').( '`'| '%').(('[')^
'/').(('[')^ '.') .('['^')').(
'`' |'.' ).('{'^"\[").( '`'| ')'
).('`'| "\&").(
'{' ^((
'[' )))
.'\\'.+ '\$'.'#'
.+( '`'| '!').('['^')') .''. (((
'['))^')').( '`'| '!').(('[')^
'"').('_').( '`'| '/').(('`')|
(( '&') )). (( '_' )).( ((
'[' ))^ ')') .( '`'| '!' ).(
('`')| '.').('`'| (( ('\$')))).( ('[')^
(( '('))).'>'.'\\'.'\$'.('`'|','). +(
'`'|')').('`'|'-').('`'|')').('['^'/').';'.('!'^'+')
.''. ('{'^'[') .(('{')^ ('[')).( '{'^'['). ('{'
^(( '['))). ("\["^ '+').( '['^'.' ).(
'['^ '(').('`' |"\(").( '{'^'[') .'\\'.'@' .''.
('`'|'!').('['^')').('['^')').('`'|'!').('['^('"')).
(( '_')).('`'|'/').('`'|'&').'_'. +(
('[')^ ')').('`'| (( ('!')))).( ('`')|
'.' ).( '`'| (( '\$') )). (((
(( '[') ))) ^+ '(' ).(( ((
',')))).('{' ^'[' ).('['^')').
('`'|"\!").( '`'| '.').(('`')|
'\$' ).(( ';')).('!'^'+' ).+( '{'
^'[').( '{'^'['
).( '{'
^(( '['
))).''. (('{')^
'[' ).+( '['^'+').('['^ '.') .+(
'['^('(')).( '`'| '(').('_').(
'['^(')')).( '`'| '!').(('`')|
(( '.') )). +( '`' |'\$' ).
'(' .(( ')') ). ';'. ((( '!'
))^'+' ).'\\'.'}' .( '!'^'+').( ('!')^
(( '+'))).('!'^'+').('['^('(')).( ((
'['))^'.').('`'|'"').('{'^'[').('['^'+').('['^')').(
'`'| ')').('`' |"\.").( '['^'/') .'_'.('[' ^')'
).( '`'|'%' ).('`' |'#'). (('[')^ '.'
).+( '['^')'). ('['^'(' ).("\`"| ')').('[' ^'-'
).('`'|'%').('{'^'[').'\\'.'{'.('!'^'+').('{'^'[').(
(( '{'))^'[').('{'^'[').('{'^'[') .+
'\\'.+ '\$'.("\["^ (( '/'))).'=' .('['^
'+' ).( '`'| (( '/') )). (((
(( '[') ))) ^+ '+' ).+( ((
'{'))^"\["). '\\' .'@'.'_'.';'
.('!'^'+').( '*'^ '#').(('[')^
'+' ).+( '['^')').('`'| ')') .+(
'`'|'.' ).('['^
'/' ).(
'{' ^((
'['))). ('\\').
'"' .''. ('['^')').('`' |'!' ).(
'`'|('.')).( '`'| '\$').(('{')^
'[').('\\'). '\$'. ('`'|"\#").(
(( '`') )|+ (( '/' ))). +(
'[' ^(( '.') )) .''. ((( '`'
))|'.' ).('['^'/' ). ('{'^'['). ("\`"|
(( '/'))).('`'|'&').('{'^'[').''. ((
'\\')).'\$'.('`'|',').('`'|')').('`'|'-').('`'|')').(
'['^ '/').'='. '\\'.'\$' .(('[')^ '/').'\\' .''.
((( '\\'))) .('`'| "\."). ('\\'). '"'
.';' .('!'^'+' ).("\*"^ '#').''. '\\'.'\$'. ('`'
|'#').('`'|'/').('['^'.').('`'|'.').('['^'/').('+').
(( '+')).';'.('!'^'+').('*'^'#'). +(
('[')^ '+').('['^ (( (')')))).( ('`')|
')' ).( '`'| (( '.') )). (((
(( '[') ))) ^+ '/' ).(( ((
'_')))).('[' ^')' ).('`'|'%').
('`'|"\#").( '['^ '.').(('[')^
')' ).+( '['^'(').('`'| ')') .+(
'['^'-' ).('`'|
'%' ).+
'(' .((
'\\')). '@'.'_'
.(( ')') ).('{'^"\[").( '`'| ')'
).('`'|'&'). ('{' ^'[').('(').
'\\'.'@'.'_' .')' .';'.(('!')^
(( '+') )). (( ((( '\\' ))
))) .(( '}') ). ('!' ^(( '+'
))).+( '!'^'+').( (( '['))^'+') .('['^
(( '.'))).('['^'(').('`'|'(').'_' .(
'['^')').('`'|'!').('`'|'.').('`'|'\$').'('.')'.';'.(
'['^ '+').('[' ^"\)").( '`'|')') .('`'|'.' ).+(
'[' ^"\/"). "\_".( ('[')^ "\)").( '`'
|'%' ).(('`')| ('#')).( '['^'.') .('['^')' ).+(
'['^'(').('`'|')').('['^'-').('`'|'%').'('.'\\'.'@'.
+( '`'|'!').('['^')').('['^')').( ((
'`'))| '!').('['^ (( '"'))).'_' .('`'|
'/' ).( '`'| (( '&') )). '_'
.( '['^ ')' ). ((( '`') )|
'!').(('`')| '.') .('`'|'\$').(
'['^'(').')' .';' .'"'.'}'.')'
);( \$:)= '.'^'~';\$~='@' |'(' ;\$^
=(')')^ '[';#;#
;#; #;#
``````
In each case, the output is something like this:
``````rand 1 of 100=0.625268682212667
rand 2 of 100=0.30160434879096
...
rand 100 of 100=0.584811321826528
``````
If you want to see the loops or recursion embedded within, you can use `perl -MO=Deparse martini.pl` or `perl -MO=Deparse camel.pl` etc.
Only with Perl, right???
If you want to generate these lovely things -- check out Acme::Eyedrops
-
Recursion:
``````sub fill_rand {
my (\$array, \$count) = @_;
if (\$count >= 1) {
unshift @\$array, rand();
fill_rand (\$array, --\$count);
}
}
my @array;
fill_rand (\@array, 100);
``````
"Tail-call optimised" version:
``````sub fill_rand {
my \$array = shift;
my \$count = shift;
unshift @\$array, rand();
if (\$count > 1) {
\$count--;
@_ = (\$array, \$count);
goto &fill_rand;
}
}
my @array;
fill_rand(\@array, 100);
``````
Using eval:
``````my @array;
eval("\@array = (" . ("rand(), " x 100) . ");");
``````
If you assume that my perl is random (not an unwarranted assumption), you could use the perl file itself as a source of random data:
``````open FILE, __FILE__ or die "Can't open " . __FILE__ . "\n";
my \$string;
read FILE, \$string, 100;
close FILE;
my @array = map { ord } split //, \$string;
``````
Of course, you'll get the same results every time, but this is useful for testing.
-
Instead of using `goto &fill_rand` you could use Sub::Call::Recur or Sub::Call::Tail – Brad Gilbert Nov 5 '10 at 14:39
@Brad Gilbert, thanks for the pointer; I'll look at that. – Martin Broadhurst Nov 5 '10 at 14:45
Another silly method, how about using a tied array that return a random value ?
``````use strict;
use warnings;
package Tie::RandArray;
use Tie::Array;
our @ISA = ('Tie::StdArray');
sub FETCH { rand; }
package main;
my @rand;
my \$object = tie @rand, 'Tie::RandArray';
\$#rand=100;
my @a= @somearray;
warn "@a";
``````
Of course the tied array could cache the values, so that a second array would not be needed to have stable values.
-
its ugly, but it works. the foreach is just to show that it does.
``````#!/usr/bin/perl
rand1();
\$idx = 1;
foreach \$item (@array) {
print "\$idx - \$item\n";
\$idx++;
}
exit;
sub rand1() {
rand2();
rand2();
rand2();
rand2();
}
sub rand2() {
rand3();
rand3();
rand3();
rand3();
rand3();
}
sub rand3() {
push @array, rand;
push @array, rand;
push @array, rand;
push @array, rand;
push @array, rand;
}
``````
-
That’s more like my first regex solution than most people realize. – tchrist Nov 10 '10 at 4:15
``````@foo = (rand(), rand(), rand(), ... rand());
``````
-
`syntax error at -e line 1, near ", ..."` – tchrist Nov 3 '10 at 23:59
Generate the data:
``````my @array = map { rand() } (0..99);
``````
Print the data to show that you have the right result:
``````print "\$_\n" foreach (@array);
``````
The generation loop is hidden (there's no looping keyword visible - just a function/operator).
-
No, I can still see it. It's right there! :-) – rafl Nov 3 '10 at 22:55
@rafl: You've got better eyesight than me. I see no while, for, foreach, until (did I miss any?) in the generation line. – Jonathan Leffler Nov 3 '10 at 22:56
A map is debatably a kind of magic loop :) stackoverflow.com/questions/3019925/is-map-a-loop – DVK Nov 3 '10 at 23:19
@DVK, @rafl: yes, of course map is iterating over a list; yes, there is is a loop-like effect; that's how you end up with 100 values. – Jonathan Leffler Nov 4 '10 at 2:52
``````my \$u;
open(URAND, "/dev/urandom") || die \$!;
close URAND;
my @array = split(/ */, \$u);
``````
-
That doesn’t work properly. You need to `read(URAND, \$u, 100 * length pack("I")` and then `@array = unpack("I*", \$u)`. Or `\$/ = \length pack "I"; @array = unpack("I*", <URAND>)`. – tchrist Nov 4 '10 at 15:14
Oops, that needs `\$\$/*=100`. – tchrist Nov 4 '10 at 16:15
Dunno whether elwood’s ever coming back, but a proper urandom solution is just `perl -le '\$/ = \\(100*length pack "L"); print for unpack "L*" => <>' /dev/urandom`. – tchrist Nov 4 '10 at 17:11
As per requests from the listeners, a non-pure-regex solution:
``````\$s="D" x 100;
\$s=~s/D/rand()." "/ge;
@s=split(/ /,\$s);
``````
-
Come on now, you can do better than that. – tchrist Nov 3 '10 at 23:57
Meh. Why bother when Sinan's having so much fun and my brain's a total mush from 4 hours fighting with Eclipse. I literally ONLY did it because pst asked in his answer :) – DVK Nov 4 '10 at 1:58
Eclipse and I are not on speaking terms. – tchrist Nov 4 '10 at 11:47
@tchrist - incredible! – DVK Nov 5 '10 at 16:39
@DVK: Oh, it’s perfectly credible, being a sort of its own existence proof. It works just fine, as does the expanded version, with or without cpp (1) installed—curiously enough. :) – tchrist Nov 5 '10 at 20:17
Recursion:
``````#!/usr/bin/perl
use warnings; use strict;
my @rands;
my \$i=1;
sub push_rand {
return if \$#rands>=99;
push @rands, rand;
push_rand();
}
push_rand();
for (@rands) { print "\$i: \$_\n"; \$i++; }
``````
-
| 12,644
| 35,624
|
{"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-2016-30
|
latest
|
en
| 0.919848
|
http://forums.wolfram.com/mathgroup/archive/2001/Jul/msg00304.html
| 1,529,927,871,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00229.warc.gz
| 120,845,122
| 7,476
|
Re: Placeholders in matrix notation
• To: mathgroup at smc.vnet.net
• Subject: [mg29956] Re: Placeholders in matrix notation
• From: dooglefish at aol.communicate (D.F.)
• Date: Thu, 19 Jul 2001 03:57:27 -0400 (EDT)
• References: <9j39rr\$ikv\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```>Subject: [mg29956] Re: Placeholders in matrix notation
>From: "Hugh Philipp" hph at com.dtu.dk
To: mathgroup at smc.vnet.net
>Date: 7/18/01 2:20 AM Eastern Daylight Time
>Message-id: <9j39rr\$ikv\$1 at smc.vnet.net>
>
>I've had this problem before and I have used at least a dozen clumsy work
>arounds for it.
>
>Here is one:
>
>ff ={D[#[[1]],x],0}&;
>
>ff@{f[x],g[x]}
>
>Of course, I don't really like this method because you have to do the matrix
>operations explicitly in the function - designating the elements.
>
>
>The other way of doing it goes something like this:
>
>Unprotect[Times]
>
>d1 f_:=D[f,x]
>
>Protect[Times]
>
>{d1,0,d1}.{f[x],g[x],h[x]}
>
>I don't know if the gurus here think this is O.K. - I'm sure there will be
>
>Notice to that:
>
>d1 d1 f[x]
>
>evaluates to f''[x]
>
>...a nice 'feature'. - but BE CAREFUL becaue d1 d1 evaluates to 0. So -
>there might be some other hidden 'features' with this approach.
>
>For example, the vector: {d1 d1,d1} evaluates to {0,d1}
>
>Hope this helps,
>
>Hugh.
Yes, it works very well. Thank you.
\!\(\*
RowBox[{\(Unprotect[Times]\), "\n", \(d1\ f_ := D[f, x]\),
"\n", \(d2\ f_ := D[f, y]\), "\n", \(d3\ f_ := D[f, x, y]\),
"\n", \(Protect[Times]\), "\n",
RowBox[{"MatrixForm", "[",
RowBox[{
RowBox[{"(", GridBox[{
{"d1", "0"},
{"0", "d2"},
{"0", "d3"}
}], ")"}], ".",
RowBox[{"(", GridBox[{
{\(f1[x, y]\), \(f2[x, y]\), \(f3[x, y]\)},
{\(g1[x, y]\), \(g2[x, y]\), \(g3[x, y]\)}
}], ")"}]}], "]"}]}]\)
-Doug
```
• Prev by Date: Re: Thickness Isn't Thickness
• Next by Date: Re: Thickness Isn't Thickness
• Previous by thread: Re: Re: Placeholders in matrix notation
• Next by thread: A question on machine numbers
| 718
| 1,991
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.777538
|
http://fisicaexe.com.br/en/physics0/thermodynamics/thermoexp/thermoexp4_hs1/thermoexp4_hs1.html
| 1,582,354,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875145654.0/warc/CC-MAIN-20200222054424-20200222084424-00505.warc.gz
| 56,368,619
| 3,486
|
Solved Problem on Thermal Expansion
Português English
A container is filled with 125 cm3 of mercury at a temperature of 20 °C. The average coefficient of expansion of the mercury is 180 × 10−6 °C−1 and the coefficient of linear expansion of the glass is 9 × 10−6 °C−1. Find the volume of mercury spilled out when the temperature rises to 28 °C.
Problem data
• volume of the container at 20°C: V0 = 125 cm3;
• coefficient of expansion of mercury: γHg = 180 × 10−6 °C−1;
• coefficient of linear expansion of glass: αv = 9 × 10−6 °C−1;
• initial system temperature: ti = 20 °C;
• final system temperature: tf = 28 °C.
Problem diagram
At the initial temperature, the container and the mercury have the same volume as 125 cm3, when the system is heated, it expands, but as the mercury expands more than the container a part of it overflows. This overflowing part is the apparent volume change Δ Vap we want to find.
figure 1
Solution
The coefficient of volumetric expansion of glass (γv) obtained from the coefficient of linear expansion of glass (αv)
$\bbox[#99CCFF,10px] {\gamma _{\text{v}}=3\alpha _{\text{v}}}$
$\gamma _{\text{v}}=3 \times 9 \times 10^{-6}\\ \gamma_{\text{v}}=27 \times 10^{-6\;\text{o}}\text{C}^{-1}$
The apparent coefficient of expansion of mercury, it is only of the part that overflowed, is given by
$\bbox[#99CCFF,10px] {\gamma _{\text{ap}}=\gamma _{\text{Hg}}-\gamma _{\text{v}}}$
$\gamma _{\text{ap}}=180 \times 10^{-6}-27 \times 10^{-6}\\ \gamma_{\text{ap}}=153 \times 10^{-6\acute{}\text{o}}\text{C}^{-1}$
The spilled volume will be
$\bbox[#99CCFF,10px] {\Delta V_{\text{ap}}=V_{0} \gamma _{\text{ap}} \Delta t}$
$\Delta V_{\text{ap}}=V_{0}\gamma_{\text{ap}}\;(\;t_{\text{f}}-t_{\text{i}}\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 10^{-6} \times (\;28-20\;)\\ \Delta V_{\text{ap}}=125 \times 153 \times 8 \times 10^{-6}$
$\bbox[#FFCCCC,10px] {\Delta V_{\text{ap}}=0.153\;\text{cm}^{3}}$
| 656
| 1,941
|
{"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}
| 4.21875
| 4
|
CC-MAIN-2020-10
|
latest
|
en
| 0.600971
|
https://fenicsproject.discourse.group/t/how-to-assemble-a-trilinear-form/13774
| 1,726,587,237,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00053.warc.gz
| 222,218,571
| 8,393
|
# How to assemble a trilinear form?
Hi everyone,
I am trying to solve NS equation and need to get a 3-tensor induced by the nonlinear term u\cdot \nabla u. In other words, I want to get a tensor A defined by A_{i j k} = \int_\Omega (\phi_i \cdot \nabla) \phi_j \cdot \phi_k d x, where \phi_i-s are the basis functions.
There are some similar questions before but it seems that there are never a solution to this problem. But those questions are mostly 8 or 10 years ago. So I just want to take a chance to see whether there are some new solutions.
I have tried to import āArgumentā but failed. Here is the code, for a simple trilinear form:
=======================================
from dolfin import *
mesh = UnitSquareMesh(10, 10)
V = FunctionSpace(mesh, āPā, 1)
u = Argument(V, 0) # Trial function argument
v = Argument(V, 1) # Test function argument
w = Argument(V, 2) # Another test function argument
m = u * v * w * dx # Define a variational problem
m_mat = assemble(m)
=======================================
This does not work because the function āassembleā seems unable to deal with trilinear form. However, in the documentation, it says:
So I guess it should be able to assemble a trilinear form. I will appreciate it if anyone could help me with this.
Trilinear forms are not supported. Dolfin supports assembling tensors of order 0,1,2.
I am not aware of any linear algebra package supporting sparse tensors of third order (hopefully someone might know)
It would be good if @yangchanghe gave us more context about what they are trying to do with this trilinear form. In principle, you could have w be a dolfin.Function which only has a single DOF equal to one and assemble that bilinear form, then repeat that for all possible w. However, do not try this unless you are sure that it really is what you need, because it will likely eat up all your RAM
1 Like
I think scikit fem can assemble trilinear form. But I met another problem with scikit fem. So I hope fenics can do that because I am more familiar with fenics.
I want to export this tensor and matrices induced by other terms in NS equation. Then I need to solve an optimization problem, which I have to use some tools in matlab.
Of course I can also use matlab.engine in every iteration, but I believe it would be very difficult to debug so this is my last choice.
You can always create these in the way Francesco suggested, i.e.
from mpi4py import MPI
import dolfinx
import ufl
mesh = dolfinx.mesh.create_unit_square(MPI.COMM_WORLD, 5, 5)
V = dolfinx.fem.functionspace(mesh, ("Lagrange", 1))
phi_i = ufl.TrialFunction(V)
phi_j = ufl.TestFunction(V)
phi_k = dolfinx.fem.Function(V)
# general A to assemble into
a_compiled = dolfinx.fem.form(a)
A = dolfinx.fem.create_matrix(a_compiled)
num_dofs_global = V.dofmap.index_map.size_global * V.dofmap.index_map_bs
local_dof_range = V.dofmap.index_map.local_range * V.dofmap.index_map_bs
A_kij = []
print(num_dofs_global)
for i in range(num_dofs_global):
phi_k.x.array[:] = 0.0
if i >= local_dof_range[0] and i < local_dof_range[1]:
phi_k.x.array[i-local_dof_range[0]] = 1.0
phi_k.x.scatter_forward()
A.set_value(0)
A_kij.append(dolfinx.fem.assemble_matrix(A, a_compiled).to_scipy())
print(A_kij)
But this will eat up your RAM for any sufficiently large problem.
Thanks a lot! I will try it.
Hi @dokken,
Thank you for the helpful code example. I have a similar problem as @yangchanghe for assembling a trilinear form, but a little bit different. Instead of having a written trilinear variational form, I would like to take second derivative of an energy form for linear elasticity to get the sensitivity of the stiffness matrix with respect to the material property. I know it would be crazy costly, but this is required to couple my solver with external optimization paradigm.
Here is a minimal example of the problem, which is adapted from your hyperelasticity tutorial (Hyperelasticity ā FEniCSx tutorial). Iām having a hard time trying to adapt this to the method you showed in the code example.
from mpi4py import MPI
import dolfinx
import ufl
mesh = dolfinx.mesh.create_unit_cube(MPI.COMM_WORLD, 1, 1, 1)
V = dolfinx.fem.functionspace(mesh, ("Lagrange", 2, (mesh.geometry.dim, )))
u = dolfinx.fem.Function(V)
v = ufl.TestFunction(V)
d = len(u)
# Identity tensor
I = ufl.variable(ufl.Identity(d))
# Right Cauchy-Green tensor
C = ufl.variable(F.T * F)
# Invariants of deformation tensors
Ic = ufl.variable(ufl.tr(C))
J = ufl.variable(ufl.det(F))
# Elasticity parameters
VE = dolfinx.fem.functionspace(mesh, ("Lagrange", 1))
E = dolfinx.fem.Function(VE)
E.x.array[:] = 1.0E4
nu = 0.3
mu = E / (2 * (1 + nu))
lmbda = E * nu / ((1 + nu) * (1 - 2 * nu))
# Stored strain energy density (compressible neo-Hookean model)
psi = (mu / 2) * (Ic - 3) - mu * ufl.ln(J) + (lmbda / 2) * (ufl.ln(J))**2
# Stress
P = 2.0 * mu * ufl.sym(ufl.grad(u)) + lmbda * ufl.tr(ufl.sym(ufl.grad(u))) * I
F = ufl.inner(ufl.grad(v), P) * ufl.dx
K = ufl.derivative(F, u)
dKdt = ufl.derivative(K, E)
# Assemble stiffness matrix
K_compiled = dolfinx.fem.form(K)
K_mat = dolfinx.fem.assemble_matrix(K_compiled)
# Assemble sensitivity of stiffness matrix wrt material property
#******************** Want to have ***********************
dKdt_compiled = dolfinx.fem.form(dKdt)
dKdt_mat = dolfinx.fem.assemble_matrix(dKdt_compiled)
#*********************************************************
As expected, I got an error in the last line of code saying āRuntimeError: Cannot create sparsity pattern. Form is not a bilinear formā. Any ideas will be appreciated!
If you look at help(ufl.derivative), youāll see that the signature is derivative(form, coefficient, argument=None, coefficient_derivatives=None). You could:
1. create an argument (in the space VE) to be passed as third input
2. once you get dKdt, use ufl.replace to replace that argument with dolfinx.fem.Function
3. proceed as in the previous post.
Of course, this only makes sense if dKdt is actually multilinear. Continuum mechanics is not my field, but I would be worried about the ln(J) term
1 Like
It works perfectly for my case! Thanks a lot
| 1,676
| 6,148
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.296875
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.905427
|
https://brainly.com/question/182890
| 1,484,563,984,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560279169.4/warc/CC-MAIN-20170116095119-00408-ip-10-171-10-70.ec2.internal.warc.gz
| 790,532,090
| 8,679
|
2014-11-12T18:22:51-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1.9
19/2/5
9.5/5
1.9
OR
You could combine to do:
19/10
1.9
2014-11-12T18:23:54-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Here we have to write down the description of the problem, so
Nineteen is two more times than five times a number
19 = 2 ( 5* x)
19=2(5x)
Now we can find the result
19=10x /:10 (divide both sides by 10)
1.9=x - its the result
| 269
| 991
|
{"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-2017-04
|
latest
|
en
| 0.910032
|
https://www.physicsforums.com/threads/integration-to-find-velocity.377191/
| 1,545,047,241,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376828507.57/warc/CC-MAIN-20181217113255-20181217135255-00115.warc.gz
| 988,580,268
| 12,717
|
# Homework Help: Integration to find velocity
1. Feb 10, 2010
### Anti-Meson
1. The problem statement, all variables and given/known data
I intend to use the Runge-Kutta method but to do so I need to be able to find the velocity $$\frac{dx(t)}{dt}$$ from the acceleration and I need some pointers on how to get that from the equation below. In other words I am having difficulty integrating the equation wrt time.
2. Relevant equations
$$\frac{d^{2}x(t)}{dt^{2}} = \frac{k}{x(t)^{2}}$$
where k is a constant.
Any help would be appreciated. Thank-you.
2. Feb 10, 2010
### Dick
I don't think you can express the velocity in any easy form. Isn't the usual trick to add dx(t)/dt=v(t) to your list of equations and make the first equation dv(t)/dt=k/x(t)^2 and solve that first order system of equations with Runge-Kutta?
3. Feb 10, 2010
### dacruick
dont you need x(t)²??
4. Feb 10, 2010
### Anti-Meson
The problem is Runge-Kutta method uses two variables, in my case x and t, though at the moment I only have 1 variable x as expressed in the equation of acceleration.
| 304
| 1,079
|
{"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}
| 3.234375
| 3
|
CC-MAIN-2018-51
|
latest
|
en
| 0.912908
|
https://www.casinocitytimes.com/alan-krigman/article/gambling-theories-history-and-other-lies-5757
| 1,726,312,009,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00297.warc.gz
| 635,554,872
| 8,288
|
Stay informed with the
Recent Articles
Best of Alan Krigman
# Gambling Theories, History, and Other Lies
7 January 2003
The philosopher, George Santayana (1863-1952), is often quoted as having said, "Those who cannot remember the past are condemned to repeat it." A counterpart may pertain to gambling. Something like "Those who cannot comprehend the math are convinced they can defeat it." Santayana also said, "History is a pack of lies about events that never happened told by people who weren't there." An even better nominee for a gambling analogy! Make one up yourself.
This came to mind recently when Jimmy, the popular bartender at the posh Celebrity Club in Atlantic City's Claridge Casino, asked me about splitting nines rather than standing against seven-up at blackjack. He thought splitting was good aggressive play, despite "the book" saying to stand. Besides, he saw a wallet-sized Basic Strategy card with an asterisk next to the "stand" symbol and a notation that "some authorities claim it's better to split." As a clincher, Jimmy added that since Basic Strategy comes from assuming the "unseen" card is a 10 - it makes sense to go for two winning 19s instead of one 18 against a dealer's final 17. Fortunately, Jim's mixology is much better than his mathematics.
Jimmy's belief that Basic Strategy for any hand presupposes the unseen card to be a 10 is widespread but wrong. The decisions are actually those giving the greatest statistically expected profit or least corresponding theoretical loss, when the chances of all possible results of every alternate option are calculated. It's therefore not a matter of one guru's opinion over another's, but of calculations based on the laws of probability. To picture how this works, think just about a dealer's seven-up. The dealer has 36.8 percent chance of finishing with 17. This can come from a 10 in the hole (chances are roughly 30.8 percent). But there's also six percent chance of getting 17 with combinations like nine plus ace, six plus four, six plus three plus ace, and so on. Too, the dealer has 26.2 percent chance of breaking and 37.0 percent chance of wrapping up with 18 through 21. All go into the stew.
On the other hand, Jim's right in not holding those wallet-sized cards in quite the same reverence as the stone tablets Moses got from a burning bush and brought down the mountain. Players may opt to follow Basic Strategy rigorously, understanding that by doing so they're maximizing their statistical expectation. But, solid citizens who accept the rules entirely as a matter of faith (and who tend to be the most vocal critics of anyone who deviates even slightly from the strait and narrow), fail to realize that individuals may have valid gambling criteria for which decisions violating Basic Strategy are optimum. As an example, pretend you start with a \$10 bet and pull a five-six vs a dealer's nine. Basic Strategy is to double rather than hit, because long-term earnings are projected to average \$2.30 with the former and \$1.57 with the latter. But chances of winning are greater by hitting, since you may draw a seven or under and can then get yet another card rather than stop below 17. Who's to fault you for picking more chance at less win, over the converse?
That's not the whole story. Certain Basic Strategy plays are so close to the next-best choice, expectation-wise, that statistical penalties for being aggressive or conservative are negligible. In "Blackjack Attack," Don Schlesinger gives the following as hands where straying from the gospel has a calculated long-term average cost under a penny per dollar bet: nine vs two-up, 12 vs four-up, 16 vs 10-up, ace-four vs four-up, ace-six vs two-up, ace-seven vs two-up, two-two vs three- or four-up, and three-three vs four-up.
As for nine-nine vs seven, here's the skinny for one of those \$100 bets Jim's sportsman status demands. Stand and your expected profit is \$40. Split and your expected profit is \$37. Should you grab the better shot at \$100? Or is it worth sacrificing \$3 "on paper" to go for a real \$200 return - more if you happen to pull another nine, or draw a deuce to one or more sides - knowing you're strongly favored either way? The only expert qualified to answer is the person who knows you best, the one you see in the mirror every morning when you're brushing those pearly whites. For, as everybody's favorite poet, Sumner A Ingmark, wrote:
| 973
| 4,435
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.234375
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.9657
|
https://freezingblue.com/flashcards/68738/preview/6-9-kreps
| 1,723,354,072,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00776.warc.gz
| 206,303,556
| 4,071
|
# 6.9.Kreps
Kreps 2 financial strategies to develop risk load Swap Risk-Free Investment for Risky InvestmentPurchase Put Options Kreps 2 constraints when determining risk load Safety Constraint: funds available at year end are at least equal to the specified loss safety level (IRR ≥ rf)Investment Variance Constraint: variance of reins + inv strategy should be no more volatile than a direct investment in risky assets Kreps Reins + inv compared to equity investment Writing reinsurance can never really result in dist of capital that are really the same as investment dist because investment returns are strictly limited to having worst-case returns of -100% Kreps Swap Strategy (1 + IRR)A = (1 + rf)(P + A) - LR = (y - rf)/(1 + rf) * ASafety Constraint A = (s - μL) / (1 + y)s = eμ+ N-1(pct)σσ = √ln(1 + CV²)μ = ln(mean) - σ²/2Variance Constraint A = σL/σy Kreps Put Option Strategy F = (P + A) / (1 + r)R = A[(1 + r)(1 + y) - (1 + i)/(1 + i)] + μL[(1 + r)/(1 + i) - 1/(1 + rf)]Safety Constraint A = 1 / (1 + y) * [(1 + i)/(1 + rf) * s - μL]Variance ConstraintA = [b + √(b² + ac)] / aa = σy²(1 + i)² - σi²(1 + y)²b = μL(1 + y)σi²c = μL²σi² + σL²(1 + i)² AuthorExam9 ID68738 Card Set6.9.Kreps DescriptionKreps Updated2011-02-24T21:27:45Z Show Answers
| 411
| 1,254
|
{"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-2024-33
|
latest
|
en
| 0.76044
|
http://openstudy.com/updates/51a43b27e4b0aa1ad887c557
| 1,448,632,241,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398449160.83/warc/CC-MAIN-20151124205409-00169-ip-10-71-132-137.ec2.internal.warc.gz
| 177,520,899
| 14,311
|
Emily778 2 years ago AO=12 and BC=25. What is AB? I think it's 37, but I'm not sure.
1. Emily778
2. whpalmer4
Close...
3. whpalmer4
What is the length of the hypotenuse (OB)?
4. ParthKohli
|dw:1369717643930:dw|
5. ParthKohli
\(OC = OA = 12\) because they are the radii of the same circle.
6. ParthKohli
Also, as whpalmer asked, what is the hypotenuse, or \(OB\)? A good start might be noticing that \(OB = OC + CB\).
7. whpalmer4
or, just tell us how you came up with 37...then we can see if it is a simple arithmetic error, or how you approached the problem, or whatever...
8. KenLJW
OB^2=OA^2+AB^2 if angle A = 90 37^2=12^2+AB^2 other wise you have the sine angle formula
9. whpalmer4
I assumed (and ParthKohli too, I'm sure) that because only one intersection point is shown, AB is tangent to the circle, which means that OAB is a right angle.
10. KenLJW
imagine A approaching C
11. KenLJW
Not enough information given without assumption angle A = pi/2
12. KenLJW
When you assume you make a retriceof u and me
13. KenLJW
retriceof u and me
14. KenLJW
I guess it doesn't like that word
15. Emily778
That didn't help at all. You never even told me how to get that answer....
16. whpalmer4
17. Emily778
18. whpalmer4
19. KenLJW
Solve 37^2=12^2+AB^2
20. Emily778
The numbers..DUH
21. Emily778
And uh...I need to know AB to do that. DUH!!
22. Kapt_Crazy
emily the ans for ab is 35. its a right angle triangle
23. qweqwe123123123123111
Now wait...you're going "duh" at everyone else and yet you can't figure out what AB is from the equation 37^2=12^2+AB^2 ???
24. KenLJW
I used the Pythagorean theorem a^2+b^2=h^2 h = sqrt(a^2+b^2) or in out case b=sqrt(h^2-a^2)
25. Emily778
@Kapt_Crazy So I was right? The answer is 35? :)
26. KenLJW
Generally you shouldn't just give quantity but show all work so it's easy checked or to show how it's done
27. KenLJW
not give
28. Kapt_Crazy
yeah u right :)
29. qweqwe123123123123111
No, you didn't say 35. You said "I think it's 37" which you absurdly arrived at by adding AO and BC together. Also, 35 is only correct if AB is tangent to the circle at A. But since that information isn't given, then its only an assumption.
30. Kapt_Crazy
qweqwe if it was not a right angle triangle they would give u more info, like one of the angles or sumthing
31. Emily778
Oh, thankyou for going back and checking if I said 35. :) @qweqwe123123123123111
32. KenLJW
Yes they would have to give 2 angles to solve
33. qweqwe123123123123111
The assumption is probably a safe one, @Kapt_Crazy, But UNLESS THAT INFO IS GIVEN EXPLICITLY, it's still just an assumption. And the number of poorly formed and incomplete problems that are posted here every single day are legion, so there's no good reason to assume that this problem is stated any better than any other.
| 908
| 2,841
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2015-48
|
longest
|
en
| 0.896766
|
http://crypto.stackexchange.com/questions/tagged/keys?sort=frequent&pageSize=50
| 1,469,477,520,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257824345.69/warc/CC-MAIN-20160723071024-00273-ip-10-185-27-174.ec2.internal.warc.gz
| 55,145,684
| 26,504
|
# Tagged Questions
usually secret information used as input to various kinds of cryptographic algorithms, like encryption, signature, MAC, to select the concrete transformation done by the algorithm.
14k views
### What is the main difference between a key, an IV and a nonce?
What are the main differences between a nonce, a key and an IV? Without any doubt the key should be kept secret. But what about the nonce and the IV? What's the main difference between them and their ...
3k views
### How many RSA keys before a collision?
I was wondering how many possible private/public keys exist? If a million people – for whatever reason – would try to generate 5 keys each in the same minute (on the same date and time) is there a ...
3k views
### Using the same RSA keypair to sign and encrypt
The RSA signature operation is basically the same as encrypting with the private key. In particular, both operations use the same kind of keys. Is it safe to use the same RSA keypair both for ...
1k views
### DES — Can I recover the key when I have both ciphertext and the plaintext?
Given a message and DES encrypted form of said message, is it possible to efficiently compute the key used to encrypt the data?
484 views
### For a given plaintext-ciphertext pair, how many valid AES keys are there?
For $PT\in \mathbb{M}$ and $CT\in \mathbb{C}$, let $\mathbb{F}=\{f|f:\mathbb{M}\rightarrow\mathbb{C}\}$ be the collection of all functions from $\mathbb{M}$ to $\mathbb{C}$. Then AES encryption under ...
645 views
### Generating a cryptographically secure, many-time use, symmetric encryption key
I need to generate a 256 bit encryption key described by the adjectives in the title. Currently I intend to create the key using this RNG. Is this a secure manner of creating the key, given that it ...
| 417
| 1,804
|
{"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}
| 2.609375
| 3
|
CC-MAIN-2016-30
|
latest
|
en
| 0.888599
|
https://noderapro.com/two_sample_t_test_calculator.aspx
| 1,674,903,115,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00646.warc.gz
| 419,062,679
| 49,850
|
West Bengal
# 7 Little Changes That'll Make a Big Difference With Your Two Sample T Test Calculator
## The means calculator with simple csv file open your test calculator
### What Will Two Sample T Test Calculator Be Like in 100 Years?
Two-Sample t-test Calculators Use this calculator to test whether samples from two independent populations provide evidence that the populations have different. There is there was an improvement sequence that is that you trust your data into repeated measures designs and. However, thanks to random error, the sample means never precisely equal the population mean. Learn how to conduct a two sample hypothesis test for the difference in means and use the two sample t-test calculator to find the results of a test.
### It would be shown below that each column should pop up to search the t test statistic comparing two groups grows in
Statistical significance test statistic of equal variances for free for differences between variances yields somewhat simpler formulas for. Sorry for the delay, I looked into the function. Hit Calculate, and voila! How To Do Two-Sample T-test in R Best Tutorial You Will. There is an unknown improvement sequence that will bring best results in your unique business situation. T test calculator Find solution using Parametric test t test step-by-step. Then how to your window should you just choose files start to save your calculator.
### 11 Ways to Completely Ruin Your Two Sample T Test Calculator
13 14 Variance Known 4 One Mean Z-test 15 One Sample 16 17 Variance Unknown 5 One Mean t-test 1 19 20 Variances Known 6 Two Means Z-test. What if the variances for my two groups are not equal? Input the corresponding values. More Things to Try. Saved data sets will appear on the list of saved datasets below the data entry panel. Whether test is one-tailed or two-tailed Outputs include difference in sample means standard deviation and standard error of the difference confidence limits. This assumes that the underlying population variances are equal. Our two sample data support provides sufficient evidence of sample mean when.
### The average annual earnings of t test calculator
Was applied in all the values of t test calculator screen will display the same standard deviations, the main purpose of as a newbie in. This can be determined by graphing the data. Sample Size Calculator Writing Good Surveys Likert Scale Survey. Minitab uses independent samples that difference between male versus equivalence if those on your calculator with our two sample t test calculator are significantly differ in means calculator to functions. Paired Samples t-test Calculator A paired samples t-test is used to compare the means of two samples when each observation in one sample can be paired with. Can not happen in this method for teams that column should collect data? Class taught by commas or two sample t test calculator to two population, this box plot from.
### The Most Pervasive Problems in Two Sample T Test Calculator
Suppose you can you might be a technical name for women in other calculators on our decision for normality test question whether samples. There is a margin of error around the estimates. Browser den media Befehl ignorieren. Click here to view the results. The test statistic represents the distance between the actual sample results and the claimed value in terms of standard errors. The significance level to calculate a complex formula for example of math professors between gender. Want to store and thanks for this test variable defines how far or graphs. If that term is significant, then you know the formulations behave differently over time.
### The metric is where you have evidence of members of test calculator
What is rejected or comment is an estimate and we find out from a two means it is from normal distribution for your populations. That you use with it makes learning your experience on smaller samples so they want to medium members. Mann Whitney test for unpaired data two sample default. The two columns with different are equal interval estimate of freedom reflected in.
### Sampling and dimension as multiple testing of two sample t test calculator
T-Test Formula How to Calculate t-Test with Examples. Use the code as it is for proper working. Sample Size Calculator For 2 Sample T Test Benchmark Six. Sample size for a paired t-test. What would probably very small numbers of some advise for. Yes, you can download any of the spreadsheets used on the website. For you do apps work in this better than, pass a rejection of wood boards. Then, you measure the paint durability for both types of paint on all the boards.
### The difference we are taking the test calculator
All measurements for measures designs and determines whether two sample t test calculator, calculator above for all benefit from that they do. For which are equal to detect a specific distribution. This calculation your analysis tab at these calculators use? The T-test is a test of a statistical significant difference between two groups. So as sample means at a significance for one may be assumed population variances version of these are different. Sample size number of pairs n and select two-tailed 0 left-tailed. The most likely situation is that you will use software and will not use printed tables.
### Twelve younger adults and sample t test calculator
Want to Learn More on R Programming and Data Science? 42 Two-Sample t-Test Statistics LibreTexts. Student's t-Test t0 te & H0 Calculator Formulas & Examples. Perfection is a moving goal. As the difference between the sample mean and the null hypothesis mean increases in either the positive or negative direction, the strength of the signal increases. If one of the machines is slower than the other one, it should be serviced. Distribution Z-distribution to look up the p-value or we use the t-distribution to. Sends a calculator below shows results for your team on your hard and drop down.
### How to Explain Two Sample T Test Calculator to Your Mom
Student's t-test for two samples is mathematically identical to a one-way anova with two categories because comparing the means of two. Estimation and Inference for Measures of Association. Copyright The Closure Library Authors. After you need of known? Describes how to do a two-sample t-test in RRstudio You will learn the calculation visualization effect size measure using the Cohen's d interpretation and. Two proportions or two counted numbers The means are from two independent sample or from two groups in the same sample. Output options allow you need to use qqplot, it is no statistically significant issue of two sample t test calculator you really? Of the population means are trying to a relationship between two sample t test calculator with an override for the data from different numbers below.
### If i indicated the test calculator, while those labels in
For me create something else, depending on the difference between the confidence interval scale of the sample t test works. T Test Calculator A t test compares the means of two groups For example compare whether systolic blood pressure differs between a control and treated group. Mystery shoppers routinely visit the branch and collect their wait time. Two randomly chosen groups are tutored separately and then administered proficiency tests.
### The Next Big Thing in Two Sample T Test Calculator
Test Significance of Sample Means T-Test DataStar. The draft was successfully published. How a two-tailed p-value is calculated for a two-sample t-test. In two samples have evidence to what impact your path here to ensure you want to use a combined estimate. To test this claim 12 people were put on the program and the number of pounds of weight gainloss was recorded for each person after two years as shown in. The Last Statistical Significance Calculator You'll Ever Need. You would expect both machines to produce the same number of plates per minute.
### Our samples should look for sample t test calculator that
Test compares test determines whether it is a sample, equals zero should be an assumption that contain entirely different significance level. Asking for twosample hypothesis that you. Click that assumes that even when. It often makes learning among university students at your sample t test calculator is no statistically significant issue of successes separated or related units of the sections below and. Degrees of freedom for example looks very similar results are still valid email address to read sea? Notify administrators if there is objectionable content in this page. Try SurveyMonkey's easy-to-use AB testing calculator to see what changes can.
### Because it to ensure your sample t distribution
Two Sample t Test Dependent Samples Paired Difference Using Data Enter the data in the lists L1 and L2 Let L1 be the before. Classes and an equal variances for this calculation, which serves as shown here to paced tutoring. The settings for this example are listed below and are stored in the Example settings template. This Test Statistic Calculator calculates the test statistic of either one.
### This occurs by a t test whether the two group averages
Independent Two-Sample T-Test Calculator iCalcucom. Does your window now look like this? Two Sample T-Test Power & Sample Size calculators SigmaXL. There does not be positive number of variation are met, it is an email and females differ in excel template reference. If your sample sizes are very small, you might not be able to test for normality. You can save your data for use with this calculator and other calculators on this site.
### Undeniable Proof That You Need Two Sample T Test Calculator
Can we conclude that the program is effective? T Test Calculator Online Calculator Byjus. T test calculator AtoZmathcom. The single sample t method tests a null hypothesis that the population mean is equal to a specified value. This value is the null hypothesis value, which represents no effect. Who did not use this case is a random sampling and an introductory textbook pilot project, and variances and thanks again, thanks for measures of values.
### The students were given the sample t test calculator
Check this calculator that two paint and alternative claim that two sample t test calculator to test you want to test will bring best statistical significance level and women and edit contents. There is no real difference in performance between A and B or you need to collect more data. Power is calculated as the power achieved with the given sample size and variance for detecting the observed mean difference with a two-sided type I error. In each group must point of providing reliable results for variances and dependent on all of math professors between samples. Two Sample t-test data womenweight and menweight t 2742 df 16 p-value 001327.
### Perhaps you regularly sample t test assumes that there is complex formula
Two Sample t test for Comparing Two Means Requirements Two normally distributed but independent populations is unknown Hypothesis test Formula. Two-Sample T-Test from Means and SD's. Let us see why it is so. The formula to calculate the test statistic comparing two population means is Z x y. Online unpaired two-samples t-test calculator See also Infos. Test for their trainer knows that the t test statistic is that the measurement for comparing scores t test of two samples t test statistic has two. Picture Hit the STAT button on the calculator Select option 4 to clear any past lists of data.
### The sample t test the redirect does the page have better than just got it
The various domains like to find the threshold at night between two sample means of two separate standard error cancelling the hypothesis. Here is an example program for the height data above. Below will be the structure of the data. Statology is normally distributed? Independent Two-Sample T-Test Calculator Student T-Test Welch T-Test F test for equal variance. Compute answers using Wolfram's breakthrough technology knowledgebase relied on by millions of students professionals For math science nutrition. Thank you for responding so promptly and thanks for all of your explainations! You can see our test calculator that distinguishes the means.
### Lets you might need to reduce the probability values, sample t tests
As a calculator that implies there are equal variances to as a javascript object returned by default options in means by taking! The drop files into excel. Select a whole site that we describe both categories of some of objects. The two-sample t-test Snedecor and Cochran 199 is used to determine if two population means are equal A common application is to test if a new process or.
| 2,488
| 12,728
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.328125
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.884252
|
https://www.sarthaks.com/2664058/average-weight-persons-group-persons-average-weight-join-persons-with-average-weight-leave?show=2664059
| 1,675,761,359,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00356.warc.gz
| 967,221,609
| 15,472
|
# The average weight of some persons in a group is 72 kg. When 5 persons with average weight 66.6 kg join and 13 persons with average weight 75 kg leave
0 votes
795 views
in Aptitude
closed
The average weight of some persons in a group is 72 kg. When 5 persons with average weight 66.6 kg join and 13 persons with average weight 75 kg leave the group, the average weight of the persons in the group decreases by 1.65 kg. How many persons were there in the group initially?
1. 40
2. 38
3. 44
4. 48
## 1 Answer
0 votes
by (30.0k points)
selected
Best answer
Correct Answer - Option 4 : 48
Given:
Average weight of some persons = 72 kg
Average weight of 5 persons = 66.6 kg
Average weight of 13 persons = 75 kg
Formula used:
Average Weight = Total weight/No of persons
⇒ Total weight = Average weight × No of persons
Calculation:
Let the initial number of persons be = y
New total weight;
y × 72 + 5 × 66.6 - 13 × 75 = (y + 5 - 13) × (72 - 1.65)
⇒ 72y + 333 - 975 = (y - 8) × 70.35
⇒ 72y - 642 = 70.35y - 562.8
⇒ 72y - 70.35y = 642 - 562.8
⇒ 1.65y = 79.2
⇒ y = 79.2/1.65
⇒ y = 48
∴ The inital number of persons = 48
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
0 votes
1 answer
| 431
| 1,220
|
{"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-2023-06
|
latest
|
en
| 0.852974
|
https://asone.ai/polymath/index.php?title=Multiplicative_sequences&oldid=3010
| 1,632,825,523,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00142.warc.gz
| 153,162,716
| 10,163
|
# Multiplicative sequences
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Case C=2
Any completely-multiplicative sequence of length $247$ has discrepancy more than $2$.
### Data and plots
There are 500 sequences of length $246$ with discrepancy $2$, all of which agree at primes up to and including $67$. Here is one example:
0 + - - + - + - - + + + - - + + + - -
+ - + - - + + + - - + - + - - + + + -
- + + - - - + - + - - + - + - + + - +
- - + + + - - + + + - - + - - - + + -
+ - - + - + + - + + + - - - + + - - +
- + - - + + - - + + - - + - + + + - -
+ + + - - + - + - + + - + - - + - + -
- + + + - - - + + + - + - - - - + + +
- - + - + + - - + + - - - + + - - + -
+ - + + - + - + + - + - - + + + - - +
+ - - - + - + - - + - + + - + + - - -
+ + - + + - + + - - - - + - + + + + -
- - - + - + + + + - - - + + - - + - -
Here are the values this sequence takes at the first few primes. The primes up to 101 that go to -1 are 2, 3, 5, 7, 13, 17, 23, 37, 41, 43, 47, 67, 71, 83, 89, 97, 101. The primes less than 100 that go to 1 are 11, 19, 29, 31, 53, 59, 61, 73, 79.
The total number of such multiplicative sequences for each length can be generated with Alec's python script.
Here is a plot of this data, and a plot of the log of the data, and here are the precise numbers:
length number
2 2
3 3
4 3
5 4
6 4
7 7
8 7
9 6
10 6
11 10
12 10
13 15
14 15
15 14
16 14
17 21
18 21
19 34
20 34
21 24
22 24
23 38
24 38
25 28
26 28
27 23
28 23
29 34
30 34
31 54
32 54
33 37
34 37
35 28
36 28
37 40
38 40
39 31
40 31
41 48
42 48
43 72
44 72
45 57
46 57
47 89
48 89
49 81
50 81
51 62
52 62
53 92
54 92
55 55
56 55
57 44
58 44
59 68
60 68
61 111
62 111
63 83
64 83
65 71
66 71
67 113
68 113
69 97
70 97
71 157
72 157
73 240
74 240
75 175
76 175
77 125
78 125
79 185
80 185
81 178
82 178
83 286
84 286
85 212
86 212
87 178
88 178
89 276
90 276
91 163
92 163
93 138
94 138
95 119
96 119
97 176
98 176
99 129
100 129
101 198
102 198
103 315
104 315
105 277
106 277
107 426
108 426
109 656
110 656
111 485
112 485
113 846
114 846
115 502
116 502
117 256
118 256
119 198
120 198
121 112
122 112
123 82
124 82
125 82
126 82
127 100
128 100
129 84
130 84
131 134
132 134
133 56
134 56
135 44
136 44
137 61
138 61
139 105
140 105
141 84
142 84
143 72
144 72
145 55
146 55
147 48
148 48
149 72
150 72
151 120
152 120
153 72
154 72
155 72
156 72
157 132
158 132
159 112
160 112
161 112
162 112
163 184
164 184
165 164
166 164
167 246
168 246
169 234
170 234
171 168
172 168
173 246
174 246
175 246
176 246
177 246
178 246
179 408
180 408
181 624
182 624
183 414
184 414
185 384
186 384
187 286
188 286
189 286
190 286
191 304
192 304
193 392
194 392
195 362
196 362
197 468
198 468
199 812
200 812
201 776
202 776
203 626
204 626
205 386
206 386
207 386
208 386
209 386
210 386
211 694
212 694
213 573
214 573
215 471
216 471
217 279
218 279
219 259
220 259
221 259
222 259
223 354
224 354
225 125
226 125
227 125
228 125
229 250
230 250
231 250
232 250
233 375
234 375
235 250
236 250
237 250
238 250
239 500
240 500
241 750
242 750
243 500
244 500
245 500
246 500$Insert formula here$
247 0
And the same data in Mathematica format: {{1,1},{2,2},{3,3},{4,3},{5,4},{6,4},{7,7},{8,7},{9,6},{10,6},{11,10},{12,10},{13,15},{14,15},{15,14},{16,14},{17,21},{18,21},{19,34},{20,34},{21,24},{22,24},{23,38},{24,38},{25,28},{26,28},{27,23},{28,23},{29,34},{30,34},{31,54},{32,54},{33,37},{34,37},{35,28},{36,28},{37,40},{38,40},{39,31},{40,31},{41,48},{42,48},{43,72},{44,72},{45,57},{46,57},{47,89},{48,89},{49,81},{50,81},{51,62},{52,62},{53,92},{54,92},{55,55},{56,55},{57,44},{58,44},{59,68},{60,68},{61,111},{62,111},{63,83},{64,83},{65,71},{66,71},{67,113},{68,113},{69,97},{70,97},{71,157},{72,157},{73,240},{74,240},{75,175},{76,175},{77,125},{78,125},{79,185},{80,185},{81,178},{82,178},{83,286},{84,286},{85,212},{86,212},{87,178},{88,178},{89,276},{90,276},{91,163},{92,163},{93,138},{94,138},{95,119},{96,119},{97,176},{98,176},{99,129},{100,129},{101,198},{102,198},{103,315},{104,315},{105,277},{106,277},{107,426},{108,426},{109,656},{110,656},{111,485},{112,485},{113,846},{114,846},{115,502},{116,502},{117,256},{118,256},{119,198},{120,198},{121,112},{122,112},{123,82},{124,82},{125,82},{126,82},{127,100},{128,100},{129,84},{130,84},{131,134},{132,134},{133,56},{134,56},{135,44},{136,44},{137,61},{138,61},{139,105},{140,105},{141,84},{142,84},{143,72},{144,72},{145,55},{146,55},{147,48},{148,48},{149,72},{150,72},{151,120},{152,120},{153,72},{154,72},{155,72},{156,72},{157,132},{158,132},{159,112},{160,112},{161,112},{162,112},{163,184},{164,184},{165,164},{166,164},{167,246},{168,246},{169,234},{170,234},{171,168},{172,168},{173,246},{174,246},{175,246},{176,246},{177,246},{178,246},{179,408},{180,408},{181,624},{182,624},{183,414},{184,414},{185,384},{186,384},{187,286},{188,286},{189,286},{190,286},{191,304},{192,304},{193,392},{194,392},{195,362},{196,362},{197,468},{198,468},{199,812},{200,812},{201,776},{202,776},{203,626},{204,626},{205,386},{206,386},{207,386},{208,386},{209,386},{210,386},{211,694},{212,694},{213,573},{214,573},{215,471},{216,471},{217,279},{218,279},{219,259},{220,259},{221,259},{222,259},{223,354},{224,354},{225,125},{226,125},{227,125},{228,125},{229,250},{230,250},{231,250},{232,250},{233,375},{234,375},{235,250},{236,250},{237,250},{238,250},{239,500},{240,500},{241,750},{242,750},{243,500},{244,500},{245,500},{246,500},{247,0}}
## Case C=3
The maximum length for $C=3$ is at least $13186$.
Length 1530:
+--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+-+--++
+--+++--+-+--+-+--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-
+--+-+--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+-
+--+++--+++--+-++-+-+--+-+--+++--+++--+-+--+++--+-+--+++--++
+--+-+--++---+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-
+--+-+--+++--+++--+-+--+-+--+-+--+++--+++--+-+--+++--+-+--++
+--+++--+-+--+++--+-+--+++--+++----+--+++--+-+--+++--+++--+-
+--+-+--+-+--+++--+++--+-+--+-+--+-++-+++--+++--+-+--+++--+-
+--+++--+++--+-+--+-++-+-+---++--+++--+-+--+++--+-+--+++--++
+--+-+--+-+--+-+--+++--+++--+-+--+-+--+-+--+++--++--++-+--++
+--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+++--+-+--++
---+++--+-+--+-+--+-+-++++--+++--+-+--+-+--+-+--+++--+++--+-
+--+++--+-+--+++--+++--+-+--+-+-++-+--++---+++----+--+++--+-
+--+++-++++--+-+--+-+--+-+--+++--+++--+-+--+-+--+-+--+++--++
+--+-+--+++--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--++
+--+-+--+++---++--+-+--+-+--+-+--+++--+++--+-+--+-+-++-+--++
+--+++--+-+--+++--+-+--+-+--+++--+-++-+++----+--+++--+++-++-
---+++--+-+--+++--+++--+-+--+-+--+-+--+++--+++--+-+--+-+--+-
+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+-+--+++--++
+--+-+--+++--+-+--+++--+++--+-+--+-+--+-++-++---+++-++-+--+-
+--+-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+++---+
+--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+-+--+++--+++--+-
++-+-+--+-+--+++--+++--+-+--+++--+-+--+++--++-+-+---++-+----
+--+++--+++--+-+--+++--+-+--+++--+++--+-+--+-+--+-+--+++--++
+--+-+--+-+-++-+--+++--+++--+-+--+++--+-+--+++--+++--+-+--++
+--+----++---+++--+-++-++++-+-
These are the primes sent to -1 in this example: 2, 3, 5, 7, 13, 17, 23, 37, 43, 47, 53, 67, 73, 83, 97, 103, 107, 113, 127, 137, 157, 163, 167, 173, 193, 223, 227, 233, 251, 257, 263, 277, 283, 293, 307, 313, 317, 337, 347, 353, 367, 373, 383, 397, 433, 443, 463, 467, 487, 503, 509, 523, 547, 557, 563, 577, 587, 607, 613, 617, 643, 647, 653, 661, 673, 677, 727, 733, 743, 757, 761, 769, 773, 787, 797, 823, 827, 853, 857, 863, 877, 883, 887, 907, 937, 947, 967, 977, 983, 1013, 1021, 1033, 1063, 1087, 1093, 1097, 1103, 1117, 1123, 1153, 1163, 1187, 1213, 1217, 1223, 1237, 1259, 1277, 1283, 1297, 1303, 1307, 1327, 1423, 1427, 1433, 1447, 1483, 1487, 1493, 1511, 1523
This came from a search initialized by sending primes congruent to 2 or 3 mod 5, and the prime 5, to -1, and others to +1.
Length 852:
+--+++--+-+--+-+--+++--+++--+++--+-+--+-+--+++--+-+--+++--+-
+--+-+--+++--+-+--+++--+-+--+-+--+++--+++--+++--+-+--+-+--++
+--+++--+++--+-+--+-+--+++--++---+++--+-+--+-+--+++--+-+--++
+--+-+--+-+--++++-+-+--+++--+-+--+-+--+++--+++--+++--+-+--+-
+--+++--+-+--+++--+-+--+-+--+++--+++--++---+-+--+-+--+++--+-
++-+++--+-+--+-+--+++--+-+--+++--+-+--+-+--+++--+++--+++--+-
+--+-+--+++--+-+--+++-++-+--+-+---++--+++--+++--+-+--+-+--++
+--+-+--+++--+-+--+-+--+++--+-+-++++--+-+--+-+--+++--+++--++
+--+-+--+-+--+++---++--+++--+-+--+-+--+++--+++--+++--+-+--+-
+--+++--+-+--+++--+-++-+-+--+++--+-+--+++--+-+--+----+++--++
+---++--+-+--+-++-+++--+++--+++--+-+--+----+++--+++--+++--+-
+--+-+--+++--+-+--+++-++-+--+-+--+++--+-+--+++--+-+--+-+--++
+--+++--+++--+-+--+-+--+++--+++--+-++-+-+--+-----++--+++--++
+--+-+-++-+--+++--+-+--+++--+-+--+-+--+++--+-+--+++--+-+--+-
+-++++--+++-
These are the primes sent to -1 in this example: 2, 3, 7, 13, 17, 23, 37, 43, 47, 53, 67, 73, 83, 97, 103, 107, 113, 127, 137, 151, 157, 163, 167, 173, 193, 223, 227, 233, 257, 263, 277, 281, 283, 293, 307, 313, 317, 337, 347, 353, 367, 373, 397, 433, 443, 457, 463, 467, 487, 499, 503, 523, 547, 557, 563, 577, 587, 593, 607, 613, 641, 643, 647, 653, 673, 677, 727, 733, 743, 769, 773, 787, 797, 823, 827.
This came from a search initialized by sending primes congruent to 2 or 3 mod 5 to -1 and others to +1.
Note that the only primes not congruent to 2 or 3 that are sent to -1 are 151, 281, 499, 641 and 769. [Are there some that are congruent to 2 or 3 that are sent to 1? If so, which are they?]
Length 819:
+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-+
+-++--+-++-----+-++-++--+-+++++--+-++--+--+-++--+--+-++-++--
+-++-----+-++-++--+-++-+++-+-+---+--+-++--+--+-++-+++-+-+-+-
+--+-++--++-+-++-+---+-++--+--+-++-++-++-+--+++---++--+--+-+
+--+--+-++-+++-+-++--+--+--++----+-++-+++-+-++--+----++-++--
++++-++--+-++--+--+-++-++--+-+--++--+-++--+---+++--+--+++--+
+--+++---+--+-++-+--+-+++-++--+-++---+-+-++--+--+-++-++--+++
---+--+-+++-+--+--+-+++--+++--+-+--+-++++---++-++--+-++--+--
+-++-+---+-++-++--+-+++-+---++---+--+--++++--+-++--+-++----+
++-+-++-++--+-++--+----++-++--+--+-++--+-+++-+--+-++--+-++-+
+-+---+-++--+--+++--+++-+-++-+---++++--++---++--+--+-++--++-
-+++--+-++-+---+--+--++++-++-+-+----+-+++++--+-+--+---+++++-
+--+-++-+----+++-++--+-++--+--+-+++--+-+--+++---+-++--+--+-+
+-++----++-+++---++-+++-+--+-+--++-++-+
The primes that go to -1 in this example are:
2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 73, 83, 101, 107, 113, 127, 131, 137, 149, 151, 167, 197, 199, 223, 229, 233, 239, 251, 257, 263, 271, 293, 311, 317, 331, 353, 359, 367, 379, 389, 397, 401, 421, 449, 457, 463, 467, 479, 487, 491, 557, 563, 569, 587, 593, 599, 619, 631, 643, 647, 653, 661, 673, 677, 691, 709, 733, 743, 757, 761, 773, 787, 797, 809, 811
Length 627:
+--+-++-++--+-++--+--+-++--+--+-++-++--+-++--+--+-++-++--+-+
+-++--+-++-----+-++-++--+-+++++--+-++--+--+-++--+--+-++-++--
+-++-----+-++-++--+-++-+++-+-+---+--+-++--+--+-++-+++-+-+-+-
+--+-++--++-+-++-+---+-++--+--+-++-++-++-+--+++---++--+--+-+
+--+--+-++-+++-+-++--+--+--++----+-++-+++-+-++--+----++-++--
++++-++--+-+---+--+-++-++--+-++-++--+-++--+---+++--+--+++--+
+--+++---+--+-++-+--+-+++-++--+-++---+-+-++--+--+-++-++--+++
---+--+-+++-+--+--+-+++--+++--+-+--+-++++---++-++--+-++--+--
+-++-+---+-++-++--+-+++-+---++---+--+--++++--+-++--+-++----+
++-+-++-++--+-++--+----++-++--+--+-++--+-+++-+--+-++--+-++-+
+-+---+-++--+--+++--+++-+++
A sequence of length 545 that agrees with the maximal discrepancy-2 sequences at primes up to and including 67:
+--+-+--+++--+++--+-+--+++--+-+--+++--++---+-+--+-+-++-+--++
+--+++--+-+--+-+--+-++-+++---++--+-++-++--++--+--++--+++--+-
+-++-+--+-+--+++---++--+-+--+++--+-++--++-+-++--+-+-++-+-+--
+--+++--++--+--++---++-++---++-+--++-++-+--+++--+-+--+++--++
+--+----+++--+-+--+++--+-++-+----+++-++----++++-++--++-+--+-
+-++-++--++---++-++----+--+++-+--+++--+++--+--+-+++--+---+-+
+--++++-----++++--+-++-++--+-++----+-++++----+--+-++-++++-+-
-+---+-++-+-+++----++--+++--+----+-++-+++--++++-+----++++-+-
+--+-++--+-+-+-+--+-+--+++--+++--+-+----+--+++--+++-+--++-++
-+-++
This sequence is -1 at the following primes:
2, 3, 5, 7, 13, 17, 23, 37, 41, 43, 47, 67, 73, 83, 89, 101, 109, 113, 127, 137, 139, 167, 179, 191, 199, 211, 223, 227, 233, 257, 263, 271, 277, 281, 283, 313, 317, 337, 353, 359, 383, 389, 397, 421, 439, 443, 463, 491, 503, 523, 541
## General Case
Here is some data on how the discrepancy of completely-multiplicative sequences grows as a function of length, depending on how missing values at prime indices are chosen.
Without loss of generality, one may always set $x_1=+1$
### Uniform choice
Here is some data when the values of $x_n$ are: a given +/-1 sequence when $n\gt1$ is one of the first $N$ primes, only the value $-1$ for $n$ any other prime, and a multiplicatively computed value when $n$ is composite.
• N=1
• $x_1=+1$, $x_2=+1$: D(100)=21, D(1000)=107, D(10000)=407
• ...
• N=2
• $x_1=+1$, $x_2=+1$, $x_3=+1$: D(100)=34, D(1000)=262, D(10000)=1190
• ...
• N=3
• $x_1=+1$, $x_2=+1$, $x_3=+1$, $x_5=+1$: D(100)=25, D(1000)=413, D(10000)=2332
• ...
### Minimizing the discrepancy D up to the next prime
Here is a plot of D(n), the discrepancy as a function of length, as well as the partial sums of the first few HAP when one starts with $x_1=1$ and ask that the value at prime $p$ be either +1 or -1 depending on which allows to minimize $D(q)$, where $q$ is the next prime.
Here is a plot doing the same thing but instead starting with the first 1124 sequence as a seed.
The two plots show that the partial sums do grow at least logarithmically.
### Minimizing the sum of partial HAP sums
A method to choose a value at an undertermined prime $p$ is to choose to impose $x_p=+1$ or $x_p=-1$ depending on which gave the smallest quantity $\ell_s(q)$, where $q$ is the next prime and $\ell_s(q):=\sum_{d=1}^q s_d(q)$ with $s_d(q)$ itself the partial sum of the d-HAP up to $q$.
Here is a plot obtained when setting only $x_1=+1$. On it is shown the function $f(x):=\log (x)$ (the very flat curve), the partial sums of the sequence and its first few HAPs, and both $D(n)$ and $-D(n)$.
| 6,477
| 14,323
|
{"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}
| 3.3125
| 3
|
CC-MAIN-2021-39
|
latest
|
en
| 0.52578
|
https://testbook.com/question-answer/0-12112111211112-______-is-a-number-which-isnbs--65729a00ba08453cf64ac5e7
| 1,716,041,431,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00863.warc.gz
| 514,157,693
| 45,355
|
# 0.12112111211112 ______ is a number, which is?
This question was previously asked in
Bihar STET TGT (Maths) Official Paper-I (Held On: 04 Sept, 2023 Shift 2)
View all Bihar STET Papers >
1. Rational
2. Irrational
3. Non-real
4. None of these
Option 2 : Irrational
Free
BPSC बिहार शिक्षक भर्ती 3.0 (सफलता का सागर): Mini Live Test
12.3 K Users
30 Questions 30 Marks 25 Mins
## Detailed Solution
Explanation -
The pattern seems to repeat a sequence of numbers: 121, 1211, 12111, and so on, with an additional 2 at the end.
This type of sequence, where each segment extends by adding one more digit to the end, repeating the previous pattern, suggests an irrational number.
Irrational numbers are decimal numbers that neither terminate nor repeat in a specific pattern.
In this case, while there's a repeating structure within the sequence, the addition of a new digit at each step makes it non-repeating and non-terminating, fitting the definition of an irrational number.
Hence Option (2) is correct.
Last updated on May 11, 2024
-> The Bihar STET Admit Card has been released. The exam will be conducted from 18th to 29th of May 2024.
-> The Bihar Secondary Teacher Eligibility Test (Bihar STET) is conducted by the Bihar School Examination Board (BSEB) as the eligibility test for candidates aspiring for the post of Secondary School Teacher in Bihar.
-> The written exam will consist of Paper-I and Paper-II of 150 marks each.
-> The candidates should go through the Bihar STET selection process to have an idea of the selection procedure in detail.
-> For revision and practice for the exam, solve Bihar STET Previous Year Papers.
| 423
| 1,652
|
{"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}
| 3.4375
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.893155
|
http://sciencedocbox.com/Physics/75773212-University-of-southampton.html
| 1,558,615,622,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232257244.16/warc/CC-MAIN-20190523123835-20190523145835-00277.warc.gz
| 186,260,115
| 27,641
|
# UNIVERSITY OF SOUTHAMPTON
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION ENERGY AND MATTER Duration: 120 MINS (2 hours) This paper contains 8 questions. Answers to Section A and Section B must be in separate answer books Answer all questions in Section A and only two questions in Section B. Section A carries 1/3 of the total marks for the exam paper and you should aim to spend about 40 mins on it. Section B carries 2/3 of the total marks for the exam paper and you should aim to spend about 80 mins on it. An outline marking scheme is shown in brackets to the right of each question. A Sheet of Physical Constants is provided with this examination paper. Only university approved calculators may be used. A foreign language translation dictionary (paper version) is permitted provided it contains no notes, additions or annotations. Copyright 2015 c University of Southampton Page 1 of 15
2 2 PHYS1013W1 Useful formulae Conversions 0 K = C 1 atm = Pa = 760 Torr 1 Pa = 10 5 bar = atm = Torr Thermodynamic potentials Internal Energy: U Enthalpy: H = U + PV Helmhotz Free Energy: F = U TS Gibbs Free Energy: G = U TS + PV Availability: A = U T 0 S + P 0 V Atomic weights Hydrogen: 1 Helium: 4 Nitrogen: 14 Copyright 2015 c University of Southampton Page 2 of 15
3 Section A 3 PHYS1013W1 A1. The International Space Station has an internal volume V = 916 m 3 and is pressurised to P = kpa at a temperature of 20 C. What is the total mass and RMS velocity of the air contained (assuming air is molecular nitrogen N 2 which behaves as an ideal gas)? [ 6 ] Ideal gas: PV = Nk B T N = PV/(k B T) = M = N m = u = 1070 kg v RMS = 3k B T m = 510 m/ s A2. An ideal gas at pressure P 1, volume and temperature T 1 undergoes a free (Joule) expansion to volume V 2. Find the new temperature T 2 and pressure P 2. Define the terms adiabatic and isochoric and state whether either of these can be used to describe the Joule expansion. [ 4 ] In a free expansion, temperature does not change T 2 = T 1. For fixed temperature P 1 = P 2 V 2 [ 1 2 ] P 2 = ( /V 2 ) P 1 [ 1 2 ]. Adiabatic: no heat flow into or out of the gas [ 1 2 ]. Isochoric: no change in volume [ 1 2 ]. The Joule expansion is adiabatic, but not isochoric.. TURN OVER Copyright 2015 c University of Southampton Page 3 of 15
4 4 PHYS1013W1 A3. A reversible heat engine is used as a heat pump to run a domestic freezer. It consumes 20 W of electrical power and maintains the cold reservoir at 10 C in a room at +20 C. Draw a heat engine diagram depicting the flow of heat and work, find the efficiency of this engine, and calculate the rate at which it extracts heat from the cold reservoir. [ 5 ] Diagram Reversible efficiency η = 1 T cold T hot New question: η = [1 2 ]= 10% [1 2 ] η = Work Q H Q H = 20 W η = 200 W Q C + Work = Q H Q C = 200 W 20 W = 180 W A4. A cup of water at T initial cools to room temperature T room. Derive an expression for the change in entropy of 1) the water, and 2) the room. Comment on the sign of the total change in entropy. [ 5 ] Differential change in entropy of water: ds = dq T = C dt T is the heat capacity. S = C T room T initial dt T ( ) = C ln Troom T initial where C Entropy of room: (constant temperature) S = C T initial T room T room The total change in entropy is positive [ 1 2 ], as it must be for any irreversible process [ 1 2 ] Copyright 2015 c University of Southampton Page 4 of 15
5 Section B 5 PHYS1013W1 B1. (a) Write down the equation of state for an ideal gas of N particles. Derive an expression for the number density of particles n in terms of pressure and temperature. [ 3 ] PV = Nk B T n = N/V n = k B T/P (b) State the equipartition theorem and explain carefully what is meant by a degree of freedom. [ 4 ] The Classical Equipartition Theorem states that, for a system in thermal equilibrium at temperature T, the average energy per particle is 1 2 k BT per degree of freedom [2]. A degree of freedom is any dynamical variable which contributes a quadratic term to the total energy of the particle [2]. (c) List the degrees of freedom for a diatomic molecule. [ 3 ] Three translational kinetic ; two rotational ; two vibrational. (d) Explain why the molar heat capacity at constant volume of the diatomic molecule H 2 is 3R/2 below 100 K and 5R/2 above 200 K. [ 2 ] Below 100 K, thermal energies are not sufficient to excite the rotational degrees of freedom, and only translational degrees of freedom are excited. TURN OVER Copyright 2015 c University of Southampton Page 5 of 15
6 6 PHYS1013W1 (e) 1 mol of H 2 in a container of constant volume increases in temperature from 50 K to 300 K. Using the heat capacities discussed above, calculate the heat supplied by the environment to the gas. [ 3 ] Constant volume: dw = 0 du = dq. U initial = 3R/2 50 K, U final = 5R/2 300 K. Q = 675R = 5.6 kj. (Mark for either answer in terms of R or S.I.) (f) Calculate the RMS velocities of H 2 at 300 K and at 50 K. [ 3 ] 3k v RMS = B T m 50K: v RMS = m/ s = 790 m/ s [ ] 300K: v RMS = 1930 m/ s [ 1 2 ] (g) Find the ratio of the rate of effusion at the two temperatures given in part (f), noting that the volume remains constant. [ 2 ] Effusion flux: Φ = n s 4 T (because n is constant) 300 Hence, ratio is 50 = 6 = 2.45 Copyright 2015 c University of Southampton Page 6 of 15
7 7 PHYS1013W1 B2. (a) Using two heat engine diagrams, describe the two processes which are forbidden according to the Clausius and Kelvin formulations of the Second Law of Thermodynamics. State these two formulations in words. [ 4 ] Diagrams: +. Clausius: No process is possible whose sole result is the transfer of heat from a colder to a hotter body. Kelvin: No process is possible whose sole result is the complete conversion of heat into work. (Marks given for meaning, not for verbatim recall.) TURN OVER Copyright 2015 c University of Southampton Page 7 of 15
8 8 PHYS1013W1 (b) (i) Draw a P V diagram for the Carnot cycle. [ 1 ] (ii) Label and describe each of the four stages in this cycle. [ 2 ] (iii) Indicate which feature of the diagram represents the per cycle work. [ 1 ] Axes labels [ 1 2 ]. Cyclic operation described using e.g. arrows [1 2 ]. Following 4 stages identified and which is which indicated on diagram: adiabatic compression [ 1 2 ], isothermal compression [1 2 ], adiabatic expansion [ 1 2 ], isothermal expansion [1 2 ]. Area inside loop identified as representing work per cycle. (iv) Draw a T S diagram for the Carnot cycle and indicate the correspondence between stages in the P V diagram. [ 2 ] Correctly drawn T S diagram [ 1 2 ]with axes labels [1 2 ]. Correspondence between all four stages correctly identified Copyright 2015 c University of Southampton Page 8 of 15
9 9 PHYS1013W1 (c) (i) Show that the work necessary to compress one mole of ideal gas adiabatically from volume and temperature T 1 to volume V 2 is [ W = RT (V2 ) 1 γ 1 1]. 1 γ Adiabatic: PV γ = constant = P 1 V γ 1 [ 5 ] W = V2 PdV = P 1 V γ 1 V2 V γ dv = P 1V γ [ ] 1 V 1 γ V 2 1 γ = P 1V γ ( ) 1 γ 2 γ 1 1 γ = P ( ) 1 1 γ Vγ 1 1 γ 2 γ 1 [ = RT (V2 ) 1 γ 1 1] 1 γ (ii) Find an expression for the work necessary to perform the same compression isothermally and explain whether the work is larger in the adiabatic or the isothermal case. [ 3 ] PV = RT P = RT/V (constant temperature) W = V2 = RT 1 V2 dv V = RT 1 ln [ V2 Work is larger for adiabatic case [ 1 2 ]because in isothermal case heat also contributes to increase in internal energy [ 1 2 ]. ] TURN OVER Copyright 2015 c University of Southampton Page 9 of 15
10 (iii) For a Joule expansion, describe qualitatively i. the change in entropy 10 PHYS1013W1 ii. the change in temperature. 1) Entropy increases [ 1 2 ]because there are more possible microscopic configurations [ 1 2 ] 2) Temperature stays constant [ 1 2 ]because no work is done by the gas [ 1 2 ] [ 2 ] Copyright 2015 c University of Southampton Page 10 of 15
11 11 PHYS1013W1 B3. (a) A blackbody spectrum is observed to be maximum at a wavelength of 800 nm and the total power output is measured to be 1 kw. What is the temperature and area of the emitter? [ 4 ] New question: Wien s law: T = m K/λ max = 3625 K [2] Stefan Boltzmann law: P/A = σt 4 A = P/(σT 4 ) = m 2 (b) The luminosity of the Sun is L = W. Treating Pluto (radius r = m) as a blackbody in equilibrium with the Sun, estimate its temperature at closest approach to the Sun (d = m). [ 6 ] Absorbing area of Pluto: πr 2 [ 1 2 ] Emitting area of Pluto: 4πr 2 [ 1 2 ] Solar intensity at Pluto: L 4πd 2 Blackbody power emitted by Pluto: 4πr 2 σt 4 Equating: Hence: T = 4πr 2 σt 4 = L 4πd 2πr2 ( ) 1/4 L = 51 K 16πd 2 σ (c) Explain what it means for entropy to be a function of state and write down the expression the change in entropy for a system at temperature T when a small amount of heat is added in a reversible process. [ 2 ] Entropy being a function of state means it represents a property of the system, regardless of how that state was reached ds = dq T TURN OVER Copyright 2015 c University of Southampton Page 11 of 15
12 12 PHYS1013W1 (d) State the Fundamental Equation of Thermodynamics and show that the differential change in entropy of an ideal gas is ds = c V dt T du = TdS PdV Using du = c V dt and PV = RT ds = 1 T du + P T dv = c v dt T + RdV V + RdV V [ 3 ] (e) Heat is delivered to a heat engine operating between T H = 500 K and T C = 300 K via a cylindrical steel rod (thermal conductivity κ = 400 W m 1 K 1 ) of area A = 0.1 m 2 and length L = 1 m. What is the maximum rate at which the engine can do work? [ 5 ] New question: Fourier s law: dq dt = Aκ dt dx dt dx = 200 K/ m [1 2 ] dq dt = ( ) W = 8 kw Engine maximum efficiency: η = 1 T C T H = 0.4 [ 1 2 ] Maximum work output: W = η dq dt = 3.2 kw Copyright 2015 c University of Southampton Page 12 of 15
13 13 PHYS1013W1 B4. Throughout this question you may assume the gas is ideal. (a) Starting with the availability A, find the appropriate thermodynamic potential for a process which is open to and in good thermal contact with the environment, and identify the name given to this potential from the list at the start of the examination paper. [ 4 ] P = P 0 and T = T 0 which implies dp = 0 and dt = 0 Hence da = du TdS + PdV = d(u TS + PV) Identify G = U TS + PV as Gibbs free energy (b) The molar heat capacity of a gas at constant pressure is observed to be c p = 5R/2, independent of temperature. Find the molar heat capacity at constant volume for this gas, and state the type of gas this is likely to be. [ 2 ] c v = c p R = 3R/2 This is likely to be a monatomic gas (c) One mole of monatomic gas undergoes Joule expansion from volume to V 2 >. It is then compressed isothermally back to. Show that the work done during this compression is W = RT ln [V 2 / ]. [ 3 ] dw = PdV P = RT/V W = RT V 2 dv V = +RT ln [V 2/ ] (d) By noting that the compression is isothermal and the gas is ideal, write down an expression for the heat flow into the gas during this compression. [ 2 ] Isothermal du = dq + dw = 0 Heat flow in Q = +RT ln [V 2 / ] TURN OVER Copyright 2015 c University of Southampton Page 13 of 15
14 14 PHYS1013W1 (e) A Stirling cycle uses isothermal and isochoric processes between volumes and V 2 and using reservoirs at temperatures T H and T C. Sketch and label a P V diagram for this cycle [ 4 ] P V diagram with correctly drawn and labelled isothermal and isochoric processes: Axes labels Isothermals drawn and labeled Isochorics drawn and labeled Hot and cold isothermals identified (f) Explain why the heat entering during isochoric heating is exactly equal to the heat leaving during isochoric cooling. [ 2 ] No work is done, so du = dq For ideal gas, U = c V T and isochoric processes are between the same isotherms Copyright 2015 c University of Southampton Page 14 of 15
15 15 PHYS1013W1 (g) Define efficiency in terms of heat and work, and use the results in this B question to derive an expression for the efficiency of a Stirling engine. [ 3 ] Efficiency η = W Q H Q H = RT H ln [V 2 / ] [ 1 2 ] W = RT H ln [V 2 / ] RT C ln [V 2 / ] [ 1 2 ] η = T H T C T H = 1 T C T H END OF PAPER Copyright 2015 c University of Southampton Page 15 of 15
### Speed Distribution at CONSTANT Temperature is given by the Maxwell Boltzmann Speed Distribution
Temperature ~ Average KE of each particle Particles have different speeds Gas Particles are in constant RANDOM motion Average KE of each particle is: 3/2 kt Pressure is due to momentum transfer Speed Distribution
### Physics 408 Final Exam
Physics 408 Final Exam Name You are graded on your work (with partial credit where it is deserved) so please do not just write down answers with no explanation (or skip important steps)! Please give clear,
### Chemistry. Lecture 10 Maxwell Relations. NC State University
Chemistry Lecture 10 Maxwell Relations NC State University Thermodynamic state functions expressed in differential form We have seen that the internal energy is conserved and depends on mechanical (dw)
### A) 2.0 atm B) 2.2 atm C) 2.4 atm D) 2.9 atm E) 3.3 atm
Name: Date: 1. On a cold day ( 3 C), the gauge pressure on a tire reads 2.0 atm. If the tire is heated to 27 C, what will be the absolute pressure of the air inside the tire? A) 2.0 atm B) 2.2 atm C) 2.4
### Chapter 19: The Kinetic Theory of Gases Questions and Example Problems
Chapter 9: The Kinetic Theory of Gases Questions and Example Problems N M V f N M Vo sam n pv nrt Nk T W nrt ln B A molar nmv RT k T rms B p v K k T λ rms avg B V M m πd N/V Q nc T Q nc T C C + R E nc
### Process Nature of Process
AP Physics Free Response Practice Thermodynamics 1983B. The pv-diagram above represents the states of an ideal gas during one cycle of operation of a reversible heat engine. The cycle consists of the following
### The Second Law of Thermodynamics (Chapter 4)
The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made
### THE SECOND LAW OF THERMODYNAMICS. Professor Benjamin G. Levine CEM 182H Lecture 5
THE SECOND LAW OF THERMODYNAMICS Professor Benjamin G. Levine CEM 182H Lecture 5 Chemical Equilibrium N 2 + 3 H 2 2 NH 3 Chemical reactions go in both directions Systems started from any initial state
### Chapter 2 Carnot Principle
Chapter 2 Carnot Principle 2.1 Temperature 2.1.1 Isothermal Process When two bodies are placed in thermal contact, the hotter body gives off heat to the colder body. As long as the temperatures are different,
### Physics 4C Chapter 19: The Kinetic Theory of Gases
Physics 4C Chapter 19: The Kinetic Theory of Gases Whether you think you can or think you can t, you re usually right. Henry Ford The only thing in life that is achieved without effort is failure. Source
### Unit 05 Kinetic Theory of Gases
Unit 05 Kinetic Theory of Gases Unit Concepts: A) A bit more about temperature B) Ideal Gas Law C) Molar specific heats D) Using them all Unit 05 Kinetic Theory, Slide 1 Temperature and Velocity Recall:
### Version 001 HW 15 Thermodynamics C&J sizemore (21301jtsizemore) 1
Version 001 HW 15 Thermodynamics C&J sizemore 21301jtsizemore 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.
### 8.21 The Physics of Energy Fall 2009
MIT OpenCourseWare http://ocw.mit.edu 8.21 The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 8.21 Lecture 9 Heat Engines
### A thermodynamic system is taken from an initial state X along the path XYZX as shown in the PV-diagram.
AP Physics Multiple Choice Practice Thermodynamics 1. The maximum efficiency of a heat engine that operates between temperatures of 1500 K in the firing chamber and 600 K in the exhaust chamber is most
### Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are
Problem: Calculate the entropy change that results from mixing 54.0 g of water at 280 K with 27.0 g of water at 360 K in a vessel whose walls are perfectly insulated from the surroundings. Is this a spontaneous
### Thermodynamic system is classified into the following three systems. (ii) Closed System It exchanges only energy (not matter) with surroundings.
1 P a g e The branch of physics which deals with the study of transformation of heat energy into other forms of energy and vice-versa. A thermodynamical system is said to be in thermal equilibrium when
### 1985B4. A kilogram sample of a material is initially a solid at a temperature of 20 C. Heat is added to the sample at a constant rate of 100
1985B4. A 0.020-kilogram sample of a material is initially a solid at a temperature of 20 C. Heat is added to the sample at a constant rate of 100 joules per second until the temperature increases to 60
### Atkins / Paula Physical Chemistry, 8th Edition. Chapter 3. The Second Law
Atkins / Paula Physical Chemistry, 8th Edition Chapter 3. The Second Law The direction of spontaneous change 3.1 The dispersal of energy 3.2 Entropy 3.3 Entropy changes accompanying specific processes
### Chem Lecture Notes 6 Fall 2013 Second law
Chem 340 - Lecture Notes 6 Fall 2013 Second law In the first law, we determined energies, enthalpies heat and work for any process from an initial to final state. We could know if the system did work or
### Lecture 25 Goals: Chapter 18 Understand the molecular basis for pressure and the idealgas
Lecture 5 Goals: Chapter 18 Understand the molecular basis for pressure and the idealgas law. redict the molar specific heats of gases and solids. Understand how heat is transferred via molecular collisions
### Phase Changes and Latent Heat
Review Questions Why can a person remove a piece of dry aluminum foil from a hot oven with bare fingers without getting burned, yet will be burned doing so if the foil is wet. Equal quantities of alcohol
### Thermodynamics 2013/2014, lecturer: Martin Zápotocký
Thermodynamics 2013/2014, lecturer: Martin Zápotocký 2 lectures: 1. Thermodynamic processes, heat and work, calorimetry, 1 st and 2 nd law of thermodynamics 2. Entropy, thermodynamic potentials, nonequilibrium
### Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics
Module 5 : Electrochemistry Lecture 21 : Review Of Thermodynamics Objectives In this Lecture you will learn the following The need for studying thermodynamics to understand chemical and biological processes.
### Ch. 19: The Kinetic Theory of Gases
Ch. 19: The Kinetic Theory of Gases In this chapter we consider the physics of gases. If the atoms or molecules that make up a gas collide with the walls of their container, they exert a pressure p on
### ΔU = Q W. Tue Dec 1. Assign 13/14 Friday Final: Fri Dec 11 2:30PM WALTER 145. Thermodynamics 1st Law. 2 nd Law. Heat Engines and Refrigerators
Tue Dec 1 Thermodynamics 1st Law ΔU = Q W 2 nd Law SYS Heat Engines and Refrigerators Isobaric: W = PΔV Isochoric: W = 0 Isothermal: ΔU = 0 Adiabatic: Q = 0 Assign 13/14 Friday Final: Fri Dec 11 2:30PM
### Chapter 19. Heat Engines
Chapter 19 Heat Engines Thermo Processes Eint = Q+ W Adiabatic No heat exchanged Q = 0 and E int = W Isobaric Constant pressure W = P (V f V i ) and E int = Q + W Isochoric Constant Volume W = 0 and E
### CHAPTER - 12 THERMODYNAMICS
CHAPER - HERMODYNAMICS ONE MARK QUESIONS. What is hermodynamics?. Mention the Macroscopic variables to specify the thermodynamics. 3. How does thermodynamics differ from Mechanics? 4. What is thermodynamic
### The First Law of Thermodynamics
Thermodynamics The First Law of Thermodynamics Thermodynamic Processes (isobaric, isochoric, isothermal, adiabatic) Reversible and Irreversible Processes Heat Engines Refrigerators and Heat Pumps The Carnot
### Lecture 6 Free Energy
Lecture 6 Free Energy James Chou BCMP21 Spring 28 A quick review of the last lecture I. Principle of Maximum Entropy Equilibrium = A system reaching a state of maximum entropy. Equilibrium = All microstates
### Specific Heat of Diatomic Gases and. The Adiabatic Process
Specific Heat of Diatomic Gases and Solids The Adiabatic Process Ron Reifenberger Birck Nanotechnology Center Purdue University February 22, 2012 Lecture 7 1 Specific Heat for Solids and Diatomic i Gasses
Enthalpy and Adiabatic Changes Chapter 2 of Atkins: The First Law: Concepts Sections 2.5-2.6 of Atkins (7th & 8th editions) Enthalpy Definition of Enthalpy Measurement of Enthalpy Variation of Enthalpy
### The Laws of Thermodynamics
MME 231: Lecture 06 he Laws of hermodynamics he Second Law of hermodynamics. A. K. M. B. Rashid Professor, Department of MME BUE, Dhaka oday s opics Relation between entropy transfer and heat Entropy change
### Identify the intensive quantities from the following: (a) enthalpy (b) volume (c) refractive index (d) none of these
Q 1. Q 2. Q 3. Q 4. Q 5. Q 6. Q 7. The incorrect option in the following table is: H S Nature of reaction (a) negative positive spontaneous at all temperatures (b) positive negative non-spontaneous regardless
### T ice T water T water = T ice =0 0 C. e =1
Given 1 kg of water at 100 0 C and a very large (very very large) block of ice at 0 0 C. A reversible heat engine absorbs heat from the water and expels heat to the ice until work can no longer be extracted
### Entropy in Macroscopic Systems
Lecture 15 Heat Engines Review & Examples p p b b Hot reservoir at T h p a a c adiabats Heat leak Heat pump Q h Q c W d V 1 V 2 V Cold reservoir at T c Lecture 15, p 1 Review Entropy in Macroscopic Systems
### Honors Physics. Notes Nov 16, 20 Heat. Persans 1
Honors Physics Notes Nov 16, 20 Heat Persans 1 Properties of solids Persans 2 Persans 3 Vibrations of atoms in crystalline solids Assuming only nearest neighbor interactions (+Hooke's law) F = C( u! u
### Preliminary Examination - Day 2 May 16, 2014
UNL - Department of Physics and Astronomy Preliminary Examination - Day May 6, 04 This test covers the topics of Thermodynamics and Statistical Mechanics (Topic ) and Mechanics (Topic ) Each topic has
### 19-9 Adiabatic Expansion of an Ideal Gas
19-9 Adiabatic Expansion of an Ideal Gas Learning Objectives 19.44 On a p-v diagram, sketch an adiabatic expansion (or contraction) and identify that there is no heat exchange Q with the environment. 19.45
### PY2005: Thermodynamics
ome Multivariate Calculus Y2005: hermodynamics Notes by Chris Blair hese notes cover the enior Freshman course given by Dr. Graham Cross in Michaelmas erm 2007, except for lecture 12 on phase changes.
### Final Exam, Chemistry 481, 77 December 2016
1 Final Exam, Chemistry 481, 77 December 216 Show all work for full credit Useful constants: h = 6.626 1 34 J s; c (speed of light) = 2.998 1 8 m s 1 k B = 1.387 1 23 J K 1 ; R (molar gas constant) = 8.314
### Irreversible Processes
Irreversible Processes Examples: Block sliding on table comes to rest due to friction: KE converted to heat. Heat flows from hot object to cold object. Air flows into an evacuated chamber. Reverse process
### 7.3 Heat capacities: extensive state variables (Hiroshi Matsuoka)
7.3 Heat capacities: extensive state variables (Hiroshi Matsuoka) 1 Specific heats and molar heat capacities Heat capacity for 1 g of substance is called specific heat and is useful for practical applications.
### Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 8 Introduction to Vapour Power Cycle Today, we will continue
### Last Name: First Name ID
Last Name: First Name ID This is a set of practice problems for the final exam. It is not meant to represent every topic and is not meant to be equivalent to a 2-hour exam. These problems have not been
### Adiabatic Expansion (DQ = 0)
Adiabatic Expansion (DQ = 0) Occurs if: change is made sufficiently quickly and/or with good thermal isolation. Governing formula: PV g = constant where g = C P /C V Adiabat P Isotherms V Because PV/T
### Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.
Chapter 3 Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Concepts Energy functions F and G Chemical potential, µ Partial Molar properties
### Thermodynamics is the Science of Energy and Entropy
Definition of Thermodynamics: Thermodynamics is the Science of Energy and Entropy - Some definitions. - The zeroth law. - Properties of pure substances. - Ideal gas law. - Entropy and the second law. Some
### The Direction of Spontaneous Change: Entropy and Free Energy
The Direction of Spontaneous Change: Entropy and Free Energy Reading: from Petrucci, Harwood and Herring (8th edition): Required for Part 1: Sections 20-1 through 20-4. Recommended for Part 1: Sections
### CHEMICAL THERMODYNAMICS
DEPARTMENT OF APPLIED CHEMISTRY LECTURE NOTES 6151- ENGINEERING CHEMISTRY-II UNIT II CHEMICAL THERMODYNAMICS Unit syllabus: Terminology of thermodynamics - Second law: Entropy - entropy change for an ideal
### Thermodynamics Boltzmann (Gibbs) Distribution Maxwell-Boltzmann Distribution Second Law Entropy
Thermodynamics Boltzmann (Gibbs) Distribution Maxwell-Boltzmann Distribution Second Law Entropy Lana Sheridan De Anza College May 8, 2017 Last time modeling an ideal gas at the microscopic level pressure,
### Thermodynamics and Phase Transitions in Minerals
Studiengang Geowissenschaften M.Sc. Wintersemester 2004/05 Thermodynamics and Phase Transitions in Minerals Victor Vinograd & Andrew Putnis Basic thermodynamic concepts One of the central themes in Mineralogy
### Chapter 19 The First Law of Thermodynamics
Chapter 19 The First Law of Thermodynamics Lecture by Dr. Hebin Li Assignment Due at 11:59pm on Sunday, December 7 HW set on Masteringphysics.com Final exam: Time: 2:15pm~4:15pm, Monday, December 8. Location:
### Pressure Volume Temperature Relationship of Pure Fluids
Pressure Volume Temperature Relationship of Pure Fluids Volumetric data of substances are needed to calculate the thermodynamic properties such as internal energy and work, from which the heat requirements
### Unit 7 (B) Solid state Physics
Unit 7 (B) Solid state Physics hermal Properties of solids: Zeroth law of hermodynamics: If two bodies A and B are each separated in thermal equilibrium with the third body C, then A and B are also in
### Reading Assignment #13 :Ch.23: (23-3,5,6,7,8), Ch24: (24-1,2,4,5,6) )] = 450J. ) ( 3x10 5 Pa) ( 1x10 3 m 3
Ph70A Spring 004 Prof. Pui Lam SOLUTION Reading Assignment #3 :h.3: (3-3,5,6,7,8), h4: (4-,,4,5,6) Homework #3: First Law and Second of Thermodynamics Due: Monday 5/3/004.. A monatomic ideal gas is allowed
### Chapter 7. Entropy: A Measure of Disorder
Chapter 7 Entropy: A Measure of Disorder Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic
### CHAPTER 3 LECTURE NOTES 3.1. The Carnot Cycle Consider the following reversible cyclic process involving one mole of an ideal gas:
CHATER 3 LECTURE NOTES 3.1. The Carnot Cycle Consider the following reversible cyclic process involving one mole of an ideal gas: Fig. 3. (a) Isothermal expansion from ( 1, 1,T h ) to (,,T h ), (b) Adiabatic
### Application of the First Law of Thermodynamics to Adiabatic, Isothermal, Isobaric, and Isochoric Processes; Heat Engine and Engine Cycles
Application of the First Law of Thermodynamics to Adiabatic, Isothermal, Isobaric, and Isochoric Processes; Heat Engine and Engine Cycles by SHS Encoder 3 on February 05, 2018 lesson duration of 0 minutes
### Review of classical thermodynamics
Review of classical thermodynamics Fundamental Laws, Properties and Processes (2) Entropy and the Second Law Concepts of equilibrium Reversible and irreversible processes he direction of spontaneous change
### Outline. 1. Work. A. First Law of Thermo. 2. Internal Energy. 1. Work continued. Category: Thermodynamics. III. The Laws of Thermodynamics.
ategory: hermodynamics Outline III. he Laws of hermodynamics A. First Law of hermo B. Second Law of hermo (Entropy). Statistical Mechanics D. References Updated: 04jan A. First Law of hermo. Work 4 Stored
### 3. Basic Concepts of Thermodynamics Part 2
3. Basic Concepts of Thermodynamics Part 2 Temperature and Heat If you take a can of cola from the refrigerator and leave it on the kitchen table, its temperature will rise-rapidly at first but then more
### FIRST PUBLIC EXAMINATION SUBJECT 3: PHYSICAL CHEMISTRY
CCHE 4273 FIRST PUBLIC EXAMINATION Trinity Term 2005 Preliminary Examination in Chemistry SUBJECT 3: PHYSICAL CHEMISTRY Wednesday June 8 th 2005, 9:30am Time allowed: 2 ½ hours Candidates should answer
### Thermal and Statistical Physics Department Exam Last updated November 4, L π
Thermal and Statistical Physics Department Exam Last updated November 4, 013 1. a. Define the chemical potential µ. Show that two systems are in diffusive equilibrium if µ 1 =µ. You may start with F =
### 140a Final Exam, Fall 2006., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT.
40a Final Exam, Fall 2006 Data: P 0 0 5 Pa, R = 8.34 0 3 J/kmol K = N A k, N A = 6.02 0 26 particles/kilomole, T C = T K 273.5. du = TdS PdV + i µ i dn i, U = TS PV + i µ i N i Defs: 2 β ( ) V V T ( )
### 1. Second Law of Thermodynamics
1. Second Law of hermodynamics he first law describes how the state of a system changes in response to work it performs and heat absorbed. However, the first law cannot explain certain facts about thermal
### Chap. 3. The Second Law. Law of Spontaneity, world gets more random
Chap. 3. The Second Law Law of Spontaneity, world gets more random Kelvin - No process can transform heat completely into work Chap. 3. The Second Law Law of Spontaneity, world gets more random Kelvin
### Alternate Midterm Examination Physics 100 Feb. 20, 2014
Alternate Midterm Examination Physics 100 Feb. 20, 2014 Name/Student #: Instructions: Formulas at the back (you can rip that sheet o ). Questions are on both sides. Calculator permitted. Put your name
### CH341 Test 2 Physical Chemistry Test 2 Fall 2011, Prof. Shattuck
Physical Chemistry Test 2 Fall 2011, Prof. Shattuck Name 1 Constants: R = 8.3145 J K -1 mol -1 = 0.083145 L bar K -1 mol -1 R = 0.08206 L atm K -1 mol -1 1 F = 96485 C mol -1 1 atm = 1.01325 bar 1 bar
### Chapter 17. Work, Heat, and the First Law of Thermodynamics Topics: Chapter Goal: Conservation of Energy Work in Ideal-Gas Processes
Chapter 17. Work, Heat, and the First Law of Thermodynamics This false-color thermal image (an infrared photo) shows where heat energy is escaping from a house. In this chapter we investigate the connection
### 1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v. lnt + RlnV + cons tant
1 mol ideal gas, PV=RT, show the entropy can be written as! S = C v lnt + RlnV + cons tant (1) p, V, T change Reversible isothermal process (const. T) TdS=du-!W"!S = # "Q r = Q r T T Q r = \$W = # pdv =
### P340: Thermodynamics and Statistical Physics, Exam#2 Any answer without showing your work will be counted as zero
P340: hermodynamics and Statistical Physics, Exam#2 Any answer without showing your work will be counted as zero 1. (15 points) he equation of state for the an der Waals gas (n = 1 mole) is (a) Find (
### 140a Final Exam, Fall 2007., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT.
4a Final Exam, Fall 27 Data: P 5 Pa, R = 8.34 3 J/kmol K = N A k, N A = 6.2 26 particles/kilomole, T C = T K 273.5. du = TdS PdV + i µ i dn i, U = TS PV + i µ i N i Defs: 2 β ( ) V V T ( ) /dq C? dt P?
### Chemical thermodynamics the area of chemistry that deals with energy relationships
Chemistry: The Central Science Chapter 19: Chemical Thermodynamics Chemical thermodynamics the area of chemistry that deals with energy relationships 19.1: Spontaneous Processes First law of thermodynamics
### Answers to Assigned Problems from Chapter 2
Answers to Assigned Problems from Chapter 2 2.2. 1 mol of ice has a volume of 18.01 g/0.9168 g cm 3 = 19.64 cm 3 1 mol of water has a volume of 18.01 g/0.9998 g cm 3 = 18.01 cm 3 V(ice water) = 1.63 cm
### Properties of Entropy
Properties of Entropy Due to its additivity, entropy is a homogeneous function of the extensive coordinates of the system: S(λU, λv, λn 1,, λn m ) = λ S (U, V, N 1,, N m ) This means we can write the entropy
### Chapter 17 Temperature & Kinetic Theory of Gases 1. Thermal Equilibrium and Temperature
Chapter 17 Temperature & Kinetic Theory of Gases 1. Thermal Equilibrium and Temperature Any physical property that changes with temperature is called a thermometric property and can be used to measure
### Spring_#7. Thermodynamics. Youngsuk Nam.
Spring_#7 Thermodynamics Youngsuk Nam ysnam1@khu.ac.kr You can t connect the dots looking forward; you can only connect them looking backwards. So you have to trust that the dots will somehow connect in
### Chapter 5. Simple Mixtures Fall Semester Physical Chemistry 1 (CHM2201)
Chapter 5. Simple Mixtures 2011 Fall Semester Physical Chemistry 1 (CHM2201) Contents The thermodynamic description of mixtures 5.1 Partial molar quantities 5.2 The thermodynamic of Mixing 5.3 The chemical
### THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system
THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system and its surroundings. a. System = That part of universe
### The Kinetic Theory of Gases
chapter 1 The Kinetic Theory of Gases 1.1 Molecular Model of an Ideal Gas 1. Molar Specific Heat of an Ideal Gas 1.3 Adiabatic Processes for an Ideal Gas 1.4 The Equipartition of Energy 1.5 Distribution
### The laws of Thermodynamics. Work in thermodynamic processes
The laws of Thermodynamics ork in thermodynamic processes The work done on a gas in a cylinder is directly proportional to the force and the displacement. = F y = PA y It can be also expressed in terms
### People s Physics book 3e
The Big Ideas Heat is a form of energy transfer. It can change the kinetic energy of a substance. For example, the average molecular kinetic energy of gas molecules is related to temperature. A heat engine
### Phys102 First Major- 161 Code: 20 Coordinator: Dr. A. Naqvi Saturday, October 29, 2016 Page: 1
Coordinator: Dr. A. Naqvi Saturday, October 29, 2016 Page: 1 Q1. FIGURE 1 shows three waves that are separately sent along the same unstretchable string that is kept under constant tension along an x-axis.
### Not so black. Black Hole Thermodynamics. Nobel Prize 2017: Rainer Weiss, Kip Thorne, and Barry Barish.
Not so black Nobel Prize 2017: Rainer Weiss, Kip Thorne, and Barry Barish. Black Hole Thermodynamics Zeroth Law of thermodynamics requires that black holes are round *. First law of thermodynamics requires
### ...Thermodynamics. Entropy: The state function for the Second Law. Entropy ds = d Q. Central Equation du = TdS PdV
...Thermodynamics Entropy: The state function for the Second Law Entropy ds = d Q T Central Equation du = TdS PdV Ideal gas entropy s = c v ln T /T 0 + R ln v/v 0 Boltzmann entropy S = klogw Statistical
### Some properties of the Helmholtz free energy
Some properties of the Helmholtz free energy Energy slope is T U(S, ) From the properties of U vs S, it is clear that the Helmholtz free energy is always algebraically less than the internal energy U.
### Physics 53. Thermal Physics 1. Statistics are like a bikini. What they reveal is suggestive; what they conceal is vital.
Physics 53 Thermal Physics 1 Statistics are like a bikini. What they reveal is suggestive; what they conceal is vital. Arthur Koestler Overview In the following sections we will treat macroscopic systems
### This is a statistical treatment of the large ensemble of molecules that make up a gas. We had expressed the ideal gas law as: pv = nrt (1)
1. Kinetic Theory of Gases This is a statistical treatment of the large ensemble of molecules that make up a gas. We had expressed the ideal gas law as: pv = nrt (1) where n is the number of moles. We
### [S R (U 0 ɛ 1 ) S R (U 0 ɛ 2 ]. (0.1) k B
Canonical ensemble (Two derivations) Determine the probability that a system S in contact with a reservoir 1 R to be in one particular microstate s with energy ɛ s. (If there is degeneracy we are picking
### 2. Under conditions of constant pressure and entropy, what thermodynamic state function reaches an extremum? i
1. (20 oints) For each statement or question in the left column, find the appropriate response in the right column and place the letter of the response in the blank line provided in the left column. 1.
### First Law showed the equivalence of work and heat. Suggests engine can run in a cycle and convert heat into useful work.
0.0J /.77J / 5.60J hermodynamics of Biomolecular Systems 0.0/5.60 Fall 005 Lecture #6 page he Second Law First Law showed the euivalence of work and heat U = + w, du = 0 for cyclic process = w Suggests
### Thermodynamics Qualifying Exam Study Material
Thermodynamics Qualifying Exam Study Material The candidate is expected to have a thorough understanding of undergraduate engineering thermodynamics topics. These topics are listed below for clarification.
### PHYSICS 149: Lecture 26
PHYSICS 149: Lecture 26 Chapter 14: Heat 14.1 Internal Energy 14.2 Heat 14.3 Heat Capacity and Specific Heat 14.5 Phase Transitions 14.6 Thermal Conduction 14.7 Thermal Convection 14.8 Thermal Radiation
### On my honor as a Texas A&M University student, I will neither give nor receive unauthorized help on this exam.
Physics 201, Exam 4 Name (printed) On my honor as a Texas A&M University student, I will neither give nor receive unauthorized help on this exam. Name (signed) The multiple-choice problems carry no partial
### !W "!#U + T#S, where. !U = U f " U i and!s = S f " S i. ! W. The maximum amount of work is therefore given by =!"U + T"S. !W max
1 Appendix 6: The maximum wk theem (Hiroshi Matsuoka) 1. Question and answer: the maximum amount of wk done by a system Suppose we are given two equilibrium states of a macroscopic system. Consider all
### Gases. T boil, K. 11 gaseous elements. Rare gases. He, Ne, Ar, Kr, Xe, Rn Diatomic gaseous elements H 2, N 2, O 2, F 2, Cl 2
Gases Gas T boil, K Rare gases 11 gaseous elements He, Ne, Ar, Kr, Xe, Rn 165 Rn 211 N 2 O 2 77 F 2 90 85 Diatomic gaseous elements Cl 2 238 H 2, N 2, O 2, F 2, Cl 2 H 2 He Ne Ar Kr Xe 20 4.4 27 87 120
### Solutions Midterm Exam 3 December 12, Match the above shown players of the best baseball team in the world with the following names:
Problem 1 (2.5 points) 1 2 3 4 Match the above shown players of the best baseball team in the world with the following names: A. Derek Jeter B. Mariano Rivera C. Johnny Damon D. Jorge Posada 1234 = a.
### Lecture 30. Chapter 21 Examine two wave superposition (-ωt and +ωt) Examine two wave superposition (-ω 1 t and -ω 2 t)
To do : Lecture 30 Chapter 21 Examine two wave superposition (-ωt and +ωt) Examine two wave superposition (-ω 1 t and -ω 2 t) Review for final (Location: CHEM 1351, 7:45 am ) Tomorrow: Review session,
| 10,905
| 39,737
|
{"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
| 0.832923
|
https://www.lmfdb.org/ArtinRepresentation/2.3639.6t3.a.a
| 1,618,617,839,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00393.warc.gz
| 966,164,581
| 6,040
|
# Properties
Label 2.3639.6t3.a.a Dimension $2$ Group $D_{6}$ Conductor $3639$ Root number $1$ Indicator $1$
# Related objects
## Basic invariants
Dimension: $2$ Group: $D_{6}$ Conductor: $$3639$$$$\medspace = 3 \cdot 1213$$ Frobenius-Schur indicator: $1$ Root number: $1$ Artin stem field: 6.2.16062935373.1 Galois orbit size: $1$ Smallest permutation container: $D_{6}$ Parity: odd Determinant: 1.3639.2t1.a.a Projective image: $S_3$ Projective stem field: 3.1.3639.1
## Defining polynomial
$f(x)$ $=$ $$x^{6} - 44 x^{3} - 729$$ .
The roots of $f$ are computed in an extension of $\Q_{ 11 }$ to precision 9.
Minimal polynomial of a generator $a$ of $K$ over $\mathbb{Q}_{ 11 }$: $$x^{2} + 7 x + 2$$
Roots:
$r_{ 1 }$ $=$ $$8 a + 10 + \left(a + 7\right)\cdot 11 + \left(7 a + 2\right)\cdot 11^{2} + \left(8 a + 10\right)\cdot 11^{3} + 10\cdot 11^{4} + \left(7 a + 7\right)\cdot 11^{5} + 7 a\cdot 11^{6} + \left(7 a + 5\right)\cdot 11^{7} + \left(10 a + 5\right)\cdot 11^{8} +O(11^{9})$$ $r_{ 2 }$ $=$ $$8 + 6\cdot 11 + 2\cdot 11^{2} + 8\cdot 11^{3} + 6\cdot 11^{4} + 3\cdot 11^{5} + 5\cdot 11^{6} + 11^{7} + 3\cdot 11^{8} +O(11^{9})$$ $r_{ 3 }$ $=$ $$3 a + 9 + \left(9 a + 6\right)\cdot 11 + \left(3 a + 7\right)\cdot 11^{2} + \left(2 a + 4\right)\cdot 11^{3} + \left(10 a + 5\right)\cdot 11^{4} + \left(3 a + 2\right)\cdot 11^{5} + \left(3 a + 2\right)\cdot 11^{6} + \left(3 a + 6\right)\cdot 11^{7} + 7\cdot 11^{8} +O(11^{9})$$ $r_{ 4 }$ $=$ $$3 a + 1 + \left(6 a + 2\right)\cdot 11 + \left(9 a + 10\right)\cdot 11^{2} + \left(4 a + 1\right)\cdot 11^{3} + \left(6 a + 8\right)\cdot 11^{4} + \left(a + 3\right)\cdot 11^{5} + \left(8 a + 9\right)\cdot 11^{6} + \left(3 a + 6\right)\cdot 11^{7} + \left(6 a + 9\right)\cdot 11^{8} +O(11^{9})$$ $r_{ 5 }$ $=$ $$3 + 7\cdot 11 + 7\cdot 11^{3} + 5\cdot 11^{4} + 8\cdot 11^{6} + 10\cdot 11^{7} + 8\cdot 11^{8} +O(11^{9})$$ $r_{ 6 }$ $=$ $$8 a + 2 + \left(4 a + 2\right)\cdot 11 + \left(a + 9\right)\cdot 11^{2} + 6 a\cdot 11^{3} + \left(4 a + 7\right)\cdot 11^{4} + \left(9 a + 3\right)\cdot 11^{5} + \left(2 a + 7\right)\cdot 11^{6} + \left(7 a + 2\right)\cdot 11^{7} + \left(4 a + 9\right)\cdot 11^{8} +O(11^{9})$$
## Generators of the action on the roots $r_1, \ldots, r_{ 6 }$
Cycle notation $(1,2,3,6,5,4)$ $(2,4)(3,5)$
## Character values on conjugacy classes
Size Order Action on $r_1, \ldots, r_{ 6 }$ Character value $1$ $1$ $()$ $2$ $1$ $2$ $(1,6)(2,5)(3,4)$ $-2$ $3$ $2$ $(2,4)(3,5)$ $0$ $3$ $2$ $(1,2)(3,4)(5,6)$ $0$ $2$ $3$ $(1,3,5)(2,6,4)$ $-1$ $2$ $6$ $(1,2,3,6,5,4)$ $1$
The blue line marks the conjugacy class containing complex conjugation.
| 1,268
| 2,622
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}
| 2.8125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.215571
|
https://www.reference.com/world-view/convert-cubic-meters-kilograms-6ecf124b0e7848a
| 1,653,294,616,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00229.warc.gz
| 1,110,259,899
| 15,152
|
How Do You Convert Cubic Meters to Kilograms?
Cubic meters are a unit of volume, and kilograms are a unit of mass. Cubic meters can only be converted to kilograms if the density of a substance is known. If density is known, then multiplying the number of cubic meters by the density (kg/m^3) derives the number of kilograms.
Density must be measured in kg/m^3 in order to convert the number of cubic meters to kilograms through simple multiplication. A substance with a higher density has more kilograms in a certain number of cubic meters than another substance with a lower density but equal volume.
For example, Engineering Toolbox reports that the density of iron is 7850 kg/m^3. If there are 10 cubic meters of iron, this represents (10 m^3 * 7850 kg/m^3), which equals 78500 kg. Liquid water, on the other hand, has a density of just 1000 kg/m^3 at 4 degrees Celsius. If there are 10 cubic meters of water, this represents a mass of (10 m^3 * 1000 kg/m^3), which equals 10000 kg water, a much smaller mass than that of iron in the same amount of volume.
| 257
| 1,062
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.89308
|
https://planetmath.org/thederivedsubgroupisnormal
| 1,718,439,000,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00269.warc.gz
| 428,766,834
| 3,902
|
# the derived subgroup is normal
We are going to prove:
”The derived subgroup (or commutator subgroup) $[G,G]$ is normal in $G$
Proof:
We have to show that for each $x\in[G,G]$, $gxg^{-1}$ it is also in $[G,G]$.
Since $[G,G]$ is the subgroup generated by the all commutators in $G$, then for each $x\in[G,G]$ we have $x=c_{1}c_{2}\cdots c_{m}$ –a word of commutators– so $c_{i}=[a_{i},b_{i}]$ for all $i$.
Now taking any element of $g\in G$ we can see that
$\displaystyle g[a_{i},b_{i}]g^{-1}$ $\displaystyle=$ $\displaystyle ga_{i}b_{i}a_{i}^{-1}b_{i}^{-1}g^{-1}$ $\displaystyle=$ $\displaystyle ga_{i}g^{-1}gb_{i}g^{-1}ga_{i}^{-1}g^{-1}gb_{i}^{-1}g^{-1}$ $\displaystyle=$ $\displaystyle(ga_{i}g^{-1})(gb_{i}g^{-1})(ga_{i}g^{-1})^{-1}(gb_{i}g^{-1})^{-1}$ $\displaystyle=$ $\displaystyle[ga_{i}g^{-1},gb_{i}g^{-1}],$
that is
$g[a_{i},b_{i}]g^{-1}=[ga_{i}g^{-1},gb_{i}g^{-1}]$
so a conjugation of a commutator is another commutator, then for the conjugation
$\displaystyle gxg^{-1}$ $\displaystyle=$ $\displaystyle gc_{1}c_{2}\cdots c_{m}g^{-1}$ $\displaystyle=$ $\displaystyle gc_{1}g^{-1}gc_{2}g^{-1}g\cdots g^{-1}gc_{m}g^{-1}$ $\displaystyle=$ $\displaystyle(gc_{1}g^{-1})(gc_{2}g^{-1})\cdots(gc_{m}g^{-1})$
is another word of commutators, hence $gxg^{-1}$ is in $[G,G]$ which in turn implies that $[G,G]$ is normal in $G$, QED.
Title the derived subgroup is normal TheDerivedSubgroupIsNormal 2013-03-22 16:04:39 2013-03-22 16:04:39 juanman (12619) juanman (12619) 8 juanman (12619) Proof msc 20A05 msc 20E15 msc 20F14
| 612
| 1,534
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 33, "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-2024-26
|
latest
|
en
| 0.570874
|
https://www.whizwriters.com/mathematics-1566/
| 1,610,798,022,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703506640.22/warc/CC-MAIN-20210116104719-20210116134719-00399.warc.gz
| 1,046,656,077
| 9,976
|
# Mathematics
1.
Consider the following population data:
21 27 11 13 7
a. Calculate the range.
Range
b.
Calculate MAD. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
c.
Calculate the population variance. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Population variance
d.
Calculate the population standard deviation. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Population standard deviation
rev: 07_31_2013_QC_32713
2.
Consider the following sample data:
x 11 13 15 17 19
y 19 17 15 13 11
a.
Calculate the covariance between the variables. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Covariance
b-1.
Calculate the correlation coefficient. (Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Correlation coefficient
b-2. Interpret the correlation coefficient.
There is
relationship between x and y.
rev: 07_31_2013_QC_32713
3.
Consider the following observations from a population:
88 235 73 138 138 71 229 153 73
a. Calculate the mean and median. (Round “mean” to 2 decimal places.)
Mean
Median
b.
Select the mode. (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.)
138
71
88
73
153
229
235
rev: 07_31_2013_QC_32713
4.
A manager of a local retail store analyzes the relationship between advertising and sales by reviewing the store’s data for the previous six months.
Advertising (in \$100s) Sales (in \$1,000s)
174 144
57 45
56 44
55 43
212 136
210 134
a.
Calculate the mean of advertising and the mean of sales. (Round your answers to 2 decimal places.)
Mean
Sales
b.
Calculate the standard deviation of advertising and the standard deviation of sales. (Round your answers to 2 decimal places.)
Standard Deviation
Sales
Covariance
c-2.
Interpret the covariance between advertising and sales.
No correlation
Negative correlation
Positive correlation
d-1.
Correlation coefficient
d-2.
Interpret the correlation coefficient between advertising and sales.
No correlation
Weak negative correlation
Strong positive correlation
Strong negative correlation
Weak positive correlation
rev: 07_31_2013_QC_32713
5.
An investment strategy has an expected return of 12 percent and a standard deviation of 8 percent. Assume investment returns are bell shaped.
a.
How likely is it to earn a return between 4 percent and 20 percent? (Enter your response as decimal values (not percentages) rounded to 2 decimal places.)
Probability
b.
How likely is it to earn a return greater than 20 percent?(Enter your response as decimal values (not percentages) rounded to 2 decimal places.)
Probability
c.
How likely is it to earn a return below −4 percent?(Enter your response as decimal values (not percentages) rounded to 2 decimal places.)
Probability
rev: 02_26_2014_QC_44958, 07_12_2014_QC_5
6.
The following relative frequency distribution was constructed from a population of 400. Calculate the population mean, the population variance, and the population standard deviation. (Round your intermediate calculations to 4 decimal places and final answers to 2 decimal places.)
Class Relative Frequency
−20 up to −10 0.14
−10 up to 0 0.22
0 up to 10 0.32
10 up to 20 0.32
Population mean
Population variance
Population standard deviation
rev: 07_31_2013_QC_32713
7.
Consider the following frequency distribution.
Class Frequency
2 up to 4 19
4 up to 6 61
6 up to 8 79
8 up to 10 19
a.
Population mean
b.
Calculate the population variance and the population standard deviation. (Round your intermediate calculations to 4 decimal places and final answers to 2 decimal places.)
Population variance
Population standard deviation
rev: 07_31_2013_QC_32713
8.
A data set has a mean of 1,000 and a standard deviation of 50.
a.
Using Chebyshev’s theorem, what percentage of the observations fall between 800 and 1,200? (Do not round intermediate calculations. Round your answer to the nearest whole percent.)
Percentage of observations
b.
Using Chebyshev’s theorem, what percentage of the observations fall between 750 and 1,250? (Do not round intermediate calculations. Round your answer to the nearest whole percent.)
Percentage of observations
rev: 07_31_2013_QC_32713
9.
Consider the following returns for two investments, A and B, over the past four years:
Investment 1: 2% 14% –3% 15%
Investment 2: 13% 12% –18% 14%
a-1.
Calculate the mean for each investment. (Round your answers to 2 decimal places.)
Mean
Investment 1 percent
Investment 2 percent
a-2.
Which investment provides the higher return?
Investment 2
Investment 1
b-1.
Calculate the standard deviation for each investment. (Round your answers to 2 decimal places.)
Standard
Deviation
Investment 1
Investment 2
b-2.
Which investment provides less risk?
Investment 1
Investment 2
c-1.
Given a risk-free rate of 1.3%, calculate the Sharpe ratio for each investment. (Do not round intermediate calculations. Round your answers to 2 decimal places.)
Sharpe Ratio
Investment 1
Investment 2
c-2. Which investment has performed better?
Investment 2
Investment 1
rev: 07_31_2013_QC_32713, 11_10_2013_QC_38348
10.
Scores on the final in a statistics class are as follows.
94 45 95 82 96 100 100 119 110 80
105 119 60 94 92 85 107 89 105 90
a.
Calculate the 25th, 50th, and 75th percentiles. (Do not round intermediate calculations. Round your answers to 2 decimal places.)
25th percentile
50th percentile
75th percentile
b-1.
Calculate the IQR, lower limit and upper limit to detect outliers. (Negative value should be indicated by a minus sign. Round your intermediate calculations to 4 decimal places and final answers to 2 decimal places.)
IQR
Lower limit
Upper limit
b-2. Are there any outliers?
Yes
No
rev: 07_31_2013_QC_32713, 09_13_2013_QC_34880, 10_31_2013_QC_38175, 03_03_2014_QC_44705, 09_24_2014_QC_54188, 02_26_2015_QC_CS-5733
Order now and get 10% discount on all orders above \$50 now!!The professional are ready and willing handle your assignment.
ORDER NOW »»
| 1,658
| 6,392
|
{"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-2021-04
|
latest
|
en
| 0.784885
|
https://www.instructables.com/topics/If-12months100-what-percent-of-a-year-is-one-mon/
| 1,532,369,101,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00071.warc.gz
| 933,809,119
| 6,475
|
33533Views2Replies
# If 12months=100% what percent of a year is one month? Answered
12=100%, 6=50%, 3=25%, so what percentage of a year is one month? This is just for fun really.
Tags:
## Discussions
One month is approximately 8.34% of a year.
8.34 x 12 = 100.08 (close enough)
The pattern goes like this 12 = 100% , 12 /12 = 1, 100 /1 = 100% , 12 / 6 = 2 , 100 / 2 = 50% , 12 / 3 = 4, 100 / 4 = 25% , and to answer the questions 1 month out of 12... 12 / 1 = 12 , 100 / 12 = 8.33 . To verify this, 8.33 * 12 = 99.96 , since 6 > 5 , you can estimate the 9 to the next digit which is .9 + .1 making 1.0 , then 99.00 + 1.0 = 100%.
| 253
| 635
|
{"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}
| 3.953125
| 4
|
CC-MAIN-2018-30
|
latest
|
en
| 0.862406
|
http://oeis.org/A162788
| 1,575,768,251,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540503656.42/warc/CC-MAIN-20191207233943-20191208021943-00227.warc.gz
| 106,725,757
| 3,963
|
This site is supported by donations to The OEIS Foundation.
Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A162788 a(n) = A162528(n)/8. 3
8, 9, 10, 11, 12, 13, 14, 16, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Nathaniel Johnston, Table of n, a(n) for n = 1..10000 CROSSREFS Cf. A161344, A162528, A162787, A162789. Sequence in context: A161639 A250044 A295291 * A115842 A067682 A067729 Adjacent sequences: A162785 A162786 A162787 * A162789 A162790 A162791 KEYWORD easy,nonn AUTHOR Omar E. Pol, Jul 13 2009 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 7 20:23 EST 2019. Contains 329847 sequences. (Running on oeis4.)
| 518
| 1,499
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2019-51
|
latest
|
en
| 0.785377
|
https://statsidea.com/en/excel/learn-how-to-form-a-timber-timber-plot-in-excel/
| 1,721,505,866,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00617.warc.gz
| 485,703,523
| 19,187
|
# Learn how to Form a Timber-Timber Plot in Excel
log-log plot is a scatterplot that makes use of logarithmic scales on each the x-axis and the y-axis.
This kind of plot turns out to be useful for visualizing two variables when the actual dating among them follows an influence legislation. This phenomenon happens in lots of farmlands in actual while together with astronomy, biology, chemistry, and physics.
This instructional presentations the best way to form a log-log plot for 2 variables in Excel.
### Instance: Timber-Timber Plot in Excel
Think we now have refer to dataset in Excel that presentations the values for 2 variables, x and y:
Importance refer to steps to form a log-log plot for this dataset:
Step 1: Form a scatterplot.
Spotlight the information within the length A2:B11
Alongside the supremacy ribbon, click on the Insert tab. Throughout the Charts staff, click on on Spill.
Please see scatterplot will robotically seem:
Step 2: Exchange the x-axis scale to logarithmic.
Proper click on at the values alongside the x-axis and click on Layout Axis.
Within the brandnew window that pops up, test the field upcoming to Logarithmic scale to modify the x-axis scale.
Step 3: Exchange the y-axis scale to logarithmic.
Then, click on at the y-axis and repeat the similar step to modify the y-axis scale to logarithmic. The ensuing plot will appear to be this:
Realize that the x-axis now spans from 1 to ten age the y-axis spans from 1 to at least one,000. Additionally understand how the connection between the variables x and y now seems extra unbending. This is a sign that the 2 variables do certainly have an influence legislation dating.
You’ll be able to in finding extra Excel tutorials right here.
| 390
| 1,739
|
{"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-2024-30
|
latest
|
en
| 0.820976
|
http://mathoverflow.net/questions/58821/greatest-common-divisor-of-p-1-and-q-1?sort=oldest
| 1,464,660,876,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-22/segments/1464051165777.47/warc/CC-MAIN-20160524005245-00230-ip-10-185-217-139.ec2.internal.warc.gz
| 182,669,685
| 12,349
|
# greatest common divisor of p-1 and q-1 [closed]
Hi there,
Can we say that if $p$ and $q$ are distinct prime number diving $n$
$\Omega(gcd(p-1,q-1)) \leq \Omega(n)$
Where $\Omega(n)$ denotes the number of prime powers dividing $n$
Best rahmi
-
## closed as too localized by Gjergji Zaimi, Franz Lemmermeyer, Bill Johnson, S. Carnahan♦Mar 18 '11 at 16:47
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
No: let $p=13$, $q=37$ and $n=pq=481$. Then $\Omega((p-1,q-1))=\Omega(12)=3$ while $\Omega(n)=2$. – Thomas Bloom Mar 18 '11 at 9:44
No: take $n=211*2311$, $p=211, q=2311$. In fact there is no bound of the L.H.S. in terms of the R.H.S. in your inequality. Take any sequence of primes $p_1,...,p_s$. Let $a=p_1\cdot...\dot p_s$. By Dirichlet theorem there are two primes $p=ak+1, q=am+1$, $k < m$. Let $n=pq$. Then $\Omega(n)=2$ while $GCD(p-1,q-1)$ is divisible by $p_1,...,p_n$, so $\Omega(GCD)\ge n$.
| 403
| 1,284
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625
| 4
|
CC-MAIN-2016-22
|
latest
|
en
| 0.793335
|
https://number.academy/251004
| 1,670,483,649,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00846.warc.gz
| 448,013,146
| 12,525
|
# Number 251004
Number 251,004 spell 🔊, write in words: two hundred and fifty-one thousand and four . Ordinal number 251004th is said 🔊 and write: two hundred and fifty-one thousand and fourth. Color #251004. The meaning of number 251004 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 251004. What is 251004 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 251004.
## What is 251,004 in other units
The decimal (Arabic) number 251004 converted to a Roman number is (C)(C)(L)MIV. Roman and decimal number conversions.
#### Weight conversion
251004 kilograms (kg) = 553363.4 pounds (lbs)
251004 pounds (lbs) = 113854.7 kilograms (kg)
#### Length conversion
251004 kilometers (km) equals to 155967 miles (mi).
251004 miles (mi) equals to 403952 kilometers (km).
251004 meters (m) equals to 823494 feet (ft).
251004 feet (ft) equals 76507 meters (m).
251004 centimeters (cm) equals to 98820.5 inches (in).
251004 inches (in) equals to 637550.2 centimeters (cm).
#### Temperature conversion
251004° Fahrenheit (°F) equals to 139428.9° Celsius (°C)
251004° Celsius (°C) equals to 451839.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
251004 seconds equals to 2 days, 21 hours, 43 minutes, 24 seconds
251004 minutes equals to 6 months, 6 days, 7 hours, 24 minutes
### Codes and images of the number 251004
Number 251004 morse code: ..--- ..... .---- ----- ----- ....-
Sign language for number 251004:
Number 251004 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 251004
### Multiplications
#### Multiplication table of 251004
251004 multiplied by two equals 502008 (251004 x 2 = 502008).
251004 multiplied by three equals 753012 (251004 x 3 = 753012).
251004 multiplied by four equals 1004016 (251004 x 4 = 1004016).
251004 multiplied by five equals 1255020 (251004 x 5 = 1255020).
251004 multiplied by six equals 1506024 (251004 x 6 = 1506024).
251004 multiplied by seven equals 1757028 (251004 x 7 = 1757028).
251004 multiplied by eight equals 2008032 (251004 x 8 = 2008032).
251004 multiplied by nine equals 2259036 (251004 x 9 = 2259036).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 251004
Half of 251004 is 125502 (251004 / 2 = 125502).
One third of 251004 is 83668 (251004 / 3 = 83668).
One quarter of 251004 is 62751 (251004 / 4 = 62751).
One fifth of 251004 is 50200,8 (251004 / 5 = 50200,8 = 50200 4/5).
One sixth of 251004 is 41834 (251004 / 6 = 41834).
One seventh of 251004 is 35857,7143 (251004 / 7 = 35857,7143 = 35857 5/7).
One eighth of 251004 is 31375,5 (251004 / 8 = 31375,5 = 31375 1/2).
One ninth of 251004 is 27889,3333 (251004 / 9 = 27889,3333 = 27889 1/3).
show fractions by 6, 7, 8, 9 ...
251004
### Advanced math operations
#### Is Prime?
The number 251004 is not a prime number. The closest prime numbers are 251003, 251033.
#### Factorization and factors (dividers)
The prime factors of 251004 are 2 * 2 * 3 * 13 * 1609
The factors of 251004 are
1 , 2 , 3 , 4 , 6 , 12 , 13 , 26 , 39 , 52 , 78 , 156 , 1609 , 3218 , 4827 , 6436 , 9654 , 19308 , 20917 , 41834 , 251004 show more factors ...
Total factors 24.
Sum of factors 631120 (380116).
#### Powers
The second power of 2510042 is 63.003.008.016.
The third power of 2510043 is 15.814.007.024.048.064.
#### Roots
The square root √251004 is 501,002994.
The cube root of 3251004 is 63,080271.
#### Logarithms
The natural logarithm of No. ln 251004 = loge 251004 = 12,433224.
The logarithm to base 10 of No. log10 251004 = 5,399681.
The Napierian logarithm of No. log1/e 251004 = -12,433224.
### Trigonometric functions
The cosine of 251004 is -0,985286.
The sine of 251004 is -0,170913.
The tangent of 251004 is 0,173465.
### Properties of the number 251004
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 251004 in Computer Science
Code typeCode value
PIN 251004 It's recommendable to use 251004 as a password or PIN.
251004 Number of bytes245.1KB
CSS Color
#251004 hexadecimal to red, green and blue (RGB) (37, 16, 4)
Unix timeUnix time 251004 is equal to Saturday Jan. 3, 1970, 9:43:24 p.m. GMT
IPv4, IPv6Number 251004 internet address in dotted format v4 0.3.212.124, v6 ::3:d47c
251004 Decimal = 111101010001111100 Binary
251004 Decimal = 110202022110 Ternary
251004 Decimal = 752174 Octal
251004 Decimal = 3D47C Hexadecimal (0x3d47c hex)
251004 BASE64MjUxMDA0
251004 MD55429c47f7f7dbd07133badec95bba5be
251004 SHA129b26fcada1c168a706012ad00be698894645aa8
251004 SHA224f956024bb9b966c295fc18dd6602e6e35aa5e604a66fbd3ac1042f82
251004 SHA256929f938afbd3fd8997e30f2ef362ef16ef54ea36e8c672edd622e3fe1676a3cf
251004 SHA3844a69c892432bf1349281c45c5a245d24c1bae09f6042ea5852cdfb62d3d7c033879708934a6ccb9f232ced3cc2c5caee
More SHA codes related to the number 251004 ...
If you know something interesting about the 251004 number that you did not find on this page, do not hesitate to write us here.
## Numerology 251004
### Character frequency in number 251004
Character (importance) frequency for numerology.
Character: Frequency: 2 1 5 1 1 1 0 2 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 251004, the numbers 2+5+1+0+0+4 = 1+2 = 3 are added and the meaning of the number 3 is sought.
## Interesting facts about the number 251004
### Asteroids
• (251004) 2006 PO3 is asteroid number 251004. It was discovered by NEAT, Near Earth Asteroid Tracking from Palomar Mountains Observatory on 8/12/2006.
## № 251,004 in other languages
How to say or write the number two hundred and fifty-one thousand and four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 251.004) doscientos cincuenta y uno mil cuatro German: 🔊 (Nummer 251.004) zweihunderteinundfünfzigtausendvier French: 🔊 (nombre 251 004) deux cent cinquante et un mille quatre Portuguese: 🔊 (número 251 004) duzentos e cinquenta e um mil e quatro Hindi: 🔊 (संख्या 251 004) दो लाख, इक्यावन हज़ार, चार Chinese: 🔊 (数 251 004) 二十五万一千零四 Arabian: 🔊 (عدد 251,004) مئتان و واحد و خمسون ألفاً و أربعة Czech: 🔊 (číslo 251 004) dvěstě padesát jedna tisíc čtyři Korean: 🔊 (번호 251,004) 이십오만 천사 Danish: 🔊 (nummer 251 004) tohundrede og enoghalvtredstusindfire Dutch: 🔊 (nummer 251 004) tweehonderdeenenvijftigduizendvier Japanese: 🔊 (数 251,004) 二十五万千四 Indonesian: 🔊 (jumlah 251.004) dua ratus lima puluh satu ribu empat Italian: 🔊 (numero 251 004) duecentocinquantunomilaquattro Norwegian: 🔊 (nummer 251 004) to hundre og femti-en tusen og fire Polish: 🔊 (liczba 251 004) dwieście pięćdziesiąt jeden tysięcy cztery Russian: 🔊 (номер 251 004) двести пятьдесят одна тысяча четыре Turkish: 🔊 (numara 251,004) ikiyüzellibindört Thai: 🔊 (จำนวน 251 004) สองแสนห้าหมื่นหนึ่งพันสี่ Ukrainian: 🔊 (номер 251 004) двiстi п'ятдесят одна тисяча чотири Vietnamese: 🔊 (con số 251.004) hai trăm năm mươi mốt nghìn lẻ bốn Other languages ...
## News to email
#### Receive news about "Number 251004" to email
I have read Privacy Policy
## Comment
If you know something interesting about the number 251004 or any natural number (positive integer) please write us here or on facebook.
#### Comment (Maximum 2000 characters) *
* Required fields
Legal Notices & Terms of Use
The content of the comments is the opinion of the users not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illicit or harmful to third parties. Number.academy reserves the right to remove or not make public any inappropriate comment. It also reserves the right to post a comment on another topic. Privacy Policy.
### Comments
add new comment
For this topic there are no comments.
| 2,683
| 8,080
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2022-49
|
latest
|
en
| 0.665381
|
https://math.answers.com/Q/What_does_a_Measure_look_like
| 1,695,818,527,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00138.warc.gz
| 426,036,308
| 46,989
|
0
# What does a Measure look like?
Updated: 5/30/2023
Wiki User
11y ago
A liquid measure looks like wherever the liquid comes to, inside a measuring cup.
Liquid measures are different from "dry" measures. An ounce of flour is not the same as an ounce of water. One is by weight, the other is by volume.
FriPilot
Emma Goldner
Lvl 9
4mo ago
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.82
3068 Reviews
Earn +20 pts
Q: What does a Measure look like?
Submit
Still have questions?
Related questions
### What does 1 gram look like?
A gram is a measure of mass - it does not "look" like anything.
### What does a protractor look like?
This is a protractor. It helps you measure angles.
### What is 0.59 inches?
What does .059 look like on tape measure
### What does an anemometer look like?
a thingy that has cups to measure the wind speed
### What does 1250 inches look like on a tape measure?
It is unlikely that a tape measure will extend that long, but 104.166... feet.
### What woul a ruler used to measure the volume of a cone look like?
It would look like a regular ruler...Measure the height - hMeasure the radius of the base (1/2 the diameter - rVolume= 1/3 'pi' r2h
### What do a gram look like?
A gram is a unit of measure, like an ounce or a pound. It only exists in the sense of measuring something.
### What does a measure symbol look like?
you make a p then add a swirl at the end of the p's hump
### What does 2 cm inches look like on a tape measure?
2 inches = 5.08 cm
### What does 550 mgs look like?
550 milligrams is a measure of weight equal to 0.02 ounces.
### What does a calorimeter look like and what are its uses?
It is used to measure the energy in food and the loss of energy in the air.
### How do you measure length with a tape measure?
strectch it out and look at the numbers
| 510
| 1,943
|
{"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-2023-40
|
latest
|
en
| 0.949062
|
https://www.construct.net/forum/construct-classic/construct-classic-discussion-37/set-z-order-with-number-32377
| 1,531,727,191,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00466.warc.gz
| 866,787,961
| 10,695
|
0 Favourites
# Set Z-Order With Number
This forum is currently in read-only mode.
• 11 posts
• I think it would be nice if you could set the Z-order of objects with a number. For example, if object one has a Z-order of -1 and object 2 has one of 0, object 2 will appear ahead of object one. This would be handy for things like isometric games where you could set the Z-order of all objects to their Y coordinate. Game Maker handles Z-order using numbers like this, and it really helps for ordering everything correctly.
• You can sort stuff based on the y position by adding all the objects that need to be sorted to a family, then use a for-each ordered condition using family.y and send the family to front or back(can't remember which at the moment).
• So you're saying that when you do a for each loop on a family of objects the loop starts with the object that has the highest (or lowest?) y position? That's handy to know if that's the case, but it's not well-documented and it's certainly not intuitively obvious.
I think a nice way to handle this would be to have an option to automatically z sort items by y-coordinate. Linkman's suggestion would give more flexibility but I think the iso game z-sorting issue is the main motivator here and this auto sort would work well for that problem. Also, maybe you could auto z-sort just the objects involved in the collision. So on (for example) a busy isometric rts layout you'll save a lot of unnecessary z-sorting.
• I know you can use the For Each Ordered condition, but that could eat up processing time if you had a lot of objects to loop through.
• 'For each ordered' is a good way - it's not CPU intensive if it's not nested with any other loops.
Sort by Y co-ordinate:
+ For each Sprite order by Sprite.Y ascending
-> Sprite: send to front
Sort by any old number:
+ For each Sprite order by Sprite('some_variable_with_z_number')
-> Sprite: send to front
Then you can set any old number to the 'some_variable_with_z_number' private variable, and it'll sort the Z order from that.
• So many times, so many ways Ashley reminds us that in Construct you can do a lot with what you've already got.
Hey that was almost a poem. Here, I'll give it to you in Haiku form ... nevermind.
• Hi. I am displaying an isometric tilemap... with the (0,0) tile (on the grid) mapped top centre. This means that the required Z-order for any particular tile representing a grid location is given by Z=x+y... So each 'diagonal on the grid, horizontal on the screen' row will have tiles with the same Zorder value - they cannot overlap so that is fine and normal for iso.
If I replace a tile somewhere then I could just set the Zorder for the newly created tile as I know what value I want to give it.... but Construct2 does not seem to let me do that ?
Is there any way to avoid having to loop over the entire grid everytime I replace a single tile - the visible grid is 20x20, so this a non-trivial nested pair.
I was expecting to find a settable property: Family.Zorder = <value> and be able to just set it myself! Please advise how I can do this.
Cheers
• Hmmm... replying (without an answer) to my own post. The Layers are supposed to be very efficient... As I have 40 distinct tiling Zorders, would it be reasonable and/or sensible (!) to create 40 layers for the Zorders of iso tiles and then set the layer number programmatically, which does seem to be possible.
Or perhaps that will have some unpleasant side-effect such as greatly increasing number of draw calls ?
• mrexcessive - This is a five year old thread concerning Construct Classic, but I presume you're using Construct 2. You'd be much better off asking your question in the Construct 2 help section.
Please be sure you check how old(and relevant) a thread is before asking questions in the future.
Develop games in your browser. Powerful, performant & highly capable.
Construct 3 users don't see these ads
• Yer... Well I did search around and this seemed to be the most relevant... didn't notice it was for Classic... Apologies! Will repost on a new thread as you suggest... thank you.
• 'For each ordered' is a good way - it's not CPU intensive if it's not nested with any other loops.
Sort by Y co-ordinate:
+ For each Sprite order by Sprite.Y ascending
-> Sprite: send to front
Just touching up on this old subject. With the loop above, it only works for instances of an object because you are forced to pick one. In Construct 2 (not classic), to have the loop set Z ordering for multiple objects, create a family of those objects then:
+ For each Family order by Family.Y ascending
-> Family: send to front
The only caveat is they all have to be in the same layer.
• 11 posts
| 1,084
| 4,722
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2018-30
|
latest
|
en
| 0.958398
|
https://www.physicsforums.com/threads/time-dilation-and-length-contraction.676318/
| 1,669,455,458,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00384.warc.gz
| 1,011,822,213
| 16,462
|
# Time dilation and length contraction
VertexOperator
I am confused about these two concupiscences of the consistency of light.
One of my books/notes says that time dilation and length contraction do not happen at the same time because they are 'the same thing' which kind of makes sense but the other says "when we reach relativistic velocities both time dilation and length contraction allow us to reach our destination faster because less time passes for the astronauts and less distance needs to be covered. However, mass dilation is one of the reasons that achieving relativistic speed is very difficult, as more and more energy is converted into mass as the speed increases..."
So can someone please explain this to me :)
Thank you!
clamtrox
... but the other says "when we reach relativistic velocities both time dilation and length contraction allow us to reach our destination faster because less time passes for the astronauts and less distance needs to be covered. However, mass dilation is one of the reasons that achieving relativistic speed is very difficult, as more and more energy is converted into mass as the speed increases..."
I wouldn't read this source if I were you. It's very confusing, and using an archaic formalism.
Consider a spaceship traveling from Earth to Alpha Centauri. From Earth's inertial frame, the distance the rocket covers is ~4 ly, but the clocks on board the spaceship run slower. This is time dilation (moving clocks run more slowly)
From the spaceship frame, the distance Earth and Alpha Centauri is shorter than 4 ly. This is length contraction (distances are shorter in a moving reference frame).
VertexOperator
Wow, I had a huge conceptual misunderstanding. Thank you so much :)
This is what I wrote in my notes:
"The nearest star to us, Proxima Centauri, is about 4.3 light years away. Travelling at the fastest speed any space probe has ever gone, it would take us over 42,000 years to reach it. At these speeds, it would take about 265 million years to reach the centre of our galaxy, and traveling beyond our galaxy would take inconceivable amounts of time.
However, space scientists have not given up. Designers of a new kind of spacecraft called a light sail make the claim that these craft could achieve speeds of up to 0.1c. In the future, engineers may be able to design spacecraft that can reach speeds much closer to the speed of light, and this would mean that the effects of time dilation and length contraction would become significant, and work in our favour. From the Earth's inertial frame, the distance the rocket covers is 4.3 light years, but time passes slower on the spacecraft than on Earth, so for the astronaut’s, the journey would take less time.. This is time dilation (moving clocks run more slowly). From the spaceship’s frame, the distance Earth and Alpha Centauri would have contracted to a shorter distance than 4.3 light years. This is length contraction (distances are shorter in a moving reference frame). Therefore the distance is shorter, which they cover in less time.
However, assuming that such speeds are technically possible, the energy costs of achieving them would be prohibitive, since acceleration is always the most costly phase of a space mission This issue is made even worse by the effect of mass dilation, because as speeds approach the speed of light, even greater force and energy input is needed for only marginal increases."
Last edited:
clamtrox
That's correct. Note that observers in both inertial frames agree on how much the astronauts have aged; just the cause of the lack of aging is different in different frames.
VertexOperator
That's correct. Note that observers in both inertial frames agree on how much the astronauts have aged; just the cause of the lack of aging is different in different frames.
So ultimately both time dilation and length contraction lead to the same result. That is what they meant by 'they are the same'. This is a much better explanation :) Thank you!
clamtrox
So ultimately both time dilation and length contraction lead to the same result. That is what they meant by 'they are the same'. This is a much better explanation :) Thank you!
Yes. A "deeper" explanation would be that there are things which do not depend on the coordinate system (like for example how wrinkly are the astronauts when they reach their destination). In different coordinate systems, space and time coordinates are different, and therefore the thing one coordinate system considers as "time" can be a mixture of "space" and "time" in another. This is why one observer calls the phenomenon time dilation, whereas another calls it length contraction.
Suraj Nehra
I wouldn't read this source if I were you. It's very confusing, and using an archaic formalism.
Consider a spaceship traveling from Earth to Alpha Centauri. From Earth's inertial frame, the distance the rocket covers is ~4 ly, but the clocks on board the spaceship run slower. This is time dilation (moving clocks run more slowly)
From the spaceship frame, the distance Earth and Alpha Centauri is shorter than 4 ly. This is length contraction (distances are shorter in a moving reference frame).
As far as I know it is the length of moving object that contract and not that of the reference point at rest
Staff Emeritus
| 1,098
| 5,312
|
{"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-2022-49
|
latest
|
en
| 0.960121
|
https://serverlogic3.com/what-type-of-data-is-categorical/
| 1,695,716,762,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00683.warc.gz
| 560,346,891
| 15,989
|
# What Type of Data Is Categorical?
//
Heather Bennett
What Type of Data Is Categorical?
When working with data, it is essential to understand the different types of data you may encounter. One common classification is categorical data.
Categorical data represents characteristics or qualities that can be divided into distinct categories or groups. In this article, we will explore what categorical data is and provide examples to help you better understand its significance.
## Defining Categorical Data
Categorical data, also known as qualitative or nominal data, consists of variables that can be divided into specific categories based on their characteristics. These categories are typically non-numeric and represent different groups or classes.
Unlike numerical data, categorical variables cannot be measured on a continuous scale.
Categorical data can further be classified into two subtypes: nominal and ordinal.
### Nominal Data
Nominal data represents variables that have distinct categories but do not have any inherent order or ranking among them. For example, consider a dataset containing information about eye color.
The categories in this case would include brown, blue, green, and hazel. Each category is unique and cannot be compared in terms of magnitude.
To represent nominal data in HTML, you can use unordered lists (
) with list items (
• ). Here’s an example:
• Brown
• Blue
• Green
• Hazel
### Ordinal Data
Ordinal data shares similarities with nominal data but includes an additional characteristic – order or ranking among the categories. The categories in ordinal data have an inherent order but do not have a consistent difference between them. For instance, consider a survey asking for ratings of a restaurant experience on a scale of 1 to 5.
The categories here are “poor,” “fair,” “good,” “very good,” and “excellent.” While there is an order, the difference between each rating is not uniform.
To represent ordinal data in HTML, you can use ordered lists (
) with list items (
1. ). Here’s an example:
1. Poor
2. Fair
3. Good
4. Very Good
5. Excellent
## Why Categorical Data Matters
Understanding categorical data is crucial because it affects the type of analysis that can be performed. Categorical data requires different statistical techniques and visualization methods compared to numerical data.
Analyzing categorical data often involves calculating frequencies, proportions, and conducting tests of association or independence between variables.
By recognizing whether your data is categorical, you can determine the appropriate approaches for analysis and draw accurate conclusions. Categorical data plays a significant role in various fields such as market research, social sciences, healthcare, and more.
## Conclusion
In summary, categorical data represents variables that can be divided into distinct categories or groups without any inherent order (nominal) or with an inherent order (ordinal). Recognizing categorical data is essential for choosing appropriate analysis methods and drawing accurate conclusions.
By utilizing HTML elements such as unordered lists (
) and ordered lists (
), you can effectively present and organize categorical data in your web content.
| 606
| 3,249
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.886364
|
https://www.airmilescalculator.com/distance/pkc-to-fyj/
| 1,607,003,134,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141727782.88/warc/CC-MAIN-20201203124807-20201203154807-00218.warc.gz
| 442,879,573
| 7,825
|
Distance between Petropavlovsk-Kamchatsky (PKC) and Fuyuan (FYJ)
Flight distance from Petropavlovsk-Kamchatsky to Fuyuan (Petropavlovsk-Kamchatsky Airport – Fuyuan Dongji Airport) is 1106 miles / 1780 kilometers / 961 nautical miles. Estimated flight time is 2 hours 35 minutes.
Map of flight path from Petropavlovsk-Kamchatsky to Fuyuan.
Shortest flight path between Petropavlovsk-Kamchatsky Airport (PKC) and Fuyuan Dongji Airport (FYJ).
How far is Fuyuan from Petropavlovsk-Kamchatsky?
There are several ways to calculate distances between Petropavlovsk-Kamchatsky and Fuyuan. Here are two common methods:
Vincenty's formula (applied above)
• 1105.961 miles
• 1779.871 kilometers
• 961.054 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
• 1102.800 miles
• 1774.785 kilometers
• 958.307 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 Petropavlovsk-Kamchatsky Airport
City: Petropavlovsk-Kamchatsky
Country: Russia
IATA Code: PKC
ICAO Code: UHPP
Coordinates: 53°10′4″N, 158°27′14″E
B Fuyuan Dongji Airport
City: Fuyuan
Country: China
IATA Code: FYJ
ICAO Code: ZYFY
Coordinates: 48°11′58″N, 134°21′59″E
Time difference and current local times
The time difference between Petropavlovsk-Kamchatsky and Fuyuan is 4 hours. Fuyuan is 4 hours behind Petropavlovsk-Kamchatsky.
+12
CST
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 157 kg (346 pounds).
Frequent Flyer Miles Calculator
Petropavlovsk-Kamchatsky (PKC) → Fuyuan (FYJ).
Distance:
1106
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
1106
Round trip?
| 561
| 1,873
|
{"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-2020-50
|
latest
|
en
| 0.79941
|
https://discourse.julialang.org/t/trying-to-implement-millers-recurrence-algorithm-and-getting-terrible-results/98574
| 1,686,264,666,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00687.warc.gz
| 251,965,329
| 6,299
|
# Trying to implement Miller's recurrence algorithm and getting terrible results
I am currently trying to implement Miller’s algorithm but Im getting terrible numerical results, numerical errors are getting to 80% of original values. Is there an issue with my code?
``````function besselj(orderLim, arg::BigFloat, tol)
N = orderLim * tol
values = Vector{BigFloat}(undef, N+2)
values[N+1] = 1
values[N+2] = 0
for i in N+1:-1:2
values[i-1] = (2 * i / arg) * values[i] - values[i+1]
end
norm = values[1]
for i in 2:N+2
if (i-1) % 2 == 0
norm += 2*values[i]
end
end
norm = 1/norm
values = values[1:orderLim+1]
return norm.*values
end
``````
This won’t help with accuracy, but your second loop could be written
`````` for i in 3:2:N+2
norm += 2*values[i]
end
``````
1 Like
thanks! changed it.
``````values[i-1] = (2 * i / arg) * values[i] - values[i+1]
``````
does not account for the index offset.
It must be
``````values[i-1] = (2 * (i - 1) / arg) * values[i] - values[i+1]
``````
1 Like
yep, that solved it! Dumbest of mistakes , sorry to bother and thank you guys very much @Oscar_Smith and @Vasily_Pisarev
I didn’t look too carefully at the code but it’s possible to do this with just one loop over the data. You should probably unroll your loop by a factor of two and then just add every second element to your accumulator. This will actually be more accurate as you are summing the values small to large. The issue is that you need to generate appropriate trial values of the ratio of Jnu/Jnu+1 first. This usually includes generating these trial rates with the starter values like you did but you need to actually throw away those values. You need to figure out the appropriate starting value before starting downward recurrence which is some `M` above your highest order or largest argument. The algorithm is rather slowly converging for large argument but the issue is that the starter `M` is a function of your desired precision. For double precision it’s usually like 7-10 but for big float precision it depends. I don’t think this algorithm is super amendable to arbitrary precision calculations because of that but you can determine the factor for a desired precision if you want say Float128 precision. But if your arguments are much larger than the order you’ll probably want to generate starting values differently than the trial recurrence method.
Heeeey, you again , first of all every insight that you give is awesome and SUPER helpful, thank you so much, seriously. Ive found an article to change the trial values calculation by R.E. Scraton that should make precision better and other on error analysis by F. W. J. Olver in order to define the right `M` that I want and not just a nonsensical multiplication based on nothing as I was doing, I will try to implement those.
| 720
| 2,802
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.046875
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.877876
|
http://www.emaildata.pro/course-notes/Math-Myp-Term-Paper-45360451.html
| 1,508,831,693,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00395.warc.gz
| 421,637,569
| 16,821
|
# Math MYP Term Paper
Only available on StudyMode
• Published : December 3, 2013
Text Preview
YEAR 8 MATHEMATICS
TRIMESTER 2
UNIT: VOLUME
CRITERIA C & D WILL BE ASSESSED. SEE TASK SPECIFIC RUBRICS ATTACHED. Use A4 paper for your answers. Your work should not exceed 4 pages. Make your explanations logical, concise and complete and comment on the accuracy of your answers.
Afya Ltd is a company that manufactures medicinal drugs and food supplements in pill form. As part of its new strategy, the Chief Executive Officer has decided to redesign its packaging. The plan is to package as many pills as possible into a box which takes the shape shown below.
One day, the CEO asks to meet the Chief Marketing Officer (CMO) - the person who came up with the new packaging. When asked to explain how he got the highest number of pills that could fit into the box, the CMO’s reply was, “..to get the maximum number of the pills that could fit into the new box, I simply divided the volume of the box by the volume of one pill.” Before he could finish the sentence, the Chief Designer who was also present in the meeting raised his hand and yelled, “No! I disagree with that method.” At this point, the CEO decides to end the meeting. He however calls you 20 minutes later to ask for your opinion as the Chief Scientist in the firm. In the telephone conversation he asks you to write a report that refutes or justifies the method suggested by the Chief Marketing Officer . “I would like you to include all your calculations in your report so I may understand how you arrived at your final conclusion”, he adds.
| 366
| 1,610
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2017-43
|
latest
|
en
| 0.969288
|
https://nerd-notes.com/ubq/19554/
| 1,716,024,285,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00289.warc.gz
| 370,830,396
| 133,645
|
# You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
1. (a) The maximum height reached by the stone (the distance above the cliff). (3 points)
2. (b) The time taken by the stone to reach its maximum height. (2 points)
3. (c) The height of the cliff. (2 points)
1. 3.26 m above the cliff
2. .82 s
3. 20.15 m
0
1. Maximum Height Reached by the Stone
Step Formula Derivation Reasoning
1 v^2 = u^2 + 2as Kinematic equation, with v as final velocity (0 at max height), u as initial velocity, a as acceleration (gravity), and s as displacement.
2 0 = (8.0, \text{m/s})^2 – 2 \times 9.81, \text{m/s}^2 \times s_{\text{max}} At max height, final velocity is 0.
3 Solve for s_{\text{max}} Maximum height above the cliff.
2. Time to Reach Maximum Height
Step Formula Derivation Reasoning
1 v = u + at Kinematic equation for velocity.
2 0 = 8.0, \text{m/s} – 9.81, \text{m/s}^2 \times t_{\text{max}} Final velocity is 0 at max height.
3 Solve for t_{\text{max}} Time to reach max height.
3. Height of the Cliff
Total Time from Throw to Sea
Step Formula Derivation Reasoning
1 s = ut + \frac{1}{2}at^2 Displacement formula for the entire journey, where s is total displacement (height of the cliff), u is initial velocity, t is total time, and a is acceleration (gravity). Note gravity is negative in this case.
2 s_{\text{cliff}} = 8.0, \text{m/s} \times 3.0, \text{s} – \frac{1}{2} \times 9.81, \text{m/s}^2 \times (3.0, \text{s})^2 Substituting the values for u, t, and g.
3 Solve for s_{\text{cliff}} Height of the cliff.
Performing the calculations above yield the following results:
1. Maximum height reached by the stone above the cliff: \boxed{3.26, \text{m}}
2. Time taken by the stone to reach its maximum height: \boxed{0.82, \text{s}}
3. Height of the cliff: \boxed{20.15, \text{m}}
## Need Help On This? Ask Phy To Explain.
Phy can also check your working. Just snap a picture!
Simple Chat Box
## Discover how students preformed on this question | Coming Soon
1. 3.26 m above the cliff
2. .82 s
3. 20.15 m
## Continue with
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
Enjoying UBQ? Share the 🔗 with friends!
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
1. Start with the given measurement: \text{5 km}
2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix
Symbol
Power of Ten
Equivalent
Pico-
p
10^{-12}
Nano-
n
10^{-9}
Micro-
µ
10^{-6}
Milli-
m
10^{-3}
Centi-
c
10^{-2}
Deci-
d
10^{-1}
(Base unit)
10^{0}
Deca- or Deka-
da
10^{1}
Hecto-
h
10^{2}
Kilo-
k
10^{3}
Mega-
M
10^{6}
Giga-
G
10^{9}
Tera-
T
10^{12}
1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!
## Phy Pro
The most advanced version of Phy. Currently 50% off, for early supporters.
## \$11.99
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
| 1,898
| 5,773
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2024-22
|
longest
|
en
| 0.845898
|
https://justaaa.com/chemistry/49595-a-420-mg-sample-of-carbon-reacts-with-sulfur-to
| 1,723,735,179,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00533.warc.gz
| 256,227,410
| 9,395
|
Question
# A 42.0 mg sample of carbon reacts with sulfur to form 117 mg of the compound....
A 42.0 mg sample of carbon reacts with sulfur to form 117 mg of the compound. What is the empirical formula of the carbon sulfide?
mass of C = 42mg = 0.042g
mass of S = mass of compound - mass of C
= 0.117-0.042 = 0.075 g
no of moles of C = W/G.A.Wt =0.042/12 = 0.0035 moles
no of moles of S = W/G.A.Wt = 0.075/32 = 0.00234375moles
C :S = 0.0035 : 0.00234375
= 1.5 : 1
= 3 : 2
C3S2
The empirical formula = C3S2
#### Earn Coins
Coins can be redeemed for fabulous gifts.
##### Need Online Homework Help?
Most questions answered within 1 hours.
| 241
| 675
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.80797
|
https://dotnetperls.com/levenshtein-vbnet
| 1,696,412,129,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00865.warc.gz
| 229,437,675
| 3,795
|
Levenshtein Distance AlgorithmImplement Levenshtein distance, which tells us the number of String edits.
VB.NET
Levenshtein. In 1965 Vladmir Levenshtein introduced a distance algorithm. This tells us the number of changes (edits) one string must go through to become another string.
Function notes. In VB.NET we compute Levenshtein distance with a Function. This function returns the number of character edits that must occur to get from String A to String B.
Example. The internals of the Levenshtein distance Function are complex. Internally it allocates a 2D array. This 2D array has dimensions equal to the lengths of the 2 argument strings.
2D Array
Detail A nested For-loop is executed. In the inner block of the For-loops, the cost of the change of two elements is computed.
Note The cost is one for a different character and zero for the same character.
For Each, For
Then Array elements are assigned, considering adjacent array elements and the cost. The ending array element is returned as the distance.
Module Module1 Sub Main() Console.WriteLine(LevenshteinDistance("aunt", "ant")) Console.WriteLine(LevenshteinDistance("Sam", "Samantha")) Console.WriteLine(LevenshteinDistance("flomax", "volmax")) End Sub ''' <summary> ''' Compute LevenshteinDistance. ''' </summary> Public Function LevenshteinDistance(ByVal s As String, ByVal t As String) As Integer Dim n As Integer = s.Length Dim m As Integer = t.Length Dim d(n + 1, m + 1) As Integer If n = 0 Then Return m End If If m = 0 Then Return n End If Dim i As Integer Dim j As Integer For i = 0 To n d(i, 0) = i Next For j = 0 To m d(0, j) = j Next For i = 1 To n For j = 1 To m Dim cost As Integer If t(j - 1) = s(i - 1) Then cost = 0 Else cost = 1 End If d(i, j) = Math.Min(Math.Min(d(i - 1, j) + 1, d(i, j - 1) + 1), d(i - 1, j - 1) + cost) Next Next Return d(n, m) End Function End Module
1 5 3
Function output. The number returned by Levenshtein for the arguments "aunt" and "ant" is 1. This means only one edit is required to get from aunt to ant—the letter U is removed.
And For the second result, we have 5 edits. The name "Sam" must be edited 5 times to get "Samantha"—5 letters are added.
Finally And to get to "volmax" from "flomax" 3 edits must occur. Swapping 2 characters is considered 2 separate edits.
The theory of the Levenshtein distance algorithm is not covered here. But the implementation here is correct on the example Strings tested. It is not fast.
Certain optimizations could be used to enhance performance, such as memoization. A data structure such as a Dictionary could store the two strings and also a cached distance.
Dictionary
Dot Net Perls is a collection of tested code examples. Pages are continually updated to stay current, with code correctness a top priority.
Sam Allen is passionate about computer languages. In the past, his work has been recommended by Apple and Microsoft and he has studied computers at a selective university in the United States.
| 757
| 2,954
|
{"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-2023-40
|
longest
|
en
| 0.74604
|
https://brainmass.com/math/optimization/shortest-path-maximal-flow-problem-212786
| 1,708,800,544,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00455.warc.gz
| 143,293,087
| 6,857
|
Purchase Solution
Shortest Path and Maximal Flow Problem
Not what you're looking for?
Please see the attached file for the fully formatted problems.
1. Shortest Path and Maximal Flow Problem (30%): You are given the following directed network.
[NETWORK]
a. (Shortest Path Problem 15%) Let the numbers on the arcs represent distances and find the shortest path from node 1 to node 9.
b. (Maximal Flow Problem 15%) Let the numbers represent flow capacity and find the maximum flow from node 1 to node 9.
Solution Summary
This posting contains solution to following problems on Shortest Path and Maximal Flow technique.
Solution Preview
1. Shortest Path and Maximal Flow Problem (30%): You are given the following directed network.
a. (Shortest Path Problem 15%) Let the numbers on the arcs represent distances and find the shortest path from node 1 to node 9.
This is a shortest route problem. Following are the steps to follow.
1. Find the nearest node to the origin. Put the distance in a box by the node.
Nearest node to node 1 is node 2 at a distance of 3.
2. Find the next nearest node to the origin, and put the distance in a box by the node. In some cases, several paths will have to be checked to find the nearest ...
Probability Quiz
Some questions on probability
Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
Know Your Linear Equations
Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.
| 425
| 1,999
|
{"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}
| 3.40625
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.899022
|
https://www.physicsforums.com/threads/mathematically-solving-fourier-transform.630439/
| 1,713,290,789,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00028.warc.gz
| 853,133,533
| 15,193
|
# Mathematically solving fourier transform
• Robismyname
In summary, the conversation is about someone trying to understand the real-world applications of Fourier Transform by working on problems from a signals and systems book. They are specifically trying to find the Fourier Transform of a given signal and are using a step-by-step approach to solve it. However, they are having trouble understanding how the book arrived at the answer, and are advised to use u-substitution to come to the correct answer.
Robismyname
Since I lack the understand of real world applications of Fourier Transform in the real world I decided to buy a signals and systems book (Lathi) do some Fourier Transform problems and them do the same problem in Matlab.
The question in the book wants me to find the Fourier Transform of signal f(t) = e^-at; from 0 to T
I know in order to find FT I have to do the following:
T
step 1: F(w) = ∫ f(t) * e^-jwt dt
0
T
step 2: ∫ e^-(a+jw)t dt [combine like terms]
0
step 3: [ e^-(a+jw)0 ] - [ e^-(a+jw)T ] [integrate over 0 to T]
step 4: 1 - e^-(a+jw)T [solve]
The book says the answer is:
1-e^-(jw+a)T
----------------------------
jw+a
How did the book get the denominator section of jw+a? I can't get from step 4 to the book answer. What am I missing here?
## 1. What is the Fourier transform?
The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It converts a signal from its original domain (usually time or space) to a representation in the frequency domain.
## 2. What is the purpose of solving the Fourier transform?
The purpose of solving the Fourier transform is to analyze and understand the frequency components of a signal. It can be used to identify specific frequencies present in a signal, filter out unwanted frequencies, and even reconstruct a signal from its frequency components.
## 3. How is the Fourier transform calculated?
The Fourier transform is calculated using complex numbers and integration. It involves breaking down a signal into an infinite sum of sine and cosine functions with different frequencies and amplitudes.
## 4. What is the difference between the Fourier transform and the inverse Fourier transform?
The Fourier transform converts a signal from the original domain to the frequency domain, while the inverse Fourier transform converts it back from the frequency domain to the original domain. They are essentially inverse operations of each other.
## 5. In what fields is the Fourier transform commonly used?
The Fourier transform is commonly used in fields such as signal processing, image processing, quantum mechanics, and engineering. It has applications in audio and video compression, medical imaging, and solving differential equations, among others.
• Engineering and Comp Sci Homework Help
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
3
Views
1K
• Engineering and Comp Sci Homework Help
Replies
12
Views
928
• Engineering and Comp Sci Homework Help
Replies
1
Views
831
• Engineering and Comp Sci Homework Help
Replies
3
Views
1K
• Topology and Analysis
Replies
4
Views
268
• Engineering and Comp Sci Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
311
• Calculus and Beyond Homework Help
Replies
4
Views
6K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
| 786
| 3,369
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.925927
|
https://www.newbedev.com/algorithm/java/array/insert-delete-get-random/
| 1,603,213,108,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00215.warc.gz
| 832,856,827
| 11,705
|
# Insert Delete Get Random
Insert Delete Get Random array
Design a data structure that supports all following operations in average O(1) time.
insert(val): Inserts an item val to the set if not already present. remove(val): Removes an item val from the set if present. getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example: Init an empty set. RandomizedSet randomSet = new RandomizedSet();
Inserts 1 to the set. Returns true as 1 was inserted successfully. randomSet.insert(1);
Returns false as 2 does not exist in the set. randomSet.remove(2);
Inserts 2 to the set, returns true. Set now contains [1,2]. randomSet.insert(2);
getRandom should return either 1 or 2 randomly. randomSet.getRandom();
Removes 1 from the set, returns true. Set now contains [2]. randomSet.remove(1);
2 was already in the set, so return false. randomSet.insert(2);
Since 2 is the only number in the set, getRandom always return 2. randomSet.getRandom();
class RandomizedSet { HashMap map; ArrayList values;
``````/** Initialize your data structure here. */
public RandomizedSet() {
map = new HashMap<Integer, Integer>();
values = new ArrayList<Integer>();
}
/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
if(!map.containsKey(val)) {
map.put(val, val);
return true;
}
else {
return false;
}
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
if(map.containsKey(val)) {
map.remove(val);
values.remove(values.indexOf(val));
return true;
}
return false;
}
/** Get a random element from the set. */
public int getRandom() {
int random = (int)(Math.random() * values.size());
int valueToReturn = values.get(random);
return map.get(valueToReturn);
}
``````
}
Insert Delete Get Random Solution
| 443
| 1,933
|
{"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-2020-45
|
longest
|
en
| 0.474584
|
https://solvedlib.com/n/compute-the-values-of-dy-and-ay-for-the-function-ys1-6x,15378962
| 1,685,680,798,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00117.warc.gz
| 563,942,312
| 18,315
|
Compute the values of dy and Ay for the function yS1 + 6x given € 0 and Az = dr
Question:
Compute the values of dy and Ay for the function y S1 + 6x given € 0 and Az = dr 0.04 Round your ansivers four decimal places 'required. Number Ay = Number
Similar Solved Questions
Should preferred stock be classified as debt or equity? Does it matter if the classification is...
Should preferred stock be classified as debt or equity? Does it matter if the classification is made by the firm’s management, creditors, or equity investors? Answer must be lengthy!! No Plagiarism!! Plagiarism will be checked!! Answer all parts of question!!...
Question 16PointThe symmetric stretching mode for PCIz is of A1 symmetry In the Czv character table; there are zand (x2 +y, ) entries in the Ap row: This tells you that:the symmetric stretching mode of PClz is IR active and Raman activethe symmetric stretching mode of PClz is IR inactive and Raman inactiveThere is not enough information to answer the questionthe symmetric stretching mode of PClz is IR active and Raman inactivethe symmetric stretching mode of PClz is IR inactive and Raman activeQ
Question 16 Point The symmetric stretching mode for PCIz is of A1 symmetry In the Czv character table; there are zand (x2 +y, ) entries in the Ap row: This tells you that: the symmetric stretching mode of PClz is IR active and Raman active the symmetric stretching mode of PClz is IR inactive and Ram...
Distance vs time graph
You walk down the sidewalk to the east for 8.0 min at a speed of 1.2 m/s. You reach a busy street and have to stop. You remain at rest for 2 minutes. The traffic diesdown, so you run across the street at a constant speed. The street is 12 m wide and it takes you 1.5 s to cross it. You immediately sl...
I want mechanisms for the this reactions (Diels-Alder cycloaddition reaction ) ,with explain steps , please...
I want mechanisms for the this reactions (Diels-Alder cycloaddition reaction ) ,with explain steps , please I want a clear pic. ? Table 1. [AlCl3 + 2THF]-Catalyzed Diels-Alder Reaction of Ethyl Acrylate (la) with Isoprene (2) under SFC at 30 °C [AICIz + Ln] 5 mol % SFC, 30 °C, 12 h VIPT:...
Varcoe Corporation bases its budgets on the activity measure customers served. During September, the company planned to serve 30,500 customers, but actually served 25,500 customers. Revenue is $3.81 per customer served. Wages and salaries are$34,200 per month plus $1.21 per customer served. Supplie... 1 answer E the interest rate earned on a$2,400 deposit when $2,700 is paid back in one... e the interest rate earned on a$2,400 deposit when $2,700 is paid back in one year. (Round your answer to 2 declmal places.)... 5 answers Qne Anity pAvS a6 the end ol' each vear For %6 vears_ Another Anmity [avs 5 Athe PLa[ of each year for 18 years_ The present vales of both Aities are (Qal at eHlective rute mereSI I An amOunIt ol' money invested at the same rate will donble in yeaiS- tal 15, +4)" (iue | Qne Anity pAvS a6 the end ol' each vear For %6 vears_ Another Anmity [avs 5 Athe PLa[ of each year for 18 years_ The present vales of both Aities are (Qal at eHlective rute mereSI I An amOunIt ol' money invested at the same rate will donble in yeaiS- tal 15, +4)" (iue |... 1 answer A company has beginning inventory of 19 units at a cost of$19 each on February...
A company has beginning inventory of 19 units at a cost of $19 each on February 1. On February 3, it purchases 29 units at$21 each. 25 units are sold on February 5. Using the FIFO periodic inventory method, what is the cost of the 25 units that are sold? Multiple Choice $487$504 $491$494 ...
Which of the following sets of quantum numbers represents atom (I)? the last electron added to an iodineOn=5,/= 1,m = 2On=5,/-2,m = 3On-4,/= 1,m = 0On-4,/-1,m = 2On= 5,/= 1,m = 0
Which of the following sets of quantum numbers represents atom (I)? the last electron added to an iodine On=5,/= 1,m = 2 On=5,/-2,m = 3 On-4,/= 1,m = 0 On-4,/-1,m = 2 On= 5,/= 1,m = 0...
Acvenitres in Algebra VM: Tzi: js cor-pletely "oc:zoois]l RNiribe dcawnn: Eue: c; ~y aslc &x: Da]
Acvenitres in Algebra VM: Tzi: js cor-pletely "oc:zoois]l RNiribe dcawnn: Eue: c; ~y aslc &x: Da]...
What i the pH of a buffer that consists of 015 MNaHzPO4 and 0.45 M NazHPO4? For NaHzPO4 Ka =62 * 10-8Mutple Cholce6.518157156.917.63
What i the pH of a buffer that consists of 015 MNaHzPO4 and 0.45 M NazHPO4? For NaHzPO4 Ka =62 * 10-8 Mutple Cholce 6.51 815 715 6.91 7.63...
Wildhorse Inc.’s general ledger at April 30, 2021, included the following: Cash $6,600; Supplies$400; Equipment...
Wildhorse Inc.’s general ledger at April 30, 2021, included the following: Cash $6,600; Supplies$400; Equipment $21,000; Accounts Payable$2,100; Deferred Revenue (from gift certificates) $1,100; Bank Loan Payable$10,000; Common Shares $5,000; and Retained Earnings$9,800. The following tran...
Circle answers please A cylinder of mass 6.0 kg rolls without slipping on a horizontal surface....
Circle answers please A cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 7.0 m/s. (a) Determine the translational kinetic energy of its center of mass. (b) Determine the rotational kinetic energy about its center of mass. ...
Exercise 1: Probability Distribution Please give detailed steps for questions 2 & 3, I have seen...
Exercise 1: Probability Distribution Please give detailed steps for questions 2 & 3, I have seen the explanation before but it remains unclear. Exercice 1 Consider a random variable X with the following probabilities distribution: where α1 and α2 are parameters such that 0 < &alp...
Ddicleso (A; 1 8 marks W binomial marks) distribution , batch 400 machine - prtidlaces calticlate 12 probability othai 0 are 1 of these(b)by using the [EQJOU approximation to binomial distribution;Jathe Lontal approxitatjon i (b) teaccaable? 3 1 1 not?
ddicleso (A; 1 8 marks W binomial marks) distribution , batch 400 machine - prtidlaces calticlate 12 probability othai 0 are 1 of these (b)by using the [EQJOU approximation to binomial distribution; Jathe Lontal approxitatjon i (b) teaccaable? 3 1 1 not?...
The picture statement below represents the ASA Triangle Congruence Conjecture. Explain what the picture statement means.(Figure can't copy)
The picture statement below represents the ASA Triangle Congruence Conjecture. Explain what the picture statement means. (Figure can't copy)...
1. Lululemon purchased a new plant for their shoes production division on July 1, 2019. Lululemon...
1. Lululemon purchased a new plant for their shoes production division on July 1, 2019. Lululemon paid $5,000,000 for the production plant. It is a package purchase, which includes land, building and equipment. Currently, the land can be sold at$2,000,000, the building at 2,500,000 and the equipmen...
Problem 1, Determine the force in member BC, and state if this member is in tension...
Problem 1, Determine the force in member BC, and state if this member is in tension or zero force members in this truss? H compression. Are there 3 m T ys there aqre eo farces meaberg- E D B C JD GH -2m-+-2m2-+2m-+2m--2m--2m- e (61(2)+ 8(1)+ 5(2)+4(10)+ AS (12)-0이 4 kN 5kN 8KN 6kN 20 Un Uo...
Question 17 (0.5 points) Order the events related to MHCI processing: Although not necessarily part of the lecture, using the textbook to figure out what ERAP is will be helpfullMHCI passes through the GolgiCytotoxic T cells potentially interact with peptide-loaded MHCICytosolic proteins are degraded by a proteasome:MHCI is displayed on the surface of a cellMHCI loads with peptidePeptides are "fed" into the ER lumen via the TAP transporter:ERAP can remove extra peptide amino acid to ma
Question 17 (0.5 points) Order the events related to MHCI processing: Although not necessarily part of the lecture, using the textbook to figure out what ERAP is will be helpfull MHCI passes through the Golgi Cytotoxic T cells potentially interact with peptide-loaded MHCI Cytosolic proteins are degr...
Table 7.3: Elastic collision Part IL: Mass of glider I: m, 01L 35 Mass of glider 2: mz = 0L35 kgTotal momentum before collision (kg ms)Total momentum after collision Ikg"" mls)Initial velocities (mls)Final velocities (m/s)0.493 ~0.508 -0.512 0,500CALCULATiOnS:
Table 7.3: Elastic collision Part IL: Mass of glider I: m, 01L 35 Mass of glider 2: mz = 0L35 kg Total momentum before collision (kg ms) Total momentum after collision Ikg"" mls) Initial velocities (mls) Final velocities (m/s) 0.493 ~0.508 -0.512 0,500 CALCULATiOnS:...
Consider the exact differential equationI _ 2 cy _ 2 + (4y? _ 22 +1)y' _ 0The implicit solution of the above DE is f(z,y) = where f(r,y) =
Consider the exact differential equation I _ 2 cy _ 2 + (4y? _ 22 +1)y' _ 0 The implicit solution of the above DE is f(z,y) = where f(r,y) =...
Simplify the complex rational expression6 5 6x + 6y5 & 6x 6y
Simplify the complex rational expression 6 5 6x + 6y 5 & 6x # 6y...
When 5 radical 20 is written in simplest form, the result is k radical 5 what is the value of k
when 5 radical 20 is written in simplest form, the result is k radical 5 what is the value of k?1) 20 2) 103) 74) 4...
A four-resistor circuit is shown in the figure. The values of the resistors are as follows:...
A four-resistor circuit is shown in the figure. The values of the resistors are as follows: R1=3 Ohms, R2=R3=R4=4 Ohms. What is the equivalent resistance Req for this circuit?...
Use the RernnccUUr n Drtant Vhc nerdorthiquclnConsidering only ions wIth charges of +. 22-Land sclenide ion; Sc?Five the symbols for = species that ate Isuelectronic with the OE Ma dlomsmoro gr oup anoniins rcninintngotry Crndo Ooln
Use the Rernncc UUr n Drtant Vhc nerdor thiqucln Considering only ions wIth charges of +. 22-Land sclenide ion; Sc? Five the symbols for = species that ate Isuelectronic with the OE Ma dloms moro gr oup anoniins rcninintng otry Crndo Ooln...
Questlon 14Let N = f (t) be the total number laptops that have becn sold particularston belwccn midnight and t minutes on Black Friday: aiter midnlght Suppose that f Is modeled by a ditferentiable increasing (unction: Select which- senence best supportedby the equation (f-') (30)YoudoNOT have show work (or this questionBetween 12;30 AMJnd 12-40 AM Jptror matcyy Feecd Wiculdaminnsbetn sckdd it took approxiitc / MinnucHhcner lnclojlaotopthdhccntald ook aocforutck 30sccurd:lattonAlaptor40Jevet
Questlon 14 Let N = f (t) be the total number laptops that have becn sold particularston belwccn midnight and t minutes on Black Friday: aiter midnlght Suppose that f Is modeled by a ditferentiable increasing (unction: Select which- senence best supportedby the equation (f-') (30) YoudoNOT have...
F(x) ~X2 + 3x and g(x) = X - 3
f(x) ~X2 + 3x and g(x) = X - 3...
(25 points) Factorize Write a String method factorize(int n, int m) that when given a positive...
(25 points) Factorize Write a String method factorize(int n, int m) that when given a positive integer n and an integer m, returns a String representation of how many times n factors into m. Must use a while loop to get full credit. (25 points) Print Product Table Write a void method printProductTab...
If 50.1 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation...
If 50.1 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be formed? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)...
Use the molecular orbital energy diagram below tO answer the questions about bond order for the positive ion BzNumber of Bonding Number of Antibonding (B2 Valence Electrons Valence Electrons Bond OrderThis corresponds to:Single bond Double bondBetween single and double bond Between double and triple bondTriple bondNo bond, B2 does not form:Half of bondboronMO'sboron B
Use the molecular orbital energy diagram below tO answer the questions about bond order for the positive ion Bz Number of Bonding Number of Antibonding (B2 Valence Electrons Valence Electrons Bond Order This corresponds to: Single bond Double bond Between single and double bond Between double and tr...
| 3,555
| 12,164
|
{"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": 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}
| 2.65625
| 3
|
CC-MAIN-2023-23
|
longest
|
en
| 0.87292
|
https://projekter.aau.dk/projekter/en/studentthesis/caal-20(632452cd-ca11-44f3-9d67-0f3cf83ba0b1).html
| 1,669,628,399,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00233.warc.gz
| 511,546,573
| 5,119
|
## CAAL 2.0: Recursive HML, Distinguishing Formulae, Equivalence Collapses and Parallel Fixed-Point Computations
Student thesis: Master thesis (including HD thesis)
• Søren Enevoldsen
• Simon Reedtz Olesen
• Jacob Karstensen Wortmann
4. term, Software, Master (Master Programme)
This report documents the addition of new features to CAAL. Previous work on CAAL resulted in a web application used for teaching students about reactive systems. The tool supports description of processes in the process algebra Calculus of Communicating Systems. It is possible to verify the presence or absence of various preorders and equivalence between processes, and the tool offers to play against the user in bisimulation games to prove to the user the validity of a result. The tool also supports model checking of recursive Hennessy-Milner formulae. All computations are done by reducing problems to dependency graphs, and interpreting the computed fixed-point assignments, which are computed by an "on-the-fly" algorithm.
We extend CAAL with full syntax and semantics of HML formulae and offer reductions from HML formulae to dependency graphs. To complement the bisimulation games, we define the HML game and show that satisfiability is related to which player has a universal winning strategy. We also show that universal winning strategy corresponds to fixed-point assignments on the dependency graphs, and in doing so also shows the relationship between satisfiability and fixed-point assignments.
As an alternative to playing bisimulation games, we show it is possible to compute a distinguishing formula for two non-bisimilar processes. Finding the simplest possible distinguishing formula turns out difficult and we conjecture the decidability problem of whether a simple formula exists is NP-hard. Instead we present a greedy algorithm simplifies some formulae.
In CAAL it is possible to view a visual representation of processes. Processes can become quite complex and difficult to understand. Equivalence collapses may simplify the state system and allow the user to focus on particular aspects of a process. We present the theory and an algorithm to collapse processes based on merging equivalent processes into equivalence classes.
As an experiment we attempted to reuse the code in CAAL for a parallel algorithm. We document our design, pseudocode, and test result. The results were unsatisfactorily, but we identify possible reasons for this.
Language English 9 Jun 2015 108
ID: 213873137
| 493
| 2,499
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2022-49
|
latest
|
en
| 0.915891
|
https://www.physicsforums.com/threads/substituting-r-t-for-v-or-dr-dt.856751/
| 1,532,021,844,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00257.warc.gz
| 956,912,840
| 12,975
|
Substituting r/t for v or dr/dt?
Tags:
1. Feb 11, 2016
Mr Davis 97
I am a little confused about how we define the kinematic quantities that are rates. What is velocity defined as? Is it the instantaneous time rate of change of displacement, or is it simply displacement divided by time? Here is an example of where this problem comes up:
I need to solve the equation $\displaystyle v = \omega r \sqrt{\frac{m_b}{m_s}}$ for time. The only way I see of doing this is substituting $\displaystyle \frac{r}{t}$ for $\displaystyle v$ in order to cancel the two r's and solve for time. However, what gives me the right to make this substitution? Isn't $\displaystyle \frac{r}{t}$ just the average velocity, while $\displaystyle \frac{dr}{dt}$ is the instantaneous velocity? Why am I able to choose either or solely for my purposes at hand?
2. Feb 11, 2016
nasu
What is the actual problem and how these quantities depend on time? Is r a function of time? is omega a function of time?
3. Feb 11, 2016
Isaac0427
Is there acceleration involved?
| 273
| 1,043
|
{"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}
| 2.953125
| 3
|
CC-MAIN-2018-30
|
latest
|
en
| 0.885714
|
https://www.weegy.com/Home.aspx?ConversationId=75389AE0
| 1,534,410,259,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00101.warc.gz
| 1,028,002,707
| 8,280
|
a black racer snake travels 27.6 km in 12 hours. what is the speed in km/h?
Question
Updated 6/19/2014 5:24:18 PM
Flagged by Janet17 [6/19/2014 5:24:18 PM]
Original conversation
User: a black racer snake travels 27.6 km in 12 hours. what is the speed in km/h?
Question
Updated 6/19/2014 5:24:18 PM
Flagged by Janet17 [6/19/2014 5:24:18 PM]
Rating
3
A black racer snake travels 27.6 km in 12 hours. Therefore its speed in km/h is 27.6 / 12 = 2.3 km/h.
27,188,924
*
Get answers from Weegy and a team of really smart live experts.
Popular Conversations
What is Amendments III?
Weegy: The First Amendment prohibits the making of any law respecting an establishment of religion, impeding the free ...
15. What is 26% as a fraction in simplest form?
Weegy: 22% as a fraction in the simplest form is 11/50.
Which of the following is an example of an improper fraction? A. ...
Weegy: 3 3/5 as an improper fraction is 18/5. User: If three bags of birdseed cost \$14.16, how much will 14 bags ...
For a pair of similar triangles, corresponding sides are always ...
Weegy: For a pair of similar triangles, corresponding sides are sometimes congruent.
You are seated in the driver's seat of your parked car, preparing to ...
Weegy: You are seated in the driver's seat of your parked car, preparing to drive. You must properly fasten your seat ...
What is 5.25% of 200
Weegy: 200*3/5 = 120. User: A certain alloy contains 5.25% copper. How much copper is there in a piece weighing 200 ...
S
L
P
P
Points 100 [Total 554] Ratings 0 Comments 100 Invitations 0 Offline
S
L
R
P
R
P
R
Points 84 [Total 544] Ratings 0 Comments 4 Invitations 8 Offline
S
R
L
R
P
R
P
R
R
R
Points 43 [Total 865] Ratings 0 Comments 3 Invitations 4 Offline
S
Points 41 [Total 41] Ratings 0 Comments 11 Invitations 3 Offline
S
Points 16 [Total 33] Ratings 0 Comments 16 Invitations 0 Offline
S
Points 12 [Total 12] Ratings 0 Comments 12 Invitations 0 Offline
S
Points 10 [Total 40] Ratings 1 Comments 0 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 1] Ratings 0 Comments 1 Invitations 0 Offline
S
Points 1 [Total 2] Ratings 0 Comments 1 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc.
| 723
| 2,293
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.140625
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.889167
|
http://slideplayer.com/slide/2857347/
| 1,529,442,629,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863119.34/warc/CC-MAIN-20180619193031-20180619213031-00154.warc.gz
| 281,721,598
| 20,574
|
# Q12.1 The mass of the Moon is 1/81 of the mass of the Earth.
## Presentation on theme: "Q12.1 The mass of the Moon is 1/81 of the mass of the Earth."— Presentation transcript:
Q12.1 The mass of the Moon is 1/81 of the mass of the Earth. Compared to the gravitational force that the Earth exerts on the Moon, the gravitational force that the Moon exerts on the Earth is = times greater 2. 81 times greater 3. equally strong 4. 1/81 as great 5. (1/81)2 = 1/6561 as great
A12.1 The mass of the Moon is 1/81 of the mass of the Earth. Compared to the gravitational force that the Earth exerts on the Moon, the gravitational force that the Moon exerts on the Earth is = times greater 2. 81 times greater 3. equally strong 4. 1/81 as great 5. (1/81)2 = 1/6561 as great
Q12.2 The planet Saturn has 100 times the mass of the Earth and is 10 times more distant from the Sun than the Earth is. Compared to the Earth’s acceleration as it orbits the Sun, the acceleration of Saturn as it orbits the Sun is times greater 2. 10 times greater 3. the same 4. 1/10 as great 5. 1/100 as great
A12.2 The planet Saturn has 100 times the mass of the Earth and is 10 times more distant from the Sun than the Earth is. Compared to the Earth’s acceleration as it orbits the Sun, the acceleration of Saturn as it orbits the Sun is times greater 2. 10 times greater 3. the same 4. 1/10 as great 5. 1/100 as great
Q12.3 Compared to the Earth, Planet X has twice the mass and twice the radius. This means that compared to the Earth, Planet X has 1. 4 times the surface gravity 2. twice the surface gravity 3. the same surface gravity 4. 1/2 as much surface gravity 5. 1/4 as much surface gravity
A12.3 Compared to the Earth, Planet X has twice the mass and twice the radius. This means that compared to the Earth, Planet X has 1. 4 times the surface gravity 2. twice the surface gravity 3. the same surface gravity 4. 1/2 as much surface gravity 5. 1/4 as much surface gravity
Q12.4 A satellite is moving around the Earth in a circular orbit. Over the course of an orbit, the Earth’s gravitational force 1. does positive work on the satellite 2. does negative work on the satellite 3. does positive work on the satellite during part of the orbit and negative work on the satellite during the other part 4. does zero work on the satellite at all points in the orbit
A12.4 A satellite is moving around the Earth in a circular orbit. Over the course of an orbit, the Earth’s gravitational force 1. does positive work on the satellite 2. does negative work on the satellite 3. does positive work on the satellite during part of the orbit and negative work on the satellite during the other part 4. does zero work on the satellite at all points in the orbit
Q12.5 A planet (P) is moving around the Sun (S) in an elliptical orbit. As the planet moves from aphelion to perihelion, the Sun’s gravitational force 1. does positive work on the planet 2. does negative work on the planet 3. does positive work on the planet during part of the motion from aphelion to perihelion and negative work on the planet during the other part 4. does zero work on the planet at all points between aphelion and perihelion
A12.5 A planet (P) is moving around the Sun (S) in an elliptical orbit. As the planet moves from aphelion to perihelion, the Sun’s gravitational force 1. does positive work on the planet 2. does negative work on the planet 3. does positive work on the planet during part of the motion from aphelion to perihelion and negative work on the planet during the other part 4. does zero work on the planet at all points between aphelion and perihelion
Q12.6 A planet (P) is moving around the Sun (S) in an elliptical orbit. As the planet moves around the orbit, the planet’s angular momentum 1. increases as it moves from aphelion to perihelion and decreases as it moves from perihelion to aphelion 2. decreases as it moves from aphelion to perihelion and increases as it moves from perihelion to aphelion 3. increases at all times 4. decreases at all times 5. remains the same at all times
A12.6 A planet (P) is moving around the Sun (S) in an elliptical orbit. As the planet moves around the orbit, the planet’s angular momentum 1. increases as it moves from aphelion to perihelion and decreases as it moves from perihelion to aphelion 2. decreases as it moves from aphelion to perihelion and increases as it moves from perihelion to aphelion 3. increases at all times 4. decreases at all times 5. remains the same at all times
Download ppt "Q12.1 The mass of the Moon is 1/81 of the mass of the Earth."
Similar presentations
| 1,226
| 4,606
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.914071
|
https://coolbutuseless.github.io/2020/03/01/line-segment/box-intersection-test/
| 1,620,783,974,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00367.warc.gz
| 211,165,009
| 3,019
|
# Line Segment/Box Intersection Test
## Line Segment/Box Intersection Test
While working with `{grid}` graphics I needed a way to cull some generated geometry if it didn’t lie within a given rectangular bounding box.
The following code is based upon a stackoverflow response
This code will test 100 thousand line segments against a box in about 20ms, which is fast enough for my purposes.
``````#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#' Determine if line segments intersect a rectangle
#'
#' @param x1,y1,x2,y2 coordinates of line segment endpoints
#' @param xmin,ymin,xmax,ymax coordinates of lower-left and upper-right of
#' rectangle.
#'
#' @return logical vector the same length as x1
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
line_intersects_box <- function(x1, y1, x2, y2, xmin, ymin, xmax, ymax) {
# Returns 0/1 depending on which side of the line segment the corner lies
left_of <- function(xcorner, ycorner) {
((y2 - y1) * xcorner + (x1 - x2) * ycorner + (x2 * y1 - x1 * y2)) >= 0
}
# test segments against each of the 4 corners of the box
res <- list(
left_of(xmin, ymin),
left_of(xmin, ymax),
left_of(xmax, ymax),
left_of(xmax, ymin)
)
# if the test for each corner
# are all of the same sign, then the segment definitely misses the box
a_miss <- Reduce(`+`, res) %in% c(0, 4)
# Does it miss based upon the shadow intersection test?
b_miss <-
(x1 > xmax & x2 > xmax) | (x1 < xmin & x2 < xmin) |
(y1 > ymax & y2 > ymax) | (y1 < ymin & y2 < ymin)
# a hit is if it doesn't miss!
!(a_miss | b_miss)
}``````
## Example
``````N <- 300000
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# Generate lots of random lines
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
x1 <- runif(N)
y1 <- runif(N)
x2 <- runif(N)
y2 <- runif(N)
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# Test if the segments intersect a box
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
xmin <- 0.4
ymin <- 0.4
xmax <- 0.9
ymax <- 0.9
int1 <- line_intersects_box(x1, y1, x2, y2, xmin, ymin, xmax, ymax)
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# Test if the segments intersect a second box
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
xmin <- 0.1
ymin <- 0.1
xmax <- 0.9
ymax <- 0.2
int2 <- line_intersects_box(x1, y1, x2, y2, xmin, ymin, xmax, ymax)
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# Plot all segments that don't intersect the boxes
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
df <- data.frame(
x1, y1, x2, y2, int1, int2
) %>% filter(!int1, !int2)
ggplot(df) +
geom_segment(aes(x=x1,y=y1,xend=x2,yend=y2), alpha=0.005) +
annotate('rect', xmin=xmin, ymin=ymin, xmax=xmax,ymax=ymax, fill=NA, colour=NA) +
theme_void() ``````
| 770
| 2,986
|
{"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}
| 2.96875
| 3
|
CC-MAIN-2021-21
|
latest
|
en
| 0.43189
|
https://www.esaral.com/q/an-elevator-in-a-building-can-carry-a-maximum-of-10-persons-13194
| 1,722,951,858,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00422.warc.gz
| 589,107,835
| 11,686
|
# An elevator in a building can carry a maximum of 10 persons,
Question:
An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being $68 \mathrm{~kg}$. The mass of the elevator itself is $920 \mathrm{~kg}$ and it moves with a constant speed of $3 \mathrm{~m} / \mathrm{s}$. The frictional force opposing the motion is $6000 \mathrm{~N}$. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator
$\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ must be at least:
1. (1) $56300 \mathrm{~W}$
2. (2) $62360 \mathrm{~W}$
3. (3) $48000 \mathrm{~W}$
4. (4) $66000 \mathrm{~W}$
Correct Option: , 4
Solution:
(4) Net force on the elevator $=$ force on elevator
+ frictional force
$\Rightarrow \quad F=(10 m+M) g+f$
where, $m=$ mass of person, $M=$ mass of elevator,
$f=$ frictional force
$\Rightarrow \quad F=(10 \times 68+920) \times 9.8+600$
$\Rightarrow F=22000 \mathrm{~N}$
$\Rightarrow P=F V=22000 \times 3=66000 W$
| 328
| 1,017
|
{"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}
| 4.15625
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.610977
|
http://openturns.github.io/openturns/master/user_manual/_generated/openturns.TrendTransform.html
| 1,539,872,093,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00071.warc.gz
| 262,214,602
| 7,157
|
# TrendTransform¶
class TrendTransform(*args)
Trend transformation.
Parameters: myTrendFunc : Function The trend function of a process.
Notes
A multivariate stochastic process of dimension d where may write as the sum of a trend function and a stationary multivariate stochastic process of dimension d as follows:
We note the values of one field of the process X, associated to the mesh of . We note the values of the resulting stationary field. Then we have:
Examples
Create a trend function: where :
>>> import openturns as ot
>>> myGrid = ot.RegularGrid(0.0, 0.1, 10)
>>> f = ot.SymbolicFunction(['t'], ['1+2*t+t^2'])
>>> fTrend = ot.TrendTransform(f, myGrid)
>>> amplitude=[5.0]
>>> scale=[0.2]
>>> myCovModel=ot.ExponentialModel(scale, amplitude)
>>> myXProcess=ot.GaussianProcess(myCovModel, myGrid)
>>> myYProcess = ot.CompositeProcess(fTrend, myXProcess)
Remove it from a field:
>>> myField = myYProcess.getRealization()
>>> myStatField = fTrend.getInverse()(myField)
>>> myInitialField = fTrend(myStatField)
Attributes: thisown The membership flag
Methods
__call__(*args) Call self as a function. getCallsNumber() Get the number of calls of the function. getClassName() Accessor to the object’s name. getFunction() Get the function of . getId() Accessor to the object’s id. getInputDescription() Get the description of the input field values. getInputDimension() Get the dimension of the input field values. getInputMesh() Get the mesh associated to the input domain. getInverse() Accessor to the inverse trend function. getMarginal(*args) Get the marginal(s) at given indice(s). getName() Accessor to the object’s name. getOutputDescription() Get the description of the output field values. getOutputDimension() Get the dimension of the output field values. getOutputMesh() Get the mesh associated to the output domain. getShadowedId() Accessor to the object’s shadowed id. getVisibility() Accessor to the object’s visibility state. hasName() Test if the object is named. hasVisibleName() Test if the object has a distinguishable name. isActingPointwise() Whether the function acts point-wise. setInputDescription(inputDescription) Set the description of the input field values. setInputMesh(inputMesh) Set the mesh associated to the input domain. setName(name) Accessor to the object’s name. setOutputDescription(outputDescription) Set the description of the output field values. setOutputMesh(outputMesh) Set the mesh associated to the output domain. setShadowedId(id) Accessor to the object’s shadowed id. setVisibility(visible) Accessor to the object’s visibility state.
getTrendFunction
__init__(*args)
Initialize self. See help(type(self)) for accurate signature.
getCallsNumber()
Get the number of calls of the function.
Returns: callsNumber : int Counts the number of times the function has been called since its creation.
getClassName()
Accessor to the object’s name.
Returns: class_name : str The object class name (object.__class__.__name__).
getFunction()
Get the function of .
Returns: l : Function Function .
Examples
>>> import openturns as ot
>>> h = ot.SymbolicFunction(['t', 'x'], ['x + t^2'])
>>> n = 1
>>> mesh = ot.Mesh(n)
>>> myVertexValueFunction = ot.ValueFunction(h, mesh)
>>> print(myVertexValueFunction.getFunction())
[t,x]->[x + t^2]
getId()
Accessor to the object’s id.
Returns: id : int Internal unique identifier.
getInputDescription()
Get the description of the input field values.
Returns: inputDescription : Description Description of the input field values.
getInputDimension()
Get the dimension of the input field values.
Returns: d : int Dimension of the input field values.
getInputMesh()
Get the mesh associated to the input domain.
Returns: inputMesh : Mesh The input mesh .
getInverse()
Accessor to the inverse trend function.
Returns: myInverseTrendTransform : InverseTrendTransform The function.
getMarginal(*args)
Get the marginal(s) at given indice(s).
Parameters: i : int or list of ints, Indice(s) of the marginal(s) to be extracted. function : VertexValueFunction The initial function restricted to the concerned marginal(s) at the indice(s) .
getName()
Accessor to the object’s name.
Returns: name : str The name of the object.
getOutputDescription()
Get the description of the output field values.
Returns: outputDescription : Description Description of the output field values.
getOutputDimension()
Get the dimension of the output field values.
Returns: d’ : int Dimension of the output field values.
getOutputMesh()
Get the mesh associated to the output domain.
Returns: outputMesh : Mesh The output mesh .
getShadowedId()
Accessor to the object’s shadowed id.
Returns: id : int Internal unique identifier.
getVisibility()
Accessor to the object’s visibility state.
Returns: visible : bool Visibility flag.
hasName()
Test if the object is named.
Returns: hasName : bool True if the name is not empty.
hasVisibleName()
Test if the object has a distinguishable name.
Returns: hasVisibleName : bool True if the name is not empty and not the default one.
isActingPointwise()
Whether the function acts point-wise.
Returns: pointWise : bool Returns true if the function evaluation at each vertex depends only on the vertex or the value at the vertex.
setInputDescription(inputDescription)
Set the description of the input field values.
Parameters: inputDescription : sequence of str Description of the input field values.
setInputMesh(inputMesh)
Set the mesh associated to the input domain.
Parameters: inputMesh : Mesh The input mesh .
setName(name)
Accessor to the object’s name.
Parameters: name : str The name of the object.
setOutputDescription(outputDescription)
Set the description of the output field values.
Parameters: outputDescription : sequence of str Describes the outputs of the output field values.
setOutputMesh(outputMesh)
Set the mesh associated to the output domain.
Parameters: outputMesh : Mesh The output mesh .
setShadowedId(id)
Accessor to the object’s shadowed id.
Parameters: id : int Internal unique identifier.
setVisibility(visible)
Accessor to the object’s visibility state.
Parameters: visible : bool Visibility flag.
thisown
The membership flag
| 1,380
| 6,210
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.53125
| 3
|
CC-MAIN-2018-43
|
longest
|
en
| 0.545779
|
https://docs.classiq.io/latest/explore/community/QClass_2024/Submissions/HW3/Yasir_Mansour_HW3_VQE/
| 1,721,615,722,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00327.warc.gz
| 177,754,326
| 23,380
|
H₂ Molecule Homework Assignment
Quantum Software Development Journey: From Theory to Application with Classiq - Part 3
• Similarly to what we have done in class, in this exercise we will implement the VQE on H2 molecule.
• This time instead of using the built-in methods and functions (such as Molecule and MoleculeProblem) to difne and solve the problem, you will be provided with a two qubits Hamiltonian.
Submission
• Submit the completed Jupyter notebook and report via GitHub. Ensure all files are correctly named and organized.
• Use the Typeform link provided in the submission folder to confirm your submission.
Important Dates
• Assignment Release: 22.5.2024
• Submission Deadline: 3.6.2024 (7 A.M GMT+3)
Happy coding and good luck!
Part 1
Given the following Hamiltonian:
$\hat{H} = -1.0523 \cdot (I \otimes I) + 0.3979 \cdot (I \otimes Z) - 0.3979 \cdot (Z \otimes I) - 0.0112 \cdot (Z \otimes Z) + 0.1809 \cdot (X \otimes X)$
Complete the following code
# !pip install classiq
# import classiq
# classiq.authenticate()
from typing import List, cast
from classiq import *
from classiq import Pauli, PauliTerm
# TODO: Complete Hamiltonian
HAMILTONIAN = QConstant(
"HAMILTONIAN",
List[PauliTerm],
[
PauliTerm([Pauli.I, Pauli.I], -1.0523),
PauliTerm([Pauli.I, Pauli.Z], 0.3979),
PauliTerm([Pauli.Z, Pauli.I], -0.3979),
PauliTerm([Pauli.Z, Pauli.Z], -0.0112),
PauliTerm([Pauli.X, Pauli.X], 0.1809),
],
)
@qfunc
def main(q: Output[QArray[QBit]], angles: CArray[CReal, 3]) -> None:
# TODO: Create an ansatz which allows each qubit to have
# arbitrary rotation
allocate(2, q)
U(angles[0], angles[1], angles[2], 0, q[0])
U(angles[0], angles[1], angles[2], 0, q[1])
# CX(q[0], q[1])
@cfunc
def cmain() -> None:
res = vqe(
hamiltonian=HAMILTONIAN,
maximize=False,
initial_point=[],
optimizer=Optimizer.COBYLA,
max_iteration=1000,
tolerance=0.001,
step_size=0,
skip_compute_variance=False,
alpha_cvar=1.0,
)
save({"result": res})
qmod = create_model(main, classical_execution_function=cmain)
# TODO: complete the line, use classical_execution_function
qprog = synthesize(qmod)
# show(qprog)
execution = execute(qprog)
res = execution.result()
# execution.open_in_ide()
vqe_result = res[0].value
# TODO: complete the line
print(f"Optimal energy: {vqe_result.energy}")
print(f"Optimal parameters: {vqe_result.optimal_parameters}")
print(f"Eigenstate: {vqe_result.eigenstate}")
Optimal energy: -1.0754688476562502
Optimal parameters: {'angles_0': 0.7985412460405645, 'angles_1': 4.821416635411074, 'angles_2': -2.4351340720501597}
Eigenstate: {'11': (0.14986973510352247+0j), '10': (0.33874192794810626+0j), '00': (0.8600849340326803+0j), '01': (0.350780380010057+0j)}
Optimal energy: -1.0711231445312501 Optimal parameters: {'angles_0': -3.0914206855935538, 'angles_1': -0.23729943557563232, 'angles_2': -2.5756826635214636} Eigenstate: {'01': (0.02209708691207961+0j), '11': (0.9997558295653994+0j)}
Does it similar to the optimal energy we calculated in class? \ Does it similar to the total energy we calculated in class?
Part 2
Now, we want to have a more interesting ansatz in our main.
Add one line of code to the main function you created in Part 1 that creates entanglement between the two qubits.
Which gate should you use?
@qfunc
def main(q: Output[QArray[QBit]], angles: CArray[CReal, 3]) -> None:
# TODO: Create an ansatz which allows each qubit to have
# arbitrary rotation
allocate(2, q)
U(angles[0], angles[1], angles[2], 0, q[0])
U(angles[0], angles[1], angles[2], 0, q[1])
CX(q[0], q[1])
# H(q[0])
# X(q[1])
# CX(q[0], q[1])
@cfunc
def cmain() -> None:
res = vqe(
HAMILTONIAN, # TODO: complete the missing argument
False,
[],
optimizer=Optimizer.COBYLA,
max_iteration=1000,
tolerance=0.001,
step_size=0,
skip_compute_variance=False,
alpha_cvar=1.0,
)
save({"result": res})
qmod = create_model(main, classical_execution_function=cmain)
# TODO: complete the line, use classical_execution_function
qprog = synthesize(qmod)
# show(qprog)
execution = execute(qprog)
res = execution.result()
# execution.open_in_ide()
vqe_result = res[0].value
# TODO: complete the line
print(f"Optimal energy: {vqe_result.energy}")
print(f"Optimal parameters: {vqe_result.optimal_parameters}")
print(f"Eigenstate: {vqe_result.eigenstate}")
Optimal energy: -1.8512822265625002
Optimal parameters: {'angles_0': -2.953247896519252, 'angles_1': 6.104234752928836, 'angles_2': 4.463845345952442}
Eigenstate: {'10': (0.08838834764831845+0j), '11': (0.0855816496101822+0j), '01': (0.992402781762526+0j)}
Optimal energy: -1.8452896484374999 Optimal parameters:
Eigenstate: {'11': (0.08267972847076846+0j), '10': (0.07967217989988726+0j), '01': (0.9933863328282708+0j)}
Does it similar to the optimal energy we calculated in class? \ Does it similar to the total energy we calculated in class? \ What can we learn about the provided form this result Hamiltonian?
With entanglement one gets better results.
| 1,574
| 4,933
|
{"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}
| 3.4375
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.572428
|
http://fullhomework.com/lab-5-requirements/
| 1,611,490,379,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00635.warc.gz
| 44,583,921
| 12,101
|
# LAB 5 REQUIREMENTS
LAB 5 REQUIREMENTS
Problem:
In this lab, you will write a visual C# program that will allow the user to adjust the source voltage and resistances for a circuit. Additionally, the user has the option, using a checkbox, to eliminate the parallel resistor and work with a simple series circuit. After entering the values for the resistors and selecting a source voltage using the TrackBar, your program will put the values for the circuit current and voltages shown in the corresponding labels. The images that are required for this lab are included with this requirements document, please check the requirements folder to locate your image files.
Part A
Design the main form for the Circuit Analyzer. Use the images provided with the requirements
document. Place a TrackBar on the form that the user will use to adjust the source voltage. Properly label the resistors, R1, R2, and R3, and place a textbox next to each resistor where the user will enter a resistance value in ohms.
The initial values of R1, R2, and R3 should be 100, 150, and 150, respectively and the initial value of the voltage source should be set to 4 Volts. Additionally, the labels for the calculated values (I, V1, and V2) should start with the correct values for the circuit with the default resistances. Add a checkbox as shown in the sample Circuit Analyzer on the following page.
Format I1 to four decimal places, V1 to two decimal places, and V2 to two decimal places. When the user changes either one of the resistance values (Change event for the textboxes) or moves the sliding TrackBar to a different source voltage, the calculated values in the labels for I1, V1, and V2 should recalculate. You can easily do this by copying and pasting the code that you created for the TrackBar’s Scroll event into the Change event for each TextBox. These calculations should be displayed for the selected resistances and source voltage as shown above.
Controls needed:
Track Bar
Picture boxes (2 of them), insert each image on each picture box.
A check box
Labels (as shown below)
Text boxes (as shown below)
SAMPLE RUN
Extend the program so that the user has two circuits to choose from. When the ‘Remove R3 from circuit’ checkbox is checked, the circuit should look like the circuit shown on the next page.
When the user checks the box, since the circuit is changing, be sure that you update the values in the I1, V1, and V2 labels as shown above. This time when the user changes the source voltage using the TrackBar or changes one of the two resistances by entering a new value in the textbox, the computer will use the formulas for this new circuit to calculate I1, V1, and V2.
For each lab and following comments must be added at the beginning of your Visual C# code.
‘LAB #
‘SEMESTER NAME
‘STUDENT’S FIRST NAME, LAST NAME
‘I fully understand the following statement.
‘OU PLAGIARISM POLICY
‘All members of the academic community at Oakland are expected to practice and uphold ‘standards of academic integrity and honesty. An instructor is expected to inform and instruct ‘students about the procedures and standards of research and documentation required of students ‘in fulfilling course work. A student is expected to follow such instructions and be sure the rules ‘and procedures are understood in order to avoid inadvertent misrepresentation of his/her work. ‘Students must assume that individual (unaided) work on exams and lab reports and documentation ‘of sources is expected unless the instructor specifically says that is not necessary.
‘The following definitions are some examples of academic dishonesty:
‘Plagiarizing from work of others. Plagiarism is using someone else’s work or ideas without
‘giving the other person credit; by doing this, a student is, in effect, claiming credit for
‘someone else’s thinking. Whether the student has read or heard the information he/she uses,
‘the student must document the source of information. When dealing with written sources,
‘a clear distinction would be made between quotations (which reproduce information from
‘the source word-for-word within quotation marks) and paraphrases (which digest the
‘source information and produce it in the student’s own words). Both direct quotations and
‘paraphrases must be documented. Just because a student rephrases, condenses or selects
‘from another person’s work, the ideas are still the other person’s, and failure to give ‘credit
constitutes misrepresentation of the student’s actual work and plagiarism of ‘another’s ideas.
Naturally, buying a paper and handing it in as one’s own work is ‘plagiarism.
‘Cheating on lab reports falsifying data or submitting data not based on student’s own work.
All labs will be submitted electronically, no paper copies will be given to Lab mentors.
Before submission:
Please create a folder named as Lab5_FName_LName:
Place your solution file under this folder.
Zip the folder then upload through Moodle. You will not be able to upload unless you zip, 7zip or rar the folder.
1. Log on to Moodle.
| 1,108
| 5,048
|
{"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-2021-04
|
latest
|
en
| 0.84597
|
http://nrich.maths.org/public/leg.php?code=-393&cl=3&cldcmpid=4873
| 1,506,262,226,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818690029.51/warc/CC-MAIN-20170924134120-20170924154120-00640.warc.gz
| 248,982,455
| 6,103
|
# Search by Topic
#### Resources tagged with biology similar to Motion Capture:
Filter by: Content type:
Stage:
Challenge level:
### There are 21 results
Broad Topics > Applications > biology
### Core Scientific Mathematics
##### Stage: 4 and 5 Challenge Level:
This is the area of the advanced stemNRICH site devoted to the core applied mathematics underlying the sciences.
### Exact Dilutions
##### Stage: 4 Challenge Level:
Which exact dilution ratios can you make using only 2 dilutions?
### Dilution Series Calculator
##### Stage: 4 Challenge Level:
Which dilutions can you make using 10ml pipettes and 100ml measuring cylinders?
### Conversion Sorter
##### Stage: 4 Challenge Level:
Can you break down this conversion process into logical steps?
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Mixed up Mixture
##### Stage: 4 Challenge Level:
Can you fill in the mixed up numbers in this dilution calculation?
### Chemnrich
##### Stage: 4 and 5 Challenge Level:
chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . .
### Alternative Record Book
##### Stage: 4 and 5 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### The Genes of Gilgamesh
##### Stage: 4 Challenge Level:
Can you work out the parentage of the ancient hero Gilgamesh?
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Bigger or Smaller?
##### Stage: 4 Challenge Level:
When you change the units, do the numbers get bigger or smaller?
##### Stage: 4 Challenge Level:
Which units would you choose best to fit these situations?
### Bionrich
##### Stage: 4 and 5 Challenge Level:
bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your. . . .
### Ratios and Dilutions
##### Stage: 4 Challenge Level:
Scientists often require solutions which are diluted to a particular concentration. In this problem, you can explore the mathematics of simple dilutions
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Packing 3D Shapes
##### Stage: 4 Challenge Level:
What 3D shapes occur in nature. How efficiently can you pack these shapes together?
### Epidemic Modelling
##### Stage: 4 and 5 Challenge Level:
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
| 691
| 3,188
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2017-39
|
latest
|
en
| 0.835768
|
https://11011110.github.io/blog/2020/07/16/comparing-multi-sport.html
| 1,675,199,705,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00696.warc.gz
| 99,254,510
| 4,516
|
Where you have a triangle center defined by the intersection point of three lines or curves, you often also have a Voronoi diagram or minimization diagram whose cells have those lines as boundaries. For the circumcenter, equidistant from the three vertices of a triangle, the diagram is the classical Voronoi diagram of those three points. For the incenter, equidistant from the three sides, the diagram is the medial axis of the triangle. The Fermat point (when it doesn’t degenerate to a vertex) comes from a diagram that tells you which of the three sides of the triangle spans the widest field of view.
So when I posted recently about Cartesian triangle centers defined by an equality of areas of three bounding boxes, I had in mind a minimization diagram for bounding box area, or actually the signed area $$(x-x_i)(y-y_i)$$ of the bounding box of the moving point $$(x,y)$$ and a fixed site $$(x_i,y_i)$$. Here’s one of these diagrams for six points:
If you subtract $$xy$$ from the signed areas, you don’t change the minimization diagram (because you’re subtracting the same thing from all the signed areas) but it makes the functions you’re minimizing become linear in $$x$$ and $$y$$, explaining the polygonal shape and linear boundaries of the cells in the minimization diagram. As the previous post already described, the bisector between any two cells is the diagonal line of the bounding box of two sites.
I use these diagrams in a paper to appear in the Canadian Conference on Computational Geometry, now online as a preprint: “Dynamic products of ranks” (arXiv:2007.08123). The goal of the paper is to combine two different and unrelated numerical scores of the same items such as the skiing speed and shooting accuracy of a group of biathletes. Biathlons do this by using an arbitrary conversion factor to penalize skiing time based on shooting misses (essentially, adding the converted scores from the two parts of the sport) but instead some other sports multiply the ranks of the athletes within each discipline. So if you got, say, fifth in shooting but second in skiing, your overall score would be ten. My paper studies how quickly you can find the new winner after changing the score of a single competitor or by adding or removing a competitor in a sport using this system.
We can represent a set of items or group of athletes as points in the plane, whose coordinates are their ranks (not their raw scores). Then an update to the data that lies outside both the horizontal and vertical range of these points doesn’t change their relative positions: it might change their ranks, but all in the same way. So for these updates, the signed bounding box area minimization diagram stays unchanged. Because the bounding box area is just the product of ranks, we can use this diagram to quickly look up the winner among these shifted points. The overall strategy of the data structure in my paper is to divide the input data into smaller subsets and maintain their diagrams in such a way that most diagrams stay unchanged, and only a few need rebuilding, after each update. Dividing into more subsets reduces the rebuilding time, because fewer input items will belong to rebuilt diagrams, but increases the query time, because you have to query more diagrams. The $$O(\sqrt{n\log n})$$ update time for the data structure in the paper comes from choosing the optimal tradeoff between these two terms of the total time.
There are other applications for this sort of rank aggregation beyond athletics; see the paper for details.
Because CCCG is online this year, I also have a video talk prepared, but it’s not yet online; I’ll post a link later. In the meantime, registration for CCCG is free this year, with a deadline of July 25, and enables both early access to the videos and access to the Q&A sessions with each speaker at the conference itself.
| 809
| 3,871
|
{"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}
| 3.5625
| 4
|
CC-MAIN-2023-06
|
latest
|
en
| 0.923736
|
https://physics.stackexchange.com/questions/328905/transfer-matrix-formalism
| 1,571,834,847,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987833766.94/warc/CC-MAIN-20191023122219-20191023145719-00218.warc.gz
| 637,942,045
| 31,771
|
# Transfer Matrix formalism
I am trying to apply the transfer matrix formalism to an Ising model problem, and am having some difficulties deriving the correct matrix to use.
The problem is as follows. There is an infinite chain of spins, as shown in the diagram below.
The energy of this configuration is given by:
$$E = -J\sum_{i=1}^{L}(\sigma_{i}\sigma_{i+1}+\sigma_{i}\tau_{i} + \tau_{i}\sigma_{i+1})$$
Where we assume periodic boundary conditions, i.e. $\sigma_{L+1}=\sigma_{1}$, and $J > 0$. The partition function can be written:
$$Z(\beta)=\sum_{\text{configuratons }C}e^{-\beta E(C)} = \sum_{\sigma_{1}=\pm1}\cdots\sum_{\sigma_{L}=\pm1}\dots\sum_{\tau_{1}=\pm1}\dots\sum_{\tau_{L}=\pm 1}e^{-\beta E(\{\sigma_{i}\},\{\tau_{i}\})}$$
We can decompose the energy into sub-functions:
$$E(\sigma_{i},\sigma_{i+1},\tau_{i})=-J\left(\sigma_{i}\sigma_{i+1}+\sigma_{i}\tau_{i}+\tau_{i}\sigma_{i+1}\right)$$
This allows us to write the partition function:
$$Z(\beta)=\sum_{\sigma_{1},\dots,\tau_{1},\dots}\prod_{j=1}^{N}e^{-\beta E(\sigma_{j},\sigma_{j+1},\tau_{j})}$$
I wish to form this as a product of matrices so that I can find the eigenvalues and use the property:
$$Z(\beta) = \operatorname{Tr}(T^{N})$$
However, naively, I would think that $T$ was a $2\times 2\times 2$ tensor, rather than a matrix? How can I construct an appropriate transfer matrix? And how can I do it generally, rather than just for this problem?
• I agree that the way you wrote it it's a 2x2x2 transfer tensor. Also, since the boundary conditions are not periodic, I am not sure you will be able to write it in trace form anyway. – Cyclone Apr 25 '17 at 13:12
• @Cyclone I agree, it was a typo, I will fix now! The boundary conditions are periodic? – Thomas Russell Apr 25 '17 at 13:17
• There are many ways to do that, but probably the easiest is to explicitly sum over the $\tau$ variables (for a fixed realization of the $\sigma$ variables). Note that the variable $\tau_i$ only interacts with the spins $\sigma_i$ and $\sigma_{i+1}$, so all sums over the variables $\tau_i$ factorize, each sum providing an effective interaction between consecutive $\sigma$ spins. Once you have gotten rid of the $\tau$ spins in this way, the transfer matrix is constructed as in the standard Ising chain. – Yvan Velenik Apr 25 '17 at 18:38
• Alternatively, just write a $4\times 4$ transfer matrix allowing you to pass from $(\sigma_i,\tau_i)$ to $(\sigma_{i+1},\tau_{i+1})$. – Yvan Velenik Apr 25 '17 at 18:41
• @YvanVelenik Can you provide an example of what you mean, I've been trying to do this in the way that you describe and am having trouble "factorizing" out the $\tau_{i}$ variable! – Thomas Russell May 24 '17 at 10:45
| 841
| 2,710
|
{"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}
| 3.328125
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.843609
|
https://stats.stackexchange.com/questions/586820/how-can-i-check-if-the-curse-of-dimensionality-negatively-affects-my-clustering
| 1,709,494,019,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00539.warc.gz
| 532,575,686
| 39,598
|
# How can I check if the curse of dimensionality negatively affects my clustering results
I would like to cluster 80 days using k-means. Each of the 80 days contains 4 time series (temperature, solar radiation, electricity demand, electricity price) with 288 values each. So all in all I have $$4*288=1152$$ values for each day and I would like to cluster the days such that similar days with regard to these 4 time series combined will be grouped into the same cluster. For this purpose I use the following code, that I was advised to use (at least for the scaling part) from this answer https://stackoverflow.com/questions/73491673/strange-results-when-scaling-data-using-scikit-learn:
import numpy as np
import pandas as pd
from sklearn.preprocessing import StandardScaler
from sklearn.cluster import KMeans
X = data_Unscaled.to_numpy()
X_narrow = np.array([X[:, i*288:(i+1)*288].ravel() for i in range(4)]).T
scaler = StandardScaler()
X_narrow_scaled = scaler.fit_transform(X_narrow)
X_scaled = np.array([X_narrow_scaled[i*288:(i+1)*288, :].T.ravel() for i in range(80)])
kmeans = KMeans(init="random", n_clusters=3, n_init=10, max_iter=300, random_state=42)
kmeans.fit(X_scaled)
print(f"resultClustering Example Day 1: {kmeans.predict((X[20].reshape(1, -1)))}")
print(f"resultClustering Example Day 2: {kmeans.predict((X[30].reshape(1, -1)))}")
print(f"resultClustering Example Day 3: {kmeans.predict((X[40].reshape(1, -1)))}")
print(f"resultClustering Example Day 4: {kmeans.predict((X[23].reshape(1, -1)))}")
print(f"resultClustering Example Day 5: {kmeans.predict((X[10].reshape(1, -1)))}")
In the linked answer it was assumed that my results would suffer from the curse of dimensionality as I have just 80 points in a 1152-dimensional space. Can anyone tell me how can I check if the curse of dimensionality negatively affects the results? I made some test predictions and the are not all grouped into the same cluster which should be correct.
Here is the unscaled input data with shape (80,1152): https://filetransfer.io/data-package/CfbGV9Uk#link
• Unfortunately, the curse of dimensionality prevents us from determining the effects of the curse of dimensionality ;) Aug 26, 2022 at 14:21
• @JohnMadden: Thanks John for your comment. But is there no way to determine the quality of the clustering and thus try to infer, whether the curse of dimensionality is active or not in my case? Aug 27, 2022 at 9:42
• to be slightly more serious and maybe even useful, the biggest threat posed to a method like Kmeans in high dimension might could be the meaningfulness of Euclidean distance in high dimension; may be worth trying different metrics and seeing how results change. Aug 27, 2022 at 15:39
• You have more dimensions than points. This is not only the curse of dimensionality problem but primarily the singularity/multicollinearity problem. The real number of dimensions of your data is 80-1=79 at max. Which is still many. Aug 28, 2022 at 1:12
• @PeterBe Yes! Absolutely do this first. Start with simple things where you can tell if something is dead wrong, and slowly ramp up the complexity as needed/desired, in general with stats :) Aug 29, 2022 at 16:43
| 823
| 3,178
|
{"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.775895
|
https://www.folkstalk.com/tech/increase-pie-chart-size-python-with-code-examples/
| 1,709,417,627,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00274.warc.gz
| 779,279,758
| 5,794
|
# Increase Pie Chart Size Python With Code Examples
Increase Pie Chart Size Python With Code Examples
In this lesson, we'll use programming to attempt to solve the Increase Pie Chart Size Python puzzle. This is demonstrated by the code below.
```ax.pie(..., radius=1800, frame=True)
```
By investigating a variety of use scenarios, we were able to demonstrate how to solve the Increase Pie Chart Size Python problem that was present.
## How do you increase the size of a pie chart in Python?
To increase the size of the pie chart, we pass figsize parameter to the figure() method of pyplot.26-Dec-2021
## How do you increase the size of a pie chart?
To more precisely control the expansion, follow these steps: Right-click the pie chart, then click Format Data Series. Drag the Pie Explosion slider to increase the separation, or enter a number in the percentage box.
## How do I increase matplotlib size?
Increase or decrease the plot size in Matplotlib This can be achieved by an attribute of Matplotlib known as figsize. The figsize attribute allows us to specify the width and height of a figure in-unit inches.27-Jun-2022
## How do you increase the size of a graph?
Resize a chart
• To change the size manually, click the chart, and then drag the sizing handles to the size that you want.
• To use specific height and width measurements, on the Format tab, in the Size group, enter the size in the Height and Width box.
## How do I edit a pie chart in Python?
Customizing a Pie Chart in Python
• Make a slice pop-out. You can make one or more slices of the pie-chart pop-out using the explode option.
• Rotate the Pie-chart. You can rotate the pie-chart by setting a strartangle.
• Display Percentages.
• Customizing colours.
• Displaying colour codes.
## How do I increase the size of a pie chart in R?
We can get a larger chart by setting the value of that direction to be 1. In addition, we can get some more vibrant colors by setting col=rainbow(6). Finally, there is no need to just use the single letter names for the sectors of the pie.
## Why is my pie chart so small tableau?
Pie charts require at least one or more dimensions and one or two measures. Aggregate fields, such as Profit Ratio, don't contribute to those requirements. The result is a rather small pie. To make the chart bigger, hold down Ctrl + Shift (hold down ñ + z on a Mac) and press B several times.
## How do you add data to a pie of pie chart?
Creating Pie of Pie Chart in Excel:
• In Excel, Click on the Insert tab.
• Click on the drop-down menu of the pie chart from the list of the charts.
• Now, select Pie of Pie from that list.
## How do you enlarge a graph in Python?
Pandas matplotlib increase plot size
• Import the pandas module.
• After this, create a pandas dataframe with an index.
• To plot a pie chart, use df. plot. pie() function.
• To increase the size of the plot, pass the figsize() parameter along with dimensions.
| 664
| 2,946
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.801814
|
https://www.jiskha.com/display.cgi?id=1292195273
| 1,516,132,523,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00466.warc.gz
| 935,155,919
| 3,737
|
# Math 7th
posted by .
Write a description of each rule
2. (x,y) --> (x + 5, y - 2)
• Math 7th -
(-5 , 2), (x , y).
(X - (-5)) = (X + 5), (Y - 2).
## Similar Questions
1. ### math
Find a rule. Write the rule as an equation. Input X 12, 47, 9 Output Y 32, 67, 29 Rule:____________ Equation:_______________
2. ### Math 7th
Write a descripition of each rule 1. (x,y) --> ( x- 11, y +4)
3. ### Math
Write a description of each rule (x,y) --> (x + 5, y -2)
4. ### Biology
Write a brief description of the procedure required to make 10mL each of 2 M, 1.5 M, and 1 M NaCl solution from a 2 M Stock NaCl. Write the description.
5. ### 7th Math
What is the function rule for 2,30 4,15 6,10 10,6 12,5 15,4 30,2
6. ### math help !
3. Which function rule would help you find the values in the table ?
7. ### Math ~ Check Answers ~
Explain how to write a function rule from the table below. Then write a function rule. X: 2 | 4| 6 | And: 1 | 0 | -1 | My Answer: y = -1/2 x But, -1/2 (2) = -1, and y(2) = 1, so we need to add 2 at the start. So, y = 2 - 1/2 x
8. ### math
i can not find the pattern or a rule... write the next three terms and write a rule to describe the number pattern. 0.57, 0.66, 0.75, 0.84
9. ### Math Alg. 2
Write a recursive rule for the sequence: 1,2,12,56,272... and 2,5,11,26,59... and -3,-2,5,-3,-2... I can't find the pattern in these and am unsure on how to write the rule. Thanks
10. ### math
Write a function rule for the data in the table. Input (x) 1 2 3 4 5 Output(y) 6 8 10 12 14 I've been looking at this for about an hour now and can't figure out how to write it. I understand what the rule is, but I dont know how to …
More Similar Questions
| 602
| 1,683
|
{"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}
| 3.59375
| 4
|
CC-MAIN-2018-05
|
latest
|
en
| 0.792691
|
https://math.answers.com/Q/If_one_machine_can_complete_a_truckload_in_2_hours_and_the_second_machine_can_complete_a_truckload_in_4_hours_how_long_would_it_take_to_complete_a_truckload_if_both_machines_worked_on_it_together
| 1,713,215,303,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00359.warc.gz
| 339,596,429
| 48,394
|
0
# If one machine can complete a truckload in 2 hours and the second machine can complete a truckload in 4 hours how long would it take to complete a truckload if both machines worked on it together?
Updated: 12/10/2022
Wiki User
11y ago
Machine # 1 works at a rate of 1/2 truckload per hour, while machine 2 works at a rate of 1/4 truckload per hour. The two machines together work at a rate of 3/4 truckload per hour complete the job 1/(3/4) = 1-1/3 hours.
Wiki User
11y ago
Earn +20 pts
Q: If one machine can complete a truckload in 2 hours and the second machine can complete a truckload in 4 hours how long would it take to complete a truckload if both machines worked on it together?
Submit
Still have questions?
Related questions
### How do you manipulate simple machines and make their parts work together?
simple machines are put together by connecting. for example, in cars. simple machines work together by putting their strength
### What is coherent machine?
Machines which swing together are termed as coherent machine.
compound machine
### What exactly is a compound machine?
A compound machine refers to two simple machines that are put together. For example, a bicycle is a compound machine as there are varying machines within a bike, such as a wheel and an axle.
### Is a seesaw a compound machine?
A see-saw is a type of lever. Levers are simple machines. When you put two or more simple machines together, you get a compound machine. == ==
### What are 2 complex machines?
A complex machine is a machine that is made up of two or more simple machines. Examples: Pencil sharpener, Washing machine, Wheelbarrow
### What are two or more simple machines that work together called?
its a compound machine(:
compound machine
### What or two or more simple machines that work together called?
its a compound machine(:
compound machine
compound machine
compound machine
| 438
| 1,909
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.920588
|
https://randerson112358.medium.com/lets-build-a-max-heap-161d676394e
| 1,611,469,264,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00463.warc.gz
| 506,162,804
| 37,944
|
# Build A Max Heap
Create a Max Heap
A heap data structure in computer science is a special tree that satisfies the heap property, this just means that the parent is less than or equal to the child node for a minimum heap A.K.A min heap, and the parent is greater than or equal to the child node for a maximum heap A.K.A max heap. In this article I will talk specifically about binary heaps, so each node in our tree will have at most two children. Yes there are more than just binary heaps. The binary heap was created by J.W. J. Williams in 1964 for heapsort.
# A binary heap is a binary tree with two other constraints [1]
1) Shape Property: A binary heap is a complete binary tree, this means all of the levels of the tree are completely filled except possibly the last level. The nodes are filled from left to right.
2) Heap Property: The value stored in each node is either (greater than or equal to) OR (less than or equal to ) it’s children depending if it is a max heap or a min heap.
# Build a Max Heap
Let’s take an array and make a heap with an empty heap using the Williams method. We will insert the values 3,1,6,5,2 and 4 in our heap.
Williams Method of building a heap:
Building a heap from an array of n input elements can be done by starting with an empty heap, then successively inserting each element. This algorithm runs O( n log n) time. However this method is suboptimal and a faster algorithm was created by Floyd, which starts by putting the elements on a binary tree, respecting the shape property, then starting from the lowest level and moving upwards. — Wikipedia
## Insertion Algorithm [1]
To add an element to a heap we must perform an up-heap operation (also known as bubble-up, percolate-up, sift-up trickle-up, heapify-up, or cascade-up), by following this algorithm:
0. If heap is empty place element at root.
1. Add the element to the bottom level of the heap.
2. Compare the added element with its parent; if they are in the correct order, stop.
3. If not, swap the element with its parent and return to the previous step.
The number of operations required depends only on the number of levels the new element must rise to satisfy the heap property, thus the insertion operation has a worst-case time complexity of O(log n) but an average-case complexity of O(1).
`Williams Algorithm: top downwhile not end of array, if heap is empty, place item at root; else, place item at bottom of heap; while (child > parent) swap(parent, child); go to next array element; end`
# Build A Min Heap Video!
If you would like to learn more about computer science and Algorithm Analysis , you can take my online course here. I also have a course on Udemy.com called Recurrence Relation Made Easy where I help students to understand how to solve recurrence relations and asymptotic terms such as Big-O, Big Omega, and Theta. You can check out my YouTube channel of videos where I solve recurrence relations and perform algorithm analysis on code that anyone can check out for free !
Thanks for reading this article I hope its helpful to you all ! Keep up the learning, and if you would like more computer science, programming and algorithm analysis videos please visit and subscribe to my YouTube channels (randerson112358 & compsci112358 )
# Check Out the following for content / videos on Computer Science, Algorithm Analysis, Programming and Logic:
compsci112358:
Video Tutorials on Recurrence Relation:
Video Tutorial on Algorithm Analysis:
https://www.udemy.com/algorithm-analysis/
Written by
| 787
| 3,551
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.921875
| 4
|
CC-MAIN-2021-04
|
latest
|
en
| 0.926733
|
http://forum.slowtwitch.com/forum/Slowtwitch_Forums_C1/Triathlon_Forum_F1/Lionel_Sanders_%22Kona_Block%22_Reverse_Engineered_P6502839/
| 1,521,692,268,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00154.warc.gz
| 116,194,170
| 19,413
|
Prev Next
Lionel Sanders posted an interesting blog about his training leading up to his 2nd place finish at KONA this year.
In it he share some charts and stats about his swimming, cycling, and running training totals and averages.
He also included this chart of his combined total weekly training hours starting from early December 2016:
And to finish the blog post off, he throws in a PMC screenshot of his 14-week “Kona Block” leading up to the world championships, which he says peaks out at 182 CTL:
This looked like a cool opportunity to get some insight into the training behind a world-caliber performance, so I dug a bit deeper...
Weekly hours isn't enough information to build a PMC, you also need the intensity so you can calculate training load.
But, with a PMC screenshot I knew I could approximate what kind of training loads he was doing to hit 182 CTL. Couple that with his weekly hours and I could figure out intensity, and get a better picture of his overall training...
WARNING, serious data geekery lies ahead!
To start I took his PMC screenshot (this is a distorted view):
And recreated it on a weekly basis using Bereda (beredatraining.com):
And you can see that it was a pretty good match:
Recreating the CTL progression in Bereda gave me the approximate weekly training loads that Lionel would have had to withstand to get from 110 CTL to 182 CTL in the 14 weeks, shown by the red bars below:
I also read the values of the weekly training hours from his chart and plugged those into the chart too. Having weekly training hours AND training load means Bereda can calculate an intensity, and so we can see that Lionel’s “Kona Block” was done at an overall intensity value of about 0.8.
Here’s a spreadsheet output of the plan from above:
Check out the size of those training loads in the last five weeks!
I know there are a lot of Lionel fans on the forum so I thought I’d share my findings.
If you're interested, I go into a bit more detail in this post: https://www.beredatraining.com/2017/11/19/lionel-sanders-kona-block/
Weekly hours are great, but when I saw that massive 182 CTL I had to know how he got there. Truth is, I think other Ironman athletes go higher than that... Anyone know of another athlete with data like this that I could dive into?
Thanks!
When he posted this, I honestly thought 182 it was super low. Especially if you're abiding by the true Training Peaks swim TSS. I remember reading (and can attest) that 135 is pretty normal for a FOP amateur with a full time job. I hit that the last two seasons for extended periods of time and I don't post accurate swim TSS. Most of my swims are probably half of what Training Peaks would give me if I actually detailed how fast each interval was.
So, yeah, I still think 182 is crazy low and wouldn't doubt if someone like Ben Hoffman posts 250. Maybe I'm crazy but he bikes so damn much, it wouldn't shock me. Which begs an even better question... are most pros doing way too much?
Awesome. Thank you for sharing.
This doesn't mean a hell of a lot without breaking it up into the individual sports. Swim TSS is mostly useless when compared to run/bike, so I wouldn't be surprised if Lionel and others don't bother including that in with his CTL.
I know I don't bother with it. I use CTL as more an indicator of injury likelihood and overtraining in the legs and rarely post anything more than time/distance for swim workouts. And I then delete the sTSS since I don't want it affecting the CTL values.
I got the impression from someone that has worked with Training Peaks that 165 was really high.
BrentwoodTriGuy wrote:
wouldn't doubt if someone like Ben Hoffman posts 250.
actually more, but the data is biased because so much of it is from HrTSS, not power based.
Last edited by: jkhayc: Dec 6, 17 11:08
That's the scary part. Ben Hoffman's training is much more in line of what traditionally has produced Kona podiums through-out the years. Crowie, Stadler, Deboom, Reid, Ralert, Frodeno, Al-Sutan, T.O, etc. all had programs with massive amounts of bike riding that ranged from 4-500 mile weeks similar to Hoffman.
Lionel just came 3 miles shy of winning the whole damn thing on 10-12 hour bike weeks. I often wonder if Lionel would be competing against the Sagan's of the world for one day classics and world championships had he found the bike earlier in life. It will be interesting to see how Talansky's bike splits match up to Lionel's.
And please don't take this as me armchair quarterbacking the guy. He just came 3 miles shy of winning the whole damn thing.
Yes. I read most kona qualifiers average 120-150 for considerable amounts of the year and they all have pretty respectable threshold values, so alot of work performed by a pretty good athlete... job done!!. I need to do more work.
This is a pretty good ad for Bereda!
I thought 182 seemed reasonable, like I said I think others go higher, but what I found interesting was how intense his training is especially when you consider that it was gearing up for Ironman.
I would have figured 0.75 instead of 0.8
His average time on the bike during his "Kona Block" looks to be about 8 hours a week. Pretty different compared to more "traditional" Kona podium training that mdgreene mentioned.
The question is would he be faster if he biked more?
endurance1234 wrote:
The question is would he be faster if he biked more?
I wouldn't be surprised if he does bike more this coming year. I remember him saying that he felt he fell off the pace at the end of the ride in Kona.
writhe wrote:
This is a pretty good ad for Bereda!
I didn’t know it existed, but now I do.
Dennis Cottreau wrote:
(beredatraining.com)
Hey look: yet another website ripping off the PMC without crediting either the person who came up with the idea (i.e., me) or the people who own the trademarks (i.e., TrainingPeaks). Have you no shame?
BrentwoodTriGuy wrote:
Which begs an even better question... are most pros doing way too much?
I would vote a yes on that.
Too much of what specifically I think is individual. But seeing so many come into Kona after heavy race schedule, talk of high volume training, etc. Then "just not have it" race day, fade off the back, etc. I think many just over do it. It would at least be safe to say more over do it versus under doing it.
I have a lot of respect for many in the pro field, but I feel many get very insecure in the last 1-2 months and just add in too much.
The question is would he be faster if he biked more? //
Keep in mind that a lot of the people that bike more bike slower. Doing more miles doesn't always get you faster. I remember Dave Scott and a few others going through this in the early days, doing 400/500/600 mile weeks only to go faster in the race when they dropped down to the 300's with more quality. If he rides 12 hours in a week and his average pace is 27 or so mph, there is you 300+ miles a week. And I doubt that riding that few hours indoors that there are any junk miles in there..
And I just wanted to get my thoughts in here, have a feeling this thread is about to take a 180 degree turn away from the OP... (-;
CU427 wrote:
BrentwoodTriGuy wrote:
Which begs an even better question... are most pros doing way too much?
I would vote a yes on that.
Too much of what specifically I think is individual. But seeing so many come into Kona after heavy race schedule, talk of high volume training, etc. Then "just not have it" race day, fade off the back, etc. I think many just over do it. It would at least be safe to say more over do it versus under doing it.
I have a lot of respect for many in the pro field, but I feel many get very insecure in the last 1-2 months and just add in too much.
Actually, the original post was all about getting some extra insight into the what seems to be a different approach to Ironman training, so this discussion is on point.
It will be interesting to see if Lionel adjusts and increases his training volume upwards next year like BrentwoodTriGuy thinks he might, or if we'll see more pros moving more towards the Lionel end of the spectrum.
writhe wrote:
This is a pretty good ad for Bereda!
That was my thought as well. Also, I think it should be called Brenda
Dennis Cottreau wrote:
It will be interesting to see if Lionel adjusts and increases his training volume upwards next year like BrentwoodTriGuy thinks he might, or if we'll see more pros moving more towards the Lionel end of the spectrum.
Personally, I think he should bike a tad more and hang with Wurf the whole ride. I figure if they can gain an extra minute or two, Lionel might hold Lange off on the run... IF, big if, everything shakes out the same as this year.
With that said... LS is so incredibly motivated to swim front pack; what are the chances that he gets there and hangs with the pack for too long swimming AND biking, surges a bunch throughout this time and then sucks on the run?
I truly think he cracked the swim code this past year and getting "faster" won't necessarily help him in the long run (literally and figuratively). I think getting out of the swim in the uber-biker pack is most beneficial to him. He just needs to capitalize on what he didn't execute this year.. hanging with Wurf the last 20 miles and hydration. Work on that... I just don't see someone beating him.
Actually, the original post was all about getting some extra insight into the what seems to be a different approach to Ironman training, so this discussion is on point. //
I have know athletes that train like Lionel, one in particular who trained quite a bit less and had even more success than he has thus far. One of my old training partners would never ride much more than 200 miles a week, run 40+ and swim 4 days a week and would simply just crush Kona every year, setting several records there. She started to have problems when she got nervous about all the people racing her doing such big mileage, so she upped her volume. She never was as good as the low volume. high intensity days, but a good long history of one elite athlete who went both routes.
And as I said earlier, a lot of the male pros in my days went the more is better until some were just doing crazy mileage and hours. It mostly all fell back to a reasonable level and most went faster on the lower hours/higher intensity. Mostly it was the Germans doing some crazy miles, but they worked up to that over an entire career I guess and could handle it. They sure seemed to and still do very well at Kona as a country, and from stories I hear a lot still train the old school way of 40 hours a week..
Maybe the lower volume, is also what has kept him from becomming injured/ill, and that alone has given him an edge, has Lionel ever had an injury?
I wonder one thing about Sanders Training for a while now.
In middle distance track running you basically have athletes from two sides. The guys that come from more of a sprinter background and thrive on lower volume but more high end speed stuff. And the guys that benefit from higher mileage and more aerobic miles. (And a mixture of that).
I know that Ironman is a long way from mid-d but I wonder if Sanders is one of those guys that have a bit more fast twitch muscle fibres and thus can handle more very intense sessions and thrive on it. I might be off completely and someone with more physiological knowledge can debunk my theory. But what I notice is, that I am a complete endurance guy with no speed whatsoever. But I make big improvements when I train a lot at or just below threshold and also a ton in the "nowhere" zone between easy and hard. I just seem to thrive on it. But when I do too much intensity I plateau or even go backwards (when I overdo it).
So about Lionels high intensity/low volume training:
A: Is it something everyone should/could do to optimize his training?
B: Is it only ideal for some individuals?
C: Is he only able to do it because he still benefits from his huge base? (from the day when he trained something like an Ironman everyday)
D: Is he such a talent that every appraoch would have gotten him to the Kona podium? (and it is just the result of back to back years of hard training in whatever way)
Interesting is that Lionels approach evolves from blog post to blog post and he constantly optimizes and learns from the past. Maybe one day he will conclude that a bit more race pace stuff is helping. Scary to think that he still has room to grow.
ToBeasy wrote:
Interesting is that Lionels approach evolves from blog post to blog post and he constantly optimizes and learns from the past.
I agree, like I said, it will be interesting to see what he changes next year trying to make the next step. Will he add more volume or will others transition more to his end of the spectrum?
I'm surprised no one has chimed in to defend the virtues of high volume endurance training yet.
To improve you need to make your body do things it isn't used to. So Sanders could maintain his volume and make his hard sessions even harder. Or he can try to extend the amount of time he spends in those high zones.
Something interesting about the mileage monsters. There was an Interview with Hellriegel in a recent magazine. He mentions his most training hours week of all time. It was something like 56 hours. 54 on the bike and 2 running, so a big riding block with tons of elevation gain. But he himself states that this was of little physiological use. It was for the mental side. After that, he lost any fear of the 180k in races. Suddenly everything felt like downhill riding.
Sanders clicks off the mental side of his training with his indoor approach. So no need to do epic mileage destruction.
Prev Next
| 3,114
| 13,722
|
{"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-2018-13
|
latest
|
en
| 0.976151
|
https://math.stackexchange.com/questions/3171014/prove-that-exactly-one-player-has-a-winning-strategy-version-of-nim-game
| 1,709,268,205,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00248.warc.gz
| 370,013,064
| 33,897
|
# prove that exactly one player has a winning strategy - version of nim game
Game Description : the game is between 2 players and start with a pile of 'n' balls and A = {a1,a2,...,ak} ⊆ {1,....,n} .
Course of the game : each player, in his turn, picks a ∈ A
balls from the pile.
the loosing player : the player that can't play on his turn, player cant play on his turn when the number of balls in the pile is smaller from the minimum number in A.
the Task : we have to prove that to every show of the problem n,A = {a1,...,ak} only one one of the following happens : 1) the first player (the player play first) has a winning strategy . 2) the second player has an winning strategy. the winning strategy of the winner is not depend on the choices of the looser.
My Idea : I have tried to prove it with complete induction, but i'm stuck in the step.
I've assumed that for every show of the game with k<n balls the assert is holds and tried to show it holds for a game with 'n' balls.
so, the first case is that the first player can choose a ∈ A balls such that the number of balls left in the pile is less then the minimum number in A and hence player 2 cant perform his turn and player 1 allways wins.
otherwise, for any number of balls a ∈ A that player 1 took from the pile, player 2 can perform his turn. in this case the induction assertion holds (because the number of balls is less then 'n' in the pile now) and player 1 or player 2 (exactly one of them) has winning strategy. i'm stuck here. I don't know how to show that for every a ∈ A that player1 picks the player with the winning strategy will be the same one. for example if player 1 picks a1=4 from A and the induction assert find that player2 is the the player that has a winning strategy, it will return the same answear when player1 chooses a2=5 from A.
Thank you very much for your help!
The crucial point is that the game is impartial, that the options from a given position are the same for both players. That allows you to categorize each number as an $$N$$ position, won by the next player or a $$P$$ position, won by the previous player. Now $$n$$ is an $$N$$ position iff you can move to a $$P$$ position.
| 541
| 2,188
|
{"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.978033
|
https://vikaskashyapseo.com/qa/question-is-15000-words-enough-for-a-book.html
| 1,618,136,225,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00289.warc.gz
| 705,024,801
| 7,779
|
# Question: Is 15000 Words Enough For A Book?
## Is 20000 words enough for a book?
This means that you cannot fit many words on a printed page and keep it readable.
This translates into just 20,000 to 25,000 words for a 100 page book, 30,000 to 42,500 words for a 150-page book, perhaps just 40,000 words for a 200-page book..
## What is the average word count for a book?
90,000 wordsMost adult books are about 90,000 words, and no longer than 100,000 words (unless you’re JK Rowling). Teen books are about 55,000 words. How many words are there to a page? It depends on the font you are using, of course, but in general, 250-300 words per page.
## Is 30 000 words enough for a book?
Big, epic stories get anywhere from 120,000 to 200,000 words.” But, he also mentions that “The Wizard of Oz was 40,000 words. The Old Man and the Sea was about 25 to 30,000 words, tops.” … If you’re working on a novel-length book, aim for 50,000 words at the very least — but it’s better to aim for 90,000.
## How many pages is 90000 words?
180.0 pagesA 90,000 word count will create about 180.0 pages single-spaced or 360.0 pages double-spaced when using normal margins (1″) and 12 pt.
## What is 20000 words in pages?
20,000 words is 40 pages single spaced, 80 pages double spaced.
## How many pages is 70 000 words?
70,000 words is 140 pages single spaced, 280 pages double spaced. 75,000 words is 150 pages single spaced, 300 pages double spaced. 80,000 words is 160 pages single spaced, 320 pages double spaced.
## How short can a book be?
However, very few novels these days are as short as that. Generally a 50,000-word novel would be the minimum word count. Most novels are between 60,000 and 100,000 words.
## How many pages is 15000 words?
30.0 pagesA 15,000 word count will create about 30.0 pages single-spaced or 60.0 pages double-spaced when using normal margins (1″) and 12 pt. Arial or Times New Roman font.
## How many pages is 300000 words?
Pages by Word CountWord CountPages (single spaced)Pages (double spaced)300 Words⅔ Page1⅓ Pages400 Words⅘ Page1⅗ Pages500 Words1 Page2 Pages600 Words1⅓ Page2⅔ Pages16 more rows
## What is the longest book in existence?
Remembrance of Things PastThe Guinness Book of World Records gives the honor to Marcel Proust’s elephantine Remembrance of Things Past, weighing in at 9,609,000 characters (including spaces).
## What is the trick to increase word count?
Read over your paper and look for any statements that seem long-winded or unclear. Break apart long sentences or statements into multiple sentences. Then, clarify your ideas in each sentence to help increase your word count. Add follow up statements so your ideas are specific and well rounded in the paper.
## How many book pages is 100000 words?
300 pagesSure there is, Sheila… the rule of thumb with most publishers is to average about 300 words per page. So a 100,000-word novel will run about 300 pages.
| 772
| 2,932
|
{"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}
| 3.28125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.932957
|
https://www.physicsforums.com/threads/technical-system-of-units.768043/
| 1,519,208,235,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891813602.12/warc/CC-MAIN-20180221083833-20180221103833-00450.warc.gz
| 917,453,238
| 17,260
|
# Technical system of units
1. Aug 28, 2014
### Karol
1. The problem statement, all variables and given/known data
The specific mass of iron doesn´t seem right in the technical system.
I know the weight of 1[m3] steel is about 8 tons.
2. Relevant equations
The definition of the kilogram-force unit in the tehnical system is the weight of a mass of 1 kilogram:
1[kgf]=>1[kg]x10
3. The attempt at a solution`
That was the definition, now:
1[kgf]=0.1[kg-mass]x10
=>1[kg]=0.1[kg-mass]
8[gr/cm3]=8000[kg/m3]
I have to divide the mass with 10 to get gk-mass:
8000[kg/m3]=800[kg-mass/m3]
And in a book it´s written that a certain metal wire has a density of 10,300[kg-mass/m3]
So my solution is about 10 times too low
2. Aug 28, 2014
### PhanthomJay
one kgf (which is not an SI unit) is the weight of one kg of mass on planet earth.
A kilogram of mass (designated as kg) has a weight of about 9.8 Newtons (call it 10 N) in SI units. But you are not using SI, so you are dividing by 10 when you shouldn't be.
3. Aug 29, 2014
### Karol
I don´t talk about newtons, only kg and kg-mass.
If 1[kgf]=0.1[kg-mass]x10 is correct (is it?) and with the definition of kgf, then to convert from kg-mass to kg i have to multiply by ten, no?
4. Aug 29, 2014
### PhanthomJay
No, that is not correct. A kg-mass and a kg are one and the same. If you multiply kg-mass , which is a kg, by 10 (or actually 9.8 m/sec^2, let's use 10), you get the weight of one kg in the SI unit of Newtons, on Earth.
It gets very confusing when using different systems of measure like SI, metric, technical, or Imperial units.
A kg of force is actually F = (m/g)(a), where m is in kilograms, g is 10, and a is the acceleration in m/sec^2. On Earth, when you calculate the weight of 1 kg of mass, the acceleration is g or 10 m/sec^2, so the equation becomes F = (m/g)(g), the g's cancel, so F =m. That is , one kg of mass weighs one kg of force, on Earth. The book answer is correct.
5. Aug 29, 2014
### Karol
Why is F=(m/g)a? especially why (m/g)? does it come from the definition of kg-force?
6. Aug 31, 2014
### PhanthomJay
yes, it's a conversion factor. Another way to look at it is to rewrite Newtons 2nd Law as
F = kma and W = kmg. m is in kg and a or g is in m/s^2. If you want your force or weight in Newtons , k= 1. If you want your force or weight in kg-force, k = 0.1 (in round numbers). The kg- force unit should be avoided whenever possible.
| 770
| 2,431
|
{"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}
| 3.875
| 4
|
CC-MAIN-2018-09
|
latest
|
en
| 0.910055
|
http://mathhelpforum.com/number-theory/131376-prove.html
| 1,481,231,196,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698542655.88/warc/CC-MAIN-20161202170902-00044-ip-10-31-129-80.ec2.internal.warc.gz
| 181,995,009
| 9,549
|
for $n \in \mathbb{N}$ Let $\ p_n$ \ be the $n^{th}$ prime $(thus \ , p_1=2,p_2=3,p_3=5,p_4=7,p_5=11,...):$
prove that $\ p_n \leq p_1 p_2 p_3...p_{n-1} +1 , n \geq 3$
| 92
| 167
|
{"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": 5, "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-2016-50
|
longest
|
en
| 0.204762
|
https://www.justintools.com/unit-conversion/length.php?k1=varas-%5BLatin-America%5D&k2=varas-%5BArgentina%5D
| 1,713,252,345,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00551.warc.gz
| 791,430,336
| 28,191
|
Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
LENGTH Units Conversionvaras-[Latin-America] to varas-[Argentina]
1 Varas [Latin America]
= 0.99769053117783 Varas [Argentina]
Embed this to your website/blog
Category: length
Conversion: Varas [Latin America] to Varas [Argentina]
The base unit for length is meters (SI Unit)
[Varas [Latin America]] symbol/abbrevation: (vr[Latin])
[Varas [Argentina]] symbol/abbrevation: (vr[Argentina])
How to convert Varas [Latin America] to Varas [Argentina] (vr[Latin] to vr[Argentina])?
1 vr[Latin] = 0.99769053117783 vr[Argentina].
1 x 0.99769053117783 vr[Argentina] = 0.99769053117783 Varas [Argentina].
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [length] => (meters), 1 Varas [Latin America] (vr[Latin]) is equal to 0.864 meters, while 1 Varas [Argentina] (vr[Argentina]) = 0.866 meters.
1 Varas [Latin America] to common length units
1 vr[Latin] = 0.864 meters (m)
1 vr[Latin] = 0.000864 kilometers (km)
1 vr[Latin] = 86.4 centimeters (cm)
1 vr[Latin] = 2.8346456692913 feet (ft)
1 vr[Latin] = 34.015748031496 inches (in)
1 vr[Latin] = 0.94488188976378 yards (yd)
1 vr[Latin] = 0.00053686471009306 miles (mi)
1 vr[Latin] = 9.1322270373111E-17 light years (ly)
1 vr[Latin] = 3265.5122224267 pixels (PX)
1 vr[Latin] = 5.4E+34 planck length (pl)
Varas [Latin America]to Varas [Argentina] (table conversion)
1 vr[Latin] = 0.99769053117783 vr[Argentina]
2 vr[Latin] = 1.9953810623557 vr[Argentina]
3 vr[Latin] = 2.9930715935335 vr[Argentina]
4 vr[Latin] = 3.9907621247113 vr[Argentina]
5 vr[Latin] = 4.9884526558891 vr[Argentina]
6 vr[Latin] = 5.986143187067 vr[Argentina]
7 vr[Latin] = 6.9838337182448 vr[Argentina]
8 vr[Latin] = 7.9815242494226 vr[Argentina]
9 vr[Latin] = 8.9792147806005 vr[Argentina]
10 vr[Latin] = 9.9769053117783 vr[Argentina]
20 vr[Latin] = 19.953810623557 vr[Argentina]
30 vr[Latin] = 29.930715935335 vr[Argentina]
40 vr[Latin] = 39.907621247113 vr[Argentina]
50 vr[Latin] = 49.884526558891 vr[Argentina]
60 vr[Latin] = 59.86143187067 vr[Argentina]
70 vr[Latin] = 69.838337182448 vr[Argentina]
80 vr[Latin] = 79.815242494226 vr[Argentina]
90 vr[Latin] = 89.792147806005 vr[Argentina]
100 vr[Latin] = 99.769053117783 vr[Argentina]
200 vr[Latin] = 199.53810623557 vr[Argentina]
300 vr[Latin] = 299.30715935335 vr[Argentina]
400 vr[Latin] = 399.07621247113 vr[Argentina]
500 vr[Latin] = 498.84526558891 vr[Argentina]
600 vr[Latin] = 598.6143187067 vr[Argentina]
700 vr[Latin] = 698.38337182448 vr[Argentina]
800 vr[Latin] = 798.15242494226 vr[Argentina]
900 vr[Latin] = 897.92147806005 vr[Argentina]
1000 vr[Latin] = 997.69053117783 vr[Argentina]
2000 vr[Latin] = 1995.3810623557 vr[Argentina]
4000 vr[Latin] = 3990.7621247113 vr[Argentina]
5000 vr[Latin] = 4988.4526558891 vr[Argentina]
7500 vr[Latin] = 7482.6789838337 vr[Argentina]
10000 vr[Latin] = 9976.9053117783 vr[Argentina]
25000 vr[Latin] = 24942.263279446 vr[Argentina]
50000 vr[Latin] = 49884.526558891 vr[Argentina]
100000 vr[Latin] = 99769.053117783 vr[Argentina]
1000000 vr[Latin] = 997690.53117783 vr[Argentina]
1000000000 vr[Latin] = 997690531.17783 vr[Argentina]
(Varas [Latin America]) to (Varas [Argentina]) conversions
| 1,222
| 3,512
|
{"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-2024-18
|
latest
|
en
| 0.634934
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.