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https://oeis.org/A194405 | 1,718,437,771,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00665.warc.gz | 373,567,209 | 3,995 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A194405 Numbers m such that Sum_{k=1..m} (<1/2 + k*e> - ) < 0 where < > denotes fractional part. 4
1, 5, 33, 37, 65, 69, 71, 72, 73, 75, 76, 77, 79, 83, 97, 101, 103, 104, 105, 107, 108, 109, 111, 115, 129, 133, 135, 136, 137, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 153, 154, 155, 157, 161, 165, 167, 168, 169, 171, 172, 173 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS See A194368. LINKS Table of n, a(n) for n=1..54. MATHEMATICA r = E; c = 1/2; x[n_] := Sum[FractionalPart[k*r], {k, 1, n}] y[n_] := Sum[FractionalPart[c + k*r], {k, 1, n}] t1 = Table[If[y[n] < x[n], 1, 0], {n, 1, 200}]; Flatten[Position[t1, 1]] (* A194405 *) t2 = Table[If[y[n] == x[n], 1, 0], {n, 1, 200}]; Flatten[Position[t2, 1]] (* A194406 *) t3 = Table[If[y[n] > x[n], 1, 0], {n, 1, 200}]; Flatten[Position[t3, 1]] (* A194407 *) CROSSREFS Cf. A194368. Sequence in context: A275233 A359971 A043068 * A299518 A338342 A135111 Adjacent sequences: A194402 A194403 A194404 * A194406 A194407 A194408 KEYWORD nonn AUTHOR Clark Kimberling, Aug 24 2011 STATUS approved
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Last modified June 15 03:12 EDT 2024. Contains 373402 sequences. (Running on oeis4.) | 657 | 1,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.63055 |
https://www.soughttech.com/front/article/19122/viewArticle | 1,680,342,994,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00199.warc.gz | 1,126,771,057 | 7,589 | ## How to find the intersection between two Numpy arrays?
In this problem, we will find the intersection between two numpy arrays. The intersection of two arrays is an array that has elements in common in both original arrays
## algorithm
```Step 1 : Import numpy.
Step 2 : Define two numpy arrays.
Step 3 : Find intersection between the arrays using the numpy.intersect1d() function.
Step 4 : Print the array of intersecting elements.```
## sample code
```import numpy as np
array_1 = np.array([1,2,3,4,5])
print( "Array 1:\n" , array_1)
array_2 = np.array([2,4,6,8,10])
print( "\nArray 2:\n" , array_2)
intersection = np.intersect1d(array_1, array_2)
print( "\nThe intersection between the two arrays is:\n" , intersection)```
output result
```Array 1:
[1 2 3 4 5 ]
Array 2:
[2 4 6 8 10 ]
The intersection between the two arrays is:
[twenty four ]```
#### Technical otaku
Sought technology together | 348 | 937 | {"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-2023-14 | latest | en | 0.232093 |
https://rdrr.io/cran/BayesS5/src/R/result.R | 1,709,536,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00204.warc.gz | 470,283,106 | 7,162 | # R/result.R In BayesS5: Bayesian Variable Selection Using Simplified Shotgun Stochastic Search with Screening (S5)
#### Documented in result
```result <-
function(fit){
#GAM = fit[-1,]; OBJ = fit[1,]
GAM = fit\$GAM; OBJ = fit\$OBJ; tuning= fit\$tuning
p = nrow(GAM)
marg.gam = rep(0,p)
for(u in 1:ncol(GAM)){
marg.gam = marg.gam + GAM[,u]*exp(OBJ[u]-max(OBJ))
}
marg.gam = marg.gam / sum(exp(OBJ-max(OBJ)))
gam0 = GAM[,which.max(OBJ)]
ind2 = which(gam0==1)
post = exp(OBJ-max(OBJ))/sum(exp(OBJ-max(OBJ)))
hppm = 1/sum(exp(OBJ-max(OBJ)))
print("# of Searched Models by S5");print(length(OBJ))
print("The MAP model is ")
print(which(gam0==1))
print(paste("with posterior probability",round(hppm,3) ))
return(list(hppm = which(gam0==1), hppm.prob = hppm, marg.prob = marg.gam,gam = GAM, obj = OBJ, post = post, tuning = tuning) )
}
```
## Try the BayesS5 package in your browser
Any scripts or data that you put into this service are public.
BayesS5 documentation built on March 26, 2020, 7:14 p.m. | 340 | 1,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.63002 |
https://www.convertunits.com/from/sthene/to/megapond | 1,600,627,347,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00269.warc.gz | 818,470,459 | 11,686 | ## ››Convert sthene to megapond
sthene megapond
## ››More information from the unit converter
How many sthene in 1 megapond? The answer is 9.80665.
We assume you are converting between sthene and megapond.
You can view more details on each measurement unit:
sthene or megapond
The SI derived unit for force is the newton.
1 newton is equal to 0.001 sthene, or 0.00010197162129779 megapond.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between sthenes and megaponds.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of sthene to megapond
1 sthene to megapond = 0.10197 megapond
5 sthene to megapond = 0.50986 megapond
10 sthene to megapond = 1.01972 megapond
20 sthene to megapond = 2.03943 megapond
30 sthene to megapond = 3.05915 megapond
40 sthene to megapond = 4.07886 megapond
50 sthene to megapond = 5.09858 megapond
75 sthene to megapond = 7.64787 megapond
100 sthene to megapond = 10.19716 megapond
## ››Want other units?
You can do the reverse unit conversion from megapond to sthene, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Sthene
The sthene is the unit of force in the former Soviet mts system, 1933-1955. The symbol is sn.
## ››Definition: Megapond
The SI prefix "mega" represents a factor of 106, or in exponential notation, 1E6.
So 1 megapond = 106 ponds.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 578 | 1,921 | {"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} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.824118 |
https://journalofcosmology.com/what-causes-gravity-in-the-universe/ | 1,696,205,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00883.warc.gz | 356,158,054 | 19,700 | What Causes Gravity in the Universe?
Gravity is a fundamental force of nature that governs the motion of objects in the universe. It is the force that keeps planets in orbit around stars and holds stars together in galaxies. Despite its importance, the nature of gravity is still not well understood. In this essay, we will explore what causes gravity in the universe.
Gravity is a fundamental force in the universe that affects all objects with mass. Despite its universal presence, the exact cause of gravity is still a mystery to scientists. In this discussion, we will explore some of the theories behind what causes gravity in the universe.
The Theory of General Relativity
The most widely accepted theory of gravity is the theory of general relativity, developed by Albert Einstein in 1915. According to this theory, gravity is not a force between masses, as described by Isaac Newton’s theory of gravity. Instead, gravity is a curvature of spacetime that is caused by the presence of mass and energy.
Spacetime
Spacetime is the four-dimensional fabric of the universe, consisting of three dimensions of space and one dimension of time. According to general relativity, the presence of mass and energy curves spacetime, causing objects to move along curved paths. The more massive an object is, the more it curves spacetime.
The Equivalence Principle
One of the key principles of general relativity is the equivalence principle, which states that gravitational and inertial forces are indistinguishable. In other words, an object that is accelerating due to a force (such as gravity) experiences the same physical effects as an object that is stationary in a gravitational field.
Gravity and Quantum Mechanics
Despite the success of general relativity in describing gravity on a macroscopic scale, it is incompatible with quantum mechanics, which describes the behavior of subatomic particles. One of the major challenges of modern physics is to develop a theory of quantum gravity that reconciles these two theories.
One key takeaway from this text is that the most widely accepted theory of gravity is the theory of general relativity, which states that gravity is a curvature of spacetime caused by the presence of mass and energy. Despite its success in describing gravity on a macroscopic scale, general relativity is incompatible with quantum mechanics, and developing a theory of quantum gravity that reconciles these two theories is a major challenge for modern physics. Other theories of gravity, such as Modified Newtonian Dynamics and Emergent Gravity, attempt to explain the nature of gravity in different ways. The existence of gravitational waves, another prediction of general relativity, was confirmed by the Laser Interferometer Gravitational-Wave Observatory (LIGO) in 2015.
The Graviton
One possible candidate for a quantum theory of gravity is the graviton, a hypothetical particle that would mediate the gravitational force. However, the graviton has not yet been observed, and its existence is still a topic of debate among physicists.
String Theory
Another approach to quantum gravity is string theory, which posits that the fundamental building blocks of the universe are one-dimensional strings rather than pointlike particles. In string theory, gravity arises from the vibrational modes of these strings.
Other Theories of Gravity
While general relativity is the most widely accepted theory of gravity, there are other theories that attempt to explain the nature of gravity.
Modified Newtonian Dynamics
Modified Newtonian Dynamics (MOND) is a theory that attempts to explain the observed behavior of galaxies without the need for dark matter. According to MOND, the gravitational force between two masses is stronger than predicted by Newton’s law of gravity at low accelerations.
Emergent Gravity
Emergent gravity is a theory that proposes that gravity is not a fundamental force, but rather an emergent property of spacetime. According to this theory, gravity arises from the collective behavior of many small degrees of freedom in spacetime.
Gravitational Waves
Another prediction of general relativity is the existence of gravitational waves, ripples in the fabric of spacetime that are caused by the motion of massive objects. In 2015, the Laser Interferometer Gravitational-Wave Observatory (LIGO) detected the first direct evidence of gravitational waves, confirming a key prediction of general relativity.
FAQs – What Causes Gravity in the Universe?
What is gravity?
Gravity is a force that exists between any two objects in the universe. It is responsible for pulling objects towards each other. The larger the objects, the stronger the force of gravity.
What causes gravity?
Gravity is caused by the mass and energy of objects. The more mass an object has, the stronger its gravitational pull. This is why the Earth has a stronger gravitational pull than the Moon.
How does gravity work?
Gravity works by creating a force field around an object that attracts other objects towards it. This force field is strongest at the center of the object and weakens as you move further away from it. The force of gravity also decreases with distance, which is why objects that are farther apart have weaker gravitational forces.
What is the theory of General Relativity?
The theory of General Relativity, developed by Albert Einstein in the early 1900s, explains gravity as the curvature of spacetime. In this theory, objects with mass or energy cause spacetime to bend, creating a gravitational force. This theory has been extremely successful in explaining the behavior of objects in space and has been confirmed by numerous experiments.
Can gravity be explained by other theories?
While General Relativity is currently the leading theory explaining gravity in the universe, there are other theories that attempt to explain it. Some of these theories include Modified Newtonian Dynamics (MOND) and Quantum Gravity. However, there is currently no other theory that has been able to match the predictive power and accuracy of General Relativity. | 1,163 | 6,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.933386 |
https://www.jiskha.com/questions/668903/The-terminal-side-of-lies-on-a-given-line-in-the-specified-quadrant-Find-the | 1,540,021,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512592.60/warc/CC-MAIN-20181020055317-20181020080817-00429.warc.gz | 991,530,330 | 4,842 | # trig
The terminal side of θ lies on a given line in the specified quadrant. Find the values of the six trigonometric functions of θ by finding a point on the line. Line: (55/48)x Quadrant:III
sin(θ)=
cos(θ)=
tan(θ)=
csc(θ)=
sec(θ)=
cot(θ)=
1. 0
2. 14
1. I will assume you are saying that the line is
y = (55/48)x
or one point on the line in quadrant III would be
(-48,-55)
then r^2 = (-48)^2 + (-55)^2
r = √5329 = 73
sinØ = y/r = -55/73
cosØ = x/r = -48/73
etc.
posted by Reiny
2. Line: 4x+3y=0 IV
posted by Anonymous
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More Similar Questions | 677 | 2,342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-43 | latest | en | 0.76524 |
https://infinitylearn.com/surge/maths/dilation-in-maths/ | 1,726,816,483,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00418.warc.gz | 286,563,604 | 25,579 | Dilation
Dilation
Explain in Detail :Dilation and its Scope in Mathematics
In mathematics, dilation is a transformation that enlarges or shrinks objects by a scale factor. The scale factor may be positive or negative, and the result is a new object that is similar to the original. Dilations can be described by pointing out a center of dilation and a scale factor.
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What is a Scalar Factor?
A scalar factor is a number that is multiplied by each element of a vector, or a number that is multiplied by each term of a series.
A Few Important Properties Of Dilation
Some of the important properties of dilation are as follows:
1. Dilation is a linear transformation.
2. The magnitude of the dilation is proportional to the magnitude of the original object.
3. The direction of the dilation is unchanged.
4. Dilated objects are always similar to the original object.
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https://www.sanfoundry.com/differential-integral-calculus-questions-answers-taylors-theorem-two-variables/ | 1,713,825,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00167.warc.gz | 849,951,115 | 22,852 | # Differential and Integral Calculus Questions and Answers – Taylor’s Theorem Two Variables
This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Taylor’s Theorem Two Variables”.
1. Among the following which is the correct expression for Taylor’s theorem in two variables for the function f (x, y) near (a, b) where h=x-a & k=y-b upto second degree?
a) $$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ +2 \frac{(x-a)(y-b)}{4!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
b) $$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ + \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
c) $$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
d) $$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ +2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
Explanation: By definition
$$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ +2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
here we can observe that second degree is of the form (p+q)2 similarly Taylor’s theorem is expanded to third degree which is of the form (p+q)3 & f (a+ h, b+ k) = f (x, y)
where$$(f_x=\frac{∂f (x,y)}{∂x}, f_y=\frac{∂f (x,y)}{∂y}, f_{xx}=\frac{∂}{∂x}(\frac{∂f(x,y)}{∂x}), f_{yy}=\frac{∂}{∂y} (\frac{∂f (x,y)}{∂y}), \\ f_{xy}=\frac{∂}{∂x}(\frac{∂f (x,y)}{∂y})).$$
2. Given f (x,y)=ex cosy, what is the value of the fifth term in Taylor’s series near (1,$$\frac{π}{4}$$) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) $$\frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}$$
b) $$-\sqrt{2} e(x-1)(y-\frac{π}{4})$$
c) $$\frac{e(x-1)^2}{\sqrt{2}}$$
d) $$\frac{e(y-\frac{π}{4})^2}{\sqrt{2}}$$
Explanation: Taylor’s series expansion is given by
$$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ +2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
Thus fifth term is given by $$2 \frac{(x-a)(x-b)}{2!} f_{xy} (a,b)$$..(1) where a=1, b=π/4 & $$f_{xy}=\frac{∂}{∂x}(\frac{∂f(x,y)}{∂x}) = \frac{∂}{∂x}(\frac{∂e^x cosy}{∂y}) =- e^x \,siny$$ at (1, $$\frac{π}{4}), f_{xy}=\frac{-e}{\sqrt{2}}$$ substituting in (1)
We get fifth term as $$2 \frac{(x-1)(x-π/4)}{2!} \frac{-e}{\sqrt{2}} = \frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}$$.
3. Given f (x,y)=sinxy, what is the value of the third degree first term in Taylor’s series near (1,-$$\frac{π}{2}$$) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) $$\frac{π^3}{8}$$
b) $$\frac{π^3}{8} \frac{(x-1)(y+\frac{π}{2})}{3!}$$
c) 0
d) $$-\frac{π^3}{8} \frac{(x-1)^3}{3!}$$
Explanation: Third degree first term in Taylor’s series is given by $$\frac{(x-a)^3 f_{xxx} (x,y)}{3!}$$ Where a=1 $$b=-\frac{π}{2}, f_{xxx} (x,y)=\frac{∂^3 f(x,y)}{∂x^3} \,i.e\, \frac{∂^3 sinxy}{∂x^3} = -y^3 cosxy$$…… (partial differentiating f (x,y) w.r.t x only)
at $$a=1, b=-\frac{π}{2}, \frac{∂^3 sinxy}{∂x^3} = -\frac{π^3 cos-\frac{π}{2}}{8}=0$$ hence third degree first term is given by $$-\frac{π^3}{8} \frac{(x-1)^3}{3!}.0 = 0.$$
4. Taylor’s theorem is mainly used in expressing the function as sum with infinite terms.
a) True
b) False
Explanation: Taylor’s theorem helps in expanding a function into infinite terms however, it can be applied to functions that can be expressed finitely.
5. Expansion of $$f (x,y) = tan^{-1} \frac{y}{x}$$ upto first degree containing (x+1) & (y-1) is __________
a) $$\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{2} + \frac{(y-1)^2}{2!} \frac{1}{2}$$
b) $$\frac{π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}$$
c) $$\frac{5π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}$$
d) $$\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}$$
Explanation: We can expand the given function according to Taylor’s theorem
$$f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\ +2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)$$
Given a=-1 & b=1, f(-1,1)=tan-1-1 = $$\frac{3π}{4}$$
$$f_x = \frac{-y}{x^2+y^2} \,at\, (-1,1) = \frac{-1}{2}$$
$$f_y = \frac{x}{x^2+y^2} \,at\, (-1,1) = \frac{-1}{2}$$
$$f_{xy} = \frac{(x^2+y^2)-2x^2}{(x^2+y^2)^2}$$ at (-1,1)=0
$$f_{xx} = \frac{2yx}{(x^2+y^2)^2} \,at\, (-1,1)=\frac{-2}{4} = \frac{-1}{2}$$
$$f_{yy} = \frac{-2yx}{(x^2+y^2)^2} \,at\, (-1,1)=\frac{2}{4} = \frac{1}{2}$$ thus the series is given by
$$\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{2} + \frac{(y-1)^2}{2!} \frac{1}{2}$$.
Sanfoundry Global Education & Learning Series – Differential and Integral Calculus.
To practice all areas of Differential and Integral Calculus, here is complete set of 1000+ Multiple Choice Questions and Answers. | 2,640 | 5,499 | {"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} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.541029 |
https://www.jiskha.com/questions/1081835/how-many-gallons-of-a-30-alcohol-solution-must-be-mixed-with-60-gallons-of-a-14-solution | 1,604,149,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107918164.98/warc/CC-MAIN-20201031121940-20201031151940-00479.warc.gz | 702,367,583 | 5,691 | # college algebra
How many gallons of a 30% alcohol solution must be mixed with 60 gallons of a 14%
solution to obtain a solution that is 20% alcohol
1. 👍 0
2. 👎 0
3. 👁 764
1. number of gallons of the 30% stuff --- x
.3x + .14(60) = .2(60+x)
30x + 14(60) = 20(60+x)
30x + 840 = 1200 +20x
10x = 360
x = 36
1. 👍 1
2. 👎 0
2. .3*V+.14*60=.2*(V+60)
solve for V
1. 👍 0
2. 👎 0
👨🏫
bobpursley
3. (2x2y2-3xy+4)(4xy2)
1. 👍 0
2. 👎 0
4. A mixture of alcohol and water contains a total of
52
52 oz of liquid. There are
13
13 oz of pure alcohol in the mixture. What percent of the mixture is water? What percent is alcohol?
1. 👍 0
2. 👎 0
5. A mixture of alcohol and water contains a total of
52 oz of liquid. There are
13 oz of pure alcohol in the mixture. What percent of the mixture is water? What percent is alcohol?
1. 👍 0
2. 👎 0
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# Course project | Numerical analysis homework help
Housing Data File
This data set contains 25 variables that are described in the data dictionary tab and include
several categorical variables, some binary variables, and numerical data. You can approach this
data from a variety of perspectives using the techniques you learned in class to answer
questions. Use the tool of your choice but make sure you know how to use it correctly. There
are 2930 observations, more than enough to provide valid and reliable statistical analysis.
You have been hired by the local real estate broker to analyze activity in the local housing
market. You must conduct three ANOVA analyses, a correlation analysis, and three Regression
analyses to answer six questions you believe will help the broker provide the best guidance to
both buyers and sellers. Categorical variables include building type, neighborhood, and house
style. If you think it is important to understand the age of the house when it sold, you will need to
create a new variable (year built and year sold are two variables provided). As with any project,
you will start with EDA to get a sense of your data. For categorical variables, using a pivot table
to get counts and proportions is an excellent way to get a better understanding of those
variables. Example questions could focus on house prices in different neighborhoods, those
sold in different years, impact of lot size or house square footage on price, etc.
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Deliverables
Perform EDA and include information about the data in the report that helps the reader get an
understanding of the data set (useful graphics should be included). Develop six research
questions (ideas include regression model to predict price, ANOVA to compare a numerical
variables based upon categorical variable groups, etc.). Perform the analysis and write a
detailed description of the results and what they mean (how you would use them). Be sure to
include appropriate graphics.
Create a 4-5 slide presentation that would support a very brief presentation (3 minutes) of your
analysis.
below is a sample paper in the files uploades. paper does not need to be 24 pages, but I have include this so you can get a sense of the overall flow of the document and what you should include.
the attachments consists of the required dataset and a sample paper and please use excel with data analytics package for the assignment and please provide the excel files also
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Order your essay today and save 15% with the discount code DISCOUNTS2023 | 664 | 3,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | latest | en | 0.913567 |
https://www.coursehero.com/file/6334026/hw3/ | 1,513,616,578,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948618633.95/warc/CC-MAIN-20171218161254-20171218183254-00727.warc.gz | 741,858,219 | 49,855 | # hw3 - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...
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AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 3 Due at the beginning of class on Thursday, September 30, 2010. Reminder: Show your reasoning! Read: Ross, Chapter 4, Sections 4.1–4.5. You may skip Example 1e, Example 4c (1). (30 points) In your pocket, you have 2 dimes, 2 nickels, and 1 penny. You select 3 coins at random (without replacement). Let X represent the amount (in cents) that you select from your pocket. (a). Give (explicitly) the probability mass function for X . Also show a plot of it. (b). Give (explicitly) the cdf, F ( x ), for X . Also show a plot of it. (c). How much money do you expect to draw from your pocket? (2). (25 points) Consider a random variable X whose distribution function (cdf) is given by F X ( x ) = 0 if x < - 1 0 . 2 if - 1 x < 1 . 2 0 . 3 if 1 . 2 x < 2 0 . 5 if 2 x < 3 0 . 6 if 3 x < 4 1 if x 4 (a). (6 points) Give the probability mass function, p ( x ), of X , explicitly. (b). (3 points) Compute
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https://answers.opencv.org/question/209492/how-to-get-the-number-of-pixels-in-an-roi-c/?answer=209607 | 1,571,096,535,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00500.warc.gz | 379,084,213 | 14,564 | Ask Your Question
# How to get the number of pixels in an ROI c++?
Hello, I have a sample image, my goal is to find the number of red pixels (black if turned to binary) per circle. A sample output like:
Circle 1: 900
Circle 2: 400
Circle 3: 300
Here is the sample image:
I have been successful in using the countNonZero function for individually cropped circle and using an area of about 30*30 to get the results. However I want to use the base image and get ROIs for each circle then calculate the pixels. Thanks!
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## 2 answers
Sort by ยป oldest newest most voted
If I've understand the question correctly, you need first to find circles in your images, you can do it using the Hough Circle Transform, here a tutorial in C++
Setting the function with the proper parameters (as I can see, you know the size of the circles, more or less..) you'll obtain the circles center coordinates back from the function:
HoughCircles( src_gray, circles, HOUGH_GRADIENT, 1, src_gray.rows/8, 200, 100, 0, 0 );
circles: A vector that stores sets of 3 values: Xc, Yc, r for each detected circle.
so using that you'll be able to get the proper ROIs for each circle and then apply the function that @ducvd wrote.
The function give to you also the radius, so what you get at the end is the center of the ROI. Is pretty easy to understand that getting the rectangle coordinate is easy. For each circles (ROI), the top left corner of the rectangle will be (Xc-r-thresh, Yc-r-thresh), while the bottom right corner will be (Xc+r+thres, Yc+r+thresh) where Xc and Yc are the circle center gave back from the function, r is the radius( again given back from the function) and thresh is a parameter that you can set to handle the uncertainty regarding the radius of the circle.
Hope that help
more
If you are able to find ROI, the remains problem is counting the Read-pixels inside ROI rectangle.
You could do it by enumerate every pixels in the images then count (please do check whether pixels is inside ROI yourself):
const int channels = I.channels();
count = 0; // count is the number of Red pixels
switch(channels)
{
case 1:
{
MatIterator_<uchar> it, end;
for( it = I.begin<uchar>(), end = I.end<uchar>(); it != end; ++it)
if (*it == 0) // black
{
count++;
}
break;
}
case 3:
{
MatIterator_<Vec3b> it, end;
for( it = I.begin<Vec3b>(), end = I.end<Vec3b>(); it != end; ++it)
{
if ((*it[2])==255) // Red
{
count++;
}
}
}
}
more
## Comments
1
I think that getting the ROI areas is the question. Count it is more than easy. If I have understand correctly, the question is: how can he detect the circles so that he can count the red values, given an arbitrary image as input? So in that case, the algorithm should be able to identify the 6 circles, crop them (getting the ROI area) automatically and then count the red pixels. I think that is what the OP was looking for. In that case, you can take a look at THIS tutorial.
( 2019-02-28 07:48:30 -0500 )edit
1
Thanks you for your idea!
( 2019-03-01 02:36:43 -0500 )edit
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https://us.sofatutor.com/mathematics/videos/the-volume-of-a-rectangular-prism-using-the-area-of-the-base | 1,657,114,651,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00655.warc.gz | 626,233,554 | 37,635 | # The Volume of a Rectangular Prism Using the Area of the Base
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## Basics on the topicThe Volume of a Rectangular Prism Using the Area of the Base
After this lesson you will be able to develop and use the formula V=Bh.
The lesson begins with the formula for a rectangular prism, V=lwh. It leads to calculating the volume of rectangular prisms with the same area of base and different heights. It concludes with a new formula for finding the volume of a rectangular prism, V=Bh, where B is the area of the base.
Learn how to use the area of base to find the volume of a rectangular prism by helping Rocky fit all of his stones in storage!
This video includes key concepts, notation, and vocabulary such as: area of base (the base of a rectangular prism is an arbitrarily chosen rectangular face); height (the height is the remaining dimension after we have calculated the area of the base); and volume of a rectangular prism (Volume equals the area of the base times the height, V=Bh).
Before watching this video, you should already be familiar with fractions, calculating area for rectangles, and calculating volume by filling with cubes or using the formula, V=lwh.
After watching this video, you will be prepared to learn that volume changes when we double or triple the dimensions of the prism.
Common Core Standard(s) in focus: 6.G.A.2 A video intended for math students in the 6th grade Recommended for students who are 11-12 years old
### TranscriptThe Volume of a Rectangular Prism Using the Area of the Base
Gem is reading the latest issue of Better Caves and Firepits magazine. Stone blocks used to be popular decoration, but the world has moved on! Unless they want to be cultural dinosaurs, Gem's husband Rocky must remove his stone collection. Rocky is attached to his stones but he has an idea! He can try to fit all his stones in his storage cave. If Rocky can calculate the volume of a rectangular prism using the area of the base he just might fit all his stone blocks in storage. Rocky has 3 different sizes of stone blocks. How can Rocky calculate the volume of each? The formula for the volume of a rectangular prism is length times width times height. The volume of the first block is the length, 5 times the width, 2 times the height which is also 2. That's 10 times 2 or 20 cubic inches. The volume of the second is 5 times 2 times 7. That's 10 times 7 or 70 cubic inches. The volume of the third is 5 times 2 times 11. That's 10 times 11 or 110 cubic inches. Notice that each prism has the same length and width. In each case, we first multiplied 5 times 2 to get 10. This leads to another way of thinking about the volume of a rectangular prism. Each prism is composed of 6 rectangular faces. The base of each prism is the face on the bottom. And so the height is the side length perpendicular to the base. The area of the base.. of each block is 5 times 2 or 10 square inches. To find the total volume, we multiply 10 times the height of each prism. In general, we can calculate the volume of a prism by finding the area of the base times the height. We can use this equation anytime we are calculating the volume of a prism. Rocky tries to neatly fit his beloved blocks into his storage cave. But some of the blocks will need to be flipped on their sides so they can fit properly. Rocky is a bit concerned--will the volumes still be the same if a different face is on the bottom? Let's take the 5 by 2 by 7 block as an example. With the 5 by 2 face as the base the area of the base is 10 the volume is the base times the height, 7 giving us a volume of 70 cubic inches. Let's flip the block on its side. Now, the seven by 2 face is on the bottom. It's the new base of the prism, and the area of the base is 7 times 2 or 14 square inches. Multiplying 14 times the new height of 5 gives us a volume of 70 cubic inches, as before. Let's flip the block one more time so that the 7 by five face is the base. The area of the base is 7 times 5 or 35 square inches. Multiplying 35 times the new height of 2 confirms the volume of 70 cubic inches. With a rectangular prism, any face can be the base and the volume can be computed by calculating the area of the base times the height. Rocky can flip his blocks freely without changing the volume--that makes his work easier! Let's review our tools for calculating the volume of a rectangular prism. We can calculate the volume as length times width times height. We can also select any face of a rectangular prism as the base. We then calculate the area of the base, and multiply by the height. Rocky has efficiently stored all his stones, and Gem is very grateful. She tells him that he can keep his stones in storage there for as long as he likes. Mmm if only they knew it was just an ordinary caveman cleaning his cave. | 1,119 | 4,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-27 | latest | en | 0.921684 |
https://en.wikipedia.org/wiki/Conditional_entropy | 1,686,407,306,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00656.warc.gz | 265,752,480 | 29,628 | # Conditional entropy
Venn diagram showing additive and subtractive relationships various information measures associated with correlated variables ${\displaystyle X}$ and ${\displaystyle Y}$. The area contained by both circles is the joint entropy ${\displaystyle \mathrm {H} (X,Y)}$. The circle on the left (red and violet) is the individual entropy ${\displaystyle \mathrm {H} (X)}$, with the red being the conditional entropy ${\displaystyle \mathrm {H} (X|Y)}$. The circle on the right (blue and violet) is ${\displaystyle \mathrm {H} (Y)}$, with the blue being ${\displaystyle \mathrm {H} (Y|X)}$. The violet is the mutual information ${\displaystyle \operatorname {I} (X;Y)}$.
In information theory, the conditional entropy quantifies the amount of information needed to describe the outcome of a random variable ${\displaystyle Y}$ given that the value of another random variable ${\displaystyle X}$ is known. Here, information is measured in shannons, nats, or hartleys. The entropy of ${\displaystyle Y}$ conditioned on ${\displaystyle X}$ is written as ${\displaystyle \mathrm {H} (Y|X)}$.
## Definition
The conditional entropy of ${\displaystyle Y}$ given ${\displaystyle X}$ is defined as
${\displaystyle \mathrm {H} (Y|X)\ =-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log {\frac {p(x,y)}{p(x)}}}$
(Eq.1)
where ${\displaystyle {\mathcal {X}}}$ and ${\displaystyle {\mathcal {Y}}}$ denote the support sets of ${\displaystyle X}$ and ${\displaystyle Y}$.
Note: Here, the convention is that the expression ${\displaystyle 0\log 0}$ should be treated as being equal to zero. This is because ${\displaystyle \lim _{\theta \to 0^{+}}\theta \,\log \theta =0}$.[1]
Intuitively, notice that by definition of expected value and of conditional probability, ${\displaystyle \displaystyle H(Y|X)}$ can be written as ${\displaystyle H(Y|X)=\mathbb {E} [f(X,Y)]}$, where ${\displaystyle f}$ is defined as ${\displaystyle \displaystyle f(x,y):=-\log \left({\frac {p(x,y)}{p(x)}}\right)=-\log(p(y|x))}$. One can think of ${\displaystyle \displaystyle f}$ as associating each pair ${\displaystyle \displaystyle (x,y)}$ with a quantity measuring the information content of ${\displaystyle \displaystyle (Y=y)}$ given ${\displaystyle \displaystyle (X=x)}$. This quantity is directly related to the amount of information needed to describe the event ${\displaystyle \displaystyle (Y=y)}$ given ${\displaystyle (X=x)}$. Hence by computing the expected value of ${\displaystyle \displaystyle f}$ over all pairs of values ${\displaystyle (x,y)\in {\mathcal {X}}\times {\mathcal {Y}}}$, the conditional entropy ${\displaystyle \displaystyle H(Y|X)}$ measures how much information, on average, the variable ${\displaystyle X}$ encodes about ${\displaystyle Y}$.
## Motivation
Let ${\displaystyle \mathrm {H} (Y|X=x)}$ be the entropy of the discrete random variable ${\displaystyle Y}$ conditioned on the discrete random variable ${\displaystyle X}$ taking a certain value ${\displaystyle x}$. Denote the support sets of ${\displaystyle X}$ and ${\displaystyle Y}$ by ${\displaystyle {\mathcal {X}}}$ and ${\displaystyle {\mathcal {Y}}}$. Let ${\displaystyle Y}$ have probability mass function ${\displaystyle p_{Y}{(y)}}$. The unconditional entropy of ${\displaystyle Y}$ is calculated as ${\displaystyle \mathrm {H} (Y):=\mathbb {E} [\operatorname {I} (Y)]}$, i.e.
${\displaystyle \mathrm {H} (Y)=\sum _{y\in {\mathcal {Y}}}{\mathrm {Pr} (Y=y)\,\mathrm {I} (y)}=-\sum _{y\in {\mathcal {Y}}}{p_{Y}(y)\log _{2}{p_{Y}(y)}},}$
where ${\displaystyle \operatorname {I} (y_{i})}$ is the information content of the outcome of ${\displaystyle Y}$ taking the value ${\displaystyle y_{i}}$. The entropy of ${\displaystyle Y}$ conditioned on ${\displaystyle X}$ taking the value ${\displaystyle x}$ is defined analogously by conditional expectation:
${\displaystyle \mathrm {H} (Y|X=x)=-\sum _{y\in {\mathcal {Y}}}{\Pr(Y=y|X=x)\log _{2}{\Pr(Y=y|X=x)}}.}$
Note that ${\displaystyle \mathrm {H} (Y|X)}$ is the result of averaging ${\displaystyle \mathrm {H} (Y|X=x)}$ over all possible values ${\displaystyle x}$ that ${\displaystyle X}$ may take. Also, if the above sum is taken over a sample ${\displaystyle y_{1},\dots ,y_{n}}$, the expected value ${\displaystyle E_{X}[\mathrm {H} (y_{1},\dots ,y_{n}\mid X=x)]}$ is known in some domains as equivocation.[2]
Given discrete random variables ${\displaystyle X}$ with image ${\displaystyle {\mathcal {X}}}$ and ${\displaystyle Y}$ with image ${\displaystyle {\mathcal {Y}}}$, the conditional entropy of ${\displaystyle Y}$ given ${\displaystyle X}$ is defined as the weighted sum of ${\displaystyle \mathrm {H} (Y|X=x)}$ for each possible value of ${\displaystyle x}$, using ${\displaystyle p(x)}$ as the weights:[3]: 15
{\displaystyle {\begin{aligned}\mathrm {H} (Y|X)\ &\equiv \sum _{x\in {\mathcal {X}}}\,p(x)\,\mathrm {H} (Y|X=x)\\&=-\sum _{x\in {\mathcal {X}}}p(x)\sum _{y\in {\mathcal {Y}}}\,p(y|x)\,\log _{2}\,p(y|x)\\&=-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}\,p(x)p(y|x)\,\log _{2}\,p(y|x)\\&=-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log _{2}{\frac {p(x,y)}{p(x)}}.\end{aligned}}}
## Properties
### Conditional entropy equals zero
${\displaystyle \mathrm {H} (Y|X)=0}$ if and only if the value of ${\displaystyle Y}$ is completely determined by the value of ${\displaystyle X}$.
### Conditional entropy of independent random variables
Conversely, ${\displaystyle \mathrm {H} (Y|X)=\mathrm {H} (Y)}$ if and only if ${\displaystyle Y}$ and ${\displaystyle X}$ are independent random variables.
### Chain rule
Assume that the combined system determined by two random variables ${\displaystyle X}$ and ${\displaystyle Y}$ has joint entropy ${\displaystyle \mathrm {H} (X,Y)}$, that is, we need ${\displaystyle \mathrm {H} (X,Y)}$ bits of information on average to describe its exact state. Now if we first learn the value of ${\displaystyle X}$, we have gained ${\displaystyle \mathrm {H} (X)}$ bits of information. Once ${\displaystyle X}$ is known, we only need ${\displaystyle \mathrm {H} (X,Y)-\mathrm {H} (X)}$ bits to describe the state of the whole system. This quantity is exactly ${\displaystyle \mathrm {H} (Y|X)}$, which gives the chain rule of conditional entropy:
${\displaystyle \mathrm {H} (Y|X)\,=\,\mathrm {H} (X,Y)-\mathrm {H} (X).}$[3]: 17
The chain rule follows from the above definition of conditional entropy:
{\displaystyle {\begin{aligned}\mathrm {H} (Y|X)&=\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log \left({\frac {p(x)}{p(x,y)}}\right)\\[4pt]&=\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)(\log(p(x))-\log(p(x,y)))\\[4pt]&=-\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}p(x,y)\log(p(x,y))+\sum _{x\in {\mathcal {X}},y\in {\mathcal {Y}}}{p(x,y)\log(p(x))}\\[4pt]&=\mathrm {H} (X,Y)+\sum _{x\in {\mathcal {X}}}p(x)\log(p(x))\\[4pt]&=\mathrm {H} (X,Y)-\mathrm {H} (X).\end{aligned}}}
In general, a chain rule for multiple random variables holds:
${\displaystyle \mathrm {H} (X_{1},X_{2},\ldots ,X_{n})=\sum _{i=1}^{n}\mathrm {H} (X_{i}|X_{1},\ldots ,X_{i-1})}$[3]: 22
It has a similar form to chain rule in probability theory, except that addition instead of multiplication is used.
### Bayes' rule
Bayes' rule for conditional entropy states
${\displaystyle \mathrm {H} (Y|X)\,=\,\mathrm {H} (X|Y)-\mathrm {H} (X)+\mathrm {H} (Y).}$
Proof. ${\displaystyle \mathrm {H} (Y|X)=\mathrm {H} (X,Y)-\mathrm {H} (X)}$ and ${\displaystyle \mathrm {H} (X|Y)=\mathrm {H} (Y,X)-\mathrm {H} (Y)}$. Symmetry entails ${\displaystyle \mathrm {H} (X,Y)=\mathrm {H} (Y,X)}$. Subtracting the two equations implies Bayes' rule.
If ${\displaystyle Y}$ is conditionally independent of ${\displaystyle Z}$ given ${\displaystyle X}$ we have:
${\displaystyle \mathrm {H} (Y|X,Z)\,=\,\mathrm {H} (Y|X).}$
### Other properties
For any ${\displaystyle X}$ and ${\displaystyle Y}$:
{\displaystyle {\begin{aligned}\mathrm {H} (Y|X)&\leq \mathrm {H} (Y)\,\\\mathrm {H} (X,Y)&=\mathrm {H} (X|Y)+\mathrm {H} (Y|X)+\operatorname {I} (X;Y),\qquad \\\mathrm {H} (X,Y)&=\mathrm {H} (X)+\mathrm {H} (Y)-\operatorname {I} (X;Y),\,\\\operatorname {I} (X;Y)&\leq \mathrm {H} (X),\,\end{aligned}}}
where ${\displaystyle \operatorname {I} (X;Y)}$ is the mutual information between ${\displaystyle X}$ and ${\displaystyle Y}$.
For independent ${\displaystyle X}$ and ${\displaystyle Y}$:
${\displaystyle \mathrm {H} (Y|X)=\mathrm {H} (Y)}$ and ${\displaystyle \mathrm {H} (X|Y)=\mathrm {H} (X)\,}$
Although the specific-conditional entropy ${\displaystyle \mathrm {H} (X|Y=y)}$ can be either less or greater than ${\displaystyle \mathrm {H} (X)}$ for a given random variate ${\displaystyle y}$ of ${\displaystyle Y}$, ${\displaystyle \mathrm {H} (X|Y)}$ can never exceed ${\displaystyle \mathrm {H} (X)}$.
## Conditional differential entropy
### Definition
The above definition is for discrete random variables. The continuous version of discrete conditional entropy is called conditional differential (or continuous) entropy. Let ${\displaystyle X}$ and ${\displaystyle Y}$ be a continuous random variables with a joint probability density function ${\displaystyle f(x,y)}$. The differential conditional entropy ${\displaystyle h(X|Y)}$ is defined as[3]: 249
${\displaystyle h(X|Y)=-\int _{{\mathcal {X}},{\mathcal {Y}}}f(x,y)\log f(x|y)\,dxdy}$
(Eq.2)
### Properties
In contrast to the conditional entropy for discrete random variables, the conditional differential entropy may be negative.
As in the discrete case there is a chain rule for differential entropy:
${\displaystyle h(Y|X)\,=\,h(X,Y)-h(X)}$[3]: 253
Notice however that this rule may not be true if the involved differential entropies do not exist or are infinite.
Joint differential entropy is also used in the definition of the mutual information between continuous random variables:
${\displaystyle \operatorname {I} (X,Y)=h(X)-h(X|Y)=h(Y)-h(Y|X)}$
${\displaystyle h(X|Y)\leq h(X)}$ with equality if and only if ${\displaystyle X}$ and ${\displaystyle Y}$ are independent.[3]: 253
### Relation to estimator error
The conditional differential entropy yields a lower bound on the expected squared error of an estimator. For any random variable ${\displaystyle X}$, observation ${\displaystyle Y}$ and estimator ${\displaystyle {\widehat {X}}}$ the following holds:[3]: 255
${\displaystyle \mathbb {E} \left[{\bigl (}X-{\widehat {X}}{(Y)}{\bigr )}^{2}\right]\geq {\frac {1}{2\pi e}}e^{2h(X|Y)}}$
This is related to the uncertainty principle from quantum mechanics.
## Generalization to quantum theory
In quantum information theory, the conditional entropy is generalized to the conditional quantum entropy. The latter can take negative values, unlike its classical counterpart. | 3,282 | 10,768 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 129, "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-2023-23 | latest | en | 0.803702 |
analysis.yellowcouch.org | 1,685,985,101,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00645.warc.gz | 2,796,703 | 3,067 | Scientific
# 2D Correlation Analysis
In 2005, I developed the 2D gel correlation technique because it was the only viable method to analyze the data we received at that time. The company I worked for tried to patent this method (which was a silly thing to do anyway, its software) and received feedback from the UK patent office that the method was already 'invented' in 1975. Of course, since the 70's improvements in computer processing power makes that the method can now produce larger volume results. If you are interested in a demonstration dataset then please have a look here
# Overview
The method correlates the content of multiple 2D gels with some external parameter (survival rate for instance) and will report the correlation using a hue color scheme. There is a movie available that summarizes the following steps. In the examples below, nothing of the biological information is correct. The examples give an overview of how the method work and don't make any biological sense. Even the reference to the 'age' parameter is wrong.
First one starts with a large collection of 2D gels (for instance of AML/ALL patients). Each of these gels is associated with an extra parameter, such as survival rate, age, disease type and so on. The question one now asks is whether there is a relation between spots on the 2D gels and that specific parameter. E.g: does specific spots correlate towards the age ?
## Step 1: Align all images
The first step in the process concerns itself with aligning all the images such that each pixel on each aligned image reflects the same pI (horizontally) and protein mass (vertically). This step might require rotation, zooming and translation of each individual image. Important here is to realize that cumulative superposition schemes do not work properly. The image above shows the superposition of 128 aligned images.
## Step 2: Normalize all images
Once the images are aligned, we need to make sure that we can compare the data between the different images. Often this might require contrast and brightness alterations. E.g: In the images above we clearly see that the background is different on the three images. Normalization will remove these differences.
## Step 3: Correlate
The correlation image can then be computed using for instance the Spearman Rank order correlation. This is done on an individual pixel basis and leads to images such as the following:
The problem with the above image is that it shows some useless information. Areas on the gel where we have a correlation but where the gel content does not alter sufficiently are not useful. Neither are areas where the probability that that specific correlation might occur by chance. To mask these areas we create two new images. One for the variability on the gel and one for the correlation significance. See figures below:
When these two masks are combined into the correlation image we get the much more understandable correlation map:
In this image, the areas where the image is red reflect the areas on the gels where we will tend to find protein for elderly patients. Where the image is blue we will find that older patients have less protein expression.
## Step 5: Creating the 3D volumetric map
To have a better understanding of which spots correlate with the external parameter it is useful to map this 2D gel in 3 dimensions. The 3th dimension is then used to reflect the correlation strength. Such volumetric maps can even be rotated:
Rotating the 3D volumetric view of the 2D correlation map
# Ordering
If you are interested in this analysis feel free to contact werner@yellowcouch.org with the following tentative information: checklist
- http://analysis.yellowcouch.org/ | 766 | 3,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-23 | longest | en | 0.92469 |
https://stats.stackexchange.com/questions/606187/why-is-median-not-a-sufficient-statistic | 1,722,661,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00481.warc.gz | 433,776,804 | 43,870 | # Why is median not a sufficient statistic? [duplicate]
Suppose a random sample of $$n$$ variables from $$N(\mu,1)$$, $$n$$ odd. The sample median is $$M=X_{(n+1)/2}$$, the order statistic of the middle of the distribution.
How to prove that sample median is not a sufficient statistic of $$\mu$$? Do we need the probability density function of M?
I would like any help.
Note 1: I'm studying for the final exam of my course, and take previous exams to practice. This question appeared in a 2015 exam, that was given by other professor than mine, so it is possible that covers things that wasn't given by my professor. This question includes others two exercises: (1) Is $$M-\bar{X}$$ an ancillary statistic for $$\mu$$? (2) Prove that if $$m$$ is a unique point, such as $$F_X(m)=1/2$$, then $$M \overset{p}{\rightarrow}m$$. I couldn't answer these two exercises, because ancillary statistics were not covered by my professor.
So, to see wheter $$M$$ is a sufficient statistic for $$\mu$$, I derived the probability density (pdf) function of $$M$$, that I bring below. But I think there is a more inteligent way to do, without getting the pdf first. So this is the reason I'm asking this question, to see if there is a better and faster way of solving this question.
Note 2: things I know about sufficient statistics:
$$T$$ is a sufficient statistic if $$\frac{f_X(x|\theta)}{q(t|\theta)}$$ does not depend on $$\theta$$, where $$f_X$$ is the probability density function of $$X$$ and $$q$$ is the probability density function of $$T$$. Also, $$T$$ is a sufficient statistic if, and only if, there exist functions $$g(t|\theta)$$ and $$h(x)$$ such that $$f_X(x|\theta)=g(T(x)|\theta)h(x)$$ for all points of $$\theta$$
Note 3: The probability density function of M is
$$f_{M}(m)=\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}f_{X}(m)[F_X(m)]^{\frac{n-1}{2}}$$
I divided the probability density functions of $$(X_1,...,X_n)$$ and $$M$$: $$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{Xn}(x_{n}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}f_{X}(m)[F_X(m)]^{\frac{n-1}{2}}}$$
So we have (I don't know if is wright): $$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{X_{n-1}}(x_{n-1}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}[F_X(m)]^{\frac{n-1}{2}}}$$ It seems that this expression depends of $$\mu.$$
• 1. This appears to be a homework-like question; please see stats.stackexchange.com/tags/self-study/info . 2. What things do you know about sufficient statistics? e.g. definition/facts/results/theorems etc? 3. What have you tried in order to use the things you know to answer the question? Commented Feb 22, 2023 at 1:53
• $T$ is a sufficient statistic if $$\frac{f_X(x|\theta)}{q(t|\theta)}$$ does not depend on $\theta$, where $f_X$ is the probability density function of $X$ and $q$ is the probability density function of $T$ Also, $T$ is a sufficient statistic if, and only if, there exist functions $g(t|\theta)$ and $h(x)$ such that $$f_X(x|\theta)=g(T(x)|\theta)h(x)$$ for all points of $\theta$ Commented Feb 22, 2023 at 2:00
• That's definitely a good thing to know. Did you try to use it? However, is that the only thing you have? 4. You should consider how $M$ is related to the parameter of the distribution of the $X_i$, $\mu$. i.e. note that $F$ and $f$ involve $\mu$ Commented Feb 22, 2023 at 2:01
• I'm not saying you need to use that fact, but if you just write it in terms of F and f without thinking about how those things relate to $\mu$ you may not even see that $\mu$ is present there and trick yourself into thinking it isn't. Commented Feb 22, 2023 at 2:07
• I divided the probability density functions of $(X_1,...,X_n)$ and $M$: $$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{Xn}(x_{n}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}f_{X}(m)[F_X(m)]^{\frac{n-1}{2}}}$$ So we have (I don't know if is wright): $$\frac{f_{X}(x|\mu)}{f_{M}(m)}=\frac{f_{X1}(x_{1}|\mu)\cdot ...\cdot f_{X_{n-1}}(x_{n-1}|\mu) }{\frac{n!}{\left ( \frac{n-1}{2} \right) ! \left ( \frac{n-1}{2} \right) !}[1-F_{X}(m)]^{\frac{n-1}{2}}[F_X(m)]^{\frac{n-1}{2}}}$$ that depends of $\mu$ Commented Feb 22, 2023 at 2:16
A tricky part of this question is that the median sounds like a good estimator and mathematically it is not so clear what the mean is gonna improve.
So imagine the following simpler case:
• Say we have a sample of size $$X_1, X_2, \dots , X_n$$ where each $$X_i$$ follows (independently) a normal distribution $$X_i \sim N(\theta,1)$$ and we want to estimate $$\theta$$.
• We could make an estimate $$\hat{\theta} = X_1$$ and it is obvious that this is not a sufficient statistic. The other $$n-1$$ values can provide information about $$\theta$$ as well. The statistic $$X_1$$ is not sufficient.
• A sufficient statistic occurs when the distribution of the data is independent of the parameter $$\theta$$ conditional on the sufficient statistic. This is not true for $$X_1$$. For example, the distribution of $$\bar{X}$$ conditional on $$X_1$$ is not independent from $$\theta$$
One obvious way to check if a statistic is sufficient is to identify a minimal sufficient statistic (if it exists) and check if the minimal sufficient statistic is a function of your proposed statistic. Here we want to use the fact that a minimal sufficient statistic is a function of any sufficient statistic.
Take $$n=3$$ for example. A minimal sufficient statistic for $$\mu$$ is the sample mean $$\overline x=\frac13(x_1+x_2+x_3)$$. But $$\overline x$$ is not a function of the sample median $$x_{(2)}$$ only. You have to argue this formally.
In essence, sufficiency is concerned with data reduction (see What does it mean that a statistic $T(X)$ is sufficient for a parameter?). So $$\overline x$$ is sufficient means that all the information about $$\mu$$ in the sample can be condensed in $$\overline x$$. This happens to be the maximum possible data condensation in this model, which makes the sample mean minimal sufficient. Given $$\overline x$$, you can make inference on $$\mu$$. But knowing $$x_{(2)}$$ alone does not give you enough information about $$\mu$$. You also need to know $$x_{(1)}$$ and $$x_{(3)}$$ to avoid any loss of information. A numerical example might help to make this point clear.
This is a great technical question. First I want to point out that your argument
$$T$$ is a sufficient statistic if $$\frac{f_X(x|\theta)}{q(t|\theta)}$$ does not depend on $$\theta$$, where $$f_X$$ is the probability density function of $$X$$ and $$q$$ is the probability density function of $$T$$.
does not hold. To see it, we know that for $$T = \bar{X} \sim N(\mu, \frac{1}{n})$$, whence \begin{align} & \frac{f_\mu(x_1, \ldots, x_n)}{q_\mu(t)} = \frac{(2\pi)^{-n/2}\exp(-\frac{1}{2}\sum_{i = 1}^n(x_i - \mu)^2)} {(2\pi n)^{-1/2}\exp(-\frac{n}{2}(t - \mu)^2)} = h(x)e^{-\frac{n}{2}(\bar{x} - t)\mu}, \end{align} which depends on $$\mu$$ (you may argue that it does not depend on $$\mu$$ if substitute $$\bar{x}$$ to $$t$$, however, this is not what the proposed ratio formula says), but $$\bar{X}$$ is of course sufficient. It looks like your "criterion" tries to match the definition of sufficiency of $$T$$ (Section 1.6, Theory of Point Estimation):
A statistic $$T$$ is said to be sufficient for $$X$$, or for the family $$\mathcal{P} = \{P_\theta, \theta \in \Omega\}$$ of possible distributions of $$X$$, or for $$\theta$$, if the conditional distribution $$X$$ given $$T = t$$ is independent of $$\theta$$ for all $$t$$.
A naive interpretation of the above definition is that the conditional density of the data $$X = (X_1, \ldots, X_n)$$ given $$T = t$$ is independent of $$\theta$$, which may be formally written as (note how the numerator in $$(1)$$ differs from that in your proposed ratio) \begin{align} f_{X|T; \theta}(x|t) = \frac{f_\theta(x, t)}{q_\theta(t)} \text{ is independent of \theta.} \tag{1} \end{align} However, a technical difficulty of the interpretation $$(1)$$ is that the "joint density" $$f_\theta(x, t)$$ of $$(X, T)$$ is degenerate (i.e., it is not a probability density on $$\mathbb{R}^{n + 1}$$. In general, the conditional density defining relation $$f_{X|Y = y}(x|y) = \frac{f(x, y)}{f_Y(y)}$$ only makes sense when $$f(x, y)$$ is a valid, non-degenerate density), hence $$(1)$$ is actually impossible to check. For a relevant discussion, see this question. To solve this difficulty, some 1-1 transformation between $$(X_1, \ldots, X_n)$$ and $$(Y_1, \ldots, Y_{n - 1}, T)$$ needs to be defined and the sufficiency needs to be augmented accordingly in terms of the conditional density of $$Y$$ given $$T$$. For details, refer to Eq (1.17) -- Eq (1.19) in Section 1.9, Testing Statistical Hypotheses.
Having clarified this, one way to show that $$M$$ is not a sufficient statistic for $$\mu$$ is to find a subset $$Y$$ (possibly with transformation) of $$(X_1, \ldots, X_n)$$, such that the conditional density of $$Y$$ given $$T$$ is well-defined (i.e., non-degenerate) and does depend on $$\theta$$. For simplicity and without losing of generality, assume $$n = 3$$.
One choice of $$Y$$ is $$(X_{(1)}, X_{(3)}) = (\min(X_1, X_2, X_3), \max(X_1, X_2, X_3))$$. It is well known by the order statistic theory that the joint density of $$(Y, M) = (X_{(1)}, X_{(3)}, X_{(2)})$$ is \begin{align} f(x_{(1)}, x_{(3)}, x_{(2)}) = 6\varphi_\mu(x_{(1)})\varphi_\mu(x_{(2)})\varphi_\mu(x_{(3)}), \quad x_{(1)} < x_{(2)} < x_{(3)}, \tag{2} \end{align} where $$\varphi_\mu(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2}}$$. And the marginal density of $$X_{(2)}$$ is (where $$\Phi_\mu(x) = \int_{-\infty}^x \varphi_\mu(t)dt$$) \begin{align} f_{X_{(2)}}(x_{(2)}) = 6\Phi_\mu(x_{(2)})(1 - \Phi_\mu(x_{(2)}))\varphi_\mu(x_{(2)}). \tag{3} \end{align} $$(2)$$ and $$(3)$$ then give the conditional density of $$(X_{(1)}, X_{(3)})$$ given $$X_{(2)}$$ as \begin{align} f_{(X_{(1)}, X_{(3)})|X_{(2)} = x_{(2)}}(x_{(1)}, x_{(3)}|x_{(2)}) = \frac{\varphi_\mu(x_{(1)})\varphi_\mu(x_{(3)})} {\Phi_\mu(x_{(2)})(1 - \Phi_\mu(x_{(2)}))}, \end{align} which depends on $$\mu$$.
In contrast, you can verify that \begin{align} f_{(X_{(1)}, X_{(3)})|\bar{X} = \bar{x}}(x_{(1)}, x_{(3)}|\bar{x}) = \frac{3\sqrt{3}}{\pi}\exp\left(\frac{3}{2}\bar{x}^2 - \frac{1}{2}(x_{(1)}^2 + x_{(3)}^2 + (3\bar{x} - x_{(1)} - x_{(3)})^2)\right), \end{align} which is independent of $$\mu$$.
Let us assume that $$n = 2k+1$$ is odd, and let $$J_i$$ be a ternary variable showing whether $$X_i$$ is equal to the median $$M$$, below it or above it. That is, $$J_i = \begin{cases} 1 & \text{X_i > M} \\ 0 & \text{X_i = M} \\ -1 & \text{X_i < M} \end{cases}$$ If $$M$$ is sufficient, then $$(M,J_1,\dots,J_n)$$ has to be sufficient as well. So it is enough to show that $$(M,J_1,\dots,J_n)$$ is not sufficient. What does sufficiency mean? That the distribution of $$(X_1,\dots,X_n)$$ given $$(M,J_1,\dots,J_n))$$ is independent of the parameter, $$\mu$$ in this case. That this independence fails in this case is intuitive, but a bit cumbersome to write down.
With probability 1, exactly one of $$J_1,\dots,J_n$$ can be equal to 0, and the rest divided equally among +1 and -1 (since $$N(\mu,1)$$ is continuous w.r.t. the Lebesgue measure the chance of seeing repeated values in a finite sequence of independent draws from it is zero.) Without loss of generality, let us condition on \begin{align} A = \{M = m,\;\; J_1 = 0, \;\;&J_2 = +1, \dots, J_{k+1} = +1, \\ &J_{k+2} = -1, \dots, J_{2k+1} = -1\}, \end{align} that is, $$X_1$$ is the median, the next $$k$$ observations are below the median and the next $$k$$ after are above the median.
Then, $$X_1 = m$$ is completely determined, and by independence $$X_2,\dots,X_{k+1}$$ are i.i.d. from $$N(\mu,1)$$ truncated to $$(m,\infty)$$ which we denote as $$N(\mu,1; m, \infty)$$. Similarly, $$X_{k+2},\dots,X_{2k+1}$$ are i.i.d. from $$N(\mu,1)$$ truncated to $$(-\infty,m)$$ which we denote as $$N(\mu,1; -\infty, m)$$. Clearly, the distribution of $$(X_1,\dots,X_n)$$ given $$A$$ depends on $$\mu$$ and hence sufficiency fails.
If you want to write it down, technically, $$(X_1,\dots,X_n) \, |\, A \;\;\sim \;\; \delta_{m} \otimes \Bigl(\prod_{i=1}^k N(\mu,1; m, \infty) \Bigr) \otimes \Bigl(\prod_{i=1}^k N(\mu,1; -\infty, m) \Bigr)$$ where $$\delta_m$$ is the point mass measure at $$m$$, and $$\otimes$$ and $$\prod$$ denote products of measures. You can also consider other combinations of values for $$\{J_i\}$$ and it would be similar (you just get a different permutation of the same terms), but not necessary to establish the failure of sufficiency (one combination is enough).
You can also answer the question about ancillarity. A statistic is ancillary if its distribution does not depend on the parameter $$\mu$$. | 4,306 | 13,031 | {"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": 158, "wp-katex-eq": 0, "align": 7, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-33 | latest | en | 0.890927 |
https://1st-in-babies.com/how-many-millimeters-are-in-87-decimeters-new-update/ | 1,680,391,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00792.warc.gz | 98,550,330 | 49,012 | How Many Millimeters Are In 87 Decimeters? New Update
How Many Millimeters Are In 87 Decimeters? New Update
Let’s discuss the question: how many millimeters are in 87 decimeters. We summarize all relevant answers in section Q&A of website 1st-in-babies.com in category: Blog MMO. See more related questions in the comments below.
How many Decimeters go into millimeters?
dm/mm length conversion result
From Symbol Result
1 dm 100.00
What is the difference between Decimeters and millimeters?
1 dm = 100 mm. 1 dm = 0.1 m.
Metric Units of Length | Convert mm, cm, m and km
Metric Units of Length | Convert mm, cm, m and km
Metric Units of Length | Convert mm, cm, m and km
How many mL are in an mL?
Metric
milliliter to cubic kilometer (km³) 0.000000000000001
milliliter to cubic decimeter (dm³) 0.001
milliliter to cubic centimeter (cc) 1
milliliter to cubic millimeter (mm³) 1,000
milliliter to hectoliter (hl) 0.00001
How many mm are in a diameter?
CATEGORIES
Diameter Nominal DN (mm) Nominal Pipe Size NPS (inches) Outside diameter (OD) inches (mm)
10 3/8 0.675 in (17.15 mm)
15 1/2 0.840 in (21.34 mm)
20 3/4 1.050 in (26.67 mm)
25 1 1.315 in (33.40 mm)
How do you convert decimeters to kilometers?
Please provide values below to convert decimeter [dm] to kilometer [km], or vice versa.
Decimeter to Kilometer Conversion Table.
Decimeter [dm] Kilometer [km]
1 dm 0.0001 km
2 dm 0.0002 km
3 dm 0.0003 km
5 dm 0.0005 km
What is the ratio of 10 Decimeters to 10 centimeters?
Since a decimeter is 10^1 larger than a centimeter, it means that the conversion factor for dm to cm is 10^1. Therefore, you can multiply 10 dm by 10^1 to get 10 dm converted to cm. Here is the answer with the math showing you how to convert 10 dm to cm by multiplying 10 by the conversion factor of 10^1.
What are millimeters?
Millimeters are used to measure very small but visible-scale distances and lengths. In terms of real-world comparisons, a millimeter is roughly the size of the wire used in a standard paper clip. The metric system is based on decimals: There are 10mm in a centimeter and 1000mm in a meter.
How many Decimeters does it take to make 1 m?
1 meter is 10 decimeters.
How many millimeters are there in every CM?
Millimeters and centimeters are separated by one tens place, which means that there are 10 millimeters for every centimeter.
What is DG 84 HG?
Simply put, hg is larger than dg. In fact, a hectogram is “10 to the power of 3” larger than a decigram. Since a hectogram is 10^3 larger than a decigram, it means that the conversion factor for hg to dg is 10^3. Therefore, you can multiply 84 hg by 10^3 to get 84 hg converted to dg.
Understanding mm, cm, m, and km
Understanding mm, cm, m, and km
Understanding mm, cm, m, and km
How do I convert diameter to centimeters?
Divide the circumference by π, or 3.14 for an estimation. The result is the circle’s diameter, 3.18 centimeters. Divide the diameter by 2. And there you go, the radius of a circle with a circumference of 10 centimeters is 1.59 centimeters.
How do you convert diameter to centimeters?
Measure the diameter in centimeters. For this example, let the diameter measure 10 cm. Multiply the diameter’s length to itself to square it — 10 cm multiplied by 10 cm results in 100 cm^2. Multiply the squared diameter by pi — 100 cm^2 multiplied by pi equals approximately 314.2 cm^2.
Is ml a ml or ml?
The ml stands for milliliter. The abbreviation ml is typically pronounced M-L (saying the letters out loud) or milliliter. This one is nice to remember. When you see the little “l” just think to yourself l = liquid.
How many millimeters is 5 ml?
ENDMEMO
1 milliliter = 10 millimeter 1 millimeter =
2 milliliter = 12.5992 millimeter 2 millimeter =
3 milliliter = 14.4225 millimeter 3 millimeter =
4 milliliter = 15.874 millimeter 4 millimeter =
5 milliliter = 17.0998 millimeter 5 millimeter =
What does 1 ml ml mean?
A Metric unit of volume. Equal to 1/1,000 (one-thousandth) of a liter. Used for small amounts of liquid. The abbreviation is ml or mL. It takes about 20 drops (from a standard eye dropper) to make one mL.
How do you find the diameter of a millimeter?
In Charrière’s scale the gauge number divided by 3 gives the needle diameter in mm. That means, the bigger is the gauge number, the thicker is the needle.
What diameter is 2inch?
2 inches is equal to 5.08 centimeters or 50.8 millimeters.
How many mm are in a mm?
The millimeters unit number 1,000,000,000.00 mm converts to 1 Mm, one megameter.
Which is more Decimeters or 15 centimeters?
15 centimeter = 15 x 0.1 decimeter = 1.5 decimeter (dm)
The most common metric system of measurement is centimeter, decimeter, meter, and kilometer.
Meters, Decimeters, Centimeters, Millimeters
Meters, Decimeters, Centimeters, Millimeters
Meters, Decimeters, Centimeters, Millimeters
Is cm bigger than dm?
Cm is 10 times smaller than a dm; a dm is 10 times smaller than a m, etc. Since you are going from a smaller unit to a larger unit, divide.
How many G are in a Hectogram?
Gram to Hectogram Conversion Table
Gram [g] Hectogram [hg]
20 g 0.2 hg
50 g 0.5 hg
100 g 1 hg
1000 g 10 hg
Related searches
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• dm to mm conversion calculator
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You have just come across an article on the topic how many millimeters are in 87 decimeters. If you found this article useful, please share it. Thank you very much. | 1,660 | 5,916 | {"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-2023-14 | latest | en | 0.825443 |
http://www.metaheuristics.org/index.php%3Fmain=4&sub=45.html | 1,582,687,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00548.warc.gz | 221,782,774 | 6,415 | Vehicle Routing with Stochastic Demand
# Vehicle Routing with Stochastic Demand
edited by Leonora Bianchi, IDSIA
## About the Vehicle Routing Problem with Stochastic Demand
### Introduction
The vehicle routing problem is a classical problem in operation research and it is one of the most challenging combinatorial optimization tasks. Many variations of this problem have been formulated, the main task of the problem being the same: designing the optimal set of routes for a fleet of vehicles in order to serve a given set of customers.
The most elementary version of the vehicle routing problem is the capacitated vehicle routing problem (CVRP). The CVRP is described as follows: n customers must be served from a unique depot. Each customer asks for a quantity q i of goods (i = 1,..., n) and a vehicle of capacity Q is available to deliver goods. Since the vehicle capacity is limited, the vehicle has to periodically return to the depot for reloading. In the CVRP, it is not possible to split customer delivery. Therefore, a CVRP solution is a collection of tours where each customer is visited only once and the total tour demand is at most Q.
From a graph theoretical point of view the CVRP may be stated as follows: Let G = (C,L) be a complete graph with node set C = (c o , c 1, c2,..., cn) and arc set L = { (c i , cj) : ci, cj Î C, i ¹ j}. In this graph model, c o is the depot and the other nodes are the customers to be served. Each node is associated with a fixed quantity q i of goods to be delivered (a quantity qo = 0 is associated to the depot c o). To each arc (ci, c j) is associated a value dij representing the distance between c i and c j. The goal is to find a set of tours of minimum total distance traveled. Each tour starts from and terminates at the depot co , each node ci (i = 1,..., n) must be visited exactly once, and the quantity of goods to be delivered on a route should never exceed the vehicle capacity Q.
### The vehicle routing problem with stochastic demand
The vehicle routing problem with stochastic demand (VRPSD) is a variation of the CVRP where each customer demand is uncertain, instead of being exactly known a priori. The VRPSD arises in practice whenever a company faces the problem of deliveries to a set of customers, whose demands are uncertain. In this formulation, it is assumed that the demand q i of customer i is a discrete random variable whose probability distribution is specified by pi k , i.e., the probability that customer i asks for a quantity qi = k of goods, with k = 0,1,..., K, and K<=Q . In the VRPSD it is also assumed that the actual customer demand is known only when the vehicle arrives at the customer's location.
In the deterministic VRP routes are planned in such a way that the vehicle always has enough capacity to satisfy all customers' demands on one of its pre-planned routes. When demands are stochastic, the situation is different. In fact, suppose a vehicle visits customers in a certain pre-planned order. It is possible that at some customer location, where the actual demand is revealed, the vehicle capacity becomes attained or exceeded, and a route failure is said to occur. In such a situation, the vehicle must return to the depot to replenish. The action taken by the vehicle after a route failure depends on the routing strategy that the planner decides to adopt. For example, the simplest strategy is to go back to the customer where demand has not been fully satisfied, and than to proceed visiting other customers in the pre-planned order. In general, also the planning of vehicle routes depends on the strategy that the planner wants to adopt.
Whatever strategy is adopted for planning the vehicle routes for the VRPSD, the distance traveled by the vehicle is a random quantity, since the number and location of recurse actions is not known in advance. Indeed, the actual route followed by the vehicle might be in general different from the pre-planned one. The goal, in the VRPSD, is to determine a routing policy such that demand at each customer is met, and the expected distance traveled is minimized.
P. Toth and D.Vigo, eds., 2002, The Vehicle Routing Problem, SIAM Monographs on Discrete Mathematics and Applications.
D. Bertsimas. A vehicle routing problem with stochastic demand. Operations Research, 40(3):574--585, 1992.
D. Bertsimas, P. Chervi and M. Peterson. Computational approaches to stochastic vehicle routing problems. Transportation Sciences, 29(4):342--352, 1995.
D. Bertsimas, P. Jaillet and A. Odoni. A priori optimization. Operations Research, 38:1019--1033, 1990.
D. Bertsimas and D. Simchi-Levi. A new generation of vehicle routing research: robust algorithms, addressing uncertainty. Operations Research, 44(2):216--304, 1996.
L. Bianchi, L. M. Gambardella and M. Dorigo. An ant colony optimization approach to the probabilistic traveling salesman problem. In Proceedings of PPSN-VII, Seventh International Conference on Parallel Problem Solving from Nature, volume 2439 of Lecture Notes in Computer Science. Springer, Berlin, Germany, 2002.
L. Bianchi, L. M. Gambardella and M. Dorigo. Solving the homogeneous probabilistic traveling salesman problem by the ACO metaheuristic. In M. Dorigo, G. Di Caro and M. Sampels, editors, Proceedings of ANTS 2002 -- From Ant Colonies to Artificial Ants: Third International Workshop on Ant Algorithms, volume 2463 of Lecture Notes in Computer Science, pages 176--187. Springer, Berlin, Germany, 2002.
J. Bramel and D. Simchi-Levi. The Logic of Logistics. Springer, Berlin, Germany, 1997.
T. G. Crainic and G. Laporte. Planning models for freight transportation. European Journal of Operational Research, 97:409--438, 1997.
M.Fisher. In O. M. Ball, T. L. Magnanti, C. L. Monma, and G. L. Nemhauser, editors, Network Routing, chapter Vehicle Routing, pages 1--33. Elsevier, Amsterdam, Holland, 1995.
M. Gendreau, G. Laporte, and R. Séguin. An exact algorithm for the vehicle routing problem with stochastic demands and customers. Transportation Sciences, 29(2):143--155, 1995.
M. Gendreau, G. Laporte, and R.Séguin. A tabu search heuristic for the vehicle routing problem with stochastic demands and customers. Operations Research, 44(3), 1996.
B. L. Golden and A.A. Assad, editors. Vehicle Routing: Methods and Studies. Elsevier, Amsterdam, Holland, 1988.
M. Haimovitch and A. Rinnooy Kan. Bounds and heuristics for capacitated routing problems. Mathematics of Operations Research, 10:527--542, 1985.
P. Jaillet. Probabilistic Traveling Salesman Problems. PhD thesis, MIT, Cambridge, MA, 1985.
P. Jaillet. A priori solution of a travelling salesman problem in which a random subset of the customers are visited. Operations Research, 36(6), 1988.
P. Jaillet and A. Odoni. In B. L. Golden and A. A. Assad, editors, Vehicle Routing: Methods and Studies, chapter The probabilistic vehicle routing problems. Elsevier, Amsterdam, Holland, 1988.
G. Laporte and F. Louveaux. The integer l-shaped method for stochastic integer programs with complete recourse. Operations Research Letters , 33:133--142, 1993.
I. Or. Traveling salesman-type combinatorial problems and their relation to the logistics of blood banking . PhD thesis, Department of Industrial Engineering and Management Sciences, Nortwestern University, Evanston, IL, 1976.
N. Secomandi. Exact and heuristic dynamic programming algorithms for the vehicle routing problem with stochastic demands . PhD thesis, University of Houston, Texas, 1998.
N. Secomandi. A rollout policy for the vehicle routing problem with stochastic demands. Operations Research , 49(5):796--802, 2001.
W. Yang, K. Mathur, and R. H. Ballou. Stochastic vehicle routing problem with restocking. Transportation Science , 34(1):99--112, 2000. | 1,847 | 7,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-10 | latest | en | 0.933889 |
https://mathhelpboards.com/threads/linear-and-angular-velocity-of-2-pulleys-and-a-belt.6893/#post-31409 | 1,638,227,895,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00098.warc.gz | 469,015,383 | 14,644 | # Trigonometrylinear and angular velocity of 2 pulleys and a belt.
#### karush
##### Well-known member
A belt contects two pulleys with radii $\displaystyle5\text { in}$ and $3\text { in}$
the $\displaystyle5\text { in}$ pulley is rotating at $\displaystyle\frac{1000\text{ rev}}{\text{min}}$
What is the linear $\displaystyle\text{v}$ in $\displaystyle\frac{\text{ft}}{\text{sec}}$ of the belt?
$\displaystyle \text{v}= \frac{1000\text{rev}}{\text{min}} \cdot\frac{\text{min}}{60\text{ sec}} \cdot\frac{10\pi \text{ in}}{\text{rev}} \cdot\frac{\text{ ft}}{12 \text{in}} =\frac{125\pi\text{ ft}}{9\text{sec}} =43.63\frac{\text{ft}}{\text{sec}}$
How many revolutions per min is the $\text{3 in}$ pulley making?
so
$\displaystyle \omega_{3in} =\frac{5}{3} \cdot\frac{1000\text{rev}}{\text{min}} \approx 1667\frac{\text{rev}}{\text{min}}$
no ans in bk on this so hope ans here is perhaps it.
#### MarkFL
$$\displaystyle v=r\omega=\left(5\text{ in}\frac{1\text{ ft}}{12\text{ in}} \right)\left(1000\frac{\text{rev}}{\text{min}} \frac{2\pi\text{ rad}}{1\text{ rev}} \frac{1\text{ min}}{60\text{ s}} \right)=\frac{125}{9}\pi\frac{\text{ ft}}{\text{s}}$$
$$\displaystyle \omega_2=\frac{r_1}{r_2}\omega_1$$ | 452 | 1,208 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.455031 |
http://www.physicsforums.com/showthread.php?t=175617 | 1,394,382,605,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394009871907/warc/CC-MAIN-20140305085751-00029-ip-10-183-142-35.ec2.internal.warc.gz | 478,740,744 | 6,503 | # Temperature vs. Resistance
by sharpnova
Tags: resistance, temperature
P: 13 1. The problem statement, all variables and given/known data At 30ąC, the resistance of a segment of gold wire is 79 ohms*. When the wire is placed in a liquid bath, the resistance increases to 165 *ohms. The temperature coefficient is 0.0034/C at 20 degrees Celcius. What is the temperature of the bath? An- swer in units of C. 2. Relevant equations Resistance = R(initial)[1+ alpha (T(final) - T(initial)] 3. The attempt at a solution 165 = 79 * ( 1 + .0034 * (T_final - 30) ) T_final = 350.1787 I'm submitting this and it's saying I'm wrong. Is the homework service in error?
Admin P: 21,591 The relevant equation is correct. The answer is correct for the given inputs.
Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 1 Introductory Physics Homework 5 Introductory Physics Homework 4 General Physics 4 | 240 | 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} | 3.1875 | 3 | CC-MAIN-2014-10 | longest | en | 0.855637 |
https://nl.webqc.org/molecularweightcalculated-190203-76.html | 1,596,581,645,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735885.72/warc/CC-MAIN-20200804220455-20200805010455-00532.warc.gz | 409,510,551 | 7,398 | #### Chemical Equations Balanced on 02/03/19
Molecular weights calculated on 02/02/19 Molecular weights calculated on 02/04/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Molar mass of N2O5 is 108.0104
Molar mass of c5 is 60.0535
Molar mass of Fe(ClO)2 is 158.7498
Molar mass of CH3OH is 32.04186
Molar mass of PO2 is 62.972562
Molar mass of Fe(ClO2)3 is 258.2004
Molar mass of (NH4)3PO4 is 149.086742
Molar mass of Mn(ClO2)4 is 324.745245
Molar mass of MnF6 is 168,9284642
Molar mass of NH4Cl is 53.49146
Molar mass of S2F10 is 254.114032
Molar mass of Al2O3 is 101,9612772
Molar mass of Ca3(PO4)2 is 310.176724
Molar mass of C6H12O6 is 180,15588
Molar mass of CO is 28.0101
Molar mass of CH3OH is 32.04186
Molar mass of PO2C5H7N is 144.088342
Molar mass of C5H7N is 81.11578
Molar mass of potassium perchlorate is 138.5489
Molar mass of NaOH is 39.99710928
Molar mass of CaCO3MgCO3 is 184,4008
Molar mass of k is 39.0983
Molar mass of h2 is 2.01588
Molar mass of K2CO3 is 138.2055
Molar mass of (C2H5)2NH2cl is 109.59778
Molar mass of NaOH is 39.99710928
Molar mass of ag2o is 231.7358
Molar mass of Al2(SO4)3 is 342.1508772
Molar mass of KOH is 56,10564
Molar mass of ag2 is 215.7364
Molar mass of CH3CH2CHO is 58.07914
Molar mass of CaCO3 is 100,0869
Molar mass of MgCO3 is 84,3139
Molar mass of Ca is 40,078
Molar mass of SiO2 is 60,0843
Molar mass of Br is 79,904
Molar mass of SCI2 is 297.88464
Molar mass of KHP is 71.080002
Molar mass of Cr2(CO3)3 is 284,0189
Molar mass of F2 is 37,9968064
Molar mass of CH3COCH3 is 58,07914
Molar mass of C6H8O4 is 144.12532
Molar mass of titanium is 47.867
Molar mass of H2SO4 is 98.07848
Molar mass of CH3O3 is 63.03272
Molar mass of SCl2 is 102.971
Molar mass of CH3OH is 32.04186
Molar mass of SCl2 is 102.971
Molar mass of C7H9O5 is 173.14336
Molar mass of Na2B2(O2)2(OH)4 is 199.62849856
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Calculate molecular weight
Molecular weights calculated on 02/02/19 Molecular weights calculated on 02/04/19
Molecular masses on 01/27/19
Molecular masses on 01/04/19
Molecular masses on 02/03/18
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https://ejournal.unp.ac.id/students/index.php/mat/article/view/7194 | 1,726,430,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00828.warc.gz | 202,939,829 | 7,647 | ### Model Matematika Tendangan Pisang Sepak Pojok pada Olahraga Sepakbola
tomy aprinaldi - Jurusan Matematika Universitas Negeri Padang
media rosha - Jurusan Matematika Universitas Negeri Padang
#### Abstract
Abstract — Football is the most interest sport by the public. The problem in football is a want to many scores. Scoring on football can be from open play and set play. In this research, the authors chose to conduct a scoring research through the set play from a football corner with a banana kick technique. The purpose of this research are: Forming a mathematical modeling banana kick of corner on football, analyzing the model, and interpreting model analysis results. The banana kick math model of the football corner is a regular differential equation-shaped system. The solution of this model to use a numerical solution with the fourth order Runge-Kutta method. Then, done simulation by using matlab. Simulated results show that, with an initial velocity 29,5 m/s the ball will be goal, but with initial velocity 23 m/s and 35 m/s the ball will not goal.
Keywords — Mathematical Modeling, Banana Kick, Corner Kick.
PDF PDF
#### References
Sucipto, dkk. 2000. Sepakbola. Departemen Pendidikan dan Kebudayaan.
Joseph A, Luxbacher. 2004. Sepakbola Langkah-langkah Menuju Sukses. Jakarta: PT Raja Grafindo Persada.
Giancoli. 2001. Fisika Edisi Kelima. Jakarta: Erlangga.
Sumarjono, dkk. 2005. Fisika Dasar 1. Malang: Penerbit Universitas Negeri Malang.
Triatmodjo, Bambang. 2002. Metode Numerik. Yogyakarta: Beta Offset.
Asmah, S. 2004. Solusi Masalah Nilai Awal Secara Numerik dengan Metode Runge-Kutta. Fakultas MIPA Universitas Negeri Makassar.
Javorova, Julian. 2018. Study of Soccer Ball Flight Trajectory. MATEC Web of Conferences 145, 01002.
Bray, Ken & David. 2003. Modelling The Flight of a Soccer Ball in a Direct Free Kick. Journal of Sports Sciences, 21:2, 75-85.
DOI: http://dx.doi.org/10.24036/unpjomath.v4i3.7194 | 543 | 1,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.503219 |
http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?s=948b8fad398390e28f833aaa0fba9198&p=9176380 | 1,545,183,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830305.92/warc/CC-MAIN-20181219005231-20181219031231-00119.warc.gz | 304,977,793 | 9,555 | Actuarial Outpost Annuity with Payment in Geometric Progressions
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#1
11-28-2017, 10:57 PM
David_Mo SOA Join Date: Nov 2017 Posts: 4
Annuity with Payment in Geometric Progressions
"On January 1, 1988, Felix inherited a perpetuity with annual payments beginning in six months. The first payment was \$3,000, and after that the payments increased by 3% each year. Find the value of this perpetuity on January 1, 1995 if the annual effective interest rate was 6% from January 1, 1988 through January 1, 1996 and 4% thereafter."
I don't how to handle this kind of question, could anyone please help me out with that? the final answer is \$397,388.55
#2
11-28-2017, 11:07 PM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,015
Using formulas for geometric series, you should be able to calculate the value of the payments starting July 1, 1997 on July 1, 1997. Then discount that to January 1, 1995.
Separately, calculate the value of the payments for July 1, 1988 to July 1, 1996 on July 1, 1988. Then accumulate that to January 1, 1995.
#3
11-29-2017, 03:01 AM
David_Mo SOA Join Date: Nov 2017 Posts: 4
Is that how I calculate the value of the payments starting July 1, 1997?
3000(1.06)^7.5[1-(1.03/1.06)^8/1-(1.03/1.06)]
#4
11-30-2017, 01:27 AM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,015
I don't know what formula you are trying to apply there, but if you are just trying to calculate the value of payments after July 1, 1997 (and are trying to get the value on January 1, 1995 [you didn't say what date your value was on]), there should be almost no use of the 6% interest rate.
You can get the value on July 1, 1997 with no use of 6%, and it gets used for only one year in discounting to Jan 1, 1995.
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Page generated in 0.24043 seconds with 11 queries | 874 | 3,221 | {"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} | 3.25 | 3 | CC-MAIN-2018-51 | latest | en | 0.930474 |
http://www.generative-ebooks.com/ebooks/A-new-look-at-the-pendulum.html | 1,669,699,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00494.warc.gz | 70,622,746 | 6,068 | A first tour through examples - Chapter 5
A new look at the pendulum
Consider the motion of the following idealized pendulum: a bob of mass $m$ is attached to one end of a massless rigid rod. The other end of the rod is pivoted so that the mass may swing in a vertical plane. We neglect both the friction of the pivot and air drag.
The swinging of the pendulum is governed by
$$\ddot{\theta}=-\frac{g}{L} \sin \theta$$
where $\theta$ is the angle between the rod and the downward vertical. This equation is derived in all textbooks of classical mechanics.
Phase plane representation.
Let’s do the same trick as for the harmonic oscillator by representing the state of the pendulum as a point in the phase plane $x=\theta$, $y=\dot{\theta}$. This yields
$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2 \sin x \end{cases}$$
where $\omega^2=\frac{g}{L}$. (Notice that $\omega$ has the dimension of a frequency.)
Let’s draw the phase portrait!
We can now look at the trajectories of this system. In the following digital experiment, one can pull the mass away from its rest position, drop it (without speed) and see at once the resulting trajectory in the phase plane. One can also select a point in the phase plane and see at once the resulting motion of the pendulum. If one does not take a point on the $x$-axis, this means that one start with a nonzero velocity (we kick the mass).
Actually the equilibrium points are $(k\pi,0)$, where $k\in\mathbb{Z}$. The origin in the phase portrait corresponds to the stable equilibrium position of the pendulum hanging straight down. The points $(\pm k\pi,0)$, $k\in\mathbb{Z}\backslash \{0\}$, correspond to the unstable equilibrium position where the pendulum is straight up. Let’s restrict to the points $(-\pi,0)$, $(0,0)$, $(\pi,0)$, since the phase portrait is periodic in the $x$-direction (translating horizontally points of the phase portrait by a distance that is a multiple of $2\pi$ yields the same phase portrait).
We observe two generic behaviors which take place into two separate regions:
• closed trajectories encircling the origin, which describe the periodic motions of the pendulum swinging back and forth;
• non closed trajectories, which are periodic in the $x$-direction, and which describe motions where the mass whirls around the pivot endlessly.
These two kinds of trajectories are separated by a eye-like curve, called the separatrix. It can be (approximately) obtained by dropping the mass very close from its upside down rest position.
Notice that, mathematically, the true separatrix corresponds to a set of initial states that has area zero in the plane. It corresponds to exceptional motions such that the pendulum tends to the unstable equilibrium, which takes an infinite amount of time.
It is well-known that we have conservation of total energy (kinetic $+$ potential) because we neglect friction of the pivot and air resistance. As shown in any textbook of classical mechanics
$$\frac{\dot{\theta}^2}{2}-\omega^2(1-\cos\theta)=\text{const}$$
which means that each motion individually has its own energy which does not change in time. Translated into our phase plane language, this means that the trajectories of solutions in the $xy$-plane are level curves of the function
$$E(x,y)=\frac{y^2}{2}-\omega^2(1-\cos x).$$
The damped pendulum.
When we add to the system of the pendulum a perturbation term to represent air friction, a physically more realistic situation, energy is no more conserved : it decreases over time and we have damped oscillations. We assume that the friction is proportional to the velocity and call $\mu$ the friction coefficient. This gives
$$\ddot{\theta}=-\omega^2\sin \theta -\mu \dot{\theta}.$$
Passing to the phase plane representation we thus obtain
$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2\sin x -\mu y. \end{cases}$$
We now see that the equilibrium point $(0,0)$ is an attractor : all solutions converge toward this point. The corresponding trajectories have a spiraling behavior if $\mu$ is not too large. Of course we recover the phase portrait of the undamped pendulum when $\mu=0$.
Linear approximation.
Because of the sine function, the equation of the pendulum is nonlinear. However, if the angular deviation of the pendulum is very small, then it is tempting to approximate $\sin x$ by $x$, which leads to
$$\begin{cases} \dot{x}= y\\ \dot{y}=-\omega^2 x-\mu y. \end{cases}$$
We get the same equation as the one for the harmonic oscillator! This is in accordance with what can be seen when we compare the phase portrait of the harmonic oscillator to the one of the pendulum in a close neighborhood of $(0,0)$.
We have an instance of what a linear approximation can tell us about a nonlinear system, but only locally about an equilibrium point. We shall develop much further this approach later on. | 1,176 | 4,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.862611 |
http://en.wikipedia.org/wiki/Hermitian_hat_wavelet | 1,411,020,190,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657125654.84/warc/CC-MAIN-20140914011205-00272-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 90,787,667 | 9,077 | Hermitian hat wavelet
The Hermitian hat wavelet is a low-oscillation, complex-valued wavelet. The real and imaginary parts of this wavelet are defined to be the second and first derivatives of a Gaussian respectively:
$\Psi(t)=\frac{2}{\sqrt{5}}\pi^{-\frac{1}{4}}(1-t^{2}+it)e^{-\frac{1}{2}t^{2}}.$
The Fourier transform of this wavelet is:
$\hat{\Psi}(\omega)=\frac{2}{\sqrt{5}}\pi^{-\frac{1}{4}}\omega(1+\omega)e^{-\frac{1}{2}\omega^{2}}.$
The Hermitian hat wavelet satisfies the admissibility criterion. The prefactor $C_{\Psi}$ in the resolution of the identity of the continuous wavelet transform is:
$C_{\Psi}=\frac{16}{5}\sqrt{\pi}.$
This wavelet was formulated by Szu in 1997 for the numerical estimation of function derivatives in the presence of noise. The technique used to extract these derivative values exploits only the argument (phase) of the wavelet and, consequently, the relative weights of the real and imaginary parts are unimportant. | 273 | 962 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-41 | latest | en | 0.754416 |
https://itprospt.com/num/18812609/points-suppose-r-is-the-triangle-with-vertices-1-0-0-1-and | 1,660,291,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571597.73/warc/CC-MAIN-20220812075544-20220812105544-00225.warc.gz | 346,289,228 | 15,097 | 5
# Points) Suppose R is the triangle with vertices (~ 1,0). (0, 1), and (1,0) (a) As an iterated integral, (Sx + 3y)? dA = EI' (Sx + 3y)? dx dy with llmits of int...
## Question
###### Points) Suppose R is the triangle with vertices (~ 1,0). (0, 1), and (1,0) (a) As an iterated integral, (Sx + 3y)? dA = EI' (Sx + 3y)? dx dy with llmits of integration(b) Evaluate the integral in part (@). Hint: substitution may make the Intogral easier: Integral
points) Suppose R is the triangle with vertices (~ 1,0). (0, 1), and (1,0) (a) As an iterated integral, (Sx + 3y)? dA = EI' (Sx + 3y)? dx dy with llmits of integration (b) Evaluate the integral in part (@). Hint: substitution may make the Intogral easier: Integral
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##### Sechleduled payments of 5825 due today replaced by two payments: a $1,000 'and$3,975 due in 15 months are to be unknown payment 2 years from payment 9 months from today: Money ' today and an (Inyour rough work It may be helpfuota earns 4.8% compounded annually.
Sechleduled payments of 5825 due today replaced by two payments: a $1,000 'and$3,975 due in 15 months are to be unknown payment 2 years from payment 9 months from today: Money ' today and an (Inyour rough work It may be helpfuota earns 4.8% compounded annually.... | 2,262 | 7,359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-33 | latest | en | 0.779347 |
https://testbook.com/question-answer/any-positive-integer-is-of-the-form--60a7549115ca54d2e70a809f | 1,632,581,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00232.warc.gz | 585,348,205 | 29,244 | # Any positive integer is of the form:
This question was previously asked in
DSSSB TGT Maths Male Subject Concerned - 23 Sep 2018 Shift 2
View all DSSSB TGT Papers >
1. 6q or 6g + 2 or 6g + 4, where q is some non-negative integer.
2. 3q or 3q + 1 or 3q + 2, where q is some non-negative integer.
3. 6g + 1 or 6q + 3 or 6g + 5, where q is some non-negative integer.
4. 6q or 6g +1 or 6q + 2, where q is some non-negative integer.
Option 2 : 3q or 3q + 1 or 3q + 2, where q is some non-negative integer.
## Detailed Solution
Concept:
Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying,
a = bq + r, where, 0 ≤ r ≤ b
Calculation:
Let a be any positive integer and b = 3.
Now, applying Euclid's division lemma with a and b = 3, we have,
a = 3q + r, where 0 ≤ r ≤ 3 and q is some non-negative integer.
⇒ a = 3q + 0 or, a = 3q + 1 or, a = 3q + 2
⇒ a = 3q or, a = 3q + 1 or, a = 3q + 2 for some non-negative integer q
Hence, Any positive integer is of the form 3q or 3q + 1 or 3q + 2, where q is some non-negative integer. | 406 | 1,079 | {"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} | 4.375 | 4 | CC-MAIN-2021-39 | latest | en | 0.757706 |
https://slideplayer.com/slide/2464903/ | 1,532,079,096,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591575.49/warc/CC-MAIN-20180720080634-20180720100634-00523.warc.gz | 762,368,595 | 20,086 | # IA#2 Review – Kinematics on Inclined Planes & Static Friction February 6, 2013.
## Presentation on theme: "IA#2 Review – Kinematics on Inclined Planes & Static Friction February 6, 2013."— Presentation transcript:
IA#2 Review – Kinematics on Inclined Planes & Static Friction February 6, 2013
Do Now
Oral Drill
A man applies a 35 N force to push a 5 kg suitcase up a 30° ramp into the baggage check-in area at the airport. If k = 0.10, what is the acceleration of the suitcase up the ramp? (sin30 = 0.5, cos30 = 0.866). FAFA m θ kk a = ?
+y +x FANG SAT Fg Fgx Fgy FNFN FAFA FfFf
Setup M (do not fully solve in class but you will solve for HW) Y-direction Fnet = ma F N - F gy = ma SOLVE X- direction Fnet = ma -F A + F gx +F f = ma SOLVE
Independent Practice
STATIC FRICTION
Expectations for Paired Reading Take turns reading each paragraph Annotate (including summarizing) each paragraph
Reading Debrief What is static friction? What is the difference between static friction and kinetic friction? What equation would I use to solve for static friction?
Reading Debrief Remember back to our mu-shoe lab. We had to apply a force to a shoe on different surfaces. The force we applied was the highest amount of force we could apply before the shoe just started to move. In this case, what formula would we use to determine the static friction force? (Hint – at this point, the applied force is opposing the maximum static friction force.)
Download ppt "IA#2 Review – Kinematics on Inclined Planes & Static Friction February 6, 2013."
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Wiring Diagram Maker | Wiring Diagram – Wiring Diagram Software
Wiring Diagram Software – wiring diagram software, wiring diagram software arduino, wiring diagram software automotive, Every electrical arrangement is composed of various different components. Each component ought to be set and linked to other parts in particular manner. Otherwise, the arrangement will not function as it ought to be. In order to be certain that the electric circuit is constructed correctly, Wiring Diagram Software is necessary. How does this diagram aid with circuit building?
The diagram provides visual representation of an electrical arrangement. However, this diagram is a simplified variant of this arrangement. This makes the process of building circuit simpler. This diagram provides information of circuit’s components in addition to their own placements.
Components of Wiring Diagram Software and Some Tips
There are two things which are going to be present in any Wiring Diagram Software. The first component is symbol that indicate electrical element from the circuit. A circuit is generally composed by numerous components. The other thing you will see a circuit diagram would be lines. Lines in the diagram show exactly how every element connects to a another.
The positions of circuit elements are comparative, not accurate. The order is also not plausible, unlike wiring schematics. Diagram only shows where to place component in a spot relative to other components within the circuit. Though it is exemplary, diagram is a good foundation for anyone to construct their own circuit.
One thing that you must learn before reading a circuit diagram is the symbols. Every symbol that’s presented on the diagram shows specific circuit element. The most common components are capacitor, resistorbattery. Additionally, there are other elements such as ground, switch, motor, and inductor. All of it depends on circuit that’s being assembled.
According to previous, the lines at a Wiring Diagram Software represents wires. At times, the cables will cross. However, it does not mean link between the cables. Injunction of 2 wires is usually indicated by black dot to the intersection of two lines. There will be main lines that are represented by L1, L2, L3, etc. Colors can also be used to differentiate wires.
Ordinarily, there are two chief types of circuit connections. The first one is known as series connection. It’s the simpler type of connection as circuit’s elements are placed within a singular line. Due to the electric current in each and every part is comparable while voltage of this circuit is complete of voltage in every component.
Parallel relationship is more complicated than the series one. Unlike in series connection, the voltage of every element is similar. It is because the element is directly connected to power source. This circuit consists of branches which are passed by distinct electrical current levels. The present joins together when the branches match.
There are lots of items that an engineer should look closely at when drawing wirings diagram. To begin with, the symbols used in the diagram ought to be accurate. It should represent the specific component needed to construct an intended circuit. After the symbol is wrong or uncertain, the circuit won’t function as it’s supposed to.
It’s also highly suggested that engineer draws positive supply and damaging supply symbols for better interpretation. Normally positive supply emblem (+) is located above the line. Meanwhile, the negative supply symbol is place below it. The current flows from the left side to right.
Besides this, diagram drawer is recommended to restrict the number of line crossing. The line and part placement ought to be designed to lessen it. But if it is unavoidable, use universal symbol to indicate whether there’s a intersection or if the lines are not actually connected.
As you can see drawing and interpreting Wiring Diagram Software may be complicated job on itself. The information and ideas which have been elaborated above should be a fantastic kick start, though. Wiring Diagram Software
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https://socratic.org/questions/58f3bf4411ef6b789ed74d32 | 1,569,267,564,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514578201.99/warc/CC-MAIN-20190923193125-20190923215125-00147.warc.gz | 663,312,567 | 6,398 | # Question #74d32
Apr 16, 2017
For a pendulum of length $l$, period is:
$T = 2 \pi \sqrt{\frac{l}{g}}$
Ie $T \propto \frac{1}{\sqrt{g}}$
So, for context, you are playing on the fact that gravity's magnitude is gonna be less on Mt Everest as it is very high up.
If we take logs of the original equation:
$\ln T = \ln 2 \pi + \frac{1}{2} \left(\ln l - \ln g\right)$
Taking the differentials, assuming that $l$ stays the same:
$\frac{\mathrm{dT}}{T} = - \frac{1}{2} \frac{\mathrm{dg}}{g} q \quad \triangle$
If we next look at what is meant by $g$, we see that it is a play on Newton's Law of Gravitation:
$F = m g = \frac{G M m}{r} ^ 2 \implies g = \frac{G M}{r} ^ 2$
We can do the same thing with logs:
$\ln g = \ln G M - 2 \ln r$
$\implies \frac{\mathrm{dg}}{g} = - 2 \frac{\mathrm{dr}}{r} q \quad \square$
Combining $\triangle$ and $\square$ means that:
$\frac{\mathrm{dT}}{T} = \frac{\mathrm{dr}}{r}$
Karachi is pretty much at sea level, Mt Everest is $8 , 848 \text{ m}$ above sea level. Ergo $\mathrm{dr} = 8 , 848$.
In terms of $r$, the earth has a radius of about: $6 , 371 \times {10}^{3} \text{ m}$.
So:
$\frac{\mathrm{dT}}{T} \approx + 0.14 \text{ %}$ | 432 | 1,179 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "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-2019-39 | longest | en | 0.744037 |
https://www.wavewalkerdsp.com/2022/02/16/can-the-fourier-transform-magnitude-be-negative/ | 1,709,326,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475701.61/warc/CC-MAIN-20240301193300-20240301223300-00345.warc.gz | 1,041,213,749 | 37,551 | # Wave Walker DSP
## Trending
DSP for Beginners: Simple Explanations for Complex Numbers! The second edition includes a new chapter on complex sinusoids.
Can the Fourier Transform Magnitude Be Negative?
February 16, 2022
#### Introduction
I came across a question on DSP Stack Exchange the other day: can the Fourier Transform magnitude be negative? This is a great question! The answer is no, and let’s take a look at why.
More blogs on DSP:
#### Discrete Time Fourier Transform
The discrete-time Fourier transform (DTFT) is
(1)
The DTFT is a cross correlation of x[n] with a complex exponential . The summation over all n produces a single complex number. The DTFT is written as a real and imaginary component,
(2)
where
(3)
and
(4)
#### Magnitude of Complex Number
Rather than use the complex number in (2) a simpler representation is given by
(5)
where the complex number c has a real part a and an imaginary part jb. The magnitude of a complex number c can be written as
(6)
The magnitude can be written in terms of a and b by substituting (5) into (6),
(7)
The square of a real number and cannot be negative and the square root of a positive number is always positive, therefore the magnitude cannot be negative.
#### Plotting Magnitude of Complex Number
Conceptually the magnitude of a complex number c is the length of the vector beginning at the origin 0 + 0j and ending at c. Consider a complex number
(8)
Figure 1 plots the complex vector c in 2D complex space.
The magnitude is the distance from 0+0j to -3-7j which is
(9)
#### Fourier Transform Magnitude
Using the representation in (2) for the complex number the magnitude can be written as
(10)
which is always a positive number.
#### Conclusion
The magnitude of a complex number always has to be positive by definition, it can never be negative. The magnitude is conceptually the length of a complex vector which is written mathematically as .
More blogs on DSP:
Got a question? Drop it in the comments below or send me an email: matt@wavewalkerdsp.com
### One Response
1. I definitely agree that the magnitude of the Fourier transform, or any complex number, is non-negative. But students or newcomers to DSP might be a bit confused about this statement: “the square root of a positive number is always positive.” Because square roots of positive numbers can be negative: -2 is a valid square root of 4. I think the difference is in the definition of “magnitude” (or “modulus” or “absolute value”) which is defined as the positive square root.
When we use the square-root symbol, and there is no ‘-‘ or ‘+’ in front of it, ‘+’ is assumed.
Interested readers might look into metric spaces and inner products:
https://en.wikipedia.org/wiki/Magnitude_(mathematics)
https://en.wikipedia.org/wiki/Euclidean_space#Euclidean_norm
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1. Nov 28, 2007
### mrroboto
1. The problem statement, all variables and given/known data
Let V = {p element of R[x] | deg(p) <=3} be the vector space of all polynomials of degree 3 or less.
b) Give the matrix for d2/dx2 in the basis {1,x,x^2, x^3} for V
2. Relevant equations
3. The attempt at a solution
[1 1 1 1
0 0 2 6]
i used the coefficients to get the first row, and then took the 2nd derivative and used the coefficients for the 2nd row. is this right?
2. Nov 28, 2007
### Dick
You have this all wrong. The matrix should be 4x4. The polynomial a+bx+cx^2+dx^3 corresponds to the column vector (a,b,c,d). Whatever the second derivative does to the polynomial, the matrix should do to the vector. E.g. x^3=(0,0,0,1), the second derivative is 3x^2=(0,0,3,0). | 262 | 807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-47 | latest | en | 0.880662 |
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# Approximately. 7.6 million women who earn incomes have
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Approximately. 7.6 million women who earn incomes have [#permalink]
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14 Sep 2004, 14:03
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16. Approximately. 7.6 million women who earn incomes have preschool-age children, and approximately 6.4 million women are the role income earners' for their families. These figures indicate that there are comparatively few income-earning women who have preschool-age children but are not the sole income earners for their families.
A major flaw in the reasoning is that it
(A) relies in figures that are too imprecise to support the conclusion drawn.
(B) overlooks the possibility that there is little or no overlap between the two populations of women cited.
(C) fails to indicate whether the difference between the two figures cited will tend to remain stable over time.
(D) ignores the possibility, that families with preschool-age children might also have older children.
(E) provides no informati0n on families in which men are the sole income earners.
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14 Sep 2004, 17:06
I will go with B.
The authors argument is that 7.6 million women who earn incomes have pre-school age children and OUT OF THESE 7.6 MILLION WOMEN, 6.4 million have the additional attribute of being role income earners of the family. So, there are 1.2 million women who earn incomes, have pre-school age children but are not the role earners versus 6.4 million who have all three characteristics. The assumption is that there is an overlap, and choice B points out the flaw in this assumption.
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14 Sep 2004, 19:58
I didnt like any of the answer choices and this CR seem to be purposefully convuluted.
Only B has some reference at all.
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### Show Tags
08 Jan 2014, 01:36
venksune wrote:
I didnt like any of the answer choices and this CR seem to be purposefully convuluted.
Only B has some reference at all.
Hello,
Here are the facts
Women with preschool age children = 7.6 mill
Women who are Sole bringers of money = 6.4 mill
We do not know the proportion of sole bringers vs ones where the men also bring in money.
The probable results therefore are these.
1. 6.4 million of the sole bringers of money COULD be in fact a part of the 7.6 million women with preschool age children; i.e. MOST of the 7.6 women with children are sole bringers of money.
2. A portion of the 6.4 million of the sole bringers of money could constitute a small to substantial part of the 7.6 million
3. the 6.4 million women who are sole bringers of money may be exclusive of this set -> ie. ALL the 7.6 million women with children are with men who bring money as well.
Therefore the conclusion that "relatively few women are in families where the man brings in money too" may not necessarily be true.
Option B does exactly this.
Hope that helps.
Ajeeth
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Re: Approximately. 7.6 million women who earn incomes have [#permalink]
### Show Tags
12 Oct 2016, 21:32
I don't like the answer either. in the stimulus what's the meaning of "comparatively few", compare to what?
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Re: Approximately. 7.6 million women who earn incomes have [#permalink] 12 Oct 2016, 21:32
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# Approximately. 7.6 million women who earn incomes have
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,560 | 6,020 | {"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-2016-44 | latest | en | 0.931291 |
https://electronics.stackexchange.com/questions/262997/op-amp-adds-spectral-components-to-amplified-output-signal | 1,714,032,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00308.warc.gz | 196,361,008 | 40,005 | # Op Amp adds spectral components to amplified output signal
I'm using this simple circuit to amplify a small signal:
simulate this circuit – Schematic created using CircuitLab
Input voltage is defined by: Vin = 1.5V + 5mV*sin(wt) with w = 5 kHz / (2*Pi).
The op-amp I use is the MCP6142 (application requires very low current consumption). I use the second op-amp only to probe the output signal from first op-amp (as scope impedance - 1 MOhm - is much higher than the resistances I use).
The problem I have is that some extra spectral components are added to the output signal:
Here is a capture of the input and output signals over time:
We can see that the input signal - blue - is "pure" as it is generated by the waveform generator of the scope. However the output signal - yellow -, though it is correctly amplified, has some extra spectral components, whose frequencies are multiples of the input frequency. Do you have any idea why this happens?
• Notice the gain is also much less than you would predict. Look at the output waveform, I think you will see it's clipping, i.e. the output is hitting a supply rail (or both). How close to the supply rails is this opamp (no link to datasheet) spec'ed to go?
– user16324
Oct 11, 2016 at 21:16
• You may also be close to the slew rate limit of the op-amp. I calculate if the op-amp operates ideally, you'd have 19 V/ms slew rate at the output, while the datasheet spec is 24 V/ms, so you might be getting close enough to that to cause some distortion. Oct 11, 2016 at 21:21
• @BrianDrummond Output is not clipped, as the GBW of the op-amp is rather small (see datasheet fig 2.14 page 9). At this frequency, gain is only 30dB. Oct 11, 2016 at 21:21
• It's picking up hum, and that's causing some of your problems. The vin- terminal is a "virtual earth" sourced from very high impedance, but drawn on the schematic as a huge area. Make it as small as possible in physical area and, if necessary, put a copper foil screen over it. (If that's with the new, lower, impedances, then the hum's coming from somewhere else)
– user16324
Oct 11, 2016 at 21:55
• Did you put a guard ring around the input terminals (see page 18 of the datasheet)? You might be picking up noise from elsewhere in the circuit. Oct 11, 2016 at 22:00 | 591 | 2,279 | {"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} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.943237 |
http://engineerexperiences.com/cruise-control-fuzzy-matlab.html | 1,488,163,093,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172404.62/warc/CC-MAIN-20170219104612-00472-ip-10-171-10-108.ec2.internal.warc.gz | 73,506,210 | 37,787 | # Fuzzy Logic:
Designing a controller for non-linear systems is difficult. For conventional controller design we require mathematical model and higher order differential equations. Some times we linearize mathematical model and design controller for system by applying well developed techniques for solving linear equations. But, as we know that real life problems are non-linear in nature and their models are not easy to handle mathematically.
We solve daily life problems by if-else rules. For example, if the ball is coming fast then swing the bat early. So, fuzzy rules logic is heuristic based which takes if-else conditions to solve daily life problems. It is linguistic based technique to design controller. It takes conditions from experience and intuition. Fuzzy logic is imitations of control laws that humans use.
# Fuzzy tool in MATLAB:
MATLAB has built-in fuzzy controller designing tool. This tool allows to design rule base for controller in linguistics. Inputs and outputs are defined and membership functions of each input and outputs. This tool also allows you to select different membership functions and deffuzzification methods.
• Write fuzzy in MATLAB command window and a new window will pop-up.
• Select number of inputs from Edit>>add variables>>input/output
• Double Click the input and a new window will pop-up containing membership functions of that input. if you want to change MFs click Edit>> and find your required option for membership functions.
• click on controller to change model parameters and add rule base for fuzzy controller.
• In first window, in left bottom defuzzification methods can be changed. We will use centroid method.
# Cruise Control:
Cruise Control is for maintaining fixed sped of object (car, for example). This is a feedback closed loop control system. A reference input at which speed is to be maintained is fed to the controller as reference input and actual speed of system is taken as feedback. The difference between these two is termed as error velocity. This error velocity is one input to the controller. Figure 1
The objective is to eliminate this error. If there is some error, controller must act accordingly.
## Cruise Controller using Fuzzy:
### Block Diagram:
Following is block diagram of complete system with controller, plant (to be controlled), input to the plant (control signal from fuzzy controller) and inputs to the fuzzy controller.
### Input Membership functions:
Using Fuzzy tool in MATLB we can input membership function of desire shape. In this example we have velocity error and derivative of velocity error i.e. acceleration. The membership function of both are given below in triangular shape. we are considering three membership functions for each inputs.
For Error Velocity:
• NE: Negative error
• ZE: Zero Error
• PE: Positive Error
For Acceleration:
• NA: Negative Acceleration
• ZA: Zero Acceleration
• PA: Positive Acceleration
Base rule for controller is designed in such way that:
• if the actual speed of car is less than desired/set speed, error will be positive which means car needs to be speed up.
• if the actual speed of car is greater, error will be negative and there is need to slow down it.
Further, if speed is slow and acceleration is negative, which means that its speed is constantly decreasing and vice versa for increasing acceleration with negative error.
### Simulink Model with fuzzy controller:
Now, the fuzzy controller from above rules is implemented in a Simulink model of cruise control as shown in following figure. It has a set value (reference point of velocity) as 10. There is summer which takes this reference and feedback actual velocity and generate a velocity error. Derivative of it (acceleration) is input with velocity to fuzzy controller. And fuzzy controller has file of above designed controller. Then state space of cruise control system and finally scopes to see the output.
Now, following is graphs for velocity error, acceleration, control signal to the system and velocity (in same order). It can be seen that when velocity is decreasing, acceleration is increasing and controller generates signal accordingly and it removes the error. (or add the decreased velocity).
This output window of cruise control shows that when velocity drops (decreasing) and acceleration is increasing and becomes constant. This shows that velocity is continuously decreasing. A control signal is generated by fuzzy controller according to rule base. Velocity is added the same amount which was required. | 906 | 4,554 | {"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-2017-09 | longest | en | 0.843346 |
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VHDL - Copy - EEE 333 ASU Spring 2010 David R Allee...
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EEE 333, ASU Spring 2010, David R. Allee Homework #2 solutions The objective of this homework is to exercise your learning of VHDL syntax, basic module definitions and modeling of combinational logic. You can try to use modelsim to simulate and check your answers. 1. Data Types: Chapter 2 in Ashenden’s book 1.1 Write constant declarations for the number of bits in a 32 bit word and for the number pi(3.14159). constant number_of_bits:integer:=32; constant pi:real:=3.14159; 1.2 Given the type declaration type state is (off, standby, active1, active2) what are the values of state’pos(standby) 1 state’succ(active2) error state’leftof(off) error state’val(2) active1 state’pred(active1) standby state’rightof(off) standby 1.3 For each of the following expressions, indicate whether they are syntactically correct, and if so, determine the resulting value. 2*3+6/4 7 3 + -4 error true and x and not y or z error (assume x,y,z are bit) “cat” & character’(‘0’) cat0 B”101110” sll 3 110000 (B”100010” sra 2) & X”2C” 11_1000_0010_1100 1.4 Write a counter model with a clock input
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Jill Tulane University ‘16, Course Hero Intern | 619 | 2,403 | {"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-2018-22 | latest | en | 0.829601 |
https://mathoverflow.net/questions/225285/a-continuous-path-between-two-sobolev-functions | 1,618,291,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00345.warc.gz | 485,707,052 | 32,605 | # A continuous path between two Sobolev functions
Let $\Omega\subset \mathbb R^N$ be open bounded, smooth boundary. Let $u_1$, $u_2\in H^{1}(\Omega)$ such that $T[u_1]=T[u_2]=T[\omega]$ where $T$ stands for the trace operator and $\omega\in H^1(\Omega)$ is a fixed function.
Define $$F(u):=\inf_{v\in\mathcal V}\left\{\int_\Omega |\nabla u|^2v^2dx + \int_\Omega |\nabla v|^2+(1-v)^2dx \right\},$$ where $\mathcal V:=\{v\in H^1(\Omega),\,0\leq v\leq 1\}$.
Question: does there exist a continuous path $a(t): [0,1]\to H^1(\Omega)$ between $u_1$ and $u_2$ satisfies the following conditions?
1. $a(0)=u_1$, $a(1)=u_2$
2. $T[a(t)]=T[\omega]$ for all $t\in (0,1)$
3. $a(t)$ is continuous in $L^2$ sense, i.e., if $t\to t_0$, then $a(t)\to a(t_0)$ in $L^2$.
4. $F(a(t))\leq \max\{F(u_1),F(u_2)\}$, for all $t\in (0,1)$.
Moreover, if $F(u_1)\leq F(a(t))\leq F(u_2)$ would be great, but it is not necessary... (here we assume $F(u_1)\leq F(u_2)$)
I tried the affine connection, i.e., $a(t)=tu_2+(1-t) u_2$, but it does not seems to work....
Any help, hint, or reference would be really welcome!
• Why doesn't the affine path work? Seems to work to me... Moreover $\nabla\alpha_t=t\nabla u_2+(1-t)\nabla u_1$ and the convexity of the norm squared even gives you 4. and the additional condition you want. – Teri Dec 4 '15 at 18:09
• @Teri: here the post is updated. – JumpJump Dec 4 '15 at 19:38
• @Teri: I edit the problem a bit more. Now it is definitely a non-trivial problem... – JumpJump Dec 5 '15 at 18:28
• Note that the compactness properties of $H^1(\Omega)$ imply that the infimum is attained. Using this one can still prove that $F$ is a convex functional, i.e. $F(tu_2+(1-t)u_1)\le tF(u_2)+(1-t)F(u_1)$. Property 4. and the additional condition always follow from this convexity. – Teri Dec 6 '15 at 20:50
• @Teri Can you be more specific and write your comments as an answer? Thank you! – JumpJump Dec 10 '15 at 22:05
I've to admit that my original argument had a (stupid) mistake. However the argument below should work.
First let me relabel: call the two functions $u_1,u_2$ by $u_0,u_1$ so as to be compatible with $u_t:=tu_1+(1-t)u_0$.
Note that given any $u\in H^1(\Omega)$, the infimum in $$F(u)=\inf_{v\in \mathcal V}\{\int_\Omega|\nabla u|^2v^2 dx+\int_\Omega(|\nabla v|^2+(1-v)^2)dx\}$$ is obtained: take a minimizing sequence for $F(u)$ and, using the Rellich-Kondrachov theorem, find $v\in \mathcal V$ and a subsequence $v_k$ so that $v_k\to v$ in $L^2$, $\nabla v_k\rightharpoonup \nabla v$ weakly in $L^2$, and $v_k\to v$ pointwise almost everywhere. Then \begin{align*} F(u)\le &\int_\Omega|\nabla u|^2v^2 dx+\int_\Omega(|\nabla v|^2+(1-v)^2)dx\\ \le &\liminf_{k\to\infty}\left[\int_\Omega|\nabla u|^2v_k^2 dx+\int_\Omega(|\nabla v_k|^2+(1-v_k)^2)dx\right]=F(u). \end{align*}
Now let $w_j$ be a function in $\mathcal V$ for which the infimum $F(u_j)$ is attained, $j=0,1$, and let $w_t=tw_1+(1-t)w_0$. By the edit below $$t\mapsto w_t^2|\nabla u_t|^2$$ is convex so we may estimate \begin{align*} F(u_t)\le & \int_\Omega |\nabla u_t|^2w_t^2 dx+\int_\Omega(|\nabla w_t|^2+(1-w_t)^2)dx\\ \le & t\int_\Omega|\nabla u_1|^2w_1^2dx+(1-t)\int_\Omega |\nabla u_0|^2w_0^2dx\\ &\ \ + t\int_\Omega(|\nabla w_1|^2+(1-w_1)^2)dx+(1-t)\int_\Omega(|\nabla w_0|^2+(1-w_0)^2)dx\\ =&tF(u_1)+(1-t)F(u_0). \end{align*} Thus $F$ is convex and this gives you the properties you want.
EDIT: Denote $$f(t)=f_x(t)=|\nabla u_t(x)|^2, g(t)=g_x(t)=w_t(x)^2.$$ Both are convex functions, but this alone doesn't guarantee the convexity of their product. Instead we use the fact that they are essentially quadratic.
Let us show that $(fg)''(t)\ge 0$. Compute \begin{align*} f'(t)=&2\nabla(u_1-u_0)\cdot \nabla u_t\\ g'(t)=&2(w_1-w_0)w_t\\ f''(t)=& 2|\nabla(u_1-u_0)|^2\\ g''(t)=& 2(w_1-w_0)^2. \end{align*} Next check that \begin{align*} (fg)''(t)&=f''(t)g(t)+2f'(t)g'(t)+g''(t)f(t)\\ =&2[|\nabla(u_1-u_0)|^2w_ t^2+4w_t\nabla(u_1-u_0)\cdot (w_1-w_0)\nabla u_t+(w_1-w_0)^2|\nabla u_t|^2]\\ \ge& 2(|\nabla(u_1-u_0)|w_t-|w_1-w_0||\nabla u_t|)^2-4|w_1-w_0|w_t|\nabla(u_1-u_0)||\nabla u_t|\\ \ge & 0. \end{align*}
• you mean that $w_t^2$ and $|\nabla u_t|^2$ is nonnegative and nondecreasing so that the product is convex? Since generally the product of nonnegative convex functions may not convex. However, we have $w_t^2$ is convex but $\nabla u_t$ may not be positive always. – JumpJump Dec 11 '15 at 20:30
• I mean, why the function $|\nabla u_t|^2w_t^2$ is convex? We usually need $|\nabla u_t|^2$ and $w_t^2$ have nonnegative or nonpositive derivative at the same time. But I don't think we have in this case. – JumpJump Dec 11 '15 at 22:38
• @tankonetoone You're right, the convexity of the product needs a separate argument, it doesn't follow automatically. I've made an edit to my answer to deal with this. – Teri Dec 14 '15 at 19:31
• Sorry if this question is naive. But how you obtain the last inequality? You are basically saying that $(|a|-|b|)^2-2|ab|\geq 0$, which is not true in generally. For example, take $a=b$. – JumpJump Dec 14 '15 at 20:29
• @tankonetoone You're right, again. I'm busy with a lot of stuff currently (hence the slow replies) but I'll try to see if anything can be done about the question. It seems that the convexity of $w_t^2|\nabla u_t|^2$ does not necessarily hold.. – Teri Dec 19 '15 at 14:52 | 2,078 | 5,330 | {"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} | 3.171875 | 3 | CC-MAIN-2021-17 | longest | en | 0.831394 |
https://www.experts-exchange.com/questions/28691962/Excel-Help-with-a-formula.html | 1,544,443,449,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823322.49/warc/CC-MAIN-20181210101954-20181210123454-00226.warc.gz | 882,204,568 | 21,974 | # Excel Help with a formula
I have a sheet that I need to do some counts on based on some text values.
I can count this way - =COUNTIF(B8:B12,"=n-****"). And I can count this way - =COUNTIF(E8:E12,"=p")+COUNTIF(E8:E12,"=t")+COUNTIF(E8:E12,"=e").
What I need to do is have something like this - IF(B8:B12,"=n-****")=COUNTIF(E8:E12,"=p")+COUNTIF(E8:E12,"=t")+COUNTIF(E8:E12,"=e")
Any help would be appreciated.
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Business Systems Analyst , ex-Senior Application EngineerCommented:
I don't quite understand your problem, what is "n", "p", "t", "e", etc ?
Do you have a sample excel can upload and indicate your issue accordingly?
the n-**** is the id number
p = present
t = tardy
e = excused
so if the usersID = n-**** then count the cells that have "p" "t" & "e" in them
Here is a copy of the file
Church-Attendee-List-Plus---EE.xlsx
Business Systems Analyst , ex-Senior Application EngineerCommented:
I guess you need to use CountIFS to allow multiple criterion, like:
=COUNTIFS(\$B8:\$B14,"=n-****",D8:D14,"=p")
=COUNTIFS(\$B8:\$B14,"=n-****",D8:D14,"=t")
=COUNTIFS(\$B8:\$B14,"=n-****",D8:D14,"=e")
Church-Attendee-List-Plus---EEb.xlsx
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Commented:
You can do this by one formula by this..
=SUMPRODUCT((LEFT(\$B8:\$B14,1)="N")*((D8:D14="p")+(D8:D14="t")+(D8:D14="e")))
This will give you the same results about what you are looking for...
Saurabh...
Commented:
You could also use this formula to give a combined count
=SUM(COUNTIFS(\$B8:\$B14,"n-*",D8:D14,{"p","t","e"}))
regards, barry
Note: using "****" doesn't do anything different from using a single "*" because the * wildcard denotes any number of characters (even zero) anyway - if you specifically want to count instances which have "n-" followed by four characters you would use ? to denote a single character, e.g. "n-????"
Older than dirtCommented:
This question has been classified as abandoned and is closed as part of the Cleanup Program. See the recommendation for more details.
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posted by .
A satellite in a nearly circular orbit is 2000 km above Earth's surface. The radius of Earth is approximately 6400 km. If the satellite completes its orbit in 112 hours, calculate the speed of the satellite in kilometers per hour.
• Geometry -
d = C = 6.28*r=6.28*6400km = 40212.4km.
C = Circumference.
V = d/t = 40212.4km / 112h = 359km/h.
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# What are three numbers that have the greatest common factor of 9?
Updated: 8/20/2019
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18, 45 & 36
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Q: What are three numbers that have the greatest common factor of 9?
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Any three.
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Group theory A major branch of modern discovered by mathematician Evariste Galois, group theory is the mathematical language of symmetry. It is the concept that is most associative with Rubik’s cube due to its invariant symmetry and the interdependence of it components. In mathematics, a “group” is a set of elements and a rule to combine those elements which satisfies certain properties. The elements of a group can consist of numbers, symmetries of a shape, or the elements themselves can be the different ways of rearranging another set of elements. (“Group”) An important part of understanding group theory, a binary operation is rule for combining elements of a set that will consequent in another element of the same type. For example, addition…show more content…
Each permutation symbolizes the result of a single element of the group. Different moves of the cube could correspond to the same final permutation and therefore to seemingly the same exact element of the group. However, a single element of the Rubik’s cube group can be expressed using different sequences of moves, just as a rotation of 4050 is the same as 450. Here, we can utilize the expression “followed by” to represent the operation in this group. Thus, one move followed by another move is itself a move of the Rubik’s cube. The number of elements in the Rubik’s cube group is, as we calculated earlier, approximately 43 quintillion because there are many distinct permutations of the cube that can be reached by legal move (without the disassembly of the cube). Any two legal moves can be combined to result in one of those permutations. To verify that this is a group: - There is an identity element, namely “not performing any move”. “Not performing any move” followed by a move x equals the move…show more content…
If the puzzle were commutative, it would not matter what order the move were implemented in, as long as each face was rotated an appropriate number of times. In this way, group theory provides deductive conclusions that would be applicable to all groups. Since the Rubik’s cube is a group, anything that is true for a group in general is true for the Rubik’s cube. (“Galois”) Furthermore, group theory is helpful in deducing certain algorithms. In particular, they are useful in deducing algorithms that possess a commutator structure, namely XYX-1Y-1(X and Y are moves or move sequences and X-1 and Y-1 are their inverses), or a conjugate structure, namely XYX-1 which is usually referred to as a “setup move”.(“Rubik’s Cube”) Both these concepts will be discussed | 565 | 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} | 3.984375 | 4 | CC-MAIN-2021-17 | latest | en | 0.935121 |
https://www.unitsconverters.com/en/R/M-To-Rad/M/Utu-4589-4596 | 1,600,739,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00166.warc.gz | 1,122,583,799 | 22,491 | 1 r/m is equivalent to 6.2832 rad/m
Formula Used
The abbreviation for r/m and rad/m is revolution/month and radian/month respectively. 1 r/m is 6 times bigger than a rad/m. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including r/m to rad/m conversion.
Check our revolution/month to rad/m converter and click on formula to get the conversion factor. When you are converting angular velocity from revolution/month to rad/m, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
The formula used to convert r/m to radian/month is 1 revolution/month = 6.28318530717902 radian/month. Measurement is one of the most fundamental concepts. Note that we have revolution/second as the biggest unit for length while radian/week is the smallest one.
How to convert r/m to rad/m? Now you can do r/m to rad/m conversion with the help of this tool. In the length measurement, first choose r/m from the left dropdown and rad/m from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from rad/m to r/m? You can check our rad/m to r/m converter.
How to convert r/m to rad/m?
The formula to convert r/m to rad/m is 1 revolution/month = 6.28318530717902 radian/month. r/m is 6.2832 times Bigger than rad/m. Enter the value of r/m and hit Convert to get value in rad/m. Check our r/m to rad/m converter. Need a reverse calculation from rad/m to r/m? You can check our rad/m to r/m Converter.
How many rad/s is 1 r/m?
1 r/m is equal to 0.00218166156499271 rad/s. 1 r/m is 458.3662361047 times Smaller than 1 rad/s.
How many rad/d is 1 r/m?
1 r/m is equal to 188.495559215371 rad/d. 1 r/m is 188.495559215371 times Bigger than 1 rad/d.
How many rad/h is 1 r/m?
1 r/m is equal to 7.85398163397377 rad/h. 1 r/m is 7.85398163397377 times Bigger than 1 rad/h.
How many rad/min is 1 r/m?
1 r/m is equal to 0.130899693899563 rad/min. 1 r/m is 7.63943726841167 times Smaller than 1 rad/min. | 595 | 2,193 | {"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-2020-40 | latest | en | 0.872095 |
https://forum.vc-mp.org/?topic=240.msg1324 | 1,709,301,254,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00319.warc.gz | 265,086,826 | 7,458 | ## Does anyone know the maximum world size in in VC?
Started by ., Jan 31, 2015, 11:11 PM
#### .
I need to know the maximum size of the world in ViceCity. Basically I need to know how much can the player go in both positive and negative X, Y, Z directions. Like a box for example that simply tells how X goes from -2131.23 to 2314.5, Y goes from -1213.2 to 321.13 and Z goes from -3154.34 to 2526.65. I'll do some digging on my own but if anyone has a knowledge of these limits than please share them with me.
EDIT:
What is the maximum positive and negative x, y, z values that a an entity can reach in the game (player, object, vehicle etc.).
.
#### Sebastian
From server.conf:
`<WorldBoundaries x1="5000.0" x2="-5000.0" y1="5000.0" y2="-5000.0"/>`
And here is the syntax:
http://wiki.vc-mp.org/wiki/Scripting/Squirrel/Functions/SetWorldBounds
#### .
Oh, so this is modifiable. Ok, I'll look into it.
.
#### Fuzzie
Maybe a developer could give a more explained reasoning behind this.
Using my settings script, I set the boundaries to 5000.0 on every side (basically the values like what seby posted). However, when I went to the far top left corner of the map (Downtown), I realized that I'm only able to go to approximately y = 1950.00 (I was in a Hunter when I test this so the size of the Hunter itself could have changed the actual world limit. Can't recall what the x-axis was). I also tested with 50000.0 but I still got the same results. Any value lower than 1950.0 (Again, I only tested the maximum y-axis) works normally however.
I don't know if this is related to boundaries being set by script or if this is a VC related boundaries issue.
x < 2000.0, x > -2100.0, y < 1900.0, y > -1900.0 are the max world bounds for sync on the 0.3r2 Squirrel server.
#### Stormeus
Quote from: Fuzzie on Feb 01, 2015, 01:27 AMMaybe a developer could give a more explained reasoning behind this.
<snip>
If you found yourself running into an invisible wall, that's some trickery maxorator invented. Vice City doesn't consider anything beyond that wall valid; to prevent players from crashing other people's games by being in an invalid coordinate, VC:MP deflects vehicles and projectiles approaching the invalid area.
#### Fuzzie
What are the coordinates for this invalid area?
#### Stormeus
Quote from: Fuzzie on Feb 03, 2015, 02:18 AMWhat are the coordinates for this invalid area?
Lower bound is (-2350.0, -1950.0, -200.0)
Upper bound is (1550.0, 1950.0, 20000.0) | 689 | 2,483 | {"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} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.922838 |
http://quizlet.com/1053055/book-n-chapter-1-introduction-to-electricity-flash-cards/ | 1,386,962,874,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164981569/warc/CC-MAIN-20131204134941-00048-ip-10-33-133-15.ec2.internal.warc.gz | 142,367,440 | 19,772 | # Book N - Chapter 1: Introduction to Electricity
## 27 terms
### Alternating current
(AC) electric current in which the charges continually switch from flowing in one direction to flowing in the reverse direction
### Battery
"a device that is made of several cells and that produces an electric current by converting chemical energy into electrical energy"
### Cell
a device that produces an electric current by converting chemical energy into electrical energy
### Circuit
a complete, closed path through which electric charges flow
### Conduction (electrical)
a method of charging an object that occurs when electrons are transferred from one object to another by direct contact
### Conductor (electrical)
a material in which charges can move easily
### Current
a continuous flow of charge caused by the motion of electrons; the rate at which charge passes a given point; expressed in amperes (A)
### District Current
"(DC) electric current in which the charges always flow in the same direction"
### Electric discharge
the loss of static electricity as charges move off an object
### Electric field
"the region around a charged particle that can exert a force on another charged particle"
### Electric force
the force between charged objects
### Electric power
"the rate at which electrical energy is used to do work; expressed in watts (W)"
### Electrode
the part of a cell through which charges enter or exit
### Electrolyte
in a cell, a mixture of chemicals that carries an electric current
### Induction
"a method of charging an object that occurs when charges in an uncharged object are rearranged without direct contact with a charged object"
### Insulator (electrical)
a material in which charges cannot easily move
### Law of electric charges
"the law that states that like charges repel and opposite charges attract"
a device that uses electrical energy to do work
### Ohm's Law
"the law that states the relationship between current (I), voltage (V), and resistance (R); expressed by the equation I = V/R"
### Parallel circuit
a circuit in which different loads are on separate branches
### Photocell
the part of a solar panel that converts light into electrical energy
### Potential difference
energy per unit charge; specifically, the difference in energy per unit charge as a charge moves between two points in an electric circuit (same as voltage); expressed in volts (V)
### Resistance
the opposition to the flow of electric charge; expressed in ohms (Ω)
### Series circuit
a circuit in which all parts are connected in a single loop
### Static electricity
the buildup of electric charges on an object
### Thermocouple
a device that converts thermal energy into electrical energy
### Voltage
"the difference in energy per unit charge as a charge moves between two points in an electric circuit (same as potential difference); expressed in volts (V)" | 590 | 2,921 | {"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} | 2.8125 | 3 | CC-MAIN-2013-48 | latest | en | 0.924837 |
https://ccssmathanswers.com/180-days-of-math-for-fifth-grade-day-3-answers-key/ | 1,725,733,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00177.warc.gz | 153,084,050 | 58,014 | # 180 Days of Math for Fifth Grade Day 3 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 3 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 3
Directions Solve each problem.
Question 1.
Perform an addition operation to find the sum of the two given numbers.
19 + 8 = 27
Question 2.
5 × 7 = _____
By performing the multiplication operation for the two given numbers we can find the product of 5 and 7.
So, multiplying 5 and 7 we get 35.
Question 3.
By performing the division operation for the two given numbers we can find the quotient of 81 and 9.
81 ÷ 9 = 9
Question 4.
Round 358 to the nearest hundred.
_______________
Answer: 400 is the nearest hundred to 358.
Question 5.
0.5 of 2 is ___.
Question 6.
8 – (10 ÷ 2) = ______
Answer: 8 – 5 = 3
Question 7.
Answer: 7 × 7 = 49
Question 8.
______ minutes = 4 hours
Answer: 240 minutes = 4 hours (1 hour = 60 minutes)
Question 9.
True or false? This figure only has one line of symmetry.
_______________
Answer: False, as square has 4 lines of symmetry.
Question 10.
Find the coordinates of:
_______________ | 326 | 1,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.865938 |
http://mamicode.com/info-detail-2679273.html | 1,568,570,138,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572235.63/warc/CC-MAIN-20190915175150-20190915201150-00188.warc.gz | 124,553,958 | 6,027 | # 加减乘除法运算（函数数组指针法）
```#include<stdio.h> double add(double x, double y)//加 { return x + y; } double sub(double x, double y)//减 { return x - y; } double mul(double x, double y)//乘 { return x*y; } double div(double x, double y)//除 { if (y != 0) { return x/y; } else { printf("输入错误"); return -1; } } void menu() { printf("************************\n"); printf("***1.add 2.sub***\n"); printf("***3.mul 4.div***\n"); printf("******* 0..exit ********\n"); printf("************************\n"); } int main() { int input = 1; double a = 0; double b = 0; double ret; double(*p[5])(double a, double b) = { 0, add, sub, mul, div }; while (input) { menu(); printf("请输入需要的操作:\n"); scanf("%d",&input); if (input > 0 && input <= 4) { printf("请输入操作数:\n"); scanf("%lf%lf",&a,&b); ret = (*p[input])(a, b); printf("%.2lf\n", ret); } else if (0==input) { break; } else { printf("输入错误"); } } return 0; }```
(0)
(0) | 310 | 893 | {"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-2019-39 | latest | en | 0.14477 |
https://nrich.maths.org/public/leg.php?code=-100&cl=1&cldcmpid=246 | 1,508,523,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824293.62/warc/CC-MAIN-20171020173404-20171020193404-00321.warc.gz | 798,251,418 | 9,940 | # Search by Topic
#### Resources tagged with Trial and improvement similar to Birthday Cakes:
Filter by: Content type:
Stage:
Challenge level:
### There are 85 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### Heads and Feet
##### Stage: 1 Challenge Level:
On a farm there were some hens and sheep. Altogether there were 8 heads and 22 feet. How many hens were there?
### Eggs in Baskets
##### Stage: 1 Challenge Level:
There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket?
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Big Dog, Little Dog
##### Stage: 1 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### The Tall Tower
##### Stage: 1 Challenge Level:
As you come down the ladders of the Tall Tower you collect useful spells. Which way should you go to collect the most spells?
### Five Steps to 50
##### Stage: 1 Challenge Level:
Use five steps to count forwards or backwards in 1s or 10s to get to 50. What strategies did you use?
### Jumping Squares
##### Stage: 1 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
### Find a Path
##### Stage: 1 Challenge Level:
Can you find a path from a number at the top of this network to the bottom which goes through each number from 1 to 9 once and once only?
### Amazing Alphabet Maze
##### Stage: 1 Challenge Level:
Can you go from A to Z right through the alphabet in the hexagonal maze?
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### A Numbered Route
##### Stage: 2 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### The Brown Family
##### Stage: 1 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
### Find the Difference
##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### 4 Dom
##### Stage: 1, 2, 3 and 4 Challenge Level:
Use these four dominoes to make a square that has the same number of dots on each side.
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Gift Stacks
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Watch the Clock
##### Stage: 2 Challenge Level:
During the third hour after midnight the hands on a clock point in the same direction (so one hand is over the top of the other). At what time, to the nearest second, does this happen?
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Four-digit Targets
##### Stage: 2 Challenge Level:
You have two sets of the digits 0 – 9. Can you arrange these in the five boxes to make four-digit numbers as close to the target numbers as possible?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Numbered Cars
##### Stage: 2 Challenge Level:
I was looking at the number plate of a car parked outside. Using my special code S208VBJ adds to 65. Can you crack my code and use it to find out what both of these number plates add up to?
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Special 24
##### Stage: 2 Challenge Level:
Find another number that is one short of a square number and when you double it and add 1, the result is also a square number.
### Pizza Cut
##### Stage: 2 Challenge Level:
Using only six straight cuts, find a way to make as many pieces of pizza as possible. (The pieces can be different sizes and shapes).
### Rocco's Race
##### Stage: 2 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Let's Face It
##### Stage: 2 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Magic Matrix
##### Stage: 2 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix? | 2,252 | 9,601 | {"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} | 4.375 | 4 | CC-MAIN-2017-43 | latest | en | 0.9275 |
https://mbaessaywritingservice.com/moore-company-is-about-to-issue-a-bond-with-semiannual-coupon-payments/ | 1,632,351,374,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00624.warc.gz | 410,753,514 | 12,071 | Moore company is about to issue a bond with semiannual coupon payments
Part A: Moore Company is about to issue a bond with semiannual coupon payments, a coupon rate of 8%, and par value of \$1,000. The yield-to-maturity for this bond is 10%.
a. What is the price of the bond if the bond matures in 5, 10, 15, or 20 years?
b. What do you notice about the price of the bond in relationship to the maturity of the bond?
Part B: The Crescent Corporation just paid a dividend of \$2 per share and is expected to continue paying the same amount each year for the next 4 years. If you have a required rate of return of 13%, plan to hold the stock for 4 years, and are confident that it will sell for \$30 at the end of 4 years, how much should you offer to buy it at today?
Part C: Use the information in the following table to answer the questions below.
State of Economy Probability of State Return on A in State Return on B in State Return on C in State Boom .35 0.040 0.210 0.300 Normal .50 0.040 0.080 0.200 Recession .15 0.040 -0.010 -0.260
a. What is the expected return of each asset?
b. What is the variance of each asset?
c. What is the standard deviation of each asset? | 321 | 1,188 | {"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-2021-39 | latest | en | 0.894063 |
https://math.stackexchange.com/questions/1575006/solving-time-needed-to-travel-a-given-distance-given-simulated-not-real-physic | 1,563,564,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526337.45/warc/CC-MAIN-20190719182214-20190719204214-00390.warc.gz | 468,116,715 | 38,626 | # Solving time needed to travel a given distance, given simulated (not real physics) properties of acceleration of object
For a small personal project I'm looking at travel time of objects in a video game called EVE-Online.
I need to calculate time it will take object to travel from stand-still, constantly accelerating, until it reaches $x$ meters.
In this game objects velocity while accelerating in straight line is defined with equation:
$$v(t)= s(1-e^{-\frac{t}{a}})$$
Where $a$ and $s$ are object specific and unchanging for duration of acceleration.
The function is constructed in a way that $v(t)$ will approach $s$ (maximum speed of object), but never reaching it.
What I did to solve I would call a brute force approach: I calculated $v(t)$ for each second of simulation and summed it up until reaching $x$ (not exactly, as You will overshoot but my system will work fine with precision around one second).
Because I have to calculate this value for many thousand of objects it is impractical to perform this simulation for each and every single one due to computing time needed (I want my system to be relatively fast) and I'm looking for directly solving for $t$ needed to sum of $v(t)$ equaling to $x$.
Is there a way to solve this other than just sum up $v(t)$ at each second or fraction of a second until reaching designated goal?
• Isolate the exponential term and use logarithms. – Claude Leibovici Dec 14 '15 at 11:06
• If I understand it correctly that will only allow me to solve for time needed to reach specific V, which is easy. What I need is taking another step and calculating entire distance traveled in a specific amount of time. – Tetlanesh Dec 14 '15 at 11:08
You can work out a formula for distance travelled using an integral :
$$x(t) = \int_0^t{v(t) dt}$$
When you work out that equation you can solve it for $t$ in terms of $x$.
What you get is :
$$x(t) = s t + sa ( e^{-t/a} - e^0 )$$
Which is :
$$x(t) = st - sa( 1 - e^{-t/a} )$$
Now we cannot provide a convenient formula for $t(x)$, the time to travel a given distance. A basic numerical approach would be to use Newton's Method. Note that for that method you will need the derivative of $x(t)$ and this is simply the velocity $v(t)$
In your case you keep calculating new values of $t_n$ using :
$$t_{n+1} = t_n - \frac{x(t_n)-X_0}{v(t_n)}$$
Where $X_0$ is your target distance.
When $(t_n-t_{n-1})$ is small enough for your needs ( which should should not take many calculations ) you have your approximate answer. You can start with any value, but try $t_0 = 0$ for simplicity.
( Thanks to Claude Leibovici for spotting a silly sign error in my original post. ).
• Thank you @StephenG, I will take a good long look at Your response and try to implement what You are proposing when I'm back at home from work :) – Tetlanesh Dec 14 '15 at 12:09
• Inspired by your answer, I proposed something which could be funny to test. Cheers. – Claude Leibovici Dec 14 '15 at 16:17
• I'm yet to try Your solution @StephenG. Anyway I figured out how to approximate the result without iterations using the formula You provided . Since $v(t)$ approaches $s$, and I can (after reorganizing) use the original formula to solve for $t(v)$ (time needed to reach certain speed) to reach for example 99% of $s$, than use Your formula to calculate $x(t)$ to get distance traveled to this point, than substract $x(t)$ from $X_0$ to get distance remaining and from there just use $distance/s$ (with assumption that object moves at sped $s$). It will be approximation but close enough for my needs. – Tetlanesh Dec 16 '15 at 15:28
• Thanks for the Integral function. I used it to estimate $t(x)$ like I wrote in my previous comment. My result are accurate enough for my needs. I will play with the Newtons Method and compare results anyway, but my approximation is many times faster computation wise and I can vectorize it. – Tetlanesh Dec 17 '15 at 23:29
• For the first few examples I runned trough Newton's Method I got almost perfect result within 2/3 iterations, so its really efficient. Also as opposed to Your advice to start with t=0 I started with 1, as $v(0)=0$. Thank you very much @StephenG. – Tetlanesh Dec 17 '15 at 23:49
This is not an answer but it is too long for a comment.
As StephenG answered (I fixed a minor sign error in the solution), the equation $$X_0 = s\,t - s\,a\,\big( 1 - e^{-\frac t a} \big)$$ would require some numerical method (Newton being probably the simplest).
However, this equation has an explicit solution in terms of Lambert function $$t=\frac{X_0}{s}+a \,\Big(W\left(z\right)+1\Big)\qquad z=\exp\Big(-\big(1+\frac{X_0}{a \,s}\big)\Big)$$ The Wikipedia page gives approximation formulas for small values of the argument $z$ (you should use it).
In fact, for your curiosity, any equation which can write $A+Bx+C\log(D+Ex)=0$ shows explicit solution(s) in terms of Lambert function.
• Wikipedia has a note showing a numerical algorithm for $W(z)$ using Newton, which is sort of funny in the circumstances. – StephenG Dec 14 '15 at 17:55
• @StephenG. It is a standard problem, just be sure. Lambert is, at least to me, one of the most important functions which already has a lot of practical applications. What has really good is that you worked the problem. The remaining is ... blabla $+\epsilon$ ! Cheers and thanks again. – Claude Leibovici Dec 14 '15 at 18:24
• @StephenG. It seems that you have a small error (a sign) in the integration of the differential equation. I fixed my answer. – Claude Leibovici Dec 15 '15 at 4:19 | 1,462 | 5,561 | {"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} | 3.703125 | 4 | CC-MAIN-2019-30 | latest | en | 0.914143 |
http://stepcalculator.com/math/adding-fractions-calculator.php | 1,713,113,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00718.warc.gz | 29,538,154 | 3,925 | 1 Adding fractions calculator (+)
1
# Adding fractions calculator
Adding fractions calculator online.
Enter fractions and press the = button.
Enter simple fractions with slash (/).
For example: 1/2 + 1/3
Enter mixed numbers with space.
For example: 2 1/2 + 1 1/3
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#### Adding fractions example
1/2 + 1/3 = (1×3+1×2) / (2×3) = 5/6
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What is the consequence of changing over the condition? Completing every one of the changes recorded above can give either a condition that has indistinguishable roots from the first condition, or a condition, among the underlying foundations of which contains every one of the foundations of the first condition, however which can likewise have different roots, or a condition that does exclude every one of the roots underlying foundations of the changed condition. In the accompanying passages, we will dissect which of these changes under which conditions lead to which conditions. It is critical to know for the fruitful arrangement of conditions.
Users write Now: BluCari • 10 months ago • edited I understood why the calculator gives 2 + 2x2 8 in the example because it counts 2 + 2 first and then only 4x2, as it were, the first 2 + 2 example solves first and then the other does the answer and it turns out 8 I know it's obvious, but some don't.
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1 | 325 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-18 | longest | en | 0.919737 |
https://www.fxsolver.com/browse/?like=2113&p=82 | 1,660,199,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00208.warc.gz | 683,536,627 | 37,189 | '
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# Advice on Diagnostic Test in OG
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Joined: 28 Jul 2006
Posts: 7
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02 Sep 2006, 17:23
I was wondering if anyone has any advice on how to interpret the Diagnostic Test in OG 11 compared to GMAT performance? Do we know if the questions in there are all difficult? (They seemed so to me!) Or what the "Excellent" vs. other categories might mean for the "real" score?
I have a week to go before the GMAT so I'm trying to figure out how I'm really doing!
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02 Sep 2006, 19:14
Take the GMAPPrep CAT. That should give you a +/- 20 point score estimate of where you currently stand. There are some repeats from GMATPrep posted in this forum and in the OG11 (supplementary version), so keep in mind that your score may be slightly skewed if you have been practicing with both sources.
02 Sep 2006, 19:14
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# Advice on Diagnostic Test in OG
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 626 | 2,352 | {"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-2017-04 | latest | en | 0.89311 |
https://academicassignmentexperts.com/the-monkey-and-bananas-problem/ | 1,670,477,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00012.warc.gz | 104,980,464 | 16,213 | # The monkey-and-bananas problem
The monkey-and-bananas problem is faced by a monkey in a laboratory with some bananas hanging out of reach from the ceiling. A box is available that will enable the monkey to reach the bananas if he climbs on it. Initially, the monkey is at A, the bananas at B. and the box at C. The monkey and box have height Low, hut if the monkey climbs onto the box he will have height high, the same as the bananas. The actions available to the monkey include Go from one place to another, Push an object from one place to another, climb Up onto or Climb Down from an object, and Grasp or Un grasp an object. Grasping results in holding the object if (he monkey and object are in the same place at the same height.
a. Write down the initial state description.
b. Write down STRIPS-style definitions of the six actions.
c. Suppose the monkey wants to fool the scientists, who are off to tea, by grabbing the bananas, but leaving the box in its original place. Write this as a general goal (i.e., not assuming that the box is necessarily at C) in the language of situation calculus. Can this goal be solved by a STRIPs-style system?
d. Your axiom for pushing is probably incorrect, because if the object is too heavy, its position will remain the same when the Push operator is applied, is this an example of the ramification problem or the qualification problem? Fix your problem description to account for heavy objects.
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http://info.iut-bm.univ-fcomte.fr/pub/gitweb/simgrid.git/blob/3f0c94e70e5ee6fc659b2e1dd45c765e7afbd0fe:/tools/MSG_visualization/trace2fig.pl | 1,606,726,098,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00361.warc.gz | 41,977,228 | 11,188 | Algorithmique Numérique Distribuée Public GIT Repository
1 #!/usr/bin/perl -w
2 use strict;
3 #use Data::Dumper;
4 use XFig;
5 use POSIX;
7 my(\$grid_Y_size)=225;
8 my(\$grid_X_size)=100550; # Can be changed to improve readability in
9 # function of the total execution time
11 my(\$color_suspended)=1;
12 my(\$color_compute)=2;
13 my(\$color_wait_for_recpt)=3;
14 my(\$color_communicate)=4;
16 # Determine the order of the colors in the legend
17 my(@color_list)=(\$color_compute,\$color_communicate,\$color_wait_for_recpt,\$color_suspended);
23 my(%Cat);
24 my(\$filename)=@_;
25 my(\$line);
27 open INPUT, \$filename;
29 while (defined(\$line=<INPUT>)) {
30 chomp \$line;
31 if(\$line =~ /^7\s/) {
32 my(\$event,\$date,\$id,\$type,\$father,@name) = split(/\s+/,\$line);
34 \$Cat{\$id}{name}="@name ";
35 \$Cat{\$id}{name}=~s/\"//g;
36 \$Cat{\$id}{father}=\$father;
37 \$Cat{\$id}{type}=\$type;
38 \$Cat{\$id}{date}=\$date;
39 }
40 }
41 close INPUT;
42 return \%Cat;
43 }
47 my(\$filename,\$Cat)=@_;
48 my(\$line);
50 open INPUT, \$filename;
52 while (defined(\$line=<INPUT>)) {
53 chomp \$line;
54 if(\$line =~ /^11\s/) {
55 my(\$event,\$date,\$type,\$id,\$state) = split(/\s+/,\$line);
56 push @{\$\$Cat{\$id}{state}}, [\$date,\$state];
57 }
58 if(\$line =~ /^12\s/) {
59 my(\$event,\$date,\$type,\$id) = split(/\s+/,\$line);
60 push @{\$\$Cat{\$id}{state}}, [\$date];
61 }
62 }
63 close INPUT;
64 }
67 my(\$filename)=@_;
68 my(\$line);
71 open INPUT, \$filename;
73 while (defined(\$line=<INPUT>)) {
74 chomp \$line;
75 if(\$line =~ /^16\s/) {
76 my(\$event,\$date,\$type,\$father,\$channel,\$src,\$key,\$trash) = split(/\t+/,\$line);
77 my(\$numkey)=hex "\$key";
82 }
83 if(\$line =~ /^17\s/) {
84 my(\$event,\$date,\$type,\$father,\$channel,\$dst,\$key,\$trash) = split(/\t+/,\$line);
85 my(\$numkey)=hex "\$key";
89 }
90 }
91 close INPUT;
93 }
96 sub build_cat_tree {
97 my(\$root,\$Cat)=@_;
98 my(@childs)=();
99 my(\$cat);
101 foreach \$cat (keys %\$Cat) {
102 if(\$\$Cat{\$cat}{father} eq \$root) {
103 push @childs, build_cat_tree(\$cat,\$Cat);
104 }
105 # print "\$\$Cat{\$cat}{name}\t\t \$Cat{\$cat}{father}\n";
106 }
108 return [\$root,@childs];
109 }
111 sub build_cat_list {
112 my(\$tree,\$cat_list)=@_;
113 my(\$root) = shift @\$tree;
114 my(\$u);
116 push @\$cat_list,\$root;
118 foreach \$u (@\$tree) {
119 build_cat_list(\$u,\$cat_list);
120 }
121 unshift @\$tree, \$root;
122 }
125 sub cat_sorting_function {
126 my(\$cat1,\$cat2,\$Cat)=@_;
127 if (!defined(\$\$Cat{\$cat1}{state})) {
128 if (!defined(\$\$Cat{\$cat2}{state})) {
129 return 0;
130 } else {
131 return 1;
132 }
133 }
135 if (!defined(\$\$Cat{\$cat2}{state})) {
136 return -1;
137 }
139 my(\$comp) = \$\$Cat{\$\$Cat{\$cat1}{'father'}}{'name'} cmp \$\$Cat{\$\$Cat{\$cat2}{'father'}}{'name'};
140 if (\$comp == 0) {
141 return \$\$Cat{\$cat1}{'name'} cmp \$\$Cat{\$cat2}{'name'};
142 } else {
143 return \$comp;
144 }
145 }
147 sub update_host_Y {
148 my(\$host,\$i) = @_;
149 if (!defined(\$\$host{'Y_min_host'})) {
150 \$\$host{'Y_min_host'} = \$i;
151 } else {
152 if (\$\$host{'Y_min_host'} > \$i) {
153 \$\$host{'Y_min_host'} = \$i;
154 }
155 }
156 if (!defined(\$\$host{'Y_max_host'})) {
157 \$\$host{'Y_max_host'} = \$i+1;
158 } else {
159 if (\$\$host{'Y_max_host'} < \$i+1) {
160 \$\$host{'Y_max_host'} = \$i+1;
161 }
162 }
163 }
165 sub set_cat_position {
166 my(\$Cat,\$cat_list)=@_;
167 my(\$i)=0;
168 my(\$cat);
169 foreach \$cat (sort {cat_sorting_function(\$a,\$b,\$Cat)} @\$cat_list) {
170 if(defined(\$\$Cat{\$cat}{state})) {
171 update_host_Y(\$\$Cat{\$\$Cat{\$cat}{'father'}},\$i);
172 \$\$Cat{\$cat}{Y_min} = \$i;
173 \$\$Cat{\$cat}{Y_max} = \$i+1;
174 \$i++;
175 }
176 }
177 }
179 sub create_fig {
180 my(\$filename)=shift;
181 my(\$fig)=new XFig;
182 \$fig->{object} = 'compound'; # Compound
183 \$fig->{elements} = [];
184 \$fig->{version} = 3.2;
185 \$fig->{orientation} = 'Landscape';
186 \$fig->{justification} = 'Center';
187 \$fig->{units} = 'Metric';
188 \$fig->{papersize} = 'A4';
189 \$fig->{magnification} = '100.00';
190 \$fig->{multiplepage} = 'Single';
191 \$fig->{transparent} = '-2';
192 \$fig->{resolution} = '1200';
193 \$fig->{coordsystem} = '2';
194 \$fig->{filename} = \$filename;
195 return \$fig;
196 }
198 sub draw_cat {
201 my(\$max_string_length)=0;
202 foreach \$cat (keys %\$Cat) {
203 next unless (defined(\$\$Cat{\$cat}{Y_min}) &&
204 defined(\$\$Cat{\$cat}{Y_max}));
205 my(\$text) = new XFig ('text');
206 # \$text->{'text'} = "\$\$Cat{\$\$Cat{\$cat}{father}}{name}"."\$\$Cat{\$cat}{name}";
207 my(\$printed_name)= \$\$Cat{\$cat}{name};
208 \$printed_name =~ s/\d+ \(0\)\s*\$//;
209 if (length(\$printed_name) > \$max_string_length) {
210 \$max_string_length = length(\$printed_name);
211 }
212 \$text->{'text'} = "\$printed_name";
213 \$text->{'y'} = (\$\$Cat{\$cat}{Y_min}+\$\$Cat{\$cat}{Y_max})/2*\$grid_Y_size+68;
214 \$text->{'x'} = -100;
215 \$text->{'subtype'} = 2;
217 }
219 my(\$max_date)=0;
220 foreach \$cat (keys %\$Cat) {
221 next unless (defined(\$\$Cat{\$cat}{Y_min}) &&
222 defined(\$\$Cat{\$cat}{Y_max}));
223 my(@states)=();
224 my(\$e);
225 foreach \$e (@{\$\$Cat{\$cat}{state}}) {
226 my(\$new_date,\$state) = (\$\$e[0],\$\$e[1]);
227 if (\$new_date > \$max_date) {
228 \$max_date = \$new_date;
229 }
230 if(defined(\$state)) {
231 push @states, \$e;
232 } else {
233 my(\$old_event) = pop @states;
234 my(\$old_date) = \$\$old_event[0];
235 \$state = \$\$old_event[1];
237 # LM: I added the next line because of "undefined values"...
238 # normally, I think that this should not happen, but this part of code
239 # is a bit too cryptic to me
240 next unless (defined(\$state));
242 my(\$line) = new XFig ('polyline');
244 \$line->{'depth'} = 50; # line
245 \$line->{'subtype'} = 1; # line
246 \$line->{'points'} = [ [\$old_date*\$grid_X_size, \$\$Cat{\$cat}{Y_min}*\$grid_Y_size],
247 [\$new_date*\$grid_X_size, \$\$Cat{\$cat}{Y_min}*\$grid_Y_size],
248 [\$new_date*\$grid_X_size, \$\$Cat{\$cat}{Y_max}*\$grid_Y_size],
249 [\$old_date*\$grid_X_size, \$\$Cat{\$cat}{Y_max}*\$grid_Y_size],
250 [\$old_date*\$grid_X_size, \$\$Cat{\$cat}{Y_min}*\$grid_Y_size] ];
251 \$line->{'areafill'} = 20;
252 if(\$state eq "S") {
253 \$line->{'fillcolor'} = \$color_suspended;
254 } elsif (\$state eq "E") {
255 \$line->{'fillcolor'} = \$color_compute;
256 } elsif (\$state eq "B") {
257 \$line->{'fillcolor'} = \$color_wait_for_recpt;
258 } elsif (\$state eq "C") {
259 \$line->{'fillcolor'} = \$color_communicate;
260 }
262 }
263 }
264 }
267 my(\$line) = new XFig ('polyline');
272 \$line->{'subtype'} = 1; # line
274 print STDERR "\$link: \$src (\$src_date) -> \$dst (\$dst_date)\n";
276 print STDERR "\$\$Cat{\$src}{name} -> \$\$Cat{\$dst}{name}\n";
277 \$line->{'points'} = [ [\$src_date*\$grid_X_size,
278 (\$\$Cat{\$src}{Y_min}+\$\$Cat{\$src}{Y_max})/2*\$grid_Y_size],
279 [\$dst_date*\$grid_X_size,
280 (\$\$Cat{\$dst}{Y_min}+\$\$Cat{\$dst}{Y_max})/2*\$grid_Y_size] ];
281 \$line->{'forwardarrow'} = ['1', '1', '1.00', '60.00', '120.00'];
283 }
287 # Host visualization
289 my(\$max_Y)= 0;
291 my(\$index_fill)=0;
292 my(\$width_of_one_letter)=80;
293 my(\$min_x_for_host)=-400 - \$max_string_length*\$width_of_one_letter;
294 my(\$host_text_x)= \$min_x_for_host + 200;
296 foreach \$cat (keys %\$Cat) {
297 next unless (defined(\$\$Cat{\$cat}{Y_min_host}) &&
298 defined(\$\$Cat{\$cat}{Y_max_host}));
299 my(\$line) = new XFig ('polyline');
301 \$line->{'depth'} = 150;
302 \$line->{'subtype'} = 1; # line
303 \$line->{'points'} = [ [\$min_x_for_host, \$\$Cat{\$cat}{Y_min_host}*\$grid_Y_size],
304 [\$max_date*\$grid_X_size+150, \$\$Cat{\$cat}{Y_min_host}*\$grid_Y_size],
305 [\$max_date*\$grid_X_size+150, \$\$Cat{\$cat}{Y_max_host}*\$grid_Y_size],
306 [\$min_x_for_host, \$\$Cat{\$cat}{Y_max_host}*\$grid_Y_size] ];
307 \$line->{'areafill'} = 4+3*(\$index_fill % 2);
308 \$line->{'fillcolor'} = 0;
309 \$line->{'thickness'} = 0;
310 \$index_fill++;
313 my(\$text) = new XFig ('text');
314 \$text->{'text'} = "\$\$Cat{\$cat}{name}";
315 \$text->{'angle'} = 3.14159265/2;
316 \$text->{'x'} = \$host_text_x;
317 \$text->{'y'} = (\$\$Cat{\$cat}{Y_min_host}+\$\$Cat{\$cat}{Y_max_host})/2*\$grid_Y_size;
318 \$text->{'subtype'} = 1;
319 \$text->{'font_size'} = 30;
322 if (\$max_Y<\$\$Cat{\$cat}{Y_max_host}) {
323 \$max_Y = \$\$Cat{\$cat}{Y_max_host};
324 }
325 }
327 # Legend:
328 my(\$i)=1;
329 my(\$color);
330 foreach \$color (@color_list) {
331 my(\$min_x)=0;
332 my(\$min_Y)=(\$max_Y+1)*\$grid_Y_size;
333 my(\$width)=1700;
334 my(\$height)=\$grid_Y_size;
336 my(\$line) = new XFig ('polyline');
338 \$line->{'depth'} = 50;
339 \$line->{'subtype'} = 1; # line
340 \$line->{'points'} = [ [\$min_x,\$min_Y + (\$i-1)*\$height ],
341 [\$min_x + \$width,\$min_Y + (\$i-1)*\$height],
342 [\$min_x + \$width,\$min_Y+\$height + (\$i-1)*\$height],
343 [\$min_x,\$min_Y+\$height + (\$i-1)*\$height],
344 [\$min_x,\$min_Y+ (\$i-1)*\$height]];
345 \$line->{'areafill'} = 20;
346 \$line->{'fillcolor'} = \$color;
349 my(\$text) = new XFig ('text');
351 if (\$color==\$color_suspended) {
352 \$text->{'text'} = "Suspended";
353 }
354 if (\$color==\$color_compute) {
355 \$text->{'text'} = "Computing";
356 }
357 if (\$color==\$color_wait_for_recpt) {
358 \$text->{'text'} = "Waiting for reception";
359 }
360 if (\$color==\$color_communicate) {
361 \$text->{'text'} = "Communicating";
362 }
364 \$text->{'y'} = \$min_Y + (\$i-0.5)*\$height +68;
365 \$text->{'x'} = 50;
366 \$text->{'subtype'} = 0;
368 \$i++;
369 }
371 # Time axis
373 my(\$line) = new XFig ('polyline');
374 \$line->{'depth'} = 0;
375 \$line->{'subtype'} = 1; # line
376 \$line->{'points'} = [ [0,0],[\$max_date * \$grid_X_size+150,0] ];
377 \$line->{'forwardarrow'} = ['1', '1', '1.00', '60.00', '120.00'];
380 my(\$digits)=POSIX::floor(log(\$max_date)/log(10));
381 my(\$exponent) = 10**\$digits;
382 my(\$mantissa)= \$max_date / \$exponent;
383 my(\$incr);
384 if (\$mantissa<2) {
385 \$incr = \$exponent/10;
386 } elsif (\$mantissa<5) {
387 \$incr = \$exponent/4;
388 } else {
389 \$incr = \$exponent/2;
390 }
392 print "\$max_date \$digits \$exponent \$mantissa \$incr\n";
393 my(\$x);
394 for(\$x=0; \$x < \$max_date; \$x += \$incr) {
395 print "\$x\n";
396 \$line = new XFig ('polyline');
397 \$line->{'depth'} = 0;
398 \$line->{'subtype'} = 1; # line
399 \$line->{'points'} = [ [\$x * \$grid_X_size,0],[\$x * \$grid_X_size, -100] ];
400 \$line->{'forwardarrow'} = 0;
403 my(\$text) = new XFig ('text');
404 \$text->{'text'} = "\$x";
405 \$text->{'y'} = -200;
406 \$text->{'x'} = \$x * \$grid_X_size;
407 \$text->{'subtype'} = 1;
409 }
411 # Empty line so that the text of the time axis can be seen on the pdf
412 \$line = new XFig ('polyline');
413 \$line->{'depth'} = 999;
414 \$line->{'subtype'} = 1; # line
415 \$line->{'thickness'} = 0;
416 \$line->{'points'} = [ [0,0],[0, -400] ];
419 }
421 sub main {
423 my(\$cat_tree)=build_cat_tree("0",\$Cat);
426 # print Dumper(\$cat_tree);
427 # print Dumper(\$Cat);
428 my(\$cat_list)=[];
429 build_cat_list(\$cat_tree,\$cat_list);
430 shift @\$cat_list;
431 shift @\$cat_list;
432 # print "@\$cat_list \n";
433 set_cat_position(\$Cat,\$cat_list);
436 my(\$fig)=create_fig("toto.fig"); | 5,360 | 13,817 | {"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-2020-50 | latest | en | 0.111857 |
https://classroomsecrets.co.uk/mixed-age-year-5-and-6-decimals-and-percentages-step-5-resource-pack/ | 1,720,877,694,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00183.warc.gz | 141,332,800 | 78,860 | Mixed Age Year 5 and 6 Decimals and Percentages Step 5 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 5 and 6 Decimals and Percentages Step 5 Resource Pack
Mixed Age Year 5 and 6 Decimals and Percentages Step 5 Resource Pack
Step 5: Mixed Age Year 5 and 6 Decimals and Percentages Resource Pack
Mixed Age Year 5 and 6 Decimals and Percentages Step 5 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 5 Thousandths as Decimals for Spring Block 2.
What's included in the Pack?
This Mixed Age Year 5 and 6 Decimals and Percentages Step 5 pack includes:
• Mixed Age Year 5 and 6 Decimals and Percentages Step 5 Teaching PowerPoint with examples.
• Thousandths as Decimals Year 5 Varied Fluency with answers.
• Thousandths as Decimals Year 5 Reasoning and Problem Solving with answers.
National Curriculum Objectives
Mathematics Year 5: (5F6b) Recognise and use thousandths and relate them to tenths, hundredths and decimal equivalents
Differentiation for Year 5 Thousandths as Decimals:
Varied Fluency
Developing Questions to support using thousandths as decimals. Includes numbers smaller than 1 and some place value grids to support.
Expected Questions to support using thousandths as decimals. Includes some numbers greater than 1 and zero as a place holder.
Greater Depth Questions to support using thousandths as decimals. Includes numbers greater than 1 and improper fractions.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Reasoning)
Developing Explain which statement is correct. Includes numbers smaller than 1.
Expected Explain which statement is correct. Includes some numbers greater than 1 and zero as a place holder.
Greater Depth Explain which statement is correct. Includes numbers greater than 1, improper fractions and mixed numbers.
Questions 2, 5 and 8 (Problem Solving)
Developing Provide possibilities from the clues given. Includes numbers smaller than 1 and place value grids to support.
Expected Provide possibilities from the clues given. Includes some numbers greater than 1 and zero as a place holder.
Greater Depth Provide possibilities from the clues given. Includes numbers greater than 1, improper fractions and mixed numbers.
Questions 3, 6 and 9 (Reasoning)
Developing Explain if a fraction has been converted correctly. Includes numbers smaller than 1 and place value grids to support.
Expected Explain if a fraction has been converted correctly. Includes some numbers greater than 1 and zero as a place holder.
Greater Depth Explain if a fraction has been converted correctly. Includes numbers greater than 1, improper fractions and mixed numbers. | 588 | 2,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.879122 |
https://support.nag.com/numeric/nl/nagdoc_27.1/flhtml/f08/f08xaf.html | 1,709,256,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00839.warc.gz | 547,251,666 | 9,790 | # NAG FL Interfacef08xaf (dgges)
Note: this routine is deprecated. Replaced by f08xcf.
## ▸▿ Contents
Settings help
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FL Specification Language:
## 1Purpose
f08xaf computes the generalized eigenvalues, the generalized real Schur form $\left(S,T\right)$ and, optionally, the left and/or right generalized Schur vectors for a pair of $n×n$ real nonsymmetric matrices $\left(A,B\right)$. f08xaf is marked as deprecated by LAPACK; the replacement routine is f08xcf which makes better use of Level 3 BLAS.
## 2Specification
Fortran Interface
Subroutine f08xaf ( sort, n, a, lda, b, ldb, sdim, beta, vsl, vsr, work, info)
Integer, Intent (In) :: n, lda, ldb, ldvsl, ldvsr, lwork Integer, Intent (Out) :: sdim, info Real (Kind=nag_wp), Intent (Inout) :: a(lda,*), b(ldb,*), vsl(ldvsl,*), vsr(ldvsr,*) Real (Kind=nag_wp), Intent (Out) :: alphar(n), alphai(n), beta(n), work(max(1,lwork)) Logical, External :: selctg Logical, Intent (Inout) :: bwork(*) Character (1), Intent (In) :: jobvsl, jobvsr, sort
#include <nag.h>
void f08xaf_ (const char *jobvsl, const char *jobvsr, const char *sort, logical (NAG_CALL *selctg)(const double *ar, const double *ai, const double *b),const Integer *n, double a[], const Integer *lda, double b[], const Integer *ldb, Integer *sdim, double alphar[], double alphai[], double beta[], double vsl[], const Integer *ldvsl, double vsr[], const Integer *ldvsr, double work[], const Integer *lwork, logical bwork[], Integer *info, const Charlen length_jobvsl, const Charlen length_jobvsr, const Charlen length_sort)
The routine may be called by the names f08xaf, nagf_lapackeig_dgges or its LAPACK name dgges.
## 3Description
The generalized Schur factorization for a pair of real matrices $\left(A,B\right)$ is given by
$A = QSZT , B = QTZT ,$
where $Q$ and $Z$ are orthogonal, $T$ is upper triangular and $S$ is upper quasi-triangular with $1×1$ and $2×2$ diagonal blocks. The generalized eigenvalues, $\lambda$, of $\left(A,B\right)$ are computed from the diagonals of $S$ and $T$ and satisfy
$Az = λBz ,$
where $z$ is the corresponding generalized eigenvector. $\lambda$ is actually returned as the pair $\left(\alpha ,\beta \right)$ such that
$λ = α/β$
since $\beta$, or even both $\alpha$ and $\beta$ can be zero. The columns of $Q$ and $Z$ are the left and right generalized Schur vectors of $\left(A,B\right)$.
Optionally, f08xaf can order the generalized eigenvalues on the diagonals of $\left(S,T\right)$ so that selected eigenvalues are at the top left. The leading columns of $Q$ and $Z$ then form an orthonormal basis for the corresponding eigenspaces, the deflating subspaces.
f08xaf computes $T$ to have non-negative diagonal elements, and the $2×2$ blocks of $S$ correspond to complex conjugate pairs of generalized eigenvalues. The generalized Schur factorization, before reordering, is computed by the $QZ$ algorithm.
## 4References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia https://www.netlib.org/lapack/lug
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
## 5Arguments
1: $\mathbf{jobvsl}$Character(1) Input
On entry: if ${\mathbf{jobvsl}}=\text{'N'}$, do not compute the left Schur vectors.
If ${\mathbf{jobvsl}}=\text{'V'}$, compute the left Schur vectors.
Constraint: ${\mathbf{jobvsl}}=\text{'N'}$ or $\text{'V'}$.
2: $\mathbf{jobvsr}$Character(1) Input
On entry: if ${\mathbf{jobvsr}}=\text{'N'}$, do not compute the right Schur vectors.
If ${\mathbf{jobvsr}}=\text{'V'}$, compute the right Schur vectors.
Constraint: ${\mathbf{jobvsr}}=\text{'N'}$ or $\text{'V'}$.
3: $\mathbf{sort}$Character(1) Input
On entry: specifies whether or not to order the eigenvalues on the diagonal of the generalized Schur form.
${\mathbf{sort}}=\text{'N'}$
Eigenvalues are not ordered.
${\mathbf{sort}}=\text{'S'}$
Eigenvalues are ordered (see selctg).
Constraint: ${\mathbf{sort}}=\text{'N'}$ or $\text{'S'}$.
4: $\mathbf{selctg}$Logical Function, supplied by the user. External Procedure
If ${\mathbf{sort}}=\text{'S'}$, selctg is used to select generalized eigenvalues to be moved to the top left of the generalized Schur form.
If ${\mathbf{sort}}=\text{'N'}$, selctg is not referenced by f08xaf, and may be called with the dummy function f08xaz.
The specification of selctg is:
Fortran Interface
Function selctg ( ar, ai, b)
Logical :: selctg Real (Kind=nag_wp), Intent (In) :: ar, ai, b
Nag_Boolean selctg (const double *ar, const double *ai, const double *b)
1: $\mathbf{ar}$Real (Kind=nag_wp) Input
2: $\mathbf{ai}$Real (Kind=nag_wp) Input
3: $\mathbf{b}$Real (Kind=nag_wp) Input
On entry: an eigenvalue $\left({\mathbf{ar}}\left(j\right)+\sqrt{-1}×{\mathbf{ai}}\left(j\right)\right)/{\mathbf{b}}\left(j\right)$ is selected if ${\mathbf{selctg}}\left({\mathbf{ar}}\left(j\right),{\mathbf{ai}}\left(j\right),{\mathbf{b}}\left(j\right)\right)=\mathrm{.TRUE.}$. If either one of a complex conjugate pair is selected, then both complex generalized eigenvalues are selected.
Note that in the ill-conditioned case, a selected complex generalized eigenvalue may no longer satisfy ${\mathbf{selctg}}\left({\mathbf{ar}}\left(j\right),{\mathbf{ai}}\left(j\right),{\mathbf{b}}\left(j\right)\right)=\mathrm{.TRUE.}$ after ordering. ${\mathbf{info}}=\mathbf{n}+{\mathbf{2}}$ in this case.
selctg must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which f08xaf is called. Arguments denoted as Input must not be changed by this procedure.
5: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrices $A$ and $B$.
Constraint: ${\mathbf{n}}\ge 0$.
6: $\mathbf{a}\left({\mathbf{lda}},*\right)$Real (Kind=nag_wp) array Input/Output
Note: the second dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: the first of the pair of matrices, $A$.
On exit: a has been overwritten by its generalized Schur form $S$.
7: $\mathbf{lda}$Integer Input
On entry: the first dimension of the array a as declared in the (sub)program from which f08xaf is called.
Constraint: ${\mathbf{lda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
8: $\mathbf{b}\left({\mathbf{ldb}},*\right)$Real (Kind=nag_wp) array Input/Output
Note: the second dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: the second of the pair of matrices, $B$.
On exit: b has been overwritten by its generalized Schur form $T$.
9: $\mathbf{ldb}$Integer Input
On entry: the first dimension of the array b as declared in the (sub)program from which f08xaf is called.
Constraint: ${\mathbf{ldb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
10: $\mathbf{sdim}$Integer Output
On exit: if ${\mathbf{sort}}=\text{'N'}$, ${\mathbf{sdim}}=0$.
If ${\mathbf{sort}}=\text{'S'}$, ${\mathbf{sdim}}=\text{}$ number of eigenvalues (after sorting) for which selctg is .TRUE.. (Complex conjugate pairs for which selctg is .TRUE. for either eigenvalue count as $2$.)
11: $\mathbf{alphar}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Output
On exit: see the description of beta.
12: $\mathbf{alphai}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Output
On exit: see the description of beta.
13: $\mathbf{beta}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Output
On exit: $\left({\mathbf{alphar}}\left(\mathit{j}\right)+{\mathbf{alphai}}\left(\mathit{j}\right)×i\right)/{\mathbf{beta}}\left(\mathit{j}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{n}}$, will be the generalized eigenvalues. ${\mathbf{alphar}}\left(\mathit{j}\right)+{\mathbf{alphai}}\left(\mathit{j}\right)×i$, and ${\mathbf{beta}}\left(\mathit{j}\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{n}}$, are the diagonals of the complex Schur form $\left(S,T\right)$ that would result if the $2×2$ diagonal blocks of the real Schur form of $\left(A,B\right)$ were further reduced to triangular form using $2×2$ complex unitary transformations.
If ${\mathbf{alphai}}\left(j\right)$ is zero, then the $j$th eigenvalue is real; if positive, then the $j$th and $\left(j+1\right)$st eigenvalues are a complex conjugate pair, with ${\mathbf{alphai}}\left(j+1\right)$ negative.
Note: the quotients ${\mathbf{alphar}}\left(j\right)/{\mathbf{beta}}\left(j\right)$ and ${\mathbf{alphai}}\left(j\right)/{\mathbf{beta}}\left(j\right)$ may easily overflow or underflow, and ${\mathbf{beta}}\left(j\right)$ may even be zero. Thus, you should avoid naively computing the ratio $\alpha /\beta$. However, alphar and alphai will always be less than and usually comparable with ${‖A‖}_{2}$ in magnitude, and beta will always be less than and usually comparable with ${‖B‖}_{2}$.
14: $\mathbf{vsl}\left({\mathbf{ldvsl}},*\right)$Real (Kind=nag_wp) array Output
Note: the second dimension of the array vsl must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ if ${\mathbf{jobvsl}}=\text{'V'}$, and at least $1$ otherwise.
On exit: if ${\mathbf{jobvsl}}=\text{'V'}$, vsl will contain the left Schur vectors, $Q$.
If ${\mathbf{jobvsl}}=\text{'N'}$, vsl is not referenced.
15: $\mathbf{ldvsl}$Integer Input
On entry: the first dimension of the array vsl as declared in the (sub)program from which f08xaf is called.
Constraints:
• if ${\mathbf{jobvsl}}=\text{'V'}$, ${\mathbf{ldvsl}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
• otherwise ${\mathbf{ldvsl}}\ge 1$.
16: $\mathbf{vsr}\left({\mathbf{ldvsr}},*\right)$Real (Kind=nag_wp) array Output
Note: the second dimension of the array vsr must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ if ${\mathbf{jobvsr}}=\text{'V'}$, and at least $1$ otherwise.
On exit: if ${\mathbf{jobvsr}}=\text{'V'}$, vsr will contain the right Schur vectors, $Z$.
If ${\mathbf{jobvsr}}=\text{'N'}$, vsr is not referenced.
17: $\mathbf{ldvsr}$Integer Input
On entry: the first dimension of the array vsr as declared in the (sub)program from which f08xaf is called.
Constraints:
• if ${\mathbf{jobvsr}}=\text{'V'}$, ${\mathbf{ldvsr}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
• otherwise ${\mathbf{ldvsr}}\ge 1$.
18: $\mathbf{work}\left(\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{lwork}}\right)\right)$Real (Kind=nag_wp) array Workspace
On exit: if ${\mathbf{info}}={\mathbf{0}}$, ${\mathbf{work}}\left(1\right)$ contains the minimum value of lwork required for optimal performance.
19: $\mathbf{lwork}$Integer Input
On entry: the dimension of the array work as declared in the (sub)program from which f08xaf is called.
If ${\mathbf{lwork}}=-1$, a workspace query is assumed; the routine only calculates the optimal size of the work array, returns this value as the first entry of the work array, and no error message related to lwork is issued.
Suggested value: for optimal performance, lwork must generally be larger than the mimimum; add, say $\mathit{nb}×{\mathbf{n}}$, where $\mathit{nb}$ is the optimal block size.
Constraints:
• if ${\mathbf{n}}=0$, ${\mathbf{lwork}}\ge 1$;
• otherwise ${\mathbf{lwork}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(8×{\mathbf{n}},6×{\mathbf{n}}+16\right)$.
20: $\mathbf{bwork}\left(*\right)$Logical array Workspace
Note: the dimension of the array bwork must be at least $1$ if ${\mathbf{sort}}=\text{'N'}$, and at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ otherwise.
If ${\mathbf{sort}}=\text{'N'}$, bwork is not referenced.
21: $\mathbf{info}$Integer Output
On exit: ${\mathbf{info}}=0$ unless the routine detects an error (see Section 6).
## 6Error Indicators and Warnings
${\mathbf{info}}<0$
If ${\mathbf{info}}=-i$, argument $i$ had an illegal value. An explanatory message is output, and execution of the program is terminated.
${\mathbf{info}}=1,\dots ,{\mathbf{n}}$
The $QZ$ iteration failed. No eigenvectors have been calculated but ${\mathbf{alphar}}\left(j\right)$, ${\mathbf{alphai}}\left(j\right)$ and ${\mathbf{beta}}\left(j\right)$ should be correct from element $⟨\mathit{\text{value}}⟩$.
${\mathbf{info}}={\mathbf{n}}+1$
The $QZ$ iteration failed with an unexpected error, please contact NAG.
${\mathbf{info}}={\mathbf{n}}+2$
After reordering, roundoff changed values of some complex eigenvalues so that leading eigenvalues in the generalized Schur form no longer satisfy ${\mathbf{selctg}}=\mathrm{.TRUE.}$. This could also be caused by underflow due to scaling.
${\mathbf{info}}={\mathbf{n}}+3$
The eigenvalues could not be reordered because some eigenvalues were too close to separate (the problem is very ill-conditioned).
## 7Accuracy
The computed generalized Schur factorization satisfies
$A+E = QS ZT , B+F = QT ZT ,$
where
$‖(E,F)‖ F = O(ε) ‖(A,B)‖ F$
and $\epsilon$ is the machine precision. See Section 4.11 of Anderson et al. (1999) for further details.
## 8Parallelism and Performance
f08xaf is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
f08xaf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The total number of floating-point operations is proportional to ${n}^{3}$.
The complex analogue of this routine is f08xnf.
## 10Example
This example finds the generalized Schur factorization of the matrix pair $\left(A,B\right)$, where
$A = ( 3.9 12.5 -34.5 -0.5 4.3 21.5 -47.5 7.5 4.3 21.5 -43.5 3.5 4.4 26.0 -46.0 6.0 ) and B= ( 1.0 2.0 -3.0 1.0 1.0 3.0 -5.0 4.0 1.0 3.0 -4.0 3.0 1.0 3.0 -4.0 4.0 ) ,$
such that the real positive eigenvalues of $\left(A,B\right)$ correspond to the top left diagonal elements of the generalized Schur form, $\left(S,T\right)$.
Note that the block size (NB) of $64$ assumed in this example is not realistic for such a small problem, but should be suitable for large problems.
### 10.1Program Text
Program Text (f08xafe.f90)
### 10.2Program Data
Program Data (f08xafe.d)
### 10.3Program Results
Program Results (f08xafe.r) | 4,648 | 14,510 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 166, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.555953 |
http://ozabluda.blogspot.com/2017_04_01_archive.html | 1,601,438,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00648.warc.gz | 91,116,631 | 15,471 | Oleg Zabluda's blog
Saturday, April 01, 2017
99% matrix sparsity is generally not enough for things to be faster on GPU than simply doing dense multiplication.
99% matrix sparsity is generally not enough for things to be faster on GPU than simply doing dense multiplication. Easy-to-understand benchmark data is hard to get. For example, typically, NVIDIA will say that their cuSparse is 2x-5x faster than Intel’s MKL, but nowhere would you find how many FLOPS they get:
https://developer.nvidia.com/cusparse
Here is the secret knowledge for you. On slide 29, AMD brags how clSparse is faster than cuSparse
getting a whopping 40 GFLOP/s (geometric mean across benchmarks) on a GPU (FURY-X) with 8600 GFLOP/s (40/8600=4.5% utilization) vs 10 GFLOP/s on Titan-X Maxwell with 6144 GFLOP/s (10/6144=1.5% utilization)
https://developer.nvidia.com/cusparse
Labels:
Exploiting Linear Structure Within Convolutional Networks for Efficient Evaluation (2014) Emily Denton, Wojciech...
Exploiting Linear Structure Within Convolutional Networks for Efficient Evaluation (2014) Emily Denton, Wojciech Zaremba, Joan Bruna, Yann LeCun, Rob Fergus
"""
3.2.1 Matrix Decomposition
Matrices are 2-tensors which can be linearly compressed using the Singular Value Decomposition. If W ∈ R^m×k is a real matrix, the SVD is defined as W = USV, where U ∈ R^m×m, S ∈ R^m×k, V ∈ R^k×k. S is a diagonal matrix with the singular values on the diagonal, and U, V are orthogonal matrices. If the singular values of W decay rapidly, W can be well approximated by keeping only the t largest entries of S, resulting in the approximation W˜ = U˜S˜V˜, where U˜ ∈ R^m×t, S˜ ∈ R^t×t, V˜ ∈ R^t×k. Then, for I ∈ R^n×m, the approximation error || IW˜ − IW|| ≤ s_{t+1}|| I || , and thus is controlled by the decay along the diagonal of S.Now the computation IW˜ can be done in O(nmt + nt2 + ntk), which, for sufficiently small t is significantly smaller than O(nmk).
"""
https://arxiv.org/abs/1404.0736
https://arxiv.org/abs/1404.0736
Labels:
Obama Administration Rushed to Preserve Intelligence of Russian Election Hacking
Obama Administration Rushed to Preserve Intelligence of Russian Election Hacking
"""
In the Obama administration’s last days, some White House officials scrambled to spread information about Russian efforts to undermine the presidential election — and about possible contacts between associates of President-elect Donald J. Trump and Russians — across the government. [...] At the Obama White House, Mr. Trump’s statements stoked fears among some that intelligence could be covered up or destroyed — or its sources exposed — once power changed hands. What followed was a push to preserve the intelligence [...] It also reflected the suspicion among many in the Obama White House that the Trump campaign might have colluded with Russia on election email hacks — a suspicion that American officials say has not been confirmed. Former senior Obama administration officials said that none of the efforts were directed by Mr. Obama.
Sean Spicer, the Trump White House spokesman, said, “The only new piece of information that has come to light is that political appointees in the Obama administration have sought to create a false narrative to make an excuse for their own defeat in the election.”
[...]
At intelligence agencies, there was a push to process as much raw intelligence as possible into analyses, and to keep the reports at a relatively low classification level to ensure as wide a readership as possible across the government — and, in some cases, among European allies. This allowed the upload of as much intelligence as possible to Intellipedia, a secret wiki used by American analysts to share information.
[...]
The opposite happened with the most sensitive intelligence, including the names of sources and the identities of foreigners who were regularly monitored. Officials tightened the already small number of people who could access that information. They knew the information could not be kept from the new president or his top advisers, but wanted to narrow the number of people who might see the information
[...]
More than a half-dozen current and former officials described various aspects of the effort to preserve and distribute the intelligence, and some said they were speaking to draw attention to the material and ensure proper investigation by Congress. All spoke on the condition of anonymity because they were discussing classified information
[...]
Beyond leaving a trail for investigators, the Obama administration also wanted to help European allies combat a threat that had caught the United States off guard. American intelligence agencies made it clear in the declassified version of the intelligence assessment released in January that they believed Russia intended to use its attacks on the United States as a template for more meddling.
"""
https://www.nytimes.com/2017/03/01/us/politics/obama-trump-russia-election-hacking.html
https://www.nytimes.com/2017/03/01/us/politics/obama-trump-russia-election-hacking.html
Labels: | 1,152 | 5,057 | {"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-2020-40 | latest | en | 0.817772 |
http://mymathforum.com/math/344577-number-sequence-3me.html | 1,558,841,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00458.warc.gz | 141,371,570 | 9,888 | My Math Forum > Math Number Sequence 3mE
Math General Math Forum - For general math related discussion and news
July 19th, 2018, 05:31 PM #1 Member Joined: Nov 2015 From: New York State Posts: 41 Thanks: 0 Number Sequence 3mE I'm adding value of the digits in a number, and I treat 0 as 1. I have not graphed this. I don't know a lot about math, but I seem to have a way of systematically obtaining a sequence from "normal" numbers. It goes like this: -35 = 2 -34 = 1 -33 = 0 -32 = -1 -31 = -2 -30 = -2 -29 = 7 -28 = 6 -27 = 5 -26 = 4 -25 = 3 -24 = 2 -23 = 1 -22 = 0 -21 = -1 -20 = -1 -19 = 8 -18 = 7 -17 = 6 -16 = 5 -15 = 4 -14 = 3 -13 = 2 -12 = 1 -11 = 0 -10 = 0 -9 = -9 -8 = -8 -7 = -7 -6 = -6 -5 = -5 -4 = -4 -3 = -3 -2 = -2 -1 = -1 0 = 1 1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 6 7 = 7 8 = 8 9 = 9 10 = 2 11 = 2 12 = 3 13 = 4 14 = 5 15 = 6 16 = 7 17 = 8 18 = 9 19 = 2 20 = 3 21 = 3 22 = 4 23 = 5 24 = 6 25 = 7 26 = 8 27 = 9 28 = 2 29 = 2 30 = 4 31 = 4 32 = 5 33 = 6 34 = 7 35 = 8 Has anyone come up with this sequence before? If yes, is there a name of it, or has it been categorized?
July 19th, 2018, 07:19 PM #2 Math Team Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timey-wimey stuff. Throw us a bone here. How is -35 --> 2? Wouldn't the values of the digits 3 and 5 give 3 + 5 = 8? It looks like you are treating -35 to be (-3)(5) but that isn't how you defined the process. OEIS doesn't have it. -Dan Last edited by topsquark; July 19th, 2018 at 07:22 PM.
July 20th, 2018, 07:44 AM #3 Member Joined: Nov 2015 From: New York State Posts: 41 Thanks: 0 -35, I look at it this way -3 and 5. -3 + 5 = 2, or to make it easier 5 + (-3) = 2. Just adding the numbers that make up -35 and all other numbers that I listed.
July 20th, 2018, 08:48 AM #4 Member Joined: Nov 2015 From: New York State Posts: 41 Thanks: 0 Let me give some examples, so that you see better what I mean. Lets try number -10,000. I stated that I count 0 as 1, so 4x0 I count as 4x1 = 4 Then -1 + 4 = 3 Lets try another number, 2438. 2 + 4 + 3 + 8 = 17, and then adding 1 + 7 = 8 I guess, an application of this could be, taking a year 1984 for example. 1 + 9 + 8 + 4 = 22 so that is 2 + 2 = 4 Last edited by Alamar; July 20th, 2018 at 08:59 AM.
July 21st, 2018, 05:17 AM #5 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,417 Thanks: 1025 1234567890 = 1+2+3+4+5+6+7+8+9+0 = 45+1 = 46: 4+6 = 10 ????????
July 22nd, 2018, 05:31 PM #6
Member
Joined: Nov 2015
From: New York State
Posts: 41
Thanks: 0
Quote:
Originally Posted by Denis 1234567890 = 1+2+3+4+5+6+7+8+9+0 = 45+1 = 46: 4+6 = 10 ????????
Exactly,
1234567890 = 1+2+3+4+5+6+7+8+9+0 = 45+1 = 46: 4+6 = 10
I stated that I treat 0 as 1, so 4 + 6 = 10 = 1 + 1 = 2
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Contact - Home - Forums - Cryptocurrency Forum - Top | 1,337 | 3,224 | {"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-2019-22 | latest | en | 0.934737 |
https://earlymathcounts.org/playful-math/ | 1,726,051,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00129.warc.gz | 200,531,379 | 32,441 | Can we play with those counting things today?”
Two bubbly three-year-olds are standing in front of me with expectant smiles on their faces. I have absolutely no idea what they talking about—or what’s going on in those busy little brains, which I can almost hear whirring with activity.
“What counting things are you looking for?” I ask.
“Those pieces of wood with the holes,” responds Hissham. “But they aren’t holes, really. They’re like circles….” He trails off, hoping that I’ll know what he means.
“And we want the buttons!” Mariana chimes in.
Ah, they want the mancala boards!
We are quite intentional about setting up math environments that don’t scream MATH CENTER.
We start with authentic materials like the mancala boards—materials that encourage sensory play and exploration. Then we sneak in other materials that lead the children into deeper math investigations.
We don’t focus much on rote counting and number recognition. Instead, we foster the development of the children’s number sense because we want the children to move on from rote number identification to a more holistic understanding of number quantity
Children learn math at their own pace by using many different skill sets that don’t always develop at the same pace or in the same order. So we give them lots of options to let them learn in their own unique ways. By keeping it playful and hands-on, we set them up for math success
The new math that is being taught in schools uses ten frames a lot. So what is a ten frame?
A ten frame is a rectangle with 10 equal spaces. It has five spaces on the top and five on the bottom. You can use counters or math manipulatives to represent numbers less than or equal to 10 on the frame. Ten frames are available to download and print from various websites—or you can use a ten frame like the one pictured above.
Ten frames help children “see” numbers by providing a visual representation of a number and its value. Ten frames also enable students to physically place items, each within a single “frame,” to create a visual representation of numbers from 0-10.
Ten frames help kids develop number sense. For our youngest learners, we often provide five frames, which are arranged in a 1-by-5 design. Research suggests that five frames are a good place to start for very young children.
By working with numbers in ten frames, children can see the number of manipulatives that a number represents. Ten frames set the stage for work with larger numbers. I love ten frames because children can use them in ways that reflect their unique developmental level. Ten frames also enable children to play with patterns and group objects together. It may look like play, but ten frames help build a basic understanding of early math—and lay the foundation for later math learning.
I love wooden ten frames like the one in the photo above, but they are pricey. I have been searching for (and finding!) mancala boards at thrift shops and yard sales for years. We now have a collection of second-hand ten frames that we use for playful math.
You can buy a ten frame or download ten-frame printables from various early childhood websites, but there are endless ways to create your own ten frames. We’ve used egg cartons, ice cube trays, tape on a table (above) and chalk on the sidewalk.
Recently, I had the grand idea of taping the ends of our mancala boards to create ten frames (see photo above). I was feeling quite giddy about my clever adaptation, but then I decided that the tape would be distracting. So I went one step further and cut my mancala boards into ten frames (see photo below).
Then I realized that five frames would be even better for the two-year-olds in my program. With a five frame, younger students can use smaller set sizes that are within their developmental counting range.
Young children are always collecting something. By incorporating a five frame into their collecting activities, we help them develop a rudimentary understanding of number sense.
Pete the Cat and his oh-so-groovy buttons have been at the top of our reading and music rotation list lately, so adding button manipulatives seemed like a logical next step.
But beware: Any small manipulative is a choking hazard, so please choose your manipulatives carefully!
By now, I was a bit “saw happy,” so I also created a three-frame to use with subitizing and a two-frame, as well as a one frame for one-to-one correlation. Remember, we want these activities to be developmentally appropriate for the youngest children in our programs.
Remember, our goal is not to work on counting, but to set the stage for later learning with these early math tools.
Older preschoolers can move up to the ten frames when they are ready. I generally suggest that they start in the top left corner and fill the entire top row before moving on to the second row (just like reading).
Of course, this is way beyond the developmental level of our youngest learners. All that we want to see with these children is the ability to place one button in one circle. As their brains continue to develop and their understanding of early math concepts grows, they can graduate to ten frames.
While the children are playing with their frames, you can help foster the development of counting skills by putting a counter in each square of the frame and saying: “One, Two, Three, Four, Five—you have five buttons.
Then tell the children to count the buttons. After they have counted to five, ask: “How many buttons are there altogether?”
If they answer five without recounting, they may be exhibiting an understanding of cardinality (the knowledge that the last word in a counting sequence represents the total number of objects in a set).
Cardinality is a school standard for kindergartners, but your early learners will just think of it as a fun game to play with buttons!
You can easily adapt these Early Math Counts lesson plans for use with ten frames in your classroom:
Don’t forget to click on the CONNECT WITH FAMILIES button on the left side of each lesson page to download and customize a Family Letter to send home with your students after each lesson.
## 22 Replies to “Playful Math”
1. Michele says:
Nice…very creative
2. playful math i like this one the many different ways can explore it in a playful way
3. Brennan says:
Love the lesson using Mancala Boards. Promotes Number Sense, Number Quantity and is a Hands on Activity. Flexibility to use different objects to place on boards and count. Examples: medium to large buttons, shells, medium rocks and as an edible treat, goldfish crackers (which could them be counted if more than one crackers is placed in the slot. Promotes fine motor skills, identifying colors of buttons & counting items.
4. Kristin Haas says:
I absolutely love this idea!!
Really cool
6. Christine says:
Like that many of the activities can be made with items I already have on hand.
7. Tara says:
I really liked these ideas and have used ten frames before. I am not sure why I never thought to adapt it to smaller frames.
8. Lindsey says:
Really interesting and love the idea!
9. Lindsey says:
Really interesting!
10. Jeanette Bye says:
Great ideas
11. Danielle says:
I always like a new way to cover content.
12. D. Brennan says:
Creative, size of mancala boards can be made smaller for younger children. Wide variety of counting materials/ ideas can be used.. Or you can create your own boards using egg cartons, ice cube trays ,tape on a table or as an outdoor activity chalk squares or circles on the sidewalk…..
13. Miranda Varney says:
Diann – I’ve never thought of using mancala boards as 10 frames; that is such a cool idea! It allows for students to be hands-on with materials as they count and practice counting out numbers of objects. I will definitely be trying this with my Pre-K students after Thanksgiving!
-Miranda
14. Angela Macias says:
I also enjoy these activities. They help evaluate the child ‘s understanding and easy to challenge them as they are ready.
15. Danielle Reilly says:
I like activities like these because they can be easily replicated at home for carry over and parent involvement.
16. Jo says:
these are really good ideas
17. Tamara says:
Very creative & challenging
This is a great one to also involve patterning! Once an older child has mastered one to one, you can also involve actual mancala rule which would extend patterning further.
19. velma Haselett says:
I never thought of using the mancala board for counting. and to cut them in different ones you could put a 2 plus a 3 and have them count out 5. i love this idea.
20. Carol says:
Great ideas | 1,830 | 8,661 | {"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} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.93944 |
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Mt. Hood Furniture Company Accuracy TestACC 491: Contemporary Auditing IMt. Hood Furniture Company Accuracy TestThe objectives of this audit are to determine the accuracy of the inventory as reported by Mt. Hood Furniture Company. The goals are to determine that the inventory is accurately transcribed to the final accumulation and valuation of the inventory and that the inventory is correctly priced and accumulated. Also included are tests to determine that the proper pricing of raw materials, work in process, and finished goods are recorded.The formula for finding the sample size is N = BV x RF / TM - (AM x EF). BV stands for the book value of the total population, which is \$6,753,764. RF stands for the reliability factor, which is 1. TM stands for the tolerable misstatement, which is \$325,000. Tolerable misstatement represents the maximum misstatement that could occur before the population would be considered materially misstated. The lower the required tolerable misstatement relative to the number and dollar size of transactions the lower the needed tolerable deviation rate. This rate is inversely related to sample size. AM is the anticipated misstatement, which is \$100,000 and EF is the expansion factor for anticipated misstatement, which is 1.15. So N= \$6,753,764 x 1.00 / \$325,000 - (\$100,000 x 1.15) = 32.16. Assuming that an auditor accepts exposure to a risk of incorrect acceptance of 37% and form an opinion of the financial data based
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https://www.impetus-afea.com/support/manual/?command=LOAD_THERMAL_BODY | 1,620,373,505,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00027.warc.gz | 874,808,802 | 18,755 | Command list
• Input handling
• Solution control and techniques
• Output
• Mesh commands
• Nodes and connectivity
• Material properties
• Initial conditions
• Boundary conditions
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"Optional title"
coid, entype, enid, cid${}_1$, $sf$, $t_{beg}$, $t_{end}$, cid${}_2$
Parameter definition
VariableDescription
coid Command ID
entype Entity type
options: P, PS, ALL
enid Entity identification number
cid${}_1$ ID of a CURVE or FUNCTION $f_1$ defining the thermal body load versus time (power/unit volume)
$sf$ Curve scale factor
default: 1
$t_{beg}$ Start time
$t_{end}$ End time
default: 1.0e20
cid${}_2$ ID of an optional CURVE or FUNCTION $f_2$ defining the total thermal body load versus time (power)
Description
This command is used to apply a thermal body load (power per unit volume).
$f_2$ is an optional CURVE or FUNCTION describing the exact total power of the heat source. It is needed in situations where the heat source described by $f_1$ (power/unit volume) is partially outside the body where the energy is to be deposited. The magnitude of $f_1$ is then scaled with a factor $\alpha$ in order to match the specified total power in $f_2$.
$\displaystyle{ \alpha \int_V f_1(\mathbf{x}, t) \mathrm{d}V = f_2(t)}$
$V$ is the body volume, $\mathbf{x}$ is a spatial coordinate and $t$ is the time. Hence, the heat power per unit volume $p$ deposited at a location $\mathbf{x}$ becomes:
$p = \alpha \cdot f_1(\mathbf{x}, t)$ | 438 | 1,580 | {"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-2021-21 | latest | en | 0.562019 |
https://community.qlik.com/t5/New-to-Qlik-Sense/Color-Expression-Qlik-Sense/m-p/902393 | 1,542,888,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00119.warc.gz | 598,368,622 | 45,258 | # New to Qlik Sense
Discussion board where members can get started with Qlik Sense.
Not applicable
## Color Expression Qlik Sense
Hi Everyone,
I would require some help in color coding for pivot tables.
Attached is a sample on the pivot table I am working on.
The expression that I have used is as follows: IF(PercentageChange < 0, Red(), IF( PercentageChange > 0, Green(), Blue()))
However, from the table, Q2 which is at -2% is not color coded red as per what my expression should have done.
Please kindly advise. Thanks in advance.
1 Solution
Accepted Solutions
MVP
## Re: Color Expression Qlik Sense
Looking the values in the table, I think that percentage in the 'Q' row should be avoided, each quarter has a row with the percentage change versus the previous quarter and the same quarter of the last year so... wich % change is telling that quarter row?
To avoid showing any value in that row you can use Dimensionality(), ie, if your actual expression is:
PercentageChange
You can change this to:
If(Dimensionality()>0, PercentageChange)
And the background color to:
If Dimensionality()>0,
IF(Round(PercentageChange, 0.01) < 0, Red(), IF(Round(PercentageChange, 0.01) > 0, Green(), Blue()))
)
8 Replies
MVP
## Re: Color Expression Qlik Sense
Hi Alvin, all the Q's have more than one different value, if you use 'PercentageChange' and there are more than one value, this will return null(), so it's not <0 or >0 and the expression returns Blue().
Enclose Percentagechange in an aggregation funtion to tell QV how to calculate the expression with different values, ie:
IF(Avg(PercentageChange)< 0, Red(), IF(Avg(PercentageChange)> 0, Green(), Blue()))
Probably you don't want a simple Avg, it's just an example
or try using the column name:
IF([Average Percentage Change] < 0, Red(), IF( [Average Percentage Change] > 0, Green(), Blue()))
Not applicable
## Re: Color Expression Qlik Sense
Hi Ruben,
Thanks for your prompt reply.
I have tried your recommendation of using IF(Avg(PercentageChange)< 0, Red(), IF(Avg(PercentageChange)> 0, Green(), Blue())). It works fine except that the first label which is 0% is green instead of blue like the rest
As from the image above, the Q1 which has 0% change is labelled green while the other 0% are in blue.
I tried the second expression that you recommended, and qlik returned me an error(Bad field name)
Please kindly advise. Thanks
MVP
## Re: Color Expression Qlik Sense
Hi Alvin, seems that it's doing: (9-8+0+0)/4 wich retuns 0,25, represented as 0, but >0. You can use Round() to avoid this:
IF(Round(Avg(PercentageChange))< 0, Red(), IF(Round(Avg(PercentageChange))> 0, Green(), Blue()))
I must advise than using avg with percentages is not an optimal solution. This only returns an exact result when all the rows have the same denominator, wich doesn't seem to happen here.
Highlighted
Not applicable
## Re: Color Expression Qlik Sense
Hi Ruben,
Putting that expression resulted in it changing all to blue. I reckon that it may be due to the explanation that you provided earlier. The answer before rounding actually provided a nearer answer to what I require.
Thanks for your advise regarding using averages. I am however unsure of how i can show it since the other expressions are sum, min, max. I am still relatively new to using qlik sense.
Anyway, thanks for your help
MVP
## Re: Color Expression Qlik Sense
Hi Alvin, as those are percentages, 1% is really 0,01, wich rounded will return 0, try with:
IF(Round(Avg(PercentageChange), 0.01)< 0, Red(), IF(Round(Avg(PercentageChange), 0.01)> 0, Green(), Blue()))
Can you post the expression that calculates PercentageChange to check how to avoid the Avg()?
Not applicable
## Re: Color Expression Qlik Sense
Hi Ruben,
Thanks! that works perfectly.
At the moment I actually use excel functions to calculate percentage changes before dropping the excel sheet to qlik sense. The Avg(PercentageChange) is actually to calculate the quarter average i think.
As i am not as familiar with the functions that I can use in qlik sense except for min, max, sum, avg, I decided to do the calculations in excel even though it is not as advisable
MVP
## Re: Color Expression Qlik Sense
Looking the values in the table, I think that percentage in the 'Q' row should be avoided, each quarter has a row with the percentage change versus the previous quarter and the same quarter of the last year so... wich % change is telling that quarter row?
To avoid showing any value in that row you can use Dimensionality(), ie, if your actual expression is:
PercentageChange
You can change this to:
If(Dimensionality()>0, PercentageChange)
And the background color to:
If Dimensionality()>0,
IF(Round(PercentageChange, 0.01) < 0, Red(), IF(Round(PercentageChange, 0.01) > 0, Green(), Blue()))
)
Not applicable
## Re: Color Expression Qlik Sense
Perfect. Thats exactly what I needed. Thanks a lot for your help. Grateful | 1,207 | 4,970 | {"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-2018-47 | latest | en | 0.825621 |
http://www.sparknotes.com/physics/linearmomentum/collisions/problems/ | 1,558,318,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255536.6/warc/CC-MAIN-20190520021654-20190520043654-00199.warc.gz | 341,095,325 | 9,516 | # Linear Momentum: Collisions
## Problems
Problem :
Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. What is the final velocity of each ball, in terms of m and v?
Though we could go through the formal application of the equations of linear momentum, it is easier to think about this problem conceptually. Since the balls of equal mass are moving at equal and opposite speeds, the total linear momentum of the system is zero. For linear momentum to be conserved after the collision, both balls must rebound with the same velocity. If one ball had more speed than the other, there would be a net linear momentum and our conservation principle would be invalid. Having established that both balls rebound with the same speed, we must find what that speed is. Since the collision is elastic, kinetic energy must be conserved. If the final velocity of each ball were more, or less, than its initial velocity, kinetic energy would not be conserved. Thus we can state that the final velocity of each ball is equal in magnitude and opposite in direction to their respective initial velocities.
Problem :
Two balls, each with mass 2 kg, and velocities of 2 m/s and 3 m/s collide head on. Their final velocities are 2 m/s and 1 m/s, respectively. Is this collision elastic or inelastic?
To check for elasticity, we need to calculate the kinetic energy both before and after the collision. Before the collision, the kinetic energy is (2)(2)2 + (2)(3)2 = 13. After, the kinetic energy is (2)(2)2 + (2)(1)2 = 5. Since the kinetic energies are not equal, the collision is inelastic.
Problem :
Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity after the collision? If so, find the conditions under which this can occur.
First of all, the collision must be inelastic, as the final kinetic energy must be zero, clearly less than the initial kinetic energy. Secondly, we can state that the collision is completely inelastic, as both objects with zero velocity must remain at the site of the collision, i.e., they must stick together. The final principle we must check is that momentum is conserved. Clearly the final momentum of the system must be zero, as neither ball is moving. Thus the same value must be true before the collision. For this to happen, both masses must have equal and opposite momentum, or m1v1 = m2v2. Thus, in a completely inelastic collision in which m1v1 = m2v2, both masses will be stationary after the collision.
Problem :
A car of 500 kg, traveling at 30 m/s rear ends another car of 600 kg, traveling at 20 m/s. in the same direction The collision is great enough that the two cars stick together after they collide. How fast will both cars be going after the collision?
This is an example of a completely inelastic collision. Since the two cars stick together, they must move with a common velocity after the collision. Thus simply using the conservation of momentum is enough to solve for our one unknown variable, the velocity of the two cars after the collision. Relating the initial and final moments:
po = pf m1v1 + m2v2 = Mvf (500)(30) + (600)(20) = (1100)vf vf = 24.5m/s
Thus both cars will travel at 24.5 m/s, in the same direction as their initial travel.
Problem :
One pool ball traveling with a velocity of 5 m/s hits another ball of the same mass, which is stationary. The collision is head on and elastic. Find the final velocities of both balls.
Here we use our two conservation laws to find both final velocities. Let's call the pool ball that is initially moving ball 1, and the stationary one ball 2. Relating the kinetic energies before and after the collision,
mv1o2 + mv2o2 = mv1f2 + mv2f2 m = mv1f2 + mv2f2 Canceling the fractions and masses, 25 = v1f2 + v2f2
We also know that momentum must be conserved. The initial momentum is provided entirely by ball 1, and has a magnitude of 5m. The final momentum has contributions from both balls. Relating the two,
5m = mv1f + mv2f
Implying that
m1f + m2f = 5
Notice the similarity of the two equations we have. Though our kinetic energy equation includes the velocities squared, both equations include the sum of the velocities being equal to a constant. The systematic approach to this problem is to substitute for m1f into our first equation using our second equation. However we can use a shortcut. Let's see what happens when we square our second equation:
(m1f+m2f)2 = 25 m1f2 + m2f2 +2m1fm2f = 25
But we know from our kinetic energy equation that 25 = v1f2 + v2f2 . Substituting this in we find that
2m1fm2f = 0
Thus we know that one of the final velocities must be zero. If the final velocity of ball 2 were zero, then the collision never would have taken place. Thus we can infer that v1f = 0 and, consequently, v2f = 5. This problem states a general principle of collisions: when two bodies of the same mass collide head on in an elastic collision, they exchange velocities. | 1,198 | 5,008 | {"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} | 4.28125 | 4 | CC-MAIN-2019-22 | latest | en | 0.936764 |
https://www.omnimaga.org/ti-basic-language/pure-ti-basic-grayscale-ascii-game/?prev_next=prev | 1,607,025,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00690.warc.gz | 796,155,720 | 13,334 | ### Author Topic: For( Loop Issues (Read 3096 times)
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#### meishe91
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##### For( Loop Issues
« on: April 07, 2010, 07:46:27 pm »
Ok, so I'm creating a program and everything is working fine but then I get to two consecutive For( and for some reason it is either executing them incredibly fast with false information or is skipping right over them. It is something inside that code because when I isolated that code into another program to test it it does the same thing. Hopefully someone can help me here.
Here is the code in question:
Code: [Select]
For(E,-B+1,-B+D-1For(F,A+1,A+C-1Str1+sub("10",1+pxl-Test(E,F),1→Str1EndEnd
For testing I was just using:
A=44
B=-27
C=7
D=-7
_→Str1
Also, I just ran a little test thing and it appears that it is the second For( loop that isn't going. Does the variable "F" have issues with the For( command?
Thanks, everyone.
Spoiler For Spoiler:
For the 51st time, that is not my card! (Magic Joke)
#### Will_W
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##### Re: For( Loop Issues
« Reply #1 on: April 07, 2010, 07:49:00 pm »
For(E,-(-27)+1,-(-27)+(-7)-1
For(E,28,19
That's the problem, the start value is higher than the end.
ex de,hl
ld (hl),e
inc hl
ld (hl),d
ld a,(hl)
#### meishe91
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##### Re: For( Loop Issues
« Reply #2 on: April 07, 2010, 08:03:18 pm »
Oh, wow. I knew it'd be something dumb like that Thanks Will_W
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#### DJ Omnimaga
• Former TI programmer
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##### Re: For( Loop Issues
« Reply #3 on: April 07, 2010, 08:05:42 pm »
That happened to me often in the past ^^ sometimes took ages for me to figure out.
Fortunately, TI-BASIC and Axe parser are nice enough to not treat such For loops as endless loops, though, like other languages do, according to Will_W
In case you are wondering where I went, I left Omni back in 2015 to form CodeWalrus due to various reasons explained back then, but I stopped calc dev in 2016 and am now mostly active on the CW Discord server at https://discord.gg/cuZcfcF
#### Will_W
• LV3 Member (Next: 100)
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##### Re: For( Loop Issues
« Reply #4 on: April 07, 2010, 08:10:28 pm »
in java, its for(variable;condition;what to do to make the variable approach the condition)
usually something along the lines of for(int i=0;i>5;i++)
a > condition will behave like BASIC and Axe, but if you did i=5 and started with i=6 or something, it would behave strangely, either giving a compile time error, or incrementing until i overflows and becomes 5.
« Last Edit: April 07, 2010, 08:12:10 pm by Will_W »
ex de,hl
ld (hl),e
inc hl
ld (hl),d
ld a,(hl)
#### DJ Omnimaga
• Former TI programmer
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##### Re: For( Loop Issues
« Reply #5 on: April 08, 2010, 10:00:30 pm »
wow Java syntax seems weird
It reminds me a bit Casio calcs, except the i++ part of your code that I don't understand.
I think it was something like For 1 to 10->X but I could be wrong. There was another argument for increment rate.
Another thing with TI-83+ for( loops is to always close the parhentesis when your For loop contains only a If:Then:code:End. If the condition is false or something, the For loop takes considerably longer to process and sometimes the rest of the program will run slower, as if a massive memory leak of OVER 9000 KB of RAM occured.
(Texas Instruments should really spend their efforts into fixing their OSes than into DMCA notices and C&D letters.)
« Last Edit: April 08, 2010, 10:01:27 pm by DJ Omnimaga »
In case you are wondering where I went, I left Omni back in 2015 to form CodeWalrus due to various reasons explained back then, but I stopped calc dev in 2016 and am now mostly active on the CW Discord server at https://discord.gg/cuZcfcF
#### meishe91
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##### Re: For( Loop Issues
« Reply #6 on: April 08, 2010, 10:05:11 pm »
Ztrumpet told me it was just to close them if only an If statement occurs after a For( loop and that If/Then were fine without the closing ones, I believe. I would have to check my emails with him to be sure that's what he said, but ya.
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#### calc84maniac
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##### Re: For( Loop Issues
« Reply #7 on: April 08, 2010, 10:08:12 pm »
in java, its for(variable;condition;what to do to make the variable approach the condition)
usually something along the lines of for(int i=0;i>5;i++)
a > condition will behave like BASIC and Axe, but if you did i=5 and started with i=6 or something, it would behave strangely, either giving a compile time error, or incrementing until i overflows and becomes 5.
I think you mean a < condition, because it acts like a "while" and not "repeat"
"Most people ask, 'What does a thing do?' Hackers ask, 'What can I make it do?'" - Pablos Holman
#### DJ Omnimaga
• Former TI programmer
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##### Re: For( Loop Issues
« Reply #8 on: April 08, 2010, 10:55:04 pm »
Ztrumpet told me it was just to close them if only an If statement occurs after a For( loop and that If/Then were fine without the closing ones, I believe. I would have to check my emails with him to be sure that's what he said, but ya.
It's actually mentionned on TI-BD site:
http://tibasicdev.wikidot.com/for
In the optimizations part.
In case you are wondering where I went, I left Omni back in 2015 to form CodeWalrus due to various reasons explained back then, but I stopped calc dev in 2016 and am now mostly active on the CW Discord server at https://discord.gg/cuZcfcF
#### meishe91
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##### Re: For( Loop Issues
« Reply #9 on: April 08, 2010, 11:04:57 pm »
Ya, TIBD says its only when an If command (when the condition is false) stands alone after a For(. Unless I'm reading it wrong.
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For the 51st time, that is not my card! (Magic Joke)
#### DJ Omnimaga
• Former TI programmer
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##### Re: For( Loop Issues
« Reply #10 on: April 08, 2010, 11:31:40 pm »
yep this is what it says, and my mistake, no Then instructions
In case you are wondering where I went, I left Omni back in 2015 to form CodeWalrus due to various reasons explained back then, but I stopped calc dev in 2016 and am now mostly active on the CW Discord server at https://discord.gg/cuZcfcF
#### meishe91
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##### Re: For( Loop Issues
« Reply #11 on: April 09, 2010, 12:01:55 am »
Oh ok, no worries. Thanks for pointing it out though, I might not have remembered otherwise
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#### Deep Toaster
• So much to do, so much time, so little motivation
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##### Re: For( Loop Issues
« Reply #12 on: June 01, 2010, 08:48:46 pm »
Ya, TIBD says its only when an If command (when the condition is false) stands alone after a For(. Unless I'm reading it wrong.
Strange, I just tested it out, and the time is almost equal in all four conditions: with the If statement true or false, and with or without the closing parenthesis on the For(.
#### DJ Omnimaga
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##### Re: For( Loop Issues
« Reply #13 on: June 01, 2010, 09:25:58 pm »
Try:
0->A
For(Z,0,100
If A=1:0
End
then compare with
0->A
For(Z,0,100)
If A=1:0
End
In case you are wondering where I went, I left Omni back in 2015 to form CodeWalrus due to various reasons explained back then, but I stopped calc dev in 2016 and am now mostly active on the CW Discord server at https://discord.gg/cuZcfcF | 2,589 | 8,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-50 | latest | en | 0.863009 |
http://algebrasfriend.blogspot.com/2012/08/questions-and-more-questions.html | 1,720,925,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00023.warc.gz | 2,013,934 | 15,698 | ## Wednesday, August 29, 2012
### Questions and more questions!
Questions and questioning are at the heart of rich math dialogue. We are in our first week of school and I realize that I need to work on my questioning skills to tap into the strengths of my diverse students.
I have found a few resources to use to help me in preparing worthwhile questions.
1. At the Ontario Ministry of Education, there is an excellent article entitled, "Asking Effective Questions." In the article there are eight tips for asking effective questions. But then there are many, many questions or question stems providing an amazing framework for asking good questions.
2. I also ran across this book by Schuster and Anderson. The book is set up by strands in math. Even though I teach algebra, I find this book useful in developing questions to help my ninth graders make sense of the prerequisite math that they still struggle with. The questions are open-ended and challenging. For example: "Carla and Fiona are having a mathematical debate about the equation y = 1/2 x + 3. Carla thinks that every time y changes by one, x changes by 2. Fiona thinks that every time y changes by one-half, x changes by one. What do you think?"
3. One other book I ran across provides ideas for differentiating. It is also organized in strands. For each topic, the authors, Small and Lin, provide ideas for open questions and 2 parallel questions at different levels. One example that caught my eye is: "Another function is a lot like the given function. What might it be? Option 1: y = 3x^2 + 4 Option 2: y = 3^x. Follow up questions are provided: What happens to your function when x gets big? When x is quite small? What kinds of things might you change but still have a similar function? What would you choose not to change?
I'm on the prowl for other great resources for developing strong questioning skills. Send me a note if you know of others! | 482 | 1,954 | {"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} | 2.984375 | 3 | CC-MAIN-2024-30 | latest | en | 0.925872 |
https://apluspal.com/tag/binomial-distribution/ | 1,627,302,338,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00222.warc.gz | 116,082,863 | 14,950 | A+ » VCE » Maths Methods U3 & 4 Master Notes » A4 Probability and Statistics » Binomial Distribution
# Binomial Distribution
## 3.7 Normal Approximation to the Binomial Distribution
#### Normal Approximation to the Binomial Distribution
• Binomial distributions depend on n and p. We will observe that as n increases, if we plot the probability distribution of the binomial distribution, a ‘bell curve’ which is a symbolic shape of normal distribution, emerges. Below is the plot of the binomal distribution for n=10,\ 20,\ 50,\ 100 and p=0.2,\ 0.5,\ 0.8.
• As you can see, as n increases, a smoother bell curve is seen. As for p, it only decides if the curve is slanted (or specifically ‘skewed’) to the left or right.
Read More »3.7 Normal Approximation to the Binomial Distribution
## 2.5 Properties of the Binomial Distribution
#### Graph of the Binomial Distribution
• Recall that a binomial distribution has the format
\operatorname{Pr}(X=x)=\left(\begin{array}{l}n \\x\end{array}\right) p^{x}(1-p)^{n-x} \quad \ \ \ \ \ \ \ \ \ x=0,1,2, \ldots, n
where X \sim \operatorname{Bi}(n, p).
• We can plot the graph of (number of successes) against \operatorname{Pr}(X=x) \text { or } p(x) (corresponding probabilities) to see the graphical representation of the binomial distribution.
• The graph of the binomial distribution will generally look like a small mountain. In other words, you would expect probability of number of successes to be higher when x is roughly half the number of trials.
• However, it is obvious that if p is higher, then we would expect more successes. Therefore, the value of p will cause the graph to skew sideways. Refer to the example below to see what ‘positively/negatively skewed’ means graphically. To give a heads up, it means the graph would be slanted towards the left/right.
Read More »2.5 Properties of the Binomial Distribution
## 2.4 Bernoulli Sequence and Binomial Distribution
#### The Bernoulli Sequence
• There are many types of random variables that contributes to a discrete probability distribution. For instance, the number of ‘sixes’ obtained from 10 throws of a fair dice, and the amount of lottery winning are examples of a discrete random variables.
• For a trial that has only either success or fail with a fixed probability, we call this as a Bernoulli trial. That is, the probability distribution table is given by
where p is a constant.
• Following this logic, we can also say that Bernoulli random variables are used to model two-outcome situations with fixed probability for each outcome.
• In Bernoulli trial, we define success as x=1, and fail as x=0. Therefore, we can calculate the mean of a Bernoulli trial random variable as p, and its variance as p(1-p).
• A particular expansion of this is the Bernoulli sequence. It describes a sequence of repeated trials with the following properties:
Read More »2.4 Bernoulli Sequence and Binomial Distribution
## The Binomial Distribution [Video Tutorial]
This tutorial covers material encountered in chapter 14 of the VCE Mathematical Methods Textbook, namely: Bernoulli distribution Binomial distribution Mean and variance of a binomial… Read More »The Binomial Distribution [Video Tutorial] | 770 | 3,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-31 | latest | en | 0.843099 |
https://www.physicsforums.com/threads/negative-potential-energy-and-negative-mass.917428/ | 1,508,699,032,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00550.warc.gz | 966,938,068 | 21,236 | # B Negative potential energy and negative mass
1. Jun 13, 2017
### Vicara
My question is:
If gravitational potential energy is normally negative, and E=m•c^2, doesen't that means that negative mass could exist?
(I don't know much about general relativity so please explain as simple as posible)
2. Jun 13, 2017
### CWatters
Why do you say gravitational potential energy is normally negative?
3. Jun 13, 2017
### Vicara
Because Ep = -G•M•m/r
4. Jun 13, 2017
### vanhees71
Usually you choose the potential being negative, i.e., for a spherically symmetric mass $M$ in the origin of a coordinate system the potential of the force on a test mass $m$ is
$$V(\vec{r})=-\frac{G m M}{r},$$
where $G$ is Newton's gravitational constant. Here masses are always positive.
I've, however no clue, what this might have to do with the rest mass-energy equivalence of special relativity. For a relativistic theory of gravitation you neen general relativity.
5. Jun 13, 2017
### Staff: Mentor
No. What it does mean is that a system consisting of two widely separated objects will have a mass very slightly greater than if the same two objects were close together. The difference in mass (multiplied by $c^2$ according to $E=mc^2$) will be the energy that was needed to move them apart, which is the difference in potential energy between the two configurations.
It's important that we work with the difference between two configurations because potential energy itself is defined that way. It makes no sense to say that the potential is positive or negative until you've specified what you're calling zero. A very common convention is that we call the potential energy zero when the two objects are infinitely distant, and with this convention the potential energy will always be negative.
6. Jun 13, 2017
### Vicara
I think that I'm just mixing concepts that I don't really understand, its just that I learned this formula (before i used Ep = m•g•h ) and the fact that the other formula gives a negative result for energy confused me and I thought that negative mas could be possible by the mass-energy equivalence
7. Jun 13, 2017
### Vicara
Hmmm, that makes sense...
And in the case of maseless particles like photons?
8. Jun 13, 2017
### Staff: Mentor
Sorry, you'll have to be more specific about what you're asking.
9. Jun 13, 2017
### Vicara
I mean, if a photon gets closer to a mass, the potential energy will decrease, hence, his mass will decrease too. But it don't have mass, so what will hapend?
10. Jun 13, 2017
### PAllen
E = mc2 is a special case (for a massive body at rest) of the general relation:
E2 = p2c2 + m2c4
For light, m is zero, and energy is related solely to momentum
Also, note that as something moves from higher to lower potential with no other forces acting, its kinetic energy increases. Since the energy of light is pure kinetic energy, this means the locally measured energy of light increases as it approaches a massive body, i.e. it blue shifts.
Last edited: Jun 13, 2017
11. Jun 13, 2017
### DrStupid
I do not see how this answers the question. Why is the energy that is needed to move the objects apart always below the total energy of the separated object? As the classical gravitational potential would allow a negative total energy for sufficiently small distances, the question can be answered in the scope of general relativity only.
12. Jun 13, 2017
### Staff: Mentor
No matter how far apart I move the objects it will always require the addition of more energy to move them farther apart; so the potential energy at infinity is always greater than the energy at any finite distance, no matter how large. Thus, if you're going to make the (arbitrary, but very convenient) choice to take the energy at infinity to be the zero point, then the potential energy at any finite separation distance must be less than zero. However, we could have taken the zero point of potential to be the minimum when the two objects have approached as closely as possible and compressed, and reshaped themselves under the gravitational forces as they merge. Do this, and the energy will be finite and positive everywhere including in the limit as $r\rightarrow{\infty}$ and whether or not we further offset the zero point by an additional constant $m_1c^2+m_2c^2$ contribution from counting the rest masses.
(And do not fall into the trap of thinking that because $F=-Gm_1m_2/r^2$ blows up as $r\rightarrow{0}$ that the minimum of energy is not a finite number that we can take as the zero point. The infinity here is just telling us that the ideal point particle approximation upon which that formula is based breaks down when $r$ becomes small enough).
This is a digression from the original poster's question, so followup should happen in a new thread in the Classical Physics subforum.
13. Jun 13, 2017
### DrStupid
The question remains: Why is there a closest possible approach with a positive total energy?
The original question can't be answered in classical physics because there is no mass-energy equvalence and no lower limit for the gravitational potential.
14. Jun 13, 2017
### Vicara
I´ve seen this formula before, but I don´t understand how light could have momentum or kinetic energy, as both of them include mass in their expressions...
15. Jun 13, 2017
### PAllen
Well, for starters, there is empirical fact: how would a light sail work if light did not have momentum?
For light, momentum is E/c, not anything involving mass.
In SR, kinetic energy is simply total energy minus rest energy. Since light has no rest energy, all of its energy is kinetic. Note that mass, per se, is not involved in this definition.
Summary: for light p = E/c, and KE = E, with E being the total energy of the light.
16. Jun 13, 2017
### Staff: Mentor
You are thinking of the classical $p=mv$ and $E_k=mv^2$ or the relativistic $p=\gamma{m_0}v$ and $E_k=(\gamma-1)m_0c^2$ which do indeed include the mass. However, these formulas only apply in the special case of particles with non-zero mass and traveling at less than the speed of light, and clearly that doesn't include photons. Do note, however, that the formula @PAllen quoted is equivalent to the two relativistic ones in the $m_0=0$ case.
We also have a FAQ: https://www.physicsforums.com/threads/how-can-light-have-momentum-if-it-has-zero-mass.512541/
17. Jun 14, 2017
### Vicara
So, what i thought that light could have negative energy (or mas, by the mass-energy equivalence) is wrong because light don't have potential energy?
18. Jun 14, 2017
### Matter_Matters
You should note that the choice of sign for the potential energy is really down to convention. You could easily define it as the following:
$$V(r) = \frac{GM}{r},$$
such that the force associated with the above potential may be given by the usual
$$F = \nabla V(r),$$
which gives the usual gravitational force produced by some body of mass $M$
$$F = -\frac{GM}{r^2}.$$
In relation to it's connection the rest energy in relativity, I think you are confusing the two theories and want to consult the general theory of relativity for a relativistic description of gravity as the other comments suggest.
19. Jun 14, 2017
### Ibix
It doesn't really matter whether light can have potential energy or not. You get to choose what zero potential is, so whether something has positive potential or negative is up to you. Something you get to choose can't have physical consequences.
20. Jun 14, 2017
### Vicara
Then the energy (and the mass) depends of the frame of reference?
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Draft saved Draft deleted | 1,855 | 7,714 | {"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": 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} | 3.484375 | 3 | CC-MAIN-2017-43 | longest | en | 0.950445 |
https://logicroots.com/math-games/grade-4/ | 1,580,197,116,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251776516.99/warc/CC-MAIN-20200128060946-20200128090946-00251.warc.gz | 521,055,366 | 14,795 | # Best Math Board Games for Grade 4 kids
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## Monster Sock Factory - Multiplication Table Board Game
8 Times More Math Practice
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4 Times More Math Practice
Squeaked the parrots, Bow-wowed the cute puppies and meowed the adorable kittens. They all welcome you to Pet Wonderland. Love them, feed them, adopt them and win their hearts. View More
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14 Times More Math Practice
ICC (International Cheese Clash) tournament for the top rats is on. The rats are fighting for the delicious cheese cakes at stake. SAY CHEESE at the perfect time and help your rat take home the prize. View More
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11 Times More Math Practice
This mountain climbing game makes student practice the whole curriculum progressively. It's system is designed to adapt as per student's skill to make him practice on her current skill level.
Math Skills:
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7 Times More Math Practice
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8 Times More Math Practice
Stranded in a strange island, 4 fishermen are wanting to go back home. But wait! There is only one way. They need to catch the fish and distribute equally among the villagers of the island. View More
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## Monkey Fractions - Fractions Card Game
8 Times More Math Practice
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## Ocean Raiders - Addition Board Game
11 Times More Math Practice
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Life Skills: | 1,705 | 7,880 | {"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} | 2.6875 | 3 | CC-MAIN-2020-05 | longest | en | 0.879602 |
http://blog.etestseries.in/lighting-stores-hhgg/6cc7e9-future-value-formula-example | 1,624,582,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00303.warc.gz | 8,435,150 | 25,855 | # future value formula example
If some are, then convert text values to numbers. The future value formula is used in essentially all areas of finance. First, we will look at the simplest case where we are using the compound interest formula to calculate the value of an investment after some set amount of time. Pmt (optional argument) – This specifies the payment per period. Typically, cash in a savings account or a hold in a bond purchase earns compound interest and so has a different value in the future. Since Jan 1, 2016, the terms of the agreement have changed, and the compound interest is attributed twice a month. I love the program, and I can't imagine using Excel without it! I thank you for reading and hope to see you on our blog next week! Must be entered as a negative number. To have all calculations performed with a single formula, do the following: Please pay attention that we lock the annual interest rate ($F$2), the number of years ($F$3) and the investment amount ($F$4) references with the dollar sign ($) so they won't shift when copying down the formula. Please remember that negative numbers should be used for all outgoing payments. 0 = end of period, 1 = beginning of period. ; pmt - The payment made each period. Enter the following formula in C2 and drag it down through C6. =FV(rate,nper,pmt,[pv],[type]) This function uses the following arguments: 1. Future value of a lump sum investment is explained on the future value of a single sum page. When investing money through a series of regular savings, it often happens that you are provided with an annual interest rate and the investment term defined in years, whereas the payments are to be made weekly, monthly, quarterly or semiannually. Let's say you pay$1,000 a month in rent. The quicker method however, is to use the following formula. the future value of the investment (rounded to 2 decimal places) is $12,166.53. Future value is the value of an asset at a specific date. rate - The interest rate per period. The formula for continously compounded interest is: $$F = Pe^{rt}$$ The future value (F) equals the present value (P) times e (Euler's Number) raised to the (rate * time) exponential. FV is an Excel financial function that returns the future value of an investment based on a fixed interest rate. Luckily, Microsoft Excel provides a special function that does all the math behind the scenes based on the arguments that you specify. Put simply, FV is the future value of an asset adjusted for interest over time. Let’s calculate the future value of this amount if Kevin keeps it for 11 years: Kevin also has account which he invested$20,000 into on January 1, 2017. This tutorial looks at how to use the FV function in Excel to find the future value of a series of periodic payments and a single lump-sum payment. With the help of the future formula, her account after 15 years will be: FV = 9,000 * … The following spreadsheets show the Excel FV function, used to calculate the future value of two different investments. Furthermore, you are going to add $100 at the beginning of each month. To correctly build a FV formula in your worksheets and avoid common errors, please keep in mind these usage notes: This example shows how to use the FV function in Excel in its simplest form to calculate future value, given a periodic interest rate, the total number of periods, and a constant payment amount per period. Unable to open Outlook window" error, Outlook Quick Parts and AutoText: how to create, edit and use, Merge data from duplicate rows based on a unique column, How to compare data in two Google sheets or columns, 0 or omitted (default) - at the end of a period (regular annuity), 1 - at the beginning of a period (annuity due), For any inflows such as dividends or other earnings, use, To get the correct future value, you must be consistent with. For example, you can calculate the future value of your 401(k) in 20 years based on a 5% interest rate, annual contribution of$3,000, and amount that you have amassed in the account. This example shows how to use the FV function in Excel in its simplest form to calculate future value, given a periodic interest rate, the total number of periods, and a constant payment amount per period. Instead of building formulas or performing intricate multi-step operations, start the add-in and have any text manipulation accomplished with a mouse click. Building your personal and corporate finances requires thorough planning. Future value is a way to calculate how much that investment is worth today. The best spent money on software I've ever spent! From the example, $110 is the future value of$100 after 1 year and similarly, $100 is the present value of$110 to be received after 1 year. Example: Let's say your goal is to end up with $10,000 in 5 years, and you can get an 8% interest rate on your savings, compounded monthly. Nper – Total number of payment periods. It is assumed to be a regular annuity where all payments are made at the end of the year. Future value formula in Excel (.xlsx file), Thank you for your comment! FV = \$9{,}000 \times (1 + 2.2\%)^{11} = \$11{,}434.11, FV = \$20{,}000 \times (1 + 2.75\%)^{4} =\$22{,}292.43, FV = \$22{,}292.43 \times (1 + 0.0046\%)^{24} = \$22{,}317.05, value is based on the time value of money, Time Value of Money Solution Grid: Additional Problems, Quarterly interest rate (r) = 11/4 = 2.75%, Number of period (n) = 2 * 12 = 24 (twice a month), Twice-monthly interest rate (r) = 0.0092/2 = 0.0046%. For this, we divide an annual interest rate (C2) by 12 and multiply the number of years (C3) by 12: Where C5 is the number of compounding periods per year: To compare the amount of growth generated by various compounding periods, you need to supply different rate and nper to the FV function. It works for both a series of periodic payments and a single lump-sum payment. FV means future value; PV means present value; i is the period discount rate Best add-ins for Microsoft Outlook in one collection to reveal the full power of your inbox and improve your emailing routine: Custom email templates for teams and individuals. The calculation of future value uses 3 variables: the cash value of payments made per period, the interest rate, and the number of payments. How much money will there be in your saving account in 5 years? As such, the higher the discount rate, the higher will be the future value of the annuity. It measures the nominal future sum of money that a given sum of money is "worth" at a specified time in the future assuming a certain interest rate, or more generally, rate of return; it is the present value multiplied by the accumulation function. . If we omit this argument, we need to provide the PV argument. Future value with simple interest uses the following formula: Future Value = Present Value (1 + (Interest Rate x Number of Years)) Let’s say Bob invests$1,000 for five years with an interest rate of 10%. By the end of the third year, you would have a balance of $1,464.10 instead of a balance of$1,300 with simple annual interest which only calculates interest on the initial cash flow. FV is simply what money is expected to be worth in the future. Ablebits is a fantastic product - easy to use and so efficient. © 1999-2021 Study Finance. Let's say you are going to make a yearly $1,000 payment for 10 years with an annual interest rate of 6%. The formula to use will depend on which 3 of the 4 variables are already known. Here we learn how to calculate the FV of an annuity due using its formula along with some practical examples and a downloadable excel template. PV (optional argument) – This specifies the present value (PV) of the inves… 35+ handy options to make your text cells perfect. You can download this Future Value Template here – Future Value Template Calculate the future value of 15,000 rupees loaned at the rate of 12 percent per annum for 10 years. This article has been a guide to the Future Value of Annuity Due and its meaning. If Mrs. Smith has$9,000 in her bank account and she earns an annual interest of 4.5%. Do not waste your time on typing the same replies to repetitive emails. To convert an annual interest rate to a periodic rate, divide the annual rate by the number of periods per year: To get the total number of periods, multiply the term in years by the number of periods per year: Now, let's see how it works in practice. At the end of those ten years, the $1,000 would be worth$1,790.85. by Svetlana Cheusheva | updated on January 13, 2021 In the following spreadsheet, the Excel Fv function is used to calculate the future value of an investment of $1,000 per month for a period of 5 years. May occur if one or more arguments are non-numeric. An initial investment of$1,000 at 10% annual interest would become a balance of $1,100 in year two, which would then also earn 10% interest. It’s a useful tool for investors and financial planners to estimate how much an investment made today will be worth in the future, and this allows investors to make sound decisions. In many circumstances, the future value formula is incorporated into other formulas. In all present value and future value lump sum formulas the following symbols are used. Normally, the FV calculation is based on an anticipated growth rate, or rate of return. The formula for future value using simple annual interest is: Kevin earns an interest rate of 2.2% on a$9,000 savings account. I have enjoyed every bit of it and time am using it. Mary has $8,500 in a checking account, and she earns an annual interest rate of 2.2%. This smart package will ease many routine operations and solve complex tedious tasks in your spreadsheets. The account has an annual rate of 11% and is compounded annually. Using RATE function in Excel to calculate interest rate, Attaching files from SharePoint to Outlook email, How to attach files to Outlook email from OneDrive, LARGE IF formula in Excel: get n-th highest value with criteria, Compare 2 columns in Excel for matches and differences, CONCATENATE in Excel: combine text strings, cells and columns, Create calendar in Excel (drop-down and printable), 3 ways to remove spaces between words in Excel cells, How to fix "Cannot start Microsoft Outlook. That's how to how to calculate future value of annuity in Excel. Like many financial tools, future value is based on the time value of money concept, which states that a dollar today is worth more than a dollar at some time in the future. Must be entered as a negative number. 100,000 for 15 years. n = 12. t = 10. type - [optional] When payments are due. Basic future value formula in Excel. A$100 invested in bank @ 10% interest rate for 1 year becomes $110 after a year. Example 1. n = number of periods Compose your response just once, save it as a template and reuse whenever you want. As one example, an annuity in the form of regular deposits in an interest account would be the sum of the future value of each deposit. AbleBits suite has really helped me when I was in a crunch! So let’s say you invested$1,000 at a fixed interest rate of 6% for 10 years. No Comments. The FV of investments in stocks, bonds or other securities may be hard to calculate accurately because of a volatile rate of return. To fix the error, check if any of the numbers referenced in your formula are formatted as text. Banking, investments, corporate finance all may use the future value formula is some fashion. And then, define the arguments in this way: Putting the arguments together, we get this formula: Suppose you wish to save some money for renovating your house in 5 years. How to calculate future value in Excel - formula examples, Find future value for different compounding periods, Excel RATE function to calculate interest rate. For example: Bob again invests $1000 today at an interest rate of 5%. That is called the future value of investment, and this tutorial will teach you how to calculate it in Excel. Anyone who works with Excel is sure to find their work made easier. Example of Future Value Formula In order to have a better understanding of the concept, we will calculate the future value by using the above-mentioned formula. Also, Mary has$20,000 in another account that pays an annual interest rate of 11% compounded quarterly. An investment is made with deposits of $100 per month (made at the end of each month) at an interest rate of 5%, compounded monthly (so, 12 compounds per period). Suppose you monthly invest$200 for 3 years with an annual interest rate of 6%. i = interest rate . 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Ltd. all rights reserved once, save it as a Template and reuse whenever you want the. The annuity arrange your data like shown in the future value calculator below to work out FV! Calculated twice per month have a look at how to how to how to calculate the balance of the commonly. Excel provides a special function that returns the future value of the account an! Copyright © 2003 - 2021 4Bits Ltd. all rights reserved 1,000 payment for 10 years be in formula! Tweak it to handle a couple of different ways important that the rate and nper units be.! | 4,910 | 21,853 | {"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} | 3.921875 | 4 | CC-MAIN-2021-25 | longest | en | 0.921412 |
https://www.askiitians.com/forums/7-grade-maths/write-the-following-rational-numbers-in-ascending_270962.htm | 1,719,137,525,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00652.warc.gz | 579,011,457 | 42,330 | # Write the following rational numbers in ascending order:-1/3, -2/9, -4/3
Harshit Singh
askIITians Faculty 5963 Points
3 years ago
Dear Student
To convert the given rational numbers into like fraction we have to find LCM, LCM of 3, 9, and 3 is 9
Now,
(-1/3)= [(-1 × 3)/ (3 × 9)] = (-3/9)
(-2/9)= [(-2 × 1)/ (9 × 1)] = (-2/9)
(-4/3)= [(-4 × 3)/ (3 × 3)] = (-12/9)
Clearly,
(-12/9) < (-3/9) < (-2/9)
Hence,
(-4/3) < (-1/3) < (-2/9)
Thanks | 193 | 442 | {"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-2024-26 | latest | en | 0.808671 |
https://everipedia.org/wiki/lang_en/Natural_number | 1,726,346,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00766.warc.gz | 221,420,137 | 49,476 | Everipedia is now IQ.wiki - Join the IQ Brainlist and our Discord for early access to editing on the new platform and to participate in the beta testing.
# Natural number
In mathematics, the natural numbers are those used for counting (as in "there are six coins on the table") and ordering (as in "this is the third largest city in the country"). In common mathematical terminology, words colloquially used for counting are "cardinal numbers" and words connected to ordering represent "ordinal numbers". The natural numbers can, at times, appear as a convenient set of codes (labels or "names"); that is, as what linguists call nominal numbers, forgoing many or all of the properties of being a number in a mathematical sense.
Some definitions, including the standard ISO 80000-2,[1] begin the natural numbers with 0, corresponding to the non-negative integers 0, 1, 2, 3, …, whereas others start with 1, corresponding to the positive integers 1, 2, 3, ….[2][3][4][5] Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers, but in other writings, that term is used instead for the integers (including negative integers).[6]
The natural numbers are a basis from which many other number sets may be built by extension: the integers (Grothendieck group), by including (if not yet in) the neutral element 0 and an additive inverse (−n) for each nonzero natural number n; the rational numbers, by including a multiplicative inverse (1/n) for each nonzero integer n (and also the product of these inverses by integers); the real numbers by including with the rationals the limits of (converging) Cauchy sequences of rationals; the complex numbers, by including with the real numbers the unresolved square root of minus one (and also the sums and products thereof); and so on.[7][8] These chains of extensions make the natural numbers canonically embedded (identified) in the other number systems.
Properties of the natural numbers, such as divisibility and the distribution of prime numbers, are studied in number theory. Problems concerning counting and ordering, such as partitioning and enumerations, are studied in combinatorics.
In common language, for example in primary school, natural numbers may be called counting numbers[9] both to intuitively exclude the negative integers and zero, and also to contrast the discreteness of counting to the continuity of measurement, established by the real numbers.
## History
### Ancient roots
The most primitive method of representing a natural number is to put down a mark for each object. Later, a set of objects could be tested for equality, excess or shortage, by striking out a mark and removing an object from the set.
The first major advance in abstraction was the use of numerals to represent numbers. This allowed systems to be developed for recording large numbers. The ancient Egyptians developed a powerful system of numerals with distinct hieroglyphs for 1, 10, and all the powers of 10 up to over 1 million. A stone carving from Karnak, dating from around 1500 BC and now at the Louvre in Paris, depicts 276 as 2 hundreds, 7 tens, and 6 ones; and similarly for the number 4,622. The Babylonians had a place-value system based essentially on the numerals for 1 and 10, using base sixty, so that the symbol for sixty was the same as the symbol for one, its value being determined from context.[13]
A much later advance was the development of the idea that 0 can be considered as a number, with its own numeral. The use of a 0 digit in place-value notation (within other numbers) dates back as early as 700 BC by the Babylonians, but they omitted such a digit when it would have been the last symbol in the number.[14] The Olmec and Maya civilizations used 0 as a separate number as early as the 1st century BC, but this usage did not spread beyond Mesoamerica.[15][16] The use of a numeral 0 in modern times originated with the Indian mathematician Brahmagupta in 628. However, 0 had been used as a number in the medieval computus (the calculation of the date of Easter), beginning with Dionysius Exiguus in 525, without being denoted by a numeral (standard Roman numerals do not have a symbol for 0); instead nulla (or the genitive form nullae) from nullus, the Latin word for "none", was employed to denote a 0 value.[17]
The first systematic study of numbers as abstractions is usually credited to the Greek philosophers Pythagoras and Archimedes. Some Greek mathematicians treated the number 1 differently than larger numbers, sometimes even not as a number at all.[18]
Independent studies also occurred at around the same time in India, China, and Mesoamerica.[19]
### Modern definitions
In 19th century Europe, there was mathematical and philosophical discussion about the exact nature of the natural numbers. A school of Naturalism stated that the natural numbers were a direct consequence of the human psyche. Henri Poincaré was one of its advocates, as was Leopold Kronecker who summarized "God made the integers, all else is the work of man".[20]
In opposition to the Naturalists, the constructivists saw a need to improve the logical rigor in the foundations of mathematics.[21] In the 1860s, Hermann Grassmann suggested a recursive definition for natural numbers thus stating they were not really natural but a consequence of definitions. Later, two classes of such formal definitions were constructed; later, they were shown to be equivalent in most practical applications.
Set-theoretical definitions of natural numbers were initiated by Frege and he initially defined a natural number as the class of all sets that are in one-to-one correspondence with a particular set, but this definition turned out to lead to paradoxes including Russell's paradox. Therefore, this formalism was modified so that a natural number is defined as a particular set, and any set that can be put into one-to-one correspondence with that set is said to have that number of elements.[22]
The second class of definitions was introduced by Charles Sanders Peirce, refined by Richard Dedekind, and further explored by Giuseppe Peano; this approach is now called Peano arithmetic. It is based on an axiomatization of the properties of ordinal numbers: each natural number has a successor and every non-zero natural number has a unique predecessor. Peano arithmetic is equiconsistent with several weak systems of set theory. One such system is ZFC with the axiom of infinity replaced by its negation. Theorems that can be proved in ZFC but cannot be proved using the Peano Axioms include Goodstein's theorem.[23]
With all these definitions it is convenient to include 0 (corresponding to the empty set) as a natural number. Including 0 is now the common convention among set theorists[24] and logicians.[25] Other mathematicians also include 0,[5] e.g., Computer languages often start from zero when enumerating items like loop counters and string- or array-elements.[26][27] Many mathematicians have kept the older tradition and take 1 to be the first natural number.[28]
Since different properties are customarily associated to the tokens0and1, e.g., neutral elements for addition and multiplications, respectively, it is important to know which version of natural numbers, generically denoted byis employed in the case under consideration. This can be done by explanation in prose, by explicitly writing down the set, or by qualifying the generic identifier with a super- or subscript (see also in#Notation),[1][29] e.g., like this:
• Naturals with zero:
• Naturals without zero:
## Notation
Mathematicians use N or ℕ (an N in blackboard bold) to refer to the set of all natural numbers. Older texts have also occasionally employed J as the symbol for this set.[30]
To be unambiguous about whether 0 is included or not, sometimes a subscript (or superscript) "0" is added in the former case, and a superscript "*" or subscript ">0" is added in the latter case:[1]
0= ℕ0= ℕ ∪ {0} = {0, 1, 2, …}
= ℕ
= ℕ1= ℕ
0
= {1, 2, 3, …}
.
Alternatively, since natural numbers naturally embed in the integers, they may be referred to as the positive, or the non-negative integers, respectively.[31]
## Properties
### Infinity
The set of natural numbers is an infinite set. This kind of infinity is, by definition, called countable infinity. All sets that can be put into a bijective relation to the natural numbers are said to have this kind of infinity. This is also expressed by saying that the cardinal number of the set is aleph-naught (ℵ0).[32]
One can recursively define an addition operator on the natural numbers by setting a + 0 = a and a + S(b) = S(a + b) for all a, b. Here S should be read as "successor". This turns the natural numbers (ℕ, +) into a commutative monoid with identity element 0, the so-called free object with one generator. This monoid satisfies the cancellation property and can be embedded in a group (in the mathematical sense of the word group). The smallest group containing the natural numbers is the integers.
If 1 is defined as S(0), then b + 1 = b + S(0) = S(b + 0) = S(b). That is, b + 1 is simply the successor of b.
### Multiplication
Analogously, given that addition has been defined, a multiplication operator × can be defined via a × 0 = 0 and a × S(b) = (a × b) + a. This turns (ℕ*, ×) into a free commutative monoid with identity element 1; a generator set for this monoid is the set of prime numbers.
### Relationship between addition and multiplication
Addition and multiplication are compatible, which is expressed in the distribution law: a × (b + c) = (a × b) + (a × c). These properties of addition and multiplication make the natural numbers an instance of a commutative semiring. Semirings are an algebraic generalization of the natural numbers where multiplication is not necessarily commutative. The lack of additive inverses, which is equivalent to the fact that ℕ is not closed under subtraction (i.e., subtracting one natural from another does not always result in another natural), means that ℕ is not a ring; instead it is a semiring (also known as a rig).
If the natural numbers are taken as "excluding 0", and "starting at 1", the definitions of + and × are as above, except that they begin with a + 1 = S(a) and a × 1 = a.
### Order
In this section, juxtaposed variables such as ab indicate the product a × b, and the standard order of operations is assumed.
A total order on the natural numbers is defined by letting ab if and only if there exists another natural number c where a + c = b. This order is compatible with the arithmetical operations in the following sense: if a, b and c are natural numbers and ab, then a + cb + c and acbc.
An important property of the natural numbers is that they are well-ordered: every non-empty set of natural numbers has a least element. The rank among well-ordered sets is expressed by an ordinal number; for the natural numbers, this is denoted as ω (omega).
### Division
In this section, juxtaposed variables such as ab indicate the product a × b, and the standard order of operations is assumed.
While it is in general not possible to divide one natural number by another and get a natural number as result, the procedure of division with remainder is available as a substitute: for any two natural numbers a and b with b ≠ 0 there are natural numbers q and r such that
a = bq + r and r < b.
The number q is called the quotient and r is called the remainder of the division of a by b. The numbers q and r are uniquely determined by a and b. This Euclidean division is key to several other properties (divisibility), algorithms (such as the Euclidean algorithm), and ideas in number theory.
### Algebraic properties satisfied by the natural numbers
The addition (+) and multiplication (×) operations on natural numbers as defined above have several algebraic properties:
• Closure under addition and multiplication: for all natural numbers a and b, both a + b and a × b are natural numbers.
• Associativity: for all natural numbers a, b, and c, a + (b + c) = (a + b) + c and a × (b × c) = (a × b) × c.
• Commutativity: for all natural numbers a and b, a + b = b + a and a × b = b × a.
• Existence of identity elements: for every natural number a, a + 0 = a and a × 1 = a.
• Distributivity of multiplication over addition for all natural numbers a, b, and c, a × (b + c) = (a × b) + (a × c).
• No nonzero zero divisors: if a and b are natural numbers such that a × b = 0, then a = 0 or b = 0 (or both).
## Generalizations
Two important generalizations of natural numbers arise from the two uses of counting and ordering: cardinal numbers and ordinal numbers.
• A natural number can be used to express the size of a finite set; more precisely, a cardinal number is a measure for the size of a set, which is even suitable for infinite sets. This concept of "size" relies on maps between sets, such that two sets have the same size, exactly if there exists a bijection between them. The set of natural numbers itself, and any bijective image of it, is said to be countably infinite and to have cardinality aleph-null (ℵ0).
• Natural numbers are also used as linguistic ordinal numbers: "first", "second", "third", and so forth. This way they can be assigned to the elements of a totally ordered finite set, and also to the elements of any well-ordered countably infinite set. This assignment can be generalized to general well-orderings with a cardinality beyond countability, to yield the ordinal numbers. An ordinal number may also be used to describe the notion of "size" for a well-ordered set, in a sense different from cardinality: if there is an order isomorphism (more than a bijection!) between two well-ordered sets, they have the same ordinal number. The first ordinal number that is not a natural number is expressed as ω; this is also the ordinal number of the set of natural numbers itself.
Many well-ordered sets with cardinal number ℵ0 have an ordinal number greater than ω (the latter is the lowest possible). The least ordinal of cardinality ℵ0 (i.e., the initial ordinal) is ω.
For finite well-ordered sets, there is a one-to-one correspondence between ordinal and cardinal numbers; therefore they can both be expressed by the same natural number, the number of elements of the set. This number can also be used to describe the position of an element in a larger finite, or an infinite, sequence.
A countable non-standard model of arithmetic satisfying the Peano Arithmetic (i.e., the first-order Peano axioms) was developed by Skolem in 1933. The hypernatural numbers are an uncountable model that can be constructed from the ordinary natural numbers via the ultrapower construction.
Georges Reeb used to claim provocatively that The naïve integers don't fill up ℕ. Other generalizations are discussed in the article on numbers.
## Formal definitions
### Peano axioms
Many properties of the natural numbers can be derived from the five Peano axioms:[33][34]
1. 0 is a natural number.
2. Every natural number has a successor.
3. 0 is not the successor of any natural number.
4. If the successor of equals the successor of , then equals .
5. The axiom of induction: If a statement is true of 0, and if the truth of that statement for a number implies its truth for the successor of that number, then the statement is true for every natural number.
These are not the original axioms published by Peano, but are named in his honor. Some forms of the Peano axioms have 1 in place of 0. In ordinary arithmetic, the successor ofis. Replacing axiom 5 by an axiom schema, one obtains a (weaker) first-order theory called *Peano arithmetic*.
### Constructions based on set theory
#### Von Neumann ordinals
In the area of mathematics called set theory, a specific construction due to John von Neumann[35][36] defines the natural numbers as follows:
• Set 0 = { }, the empty set,
• Define S(a) = a ∪ {a} for every set a. S(a) is the successor of a, and S is called the successor function.
• By the axiom of infinity, there exists a set which contains 0 and is closed under the successor function. Such sets are said to be 'inductive'. The intersection of all such inductive sets is defined to be the set of natural numbers. It can be checked that the set of natural numbers satisfies the Peano axioms.
• It follows that each natural number is equal to the set of all natural numbers less than it:
• 0 = { },
• 1 = 0 ∪ {0} = {0} = {{ }},
• 2 = 1 ∪ {1} = {0, 1} = {{ }, {{ }}},
• 3 = 2 ∪ {2} = {0, 1, 2} = {{ }, {{ }}, {{ }, {{ }}}},
• n = n−1 ∪ {n−1} = {0, 1, …, n−1} = {{ }, {{ }}, …, {{ }, {{ }}, …}}, etc.
With this definition, a natural number n is a particular set with n elements, and nm if and only if n is a subset of m. The standard definition, now called definition of von Neumann ordinals, is: "each ordinal is the well-ordered set of all smaller ordinals."
Also, with this definition, different possible interpretations of notations like ℝn (n-tuples versus mappings of n into ℝ) coincide.
Even if one does not accept the axiom of infinity and therefore cannot accept that the set of all natural numbers exists, it is still possible to define any one of these sets.
#### Zermelo ordinals
Although the standard construction is useful, it is not the only possible construction. Ernst Zermelo's construction[36] goes as follows:
• Set 0 = { }
• Define S(a) = {a},
• It then follows that
• 0 = { },
• 1 = {0} = {{ }},
• 2 = {1} = {{{ }}},
• n = {n−1} = {{{…}}}, etc.
Each natural number is then equal to the set containing just the natural number preceding it. (This is the definition of Zermelo ordinals.)
• Benacerraf's identification problem
• Canonical representation of a positive integer
• Countable set
• Number#Classification for other number systems (rational, real, complex etc.)
## References
[1]
Citation Linkwww.iso.org"Standard number sets and intervals". ISO 80000-2:2009. International Organization for Standardization. p. 6.
Sep 29, 2019, 10:47 PM
[2]
Citation Linkmathworld.wolfram.comWeisstein, Eric W. "Natural Number". MathWorld.
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[3]
Citation Linkwww.merriam-webster.com"natural number", Merriam-Webster.com, Merriam-Webster, retrieved 4 October 2014
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[4]
Citation Linkopenlibrary.orgCarothers, N.L. (2000), Real analysis, Cambridge University Press, ISBN 978-0-521-49756-5 says: "ℕ is the set of natural numbers (positive integers)" (p. 3)
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[5]
Citation Linkopenlibrary.orgMac Lane, Saunders; Birkhoff, Garrett (1999), Algebra (3rd ed.), American Mathematical Society, ISBN 978-0-8218-1646-2 include zero in the natural numbers: 'Intuitively, the set ℕ = {0, 1, 2, ...} of all natural numbers may be described as follows: ℕ contains an "initial" number 0; ...'. They follow that with their version of the Peano Postulates. (p. 15)
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[6]
Citation Linkbooks.google.comJack G. Ganssle & Michael Barr (2003). Embedded Systems Dictionary. pp. 138 (integer), 247 (signed integer), & 276 (unsigned integer). ISBN 978-1-57820-120-4. integer 1. n. Any whole number.
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[7]
Citation Linkopenlibrary.orgMendelson, Elliott (2008) [1973], Number Systems and the Foundations of Analysis, Dover Publications, ISBN 978-0-486-45792-5 says: "The whole fantastic hierarchy of number systems is built up by purely set-theoretic means from a few simple assumptions about natural numbers." (Preface, p. x)
Sep 29, 2019, 10:47 PM
[8]
Citation Linkopenlibrary.orgBluman, Allan (2010), Pre-Algebra DeMYSTiFieD (Second ed.), McGraw-Hill Professional, ISBN 978-0-07-174251-1: "Numbers make up the foundation of mathematics." (p. 1)
Sep 29, 2019, 10:47 PM
[9]
Citation Linkmathworld.wolfram.comWeisstein, Eric W. "Counting Number". MathWorld.
Sep 29, 2019, 10:47 PM
[10]
Citation Linkwww.naturalsciences.be"Introduction". Royal Belgian Institute of Natural Sciences. Brussels, Belgium. Archived from the original on March 4, 2016.
Sep 29, 2019, 10:47 PM
[11]
Citation Linkishango.naturalsciences.beFlash presentation, Royal Belgian Institute of Natural Sciences, Brussels, Belgium.
Sep 29, 2019, 10:47 PM
[12]
Citation Linkwww2.astronomicalheritage.netThe Ishango Bone, Democratic Republic of the Congo, on permanent display at the Royal Belgian Institute of Natural Sciences, Brussels, Belgium. UNESCO's Portal to the Heritage of Astronomy
Sep 29, 2019, 10:47 PM
[13]
Citation Linkopenlibrary.orgGeorges Ifrah, The Universal History of Numbers, Wiley, 2000, ISBN 0-471-37568-3
Sep 29, 2019, 10:47 PM
[14]
Citation Linkwww-history.mcs.st-and.ac.uk"A history of Zero". MacTutor History of Mathematics. Retrieved 2013-01-23. a tablet found at Kish ... thought to date from around 700 BC, uses three hooks to denote an empty place in the positional notation. Other tablets dated from around the same time use a single hook for an empty place
Sep 29, 2019, 10:47 PM
[15]
Citation Linkbooks.google.comMann, Charles C. (2005), 1491: New Revelations Of The Americas Before Columbus, Knopf, p. 19, ISBN 978-1-4000-4006-3.
Sep 29, 2019, 10:47 PM
[16]
Citation Linkbooks.google.comEvans, Brian (2014), "Chapter 10. Pre-Columbian Mathematics: The Olmec, Maya, and Inca Civilizations", The Development of Mathematics Throughout the Centuries: A Brief History in a Cultural Context, John Wiley & Sons, ISBN 978-1-118-85397-9.
Sep 29, 2019, 10:47 PM
[17]
Citation Linkhbar.phys.msu.ruMichael Deckers (2003-08-25). "Cyclus Decemnovennalis Dionysii – Nineteen year cycle of Dionysius". Hbar.phys.msu.ru. Retrieved 2012-02-13.
Sep 29, 2019, 10:47 PM
[18]
Citation Linkaleph0.clarku.eduThis convention is used, for example, in Euclid's Elements, see Book VII, definitions 1 and 2.
Sep 29, 2019, 10:47 PM
[19]
Citation Linkopenlibrary.orgMorris Kline, Mathematical Thought From Ancient to Modern Times, Oxford University Press, 1990 [1972], ISBN 0-19-506135-7
Sep 29, 2019, 10:47 PM
[20]
Citation Linkwww.digizeitschriften.deThe English translation is from Gray. In a footnote, Gray attributes the German quote to: "Weber 1891/92, 19, quoting from a lecture of Kronecker's of 1886."Gray, Jeremy (2008), Plato's Ghost: The Modernist Transformation of Mathematics, Princeton University Press, p. 153, ISBN 978-1-4008-2904-0 Weber, Heinrich L. 1891-2. Kronecker. Jahresbericht der Deutschen Mathematiker-Vereinigung 2:5–23. (The quote is on p. 19.)
Sep 29, 2019, 10:47 PM | 5,650 | 22,473 | {"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-2024-38 | latest | en | 0.918858 |
https://www.physicsforums.com/threads/f-ma-question.383126/ | 1,591,342,264,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00210.warc.gz | 840,750,656 | 15,314 | # F=ma question
## Homework Statement
A person weighing 0.6 kN rides in an elevator
that has a downward acceleration of 3 m/s2.
The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the force of the
elevator floor on the person?
Answer in units of kN.
F = ma
## The Attempt at a Solution
Is the downward acceleration of the elevator and gravity in the same direction, so should the acceleration in the formula equal the sum of the acceleration of the elevator and gravity.
Or are they in opposite directions?
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
ideasrule
Homework Helper
What do you think? Do you feel lighter on an elevator accelerating downwards, or heavier?
lighter, so would is acceleration due to gravity upwards?
ideasrule
Homework Helper
Acceleration due to gravity is downwards because gravity doesn't change; it depends only on the Earth. However, the effect of the elevator's acceleration counteracts the effect of gravity.
thanks | 221 | 1,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | longest | en | 0.921417 |
http://mathhelpforum.com/pre-calculus/171752-accurate-3-significant-figures.html | 1,480,748,078,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540909.75/warc/CC-MAIN-20161202170900-00217-ip-10-31-129-80.ec2.internal.warc.gz | 182,577,236 | 9,532 | # Thread: Accurate to 3 significant figures?
1. ## Accurate to 3 significant figures?
A) square root of 11 + j5 =6 angle 56.4 degrees
B) 7 angle38 degrees = 5.52 + j4.30
A. a & B are both FALSE
B. A is TRUE and b is FALSE
C. a and b are both TRUE
D. a is FALSE and b is TRUE
WHICH is Correct and how?
2. What ideas have you had so far? | 115 | 340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-50 | longest | en | 0.938138 |
https://edurev.in/course/quiz/attempt/-1_Test-Level-2-Structure-Of-Atom-1/9f5ce526-41c2-45c3-9649-ad71b8b65b82 | 1,623,867,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00163.warc.gz | 217,241,534 | 47,020 | Courses
# Test Level 2: Structure Of Atom- 1
## 30 Questions MCQ Test Chemistry for JEE | Test Level 2: Structure Of Atom- 1
Description
This mock test of Test Level 2: Structure Of Atom- 1 for Class 11 helps you for every Class 11 entrance exam. This contains 30 Multiple Choice Questions for Class 11 Test Level 2: Structure Of Atom- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test Level 2: Structure Of Atom- 1 quiz give you a good mix of easy questions and tough questions. Class 11 students definitely take this Test Level 2: Structure Of Atom- 1 exercise for a better result in the exam. You can find other Test Level 2: Structure Of Atom- 1 extra questions, long questions & short questions for Class 11 on EduRev as well by searching above.
QUESTION: 1
Solution:
3 eV
Corite Actual
Correct option
= 8 – 5
= 3eV
∴ V0 = 3 eV
QUESTION: 2
### Increasing order of magnetic moment among the following species is __________ . Na+, Fe+3, Co2+, Cr+2
Solution:
Na+, Co+2, Cr2+, Fe+3
we get Na+, Co+2, Cr2+, Fe+3
QUESTION: 3
### If in the hydrogen atom P.E. at ∞ is choosen to be 13.6 eV then the ratio of T.E. to K.E. for 1st orbit of H-atom is __________ .
Solution:
Zero
Refrence level is Ist orbit
itself T. E. = 0
Ratio becomes zero
QUESTION: 4
The light radiations with discrete quantities of energy are called ___________ .
Solution:
A photon is the quantum of electromagnetic radiation. The term quantum is the smallest elemental unit of a quantity, or the smallest discrete amount of something. Thus, one quantum of electromagnetic energy is called a photon. The plural of quantum is quanta.
QUESTION: 5
The ratio of the energy of a photon of 2000 Å wavelength radiation to that of 4000 Å radiation is
Solution:
QUESTION: 6
The energy of electron is maximum at :
Solution:
An electron bound to the nucleus by the mutual attraction between unlike charges, has a negative energy. Its most tightly bound to the nucleus nearer it is to the nucleus. That is an electron has least energy when it is in the n=1 shell. As the electron moves away from the nucleus its energy becomes less and less negative, i.e. its energy increases. When the electron is not under the attractive Coulomb force of the nucleus its potential energy is zero and whatever energy it has now, is kinetic energy, which is always positive. So farther the electron more energy it has.
QUESTION: 7
Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon:
Solution:
Is → As it is the ground state
QUESTION: 8
The third line in Balmer series corresponds to an electronic transition between which Bohr's orbits in hydrogen :
Solution:
Balmer means transition to n = 2
1. line → 3 to 2
2. line → 4 to 2
3. line → 5 to 2
QUESTION: 9
The orbital angular momentum of an electron in 2s orbital is :
Solution:
Orbital Angular momentum is given by [√l(l+1)]×h/2π
For s orbital, we have l = 0
Therefore orbital angular momentum = 0
QUESTION: 10
Which quantum number is not related with Schrodinger equation :
Solution:
Spin quantum number do not follow Schrodinger Wave Equation because it is related to spin of the electrons.It is needed to explain Pauli's exclusion principle which sates that no two electrons can share the same state (i.e. have the same quantum numbers) at the same time.Therefore, if the quantum numbers n, l and are same then the electrons should have different spin.
QUESTION: 11
The shortest wavelength of He atom in Balmer series is x, then longest wavelength in the Paschene series of Li+2 is :
Solution:
QUESTION: 12
An electron in a hydrogen atom in its ground state absrobs energy equal to the ionisation energy of Li+2. The wavelength of the emitted electron is
Solution:
V2 = 4.32 × 1013
V2 = 43.2 × 1012
V = 6.57 × 106
QUESTION: 13
An electron, a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. What is the qualitative order of their de-Broglie wavelengths?
Solution:
Mass of a particle = 4 (mass of proton)
Mass of proton = 1840 (mass of e)
Let Mass of e = m
∴ Mass of p+ = 1840 m
and mass of a particle = 7360 m
QUESTION: 14
Given ΔH for the process Li(g) → Li+3(g) + 3e- is 19800 kJ/mole & IE1 for Li is 520 then IE2 & IE3 of Li+ are respectively (approx, value) :
Solution:
IE1 + IE2 + IE3 = 19800
IE2 + IE3 = 19800 – 520
IE2 + IE3 = 19280
QUESTION: 15
The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is :
Solution:
QUESTION: 16
If radius of second stationary orbit (in Bohr's atom) is R. Then radius of third orbit will be
Solution:
= 2.25 R
QUESTION: 17
The ratio of wave length of photon corresponding to the α-line of lyman series in H-atom and β-line of Baimer series in He+ is
Solution:
QUESTION: 18
Three energy levels P, Q, R of a certain atom are such that EP < EQ < ER. If λ1, λ2, and λ3, are the wave length of radiation corresponding to transition R → Q : Q → P and R → P respectively. The correct relationship between λ1, λ2 and λ3 is
Solution:
We know that, for wave no.
QUESTION: 19
The Value of (n2 + n1) and for He+ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted photon when electron jump from n2 to n1 is
Solution:
QUESTION: 20
Number of possible spectral lines which may be emitted in bracket series in H atom if electrons present 9th excited level returns to ground level, are
Solution:
Total line = 6
QUESTION: 21
The first use of quantum theory to explain the structure of atom was made by :
Solution:
Bohr's theory was based upon some postulations of classical physics and some postulations of the newly proposed quantum theory of Planck.
QUESTION: 22
The wavelength associated with a golf ball weighing 200g and moving at a speed of 5m/h is of the order:
Solution:
From de Broglie wavelength formula; λ = h/p where h=Planck's constant and p=momentum.
Putting all the values, we have h=6.626×10-34/(200/1000)(5/3600)
= 2.3853×10-30
Order is of 10-30
QUESTION: 23
The longest wavelength of He+ in Paschen series is "m", then shortest wavelength of Be3+ in Paschen series is (in terms of m) :
Solution:
The expression for the wavelength is :
For the longest wavelength of the He+ atom in Paschen series , the expression becomes
(1)
For the shortest wavelength of the expression becomes
(2)
Divide equation (2) with equation (1).
On solving this, we get λ=7/64m
QUESTION: 24
What is uncertainity in location of a photon of wavelength 5000 Å if wavelength is known to an accuracy of 1 pm?
Solution:
0.0199 m
QUESTION: 25
Consider the following nuclear reactions involving X & Y.
X → Y + 2He4 Y → 8O18 + 1H1
If both neutrons as well as protons in both the sides are conserved in nuclear reaction then moles of neutrons in 4.6 gm of X :
Solution:
By conservation of mass
X = 4 + 18 + 1 gm
= 23
23 gm → (2 + 10) moles neutrons.
QUESTION: 26
Electromagnetic radiations having λ = 310 Å are subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photoelectrons with maximum Kinetic energy.
Solution:
V2 = 9.56 × 1012
V = 3.09 × 106 m/sec.
QUESTION: 27
Assuming Heisenberg Uncertainity Principle to be true what could be the miniumum uncertainity in de–Broglie wavelength of a moving electron accelerated by Potential Difference of 6V whose uncertainity in position is 7/22 n.m.
Solution:
QUESTION: 28
A 1-kW radio transmitter operates at a frequency of 880 Hz. How many photons per second does it emit -
Solution:
Given :
P = 1 kW
P = 1 × 103 watt
E = 103 J/S in one sec
v = 880 Hz
∴ E = nhv
⇒ 103 x × 6.626 × 10–34 × 880
⇒ x = 1.71 × 1033 in one sec
QUESTION: 29
The value of Bohr radius of hydrogen atom is -
Solution:
= 0.529 × 10–10 m
= 0.529 × 10–8 cm
QUESTION: 30
On the basis of Bohr's model, the radius of the 3rd orbit is -
Solution: | 2,229 | 7,995 | {"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-2021-25 | latest | en | 0.857886 |
https://www.convertunits.com/from/link+%5BRamden,+engineer%5D/to/reed | 1,653,045,896,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00193.warc.gz | 812,711,321 | 13,020 | ## ››Convert link [Ramden, engineer] to reed [Israel]
Did you mean to convert link [Gunter, survey] link [Ramden, engineer] to reed
How many link [Ramden, engineer] in 1 reed? The answer is 8.7893700787402.
We assume you are converting between link [Ramden, engineer] and reed [Israel].
You can view more details on each measurement unit:
The SI base unit for length is the metre.
1 metre is equal to 3.2808398950131 link [Ramden, engineer], or 0.3732736095558 reed.
Note that rounding errors may occur, so always check the results.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of link [Ramden, engineer] to reed
1 link [Ramden, engineer] to reed = 0.11377 reed
5 link [Ramden, engineer] to reed = 0.56887 reed
10 link [Ramden, engineer] to reed = 1.13774 reed
20 link [Ramden, engineer] to reed = 2.27548 reed
30 link [Ramden, engineer] to reed = 3.41321 reed
40 link [Ramden, engineer] to reed = 4.55095 reed
50 link [Ramden, engineer] to reed = 5.68869 reed
75 link [Ramden, engineer] to reed = 8.53303 reed
100 link [Ramden, engineer] to reed = 11.37738 reed
## ››Want other units?
You can do the reverse unit conversion from reed to link [Ramden, engineer], or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 521 | 1,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.853484 |
https://exercism.io/tracks/elm/exercises/sum-of-multiples/solutions/4b05e0e5216947f289015a79af30e745 | 1,558,481,362,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256586.62/warc/CC-MAIN-20190521222812-20190522004812-00176.warc.gz | 467,676,774 | 6,059 | # syvanpera's solution
## to Sum Of Multiples in the Elm Track
Published at Apr 16 2019 · 0 comments
Instructions
Test suite
Solution
Given a number, find the sum of all the unique multiples of particular numbers up to but not including that number.
If we list all the natural numbers below 20 that are multiples of 3 or 5, we get 3, 5, 6, 9, 10, 12, 15, and 18.
The sum of these multiples is 78.
## Elm Installation
Refer to the Installing Elm page for information about installing elm.
## Writing the Code
The first time you start an exercise, you'll need to ensure you have the appropriate dependencies installed. Thankfully, Elm makes that easy for you and will install dependencies when you try to run tests or build the code.
Execute the tests with:
``````\$ elm-test
``````
Automatically run tests again when you save changes:
``````\$ elm-test --watch
``````
As you work your way through the test suite, be sure to remove the `skip <|` calls from each test until you get them all passing!
## Source
A variation on Problem 1 at Project Euler http://projecteuler.net/problem=1
## Submitting Incomplete Solutions
It is possible to submit an incomplete solution so you can see how others have completed the exercise.
### Tests.elm
``````module Tests exposing (tests)
import Expect
import SumOfMultiples exposing (sumOfMultiples)
import Test exposing (..)
tests : Test
tests =
describe "Sum Of Multiples"
[ test "[3, 5] 15" <|
\() -> Expect.equal 45 (sumOfMultiples [ 3, 5 ] 15)
, skip <|
test "[7, 13, 17] 20" <|
\() -> Expect.equal 51 (sumOfMultiples [ 7, 13, 17 ] 20)
, skip <|
test "[4, 6] 15" <|
\() -> Expect.equal 30 (sumOfMultiples [ 4, 6 ] 15)
, skip <|
test "[5, 6, 8] 150" <|
\() -> Expect.equal 4419 (sumOfMultiples [ 5, 6, 8 ] 150)
, skip <|
test "[43, 47] 10000" <|
\() -> Expect.equal 2203160 (sumOfMultiples [ 43, 47 ] 10000)
, skip <|
test "[5, 25] 51" <|
\() -> Expect.equal 275 (sumOfMultiples [ 5, 25 ] 51)
]``````
``````module SumOfMultiples exposing (sumOfMultiples)
sumOfMultiples : List Int -> Int -> Int
sumOfMultiples multiples limit =
let
numsUntilLimit =
List.range 1 (limit - 1)
isInMultiples num =
List.any (\x -> modBy x num == 0) multiples
in
numsUntilLimit
|> List.filter isInMultiples
|> List.sum`````` | 686 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-22 | latest | en | 0.714795 |
https://math.answers.com/Q/What_are_the_whole_divisors_of_1356 | 1,716,454,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00594.warc.gz | 337,688,805 | 47,328 | 0
# What are the whole divisors of 1356?
Updated: 9/25/2023
Wiki User
8y ago
Best Answer
1, 2, 3, 4, 6, 12, 113, 226, 339, 452, 678, 1356
Wiki User
8y ago
This answer is:
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Related questions
12458
### What are the whole number factors of 1356?
1, 2, 3, 4, 6, 12, 113, 226, 339, 452, 678, 1356
### What are the Whole number divisors of 512?
512,256,128,64,32,16,8,4,2,1
1 2 3 4 6.
6
15
1, 2, 43, 86.
It is 6.
1356
### What is 4601 minus 1356?
4601 - 1356 = 3245
### A whole number that has more than two divisors is?
Called a composite number.
### What is 1356.48 to the nearest cubic unti?
Assuming that figure represents cubic units, the nearest whole one is 1356 | 305 | 812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.828624 |
https://questionitnow.com/how-many-litres-of-home-heating-oil-is-used-per-hour/ | 1,670,201,991,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00294.warc.gz | 504,814,700 | 32,486 | As a very rough guide, a medium sized boiler could use around 3.5 litres of oil per hour when on full, whereas a large appliance might burn around four litres per hour.
How many litres of oil does a wrx take? 2015 wrx oil capacity.
How many Litres of oil does my boiler burn per hour?
As a very rough guide, a medium sized boiler could use around 3.5 litres of oil per hour when on full, whereas a large appliance might burn around four litres per hour.
How much oil does a oil burner use per hour?
On average, an oil-burning furnace will burn between 0.8 and 1.7 gallons per hour while in operation. So, if you have a one-bedroom home, requiring a 300-gallon tank, and you keep your furnace running for 10 hours, you’ll need to fill your rank after 17 days.
How long should 500l of oil last?
500 litres, even in winter, will last us 2-3 months.
How long will 250 Litres of oil last?
How much did 250 litres cost you? An average burner will burn around 3 ltrs getting upto temperature and then around a ltr/hr maintaining it. Lot of things come into the equation but i would guestimate around 6 weeks.
How much oil does a house use per day?
In the U.S., the average household that uses oil for heat uses about 500 gallons between October and March (182 days). If you divide the average amount of fuel used (500 gallons) by the number of days (182), the average home uses about 2.75 gallons of oil each day.
How much oil does it take to heat a house per month?
So on average, one home demands around 5.3 gallons of heating oil in a day, which means that a full tank can last for 52 days or one month and three weeks. Therefore, if you’re using a 275-gallon tank, that would most likely be enough.
How do you calculate heating oil usage?
Multiply the hours of burner operation by the gallons per hour to find out the gallons per day your furnace will use. For instance, if your oil burner uses 1.7 gallons per operating hour and it runs for 13 hours per day, it will burn 22 gallons per day.
How much electricity does an oil burner use?
Generally, it takes approximately 2,500 watts of electricity to run an oil furnace. This is significantly less than a minimum of 15,000 watts as required for electric furnaces.
How long will an eighth of a tank of oil last?
How Long Will 1/8 Tank of Heating Oil Last? Typical home oil tanks will fill to 245 gallons, so 1/8 tank will supply about 31 gallons of oil. At 40 degrees F outside temperature, 1/8 tank will last you 2-4 days ( your furnace won’t likely be running 100% of that time and, again, there are many variables involved).
How many hours should 500 Litres of heating oil last?
500 litres equates therefore to 166 to 200 hours of the boiler being on.
How much oil does my boiler use per hour UK?
How much oil does my boiler use per hour? The average home in the UK that uses an oil boiler to heat the property is thought to use around 17,000 kWh on heating. Most oil boiler systems use kerosene and one litre of kerosene equals around 10 kWh.
Why is my heating oil burning so fast?
Oil burner cycling rate – Most heating systems take a few minutes after start-up to reach peak operating efficiency. If your heating oil burner is turning on and off rapidly (short cycling) – most likely due to it being mismatched to the heating load of your home – you are wasting energy and money.
How long should my oil heating last?
Heating oil will generally last between 18 and 24 months inside a residential oil tank. As long as the tank is clean and free of impurities, the oil should flow fine and heat your living quarters efficiently.
Are oil boilers being phased out?
The government plans to phase out high-carbon fossil fuel heating systems, such as oil boilers, during the 2020s.
How efficient are oil boilers?
Usually, a new condensing oil-fired boiler will have an efficiency of 92% to 93%, compared with 85% for a new non-condensing boiler and 60% to 70% for older systems. An oil boiler purchased from a well-established company should be just as efficient as a gas powered boiler system, if not more.
How many Litres of oil does the average house use?
How much heating oil do I need? The average UK household uses around 26,999 kWh of energy per year. This roughly translates to around 1,700 – 1,800 litres of heating oil – or between two and three tanks (depending on how big your tank is).
Is it expensive to heat a home with oil?
oil heat and crunch the numbers on switching to cheaper, cleaner gas heating. Last winter, heating a house with oil cost an average of \$1,700, while natural gas averaged less than \$900, according to the US Energy Information Administration. … Since 2002, oil heat has averaged 30% to 50% more than gas every year.
Is heating a house with oil expensive?
On average, oil heating prices are 2.5 to 5.5 times lower than electric-based heating prices, and older gas furnaces can be much more expensive to run than oil-based systems. … Oil heat is an especially cost-efficient method for warming larger homes in colder climates, as oil tends to burn hotter than gas.
How much does oil heat cost per month in Massachusetts?
Heating FuelForecasted Fuel Price (\$)Estimated ExpendituresHeating Oil\$3.27/gallon\$2,343Propane\$4.18/gallon\$3,192Natural Gas\$1.87/therm\$1,243Electric Resistance Heat\$0.26/kWh\$939
How much oil does a house use per year?
A typical 2500 square foot home in the northeast will use around 880 gallons of heating oil per year. Houses with great insulation will use less fuel oil, while poorly insulated houses with drafty windows can expect to use more heating oil.
How many hours a day should heat run?
Most systems should run about 2-3 cycles per hour. How long should the furnace or AC run in a given cycle? Typically 10-15 minutes in mild weather. The closer you get to the max design temperature in heating (0 Degrees) or cooling (90 Degrees) the longer the run cycle will be.
Do oil boilers use much electricity?
Oil fired boilers are more expensive to buy and install than gas or electric. However, the running costs are significantly lower than most electricity tariffs and around the same as gas (depending on fluctuations).
Does an oil boiler use a lot of electricity?
Do boilers use electricity? Boilers only use a minimal amount of electricity, but it is still needed to make certain functions work. Display screens, back-lighting bulbs and smart thermostats will all use electricity to help keep your home heated and supplied with hot water.
How much watts does a oil boiler use?
Here’s a simple way to estimate the power consumption: The power consumption for a typical boiler is about 200 watts running. Starting watts for the burner motor will be about 3 times that, or 600 watts peak.
How can I make my heating oil last longer?
Heat savings tips: easy dos Get a heating system tune-up — it will ensure you get the maximum performance out of every drop of heating oil burned. Open shades and drapes when the sun is out to help warm your home. Close them when the sun goes down to reduce heat loss through drafty windows.
How low should you let your oil tank get?
Glatz suggests you not let the level drop below 8 inches so you won’t get caught short. If you have a regular oil provider, they may be able to supply you with a chart that estimates the amount of oil left in your tank according to the level on the dipstick.
Can I put diesel in my oil tank?
Diesel, as sold at many gas stations, is an acceptable replacement for home heating oil in virtually all furnaces. Both diesel and heating oil No. … If you’re on the verge of running out of heating oil, or have run out, pouring diesel fuel into the tank can hold you over until a delivery is made.
Is oil central heating expensive to run?
When it comes to heating oil prices, oil fired central heating is generally a little more expensive than gas for heating, but is often the cheapest option in rural areas where there is no gas supply.
How do I know what size oil boiler I need?
1. Up to 10 radiators: 24–27 kW.
2. 10 to 15 radiators: 28–34 kW.
3. 15 to 20 radiators: 32–40 kW.
How much oil is in my tank UK?
To measure your tank manually Mister C says; take the cap off the top of the tank, which is where the fuel goes in, and put the dipstick right down to the bottom of the tank. When the stick is withdrawn, the oil will leave a mark on the stick consistent with the level of oil.
How long will 1000l oil last?
1000 litres of heating oil may last one home a year, and it may last one family three months.
Is it cheaper to leave oil heating on constant?
This is a hotly debated one. According to experts at the Energy Saving Trust, the idea it’s cheaper to leave the heating on low all day is a myth. … The Energy Saving Trust says if you’re keeping the heating on all day you’re losing energy all day, so it’s better to heat your home only when you need it.
How much oil should I use a month?
A:Oil consumption per person per month should be less than 500ml. Try to keep the oil consumption of the entire family of 3 between 1000-1500 ml per month. Do not exceed beyond this.
How long should a full tank of oil last in the winter?
If your house is totally average and it’s the middle of winter then a full tank of heating oil will last approximately 3-4 weeks. Remember to always Order at a Quarter!
How long does an oil change take?
Having an oil change completed is an easy commitment. Typically, an oil change takes about 30-45 minutes. The process includes draining out the existing oil and changing the oil filter, then placing new oil into the engine. | 2,219 | 9,612 | {"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-2022-49 | latest | en | 0.947974 |
https://tophat.com/marketplace/engineering/engineering/questions/oer-fluid-flow-saylor-academy/958/ | 1,596,446,701,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735792.85/warc/CC-MAIN-20200803083123-20200803113123-00560.warc.gz | 547,610,576 | 31,053 | # Fluid Flow
Source: Saylor
Student Price: FREE
A question pack on fluid flow in thermal dynamic systems.
Fluid Flow Q42
The equation of continuity (mass conservation) reduces to $\triangledown \cdot$ v = 0 for the special case of ______________.
A
B
Constant density
C
Constant temperature
D
Constant velocity
E
Constant pressure
Fluid Flow Q41
Calculate the Reynolds number for water flowing through a 2‑inch internal diameter pipe at a flow rate of 500 gallons/hour.
A
10
B
100
C
1000
D
10,000
E
100,000
Fluid Flow Q43
Which group describes the ratio of inertial forces to viscous forces?
A
Re
B
Pr
C
Gr
D
St
E
Nu
Fluid Flow Q39
Which group quantifies the ratio of thermal transfer and momentum transfer resistance in a fluid?
A
Re
B
St
C
Gr
D
Pr
E
Sc
Fluid Flow Q40
The ratio of convective heat transfer to conductive heat transfer is commonly represented by which of the following?
A
Pr
B
St
C
Re
D
Nu
E
Gr
Fluid Flow Q38
Which of the following is a state variable?
A
Flow rate
B
Diffusivity
C
Reynolds number
D
Gibbs free energy
E
Location
Fluid Flow Q37
Which of the following is a form of anemometry?
A
Wind sock observations
B
Laser doppler
C
Particle imaging velocimetry
D
Hot wire
E
All of the above
Fluid Flow Q36
Water flowing at 2 gallons per minute passes through a contraction from 1/2 inch internal diameter pipe to a 1/4 inch internal diameter pipe. Use Bernoulli's equation to calculate the pressure difference from immediately before the contraction to in the contraction. This pressure difference is defined as the pressure before the contraction minus the pressure in the contraction.
A
11 mm Hg
B
55 mm Hg
C
1120 mm Hg
D
‑1120 mm Hg
E
‑112 mm Hg
Fluid Flow Q35
Water flowing at 10 gallons per minute passes through a contraction from 1 inch internal diameter pipe to a 1/2 inch internal diameter pipe. Use Bernoulli's equation to calculate the pressure difference from immediately before the contraction to immediately after the contraction. This pressure difference is defined as the pressure before the contraction minus the pressure after the contraction.
A
87 mm Hg
B
17.5 mm Hg
C
1750 mm Hg
D
‑17.5 mm Hg
E
‑175 mm Hg
Fluid Flow Q34
Water flowing at 200 gallons per minute passes through an expansion from 2 inch internal diameter pipe to a 2.25 inch internal diameter pipe. Use Bernoulli's equation to calculate the pressure difference from immediately before the expansion to immediately after the expansion. This pressure difference is defined as the pressure before the contraction minus the pressure after the contraction.
A
110 mm Hg
B
‑56 mm Hg
C
‑1110 mm Hg
D
1110 mm Hg
E
None of the above
Fluid Flow Q33
In studying Bernoulii's equation and the forms for the relationship between pressure drop and flow rate for an orifice plate and venturi meter, the results can be summarized by which of the following statements?
A
Q is proportional to pressure drop.
B
Q is inversely proportional to pressure drop.
C
Q is inversely proportional to the square root of pressure drop.
D
Q is proportional to the square root of pressure drop.
E
None of the above
Fluid Flow Q32
What is the lowest absolute pressure that can exist?
A
14.7 psi
B
‑14.7 psi
C
‑29.4 psi
D
0 psi
E
‑1 atm
Fluid Flow Q31
Air is flowing through a pipe with an internal diameter of 15 cm. It passes through an orifice of diameter 8 cm. The air density at the operating conditions is about 1.3 kg/m$^3$. The observed pressure difference across the orifice is approximately 100 mm Hg, and the flow coefficient for the orifice is approximately 0.7. What is the mass flow rate of air?
A
0.506 kg/s
B
0.503 kg/s
C
6.5 kg/s
D
0.05 kg/s
E
0.65 kg/s
Fluid Flow Q44
Water is flowing through a pipe with an internal diameter of 15 cm. It passes through an orifice of diameter 8 cm. The fluid density at the operating conditions is about 1000 kg/m$^3$. The observed pressure difference across the orifice is approximately 100 mm Hg, and the how coefficient for the orifice is approximately 0.7. What is the mass flow rate of water?
A
2 kg/s
B
8.1 kg/s
C
18 kg/s
D
180 kg/s
E
81 kg/s
Fluid Flow Q30
A liquid with 5.3 times the density of water is used in a manometer. A pressure difference of 4 psi corresponds to how many mm head of this liquid?
A
53
B
530
C
5.3
D
5300
E
22
Fluid Flow Q29
A liquid with a density 8 times that of water is used in a manometer. A pressure difference of 0.25 atm corresponds to how many mm head of that liquid?
A
32
B
3200
C
320
D
‑32
E
3.2
Fluid Flow Q28
Which of the following best characterizes the inner channel of a rotameter?
A
It is typically close to perfectly cylindrical.
B
It is typically tapered from top to bottom.
C
It is typically roughened to promote turbulence.
D
It is typically notched for each
E
None of the above
Fluid Flow Q27
A sharp crested, contracted, rectangular weir of length 12 ft and height 1 ft exhibits a head (h0) of 2 inches of water. Calculate the water flow in ft$^3$/s. Assume the head of velocity approach is small.
A
0.2 ft$^3$/s
B
0.4 ft$^3$/s
C
20 ft$^3$/s
D
2.6 ft$^3$/s
E
1 ft$^3$/s
Fluid Flow Q26
Which of the following statements about pump affinity laws is true?
A
Pump affinity laws describe similarity to other pumps.
B
Pump affinity laws describe cost versus performance.
C
Pump affinity laws describe relationship of geometric variables and dynamic performance.
D
Pump affinity laws describe dynamic head versus fluid viscosity.
E
None of the above
Fluid Flow Q25
When does cavitation occur?
A
When vapor appears in the suction line
B
When the net postive suction head at the pump input is less than the vapor pressure of the liquid
C
When the liquid to be pumped begins to boil in the suction line
D
All of the above
E
None of the above
Fluid Flow Q24
In considering the exits of two pumps or blowers configured in parallel, which of the following statements is true?
A
The flow rates from the two pieces of equipment are equal.
B
The total mass flow rate is the sum of the mass flow rates from each piece of equipment.
C
The temperatures of the streams from each piece of equipment are equal.
D
The pressures in the output streams from each piece of equipment are nearly equal.
E
Both A and C
F
Both B and D
Fluid Flow Q23
In considering two pumps (1 and 2) connected in series, which of the following is true?
A
The flow rates are the same in 1 and 2.
B
The inlet pressures are the same for 1 and 2.
C
The outlet pressures are the same for 1 and 2.
D
The inlet pressure of one of the pumps is equivalent to the outlet pressure of the other pump.
E
Both A and D
Fluid Flow Q22
For both pumps and fans, the system resistance varies as the ______________.
A
Square root of flow rate
B
Cube root of flow rate
C
Flow rate
D
Flow rate squared
E
Cube of flow rate
Fluid Flow Q21
Air is moved through a conduit at a volumetric rate of 100 m$^3$/min. The conditions are P=18 psi, T = 27°C. What is the mass flow rate of air in the conduit?
A
2.4 lb/min
B
2.4 g/s
C
2.4 kg/min
D
2.4 kg/s
E
2.4 kg/hr
Fluid Flow Q20
Positive displacement pumps do which of the following?
A
Produce a flow rate independent of operating speed
B
Produce a flow rate independent of discharge pressure
C
Produce a flow rate independent of suction head
D
Usually require a relief or safety valve
E
Both B and D
Fluid Flow Q19
A centrifugal pump is used to lift water at five gallons per minute against a head of 50 feet of water. The efficiency of the pump is 70%. Approximately, what is the power required by the pump?
A
0.7 kW
B
0.07 kW
C
7 kW
D
70 kW
E
750 kW
Fluid Flow Q18
A centrifugal pump is used to lift water at fifty gallons per minute against a head of 75 feet of water. The efficiency of the pump is 65%. Approximately, what is the power required by the pump?
A
0.1 kW
B
0.01 kW
C
1 kW
D
10 kW
E
100 kW
Fluid Flow Q17
A centrifugal pump is used to lift water at 75 gallons per minute against a head of 175 feet of water. The efficiency of the pump is 75%. Approximately, what is the power required by the pump?
A
0.33 kW
B
3.3 W
C
3.3 hp
D
3.3 tons
E
3.3 kw
Fluid Flow Q16
You have a pump that is using 8.5 kW of power to pump water at about 450 gallons per minute against a head of 32 psi. What is the operating efficiency of the pump?
A
93%
B
85%
C
79%
D
74%
E
65%
Fluid Flow Q15
What is the primary difference between a pump and a blower or fan?
A
Blowers have an unconstrained exit stream, and pumps do not.
B
Blowers move gases, and pumps move liquids and gases.
C
Pumps pressurize liquids, and blowers move gases.
D
Pumps operate at lower pressures than blowers.
E
Pumps operate on incompressible fluids, and blowers operate on compressible fluids.
Fluid Flow Q14
One atmosphere of pressure corresponds to a water head of how many feet?
A
33.9 ft
B
3.39 ft
C
340 ft
D
3340 ft
E
None of the above
Fluid Flow Q13
Water flows through a 1 inch ID 90‑degree elbow at 100 L/min. What is the pressure drop associated with the fitting?
A
3.3 Pa
B
2.2 Pa
C
2.2 kPa
D
3.7 kPa
E
440 kPa
Fluid Flow Q11
Water flows through a 2 inch ID 90‑degree elbow at 300 L/min. What is the pressure drop associated with the fitting?
A
1 psi
B
2.5 psi
C
0.25 psi
D
43 psi
E
10 psi
Fluid Flow Q12
A fluid with density 0.78 g/cm$^3$ and viscosity 2.3 cP flows at 1 gallon per minute through a 2 inch internal diameter pipe. The flow should be considered as which of the following?
A
Turbulent
B
Laminar
C
Transitional
D
None of the above
E
All of the above
Fluid Flow Q10
For gas flow in a pipe, as the pressure decreases downstream the gas velocity does which of the following?
A
Remains constant
B
Decreases
C
Increases
D
Doubles
E
None of the above
Fluid Flow Q9
A centrifugal pump can provide a flow of 20 L/min at a pressure of 20 psi above atmospheric pressure. What is the minimum pipe diameter $(\epsilon/D = 0.002)$ that can be used to carry water for 50 m?
A
5 cm
B
0.6 cm
C
2.0 cm
D
15 cm
E
7 cm
Fluid Flow Q8
A centrifugal pump can provide flow of 100 L/min at 1.7 atm gauge. What is the minimum pipe diameter $(\epsilon/D=0.005)$that can be used to carry water under these conditions for 100 m?
A
1.4 cm
B
2.5 cm
C
3.9 cm
D
5.0 cm
E
1.1 cm
Fluid Flow Q7
Important factors for choosing a pipe material include which of the following?
A
Presence/absence of organic chemicals in the pipe environment
B
C
Brittleness
D
Ease of corrosion
E
All of the above
Fluid Flow Q6
How are pipe diameters most commonly specified and tabulated?
A
Internal diameter
B
Internal and external diameter
C
External diameter (OD) and wall thickness
D
External Diameter
E
None of the above
Fluid Flow Q5
A piece of analytical equipment employs very narrow diameter tubing to supply solvent (density = 0.9 g/cm$^3$, viscosity = 0.78 cP) to a workstation at a flow rate of 5 ml/min. The internal diameter of the tubing is 1000 $\mu$ m and the length of the tubing is 75 cm. What is the pressure drop in the tubing?
A
0.5 atm
B
1 atm
C
0.2 atm
D
0.02
E
5 atm
Fluid Flow Q4
Water flows through a 3 inch internal diameter 75 ft long pipe at 0.1 gallon per minute. The surface roughness of the pipe is 0.02 cm. What is the pressure drop over the entire length of the pipe?
A
1.8 kPa
B
1800 Pa
C
17.6 Pa
D
1.16 Pa
E
0.18 Pa
Fluid Flow Q3
Water flows through a 5 inch internal diameter pipe 50 m long at 5000 liters /min. The roughness of the pipe is characterized by $\epsilon$ /D = 0.002. What is the pressure drop over the entire length of pipe?
A
24 kPa
B
90 Pa
C
900 Pa
D
9 kPa
E
45 kPa
Fluid Flow Q2
A fluid of density 0.83 g/cm$^3$ and viscosity 1.8 cP flows through a 1 inch internal diameter pipe 50 m long at 80 kg/min. The roughness of the pipe is characterized by $\epsilon$ /D = 0.005. What is the pressure drop over the entire length of pipe?
A
112 kPa
B
269 Pa
C
2.69 Pa
D
524 kPa
E
262 kPa
Fluid Flow Q1
Pressure drops for elements in series are ______________.
A
B
Multiplicative
C
Equal
D
Related reciprocally
CC BY 3.0 - Saylor - Steve Gibbs | 3,448 | 12,396 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-34 | latest | en | 0.846489 |
https://www.mathworks.com/matlabcentral/answers/433851-how-to-add-lines-to-histograms-plotted-by-plotmatrix?s_tid=prof_contriblnk | 1,679,856,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00066.warc.gz | 993,724,759 | 26,398 | # how to add lines to histograms plotted by plotmatrix
4 views (last 30 days)
Etsuko on 4 Dec 2018
Commented: Etsuko on 5 Dec 2018
Hi,
I want to add vertical lines to histograms plotted in the diagonal of a matrix of pair-wise scatter plots. I tried to use the graphic objects returned by plotmatrix;
[S,AX,BigAx,H,HAx] = plotmatrix(X); % X has 15 columns
for example,
subplot(15,15,1); hold on; line([1.2,1.2],get(AX(1),'YLim'),'Color','r')
AX(15); plot([1.2,1.2],[0,1.5],'Color','r')
etc. But so far none has worked out. I would appreciate any clue as I am clueless.
I use Matlab 2018a. Thanks.
Akira Agata on 4 Dec 2018
How about the following?
% Sample data
X = rand(50,3);
figure
[~,ax] = plotmatrix(X);
hold(ax(1,3),'on')
plot(ax(1,3),[0.2 0.2],ax(1,3).YLim,'r:','LineWidth',2)
hold(ax(2,1),'on')
plot(ax(2,1),[0.5 0.5],ax(2,1).YLim,'r','LineWidth',2)
Just FYI:
If you can upgrade your MATLAB to R2018b, you can do the same thing slightly easier by using newly introduced xline function, like:
figure
[~,ax] = plotmatrix(X);
xline(ax(1,3),0.2,'r:','LineWidth',2)
xline(ax(2,1),0.5,'r','LineWidth',2)
##### 1 CommentShowHide None
Etsuko on 5 Dec 2018
Thank you very much!
It needed to specify the axis of the histograms. Just an easy modification of your suggestion did what I wanted. Thank you!
[S,AX,BigAx,H,HAx] = plotmatrix(X); % X has 15 columns
hold(HAx(1,1),'on')
plot(HAx(1,1),[2.2 2.2],HAx(1,1).YLim,'r:','LineWidth',2)
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Start Hunting! | 553 | 1,638 | {"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-2023-14 | longest | en | 0.747709 |
http://planetmath.org/ProofOfEulerMaclaurinSummationFormula | 1,521,873,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00483.warc.gz | 231,190,749 | 5,521 | # proof of Euler-Maclaurin summation formula
Let $a$ and $b$ be integers such that $a, and let $f:[a,b]\to\mathbb{R}$ be continuous. We will prove by induction that for all integers $k\geq 0$, if $f$ is a $C^{k+1}$ function,
$\sum_{a (1)
where $B_{r}$ is the $r$th Bernoulli number and $B_{r}(t)$ is the $r$th Bernoulli periodic function.
To prove the formula for $k=0$, we first rewrite $\int_{n-1}^{n}f(t)\mathrm{d}t$, where $n$ is an integer, using integration by parts:
$\displaystyle\int_{n-1}^{n}f(t)\mathrm{d}t$ $\displaystyle=$ $\displaystyle\int_{n-1}^{n}\frac{\mathrm{d}}{\mathrm{d}t}(t-n+\frac{1}{2})f(t)% \mathrm{d}t$ $\displaystyle=$ $\displaystyle(t-n+\frac{1}{2})f(t)\big{|}_{n-1}^{n}-\int_{n-1}^{n}(t-n+\frac{1% }{2})f^{\prime}(t)\mathrm{d}t$ $\displaystyle=$ $\displaystyle\frac{1}{2}(f(n)+f(n-1))-\int_{n-1}^{n}(t-n+\frac{1}{2})f^{\prime% }(t)\mathrm{d}t.$
Because $t-n+\frac{1}{2}=B_{1}(t)$ on the interval $(n-1,n)$, this is equal to
$\int_{n-1}^{n}f(t)\mathrm{d}t=\frac{1}{2}(f(n)+f(n-1))-\int_{n-1}^{n}B_{1}(t)f% ^{\prime}(t)\mathrm{d}t.$
From this, we get
$f(n)=\int_{n-1}^{n}f(t)\mathrm{d}t+\frac{1}{2}(f(n)-f(n-1))+\int_{n-1}^{n}B_{1% }(t)f^{\prime}(t)\mathrm{d}t.$
Now we take the sum of this expression for $n=a+1,a+2,\ldots,b$, so that the middle term on the right telescopes away for the most part:
$\sum_{n=a+1}^{b}f(n)=\int_{a}^{b}f(t)\mathrm{d}t+\frac{1}{2}(f(b)-f(a))+\int_{% a}^{b}B_{1}(t)f^{\prime}(t)\mathrm{d}t$
which is the Euler-Maclaurin formula for $k=0$, since $B_{1}=-\frac{1}{2}$.
Suppose that $k>0$ and the formula is correct for $k-1$, that is
$\sum_{a (2)
We rewrite the last integral using integration by parts and the facts that $B_{k}$ is continuous for $k\geq 2$ and $B_{k+1}^{\prime}(t)=(k+1)B_{k}(t)$ for $k\geq 0$:
$\displaystyle\int_{a}^{b}B_{k}(t)f^{(k)}(t)\mathrm{d}t$ $\displaystyle=$ $\displaystyle\int_{a}^{b}\frac{B_{k+1}^{\prime}(t)}{k+1}f^{(k)}(t)\mathrm{d}t$ $\displaystyle=$ $\displaystyle\frac{1}{k+1}B_{k+1}(t)f^{(k)}(t)\big{|}_{a}^{b}-\frac{1}{k+1}% \int_{a}^{b}B_{k+1}(t)f^{(k+1)}(t)\mathrm{d}t.$
Using the fact that $B_{k}(n)=B_{k}$ for every integer $n$ if $k\geq 2$, we see that the last term in Eq. 2 is equal to
$\frac{(-1)^{k+1}B_{k+1}}{(k+1)!}(f^{(k)}(b)-f^{(k)}(a))+\frac{(-1)^{k}}{(k+1)!% }\int_{a}^{b}B_{k+1}(t)f^{(k+1)}(t)\mathrm{d}t.$
Substituting this and absorbing the left term into the summation yields Eq. 1, as required.
Title proof of Euler-Maclaurin summation formula ProofOfEulerMaclaurinSummationFormula 2013-03-22 13:28:41 2013-03-22 13:28:41 pbruin (1001) pbruin (1001) 5 pbruin (1001) Proof msc 65B15 | 1,144 | 2,622 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 46, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-13 | latest | en | 0.563917 |
http://www.jiskha.com/display.cgi?id=1381961728 | 1,496,010,650,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611569.86/warc/CC-MAIN-20170528220125-20170529000125-00028.warc.gz | 677,874,139 | 4,041 | # Math
posted by on .
1. The US Senate is composed of 2 senators from each of the 50 states. For a treaty to be ratified, at least two-thirds of the senators present must approve the treaty. Suppose all senators are present and 48 of them have voted in favor of a treaty. What are the possible numbers of additional senators who must vote in favor of the treaty to ratify it?
at least 12 senators
at least 18 senators
at least 19 senators
at least 20 senators
How do you solve this?
• Math - ,
2/3 * 100 = ? senators needed to pass the treaty. (Hint: round up since we don't usually see a fraction of a senator.)
• Math - ,
Ok got it thanks!
• Math - ,
You're welcome. | 169 | 678 | {"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-2017-22 | latest | en | 0.969406 |
https://www.pharmaceuticalsky.com/2020/06/sop-for-rounding-off-of-analytical-test.html | 1,679,640,016,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00553.warc.gz | 1,048,331,743 | 53,957 | # SOP for Rounding off of Analytical Result
• Rounding of the figure is a mathematical method in which replacing one number with another number to get the approximate and nearby value.
• The rounding of figures is important because it gives the final reporting results an accurate and easier form.
• Rounding of figures can be done for both rounding of decimals (e.g. 10.769 rounded to 10.78) and rounding of whole numbers (3350 rounded to 3300).
• Mostly in the pharma industry, the final results are calculated either manually or by using validated excel sheets or validated calculators which are prepared in an automated way to give the final results in rounding figures.
OBJECTIVE:
To lay down a procedure for establishing guidelines for reporting of analytical values and rounding off the digits.
SCOPE:
This Procedure is applicable to Rounding off of Analytical Results at a quality control laboratory.
RESPONSIBILITY:
Execution: Chemist and above – QC department.
Checking: Asst. Manager and above – QC department.
PROCEDURE:
• After completion of the analysis, the analytical chemist shall calculate the analytical result in the work sheet and compare with the stated limits in the specification.
• If the calculated value(s) contains more digits in the decimal place than that in the stated limit, the digits after decimal place shall be rounded off as described below.
• Rounding off shall be done by considering only one digit in the decimal place to the right of the last place in the limit expression as shown in the illustration below.
• If the digit is smaller than 5, it shall be eliminated and the preceding digit shall remain unchanged
• If the digit is greater than or equal to 5, it shall be eliminated and the preceding digit shall be increased by one.
Illustration of Rounding Numerical Values for comparison with requirements
Compilations Requirement Unrounded value Rounded results Conforms Assay limit > 98.0% 97.96% 98.0% Yes 97.92% 97.9% No 97.95% 98.0% Yes Assay limit < 101.5% 101.55% 101.6% No 101.46% 101.5% Yes 101.45% 101.5% Yes Limit test < 0.02% 0.025% 0.03% No 0.015% 0.02% Yes 0.027% 0.03% No Limit test < 3ppm 0.00035% 0.0004% No 0.00025% 0.0003% Yes 0.00028% 0.0003% Yes
• When the analytical method or test requires quantification, in such cases, the analytical result shall be expressed as observed numerical value.
• When the analytical method or test does not require quantification, in such cases the analytical result shall be expressed as “Conform” or “does not conform” as the case may be on Certificate of Analysis. However, in the work sheet, the actual observation of analysis shall be specified (for example: Less than 50ppm, colour intensity of sample solution is not more than that of standard solution “x” or formation of specific colour to the solution or formation of precipitate).
ABBREVIATION
SOP : Standard Operating Procedure
QC : Quality Control
% : Percentage
ANNEXURE
Nil
REVISION HISTORY
Nil | 716 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.895444 |
https://www.findaphd.com/phds/project/design-of-experiments-estimation-of-selected-interactions-in-factorial-experiments/?p133667 | 1,657,150,046,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104678225.97/warc/CC-MAIN-20220706212428-20220707002428-00615.warc.gz | 816,921,682 | 47,218 | # Design of Experiments: estimation of selected interactions in factorial experiments
### University of SurreyDepartment of Mathematics
Dr Janet Godolphin Friday, July 22, 2022 Competition Funded PhD Project (UK Students Only)
Experiments with factorial treatment structure are used widely in industrial experiments and have applications in many other areas, including medical research and the horticultural and agricultural industries. Their importance lies in their capacity to provide estimates of main effects and interactions with relatively small numbers of runs.
This project focuses on the construction of blocked or fractional factorial designs, to enable the estimation of all main effects and of selected interactions. Work in the literature predominantly focuses on designs with all factors at two levels. Further, designs are typically constructed under the assumption of the effect hierarchy principle, namely that “for given m, all m-factor interactions are equally important”. Thus, much of the existing work in this area uses minimum aberration criteria to maximise the number of estimable two factor interactions in designs with factors all at two levels.
In practice, knowledge of an experimental situation can mean that a subset of interactions is of interest, with the remaining interactions assumed to be negligible. It can be shown that designs constructed according to minimum aberration criteria are not necessarily the most appropriate designs in this situation. The project seeks design construction approaches with a focus on this scenario.
Two specific research directions are proposed:
(i) Use of proper vertex colouring approaches in graph theory to aid construction of bespoke factorial and fractional factorial designs, with one form of blocking, enabling estimation of main effects and selected interactions. The aim is to extend work completed in (1) to construction of designs with factors at more than two levels and to encompass three factor interactions.
(ii) Investigation of the various “folding” techniques available to de-alias effects of interest for factorial designs where sequential sets of small numbers of runs are planned. Regular and non-regular fractional factorial designs will be considered, both with and without blocking. See (2) and (3) for background reading.
The successful candidate will receive comprehensive research training related to all aspects of the research and opportunities to participate in conferences, workshops and seminars to develop professional skills and research network.
Supervisor: Dr Janet Godolphin
This is a minimum 3 year project. We are able to offer this opportunity starting in January 2022, April 2022 or October 2022.
Entry requirements
Applicants should have a minimum of a first class honours degree in mathematics, the physical sciences or engineering. Preferably applicants will hold a MMath, MPhys or MSc degree, though exceptional BSc students will be considered.
English language requirements: IELTS Academic 6.5 or above (or equivalent) with 6.0 in each individual category.
How to apply
Applications should be submitted via the Mathematics PhD programme page on the "Apply" tab.
Please state clearly the studentship project at you would like to apply for.
## Funding Notes
Full UK tuition fees and a tax-free stipend. This project is on offer in competition with a number of other projects for funding. This opportunity may be available with partial funding for overseas fees for exceptional applicants. However, funding for overseas students is limited and applicants are encouraged to find suitable funding themselves. Funded by the University of Surrey.
## References
(1) Godolphin, J.D., 2021. Construction of blocked factorial designs to estimate main effects and selected two-factor interactions, Journal of the Royal Statistical Society, Series B, 83, 5-29.
(2) Li, F., Jacroux, M., 2007. Optimal foldover plans for blocked fractional factorial designs. Journal of Statistical Planning and Inference, 137, 2439-2452.
(3) Li, W., Lin, D.K.J., 2003. Optimal foldover plans for two-level fractional factorial designs. Technometrics, 45, 142-149.
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Based on your current search criteria we thought you might be interested in these. | 849 | 4,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.925797 |
http://newarthurianeconomics.blogspot.com/2013/07/arithmetic-and-integrity.html | 1,519,324,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814249.56/warc/CC-MAIN-20180222180516-20180222200516-00073.warc.gz | 256,696,047 | 26,975 | ## Wednesday, July 31, 2013
### Arithmetic and Integrity
These are the relations between GDP, inflation-adjusted GDP, and prices:
That is a fact. So, what does the arithmetic tell us? It says
1. If you divide "nominal" GDP by "real" GDP, you get "prices".
2. If you divide "nominal" GDP by "prices", you get "real" GDP. And
3. If you multiply "real" GDP by "prices" you get "nominal" GDP.
The three series of numbers considered here are more closely tied than braided hair. You can always convert nominal GDP values to real, by dividing prices out. You can always convert real GDP values to nominal, by multiplying prices in.
Again: The relations between these series is such that if you divide NGDP by RGDP it gives you prices; if you divide NGDP by prices you get RGDP; and if you multiply RGDP by prices you get NGDP. These are the relations, and nothing can be done about it.
On Sunday I wrote:
Unit Labor Cost is Employee Comp multiplied by the price level...
They take numbers like Employee Compensation going down relative to GDP. They times it by prices to make the numbers go up. They say Look, look! Labor costs are going up! And they claim that rising labor costs are pushing prices up.
I don't know which employee cost data is used to figure Unit Labor Cost. I know I'm in the ballpark because the lines on the graph were very close. But that's not the point. The point is, they multiply the employee cost numbers by the price level to get the Unit Labor Cost numbers.
It is a point I've made before. In mine of 6 October 2012, I wrote:
Kaminska seems to think the graph shows labor cost. How does she describe it? "The labour cost attached to the production of one unit". Oh right, right: "One unit".
As I showed the other day, the Unit Labor Cost plot is almost identical to the price level plot:
Graph #2: Unit Labor Cost (blue) and the price level (red)
The similarity between ULC and the price level is so remarkable as to inspire disbelief. And well it should, for the red line is used to calculate the blue line. To calculate Unit Labor Cost, labor costs are multiplied by prices.
And then the graph is used to claim that Labor cost makes prices go up.
The first comment on that old post disputes my view:
I'm afraid you've got everything upside down. Unit labour costs are exactly as described, the cost of one unit of output. Normally when we say this, we mean nominal unit costs. Since labour costs account for most costs, as labour costs go up so do prices. It is therefore not remotely surprising that the labour cost line and the price line move together.
Labour costs are NOT multiplied by prices. They are a nominal variable which at least partially drive prices.
There is nothing about multiplying by prices involved. Nominal gdp does go up as prices go up or or as real gdp, which is nominal gdp divided by prices, goes up. It is the nominal gdp which is the directly observed measure.
The similarity between the two lines is "so remarkable as to inspire disbelief," I said. My anonymous commenter's view is that the similarity is not remotely surprising because "labour costs account for most costs". I had to laugh at the difference of opinion. I do think that someone with enough knowledge of math and economics could determine with certainty which view is closer to the truth. This one has an answer.
Meanwhile, I stand by my view. The two lines are inordinately, unjustifiably similar. More similar than can be accounted by labor's share of cost which, unlike prices, has been falling for 30 years. The similarity is artificial.
The similarity is created when labor costs are multiplied by prices.
That brings me to the second point in the comment: "There is nothing about multiplying by prices involved," my anonymous friend writes. "Nominal gdp does go up as prices go up or or as real gdp, which is nominal gdp divided by prices, goes up."
Real GDP is Nominal GDP divided by prices, he says. He's right about that. So if we are dividing something by Real GDP, we can instead divide by "Nominal GDP divided by Prices" and get the same result. The same result, and better transparency:
But we're dividing by a fraction here. Do you remember how to do that? "Invert and multiply." To divide by a fraction, invert the fraction and multiply. We can do that:
The fraction "Nominal GDP over Prices" becomes "Prices over Nominal GDP", and the "divided by" symbol gets replaced by a multiply. But now that we're multiplying, we can rearrange the calculation a bit more:
Now it is obvious that we are dividing our original number by Nominal GDP and multiplying by prices. This is the arithmetic: You can divide something by Real GDP, or you can divide it by Nominal GDP -- actual GDP -- and multiply by prices. Either way you get the same answer.
Either way, you get the same answer.
For figuring unit labor costs, the commenter says, "Labour costs are NOT multiplied by prices ... There is nothing about multiplying by prices involved." But Unit Labor Cost is total labor compensation divided by real GDP. And dividing by "real GDP" gives the same answer as dividing by actual GDP and multiplying by prices.
Actual GDP is GDP at the prices we actually paid to buy it. They call it "nominal".
If Real GDP was "the directly observed measure" then there would be nothing wrong with the Unit Labor Cost calc. But that's not the case. Real GDP is an artificial measure, created by stripping price changes out of actual GDP.
Or if they didn't multiply prices into labor cost, and compare the resulting numbers to prices (and discover an unbelievable similarity) then what they are doing might be okay. But the arithmetic is not okay, because they multiply prices into labor cost and create the similarity they pretend to discover.
When they choose to divide labor costs by "real" GDP rather than actual GDP, they are choosing to multiply labor costs by prices. So doing, they create the appearance of similarity between Unit Labor Cost and prices. To use this artifice as evidence that labor costs have been pushing prices up is an abomination.
daphne millar said...
I'm not sure what point you are trying to make with all this. Think of it at the micro level. GDP is made up of cars, roads, services and so on. Unit Labor Costs are a nominal variable. The cost of the labor component of a car is the Unit Labor Cost times the units of labor (hours, weeks, whatever.) That in turn decides the price. So if the unit labor cost goes up so does the price of the car.
There's no complicated calculation here with real gdp or something. Rising productivity is already incorporated in the ULC figure.
So what's the issue? You write as if there is some plot to fool people. I think this is a pity because it is wholly irrelevant to your really important insight, which is that private debt not public debt is the source of the problem.
The Arthurian said...
Hi Daphne, thanks for that last.
My point has to to do neither with micro nor macro, but with arithmetic. When you strip inflation out of the denominator of a fraction and leave it in the numerator, the resulting values are pushed upward in proportion to the price level.
The problem is that the resulting values are used as evidence that labor costs are responsible for inflation. As you put it: "if the unit labor cost goes up so does the price of the car."
Where are the studies of "unit profit cost" and "unit rent cost" and "unit finance cost"?? The only factor of production that gets held up for inspection is labor cost. But labor costs are going down as a share of actual GDP.
If the problem is private debt, then the problem is the *cost* of that debt. But how can I show this is the problem if corrupt arithmetic is accepted as proof that labor cost is the problem?
I majored in math. But I spent 20 years reading Free to Choose and Money Mischief before I figured out Friedman had stripped inflation from the denominator of his ratio. Is there a plot to fool people? I have no idea. People are fooled nonetheless.
The Arthurian said...
... Maybe ten years. It was a long time, I know that.
Daphne, thanks for a thoughtful critique. | 1,783 | 8,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-09 | longest | en | 0.962389 |
https://convert-dates.com/days-before/4130/2023/05/26 | 1,685,718,466,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00600.warc.gz | 209,354,060 | 4,051 | ## 4130 Days Before May 26, 2023
Want to figure out the date that is exactly four thousand one hundred thirty days before May 26, 2023 without counting?
Your starting date is May 26, 2023 so that means that 4130 days earlier would be February 3, 2012.
You can check this by using the date difference calculator to measure the number of days before Feb 3, 2012 to May 26, 2023.
February 2012
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February 3, 2012 is a Friday. It is the 34th day of the year, and in the 5th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 29 days in this month. 2012 is a leap year, so there are 366 days in this year. The short form for this date used in the United States is 02/03/2012, and almost everywhere else in the world it's 03/02/2012.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 4130 weekdays before May 26, 2023, you can count up each day skipping Saturdays and Sundays.
Start your calculation with May 26, 2023, which falls on a Friday. Counting forward, the next day would be a Monday.
To get exactly four thousand one hundred thirty weekdays before May 26, 2023, you actually need to count 5782 total days (including weekend days). That means that 4130 weekdays before May 26, 2023 would be July 27, 2007.
If you're counting business days, don't forget to adjust this date for any holidays.
July 2007
• Sunday
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July 27, 2007 is a Friday. It is the 208th day of the year, and in the 208th week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 31 days in this month. 2007 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 07/27/2007, and almost everywhere else in the world it's 27/07/2007.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 928 | 2,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.929897 |
https://indicatorchart.com/linear-regression-channel-indicator/ | 1,716,129,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00259.warc.gz | 276,634,425 | 18,887 | # Linear Regression Channel Indicator for MT4
The Linear Regression Channel Indicator is created by first calculating the linear regression line for the price data. The linear regression line is then extended to create upper and lower parallel lines.
These lines form the Cap channel within which prices are expected to move. The Linear Regression Channel Indicator can be used to trade any market, but is most commonly used in Forex trading.
## Linear Regression Channel Indicator For MT4 Trading Signals
When using the Linear Regression Channel Indicator to trade Forex, it is important to remember that price action tends to respect the upper and lower limits of the channel. As such, traders may look for buying opportunities when prices approach the lower limit Scalper Indicators of the channel and selling opportunities when prices approach the upper limit of the channel.
Linear regression is a statistical method that can be used to show the relationship between two variables, in this case, price and time. The Linear Regression Channel indicator uses this relationship to create a channel around the price action.
## Linear Regression Channel indicator Formula
The Linear Regression Channel has three lines: the center line, which is the linear regression line itself; the upper line, which is created by adding a certain number of standard deviations above the center line; and the lower line, which is created by subtracting Trend Strategy a certain number of standard deviations below the center line. These lines are then used to create a channel around the price action.
The purpose of the Linear Regression Channel indicator is to show trends and reversals in the market. When prices are trending upwards, they will tend Non-Repainting to stay within the upper and lower lines of the channel. However, when prices reverse and begin to trend downwards, they will break below the lower line of the channel.
## Some alternative indicators to linear regression channel indicator include:
• Bollinger Bands
• MACD
• RSI
• Fibonacci Retracements
• Stochastic Oscillator
### How to Trad with Linear Regression Channel MT4 Strategy
The indicator is based on the linear regression channel, which is a technical analysis tool used to predict the direction of prices. The Linear Regression Channel Indicator for MT4 is easy to use and can be added to your charts in just a few clicks.
Once you’ve added the indicator to your chart, you’ll need to set the parameters according to your preferences. You can choose the period Ichimoku Strategy of the linear regression, the price source, and the MA type.
it can be an effective addition to any trading strategy. However, like all indicators, it should be used with caution as false readings are possible. This can be used as a signal to enter into short positions. Conversely, when prices break above the upper line of the channel after trending downwards,
this can be used as a signal to enter into long positions. It should be noted that false signals can occur with this indicator, so it is important to use other indicators in conjunction with it in order to confirm any potential trades that may be generated. | 608 | 3,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.890127 |
https://www.physicsforums.com/threads/friction-impulsive-or-non-impulsive.766337/ | 1,532,037,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00551.warc.gz | 981,296,794 | 12,863 | # Friction : impulsive or non impulsive?
1. Aug 17, 2014
### swayne221b
Maximum value of friction depends upon the normal reaction and normal reaction can be impulsive, so does that imply friction can be impulsive?
2. Aug 17, 2014
### jbriggs444
Sure. A couple of examples that spring to mind are the application of "English" to the cue ball on a pool table and the behavior of a spinning rubber ball as it bounces.
Edit: And the application of topspin to a tennis shot as the racket hits the ball.
Last edited: Aug 17, 2014
3. Aug 17, 2014
### WhatIsGravity
There are a bunch of different kinds of friction. It depends on the surfaces, forces, and definitely the impulses. If you account for pressure drag as a part of friction, it's kinda like skipping a rock. | 197 | 772 | {"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-2018-30 | latest | en | 0.90551 |
https://torusacademia.com/course/pure-mathematics-p1-ial-wma11-01/86 | 1,721,056,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00275.warc.gz | 509,218,637 | 16,032 | # Pure Mathematics P1-(IAL-WMA11/01)
Edexcel International Advanced Level (IAL) Mathematics. It is a pure math unit that covers topics such as algebraic manipulation, functions, trigonometry and calculus.
Advanced Level 5(1 Ratings) 1 Students enrolled
Created by Raja Nadeem Amin Last updated Wed, 10-Jan-2024 English
What will i learn?
• This course will teach you everything you need to know about Pure Math P1 IAL. You will learn how to apply algebra, geometry, trigonometry and calculus to solve various problems. You will also develop your logical thinking and analytical skills. By the end of this course, you will be able to confidently tackle any question in the exam and achieve your desired grade. Students will be able to master the paper techniques, time management to get top grades.
Curriculum for this course
8 Topics
Algebraic Fraction
3 Sub Topics
• Surds
• Rationalising Denominators
7 Sub Topics
• Completing the Square
• Nature of Roots
• Sketching of Graphs.
• Simultaneous Equations
• Inequalities on Graphs
• Inequalities on Graphs
Graphs and Transformations
3 Sub Topics
• Sketching of Cubic Graphs
• Sketching of Reciprocal Graphs
• Transfomations
Coordinate Geometry
1 Sub Topics
• Equation of Straight Lines
Trigonometry
3 Sub Topics
• Solution of Trigonometrical Equations
• Trigonometrical Graphs
• Basic Trigonometrical Applications
1 Sub Topics
• Arc Length and Area of Sectors
Differentiation
3 Sub Topics
• Differentiating Technique
• Equation of Tangents and Normal
• 2nd derivative
Integration
1 Sub Topics
• Integration Techniques
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Title Lectures Price
Inquiry Based Session-1 lesson 1 25\$
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Description
Do you want to master Pure Math P1 IAL and ace your exams? If yes, then this online course is for you. You will learn all the concepts and skills you need to solve any problem from the book, the worksheets and the past papers. You will also get to practice with real exam questions and get feedback on your performance. This course will last one hour per class and will cover every aspect of the syllabus in depth. Don’t miss this opportunity to boost your confidence and grades in Pure Math P1 IAL. Book your course today and get ready to succeed.
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Hi, my name is Raja Nadeem Amin. I have the teaching experience of almost twenty three years in the domain of assisting Mathematics to O’ Levels (IGCSE, GCSE & Edexcel), Add Mathematics (GCSE & Edexcel), Pure Mathematics, Mechanics, Statistics to A Levels and teaching Mathematics. I will not only help you with your examinations but also with any mathematical works, presentations, queries in the friendliest way. I will be teaching you in a very flexible way, with computer animations of mathematical concepts. If you don’t know the basics of the subject, or want to improve your mathematical skills, I am excited to help achieve your goals. In the first lesson we will briefly discuss your goals and I’ll plan the easiest way for you at your own learning pace. I am very excited and looking forward for your first lesson booking!
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I engaged in discussion about topics and past papers with the teacher and the experience was truly exceptional.
Includes:
• 00:00:00 Hours On demand videos
• 8 Topics , 22 Sub Topics
• Access on mobile and tv | 953 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.810339 |
https://www.physicsforums.com/threads/pitcher-throwing-a-ball-in-protest.336448/ | 1,508,654,764,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825147.83/warc/CC-MAIN-20171022060353-20171022080353-00403.warc.gz | 963,144,578 | 17,413 | # Pitcher Throwing A Ball In Protest
1. Sep 11, 2009
### Warmacblu
1. The problem statement, all variables and given/known data
As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of 11.4 m/s. In the process, he moves his hand through a distance of 1.68 m.
Acceleration of gravity is 9.8 m/s2.
If the ball has a mass of 0.168 kg, find the force he exerts on the ball to give it this upward speed. Answer in units of N.
2. Relevant equations
F = ma
A kinematic equation, not sure which one though.
3. The attempt at a solution
F = ma
F = .168 kg * (11.4 / 1.68) = 1.14
F = ma
F = .168 kg * (1.68 / 11.4) = .025
I did look at old threads regarding this question but cannot understand the hints. I guess I need to find a proper acceleration taking into account all the forces acting upon the ball but I do not know which kinematic equation to use.
Any help is appreciated,
Thanks
Last edited: Sep 11, 2009
2. Sep 11, 2009
### Andrew Mason
Try using an energy approach. What is the kinetic energy of the ball when it leaves the pitcher's hand? How does that relate to the force applied by the pitcher over the 1.68 m. distance? Be careful to include the force of gravity and the increase in potential energy over that distance.
AM
Last edited: Sep 11, 2009
3. Sep 11, 2009
### Warmacblu
We haven't really discussed energy yet which could contribute to why I am having difficulty understanding this problem.
However, I looked ahead a bit and found that k = 1/2mv2.
I do not know if that v is initial velocity or not and I do not know how I can relate it to the force if I can't figure out an acceleration for the force equation.
4. Sep 11, 2009
### Staff: Mentor
Look for a kinematic equation that relates velocity and distance.
5. Sep 11, 2009
### Warmacblu
I believe this is the one:
Vf2 = Vi2 + 2a (Xf - Xi)
6. Sep 11, 2009
### Staff: Mentor
That's the one you need. Assume that the ball starts from rest.
7. Sep 11, 2009
### Warmacblu
Okay, here's what I did:
Vf2 = Vi2 + 2a (Xf - Xi)
0 = 11.4 + 2a (1.68)
a = -11.4 / (2 * 1.68)
a = -3.39
I don't think I can just plug this into F = ma because I haven't taken gravity into consideration, but I don't know where to take that into account.
8. Sep 11, 2009
### Staff: Mentor
You forgot to square the velocity. And you have the wrong sign because you mixed up Vi and Vf. (Vi = 0.)
Using F = ma will tell you the net force. You'll need to solve for the force the player exerts on the ball, taking into account that gravity also acts on the ball.
9. Sep 11, 2009
### Warmacblu
Okay, here's what I have now:
Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
a = 38.68 + (-9.8)
a = 28.88
F = ma
F = .168 * 28.88
F = 4.85
How does that look? Is my math and gravity assumption correct?
10. Sep 11, 2009
### Staff: Mentor
So far, so good. Solve for a.
Do not try to take shortcuts! Solve for a, find the net force, then worry about the force of gravity
Hint: Two forces act on the ball, which combine to give you the net force.
11. Sep 11, 2009
### Warmacblu
Vi2 = Vf2 + 2a (1.68)
11.42 = 0 + 2a (1.68)
129.96 = 2a (1.68)
77.36 = 2a
a = 38.68
F = ma
F = .168 * 38.68
F = 6.50
Now, to worry about the force of gravity.
F = ma
F = .168 * (-9.8) = -1.65
So ...
6.50 + (-1.65) = 4.85
However, I am unsure if gravity should be negative or not:
F = ma
F = .168 * 9.8 = 1.65
So ...
6.50 + 1.65 = 8.15
I need a little bit of reinforcement (explanation) on what sign gravity should have in this problem.
12. Sep 11, 2009
### Staff: Mentor
Good. The net force on the ball is 6.5 N. (Since that force is upward, let's call it positive.)
OK, the weight is 1.65 N downward, so negative.
Here's where you went wrong.
Think this way:
Net Force = Force of Pitcher (upward) + Force of gravity (downward)
+ 6.5 = Fp - 1.65
solve for Fp:
Fp = 6.5 + 1.65
13. Sep 11, 2009
### Warmacblu
Okay, I understand now. Thanks for all the help. | 1,294 | 3,999 | {"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} | 3.859375 | 4 | CC-MAIN-2017-43 | longest | en | 0.944348 |
https://www.branchor.com/how-to-calculate-the-gdp-deflator/ | 1,726,349,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00668.warc.gz | 643,083,767 | 12,158 | Understanding GDP Deflator: A Complete Guide to Calculation and Interpretation
I. Introduction
Gross Domestic Product (GDP) is one of the most significant indicators of a country’s economic growth. However, it is vital to adjust for inflation to ensure accurate measurements. This is where the GDP Deflator comes in. The deflator helps to adjust nominal GDP to reflect the inflation rate of an economy. This article aims to provide a comprehensive guide to understanding the GDP deflator, its calculation, interpretation, and its importance in economic policy and financial analysis.
II. Understanding the Importance of the GDP Deflator: A Comprehensive Guide to Calculation
The GDP deflator is the ratio of nominal GDP to real GDP. It measures the overall price level change of all goods and services produced by a country. The nominal GDP measures the total value of goods and services produced without adjusting for inflation, while the Real GDP is the value of goods and services produced, but adjusted for inflation.
The GDP deflator is significant in measuring economic growth and inflation. It helps to provide an accurate picture of changes in prices, which is an essential determinant of inflation, and it is a measure of overall economic performance. Therefore, computing GDP Deflators is essential in economix growth analysis and forecasting. Here is how to calculate GDP deflator:
GDP Deflator = (Nominal Gross Domestic Product / Real Gross Domestic Product) x 100.
The result obtained represents the percentage change in the GDP Deflator from one period to another. For instance, if the GDP deflator this year is 105, it means that prices have increased by 5% compared to the base year.
III. How to Use GDP Deflators to Measure Inflation: A Simple Guide
Inflation is the increase in the cost of goods and services over time. Understanding inflation and its effects are essential in making economic decisions. The GDP deflator plays a fundamental role in measuring inflation because it measures all goods and services produced in an economy. Unlike other inflation indicators that measure specific products or services, the GDP deflator measures all the products, making it a more accurate measure of inflation.
In comparison with other inflation indicators like the Consumer Price Index (CPI), the GDP Deflator includes all goods and services involved in the production of final goods and services. The CPI only includes the cost of goods and services purchased by consumers. Therefore, the GDP deflator is a more comprehensive measure of inflation.
To measure inflation using the GDP deflator, you need to calculate the percentage change in the deflator from one year to another. If the GDP Deflator is increasing over time, it suggests increasing inflation. Conversely, if the GDP is decreasing over time, it suggests reducing inflation.
IV. The Role of GDP Deflators in Economic Policy: A Beginner’s Guide
Government and economists use GDP deflators in various economic policies and decision-making processes. Since the GDP deflator measures the overall change in the price level, it plays a significant role in economic forecasting and actual performance analysis. It helps the government to have an accurate picture of economic growth, inflation, and other economic indicators that are crucial in formulating economic policies.
The GDP deflator plays a crucial role in economic forecasting. It helps policy-makers make better decisions about budget allocations, interest rates, inflation control, among others. Accurate GDP deflator data is therefore significant in creating timely and informed economic decisions. However, it is important to note that the GDP deflator has limitations, such as the fact that the goods and services whose prices are reflected in the deflator are only a subset of the total economy.
V. Why GDP Deflators Matter: Exploring the Significance of Inflation Adjustment
Adjusting for inflation is essential to ensure accurate growth measurements and comparisons. Failure to adjust GDP values can lead to erroneous conclusions about economic performance. This is because the nominal GDP values cannot reflect changes that are due to price changes, while real GDP changes take the inflation rate into account.
Using GDP deflators helps to provide a more accurate measurement of economic growth and inflation. It is essential to note that comparing GDP with and without inflation adjustment shows different results, which are essential in making informed decisions.
The GDP deflator plays a vital role in inflation adjustment by giving an accurate measure of the overall price level. Without inflation adjustment, governments may misinterpret economic performance, which can lead to wrong decisions, such as influencing interest rates, tax, and budgetary adjustments. Such erroneous decisions can have adverse effects on the economy as a whole.
VI. Maximizing Your Data: How to Calculate GDP Deflators for Accurate Economic Analysis
The accuracy of the GDP deflator depends on the quality and accuracy of the data. However, there are common errors when calculating the GDP deflator that you can avoid by following these tips and tricks:
• Ensure the nominal and real GDP values are in the same price base year.
• Keep track of the inflation rate and update your GDP deflator values regularly
• Use the appropriate GDP components to calculate the GDP Deflator.
• Use reliable and accurate data sources.
An inaccurate GDP deflator can lead to incorrect economic interpretations and wrong decisions. Therefore, it is essential to ensure that your deflator calculation is accurate and reliable.
VII. How to Interpret GDP Deflator Data: A Guide to Drawing Insights from the Numbers
Your ability to analyze and interpret the GDP deflator data can provide valuable economic insights. Here are some key points to consider:
• Identify trends and patterns emerging from the GDP deflator trend data.
• Understand the relationship between GDP levels and inflation.
• Make logical comparisons between GDP deflators from different periods to derive meaningful insights.
It is essential to note that the GDP deflator has limitations, such as not capturing the full economic activity of the country. Such knowledge is necessary to help you understand how to interpret GDP deflator data accurately.
VIII. A Step-by-Step Guide to Using GDP Deflators in Financial Analysis
The GDP deflator plays a vital role in financial analysis as it gives you a broad view of economic trends that can influence your investment decisions. Here how you can utilize GDP deflators to make informed investment decisions:
• Use GDP deflators to assess the risk levels of your investments.
• Analyze consumer price index numbers and financial markets in comparison with GDP deflator data.
• Use GDP deflator data to forecast future inflation, helping you to adjust your investments accordingly.
Using GDP deflators in financial analysis provides guidance for investors in making informed investment decisions. Therefore, it is important to incorporate GDP deflators in your portfolio management and investment decisions.
IX. Conclusion
The GDP deflator is a crucial tool in measuring economic growth, inflation, and provides valuable insights for policy-makers and financial analysts. Through a step-by-step guide, we have explored the calculation of GDP deflators, their importance in measuring inflation and economic growth, their role in economic policy, and their use in financial analysis. As such, it is essential to regularly update GDP deflator data to make informed decisions based on accurate calculations and interpretations.
Therefore, we encourage everyone to incorporate GDP deflators in their economic analysis and decision-making processes for informed, timely, and accurate economic forecasting. | 1,460 | 7,846 | {"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-2024-38 | latest | en | 0.932674 |
http://nrich.maths.org/public/leg.php?code=149&cl=1&cldcmpid=4964 | 1,506,100,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689028.2/warc/CC-MAIN-20170922164513-20170922184513-00503.warc.gz | 255,427,040 | 9,314 | Search by Topic
Resources tagged with Area similar to Wallpaper:
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There are 55 results
Broad Topics > Measures and Mensuration > Area
Wallpaper
Stage: 1 Challenge Level:
These pieces of wallpaper need to be ordered from smallest to largest. Can you find a way to do it?
Being Thoughtful - Primary Measures
Stage: 1 and 2 Challenge Level:
Measure problems at primary level that require careful consideration.
Being Collaborative - Primary Measures
Stage: 1 and 2 Challenge Level:
Measure problems for primary learners to work on with others.
A Day with Grandpa
Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
Rope Mat
Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
Overlapping Squares
Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
Being Curious - Primary Measures
Stage: 1 and 2 Challenge Level:
Measure problems for inquiring primary learners.
Being Determined - Primary Measures
Stage: 1 and 2 Challenge Level:
Measure problems at primary level that may require determination.
Circle Panes
Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
Triangle Island
Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
Different Sizes
Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
Transformations on a Pegboard
Stage: 2 Challenge Level:
How would you move the bands on the pegboard to alter these shapes?
Shape Draw
Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
Cutting it Out
Stage: 1 and 2 Challenge Level:
I cut this square into two different shapes. What can you say about the relationship between them?
Shaping It
Stage: 1 and 2 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
Lawn Border
Stage: 1 and 2 Challenge Level:
If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time?
From One Shape to Another
Stage: 2
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
Triangle Relations
Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
Fencing
Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
Tiles in the Garden
Stage: 2 Challenge Level:
How many tiles do we need to tile these patios?
Fitted
Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
Sizing Them Up
Stage: 1 Challenge Level:
Can you put these shapes in order of size? Start with the smallest.
The Big Cheese
Stage: 2 Challenge Level:
Investigate the area of 'slices' cut off this cube of cheese. What would happen if you had different-sized block of cheese to start with?
It Must Be 2000
Stage: 2 Challenge Level:
Here are many ideas for you to investigate - all linked with the number 2000.
Fit These Shapes
Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
Area and Perimeter
Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
Inside Seven Squares
Stage: 2 Challenge Level:
What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?
Cover the Tray
Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
Torn Shapes
Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
My New Patio
Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
Ribbon Squares
Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
Making Squares
Stage: 2
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
Fencing Lambs
Stage: 2 Challenge Level:
A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales.
A Square in a Circle
Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
Geoboards
Stage: 2 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
Numerically Equal
Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
Dicey Perimeter, Dicey Area
Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
Two Squared
Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
Great Squares
Stage: 2 and 3 Challenge Level:
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
Through the Window
Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
Tiles on a Patio
Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
Extending Great Squares
Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
Tiling Into Slanted Rectangles
Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
Uncanny Triangles
Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
Pebbles
Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Wrapping Presents
Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
Making Boxes
Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
Exploration Versus Calculation
Stage: 1, 2 and 3
This article, written for teachers, discusses the merits of different kinds of resources: those which involve exploration and those which centre on calculation.
Making Rectangles
Stage: 2 and 3 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding. | 1,786 | 7,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-39 | latest | en | 0.915026 |
https://sites.psu.edu/mathmug/2018/02/01/godels-incompleteness-theorems/ | 1,534,530,044,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00625.warc.gz | 791,460,270 | 8,131 | # Gödel’s Incompleteness Theorems
Godel’s Incompleteness Theorems are some of the most important results in logic, that I’ve used in many forms day to day. “any formal system (such as a system of the natural numbers), there are certain true statements about the system which cannot be proven by the system itself.” https://listverse.com/2013/05/05/10-coolest-mathematics-results/ This, although seemingly technical, means to solve a problem you need to look outside the problem itself. It isn’t necessarily the theorem itself, but the state of mind. We often think that we can find the problems and solutions of a problem on whether something is true or not, without looking past the system itself. It is like determining whether the legal system is just based only on the legal system. Which leads us to the other Incompleteness Theorem, which basically says you can’t determine if something is consistent based on itself. This comes in handy whenever you are trying to figure out how to handle an issue.
The first step when presented with an issue is to find a good goal, or achievement you want to achieve, even if it’s broad, like, betterment for society, as long as there is an outside goal, you can work backwards to determine if you are considering everything.
The other great use, is that you can never fully figure out the impact of a system by looking at the system itself, consider iphones, while we may enjoy them, is our enjoyment worth the bad conditions of the people working to make them? And what can we do about it? Math is at its heart all about problem solving, and I feel like sometimes it gets undervalued to memorization of what dead people figured out and numbers, but really it is about solving problems given axioms. And this result tells us, we never have it all figured out. (and even if we thought we were at least on the right track, we wouldn’t know it)
## One Comment
1. mfd5415
Austin, this is a very interesting take on something that too many people don’t know about. Only by looking at many different fields and systems did our classic philosophers, mathematicians, and scientists developed a deeper understanding of the world around them. Now, even our society’s “smart people” are tied to one or maybe two issue areas. While I know this isn’t always the case, it certainly seems as though society’s intellectuals are more specialized, which I would argue places them further away from their goals than trying to be the “jack of all trades.” Regardless though, the first step in figuring anything out is gathering information, which will always be fundamentally incomplete. Your post shines a light on that, and in turn results in better decision making if closely adhered to. Great work man, and great topic selection.
Mitch Dobbs | 594 | 2,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-34 | longest | en | 0.965709 |
https://www.gospel10.com/how-to-calculate-bond-order-resonance-a-step-by-step-guide/ | 1,725,738,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00165.warc.gz | 766,746,853 | 55,920 | # How to Calculate Bond Order Resonance: A Step-by-Step Guide
Bond order resonance is a computational method used to estimate the bond order between atoms in a molecule.
Understanding bond order resonance is essential for predicting the stability, reactivity, and properties of chemical compounds. Resonance structures are often used to represent the electronic structure of molecules, and bond order resonance provides a quantitative measure of the relative importance of these structures.
This article will explore the steps involved in calculating bond order resonance, including the use of resonance theory and quantum mechanics.
## How to Calculate Bond Order Resonance
Understanding the aspects of bond order resonance is fundamental to comprehend its calculation. Key aspects include:
• Resonance structures
• Resonance hybrid
• Bond order
• Electronegativity
• Molecular orbitals
• Quantum mechanics
• Chemical bonding
• Molecular stability
• Chemical reactivity
• Molecular properties
These aspects are interconnected. Resonance structures provide a qualitative description of the electronic structure of a molecule, while bond order resonance provides a quantitative measure of the relative importance of these structures. Quantum mechanics provides the theoretical framework for understanding resonance and bond order. Electronegativity, molecular orbitals, and chemical bonding are all factors that influence bond order resonance. Ultimately, understanding bond order resonance is essential for predicting the stability, reactivity, and properties of chemical compounds.
### Resonance structures
Resonance structures are a fundamental concept in chemistry, used to describe the electronic structure of molecules. They are particularly important in understanding how to calculate bond order resonance, which is a measure of the strength of the bond between two atoms.
Resonance structures arise when there are multiple possible Lewis structures for a molecule. These structures differ in the placement of the electrons, but they all have the same number of valence electrons and the same overall charge. For example, the carbonate ion (CO32-) has three resonance structures, as shown below:
O=C-O- O-C=O- O-C-O- | | | O O O
The bond order resonance between the carbon and oxygen atoms in the carbonate ion is calculated by taking the average of the bond orders in the three resonance structures. In this case, the bond order resonance is 1.33, which indicates that the bond between the carbon and oxygen atoms is stronger than a single bond but weaker than a double bond.
Understanding bond order resonance is essential for predicting the stability, reactivity, and properties of chemical compounds. For example, compounds with higher bond order resonances are generally more stable and less reactive. Bond order resonance can also be used to explain the geometry of molecules and the strength of their bonds.
### Resonance hybrid
In the context of calculating bond order resonance, resonance hybrids are crucial for understanding the electronic structure of molecules and their properties. A resonance hybrid is a hypothetical structure that represents the average of all possible resonance structures for a molecule. It combines the features of all the resonance structures, taking into account their relative contributions.
• Weighted average: A resonance hybrid is a weighted average of all resonance structures, where the weights are determined by the stability of each structure.
• Intermediate properties: The properties of a resonance hybrid are intermediate between those of the contributing resonance structures.
• Delocalized electrons: In a resonance hybrid, the electrons are delocalized over the entire molecule, rather than being localized to specific atoms or bonds.
• Stability: Resonance hybrids are generally more stable than any of the individual resonance structures.
By considering resonance hybrids, chemists can gain a more accurate understanding of the electronic structure and properties of molecules. Resonance hybrids provide a valuable tool for predicting molecular properties, such as bond lengths, bond angles, and molecular polarity. They also help to explain the reactivity and stability of molecules.
### Bond order
Bond order is a central concept in chemistry that describes the strength and type of chemical bond between two atoms. In the context of calculating bond order resonance, it plays a crucial role in determining the relative importance of different resonance structures and understanding the electronic structure of molecules.
• Bond length: Bond order is inversely related to bond length. Shorter bonds correspond to higher bond orders, indicating a stronger bond.
• Bond strength: Bond order is directly proportional to bond strength. Bonds with higher bond orders are stronger and less likely to break.
• Bond type: Bond order can be used to distinguish between different types of bonds, such as single bonds, double bonds, and triple bonds. Single bonds have a bond order of 1, double bonds have a bond order of 2, and triple bonds have a bond order of 3.
• Electron density: Bond order is related to the electron density between the bonded atoms. Higher bond orders correspond to higher electron density, indicating a greater overlap between the atomic orbitals.
Understanding bond order is essential for predicting the stability, reactivity, and properties of chemical compounds. By considering bond order resonance, chemists can gain a more accurate picture of the electronic structure of molecules and their behavior.
### Electronegativity
Electronegativity is a fundamental property of atoms that influences their ability to attract electrons, thereby shaping the electronic structure and properties of molecules. In the context of calculating bond order resonance, electronegativity plays a crucial role in determining the distribution of electrons and the strength of chemical bonds.
• Electronegative Potential: Electronegative potential quantifies the tendency of an atom to attract electrons towards itself, determining the polarity of bonds and the overall charge distribution within a molecule.
• Pauling Scale: The Pauling scale is a widely used measure of electronegativity, assigning numerical values to different elements based on their ability to attract electrons. This scale provides a relative measure of electronegativity, aiding in the prediction of bond polarity and molecular properties.
• Electronegativity and Bond Length: Electronegativity influences bond length, as more electronegative atoms tend to form shorter and stronger bonds. This is due to the increased attraction between the electronegative atom and the shared electrons.
• Electronegativity and Resonance: Electronegativity affects resonance by stabilizing certain resonance structures over others, as more electronegative atoms tend to concentrate electron density towards themselves. This can influence the overall electronic structure and properties of the molecule.
Understanding electronegativity is crucial for calculating bond order resonance accurately, as it provides insights into the distribution of electrons and the strength of chemical bonds. By considering the electronegativities of the atoms involved, chemists can gain a more precise understanding of the electronic structure and properties of molecules.
### Molecular orbitals
In the realm of calculating bond order resonance, molecular orbitals (MOs) emerge as a fundamental concept, providing a detailed account of the electronic structure of molecules. These orbitals are mathematical functions that describe the wave-like behavior of electrons within a molecule, encompassing their energy, spatial distribution, and interactions.
• Function: MOs are represented by mathematical functions () that describe the wave-like behavior of electrons, capturing their energy and spatial distribution.
• Electron Density: MOs provide insights into electron density distribution within a molecule, indicating the probability of finding electrons in specific regions.
• Bonding and Antibonding: MOs can be classified as bonding or antibonding, depending on their effect on the overlap and strength of chemical bonds.
• Molecular Symmetry: The symmetry of MOs aligns with the symmetry of the molecule, allowing for the prediction of molecular properties and spectroscopic behavior.
Comprehending molecular orbitals is crucial for accurately calculating bond order resonance. By considering the energy levels, spatial distributions, and interactions of electrons within MOs, chemists gain a deeper understanding of the electronic structure of molecules, enabling precise predictions of bond order resonance and various molecular properties.
### Quantum mechanics
Quantum mechanics plays a crucial role in understanding and calculating bond order resonance. It provides the theoretical framework for describing the electronic structure of molecules and the interactions between electrons. By applying the principles of quantum mechanics, chemists can accurately predict the bond order resonance of a molecule, which is essential for understanding its stability, reactivity, and properties.
One of the key concepts in quantum mechanics is the wave-particle duality of electrons. Electrons can behave both as particles and as waves, and their behavior is described by wave functions. The wave function of an electron provides information about its energy, momentum, and spatial distribution. In the context of bond order resonance, the wave function of an electron can be used to determine the probability of finding the electron in a particular region of space. This information is essential for calculating the bond order resonance, as it allows chemists to determine the extent to which the electrons are delocalized over the molecule.
Another important concept in quantum mechanics is the Pauli exclusion principle. This principle states that no two electrons can occupy the same quantum state. In the context of bond order resonance, the Pauli exclusion principle can be used to determine the number of electrons that can occupy a particular molecular orbital. This information is essential for calculating the bond order resonance, as it allows chemists to determine the number of bonds that can be formed between two atoms.
Quantum mechanics is a powerful tool for understanding and calculating bond order resonance. By applying the principles of quantum mechanics, chemists can gain a deeper understanding of the electronic structure of molecules and their properties.
### Chemical bonding
Chemical bonding is the interaction between atoms or molecules that enables the formation of chemical substances that contain two or more atoms. The strength of chemical bonding influences various properties of a molecule, including its stability, reactivity, and physical properties. Understanding chemical bonding is essential for accurately calculating bond order resonance, as it provides insights into the nature and strength of the bonds between atoms.
• Types of Bonds: Chemical bonds can be classified into various types, including covalent bonds, ionic bonds, and metallic bonds. Each type of bond has its own unique characteristics and properties, which affect the bond order resonance.
• Bond Strength: The strength of a chemical bond is determined by the amount of energy required to break the bond. Bond strength is directly related to bond order resonance, as stronger bonds typically have higher bond order resonance values.
• Bond Length: Bond length is the distance between the nuclei of two bonded atoms. Bond length is inversely related to bond order resonance, as shorter bonds generally have higher bond order resonance values.
• Molecular Geometry: The arrangement of atoms in a molecule is determined by the type and strength of the chemical bonds between them. Bond order resonance can influence the molecular geometry by affecting the bond lengths and bond angles.
In summary, chemical bonding plays a crucial role in determining the bond order resonance of a molecule. By understanding the different types of chemical bonds, their strengths, and their effects on bond length and molecular geometry, chemists can accurately calculate bond order resonance values, which provide valuable insights into the electronic structure and properties of molecules.
### Molecular stability
In the context of calculating bond order resonance, molecular stability plays a critical role in understanding the electronic structure and properties of molecules. Bond order resonance provides insights into the relative stability of different resonance structures, which contributes to the overall stability of the molecule.
• Bond strength: The strength of the bonds between atoms affects the stability of a molecule. Bond order resonance can help determine the strength of bonds, as stronger bonds typically have higher bond order resonance values.
• Resonance energy: The resonance energy of a molecule is the energy difference between the actual molecule and the hypothetical molecule with the lowest energy resonance structure. Higher resonance energy indicates greater stability, as the molecule is more likely to adopt resonance structures that lower its energy.
• Molecular geometry: The arrangement of atoms in a molecule can influence its stability. Bond order resonance can provide insights into the preferred molecular geometry by indicating the most stable resonance structures.
• Reactivity: The reactivity of a molecule is related to its stability. More stable molecules are generally less reactive, as they are less likely to undergo chemical reactions that would disrupt their stable electronic structure.
Understanding molecular stability is essential for accurately calculating bond order resonance. By considering the various factors that contribute to molecular stability, chemists can gain a deeper understanding of the electronic structure and properties of molecules, and make more accurate predictions about their behavior.
### Chemical reactivity
Chemical reactivity is a crucial aspect of “how to calculate bond order resonance” as it provides insights into the behavior and stability of molecules. It refers to the tendency of a molecule to undergo chemical reactions, forming new substances with different properties.
• Reaction Rates: Bond order resonance can influence the rate at which a molecule reacts with other species. Molecules with higher bond order resonance values tend to react more slowly, as the delocalized electrons are less available for bond formation with other molecules.
• Activation Energy: The activation energy of a reaction is the minimum amount of energy required for the reaction to occur. Bond order resonance can affect the activation energy by stabilizing the transition state of the reaction. A more stable transition state leads to a lower activation energy and, consequently, a faster reaction rate.
• Selectivity: Bond order resonance can influence the selectivity of a reaction, which refers to the preference of a molecule to react with one type of molecule over another. By considering the resonance structures of the reactants and products, chemists can predict the most favorable reaction pathway.
• Regioselectivity: Regioselectivity refers to the preference of a molecule to react at a specific site. Bond order resonance can provide insights into the regioselectivity of a reaction by indicating the most reactive site on a molecule.
In summary, chemical reactivity is an important factor to consider when calculating bond order resonance, as it provides valuable information about the stability, reaction rates, selectivity, and regioselectivity of molecules. Understanding chemical reactivity enables chemists to make more accurate predictions about the behavior of molecules and design molecules with desired properties.
### Molecular properties
Molecular properties play a crucial role in understanding the chemical behavior and applications of molecules. By calculating bond order resonance, chemists can gain insights into various molecular properties that determine how molecules interact with each other and their surroundings.
• Dipole moment: Dipole moment measures the polarity of a molecule, indicating the uneven distribution of electrical charge. Bond order resonance can influence the dipole moment by affecting the distribution of electrons within the molecule.
• Polarizability: Polarizability describes how easily a molecule’s electron cloud can be distorted by an external electric field. Bond order resonance can affect polarizability by influencing the delocalization of electrons.
• Magnetic susceptibility: Magnetic susceptibility measures the response of a molecule to a magnetic field. Bond order resonance can affect magnetic susceptibility by influencing the number and type of unpaired electrons in the molecule.
• Optical properties: Optical properties, such as color and absorption spectra, are influenced by the electronic structure of a molecule. Bond order resonance can affect optical properties by altering the energy levels and transitions of electrons.
In summary, understanding molecular properties is essential for accurately calculating bond order resonance. By considering the various molecular properties outlined above, chemists can gain a deeper understanding of the electronic structure and behavior of molecules.
This FAQ section addresses common questions and clarifies key aspects of “how to calculate bond order resonance”.
Question 1: What is bond order resonance?
Bond order resonance is a method for calculating the bond order between atoms in a molecule, taking into account the resonance structures of the molecule.
Question 2: Why is bond order resonance important?
Bond order resonance provides valuable insights into the electronic structure and properties of molecules, such as their stability, reactivity, and molecular properties.
Question 3: What are resonance structures?
Resonance structures are different Lewis structures for the same molecule that have the same number of valence electrons and the same overall charge. They represent the delocalization of electrons within the molecule.
Question 4: How do you calculate bond order resonance?
Bond order resonance is calculated by taking the average of the bond orders in the resonance structures of the molecule.
Question 5: What factors affect bond order resonance?
Bond order resonance is influenced by factors such as electronegativity, molecular orbitals, and quantum mechanics.
Question 6: How does bond order resonance relate to molecular properties?
Bond order resonance provides insights into various molecular properties, including dipole moment, polarizability, magnetic susceptibility, and optical properties.
In summary, understanding bond order resonance is crucial for gaining insights into the electronic structure and behavior of molecules. By considering the factors that affect bond order resonance, chemists can make accurate predictions about molecular properties and chemical reactivity.
In the next section, we will explore advanced applications of bond order resonance in chemistry.
### Tips for Calculating Bond Order Resonance
Accurately calculating bond order resonance is crucial for understanding the electronic structure and properties of molecules. Here are some valuable tips to help you master this technique:
Tip 1: Understand Resonance Structures
Identify and analyze the resonance structures of the molecule to gain insights into the delocalization of electrons.
Tip 2: Calculate Bond Orders
Determine the bond orders in each resonance structure using appropriate methods, such as the Pauling formula or molecular orbital theory.
Tip 3: Consider Electronegativity
Take into account the electronegativity of the atoms involved, as it influences the distribution of electrons and bond strength.
Tip 4: Utilize Quantum Mechanics
Apply the principles of quantum mechanics, such as wave-particle duality and the Pauli exclusion principle, to understand electron behavior and bond formation.
Tip 5: Analyze Molecular Geometry
Consider the molecular geometry and symmetry to determine the most stable resonance structures and their impact on bond order resonance.
Summary: By following these tips, you can effectively calculate bond order resonance and gain valuable insights into the electronic structure, stability, and properties of molecules.
In the concluding section of this article, we will explore advanced applications of bond order resonance in chemistry, showcasing its significance in understanding complex molecular systems.
### Conclusion
In this article, we have explored the intricacies of “how to calculate bond order resonance”. We have gained insights into the fundamental concepts, such as resonance structures, bond orders, and the influence of factors like electronegativity and molecular geometry. By understanding bond order resonance, chemists can accurately predict the stability, reactivity, and properties of molecules.
Key points to remember:
• Bond order resonance provides a quantitative measure of the relative importance of resonance structures.
• It is crucial for understanding the electronic structure, stability, and properties of molecules.
• Factors such as electronegativity, molecular orbitals, and quantum mechanics play significant roles in determining bond order resonance.
Understanding bond order resonance is not merely an academic exercise but a powerful tool for unraveling the mysteries of molecular behavior. It empowers chemists to design new materials, optimize chemical reactions, and develop innovative technologies that shape our world. As we continue to push the boundaries of chemistry, the significance of bond order resonance will only grow, enabling us to harness the power of molecules for the betterment of society. | 3,867 | 21,982 | {"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-2024-38 | latest | en | 0.880092 |
https://www.jiskha.com/display.cgi?id=1316616465 | 1,511,156,980,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00406.warc.gz | 821,161,711 | 3,599 | # Physics
posted by .
On a spacecraft two engines fire for a time of 589 s. One gives the craft an acceleration in the x direction of ax = 5.09 m/s2, while the other produces an acceleration in the y direction of ay = 7.50 m/s2. At the end of the firing period, the craft has velocity components of vx = 3781 m/s and vy = 4857 m/s. Calculate the magnitude of the initial velocity.
• Physics -
Vf=Vi+at
solve for Vi. Do this in vector form, ie, i,j components of acceleration. Then Vi will have both components.
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# What is the mean absolute deviation formula?
A:
The mean absolute deviation formula is Σ|x - μ| / N. The symbol Σ is used to denote the sum of a series of numbers, while μ represents the mean, x represents each value and N represents the total number of values. The formula takes the absolute value of the difference between a value and the mean and divides it by the number of values.
## Keep Learning
Credit: Imgorthand E+ Getty Images
The mean absolute value of a data set, sometimes abbreviated as MAD, represents the mean of the data's absolute deviations about the data's mean. In other words, the mean absolute deviation represents the absolute value of the average distance from the data set's mean. The mean of a data set is the average of all of its values.
Sources:
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posted by Lauren
I want to ensure I've done this problem correctly. Thanks!
3. In one elementary school, 200 students are tested on the subject of Math and English. The table below shows the mean and standard deviation for each subject.
Mean SD
Math 67 9.58
English 78 12.45
One student’s Math score was 70 and the same individual’s English score was 84. On which exam did the student do better?
Work:
Math Score Z-Score = 0.313152
Math Score Percentile Rank = 62.2918th
English Score Z-Score = 0.481928
English Score Percentile Rank = 68.5072th
The student did better on the English test because they ranked in a higher percantile than the Math test (68.5072th percantile versus 62.2981th percantile).
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# Quantile Function
## What is Quantile Function?
A Quantile Function is a phrase that describes a portion of a data set; it determines how many values in a distribution are above or below a given threshold. The quartile (quarter), quintile (fifth), and percentiles are special quantiles (hundredth).
The Q(a), is the inverse of the CDF; it gives us the probability that a random variable will be less than or equal to some value X = x.
## Example of Quantile Function:
Two quantile measures are:
1. One is the value of the quantile function for 50%. This is termed the median of the distribution. On average, 50% of the variable’s outcomes will be below the median, and 50% of the variable’s outcomes will be above the median. The mean and median will be equal for a symmetric distribution (skew = 0). For distributions with positive (right) skew, the median will be less than the mean but greater for distributions with a negative (left) skew.
2. The second quantile measure of interest here is the interquartile range. The interquartile range is the upper and lower value of the outcomes of a random variable that include the middle 50% of its probability distribution. The lower value is Q(25%), and the upper value is Q(75%). The lower value is the value that we expect 25% of the outcomes to be less than, and the upper value is the value that we expect 75% of the values to be less than.
## Why are quantiles important?
Quantiles provide information on the shape of a distribution, namely whether or not it is skewed. We can deduce that the distribution is skewed to the right if the higher quartile is further from the median than the lower quartile and vice versa.
Owais Siddiqui | 399 | 1,746 | {"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-2024-22 | latest | en | 0.874562 |
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## Example 1: Reassigning a Variable
``````
# Initial value
x = 5
# Reassigning the variable
x = 10
# Print the updated value
print(x)
``````
Output:
``````
10
``````
## Example 2: Incrementing a Numeric Variable
``````
# Initial value
count = 0
# Increment the variable
count += 1
# Print the updated value
print(count)
``````
Output:
``````
1
``````
## Example 3: Decrementing a Numeric Variable
``````
# Initial value
quantity = 10
# Decrement the variable
quantity -= 2
# Print the updated value
print(quantity)
``````
Output:
``````
8
``````
## Example 4: Concatenating Strings
``````
# Initial values
first_name = "John"
last_name = "Doe"
# Concatenate strings
full_name = first_name + " " + last_name
# Print the concatenated string
print(full_name)
``````
Output:
``````
John Doe
``````
## Example 5: Updating Lists
``````
# Initial list
fruits = ["apple", "banana", "cherry"]
# Update the list
fruits[1] = "orange"
# Print the updated list
print(fruits)
``````
Output:
``````
['apple', 'orange', 'cherry']
``````
## Example 6: Modifying Dictionary Values
``````
# Initial dictionary
student = {"name": "Alice", "age": 25}
# Modify a dictionary value
student["age"] = 26
# Print the modified dictionary
print(student)
``````
Output:
``````
{'name': 'Alice', 'age': 26}
``````
## Example 7: Multiplying Numeric Variables
``````
# Initial values
length = 5
width = 3
# Multiply variables
area = length * width
# Print the calculated area
print("Area:", area)
``````
Output:
``````
Area: 15
``````
## Example 8: Updating Variables Conditionally
``````
# Initial values
temperature = 25
# Check a condition and update the variable
if temperature > 30:
status = "Hot"
else:
status = "Moderate"
# Print the status
print("Temperature Status:", status)
``````
Output:
``````
Temperature Status: Moderate
``````
## Example 9: Swapping Variable Values
``````
# Initial values
a = 5
b = 10
# Swap the values of variables
a, b = b, a
# Print the swapped values
print("a:", a)
print("b:", b)
``````
Output:
``````
a: 10
b: 5
``````
## Example 10: Updating Variables Using a Function
``````
# Initial value
value = 5
# Define a function to update the value
def update_value(num):
return num * 2
# Update the value using the function
value = update_value(value)
# Print the updated value
print("Updated Value:", value)
``````
Output:
``````
Updated Value: 10
``````
## Example 11: Appending to a List
``````
# Initial list
fruits = ["apple", "banana", "cherry"]
# Append an item to the list
fruits.append("orange")
# Print the updated list
print("Fruits:", fruits)
``````
Output:
``````
Fruits: ['apple', 'banana', 'cherry', 'orange']
``````
## Example 12: Removing an Item from a List
``````
# Initial list
fruits = ["apple", "banana", "cherry"]
# Remove an item from the list
fruits.remove("banana")
# Print the updated list
print("Fruits:", fruits)
``````
Output:
``````
Fruits: ['apple', 'cherry']
``````
## Example 13: Updating Dictionary Values
``````
# Initial dictionary
person = {"name": "Alice", "age": 25}
# Update a dictionary value
person["age"] += 1
# Print the updated dictionary
print("Person:", person)
``````
Output:
``````
Person: {'name': 'Alice', 'age': 26}
``````
## Example 14: Clearing a List
``````
# Initial list
numbers = [1, 2, 3, 4, 5]
# Clear the list
numbers.clear()
# Print the empty list
print("Numbers:", numbers)
``````
Output:
``````
Numbers: []
``````
## Example 15: Deleting a Variable
``````
# Initial value
variable_to_delete = "I will be deleted"
# Delete the variable
del variable_to_delete
# Attempt to print the deleted variable (will raise an error)
try:
print(variable_to_delete)
except NameError as e:
print("Variable deleted:", e)
``````
Output:
``````
Variable deleted: name 'variable_to_delete' is not defined
`````` | 1,147 | 3,893 | {"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-2024-10 | latest | en | 0.621044 |
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