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### Capital evolution with five-year step
Posted: Tue Jul 04, 2017 2:51 pm
Dear all,
I am learning to build a global CGE model, but came across some questions.
The question is about the dynamics of capital stock. According to the standard perpetual inventory assumption,
Capital Stock(t+1) = NewInvestment(t)+ (1-depreciation rate)*Capital Stock(t).
However, when considering a five-year time step, shall I modify the depreciation part from “(1-depreciation rate)” to “(1-depreciation rate)^5”?
Moreover, how shall I deal with the new investment in period t? Shall I simply multiply it with 5? That is, shall I use the following equation instead to model the dynamics of capital stock when using a five-year time?
Capital Stock(t+1) = 5*NewInvestmentt+ (1-depreciation rate)^5*Capital Stock(t).
I wonder if anyone have experiences with the recursive dynamic CGE model could spare some time to help me solve them out.
Thanks for your help!
### Re: Capital evolution with five-year step
Posted: Sat Jul 22, 2017 7:27 pm
Hi
Take a look at "Calibration of Models with Multi-Year Periods" on http://www.mpsge.org/mainpage/mpsge.htm by Tom Rutherford.
Cheers
Renger | 300 | 1,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.883968 |
http://perplexus.info/show.php?pid=2200 | 1,606,601,349,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00062.warc.gz | 69,005,607 | 5,089 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Firing Line (Posted on 2004-08-25)
There is a group of N soldiers arranged in a straight line, standing side by side. Soldier number 1 is at the extreme left and soldier "N" is at the extreme right. Each soldier has a rifle that can be fired only once, a primitive timer, understands a finite list of commands, and can exist in a finite number of states, like a finite state machine.
Each soldier has the ability to communicate only with the two adjacent soldiers, and has no means of communication with more distant soldiers. The i-th soldier can not see or hear any signals given by the (i+2)th soldier, for example. There are no radios, cell phones, or megaphones.
Your mission as the commander is to devise an algorithm by which all soldiers fire their weapons simultaneously. Soldiers 1 and N are aware of the fact that they are "different" in that they each have only one neighbor. Other than that, however, the soldiers are initially all identical. The algorithm has to work for any value of N>2.
The primitive timers are synchronized and tick off once a second. Once a soldier receives new information, the earliest he can respond in any way is on the next tick of the clock. (I would say he/she, except that they are all identical). A soldier can give a signal to each neighbor simultaneously, based on the information he received one tick earlier. Whenever a soldier's state changes, his neighbors are aware of this one tick later. At time=0, soldier 1 is given the command to start the firing procedure
1. Devise an algorithm that results in all N soldiers firing simultaneously
2. As a function of N, how many clock ticks does this take?
See The Solution Submitted by Larry Rating: 3.7500 (4 votes)
Subject Author Date The way to do it Math Man 2020-11-27 20:54:10 Here is a freaky idea ! Ethen Hunt 2004-09-09 09:33:34 how about this? Patrick 2004-08-29 16:52:43 re(2): More Information - another question Larry 2004-08-28 10:45:29 Jacques Mazoyer's Solution Bractals 2004-08-27 23:35:38 re: More Information - another question nikki 2004-08-27 13:57:42 More Information Larry 2004-08-27 10:44:41 Spoiler Bractals 2004-08-27 02:13:33 Communication Confirmation nikki 2004-08-26 11:20:09 re: Assumption nikki 2004-08-26 09:37:15 Assumption Luke 2004-08-25 20:05:54 re: faster nikki 2004-08-25 15:18:00 faster Cory Taylor 2004-08-25 15:10:18 I think this is right nikki 2004-08-25 14:01:38 Solution! dhruv 2004-08-25 12:52:14
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Forums (0) | 703 | 2,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-50 | latest | en | 0.943345 |
https://boymamateachermama.com/2016/11/17/free-thanksgiving-addition-bump-game/ | 1,725,850,732,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00402.warc.gz | 126,928,724 | 19,442 | # FREE Thanksgiving Addition Bump Game
I have seen all these math “bump” games floating around Pinterest and TPT and had no idea what they were all about. So, I did a little investigating and learned what they are all about. Then, of course, I had to make a game or two to play with my first graders. And, since it is Thanksgiving next week and all, I figured a turkey theme would be fitting! So here it is, my very first “bump” game!
### Materials
one game board per 2 players
2 sets of 12 unifix or any interlocking cubes (1 color per player)
2 regular 6 sided dice
### How to Play
Player 1 rolls the two dice and determines the sum.
Player 1 then puts one of his colored cubes on that sum.
Player 2 rolls and determines the sum.
Player 2 puts her cube on that sum.
If that sum is already occupied by her opponents cube, she can “bump” him off and take the spot.
If that sum is already occupied by her own cube, she can attached a second cube to the first cube and “lock” the space. That space is then locked out and no longer in play.
Players continue to play until one player has covered 8 of the 11 spaces or until the allotted time is up.
That’s it! Sounds pretty fun right? Well, it is yours for FREE. | 297 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.975797 |
https://plainmath.net/advanced-physics/103527-in-the-relation-kinetic-energy | 1,679,897,956,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00571.warc.gz | 517,857,138 | 19,130 | Rocco May
2023-03-11
In the relation Kinetic Energy $={p}^{2}/2m$ (where p is momentum, m is mass ) . If Momentum is raised by 0.1 % what will be the percentage increase in Kinetic Energy
stunaire2dt
Kinetic energy E=P22m∴E∝P2
Percentage increase in kinetic energy = 2(% increase in momentum) = 2(0.1%) = 0.2%.
Do you have a similar question? | 115 | 347 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-14 | latest | en | 0.864853 |
https://wikimili.com/en/Composition_algebra | 1,627,228,987,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00256.warc.gz | 636,137,392 | 19,544 | # Composition algebra
Last updated
In mathematics, a composition algebraA over a field K is a not necessarily associative algebra over K together with a nondegenerate quadratic form N that satisfies
## Contents
${\displaystyle N(xy)=N(x)N(y)}$
for all x and y in A.
A composition algebra includes an involution called a conjugation: ${\displaystyle x\mapsto x^{*}.}$ The quadratic form ${\displaystyle N(x)\ =\ xx^{*}}$ is called the norm of the algebra.
A composition algebra (A, ∗, N) is either a division algebra or a split algebra, depending on the existence of a non-zero v in A such that N(v) = 0, called a null vector. [1] When x is not a null vector, the multiplicative inverse of x is ${\displaystyle {\frac {x^{*}}{N(x)}}\ .}$ When there is a non-zero null vector, N is an isotropic quadratic form, and "the algebra splits".
## Structure theorem
Every unital composition algebra over a field K can be obtained by repeated application of the Cayley–Dickson construction starting from K (if the characteristic of K is different from 2) or a 2-dimensional composition subalgebra (if char(K) = 2). The possible dimensions of a composition algebra are 1, 2, 4, and 8. [2] [3] [4]
• 1-dimensional composition algebras only exist when char(K) ≠ 2.
• Composition algebras of dimension 1 and 2 are commutative and associative.
• Composition algebras of dimension 2 are either quadratic field extensions of K or isomorphic to KK.
• Composition algebras of dimension 4 are called quaternion algebras. They are associative but not commutative.
• Composition algebras of dimension 8 are called octonion algebras. They are neither associative nor commutative.
For consistent terminology, algebras of dimension 1 have been called unarion, and those of dimension 2 binarion. [5]
## Instances and usage
When the field K is taken to be complex numbers C and the quadratic form z2, then four composition algebras over C are C itself, the bicomplex numbers, the biquaternions (isomorphic to the 2×2 complex matrix ring M(2, C)), and the bioctonions CO, which are also called complex octonions.
Matrix ring M(2, C) has long been an object of interest, first as biquaternions by Hamilton (1853), later in the isomorphic matrix form, and especially as Pauli algebra.
The squaring function N(x) = x2 on the real number field forms the primordial composition algebra. When the field K is taken to be real numbers R, then there are just six other real composition algebras. [3] :166 In two, four, and eight dimensions there are both a division algebra and a "split algebra":
binarions: complex numbers with quadratic form x2 + y2 and split-complex numbers with quadratic form x2y2,
quaternions and split-quaternions,
octonions and split-octonions.
Every composition algebra has an associated bilinear form B(x,y) constructed with the norm N and a polarization identity:
${\displaystyle B(x,y)\ =\ [N(x+y)-N(x)-N(y)]/2.}$ [6]
## History
The composition of sums of squares was noted by several early authors. Diophantus was aware of the identity involving the sum of two squares, now called the Brahmagupta–Fibonacci identity, which is also articulated as a property of Euclidean norms of complex numbers when multiplied. Leonhard Euler discussed the four-square identity in 1748, and it led W. R. Hamilton to construct his four-dimensional algebra of quaternions. [5] :62 In 1848 tessarines were described giving first light to bicomplex numbers.
About 1818 Danish scholar Ferdinand Degen displayed the Degen's eight-square identity, which was later connected with norms of elements of the octonion algebra:
Historically, the first non-associative algebra, the Cayley numbers ... arose in the context of the number-theoretic problem of quadratic forms permitting composition…this number-theoretic question can be transformed into one concerning certain algebraic systems, the composition algebras... [5] :61
In 1919 Leonard Dickson advanced the study of the Hurwitz problem with a survey of efforts to that date, and by exhibiting the method of doubling the quaternions to obtain Cayley numbers. He introduced a new imaginary unit e, and for quaternions q and Q writes a Cayley number q + Qe. Denoting the quaternion conjugate by q, the product of two Cayley numbers is [7]
${\displaystyle (q+Qe)(r+Re)=(qr-R'Q)+(Rq+Qr')e.}$
The conjugate of a Cayley number is q'Qe, and the quadratic form is qq′ + QQ, obtained by multiplying the number by its conjugate. The doubling method has come to be called the Cayley–Dickson construction.
In 1923 the case of real algebras with positive definite forms was delimited by the Hurwitz's theorem (composition algebras).
In 1931 Max Zorn introduced a gamma (γ) into the multiplication rule in the Dickson construction to generate split-octonions. [8] Adrian Albert also used the gamma in 1942 when he showed that Dickson doubling could be applied to any field with the squaring function to construct binarion, quaternion, and octonion algebras with their quadratic forms. [9] Nathan Jacobson described the automorphisms of composition algebras in 1958. [2]
The classical composition algebras over R and C are unital algebras. Composition algebras without a multiplicative identity were found by H.P. Petersson (Petersson algebras) and Susumu Okubo (Okubo algebras) and others. [10] :463–81
## Related Research Articles
In abstract algebra, an alternative algebra is an algebra in which multiplication need not be associative, only alternative. That is, one must have
In mathematics, a Clifford algebra is an algebra generated by a vector space with a quadratic form, and is a unital associative algebra. As K-algebras, they generalize the real numbers, complex numbers, quaternions and several other hypercomplex number systems. The theory of Clifford algebras is intimately connected with the theory of quadratic forms and orthogonal transformations. Clifford algebras have important applications in a variety of fields including geometry, theoretical physics and digital image processing. They are named after the English mathematician William Kingdon Clifford.
In mathematics, the octonions are a normed division algebra over the real numbers, meaning it is a hypercomplex number system; Octonions are usually represented by the capital letter O, using boldface O or blackboard bold . Octonions have eight dimensions; twice the number of dimensions of the quaternions, of which they are an extension. They are noncommutative and nonassociative, but satisfy a weaker form of associativity; namely, they are alternative. They are also power associative.
In mathematics, hypercomplex number is a traditional term for an element of a finite-dimensional unital algebra over the field of real numbers. The study of hypercomplex numbers in the late 19th century forms the basis of modern group representation theory.
In mathematics, the Cayley–Dickson construction, named after Arthur Cayley and Leonard Eugene Dickson, produces a sequence of algebras over the field of real numbers, each with twice the dimension of the previous one. The algebras produced by this process are known as Cayley–Dickson algebras, for example complex numbers, quaternions, and octonions. These examples are useful composition algebras frequently applied in mathematical physics.
In mathematics, and more specifically in abstract algebra, a *-algebra is a mathematical structure consisting of two involutive ringsR and A, where R is commutative and A has the structure of an associative algebra over R. Involutive algebras generalize the idea of a number system equipped with conjugation, for example the complex numbers and complex conjugation, matrices over the complex numbers and conjugate transpose, and linear operators over a Hilbert space and Hermitian adjoints. However, it may happen that an algebra admits no involution at all.
In mathematics, a square is the result of multiplying a number by itself. The verb "to square" is used to denote this operation. Squaring is the same as raising to the power 2, and is denoted by a superscript 2; for instance, the square of 3 may be written as 32, which is the number 9. In some cases when superscripts are not available, as for instance in programming languages or plain text files, the notations x^2 or x**2 may be used in place of x2.
In abstract algebra, a split complex number has two real number components x and y, and is written z = x + yj, where j2 = 1. The conjugate of z is z = xyj. Since j2 = 1, the product of a number z with its conjugate is zz = x2y2, an isotropic quadratic form, N(z) = x2y2.
In abstract algebra, the biquaternions are the numbers w + xi + yj + zk, where w, x, y, and z are complex numbers, or variants thereof, and the elements of {1, i, j, k} multiply as in the quaternion group and commute with their coefficients. There are three types of biquaternions corresponding to complex numbers and the variations thereof:
In abstract algebra, the split-quaternions or coquaternions are elements of a 4-dimensional associative algebra introduced by James Cockle in 1849 under the latter name. Like the quaternions introduced by Hamilton in 1843, they form a four dimensional real vector space equipped with a multiplicative operation. But unlike the quaternions, the split-quaternions contain nontrivial zero divisors, nilpotent elements, and idempotents. As an algebra over the real numbers, they are isomorphic to the algebra of 2 × 2 real matrices. For other names for split-quaternions see the Synonyms section below.
In mathematics, a quaternion algebra over a field F is a central simple algebra A over F that has dimension 4 over F. Every quaternion algebra becomes a matrix algebra by extending scalars, i.e. for a suitable field extension K of F, is isomorphic to the 2×2 matrix algebra over K.
In mathematics, the split-octonions are an 8-dimensional nonassociative algebra over the real numbers. Unlike the standard octonions, they contain non-zero elements which are non-invertible. Also the signatures of their quadratic forms differ: the split-octonions have a split signature (4,4) whereas the octonions have a positive-definite signature (8,0).
In abstract algebra, a bicomplex number is a pair (w, z) of complex numbers constructed by the Cayley–Dickson process that defines the bicomplex conjugate , and the product of two bicomplex numbers as
A non-associative algebra is an algebra over a field where the binary multiplication operation is not assumed to be associative. That is, an algebraic structure A is a non-associative algebra over a field K if it is a vector space over K and is equipped with a K-bilinear binary multiplication operation A × AA which may or may not be associative. Examples include Lie algebras, Jordan algebras, the octonions, and three-dimensional Euclidean space equipped with the cross product operation. Since it is not assumed that the multiplication is associative, using parentheses to indicate the order of multiplications is necessary. For example, the expressions (ab)(cd), d and a(b ) may all yield different answers.
In mathematics, an octonion algebra or Cayley algebra over a field F is an algebraic structure which is an 8-dimensional composition algebra over F. In other words, it is a unital non-associative algebra A over F with a non-degenerate quadratic form N such that
In mathematics, Hurwitz's theorem is a theorem of Adolf Hurwitz (1859–1919), published posthumously in 1923, solving the Hurwitz problem for finite-dimensional unital real non-associative algebras endowed with a positive-definite quadratic form. The theorem states that if the quadratic form defines a homomorphism into the positive real numbers on the non-zero part of the algebra, then the algebra must be isomorphic to the real numbers, the complex numbers, the quaternions, or the octonions. Such algebras, sometimes called Hurwitz algebras, are examples of composition algebras.
In the field of mathematics called abstract algebra, a division algebra is, roughly speaking, an algebra over a field in which division, except by zero, is always possible.
In mathematics, the Hurwitz problem, named after Adolf Hurwitz, is the problem of finding multiplicative relations between quadratic forms which generalise those known to exist between sums of squares in certain numbers of variables.
In algebra, an Okubo algebra or pseudo-octonion algebra is an 8-dimensional non-associative algebra similar to the one studied by Susumu Okubo. Okubo algebras are composition algebras, flexible algebras, Lie admissible algebras, and power associative, but are not associative, not alternative algebras, and do not have an identity element.
In mathematics, a bioctonion, or complex octonion, is a pair (p,q) where p and q are biquaternions.
## References
1. Springer, T. A.; F. D. Veldkamp (2000). Octonions, Jordan Algebras and Exceptional Groups. Springer-Verlag. p. 18. ISBN 3-540-66337-1.
2. Jacobson, Nathan (1958). "Composition algebras and their automorphisms". Rendiconti del Circolo Matematico di Palermo . 7: 55–80. doi:10.1007/bf02854388. Zbl 0083.02702.
3. Guy Roos (2008) "Exceptional symmetric domains", §1: Cayley algebras, in Symmetries in Complex Analysis by Bruce Gilligan & Guy Roos, volume 468 of Contemporary Mathematics, American Mathematical Society, ISBN 978-0-8218-4459-5
4. Schafer, Richard D. (1995) [1966]. An introduction to non-associative algebras. Dover Publications. pp. 72–75. ISBN 0-486-68813-5. Zbl 0145.25601.
5. Kevin McCrimmon (2004) A Taste of Jordan Algebras, Universitext, Springer ISBN 0-387-95447-3 MR 2014924
6. Arthur A. Sagle & Ralph E. Walde (1973) Introduction to Lie Groups and Lie Algebras, pages 194−200, Academic Press
7. Dickson, L. E. (1919), "On Quaternions and Their Generalization and the History of the Eight Square Theorem", Annals of Mathematics , Second Series, Annals of Mathematics, 20 (3): 155–171, doi:10.2307/1967865, ISSN 0003-486X, JSTOR 1967865
8. Max Zorn (1931) "Alternativekörper und quadratische Systeme", Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg 9(3/4): 395–402
9. Albert, Adrian (1942). "Quadratic forms permitting composition". Annals of Mathematics . 43: 161–177. doi:10.2307/1968887. Zbl 0060.04003.
10. Max-Albert Knus, Alexander Merkurjev, Markus Rost, Jean-Pierre Tignol (1998) "Composition and Triality", chapter 8 in The Book of Involutions, pp. 451–511, Colloquium Publications v 44, American Mathematical Society ISBN 0-8218-0904-0 | 3,556 | 14,523 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-31 | latest | en | 0.866503 |
https://forum.arduino.cc/t/led-speedometer-at-rest/275347 | 1,653,829,186,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00060.warc.gz | 307,343,691 | 5,767 | # LED speedometer at rest
hi!
I ve been building a LED bar speedometer on a model car, calculating the velocity with an infrared RPM sensor. It works, except for one thing: once the car brakes and comes to a stop, the velocity wont become 0.
This is because we used the attachinterrupt function and it wont update the velocity until the RPM sensor gets HIGH again. Does anyone of you have any idea as to how we can force a function to update the velocity anyway?
Here is some of the code, the attachinterrupt:
``````attachInterrupt(5, interruption, FALLING);
``````
the function it calls:
``````void interruption(){Â Â
Â
 interval = (millis()-time_last);
 time_last = millis();
Â
}
``````
interval is then used to update the velocity:
``````void velocityCalc(){
 rpsec = (1000/interval)/2;
  velocity = (0.376991*rpsec*60*3.6)/60;
}
``````
``````if (millis() - time_last > something)
{
velocity = 0;
}
``````
I have already tried that, i have put it in the interrupt function. But that still doesnt work. It only goes back to zero after you start to drive again, so during the standstill, it still displays a couple of of km/h,
and once you provide throttle again, it quickly goes back to zero and then displays the new speed.
i have put it in the interrupt function
Why when you know that the ISR will not be triggered ?
Put it or something similar in the loop() function and set time_last to millis() as part of the ISR or outside the ISR by setting a flag variable inside the ISR.
The problem is that you're only triggering an update when you get a pulse from the speed sensor.
instead of measuring how long it's been since the last pulse, what you really need to do is measure how many pulses in a given time period.
``````const unsigned long UpdateInterval = 500; // update speed twice per second.
const unsigned long SpeedConstant = (whatever it needs to be for pulsecounts/interval);
unsigned long LastUpdate = millis(); // When was the Speed last updated?
unsigned long PulseCount = 0l;
unsigned int CurrentSpeed = 0;
...
void Interruption() {
 PulseCount++;
}
void CalculateSpeed() {
CurrentSpeed = PulseCount / (millis() - LastUpdate) * SpeedConstant;
}
...
void main() {
if (millis() - LastUpdate > UpdateInterval) {
 CalculateSpeed();
 DisplaySpeed(); // whatever code you want to use to display the result of the previous calculation.
 PulseCount=0;
 LastUpdate=millis();
}
// whatever other code you want to keep running
}
``````
Something like that will update the speed at consistent intervals which will avoid the display flashing between 2 similar values many times per second. In this case, it's every 1/2 second, but you could use longer values for more meaningful results to a human being. | 664 | 2,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.902401 |
http://nrich.maths.org/public/leg.php?code=72&cl=1&cldcmpid=143 | 1,503,284,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00413.warc.gz | 319,722,977 | 9,645 | # Search by Topic
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### Odd Squares
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### Circles, Circles
##### Stage: 1 and 2 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### Stop the Clock
##### Stage: 1 Challenge Level:
This is a game for two players. Can you find out how to be the first to get to 12 o'clock?
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### Stop the Clock for Two
##### Stage: 1 Challenge Level:
Stop the Clock game for an adult and child. How can you make sure you always win this game?
### Cuisenaire Rods
##### Stage: 2 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
### Counting Counters
##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Growing Garlic
##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### Triangle Pin-down
##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### The Add and Take-away Path
##### Stage: 1 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own?
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Snake Coils
##### Stage: 2 Challenge Level:
This challenge asks you to imagine a snake coiling on itself.
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### 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.
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Sticky Triangles
##### Stage: 2 Challenge Level:
Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Walking the Squares
##### Stage: 2 Challenge Level:
Find a route from the outside to the inside of this square, stepping on as many tiles as possible.
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### 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?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### 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?
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### 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?
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Roll over the Dice
##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Unit Differences
##### Stage: 1 Challenge Level:
This challenge is about finding the difference between numbers which have the same tens digit.
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Cut it Out
##### Stage: 2 Challenge Level:
Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Up and Down Staircases
##### Stage: 2 Challenge Level:
One block is needed to make an up-and-down staircase, with one step up and one step down. How many blocks would be needed to build an up-and-down staircase with 5 steps up and 5 steps down?
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. | 2,063 | 9,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.905559 |
https://www.calculatorsconversion.com/en/electrical/page/11/ | 1,550,267,276,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479159.2/warc/CC-MAIN-20190215204316-20190215230316-00556.warc.gz | 810,448,026 | 11,661 | ## Amp to kw – Conversion, formula, chart, convert and calculator free.
With this calculator you can convert online of Amperes to kW or kW to Amperes automatically, easily, quickly and free.
In order to facilitate the calculation we explain what formula is used, how to calculation in only 2 steps, table and examples of Ampere to kW conversions.
We also show the typical power factors of different constructions, appliances and motors.
## Amp to watts – Conversion, equation, table, convert and calculator free.
With this calculator you can convert from Amp to Watts easily, quickly and free any electric power, the calculation takes into account the power factor.
The calculation takes into account the power factor of the electrical system and give the most common values.
We also show the formula that is used for the conversion and a table with the main conversions from Amp to Watts.
## Hp to Amps (Amperes) – Conversion, calculator, formula, table, chart.
With this calculator you can convert online from Hp to Amps or Amperes to Hp automatically, easily, quickly and free.
For ease we explain how to convert from Hp to Amps in only 3 step, that the formula is used for the calculation, some examples and a table with the main conversions of Hp to Amps.
We also show the most common power factors in addition to efficiency values.
## kVA to Hp – Conversion, equation, table, convert and calculator free.
With this tool you can convert online from kVA to Hp automatically, easily, quickly and free.
For greater ease we explain that formula is used for the calculation and a table with the main conversions of kVA to Hp.
We also show the most common power factors of different constructions, appliances and motors, in addition to the most common efficiency values of the latter.
## HP to kVA (kilo-volt-ampere) – Online Convert free
With this tool you can convert from HP to kVA or also kVA to HP easily, quickly and free any electric power, the calculation takes into account the power factor. For greater ease we explain that formula is used for the calculation, some examples , how to convert from Hp to kVA in only 3 steps and … | 462 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-09 | longest | en | 0.881298 |
https://www.thestudentroom.co.uk/showthread.php?t=35669 | 1,524,684,242,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947939.52/warc/CC-MAIN-20180425174229-20180425194229-00409.warc.gz | 889,308,574 | 36,972 | x Turn on thread page Beta
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# P3 Trig. watch
Announcements
1. Hi,
I'd appreciate any help on this question.
f(A)= cosA/1+sinA + 1+sinA/cosA
Prove that f(A)=2secA
I cross multiplied and attempted to put under common denominator,but i could'nt get final ans.Is this the right method?Can anyone help!!!!
Sukh.
2. (Original post by sukh_b)
Hi,
I'd appreciate any help on this question.
f(A)= cosA/1+sinA + 1+sinA/cosA
Prove that f(A)=2secA
I cross multiplied and attempted to put under common denominator,but i could'nt get final ans.Is this the right method?Can anyone help!!!!
Sukh.
Yeh it is right. Just work through it again.
3. f(A) = cosA/1+sinA + 1+sinA/cosA
f(A) = (cos^2A + 1 + 2sinA + sin^2A)/cosA(1 + sinA)
But cos^2A + sin^2A = 1, so
f(A) = (2 + 2sinA)/cosA(1 + sinA)
f(A) = 2(1 + sinA)/cosA(1 + sinA)
f(A) = 2/cosA = 2secA as required
4. Cheers people.
I made the mistake on squishy's 3rd step.I didnt simplify.
Thanks again.
Sukh.
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# 5.14. How @RISK Computes Rank-Order Correlation
Applies to: @RISK 5.5.0 and later
The RiskCorrel( ) function can return the Pearson product-moment or Spearman rank-order correlation coefficient. How is the rank-order coefficient computed?
@RISK uses the method in Numerical Recipes by Press, Flannery, Teulosky, and Vetterling (Cambridge University Press; 1986), pages 488 and following.
Each number in each of the simulated distributions is replaced with its rank within that distribution, as an integer from 1 to N (number of iterations). If the values in a distribution are all different, as they usually are with continuous distributions, then the rank numbers will all be distinct. If there are duplicate numbers within the distribution, as often happens with discrete distributions, then "it is conventional to assign to all these 'ties' the mean of the ranks that they would have had if their values had been slightly different. This [is called the] midrank" (quoting from the reference book above).
Once the ranks are obtained, the rank-order coefficient is simply the Pearson linear correlation coefficient of the ranks.
The above explains how @RISK computes rank-order correlation after a simulation is complete. @RISK also uses rank-order correlation within a simulation, when drawing numbers for correlated distributions. This page gives details: How @RISK Correlates Inputs.
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# How Much Do We Need?
Name: Meghan Anderson Time Allotted: 70 minutes Grade Level: Kindergarten Subject: Math Materials Required: Math manipulatives A bag of apples The costume bag Quiz for each student iPads
IXL Math. Retrieved December 15, 2013, from http://www.ixl.com/math/kindergarten/addition-word-problems-sums-up-to-5 ABCYA. Retrieved December 15, 2013, from http://www.abcya.com/addition.htm
Michigan Content Expectations: K.OA.2 Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Objective: The student will solve addition word problems (within the range of numbers 1-10) about human needs and will use objects representing physical needs (i.e.: mittens, Linkin-Logs, candy, etc.) as aids to represent the problems, and the student will score 80% on the assessment in order to demonstrate proficiency. (Application) Assessment: Formative: During the stations, the students will write down the answers to their story problems on page two of each of the flip books next to their name. The teacher will be able to grade these and assess student learning from the data. Summative: The students will take a quiz about story problems and needs during the independent practice. They must score 80% on the assessment in order to demonstrate proficiency. If they do not score 80%, the teacher will take the time in a later lesson to do remediation work.
Instructional Procedure: 1. Anticipatory Set: (5 minutes) a. Gather the students all into one big circle and have the following class discussion: b. We have been learning about our many different needs and wants. We have talked about how our wants are the things we would like, but do not have to have in order to survive. We have talked about how our needs can be both physical needs and emotional needs. Remember, our main physical needs are food, shelter, clothing, and safety. Our emotional needs are fun, freedom, power, and belonging. Today, during this math lesson, we are going to talk about how much of each of our needs we need in order to live! c. Can we start with our emotional needs? How much fun does someone need? How much freedom? How much power? How much belonging? d. We cannot count our emotions can we? That is because these are feelings happen in our head and our heart. They are too important to count. | 541 | 2,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.943487 |
http://nrich.maths.org/public/leg.php?code=87&cl=3&cldcmpid=6793 | 1,485,168,412,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00566-ip-10-171-10-70.ec2.internal.warc.gz | 196,130,273 | 6,624 | # Search by Topic
#### Resources tagged with Isosceles triangles similar to Isosceles Reduction:
Filter by: Content type:
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Challenge level:
### There are 17 results
Broad Topics > 2D Geometry, Shape and Space > Isosceles triangles
### Tricircle
##### Stage: 4 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### The Eyeball Theorem
##### Stage: 4 and 5 Challenge Level:
Two tangents are drawn to the other circle from the centres of a pair of circles. What can you say about the chords cut off by these tangents. Be patient - this problem may be slow to load.
### Xtra
##### Stage: 4 and 5 Challenge Level:
Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Farhan's Poor Square
##### Stage: 4 Challenge Level:
From the measurements and the clue given find the area of the square that is not covered by the triangle and the circle.
### Lighting up Time
##### Stage: 2 and 3 Challenge Level:
A very mathematical light - what can you see?
### Arrh!
##### Stage: 4 Challenge Level:
Triangle ABC is equilateral. D, the midpoint of BC, is the centre of the semi-circle whose radius is R which touches AB and AC, as well as a smaller circle with radius r which also touches AB and AC. . . .
### Three Way Split
##### Stage: 4 Challenge Level:
Take any point P inside an equilateral triangle. Draw PA, PB and PC from P perpendicular to the sides of the triangle where A, B and C are points on the sides. Prove that PA + PB + PC is a constant.
### Lens Angle
##### Stage: 4 Challenge Level:
Find the missing angle between the two secants to the circle when the two angles at the centre subtended by the arcs created by the intersections of the secants and the circle are 50 and 120 degrees.
### Are You Kidding
##### Stage: 4 Challenge Level:
If the altitude of an isosceles triangle is 8 units and the perimeter of the triangle is 32 units.... What is the area of the triangle?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Hexy-metry
##### Stage: 4 Challenge Level:
A hexagon, with sides alternately a and b units in length, is inscribed in a circle. How big is the radius of the circle?
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
##### Stage: 4 Challenge Level:
Find the area of the shaded region created by the two overlapping triangles in terms of a and b?
### Interacting with the Geometry of the Circle
##### Stage: 1, 2, 3 and 4
Jennifer Piggott and Charlie Gilderdale describe a free interactive circular geoboard environment that can lead learners to pose mathematical questions.
### Pareq Exists
##### Stage: 4 Challenge Level:
Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.
### Two Triangles in a Square
##### Stage: 4 Challenge Level:
Given that ABCD is a square, M is the mid point of AD and CP is perpendicular to MB with P on MB, prove DP = DC. | 919 | 3,811 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2017-04 | longest | en | 0.853346 |
https://www.studyadda.com/question-bank/dual-nature-of-electron_q12/1324/94569 | 1,601,215,324,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400279782.77/warc/CC-MAIN-20200927121105-20200927151105-00561.warc.gz | 1,042,827,305 | 18,384 | • question_answer What will be de-Broglie wavelength of an electron moving with a velocity of $1.2\times {{10}^{5}}\,m{{s}^{-1}}$ [MP PET 2000] A) $6.068\times {{10}^{-9}}$ B) $3.133\times {{10}^{-37}}$ C) $6.626\times {{10}^{-9}}$ D) $6.018\times {{10}^{-7}}$
$\lambda =\frac{h}{p}$, $p=mv$ $\lambda =\frac{h}{mv}$$=\frac{6.62\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times 1.2\times {{10}^{5}}}$ $\lambda =6.626\times {{10}^{-9}}m$. | 215 | 558 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-40 | latest | en | 0.361335 |
http://webwork.maa.org/w/images/2/24/PointAnswers1.txt | 1,563,709,664,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-30/segments/1563195526948.55/warc/CC-MAIN-20190721102738-20190721124738-00535.warc.gz | 170,138,067 | 1,343 | ## DESCRIPTION ## Precalculus: answers are lists of points ## ENDDESCRIPTION ## KEYWORDS('precalculus', 'list of points') ## DBsubject('WeBWorK') ## DBchapter('WeBWorK Tutorial') ## DBsection('Fort Lewis Tutorial 2011') ## Date('01/30/2011') ## Author('Paul Pearson') ## Institution('Fort Lewis College') ## TitleText1('') ## EditionText1('') ## AuthorText1('') ## Section1('') ## Problem1('') ########################### # Initialization DOCUMENT(); loadMacros( "PGstandard.pl", "MathObjects.pl", "AnswerFormatHelp.pl", "contextLimitedPoint.pl", ); TEXT(beginproblem()); ########################### # Setup Context("LimitedPoint"); \$f = Compute("x^2-1"); \$xint = List( Point("(1,0)"), Point("(-1,0)") ); \$yint = List( Point("(0,-1)") ); ########################### # Main text Context()->texStrings; BEGIN_TEXT Enter the x-intercept(s) and y-intercept(s) of \( y = \$f \). Enter a point as \( (a,b) \), including the parentheses. If there is more than one correct answer, enter a comma separated list of points. \$BR \$BR x-intercept(s): \{ ans_rule(20) \} \{ AnswerFormatHelp("points") \} \$BR y-intercept(s): \{ ans_rule(20) \} \{ AnswerFormatHelp("points") \} END_TEXT Context()->normalStrings; ############################ # Answer evaluation \$showPartialCorrectAnswers = 1; ANS( \$xint->cmp() ); ANS( \$yint->cmp() ); ############################ # Solution Context()->texStrings; BEGIN_SOLUTION \${PAR}SOLUTION:\${PAR} Solution explanation goes here. END_SOLUTION Context()->normalStrings; COMMENT('MathObject version.'); ENDDOCUMENT(); | 406 | 1,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-30 | latest | en | 0.351024 |
https://brainly.in/question/40582 | 1,485,302,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00255-ip-10-171-10-70.ec2.internal.warc.gz | 792,526,731 | 11,100 | Derive expression for drift speed by using drude Lorentz theory?
1
2014-09-23T21:49:30+05:30
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Drift speed of electrons is the average speed of electrons over the length of a conductor when a potential difference is applied to the ends of the conductor. The electrons move under the influence of electric and magnetic effects of atoms and particles inside conductor.
Let A be cross section of a conductor wire of length L. Let its resistivity be ρ. Let a current I flow through it. Let V be voltage difference across the conductor. R be the resistance of wire.
Let T be the temperature of the wire. Let α be the thermal coefficient of resistance. Let e be the charge on an electron. Let there be n electrons per unit volume of the conductor. Let m be mass of the wire. Let M be molar mass of the conductor. Letd be volume density of the conductor.
N = Avogadro number (number of atoms in a mole of the conductor).
Let us say that there are f free electrons in each atom.
I = current flowing across the wire = number of charged particles * their charge crossing a particular cross section P' of wire in one second.
Suppose an electron travels (on an average) x meters in t seconds. Then average drift speed v of an electron is x/t meters/sec.
Let us take volume x * A to one side of P'. All the electrons in the volume x * Awill cross P' in t seconds.
So the charge crossing P' in one second is = I = x*A*n * e / t
I = n A e v or v = I / (n A e)
Resistivity of a conductor = ρ = ρ
₀ (1+αT) taking into account the thermal
increase of resistance.
Resistance of a conductor = R = ρL / A = ρ
₀ (1+α T) L / A
current = I = V/R = V / [ ρ
₀ L (1+α T) L / A ] = V A / [ ρ₀ L (1+α T)]
n = electron density = N atoms * f free electrons per atom / molar volume
= N f / (M/d) = N f d / M
So, the drift velocity (speed) = v =
= I / n A e
= {V A / [ρ
₀ L (1+αT) ] } / (N f d /M) (A) e
drift speed
v = V M / N f d e [ρ
L (1+αT) ]
hope u understand that easily. just read it on. | 683 | 2,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-04 | latest | en | 0.745217 |
https://www.cfd-online.com/W/index.php?title=Introduction_to_turbulence/Homogeneous_turbulence&diff=8995&oldid=8994 | 1,503,185,963,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00644.warc.gz | 899,891,188 | 10,471 | # Introduction to turbulence/Homogeneous turbulence
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Revision as of 07:16, 12 March 2008 (view source)Michail (Talk | contribs) (→A first look at decaying turbulence)← Older edit Revision as of 09:25, 12 March 2008 (view source)Michail (Talk | contribs) Newer edit → Line 24: Line 24: [/itex] [/itex]
(3)
(3)
+ + Now you can't get any simpler than this. Yet unbelievably we still don't have enough information to solve it. Let's try. Suppose we use the extanded ideas of Kolmogorov we introduced in Chapter 3 to related the dissipation to the turbulence energy, say: + +
+ :$+ \epsilon = f \left( Re \right) \frac{u^{3}}{l} +$ + (4)
## A first look at decaying turbulence
Look, for example, at the decay of turbulence which has already been generated. If this turbulence is homogeneous and there is no mean velocity gradient to generate new turbulence, the kinetic energy equation reduces to simply:
$\frac{d}{dt} k = - \epsilon$ (1)
This is often written (especially for isotropic turbulence) as:
$\frac{d}{dt} \left[ \frac{3}{2} u^{2} \right] = - \epsilon$ (2)
where
$k \equiv \frac{3}{2} u^{2}$ (3)
Now you can't get any simpler than this. Yet unbelievably we still don't have enough information to solve it. Let's try. Suppose we use the extanded ideas of Kolmogorov we introduced in Chapter 3 to related the dissipation to the turbulence energy, say:
$\epsilon = f \left( Re \right) \frac{u^{3}}{l}$ (4) | 421 | 1,481 | {"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": 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} | 3.453125 | 3 | CC-MAIN-2017-34 | latest | en | 0.93731 |
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# If y is an integer, y=|x|+x, is y=0? 1)x<0 2)y<1
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If y is an integer, y=|x|+x, is y=0? 1)x<0 2)y<1 [#permalink]
### Show Tags
28 Dec 2006, 00:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If y is an integer, y=|x|+x, is y=0?
1)x<0
2)y<1
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### Show Tags
28 Dec 2006, 01:51
I must pick A....
If x is less than 0 then you take any value and you will get y=0
x=-0.5
0.5-0.5=0
x=-1.8
1.8-1.8=0.....
So on.....
B is obviously not enougth....
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### Show Tags
28 Dec 2006, 03:06
in fact, st2 is sufficient as well so answer is D.
st1 is sufficient as noted above:
if x<0 then |x|=-x
so y=|x|+x = -x+x = 0
suff.
st2 is sufficient as well:
from stem only we can deduce that y is non-negative:
if x<0 then y=0 as we showed above,
if x=0 then obviously y=0.
if x>0 then |x|=x and y = 2x > 0
also note that y is an integer (!!!!!!!)
st2 says y<1. the only non-negative integer less than 1 is 0.
so y must be 0.
hence sufficient.
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### Show Tags
28 Dec 2006, 04:18
I'd go with D
Given: y is integer, y = |x| + x
asked : does y = 0 ?
from the question, 0=< y =< infinity; that is, y can range between zero and any positive integer.
because |x| is always positive.
if x is positive then y = 2x
if x is negative then y = 0
(1) x<0
--------------
x is negative then y = x - x = 0 .. YES y = 0
statement 1 is sufficient
(2) y<1
--------------
the only possible value for y < 1 is zero because y can never be negative and y is an integer so it can't be some fraction less than 1 like 1/2 or so.
So y in this case has to equal zero [ y = 0 ]
statement 2 is sufficient
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### Show Tags
28 Dec 2006, 04:58
Damn, again, i somehow tend to misread this "interger" thing.... yes should be D...
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28 Dec 2006, 07:57
You are right! OA is D
Mind works better after a good night's sleep!
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### Show Tags
28 Dec 2006, 22:22
D.
y can be only >= 0. so (2) is actually Suff.
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### Show Tags
30 Dec 2006, 19:22
D for me too
From 1) Obvious
From 2)
Possible values of Y is 0 or negative..but negative is not possible because y = |x| + x
|x| + x could be either positive or 0.
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### Show Tags
31 Dec 2006, 00:52
y=|x|+x = {0 for x<0; 0 for x=0; 2x for x>0}
Both (1) and (2) are suff to arrive at a single solution, then D.
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# If y is an integer, y=|x|+x, is y=0? 1)x<0 2)y<1
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®. | 1,404 | 4,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | longest | en | 0.874887 |
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# CAT 2018 [SLOT 2] Quant Question with Solution 10
Question:
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ?lling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ?lled on Thursday if both pumps were used simultaneously all along?
1. 4:36 pm
2. 4:12 pm
3. 4:24 pm
4. 4:48 pm
Let x be the time, on a 24 hours clock, at which the tank is empty.
Time taken by pipe A alone to fill the tank is (20 – x) hrs.
Time taken by pipe B alone to fill the tank is (18 – x) hrs.
On the other day, A fill the tank for (15 – x) hrs and B for 2 hrs.
Let A and B be the rate of works of pipe A and B respectively.
$\Rightarrow(20-x) A=(18-x) B=(17-x) A+2 B$
$\Rightarrow \frac{A}{B}=\frac{2}{3}$
$\Rightarrow(20-x) 2=(18-x) 3$
$\Rightarrow(20-x) A=1$
Let $(20-x) A=1$
$A=\frac{1}{6}$
$B=\frac{1}{4}$
When both work simultaneously, time taken
$=\frac{1}{\frac{1}{6}+\frac{1}{4}}=2.4 \mathrm{hrs}=2 \mathrm{hrs} 24 \mathrm{min}$
The tank will be filled by $16 : 24 \mathrm{i.e.} .4 : 24 \mathrm{pm}$
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CAT 24 & 25 best online courses | 540 | 1,715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.812851 |
https://mathoverflow.net/questions/140819/graphs-with-graphic-imbalance-sequences | 1,674,795,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494936.89/warc/CC-MAIN-20230127033656-20230127063656-00437.warc.gz | 399,836,420 | 26,207 | Graphs with graphic imbalance sequences
Let $G$ be simple undirected graph and $e=uv\in E(G)$.
The imbalance of the edge $e$ is the value $imb(e)=|d(u)-d(v)|$.
Let $M_{G}$ denotes the imbalance sequence (or more correctly, multiset of all edge imbalances) of $G$.
I can prove that if $T$ is a tree, then $M_{T}$ is graphic.
However, in general case it isn't true. See A question on graphic sequences for simple counterexamples.
Based on these I came to the next
Imbalance Conjecture: Suppose that for all edges $e\in E(G)$ we have $imb(e)>0$. Then $M_{G}$ is graphic.
ADDED 1: Vova Skochko, a good friend of mine, verified the conjecture for all such graphs with $\leq 9$ vertices.
ADDED 2: Recall that the value $Irr(G)=\sum_{e\in E(G)}imb(e)$ is called irregularity of $G$.
In light of Erdos-Gallai theorem the Imbalance Conjecture is equivalent to the next statement
Suppose that for all edges $e\in E(G)$ we have $imb(e)>0$. Then for all $E'\subset E(G)$ it holds $$Irr(G)\leq |E'|(|E'|-1)+\sum_{e\notin E'}\left(\min\{{|E'|,imb(e)}\}+imb(e)\right).$$
I wonder does this bound on $Irr(G)$ implies other known bounds on $Irr(G)$ for such graphs?
• Sergiy, It seems like for any graph, the imbalance sequence is the degree sequence of a multigraph (repeated edges allowed). For trees this multigraph happens to be a simple graph. Perhaps the question should be to find a direct description of the associated multigraph from the original graph. Sep 1, 2013 at 22:10
• Gjergji Zaimi, I'm interested in degrees sequences of simple graphs, not multigraphs. However, it is easy to see that $\sum M_{G}$ is always even, so the imbalance sequence of any graph is the degree sequence of some pseudograph. Oh, and I'm curious how can you prove that the imbalance sequence of a tree is graphic? Because my proof is inductional. Sep 2, 2013 at 11:18
• Jernej, no it's not! For counterexample on $9$ vertices see mathoverflow.net/questions/140198/… Sep 3, 2013 at 12:00
• Obviously, I like this question :) Is there some special motivation or just curiosity? Sep 8, 2013 at 14:19
• @Gjergji Zaimi: It is known that the multiset $M$ is multigraphic if and only if $\sum M$ is even and $\sum M\geq 2\max M$. Thus for every even $m\in\mathbb{N}$ one can take any $m+1$-regular graph $H$ with $\geq 4$ vertices, such that $H$ has a matching of a size $\frac{m}{2}$ (for example $H=K_{m+1,m+1}$). Then delete this matching, add new vertex $v$ and new edges from $v$ to the vertices from this matching. Then again add a new vertex $u$ and the edge $uv$ to obtain a new graph $G$. We have, that $M_{G}=\{m\}$ is not multigraphic. Sep 11, 2013 at 11:33 | 803 | 2,647 | {"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} | 3.390625 | 3 | CC-MAIN-2023-06 | latest | en | 0.882423 |
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What does a barometer measure? During which year do humans grow the fastest? Gather your wits and measure your knowledge by taking this quiz.
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### Trains, Speed, and Tunnels
```
Date: 09/10/98 at 19:16:05
From: Matthew
Subject: Trains and Tunnels
A freight train is 500m long. It passes through a tunnel 2000m long.
Sixty seconds elapse between the time the last car enters and the time
the engine exits the tunnel. How fast is the train traveling?
I would probably be able to figure this out if I knew the time when the
engine enters and exits the tunnel, but I don't know how to calculate
the speed the way the question says. I've tried drawing a picture on
graph paper, but it doesn't help. I hope you can help me. Thank you.
```
```
Date: 09/11/98 at 17:00:25
From: Doctor Peterson
Subject: Re: Trains and Tunnels
Hi, Matthew. It is a little tricky, but when you see it right, it's
not too bad. Let's draw a sort of graph, with the distance through the
tunnel along the x-axis and time along the y-axis (which I'll draw
downward for convenience):
tunnel
in 2000m out
0+====================+---> <-- at time 0, the last car
|\ 500m \ enters the tunnel
| \ \
| \ \
| \ train \
| \ \
| \ \
| \ \
60| \ \ <-- at time 60, the engine
V last engine exits the tunnel
car
You could also just draw a picture of the train and tunnel at the
start and end of the 60 seconds, to see where it is:
___________________________ time 0
|500m tunnel |
XXXXXX
|___________________________|
2000m
___________________________ time 60
| tunnel 500m|
XXXXXX
|___________________________|
At time 0, the engine was 500 m into the tunnel, and at time 60 it was
at the end. So in those 60 seconds, the engine traveled (2000 - 500)
meters. Now can you finish the problem?
- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 615 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-26 | longest | en | 0.915153 |
http://www.brightstorm.com/tag/distance/page/2 | 1,432,517,801,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928114.23/warc/CC-MAIN-20150521113208-00255-ip-10-180-206-219.ec2.internal.warc.gz | 352,767,745 | 13,753 | • #### The Midpoint and Distance Formulas in 3D - Problem 4
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 948 | 3,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.908123 |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=719&pid=6261#pid6261 | 1,675,328,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00806.warc.gz | 410,079,302 | 7,585 | fractional iteration by schröder and by binomial-formula Gottfried Ultimate Fellow Posts: 889 Threads: 130 Joined: Aug 2007 11/23/2011, 04:45 PM (This post was last modified: 11/23/2011, 04:49 PM by Gottfried.) Just a small result from some playing around in a break; I'll look at the confirmation for a wider range of parameters later. This days I'm re-considering the differences of the fractional iterates of $b^x -1$ using the schröder-function by their carleman-matrices. At the moment I use b=2, comparing the results when I use the power series centered around fixpoint t0=0 versus that around fixpoint t1=1. The range of 01. The point x0, at which the most symmetric sinusoidal curve occurs seems to be x0~0.382160520000... (not rational!) which has a special property for base b=2 which I'll discuss in a later post. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »
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Users browsing this thread: 1 Guest(s) | 678 | 2,142 | {"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": 1, "/images/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-2023-06 | latest | en | 0.795136 |
https://likunxie.wordpress.com/2016/08/06/ch-1-solutions/comment-page-1/ | 1,606,320,644,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00559.warc.gz | 374,408,070 | 22,524 | # [Solutions]Ch 1 Special Relativity and Flat Spacetime
1-6
https://petraaxolotl.wordpress.com/chapter-1-special-relativity-and-flat-spacetime/
9. For a system of discrete point particles the energy-momentum tensor take the form
$T^{\mu \nu}=\Sigma_a \frac{p_{\mu}^{(a)} p_{\nu}^{(a)} }{p^{(a)}}\delta^{(3)}(x-x^{(a)})$,
where the index a labels the different particles. Show that, for a dense collection of particles with isotopically distributed velocities, we can smooth over the individual particle worldlines to obtain the perfect-fluid energy-momentum tensor
$T^{\mu\nu}=(\rho +p)U^{\mu}U^{\nu} + p\eta^{\mu\nu}$.
Doing average over 4-volume,
${\bf T}^{\mu\nu}=\dfrac{1}{\Delta V_4}\int_{\Delta V_4}T^{\mu\nu}d V=\dfrac{1}{\sqrt{-g} d^3x^i dx^0}\int_{\Delta V_4}T^{\mu\nu} \sqrt{-g} d^3x^i dx^0$.
Then a) in ${\bf T}^{\mu\nu}$ only delta-functions depend on x, b) metric determinant g is a macroscopic quantity, is constant over selected volume and can also be taken away from the integral.
${\bf T}^{\mu\nu}=\dfrac{1}{d^3x^i dx^0}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}\int_{\Delta V_4} \delta^{(3)}({\bf x}-{\bf x}^{(a)}) d^3 x^i dx^0 = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^\mu p_a^\nu}{p_a^0}$
In the last expression the sum is taken over the particles which have worldlines passing through $\Delta V_4$.
Consider symmetry,
${\bf T}^{0 0} = \dfrac{1}{d^3x^i}\sum_a p_a^0 \equiv \rho$
${\bf T}^{i 0} = \dfrac{1}{d^3x^i}\sum_a p_a^i$. Because of isotropy, ${\bf T}^{i 0}= 0$.
${\bf T}^{i j} = \dfrac{1}{d^3x^i}\sum_a\dfrac{p_a^i p_a^j}{p_a^0}$. The sum should produce a symmetric macroscopic 3-tensor of second order. But all 3-tensors are defined by 3 eigenvectors. Because no preferred direction exits and no preferred directions correspond to the case when the matrix has all eigenvalues equal, that is when the matrix is proportional to kronecker delta. The coefficient of proportionality is the pressure: ${\bf T}^{i j} \equiv P \delta^{ij}$.
Replace ：
$\delta ^{ij}\to \eta^{ij}+U^i U^j$
$T^{00}\to \rho U^0U^0$,
then we get the result.
10. Using the tensor transformation law applied to $F_{\mu\nu}$, show how the electric and magnetic field 3-vectors E and B transform under
(a) a rotation about the y-axis;
(b) a boost along the z-axis.
(a) $\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} 1&0&0&0 \\ 0&cos\theta&0&sin\theta\\ 0&0&1&0\\ 0&-sin\theta&0&cos\theta \end{array}\right)$
(b)$\Lambda^{\mu'}_{\nu}=\left(\begin{array}{cccc} cosh\theta&0&0&-sinh\theta \\ 0&1&0&0\\ 0&0&1&0\\ -sinh\theta&0&0&cosh\theta \end{array}\right)$
$\overline{F}^{\mu\nu}=\Lambda F \Lambda^T=\left(\begin{array}{cccc} 0& - E1*cosh(x) - B2*sinh(x)& B1*sinh(x) - E2*cosh(x)& E3*sinh(x)^2 - E3*cosh(x)^2\\ E1*cosh(x) + B2*sinh(x)& 0& B3&- B2*cosh(x) - E1*sinh(x)\\ E2*cosh(x) - B1*sinh(x)& -B3& 0& B1*cosh(x) - E2*sinh(x)\\ E3*cosh(x)^2 - E3*sinh(x)^2& B2*cosh(x) + E1*sinh(x)&E2*sinh(x) - B1*cosh(x)& 0 \end{array}\right)$.
13. Consider adding to the Lagrangian for electromagnetism an additional term of the form $\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}$
(a) Express $\mathcal{L}'$ in terms of E and B.
(b) Show that including $\mathcal{L}'$ does not affect Maxwell’s equations. Can you think of a deep reason for this?
(a) $\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=\tilde{\epsilon}_{ijk}(F^i F^{jk}-F^{i0}F^{jk}+F^{ij}F^{0k}-F^{ij}F^{k0})=4F^{0i}F^{jk}\tilde{\epsilon}^{ijk}=-8\vec{E}\vec{B}$
(b) The expression in (a) can be expressed by 4-dimension divergence:
$\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F{\rho\sigma}=4\partial_i[\tilde{\epsilon}^{ijlm}A_k (\partial_l A_m)]$
Integrate $\mathcal{L}'$ over 4-dimension space, according to Stokes theorem, we can get the additional term in Lagrangian $L$. This term vanishes when varying action $S$ | 1,470 | 3,824 | {"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": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | longest | en | 0.683637 |
https://quizzino.com/quiz-the-trickiest-math-challenge-youll-ever-take/ | 1,674,795,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494936.89/warc/CC-MAIN-20230127033656-20230127063656-00156.warc.gz | 480,796,932 | 24,496 | Quiz: The Trickiest Math Challenge You’ll Ever Take
We gave this math quiz to 890 educated adults and only 2% got a perfect score. Take on the challenge and let us know how well you did.The Trickiest Math Challenge You’ll Ever Take
What are the 7 unsolved math questions?
Clay “to increase and disseminate mathematical knowledge.” The seven problems, which were announced in 2000, are the Riemann hypothesis, P versus NP problem, Birch and Swinnerton-Dyer conjecture, Hodge conjecture, Navier-Stokes equation, Yang-Mills theory, and Poincaré conjecture.
What's the hardest math problem ever solved?
In 2019, mathematicians finally solved a math puzzle that had stumped them for decades. It's called a Diophantine Equation, and it's sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from one to 100.
What is the hardest math test in the world?
The 3 hardest exams in the world The International Mathematical Olympiad (IMO) Remember the Maths Challenge? The All Soul's College Fellowship Examination. It's tough getting into Oxford. 'Gaokao', the Chinese National College Entrance Exam.
Has 3X 1 been solved?
After that, the 3X + 1 problem has appeared in various forms. It is one of the most infamous unsolved puzzles in the word. Prizes have been offered for its solution for more than forty years, but no one has completely and successfully solved it [5].
What's the answer to x3 y3 z3 K?
In mathematics, entirely by coincidence, there exists a polynomial equation for which the answer, 42, had similarly eluded mathematicians for decades. The equation x3+y3+z3=k is known as the sum of cubes problem.
What is the 1 million dollar math problem?
The Riemann hypothesis – an unsolved problem in pure mathematics, the solution of which would have major implications in number theory and encryption – is one of the seven \$1 million Millennium Prize Problems. First proposed by Bernhard Riemann in 1859, the hypothesis relates to the distribution of prime numbers.
What is an E in math?
Euler's Number 'e' is a numerical constant used in mathematical calculations. The value of e is 2.718281828459045…so on. Just like pi(π), e is also an irrational number. It is described basically under logarithm concepts. 'e' is a mathematical constant, which is basically the base of the natural logarithm.
Who Solved Millennium Problems?
Grigori Perelman, a Russian mathematician, solved one of the world's most complicated math problems several years ago. The Poincare Conjecture was the first of the seven Millennium Prize Problems to be solved.
Who created math?
Archimedes is known as the Father of Mathematics. Mathematics is one of the ancient sciences developed in time immemorial. A major topic of discussion regarding this particular field of science is about who is the father of mathematics.
Is there an unsolvable equation?
For decades, a math puzzle has stumped the smartest mathematicians in the world. x3+y3+z3=k, with k being all the numbers from one to 100, is a Diophantine equation that's sometimes known as "summing of three cubes."
What is this pi?
Succinctly, pi—which is written as the Greek letter for p, or π—is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle's size, this ratio will always equal pi. In decimal form, the value of pi is approximately 3.14.
What is the easiest math problem in the world?
The Collatz Conjecture is the simplest math problem no one can solve — it is easy enough for almost anyone to understand but notoriously difficult to solve. So what is the Collatz Conjecture and what makes it so difficult? Veritasium investigates.
What is the hardest exam to pass?
Top 10 Toughest Exams in the World UMSLE (The United States Medical Licensing Examination) CA (Chartered Accountant) California Bar Exam. LNAT (Law National Aptitude Test) UMSLE (The United States Medical Licensing Examination) CA (Chartered Accountant) California Bar Exam. IES (Indian Engineering Services)
What country has the hardest exams?
China ✅ Gaokao (China) Each year, millions of students in China take the gaokao, taken by students in their third and final year of high school typically from June 7 to June 8 or 9. It's China's version of the American SAT and British A-level tests, and is known as one of the hardest exams in the world.
Why is 3x1 impossible?
Multiply by 3 and add 1. From the resulting even number, divide away the highest power of 2 to get a new odd number T(x). If you keep repeating this operation do you eventually hit 1, no matter what odd number you began with? Simple to state, this problem remains unsolved.
Is 3x 1 possible?
The 3x+1 problem concerns an iterated function and the question of whether it always reaches 1 when starting from any positive integer. It is also known as the Collatz problem or the hailstone problem. . This leads to the sequence 3, 10, 5, 16, 4, 2, 1, 4, 2, 1, which indeed reaches 1.
Is 0 an even number?
When 0 is divided by 2, the resulting quotient turns out to also be 0—an integer, thereby classifying it as an even number. Though many are quick to denounce zero as not a number at all, some quick arithmetic clears up the confusion surrounding the number, an even number at that.
How do you solve x3 y3 z3?
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx).
What is the longest equation?
the Boolean Pythagorean Triples problem What is the longest equation in the world? According to Sciencealert, the longest math equation contains around 200 terabytes of text. Called the Boolean Pythagorean Triples problem, it was first proposed by California-based mathematician Ronald Graham, back in the 1980s.
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https://computinglearner.com/using-the-stack-data-structure-in-java-with-examples/ | 1,712,986,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00046.warc.gz | 165,870,997 | 29,456 | # Using the Stack data structure in Java (with examples)
The Stack data structure has many applications in programming. In this post, you will learn what are the Stack operations and how to use them to solve real-life problems.
## Stacks in the real world
Programming is about modelling real-life situations. The use of stacks is not the exception.
The Tower of Hanoi is a game “consisting of three rods and a number of disks of various diameters, which can slide onto any rod. The puzzle begins with the disks stacked on one rod in order of decreasing size, the smallest at the top, thus approximating a conical shape. The objective of the puzzle is to move the entire stack to the last rod, obeying the following rules:
1. Only one disk may be moved at a time.
2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack or on an empty rod.
3. No disk may be placed on top of a disk that is smaller than it.
“. Source Wikipedia.
The rules of the game, states that you can only move one disk at a time, and it must be the disc on top of any of three rods.
That behaviour, of removing only the element at the top each time, and adding a new element only to the top, is known as a stack in Computer Science. Also known as the Last In, First Out (LIFO) principle.
Notice that if you remove more than a disc at a time (something you can do) while playing the game, you will be breaking the rules.
Let’s now see the basic operations of the Stack Abstract Data Type (ADT).
## The Stack Abstract Data Type
Find below the basic operations of the Stack ADT and an explanation.
• Push: adds an element to the top of the stack
• Pop: retrieve and removes the element at the top
• Top: retrieve the element at the top without removing it
• isEmpty: return true if the stack is empty, otherwise it returns false.
In different programming languages, you can find extra operations and different names for the operations above. But you will always find an operation that will do the action of each of the four operations mentioned above.
## Stack data structure examples
One example is to find the solution to the Tower of Hanoi problem given the number of disks. However, this solution is difficult to understand at a beginner level. So, I’ll show you examples that are easier to understand, so later you can solve the Tower of Hanoi problem on your own.
### Problem 1: Invert an array of integers
Let’s create a program that takes an array as input and print the inverted array.
We can use a stack to solve the previous problem. Notice that in a stack, the last in is the first out. So, if we add all the array elements to a stack, when we take them out, they will be in reverse order.
Let’s see the implementation.
`````` public static int [] invertArrray(int arr[]){
Stack<Integer> stack = new Stack<>();
for (int i=0; i<arr.length; i++){
stack.push(arr[i]);
}
int index = 0;
while(!stack.isEmpty()){
arr[index] = stack.pop();
index++;
}
return arr;
}``````
All we did in the previous code was to add the elements of the array to the stack and remove the elements from stacks and add them to the array. That’s it! now the array is inverted.
When working with stacks, we use a while loop to go through all the elements in a similar to the example of the Queue data structure. In this case, we don’t need an additional stack because we are storing the elements in the array.
You can also add a method to print an array so you can test the code below. Such a method can be like this:
```public static void printArray(int arr[]){
System.out.println();
for (int i=0; i<arr.length; i++)
System.out.print(arr[i] + ", ");
}```
You are just printing all the elements in the array to confirm if the array is inverted or not.
The main method of the console application can look like this:
```public static void main(String[] args) {
int array[] = {1,2,3,4,5,6,7};
printArray(array);
//Invert the array
array = invertArrray(array);
// Print the inverted array.
printArray(array);
}```
## Summary
The Stack data structure follows the LIFO principle. It means that the last element to go in the stack is the first one to go out.
Many problems can be solved using stacks. However, in all cases, you won’t see the word stack as part of the problem description. Like you saw in the example of inverting an array.
The key to know when to use stacks is to think about the process you can follow to give the solution to the problem. If it follows the LIFO principle, then you can use stacks.
Some programming languages implement extra operations in the Stack data structure. As a word of caution, if you have to use more operations than the basic ones, you should evaluate if stacks are the right data structure to use in the solution to that particular problem.
H@ppy coding!
1- Using the List data in Java
2- Using the Queue data structure in Java | 1,114 | 4,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.911746 |
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Piping elbows and bends are very important fitting which are used very frequently for changing
direction
in piping system. Piping Elbow and Piping bend are not the same, even though sometimes these two terms
are interchangeably used. A BEND is simply a generic term in piping for an “offset” – a change in direction
of the piping. It signifies that there is a “bend” i.e, a change in direction of the piping (usually for some
specific reason) – but it lacks specific, engineering definition as to direction and degree. Bends are usually
made by using a bending machine (hot bending and cold bending) on site and suited for a specific need.
Uses of bends are economic as it reduces number of expensive fittings. An ELBOW, on the other hand, is
a specific, standard, engineered bend pre-fabricated as a spool piece (based on ASME B 16.9) and designed
to either be screwed, flanged, or welded to the piping it is associated with. An elbow can be 45 degree or 90
degree. There can also be custom-designed elbows, although most are catagorized as either “short radius”
In short “All bends are elbows but all elbows are not bend”
Whenever the term elbow is used, it must also carry the qualifiers of type (45 or 90 degree) and radius
(short or long) – besides the nominal size.
Elbows can change direction to any angle as per requirement. An elbow angle can be defined as the angle
by which the flow direction deviates from its original flowing direction (See Fig.1 below).Even though An
elbow angle can be anything greater than 0 but less or equal to 90°But still a change in direction greater
than 90° at a single point is not desirable. Normally, a 45° and a 90° elbow combinedly used while making
piping layouts for such situations.
Fig.1 A typical elbow with elbow angle (phi)
Elbow angle can be easily calculated using simple geometrical technique of mathematics. Lets give an
example for you. Refer to Fig.2. Pipe direction is changing at point A with the help of an elbow and again
the direction is changing at the point G using another elbow.
Fig.2 Example figure for elbow angle calculation
In order to find out the elbow angle at A, it is necessary to consider a plane which contains the arms of the
elbow. If there had been no change in direction at point A, the pipe would have moved along line AD but
pipe is moving along line AG. Plane AFGD contains lines AD and AG and elbow angle (phi) is marked
which denotes the angle by which the flow is deviating from its original direction.
Considering right angle triangle AGD, tan(phi) = √( x2 + z2)/y
Similarly elbow angle at G is given by : tan (phi1)=√ (y2 +z2)/x
Elbows or bends are available in various radii for a smooth change in direction which are expressed in
terms of pipe nominal size expressed in inches. Elbows or bends are available in three radii,
a. Long radius elbows (Radius = 1.5D): used most frequently where there is a need to keep the frictional
fluid pressure loss down to a minimum, there is ample space and volume to allow for a wider turn and
generate less pressure drop.
b. Long radius elbows (Radius > 1.5D): Used sometimes for specific applications for transporting high
viscous fluids likes slurry, low polymer etc. For radius more than 1.5D pipe bends are usually used and
these can be made to any radius.However, 3D & 5D pipe bends are most commonly used
b. Short radius elbows (Radius = 1.0D): to be used only in locations where space does not permit use of
long radies elbow and there is a need to reduce the cost of elbows. In jacketed piping the short radius
elbow is used for the core pipe.
Here D is nominal pipe size in inches.
There are three major parameters which dictates the radius selection for elbow. Space availability, cost and
pressure drop.
Pipe bends are preferred where pressure drop is of a major consideration.
Use of short radius elbows should be avoided as far as possible due to abrupt change in direction causing
high pressure drop.
Minimum thickness requirement:
Whether an elbow or bend is used the minimum thickness requirement from code must be met. Code
ASME B 31.3 provides equation for calculating minimum thickness required (t) in finished form for a
given internal design pressure (P) as shown below:
Fig.3: Code equation for minimum thickness requirement calculation
Here,
R1 = bend radius of welding elbow or pipe bend
D = outside diameter of pipe
W = weld joint strength reduction factor
Y = coefficient from Code Table 304.1.1
S = stress value for material from Table A-1 at maximum temperature
E = quality factor from Table A-1A or A-1B
Add any corrosion, erosion, mechanical allowances with this calculated value to get the thickness required.
End Connections:
For connecting elbow/bend to pipe the following type of end connections are available
http://www.whatispiping.com/piping-elbows-and-bends | 1,127 | 4,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.943751 |
https://interunet.com/amortization-vs-depreciation | 1,603,972,440,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00119.warc.gz | 368,405,983 | 6,127 | # Amortization vs depreciation
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Amortization vs depreciation (examples)
What is amortization?
Amortization is an allocation of an intangible finite asset’s initial cost over its service life. To calculate the amortization of finite assets entities usually use the straight line amortization method which is the same as for depreciable assets. When calculating amortization an entity should deduct capital expenses over service life of an asset.
Formula:
Amortization expense per year = (initial cost* – residual value**) / # of years of services life.
*Initial cost of an amortizable asset is the purchase price and any other cost related to it in order to get the asset ready for its intended use.
**Residual value is remaining estimated worth of the amortizable asset (if any) at the end of its service life. It is the fair market value that the potential buyers might be willing to pay for it.
What is accounting entries to record an intangible asset’s amortization
Dr. Intangible asset amortization expense
Cr. Intangible asset (name)
(to record an annual amortization expense of an intangible asset)
The amortization computations are almost the same as they would be in case of depreciation, except that in computation of amortizable assets involved intangible finite assets, but depreciable assets include tangible assets. Therefore, the difference between depreciation and amortization is that amortization refers to intangible finite assets while depreciation relates to tangible depreciable assets.
What is depreciation?
Depreciation is the process of spreading the purchased depreciable asset’s initial cost over its useful life. Companies use the depreciation methods to have reasonable and consistent revenue / expense matching. Commonly used depreciation methods are straight line (the easiest and most used one) and accelerated depreciation methods such as sum of the years’ digits, declining balance (items such as 200% and 150%), and units of production. All of the depreciation methods have their own advantages and disadvantages and a firm depending on the specification of certain depreciable asset may choose for it the right accounting method which will benefit them the most.
Which depreciation method is better?
Typically entities choose the depreciation method depending on specifications of a depreciable asset and the goals they pursue, whether they want to show lower or higher reduction amounts. However, it is not all the time the case. Since all types of properties have their own unique characteristics even within the same company for different depreciable assets can be chosen different depreciation methods. This is due to the fact that some assets wear out faster than the other or require more maintenance cost with years of service. For example, if for an office table the straight line depreciation method might be appropriate, but for computers or some equipment it might be more appropriate to use one of accelerated depreciation methods (due to the rapidly changing technology). In addition, such depreciation method as units of production cannot be applies to all depreciable assets. It is not the right method for such assets as furniture and buildings which weariness depends usually on the years used rather than on the production levels. The firm can change the selected accounting method of depreciable assets later on if necessary, but they should reflect that in the notes to the financial statements for the period.
What accounting entries to record depreciation?
Dr. Depreciation expense
Cr. Accumulated depreciation | 675 | 3,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.927491 |
https://www.nytimes.com/interactive/projects/4thdownbot/play/20151213012482 | 1,537,810,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160620.68/warc/CC-MAIN-20180924165426-20180924185826-00187.warc.gz | 809,241,773 | 7,749 | Week 14, Chargers at Chiefs
4th-and-10, 12:13 remaining in 3rd quarter, Up by 10
The Chiefs attempted a field goal on a 4th-and-10 on the Chargers’ 28. It doesn’t matter to NYT 4th Down Bot.
If you disagree
Well, this is weird — it seems that the Chiefs’ chances of winning are about the same no matter what they do. I lean very slightly toward attempting a field goal, but I don’t have strong feelings here — just go with your gut, coach.
Here's the full breakdown of my calculations:
Option Chance of converting
Chance of winning
Before play
After play Change
Field goal try45 yard kick (est.) 69% 94% 94% -
Go for it4th and 10, opp. 28 29% 94% 93% –1%
Punt 94% 93% –1%
My decision in context
Along with some circuitry to come up with a win probability for every game situation, all you need to figure out what you should do next is an estimate of how likely you are to make a field goal or convert a first down.
My estimates for these are based on the results of thousands of similar plays, but you may think you're smarter than I am. This chart shows you how changing those estimates would change my recommendation.
What to do on 4th-and-10 on opp. 28
Up by 10 with 12:13 remaining in the 3rd quarter
Behind my field goal number
What coaches usually do
Field goal try 64% of the time Punt 32% of the time Go for it 5% of the time
Based on about 1,135 fourth downs in similar situations since 2001.
What happened
46 yard field goal attempt by Cairo Santos is NO GOOD. 46 yard field goal attempt by Cairo Santos is NO GOOD.
Where did these numbers come from?
To estimate a team’s chances of winning, I use a mathematical model that accounts for a whole lot of variables — including the difference in score, the time remaining in the game, and the number of timeouts each team has left. On top of that, I have models for the likelihood that a team makes a field goal and the likelihood that it will convert a first down.
By combining all of this information, I can come up with the best decision a team can make, according to math.
If you want even more details about the numbers behind my decisions, my full model is available on GitHub. Help make me better! | 539 | 2,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-39 | latest | en | 0.96499 |
https://fr.slideserve.com/jui/chapter-20-electricity-chapter-21-magnetism | 1,642,624,306,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301488.71/warc/CC-MAIN-20220119185232-20220119215232-00409.warc.gz | 312,951,088 | 23,118 | Chapter 20 Electricity Chapter 21 Magnetism
# Chapter 20 Electricity Chapter 21 Magnetism
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## Chapter 20 Electricity Chapter 21 Magnetism
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##### Presentation Transcript
1. Chapter 20 ElectricityChapter 21 Magnetism Prentice Hall 2006
2. Assignments for Chapter 20 • Define vocabulary terms in 20.1-3 (24) • Write the key concepts in 20.1-3 (14) • 611/1-3 Math Practice re Power • 613/5-8 • 625/1-10(complete sentences), 14,15,17,18,27,29,30 • Workbook pages: 239-248
3. 20.1 Electric Charge and Static Electricity A. Electric charge – an electrical property of matter that creates a force between objects. *based on number of electrons and protons in the objects *excess number of electrons creates a negative charge - atom gains electrons *excess number of protons creates a positive charge - atom loses electrons
4. Electric Charge and Force srikant.org/core/node8.html
5. Electric Force Fields • The strength of an electric field depends on the • Amount of charge that produced the field and • The distance from the charge. • Potential difference is the difference in electrical potential difference between 2 places in an electric field. www.stkate.edu/physics/phys112/index.html
6. 20.1 Electric Charge and Static Electricity, continued *equal number of electrons and protons creates a neutral charge *unit of charge is a coulomb >electron charge is –1.6 x 10-19 coulombs >proton charge is +1.6 x 10-19 coulombs
7. devices used to detect charges: Van deGraaf Generators Electroscopes Pith balls
8. Van de Graaf Generator
10. Charging a Pith Ball On the following diagrams; i) Show how the negative and positive charges are arranged and ii) Explain why the pith ball is first attracted then repelled from the rod Pith Balls Discussion 1) The neutral pith ball has equal number of (+) and (-) charges 2) The negative (-) charges on the pith ball are repelled away form the rod. The positive (+) charges on the pith ball are closer to the negative (-) charges on the rod than the pith balls positive (+) charges. The pith balls positive charges are therefore attracted more strongly than the pith balls negative (-) charges are repelled. Hence there is an overall attraction 3) When rod makes contact, surface negative (-) charges are able to move onto the pith ball, giving it an overall negative charge. 4) The pith ball is then repelled away from the rod.
11. Discussion of prior slide 1) The neutral pith ball has equal number of (+) and (-) charges 2) The negative (-) charges on the pith ball are repelled away form the rod. The positive (+) charges on the pith ball are closer to the negative (-) charges on the rod than the pith balls positive (+) charges. The pith balls positive charges are therefore attracted more strongly than the pith balls negative (-) charges are repelled. Hence there is an overall attraction 3) When rod makes contact, surface negative (-) charges are able to move onto the pith ball, giving it an overall negative charge. 4) The pith ball is then repelled away from the rod.
12. Static Electricity • Study of the behavior of electric charges (at rest) including how charge is transferred between objects by • Friction (like walking across a carpet) • By conduction (contact) • By induction (bringing a charged object near a neutral object)
13. Electrical Potential Energy Potential difference (measured in volts) between two points is what causes electricity to move. Lightning is a natural result of this. Batteries provide a potential difference this across their terminals and produce direct current.
14. DC circuit is produced by a battery. Potential Difference (voltage drop) is maintained across the + and - terminals of a battery. What is the difference between the circuit at left and the one above? The circuit at left is in series The circuit above is in parallel
15. 20.2 For current to flow: • You must have • Source of voltage • Complete path or circuit • Conductor with low resistance • Device to use the energy (i.e., light bulb) • Current flows from positive to negative • This is considered conventional current • Direction of positive charge movement that is equivalent to actual motion of charge in the material
16. Types of Current • Direct current (DC) - flows in one direction only • Flashlight and car batteries produce DC • Alternating current (AC) - flows back and forth • This is produced by an electrical generating power plant that sends electricity to places like businesses, schools, and homes
17. Resistances • Conductorshave low resistances and allow charges to flow easily • Samples: metals • Insulatorshave high resistances because electrons are tightly bound to its atoms • Samples: plastics, dry wood • Superconductors have little or no resistance below their critical temperature • Samples: some metals (Nb, Sn, Hg,…)
18. Ohm’s Law • V = IR • V, potential difference in volts, v • I, current in amperes, a • R, resistance in ohms, W • * Resistance is affected by a material’s thickness, length, and temperature. *Increasing the thickness of a metal wire will reduce its resistance. V I R
19. Problem samples Find the resistance in a circuit with an 8.0 volt battery and 0.2 amp flowing when the current is on. R = V I R = 8.0 v 0.2 a R = 40. W
20. Practice Problems
21. Practice Problems, answers • 1. R=V/I = 24v/0.80a = 3.0 x 101W • 2. R=V/I = 120v/0.50a = 240 W • 3. V = IR = (0.50a)(12W) = 6.0 v • 4. I = V/R = 1.5 v/ 3.5 W = 0.43a
22. 20.3 Circuits • Two main types • Series where there is one path for current flow • Parallel where there is more than one path for current to flow. Most circuits in your home are of this type - parallel. • Series Parallel
23. Sw.1 4
24. Answers to the previous 4 slides • Frame 25: no, no, yes • Frame 26: 1-abc; 2-ab; 3-a; 4-abc; 5-abc • Frame 27: 1-a; 2-c; 3-a • Frame 28: 4-d; 5-d; 6-c
25. Electrical Safety • These devices are used make electricity safer: • Circuit breaker (if too much current flows, this opens the circuit - page 609) • Fuse (if too much current flows, the wire in this melts and opens the circuit - page 612) • Ground-fault circuit interrupter (this automatically opens the circuit if it senses unequal currents - see page 613)
26. Schematic Diagrams • The figures on the prior slide are schematic diagrams where symbols represent the parts of a circuit. See page 374 in book. lamp
27. Power = Current x Voltage P = IV, unit: watt = amp volt An electric space heater requires 29 amp of 120 v current to adequately warm a room. What is the power rating of the heater? P = IV P = 29a (120v) P = 3.5 x 103 watt
28. Your assignment for 20.3 • 611/1-3 math practice • 613/5-8
29. 20.4 Electronic Devices • Electronic Signal - information sent as patterns in controlled flow of electrons through a circuit. • Analog signals are produced by continuously changing voltage or current (page 619) • Vacuum tubes can be used to change AC into DC, increase signal strength, turn current on or off. • Semiconductors are made from crystalline solids and conduct current under certain circumstances
30. Transistors are solid state components with 3 layers of semiconductors Transistors amplify a mobile phone’s incoming signal A diode maintains proper voltage levels in the circuits in a mobile phone An Integrated Circuit is a thin slice of silicon that contains many solid-state components Other electrical information
31. 21.1&2Magnetism & Electromagnetism • Define vocabulary terms in 21.1-3 (15) • Write the key concepts in 21.1-3 (10) • 651/1-10 (complete sentences), 14, 15,17, 30 • Workbook pages: 251-258
32. 21.1&2Magnetism & Electromagnetism Review: 1. Like magnetic poles _____ 2. Unlike magnetic poles ____ 3. Magnetic field lines run from ___ to ___
33. answers • Like poles repel • Unlike poles attract • Magnetic field flows from N to S
34. Earth’s Magnetic Field
35. Electricity can produce magnetism and … • Magnetism can produce electricity!!! • When electricity flows through a conductor, a magnetic field is produced • Right hand rule for a straight conductor
36. 21.3 Electrical Energy Generation and Transmission • When a coil of wire moves in a magnetic field, electricity will flow in the coil (aka: electromagnetic induction) • This is how our electricity is generated • Faraday’s Law - an electric current can be produced in a circuit by a changing magnetic field.
37. Right hand rule used in designing and understanding of generators I current flow B magnetic field direction F resulting force acting on wire
38. Faraday’s Law • Just by moving a magnet in a coil of wire or [as in fig. (c), the switch is turned on or off] electricity is produced. This is how electricity is generated.
39. Transformers These change the voltage and current of electricity by having different numbers of turns on the primary and secondary coils Before electricity gets to your house it is stepped down by a passing through a step-down transformer.
40. The stronger the magnetic force of a magnet, the more effect it has on another magnetic or on moving charges Magnetic fields are produced by moving electric charges AC and DC generators are similar except DC generators use commutators to produce direct current Heat from fossil fuels spins a turbine (coil of wire that spins in a magnetic field) that generates electrical energy More ElectroMagnetic Information | 2,317 | 9,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.720859 |
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Yesterday, I was asked how to write a code to generate a compound Poisson variables, i.e. a series of random variables $S=X_1+\cdots+X_N$ where $N$ is a counting random variable (here Poisson disributed) and where the $X_i$‘s are i.i.d (and independent of $N$), with the convention $S=0$ when $N=0$. I came up with the following algorithm, but I was wondering if it was possible to get a better one…
> rcpd=function(n,rN,rX){
+ N=rN(n)
+ X=rX(sum(N))
+ I=as.factor(rep(1:n,N))
+ S=tapply(X,I,sum)
+ V=as.numeric(S[as.character(1:n)])
+ V[is.na(V)]=0
+ return(V)}
Here, consider – to illustrate – the case where $N\sim\mathcal{P}(5)$ and $X_i\sim\mathcal{E}(2)$,
> rN.P=function(n) rpois(n,5)
> rX.E=function(n) rexp(n,2)
We can generate a sample
> S=rcpd(1000,rN=rN.P,rX=rX.E)
and check (using simulation) than $\mathbb{E}(S)=\mathbb{E}(N)\mathbb{E}(X_i)$
> mean(S)
[1] 2.547033
> mean(rN.P(1000))*mean(rX.E(1000))
[1] 2.548309
and that $\text{Var}(S)=\mathbb{E}(N)\text{Var}(X_i)+\mathbb{E}(X_i)^2\text{Var}(N)$
> var(S)
[1] 2.60393
> mean(rN.P(1000))*var(rX.E(1000))+
+ mean(rX.E(1000))^2*var(rN.P(1000))
[1] 2.621376
If anyone might think of a faster algorithm, I’d be glad to hear about it…
To leave a comment for the author, please follow the link and comment on their blog: Freakonometrics - Tag - R-english.
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Click here to close (This popup will not appear again) | 685 | 2,061 | {"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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | longest | en | 0.737461 |
https://www.brooklandsprimary.com/numeracy-8175/ | 1,723,300,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00057.warc.gz | 530,505,116 | 9,608 | # Numeracy
Data Handling: Bar Charts
I'm sure as P7's you are all very familiar with bar charts and how we can use these.
Here are a few reminders, for you to refresh your brain on this topic:
• A bar chart displays information (also called data) by using rectangular bars. These bars will be of different heights.
• A bar chart has a vertical axis (going up long ways) with numbers on it.
• These numbers can be displayed in a range of different ways. For example they may move up one number at a time- 0, 1, 2, 3 and so on... OR
• The numbers may move up with only some numbers displayed. For example 0, 2, 4, 6 and so on... or 0, 5, 10 ,15 and so on... In this case you will need to work out the value that each square represents, as some numbers may not be marked on the chart.
• A bar chart also has a horizontal axis (going from left to right), showing values of something that has been investigated.
• Some bar charts may use something called class intervals. A class interval is normally used if the horizontal axis is showing values in a number format.
For example:
A horizontal axis may show various different measurements. A class interval means a range of these measurements will be grouped together and represented by one bar within the chart. One of these class intervals could be 18cm to 20cm. This means that it includes sizes 18cm, 19cm and 20cm.
Look at the image below, showing the different types of bar charts you may come across.
First of all
We would like everyone to start off with the Mild/Medium level of this task. Here, you will have to use the graph to answer some questions about the protective clothing children must wear when cleaning up rubbish around their school.
Unfortunately you cannot complete question 2 of this level, as we would need to be in school.
Secondly
There are a total of 7 questions on the spicy level task. We would ask that you try and complete at least a few of the questions on this level.
For some of these questions you will need squared paper to work on. There is some printable graph templates below which you could work from.
Top | 481 | 2,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-33 | latest | en | 0.938696 |
https://ar.casact.org/a-numerical-bar-game/ | 1,679,446,203,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00755.warc.gz | 152,811,797 | 21,167 | # A Numerical Bar Game
Steve Mildenhall contributed this puzzle.
The nerdier regulars at the Bon Pint pub enjoy a peculiar number game. They start with an integer n and expand it in powers of 2. Then, they expand all the exponents in powers of 2, and so on, until n is written with just 1s and 2s. For example, 9 = 23 + 1 = 2(21+1) + 1. They then make a new number by replacing each 2 with 3 and subtracting 1 from the result, carrying terms (like grade school math subtraction, see example below) to ensure all the coefficients in the base 3 representation are positive. Next, they replace 3 with 4 and subtract 1, and so on. They keep going until the pub closes or the sequence stops, whichever comes first.
How often do the sequences stop? If they stop, how long do the players play?
To get you started, here’s the game for n = 2, n = 3, and the initial terms for n = 4.
• When n = 2, the game stops after 3 steps: 21 → 31 – 1 = 2 → 1 → 0.
• When n = 3, it stops after 5 steps: 21 + 1 → 31 + 1 – 1 = 31 → 41 – 1 = 3 → 2 → 1 → 0.
• When n = 4, the game starts: 22 → 33 – 1 = 2 × 32 = 2 × 3 + 2 = 26, which illustrates carrying (twice) to ensure all coefficients are positive. The sequence continues, 26 → 2 × 42 + 2 × 4 + 1 = 41 → 60 → 83 → 109 → … .
First, decide the outcome for n = 4. Then try to generalize. Show your work for partial credit.
## Follow-ups on solutions to previous puzzles
### Proof of Crypto Mining Work
This puzzle was to find a number (a “nonce”) that when appended to “Casualty Actuarial Society” results in a SHA-256 hash with at least 20 leading binary 0s (same as at least 5 leading 0s in hexadecimal representation), or equivalently smaller than 2236. Two solutions that should have been mentioned are below:
Dave Schofield should have been mentioned as also submitting the strongest nonce at previous publication time, 7180096807, which results in 9 leading hex 0s and 37 leading binary 0s in the hash value of: 0000000004e11d3163164d3485ad2588f56eda9630c71405acf23f004c9060f9.
More recently, Mike Convey submitted the nonce 1ff8640245, which results in 10 leading hex 0s and 42 leading binary 0s in the hash value of: 00000000026aa1c8ce977957f4dbf2e4b9951b61300eb996a555c0df47e8e2e.
Soon, a follow-up column is anticipated dealing with the ongoing search for a stronger (more leading binary 0s) Casualty Actuarial Society nonce.
### Questionable Odds
Jeff Subeck pointed out a subtle logical deficiency in the wording of the original puzzle, something most readers would assume but should have been explicitly stated. This assumption should have been explicitly stated — something like the following: “The question must have a fixed answer for any given individual, independent of whether or not that individual is the randomly chosen person.”
Without the clarification above, a trivial solution for part of the puzzle could be as follows:
“A question to ask to maximally improve your expected probability of correctly guessing is whether this person is in the set of persons such that, if it is guessed that they are the person who was randomly selected, the guess would be correct. The list of people having an answer of yes will be a list of the one selected person. This trivially results in 100% probability of correctly guessing this person.”
### A Game of Coins
Solutions were sent in by Shyam Bihari Agarwal, John Berglund, Olivier Guillot-Lafrance, Dave Oakden and Andrew Yuhasz. Stay tuned for the solution in an upcoming AR! | 940 | 3,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-14 | latest | en | 0.917478 |
http://canadianenergyissues.com/2013/12/10/ontario-gas-plant-carbon-the-rogers-centre-metric/ | 1,696,015,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00252.warc.gz | 7,502,094 | 24,430 | Ontario gas plant carbon: the Rogers Centre metric
How many times do Ontario gas plants emit enough carbon dioxide (CO2) to fill up the Rogers Centre with the roof closed? You can figure out how much they have emitted so far today: just divide the amount of CO2 in the Gas category in Table 2 on the left-hand sidebar by 2,877, which is how many tons of CO2 would fit into the Rogers Centre at 25 °C and one atmosphere pressure. (For an explanation of how I got that figure, see the Info Box below). As of nine a.m. today (December 10), they had emitted enough CO2 to fill Rogers Centre just over 2.6 times.
Here’s how many times Ontario’s gas plants filled up Rogers Centre per day in November 2013:
DAY IN NOVEMBER 2013CO2, TONS# OF ROGERS CENTRES FILLED
1126264
2137955
3169406
4179846
5175056
6160186
7157955
8161736
9122394
10134605
11161106
12196007
13168856
14161716
15144995
16131295
17132105
18148895
19193177
20209087
21228268
22193987
23158015
24203817
25199067
26249749
27230798
283086011
293198011
30208167
Total547274190
As you can see, Ontario gas plants filled the Rogers Centre 190 times in November. On November 29, the Ontario gas plants filled Rogers Centre with CO2 11 times. The lowest days were the 1st and 9th, when they only filled it 4 times. The average for the month was 6 times.
At 25 °C, Rogers Centre with its roof closed holds 2,877 metric tons of CO2. Here is how I got that number:
1. The mass of one mole of CO2 is 44.01 grams. (Most versions of the Periodic Table, including this one, give the mass of each element. Look up carbon and oxygen—the constituent atoms in a molecule of CO2—and note their mass. Don’t confuse the atomic mass of an element with its atomic number! Add the mass of one atom of carbon, ~12.01 atomic mass units or AMU, to that of two atoms of oxygen, ~32 AMU. The result: a molecule of CO2 has a mass of 44.01 AMU. A mole of CO2 is therefore 44.01 grams.)
2. One mole of any gas, at 25 °C and one atmosphere pressure, occupies 24.47 litres of volume. (One mole of any gas at standard temperature and pressure occupies 22.414 litres. To calculate molar volume at another temperature, let’s call it T2, with the same pressure, convert temperature to Kelvins and then multiply the ratio of the STP volume to temperature by T2; this is Charles’s Law. In this case, your T2 is 25 °C, which is 298 Kelvins: 25 + 273 = 298. Your ratio of STP volume to temperature is 22.414/273 = 0.0821. Multiply 0.0821 x 298 = 24.466 litres.)
3. One metric tonne, or one million grams, of CO2 contains 22,727 moles: divide one million by 44.
4. Multiply those 22,727 moles in a tonne of CO2 by the molar volume of a gas at 25 °C, which from point 3 is 24.47 litres.
5. Therefore one metric tonne of CO2 at the above-mentioned temperature and pressure occupies 556,136 litres.
6. One cubic meter is 1,000 litres, so a metric tonne of CO2 occupies 556.14 cubic meters (divide the 556,136 litres of CO2 that make up a metric tonne by 1,000).
7. Now you need to know the volume of Rogers Centre. According to Rogers, Rogers Centre’s volume is 1,600,000 m3. So divide that by 556.14 to get 2,877.
How many times can we as a society keep doing this? Dumping so much CO2 into the air won’t make our atmosphere thicker or more dense, as Al Gore claimed in his documentary An Inconvenient Truth. But it will make it more “opaque” to long-wave (infrared) radiation. Which is to say, it will trap more heat.
It will also increase the number of CO2 molecules making contact with surface water. This will increase the amount of CO2 being absorbed into the world’s oceans—most of the world is covered with salt water. When CO2 absorbs into water, it turns it slightly more acidic. This has been increasing since the Industrial Revolution, and scientists now estimate that half of the man-made CO2 has been absorbed in the world’s oceans. The pH of the oceans has historically been 8.2, which is slightly on the base side (neutral is 7); today it averages about 8.1. That might sound small, but it has altered the water chemistry especially of shallow areas, where a lot of ocean life is. More CO2 in the water means less of the calcium carbonate minerals which are the building blocks for the shells and skeletons of many marine animals.
Scientists also think the oceans absorb today about one-third of the man-made CO2 dumped into the air. So have another look at the table above. Divide the tons of CO2 emitted by Ontario gas plants by three. That is how much our electricity is affecting the world’s oceans.
Next time you pass the Rogers Centre, think of it as a receptable for CO2 for Ontario gas plants. In November, the provincial gas fleet dumped enough carbon to fill Rogers Centre once every four hours. A third of that will acidify the oceans, and harm marine animals. How much should we rely on these plants for our electricity, especially when we have nuclear plants that dump no CO2 at all? | 1,299 | 4,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-40 | latest | en | 0.878235 |
https://wiki.ard-site.net/index.php?title=Mercalli_intensity_scale&printable=yes | 1,618,628,404,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038098638.52/warc/CC-MAIN-20210417011815-20210417041815-00468.warc.gz | 657,315,025 | 9,347 | Mercalli intensity scale
The Mercalli intensity scale (or more precisely the Modified Mercalli intensity scale) is a scale to measure the intensity of earthquakes. Unlike with the Richter scale, the Mercalli scale does not take into account energy of an earthquake directly. Rather, they classify earthquakes by the effects they have (and the destruction they cause). When there is little damage, the scale describes how people felt the earthquake, or how many people felt it.
Very often, non-geologists use this scale, because it is easier for people to describe what damage an earthquake caused, than to do calculations to get a value on the Richter scale.
Values range from I - Instrumental to XII - Catastrophic.
Giuseppe Mercalli (1850-1914) originally developed the scale, with ten levels. In 1902, Adolfo Cancani extended the scale to include twelve levels. August Heinrich Sieberg completely rewrote the scale. For this reason, the scale is sometimes named Mercalli-Cancani-Sieberg scale, or MCS scale.
Harry O. Wood and Frank Neumann translated it into English, and published it as Mercalli–Wood–Neumann (MWN) scale. Charles Francis Richter also edited it. He also developed the Richter scale, later on.
Modified Mercalli Intensity scale
The lower degrees of the Modified Mercalli Intensity scale generally deal with the manner in which the earthquake is felt by people. The higher numbers of the scale are based on observed damage to structures
The large table gives Modified Mercalli scale intensities that are typically observed at locations near the epicenter of the earthquake.[1]
I. Not felt Not felt by humans but technology is capable of sensing it. Felt only by a few persons during sleep, especially on upper floors of buildings. Felt quite noticeably by persons indoors, especially on upper floors of buildings. Many people do not recognize it as an earthquake. Standing motor cars may rock slightly. Vibrations similar to the passing of a truck. Duration estimated. Felt indoors by many, outdoors by few during the day. At night, some awakened. Dishes, windows, doors disturbed; walls make cracking sound. Sensation like heavy truck striking building. Standing motor cars rocked noticeably. Felt by nearly everyone; many awakened. Some dishes, windows broken. Unstable objects overturned. Pendulum clocks may stop. Felt by all, many frightened. Some heavy furniture moved; a few instances of fallen plaster. Damage slight. Damage negligible in buildings of good design and construction; slight to moderate in well-built ordinary structures; considerable damage in poorly built or badly designed structures; some chimneys broken. Damage slight in specially designed structures; considerable damage in ordinary substantial buildings with partial collapse. Damage great in poorly built structures. Fall of chimneys, factory stacks, columns, monuments, walls. Heavy furniture overturned. Damage considerable in specially designed structures; well-designed frame structures thrown out of plumb. Damage great in substantial buildings, with partial collapse. Buildings shifted off foundations. Some well-built wooden structures destroyed; most masonry and frame structures destroyed with foundations. Rails bent. Few, if any, (masonry) structures remain standing. Bridges destroyed. Broad fissures in ground. Underground pipe lines completely out of service. Earth slumps and land slips in soft ground. Rails bent greatly. Damage total. Waves seen on ground surfaces. Lines of sight and level distorted. Objects thrown upward into the air.
Correlation with magnitude
Magnitude Typical MaximumModified Mercalli Intensity I II – III III – IV IV – V V – VI VI – VII VII – VIII VIII or higher
There is a correlation between the magitude and the intensity of the earthquake. Even though this correlation is there, it may be difficult to link one to the other: This correlation depends on several factors, such as the depth of the earthquake, terrain, population density, and damage. For example, on May 19, 2011, an earthquake of magnitude 0.7 in Central California, United States 4 km deep was classified as of intensity III by the United States Geological Survey (USGS) over 100 miles (160 km) away from the epicenter (and II intensity almost 300 miles (480 km) from the epicenter), while a 4.5 magnitude quake in Salta, Argentina 164 km deep was of intensity I.[2]
The small table is a rough guide to the degrees of the Modified Mercalli Intensity scale.[1][3] The colors and descriptive names shown here differ from those used on certain shake maps in other articles. However, it will not be 100% accurate.
References
1. "Magnitude / Intensity Comparison". USGS.
2. USGS: Did you feel it? for 20 May 2011
3. "Modified Mercalli Intensity Scale". Association of Bay Area Governments (ABAG). | 1,007 | 4,817 | {"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} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.948357 |
http://fug-experimentalerror.blogspot.com/2009/10/second-law-of-thermodynamics.html | 1,532,068,860,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00228.warc.gz | 156,416,057 | 16,052 | ## Thursday, October 22, 2009
### The Second Law of Thermodynamics
The Second Law clicked today. It took two hours of work at a chalk board along with conversations with a professor (who happens to be very generous with his time), but it clicked in my head, and the interpretation that helped it click was the statistical formulation of the Second Law. So, for me, the most confusing part of the second law is NOT how esoteric it is -- it's far from esoteric. It makes perfect sense and matches up with what we observe. To me, describing the Second Law
as "In spontaneous processes the entropy of the universe tends to increase, where entropy is the measure of disorder" is the confusing part. This statement makes sense, but only if you're familiar with the jargon. And even then, I was still left with wondering "So... why is this, again...?" While you can always ask why (and one ought to), the statistical interpretation satiates the confusing "Why?" for the "Hm, I wonder Why?" kind of why -- bridging the gap from frustrating unfamiliarity to curiosity.
But stating the statistical formulation takes a lot more room. I'll still take a go at it, however.
Suppose you have a chunk of energy. You split that energy into 10 equal parts to observe how it behaves, and you have two metal blocks that can absorb that energy. Placing all 10 equal parts into one of the metal blocks (We'll say so that the energy heats up the block, since I am referring to thermodynamics here) and sitting it next to the other metal block, you sit and wait to see what happens. The heat from the first block should heat up the second block until they're about the same temperature. For our purposes, this is no different than when you let your soup cool off to room temperature, or your ice melts in a glass of water, or when you cuddle up with someone when you feel cold. Eventually heat will be transferred until you reach the same temperature. At this point, heat transfer seems to stop. Ice does not later boil, the soup does not freeze, and you and your partner remain at about the same temperature (though there are some extra complications involved with cuddling, since human bodies produce their own heat, but for rough analogy and everyday experience, it works). Something stops the transfer of heat from continuing in the same direction that is initially observed. Something also stops the transfer of heat from going back to where it used to be (Hot soup, ice cubes, you stay cold). This "Something" is the Second Law of Thermodynamics. From the 10 pieces of energy analogy above:
You have two blocks of metal. However, those blocks of metal have places to store this energy -- atoms. Everything has atoms that it can store energy in. The question really becomes which atoms hold what amount of energy. This is a question that can be addressed mathematically with a concept termed "Multiplicity". Multiplicity is the number of ways you can store those 10 units of energy in however many atoms are present in the metal block. You can place all 10 in the first atom you touch, or spread them out in 10 different atoms, or put 5 in one atom and 5 in another. These are all different ways to arrange this amount of energy. Even so, if all 10 of the energy units are still in the first block, this would mean that the block is at the same temperature (if you'll recall that our energy units tell us how hot our blocks are) no matter how they are arranged within the individual atoms that make up the block. This is something called a "Macrostate" -- a mathematical description of what we observe, namely, the temperature of the block. However, the "Microstate", or the mathematical description of how the energy units are distributed amongst the individual atoms in the block, still plays a crucial role. See, if we take into consideration the second block of metal we just touched to the first block (Let's suppose that both of the blocks are the same size), we essentially double the number of atoms our 10 units of energy can spread between. We also increase the number of macrostates from the single one before (Where our block stayed at the temperature of 10 units of energy that we placed there) to 11 different possible macrostates -- 9 units of energy in the first block, 1 unit of energy in the second block, or 8 units of energy in the first block and 2 units of energy in the second block, so on and so forth.
So the question becomes: Which macrostate is the most likely one to observe? From common experience, we know that things tend to have the same temperature as one another if given enough time, such as the soup cooling off in a room example above. So we should expect that what we observe will be 5 units of energy in the first and 5 units of energy in the 2nd block, given enough time. But why? That is where the term for "Microstates" comes in. It turns out that when you have 5 in the first and 5 in the second, you have more possible ways of distributing the energy throughout the different atoms than you do with any other macrostate. So, it just becomes a statistical issue: There are more possible ways for the Macrostate 5/5 to be observed, therefore it is the one most often observed. There may be some oscillations about this point, but we still observe this more often than anything else.
Now the real kicker is that when dealing with the real world, one deals with more than 10 energy units. We deal with billions upon billions of energy units. And, as atoms are awfully small, we also deal with billions upon billions of atoms. So, with such large numbers the oscillations about the midpoint become immeasurable. So, while oscillations are dictated by probability to occur, as every possible way to arrange the energy in the atoms is just as likely as any other way, we don't notice them due to the sheer improbability of that happening. Like, much more than 10^23. I'm not sure how to express how improbable it is to feel an object heat up without anything heating it up(as it is REALLY FRIGGEN IMPROBABLE), but as you've never experienced it in your life, and I am confident in saying that, you too can feel confident that the 2nd Law is pretty sound stuff! Cool factoid: another common experience unrelated to heat, table salt dissolving in water is an entropy driven process, which is to say that without the 2nd Law of Thermodynamics, table salt wouldn't dissolve. | 1,383 | 6,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-30 | latest | en | 0.963333 |
https://usethinkscript.com/threads/2-period-roc-with-lbr-310-creating-dots.18334/ | 1,720,865,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00061.warc.gz | 497,685,621 | 20,357 | # 2 Period ROC with LBR 310 creating DOTS
#### HushPuppy
##### New member
@BenTen Good day could you assist me in coding this Linda Raschke indicator in TOS:
Indicator: 2 Period ROC with LBR 310 creating DOTS
Indicator Process: The line is the 2 period ROC. The dots are generated as follows - If the slope of the LBR 3/10 slow line, fast line, and the ROC are up, then you get a pink dot at the top of the ROC- If all three slopes are down, you get the blue dot at the bottom of the ROC line
** Note: The LBR 3/10 is Linda's Raschke oscillator. The fast line is just the SMA 3 minus the SMA 10. The slow line is the 16-period average of the fast line.
That should be everything you need to recreate it ! -- Attached is how this indicator should look:
Any help would be greatly appreciated.
Thanks
#### Attachments
• Screenshot 2024-03-08 at 10.39.06 PM.png
332.8 KB · Views: 164
Solution
this plots the LBR Osc
Code:
``````declare lower;
input price = close;
def ROC = if price[2] != 0 then (price / price[2] - 1) * 100 else 0;
plot ZeroLine = 0;
input calculationMode = {default Normal, Alternate};
plot FastLine;
switch (calculationMode) {
case Normal:
FastLine = Average(price, 3) - Average(price, 10);
case Alternate:
FastLine = Average(price - Average(price[3], 3), 2);
}
plot SlowLine = Average(FastLine, 16);
Def Con1 = if ROC >= ROC[1] then 1 else if ROC < ROC[1] then -1 else 0;
Def Con2 = If Fastline >= Fastline[1] then 1 else if Fastline < Fastline[1] then -1 else 0;
Def Con3 = If Slowline >= Slowline[1] then 1 else if Slowline < Slowline[1] then -1 else 0;
Plot Dots = If (Con1 + Con2 + Con3) == 3 then Fastline +1...``````
this plots the LBR Osc
Code:
``````declare lower;
input price = close;
def ROC = if price[2] != 0 then (price / price[2] - 1) * 100 else 0;
plot ZeroLine = 0;
input calculationMode = {default Normal, Alternate};
plot FastLine;
switch (calculationMode) {
case Normal:
FastLine = Average(price, 3) - Average(price, 10);
case Alternate:
FastLine = Average(price - Average(price[3], 3), 2);
}
plot SlowLine = Average(FastLine, 16);
Def Con1 = if ROC >= ROC[1] then 1 else if ROC < ROC[1] then -1 else 0;
Def Con2 = If Fastline >= Fastline[1] then 1 else if Fastline < Fastline[1] then -1 else 0;
Def Con3 = If Slowline >= Slowline[1] then 1 else if Slowline < Slowline[1] then -1 else 0;
Plot Dots = If (Con1 + Con2 + Con3) == 3 then Fastline +1 else if (Con1 + Con2 + Con3) == -3 then Fastline -1 else Double.Nan;
Dots.AssignValueColor(if (Con1 + Con2 + Con3) == 3 then Color.Magenta else if (Con1 + Con2 + Con3) == -3 then Color.Cyan else Color.Black);
Dots.SetPaintingStrategy(PaintingStrategy.Points);
Dots.setLineWeight(2);
plot Hist = FastLine;
FastLine.SetDefaultColor(Color.Cyan);
SlowLine.SetDefaultColor(Color.Yellow);
FastLine.SetLineWeight(2);
SlowLine.SetLineWeight(2);
Hist.DefineColor("Positiveup", Color.Green);
Hist.DefineColor("NegativeUp", Color.Light_Red);
Hist.DefineColor("PositiveDown", Color.DarK_Green);
Hist.DefineColor("NegativeDown", Color.Dark_Red);
Hist.AssignValueColor(if Hist >= 0 && Hist >= Hist[1] then Color.Green else if Hist >= 0 && Hist < Hist[1] then Color.Dark_Green else if Hist < 0 && Hist < Hist[1] then Color.Dark_Red else if Hist < 0 && Hist >= Hist[1] then Color.Light_Red else Color.Black);
Hist.SetPaintingStrategy(PaintingStrategy.HISTOGRAM);
Hist.HideTitle();
ZeroLine.SetDefaultColor(Color.Gray);``````
Last edited:
this plots the ROC
Code:
``````declare lower;
input price = close;
Plot ROC = if price[2] != 0 then (price / price[2] - 1) * 100 else 0;
plot ZeroLine = 0;
input calculationMode = {default Normal, Alternate};
def FastLine;
switch (calculationMode) {
case Normal:
FastLine = Average(price, 3) - Average(price, 10);
case Alternate:
FastLine = Average(price - Average(price[3], 3), 2);
}
Def SlowLine = Average(FastLine, 16);
Def Con1 = if ROC >= ROC[1] then 1 else if ROC < ROC[1] then -1 else 0;
Def Con2 = If Fastline >= Fastline[1] then 1 else if Fastline < Fastline[1] then -1 else 0;
Def Con3 = If Slowline >= Slowline[1] then 1 else if Slowline < Slowline[1] then -1 else 0;
Plot Dots = If (Con1 + Con2 + Con3) == 3 then ROC +1 else if (Con1 + Con2 + Con3) == -3 then ROC -1 else Double.Nan;
Dots.AssignValueColor(if (Con1 + Con2 + Con3) == 3 then Color.Magenta else if (Con1 + Con2 + Con3) == -3 then Color.Cyan else Color.Black);
Dots.SetPaintingStrategy(PaintingStrategy.Points);
Dots.setLineWeight(2);
plot Hist = ROC;
ROC.SetDefaultColor(Color.Cyan);
ROC.SetLineWeight(2);
Hist.AssignValueColor(if Hist >= 0 && Hist >= Hist[1] then Color.Green else if Hist >= 0 && Hist < Hist[1] then Color.Dark_Green else if Hist < 0 && Hist < Hist[1] then Color.Dark_Red else if Hist < 0 && Hist >= Hist[1] then Color.Light_Red else Color.Black);
Hist.SetPaintingStrategy(PaintingStrategy.HISTOGRAM);
Hist.HideTitle();
ZeroLine.SetDefaultColor(Color.Gray);``````
@henry1224 Thank you sooo much for your help on these 2 indicators. Is there any way I can have the ROC2 DOTs show on the chart as well when needed? Also what parameter controls how close or far off the candle the Dots sit?
• I tried moving the study up but the DOTS were way off from the candle.
• Then I tried changing the " declare lower " to "declare Upper" but that just made the dots disappear completely. Ahhh - Any help or guidance would be greatly appreciated.
Kind Thanks,
Hush
Last edited:
@henry1224 Thank you sooo much for your help on these 2 indicators. Is there any way I can have the ROC2 DOTs show on the chart as well when needed? Also what parameter controls how close or far off the candle the Dots sit?
• I tried moving the study up but the DOTS were way off from the candle.
• Then I tried changing the " declare lower " to "declare Upper" but that just made the dots disappear completely. Ahhh - Any help or guidance would be greatly appreciated.
Kind Thanks,
Hush
to plot the dots on top of price you first need
Code:
``````declare upper;
input price = close;
def ROC = if price[2] != 0 then (price / price[2] - 1) * 100 else 0;
def ZeroLine = 0;
input calculationMode = {default Normal, Alternate};
def FastLine;
switch (calculationMode) {
case Normal:
FastLine = Average(price, 3) - Average(price, 10);
case Alternate:
FastLine = Average(price - Average(price[3], 3), 2);
}
Def SlowLine = Average(FastLine, 16);
Def Con1 = if ROC >= ROC[1] then 1 else if ROC < ROC[1] then -1 else 0;
Def Con2 = If Fastline >= Fastline[1] then 1 else if Fastline < Fastline[1] then -1 else 0;
Def Con3 = If Slowline >= Slowline[1] then 1 else if Slowline < Slowline[1] then -1 else 0;
Plot Dots = If (Con1 + Con2 + Con3) == 3 then High +1 else if (Con1 + Con2 + Con3) == -3 then Low -1 else Double.Nan;
Dots.AssignValueColor(if (Con1 + Con2 + Con3) == 3 then Color.Magenta else if (Con1 + Con2 + Con3) == -3 then Color.Cyan else Color.Black);
Dots.SetPaintingStrategy(PaintingStrategy.Points);
Dots.setLineWeight(2);``````
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What are the benefits of VIP Membership? | 2,311 | 8,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.725487 |
https://www.fractionsweb.com/simplifying-fractions-1/ | 1,670,299,816,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00192.warc.gz | 811,841,361 | 28,856 | # Simplifying Fractions Level 1
## The 5-step plan
Here you can learn to simplify fractions with the 5-step plan. When you simplify a fraction, you make a fraction a small as possible. Divide the fraction by the greatest common factor. This way you get the simplest form of a fraction, with the smallest possible numerator an denominator. The 5 steps of the 5-step plan are as follows: At step 1 you only need to fill in the numerator (top number). At step 2 you drag the answer to the matching question. At step 3 you need to fill in the numerator and denominator (bottom number). At step 4 there are 15 multiple choice questions. And finally step 5, the diploma with 20 questions.
### Step 4 - Multiple choice
Try to answer all the 15 questions right!
### Step 5: Get your diploma
Answer all the 20 questions right to get the diploma!
## Description Simplifying Fractions 1
When you want to simplify fractions, you need to look for the greatest common factor, also called the GCF. This is the highest positive cardinal number both the numerator and denominator can be divided by, without leaving remainders. The greatest common factor of the numbers 6 and 9, for example, is 3.
Advertisement: | 274 | 1,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.879171 |
https://www.excelforum.com/excel-general/1289788-countif-for-date-counts-with-weeknumber-and-specific-month.html?s=25be95bce9476896f4f789351d821dd1 | 1,568,674,449,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00011.warc.gz | 840,857,798 | 14,697 | # Countif for Date Counts with Weeknumber and specific month
1. ## Countif for Date Counts with Weeknumber and specific month
I have an excel sheet where I want to count dates appearing in a particular month and a particular week of that month. I have to use Countifs formula. Sample sheet attached. Please assist. For example, in September, 2019, I have count of dates 16 but the count of dates in September's 37th week is 7. I want the result to be 7 through countif(s)
2. ## Re: Countif for Date Counts with Weeknumber and specific month
It's not clear what the expected results are.
I recommend including some of the desired results manually then re-uploading your sample.
3. ## Re: Countif for Date Counts with Weeknumber and specific month
Thank you so much for your quick reply. I have edited my thread along with new attachment.
4. ## Re: Countif for Date Counts with Weeknumber and specific month
Put this in B2:
=WEEKNUM(A2)
Drag the formula down column B
Then you can use this:
=COUNTIFS(A:A,">="&DATE(2019,9,1),A:A,"<"&DATE(2019,10,1),B:B,37)
See attachment.
5. ## Re: Countif for Date Counts with Weeknumber and specific month
Thanks a lot. I have the limitation that I cannot add any additional column in the sheet. What I can do that I can copy that entire sheet in a new workbook and then apply the formula you have provided. Is there any specific formula by which I can copy an entire sheet from one workbook and paste the same in another workbook.
6. ## Re: Countif for Date Counts with Weeknumber and specific month
Maybe just right-click on the Date & Result worksheet > Move or Copy > check the "Create a copy" box and tell it where to go > OK
Or just copy/paste into a new workbook.
7. ## Re: Countif for Date Counts with Weeknumber and specific month
Thank you so much. That's really a great help
8. ## Re: Countif for Date Counts with Weeknumber and specific month
You're welcome. Happy to help.
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Tables of multiplication - - - - - - - - - - Tables of division - - - - - - - - - -
Simple operations Combined operations Problems Aritmetic tables
Instructions: Press one of the buttons on the right column The correct number is put in red .
In the combined operations, you must do the first operation with the first two numbers. Then, you must do the second operation with the result and the third number.
# Simple operations
3 x 9 = 5 x 6 = 7 x 3 = 9 x11 = 3 x 11 = 6 : 2 = 24 : 4 = 66 : 6 = 40 : 8 = 110 : 10 =
1 x 4 - 3 = 6 x 6 + 5 = 8 x 5 + 8 = 4 x 11 + 6 = 18 : 6 - 2 = 20 : 5 + 1 = 2 : 1 + 5 = 80 : 8 + 9 = 60 : 6 + 11 = 80 : 10 + 7 =
# Problems
1. 2 friends receive 10 candies. They are shared out equally.
How many candies will each one receive?
a) 12 candies.
b) 5 candies.
c) 8 candies.
2. We are a group of 6 players and we want to form 2 teams.
How many players will there be on each team?
a) 8 players.
b) 4 players.
c) 3 players.
3. We shared out 90 trading cards among 10 friends.
How many trading cards will I have?
a) 100 cards.
b) 9 cards.
c) 80 cards.
4. Guadalupe has 6 candies and her sister Pilar has 9 times more.
How many candies does Pilar have?
a) 15 candies.
b) 3 candies.
c) 54 candies.
5. A girl bought 5 kilos of potatoes at 20 cents per kilo and 1 can of sardine at 40 cents.
How much money did she spend altogether?
a) 100 cents.
b) 140 cents.
c) 65 cents.
6. Pedro wants to put 24 photos in an album. If 4 photos fit in each page and 2 pages are free,
How many pages did the album have?
a) 8 pages.
b) 30 pages.
c) 4 pages.
7. My dad bought 2 canaries at 5 euros each one. If he only had 5 euros,
How much money does he need to pay for them?
a) 12 Euros.
b) 15 Euros.
c) 5 Euros.
8. 80 euros are shared out among 8 friends and one of them spends 9 euros.
How much money does he have left?
a) 19 Euros.
b) 1 Euro.
c) 97 Euros.
9. Ana bought 7 boxes with 10 colors in each one. If she loses 7 colors,
How many colors does she have left?
a) 63 colors.
b) 24 colors.
c) 77 colors.
10. If we put 77 pencils in several cases and each one of them contains 11 pencils
How many cases did we have if 11 cases are free?
a) 88 cases.
b) 22 cases.
c) 18 cases. | 739 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-09 | latest | en | 0.926852 |
http://mathhelpforum.com/algebra/156388-profit-loss-problem-print.html | 1,503,284,786,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00395.warc.gz | 267,004,272 | 3,106 | # Profit Loss Problem
• Sep 16th 2010, 04:07 AM
rn5a
Profit Loss Problem
Albert sells a cycle to Ben at a profit of 20% & Ben sells it to Cindy at a profit of 25%. If Cindy pays \$1500, what did Albert pay for it?
This is how I tried:
Let Cost Price (C.P.) of Albert be 100
Profit Percentage of Albert=20%
So Selling Price (S.P.) of Albert (to Ben)=100+(20% of 100)=120
Profit Percentage of Ben=25%
So S.P. of Ben (to Cindy)=120+(25% of 120)=150
Now how do I proceed ahead?
Thanks,
Ron
• Sep 16th 2010, 04:32 AM
Unknown008
Well, while you picked the lucky ticket but didn't know how to use it, I'd have done it using what I know. (Wink)
The CP to Cindy is \$1500.
Let the cost price of Ben be x.
To find the selling price of Ben, we need $x + (\dfrac{20}{100})x$
So, we get:
$x + (\dfrac{25}{100})x = 1500$
Find x.
Then, do the same. Let y for example be the cost price of Albert. The selling price of Albert is given by:
$y + (\dfrac{20}{100})y$
• Sep 16th 2010, 04:48 AM
rn5a
OK...this is how I tried it.....
Let Ben's CP be 100
So Ben's SP=100+(25% of 100)=125
Ben's SP=Cindy's CP=1500
If Ben's SP=125, Ben's CP=100
If Ben's SP=1500, Ben's CP=(1500X100)/125=1200
Let Albert's CP be 100
So Albert's SP=100+(20% of 100)=120
Albert's SP=Ben's CP=1200
If Albert's SP=120, Albert's CP=100
If Albert's SP=1200, Albert's CP=(1200x100)/120=1000
So Albert pays \$1000 for the cycle.
Have I done it correctly (answer is correct)? Are the steps correct?
Thanks,
Ron
• Sep 16th 2010, 04:50 AM
Unknown008
The answers are correct. As for the method, I'm not sure if it is accepted, but I understand the logic behind it. (Happy) | 552 | 1,642 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-34 | longest | en | 0.87044 |
https://documen.tv/question/on-what-factors-does-coefficient-friction-force-and-static-friction-force-depend-22687574-85/ | 1,675,628,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00617.warc.gz | 224,196,470 | 19,543 | # On what factors does coefficient friction force and static friction force depend?
Question
On what factors does coefficient friction force and static friction force depend?
in progress 0
1 year 2021-09-04T04:51:41+00:00 1 Answers 2 views 0
Coefficient of friction is defined as the ratio of the force of friction acting between two objects and the force which is pushing them together.
F = μN
F → Friction Force
N → Normal Force
μ → Coefficient of Friction
Coefficient of Friction depends on:
i) Nature of the surface
ii) Material of the object | 133 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.874395 |
https://www.queryhome.com/puzzle/29889/if-7-5-0-6-8-3-9-9-2-9-1-1-and-8-7-2-then-4-5 | 1,716,829,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00526.warc.gz | 832,648,166 | 26,743 | # If 7 + 5 = 0; 6 + 8 = 3; 9 + 9 = 2; 9 + 1 = 1 and 8 + 7 = 2 then 4 + 5 = ?
875 views
If 7 + 5 = 0; 6 + 8 = 3; 9 + 9 = 2; 9 + 1 = 1 and 8 + 7 = 2 then 4 + 5 = ?
posted Nov 19, 2018
The '+' operation between the 2 numbers here takes The number of loops in these numbers and adds them together.
So 7 + 5 = 0 as (7 => has 0 loops and so does 5)
6 + 8 = 3 as (6 => has 1 loop and 8 => has 2 loops)
In that sense
4 + 5 = 1 is the answer.
what did you mean by loop clarify plzz
Here by loop I meant holes with unbroken borders.
7+5=0, 6+8=3, 9+9=2, 9+1-1, 8+7=2, 4+5=2
Similar Puzzles
Arrange all the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 to give four different square numbers. (You make a square number when you multiply something by itself, so 49 is square because it is 7 times 7)
The digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 can be arranged into an addition sum to add up to almost any total, except that nobody has yet found a way to add up to 1984.
However 9 digits can equal 1984 by an addition sum.
Which digit is omitted? | 420 | 1,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.929903 |
https://paspolini.studio/en/what-exactly-is-pitch-in-terms-of-nuts-and-bolts/ | 1,713,159,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00758.warc.gz | 413,142,814 | 14,121 | The pitch is how steep the threads are. On US nuts and bolts, they are measured according to the number of threads per inch. So a pitch of 20 would have 20 threads per inch. For metric nuts and bolts, it’s measured in mm per thread, so a 1.50 pitch will be a thread every 1.5 mm.
## What does pitch mean for a bolt?
Pitch is the distance between screw threads and is commonly used with inch sized products and specified as threads per inch.
## What does pitch mean in thread?
The thread pitch is the distance between corresponding points on adjacent threads. Measurements must be taken parallel to the thread axis. • The major diameter or outside diameter is the diameter over the crests of the thread, measured at right angles to the thread axis. •
## What is the difference between thread and pitch?
Pitch is the distance between a screw’s threads, whereas lead is the distance a screw will travel when turned 360 degrees. As previously mentioned, all screws have threading. The way in which the threading is designed can affect both the pitch and lead. Pitch is simply a measurement of the distance between the threads.
## How is nut pitch measured?
Quote from the video:
Quote from Youtube video: So there are some tools we can use to measure the the pitch of our threads. Using a thread pitch gauge like one of these I have here. We can determine what the pitch of these threads.
## How do you determine thread pitch?
Quote from the video:
Quote from Youtube video: Use a pitch gauge. And check the thread against each form until you find a match. Some fractional and metric thread forms are very similar. So this may take a little time.
## Does thread pitch affect torque?
Thread pitch tends to be more independent of nominal diameter than the other variables. Increasing just thread pitch by 40% cuts tension 5% for a given torque.
## What is the standard pitch of a bolt?
Bolt Diameter (mm) Thread Pitch (mm)
Standard Extra or Super Fine
8 1.25
10 1.5 1
12 1.75 1.25
## How does the pitch affect a screw?
The smaller the pitch (the distance between the screw’s threads), the greater the mechanical advantage (the ratio of output to input force).
## Does screw thread pitch matter?
The thread pitch measurement is critical because if the thread pitch on the screw doesn’t match the nut that it will be used with, you won’t be able to secure the screw properly.
## What is a pitch diameter?
Pitch diameter, denoted by dm or d2, is the imaginary diameter for which the widths of the threads and the grooves are equal.
## How are TPI & thread pitch related?
TPI stands for Threads Per Inch, a count of the number of threads per inch measured along the length of a fastener. TPI is used only with American fasteners. Metric Fasteners use a thread Pitch. In general, smaller fasteners have finer threads, so the thread count is higher.
## How do you convert pitch to TPI?
The TPI is, then, four because it measures “threads per inch.” To convert the pitch to millimeters, use the conversion that 1 inch equals 25.4 millimeters. You can convert 26 TPI to inches per thread by dividing 1 by 26 to get 0.038, and then multiplying this by 25.4 to get a pitch of 0.98 millimeters.
## What is 1.25 mm thread pitch?
Table 1.
Pitch (mm) Pitch (inches) Threads per Inch
1.25 0.0492 20.32
1.5 0.0591 16.93
1.75 0.0689 14.51
2.0 0.0787 12.70 | 800 | 3,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-18 | latest | en | 0.953712 |
https://math.stackexchange.com/questions/4703010/how-to-determine-equivalence-classes | 1,718,705,691,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00865.warc.gz | 331,142,833 | 36,661 | # How to determine equivalence classes?
Let the relation $$µ$$ on $$Z$$ (set of integers) be defined by $$xµy$$ if and only if $$x^2 ≡ y^2(mod 4).$$
I have already proven that the relation $$µ$$ is an equivalence relation, but I am currently struggling with determining the equivalence classes
Do I just testing values for $$x \in Z$$?
For example:
If $$x ≡ 0 (mod 4)$$, then $$x^2 ≡ 0^2 ≡ 0 (mod 4)$$. So, the equivalence class [x] contains all integers that are congruent to $$0$$ modulo $$4$$.
If $$x ≡ 1 (mod 4)$$, then $$x^2 ≡ 1^2 ≡ 1 (mod 4).$$ The equivalence class [x] contains all integers that are congruent to $$1$$ modulo $$4$$.
Therefore, the equivalence classes of the relation $$µ$$ on $$Z$$ are:
$$[0] = \{..., -8, -4, 0, 4, 8, ...\}$$
$$[1] = \{..., -7, -3, 1, 5, 9, ...\}$$
Is this correct? But I believe that my equivalence classes need to complete $$Z$$? Not sure if I am doing the right thing. And where is $$2$$?
Actually, $$x\mathrel\mu 0\iff x^2\equiv0\pmod4\iff x\text{ is even},$$ and therefore $$[0]$$ is the set of all even integers.
And $$x\mathrel\mu 1\iff x^2\equiv1\pmod4\iff x\text{ is odd,}$$ and therefore $$[1]$$ is the set of all odd integers.
Since every integer is even or odd, you are done: these are the only equivalence classes.
Here's another way of reaching the same conclusion. If $$x,y\in\mathbb{Z}$$, then \begin{align} x\mathrel\mu y&\iff x^2\equiv y^2\pmod4\\ &\iff4\mid(x-y)(x+y). \end{align} Now, when a product of integers is a multiple of $$4$$, then either both factors are even or one of them is odd whereas the other one is a multiple of $$4$$. But the current situation the second possibility cannot occur. Indeed, if, say $$x+y$$ is odd, the $$x-y$$ is odd too, since it is equal to $$(x+y)-2y$$. So, both $$x+y$$ and $$x-y$$ are even, and this means that $$x$$ and $$y$$ have the same parity. So, \begin{align} [x]&=\{y\in\mathbb{Z}\,|\,x\text{ and }y\text{ have the same parity}\}\\ &=\begin{cases} \{\text{even integers}\}&\text{ if x is even}\\ \{\text{odd integers}\}&\text{ if x is odd.} \end{cases} \end{align}
• Thank you for your answer, just one more question, which is kinda unrelated, is μ anti symmetry? From my proof, I think so, but I am not sure. Commented May 20, 2023 at 11:44
• No, it is not: $0\mathrel\mu2$ and $2\mathrel\mu0$, but $0\ne2$. Commented May 20, 2023 at 12:05
We denote $$[\ell]_{\mathbb{Z}_4}:=\{n\in\mathbb{Z}:n\equiv \ell\mbox{ mod } 4\},\qquad\ell=0,1,2,3.$$ Then for each $$m\in[\ell]_{\mathbb{Z}_4}$$, there is $$k\in\mathbb{Z}$$ such that $$m=4k+\ell$$. Therefore $$m^2=(4k+\ell)^2=\ell^2.$$ Denote $$[p]_{\mathbb{Z}_4,2}:=\{n\in\mathbb{Z}:n^2\equiv p^2 \mbox{ mod }4\}.$$ It follows from $$0^2\equiv 2^2\equiv 0 \mbox{ mod 4}$$ and $$1^2\equiv 3^2\equiv 1 \mbox{ mod 4}$$ that $$\begin{equation*} \begin{cases} [0]_{\mathbb{Z}_4,2}=[0]_{\mathbb{Z}_4}\bigcup [2]_{\mathbb{Z}_4}\\ [1]_{\mathbb{Z}_4,2}=[1]_{\mathbb{Z}_4}\bigcup [3]_{\mathbb{Z}_4} \end{cases} \end{equation*}$$ and $$[0]_{\mathbb{Z}_4,2}\bigcup[1]_{\mathbb{Z}_4,2}=\left([0]_{\mathbb{Z}_4}\bigcup [2]_{\mathbb{Z}_4}\right)\bigcup\left([1]_{\mathbb{Z}_4}\bigcup [3]_{\mathbb{Z}_4}\right)=\mathbb{Z}.$$ | 1,225 | 3,180 | {"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": 46, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.797952 |
https://nethercraft.net/a-12-4-nc-charge-is-located-at-position-xy-1-0cm-2-0cm/ | 1,695,913,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00445.warc.gz | 464,869,744 | 12,135 | # A 12.4 nC charge is located at position (x,y)= 1.0cm, 2.0cm?
0
A 12.4 nC charge is located at position (x,y)= 1.0cm, 2.0cm
A) At what (x,y) position is the electric field -225,000i N/C?
B) At what (x,y) position is the electric field (161,000i + 80,500j) N/C?
C) At what (x,y) position is the electric field (21,600i – 28,800j) N/C?
• The general formula is |E| = kq/|r|^2 ==> |r| = sqrt(kq/|E|)
To keep the formulas uncluttered, I’m going to drop the i+j notation, so, e.g., the charge location = (1,2) cm.
Since q is positive, the signs of the components of r = those of E.
For each of the three cases:
Define a and b as the x and y components respectively of vector E (e.g., for case A, a = -225000, b = 0)
|E| = sqrt(a^2+b^2) N/C
Solve for |r| using formula above
deltar (x,y) = |r|*(a, b)/|E| cm
position (x,y) = deltar+(1, 2) cm
Be careful with signs.
Also Check This I need some help with Spanish!…again lol? | 338 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-40 | latest | en | 0.888195 |
webdev.funbridge.com | 1,653,655,063,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00220.warc.gz | 679,031,019 | 17,526 | Select a question per theme
in the purple boxes.
## How Funbridge works
### Average performance
#### What is the average performance?
Player's average performance is represented by three values: percentage (MP tournaments), IMP score (IMP tournaments) and overall value (performance taking all deals into account by converting IMP deals into a percentage). They reflect the player's level.
Example: 53%/+1 IMP/55%. It means that on average, the player scores 53% in MP tournaments and +1 IMP per deal in IMP tournaments, which is an overall average of 55%.
Here it is important to note that how your average performance is calculated also depends on the one achieved by other tournament participants. For more details, read the section "How is the average performance calculated?" (See APT, which stands for "Average Performance in the Tournament").
#### Where can I find other players' average performance and mine?
The average performance can be found in your profile and in any player's profile (click on any player's username anywhere in the app to access it).
#### How is the average performance calculated?
Player's average performance is calculated based on:
• The deals he plays in the following game modes: practice deals, tournaments of the day, series tournaments, challenges, exclusive tournaments, team tournaments, easy tournaments, Face the Elites, special tournaments and federation tournaments.
• The average performance achieved by other tournament participants and hence "APT" (Average Performance in the Tournament).
Game modes involving a partner (private tables, pairs tournaments, etc.) are not included.
The average performance in the tournament (APT) represents the average performance of all players in the tournament who have played at least 1 deal.
The APT is expressed as a percentage in MP tournaments and as an IMP score in IMP tournaments.
To determine the player's average performance on a deal, we use the following calculation formulas:
Scored by MPs
Player's performance on the deal = player's result on the deal * (100 - 2 * (50 - APT)) / 100
Example #1:
The player scores 60% in a tournament with an APT of 70%.
His average performance on the deal = 60 * (100 - 2 * (50 - 70)) / 100 = 84%.
Example #2:
The player scores 60% in a tournament with an APT of 40%.
His average performance on the deal = 60 * (100 - 2 * (50 - 40)) / 100 = 48%.
Scored by IMPs
Player's performance on the deal = player's result on the deal + APT
Example #1:
The player scores +2 in a tournament with an APT of +4.
His average performance on the deal = 2 + 4 = 6 IMPs.
Example #2:
The player scores +2 in a tournament with an APT of -2.
His average performance on the deal = 2 + (-2) = 0 IMPs.
The player's overall average performance is the average of all his average performances up to the last 1,000 deals played and is expressed as a percentage.
Therefore his IMP average performance is converted into a percentage as follows:
Player's average performance as a percentage = 50 + (player's IMP average performance * 7)
Example #1:
The player has an IMP average performance of +2.
His average performance converted into a percentage = 50 + (2 * 7) = 64%.
Example #2:
The player has an IMP average performance of 0.
His average performance converted into a percentage = 50 + (0 * 7) = 50%.
It is important to note that the deals taken into account in the calculation of the player's (IMP and MP) average performance are the last 1,000 deals played.
Example:
The player has a MP average performance of 60% for the last 900 deals played (among the last 1,000 deals played).
The player has an IMP average performance of -5 for the last 100 deals played (among the last 1,000 deals played).
To calculate the player's overall average performance, his IMP average performance is converted into a percentage: 50 + (-5 * 7) = 15%.
Then we calculate the average of his two types of average performance weighted by how the last 1,000 deals played are split:
((60 * 900) + (15 * 100)) / 1,000 = 55.5%.
#### Is my average performance impacted in case I withdraw?
The calculation of the average performance is not impacted in case you withdraw from a deal with no determined contract. In other words, if you withdraw from a deal after choosing your contract, it will be taken into account in the calculation.
#### My average performance is not changing. Is this normal?
The average performance is not updated until the end of a tournament.
We remind you that series tournaments are closed at the end of each period, i.e. 1st-15th and 16th-end of each month.
Moreover, below 100 deals played, we don't consider the average performance as representative and your values remain 50%/0 IMPs. | 1,074 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-21 | longest | en | 0.951708 |
http://www.downloadready.com/dl/download_74477.htm | 1,571,245,394,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00289.warc.gz | 246,680,918 | 9,951 | 3 logic games. Each game has 4 levels of difficulty. Tic-tac-toe is played on a 4x4 game board. One player uses X's and the other uses O's. Players move in turn, each placing their tokens on the board. If a row of four identical tokens (either X's or O's) appears somewhere on the board - horizontally, vertically or on the upper-left to lower-right diagonal - the player who started the game wins. Otherwise the second player wins. Cram will remind you of the game Tic-tac-toe but is played on a 8x8 game field. The players move in turn, putting designated stones on any vacant space. The object of the game is to create a line of three stones horizontally, vertically, or diagonally. With that, your stones and your opponent's stones are counted. In the game Super Nim there are randomly placed pieces on the board when the game begins. The players in turn remove all of the pieces from one selected column or row. If you are the player to remove the very last piece, you win! | 225 | 978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.95707 |
https://www.coursehero.com/file/5926262/ams316-hw2/ | 1,485,289,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285244.23/warc/CC-MAIN-20170116095125-00574-ip-10-171-10-70.ec2.internal.warc.gz | 884,557,817 | 21,010 | ams316_hw2
# ams316_hw2 - AMS316 HW2 Due Oct 19, 2009 1. Show that the...
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Unformatted text preview: AMS316 HW2 Due Oct 19, 2009 1. Show that the acf of teh mth-order MA process given by m Xt = k =0 Zt−k /(m + 1) is ρ(k ) = (m + 1 − k )/(m + 1) k = 0, 1, . . . , m 0 k>m 2. Show that the infinite-order MA process {Xt } defined by Xt = Zt + C (Zt−1 + Zt−2 + . . . ) where C is a constant, is non-stationary. Also show that the series of first differences {Yt } defined by Yt = Xt − Xt−1 is a first-order MA process and is stationary. Find the ACF of {Yt }. 3. Find the values of λ1 and λ2 such that the 2nd-order AR process defined by Xt = λ1 Xt−1 + λ2 Xt−2 + Zt is stationary. If λ1 = 1/3, λ2 = 2/9, show that the ACF of Xt is given by ρ(k ) = 16 2 27 3 |k | + 1 5 − 21 3 |k | , k = 0, ±1, ±2, . . . 4. Show that the ACF of the stationary 2nd-order AR process Xt = is given by ρ(k ) = 1 1 Xt−1 + Xt−2 + Zt 12 12 1 32 − 77 4 |k | 45 1 77 3 |k | + , k = 0, ±1, ±2, . . . ...
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Ask a homework question - tutors are online | 473 | 1,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.880218 |
https://www.computing.net/answers/office/countifs-and-countif-and-and/18241.html | 1,603,781,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00070.warc.gz | 675,781,110 | 10,042 | # Solved Countifs and Countif and AND
Microsoft Excel 010 - complete package
May 16, 2013 at 06:31:24
Specs: Windows XP
Why won't this formula work???***=IF(COUNTIFS(Data!E:E,A4),"MMN","")+AND(IF(COUNTIFS('Metro South'!E:E,A4),"MSNJ",""))+AND(IF(COUNTIFS('West Coast'!E:E,A4),"WC",""))))Watching the Evaluate formula, it picks up the first response as 'MMN', then continues and comes back with #Value for the next two responses instead of the blank. I've tried it without the + before the And (which Excel added) and it put in an asterick instead. I've also tried it with a Countif.Help, please!
See More: Countifs and Countif and AND
#1
May 16, 2013 at 12:39:48
You can return only ONE answer, either MMN, or MSNJ or WCSo which answer to you want to give?You could do a nested IF, something like:=IF(COUNTIFS(Data!E:E,A4),"MMN",IF(COUNTIFS('Metro South'!E:E,A4),"MSNJ",IF(COUNTIFS('West Coast'!E:E,A4),"WC",""))))Without knowing what it is your trying to do, not much more to offer.MIKEhttp://www.skeptic.com/
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#2
May 16, 2013 at 19:13:15
While I'll agree that we can't be much help without knowing what GingerLeake is trying to do, I don't necessarily agree that she can only return one answer.If she is trying to use the plus signs and/or the AND function to return multiple answers, she should be using the Concatenation operator, &.For example, this will return 23: =IF(1,2)&IF(1,3)P.S. I don't think that that is what she is trying to do, I'm merely pointing out that it's possible to return multiple answers since she is performing her IF's across multiple sheets. If the value in A4 is found in Column E of all three sheets, it's possible to return MMNMSNJWC with the &.Click Here Before Posting Data or VBA Code ---> How To Post Data or Code.
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#3
May 17, 2013 at 05:34:49
I understand that, the reason I did not mention it, is it would end up a very complicated formula:Something like this will return all three strings concatenated in one cell:=IF(COUNTIFS(E:E,A4),"MMN"&IF(COUNTIFS(F:F,A4),"MSNJ"&IF(COUNTIFS(G:G,A4),"WC","")))But it works only if the First COUNTIFS returns True,else you get a False string returned.I did not try to account for all combinations of strings. A bit more then I wish to attempt just now seeing as the OPhas not replied.MIKE
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#4
May 17, 2013 at 05:54:34
I was really hoping someone would say, "Oh you have too many ( or the ' is in the wrong place!" Ah well.And what I'm trying to do is check for an employee's ID (A4) but on all of the regional sheets (Data, Metro South, West Coast) and if found, show the regional code (MMN, MSNJ, WC.) This is on a worksheet in the same workbook called 'Reports' and it's for management to keep track of where the employee is picking up cases from.It IS possible that an employee could be on all three sheets so I'll try it with the & that DerbyDad suggested and will let you know what happens.Thanks!
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#5
May 17, 2013 at 06:44:48
Try this:=IF(COUNTIF(Data!E:E,A4),"MMN","")&IF(COUNTIF('Metro South'!E:E,A4),"MSNJ","")&IF(COUNTIF('West Coast'!E:E,A4),"WC","")Just concatenated your three =IF() functions.MIKEhttp://www.skeptic.com/
Report • | 916 | 3,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.941898 |
https://www.studiestoday.com/ncert-solution/ncert-class-11-solutions-equilibrium-202418.html | 1,571,687,952,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00361.warc.gz | 1,103,127,244 | 17,563 | # NCERT Class 11 Solutions Equilibrium
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NCERT Class 11 Solutions Equilibrium - NCERT Solutions prepared for CBSE students by the best teachers in Delhi.
Class XI Chapter 7 – Equilibrium Chemistry
Question 7.1: A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer: (a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.
Question 7.2: What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ?
Question 7.3: At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms Calculate Kpfor the equilibrium.
Question 7.9: Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.
Question 7.11: A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
Question 7.17: Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when
Students should free download the NCERT solutions and get better marks in exams. Studiestoday.com panel of teachers recommend students to practice questions in NCERT books and download NCERT solutions.
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UP government schools will follow NCERT pattern from April 2020Are you appearing for the UP exams 2020? If yes, then you have to see what major changes UP government schools have brought for 2020.The UP government is fully prepared to implement NCERT pattern for... | 1,075 | 4,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.815979 |
https://pdfcoffee.com/chapter-8-solutions-gitman-pdf-free.html | 1,720,977,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00799.warc.gz | 398,114,351 | 17,382 | # Chapter 8 Solutions Gitman
##### Citation preview
Answers to Warm-Up Exercises E8-1. Total annual return Answer: (\$0 \$12,000 \$10,000)
\$10,000
\$2,000
\$10,000
20%
Logistics, Inc. doubled the annual rate of return predicted by the analyst. The negative net income is irrelevant to the problem. E8-2. Expected return Answer: Analyst 1 2 3 4 Total
Probability
Return
Weighted Value
0.35 0.05 0.20 0.40 1.00
5% 5% 10% 3% Expected return
1.75% 0.25% 2.0% 1.2% 4.70%
E8-3. Comparing the risk of two investments Answer: CV1 0.10 0.15 0.6667 CV2 0.05
0.12
0.4167
Based solely on standard deviations, Investment 2 has lower risk than Investment 1. Based on coefficients of variation, Investment 2 is still less risky than Investment 1. Since the two investments have different expected returns, using the coefficient of variation to assess risk is better than simply comparing standard deviations because the coefficient of variation considers the relative size of the expected returns of each investment. E8-4. Computing the expected return of a portfolio Answer: rp (0.45 0.038) (0.4 0.123) (0.15 0.174) (0.0171) (0.0492) (0.0261 0.0924 9.24% The portfolio is expected to have a return of approximately 9.2%. E8-5. Calculating a portfolio beta Answer: Beta (0.20 1.15) (0.10 0.85) (0.15 1.60) (0.20 1.35) (0.35 1.85) 0.2300 0.0850 0.2400 0.2700 0.6475 1.4725 E8-6. Calculating the required rate of return Answer: a. Required return 0.05 1.8 (0.10 0.05) 0.05 0.09 0.14 b. Required return 0.05 1.8 (0.13 0.05) 0.05 0.144 0.194 c. Although the risk-free rate does not change, as the market return increases, the required return on the asset rises by 180% of the change in the market’s return.
P8-1.
Solutions to Problems Rate of return: rt =
(Pt Pt 1 Ct ) Pt 1
LG 1; Basic a.
Investment X: Return
(\$21,000 \$20,000 \$1,500) 12.50% \$20,000
Investment Y: Return
(\$55,000 \$55,000 \$6,800) 12.36% \$55,000
b. Investment X should be selected because it has a higher rate of return for the same level of risk. P8-2.
Return calculations: rt =
(Pt Pt 1 Ct ) Pt 1
LG 1; Basic Investment
P8-3.
Calculation
A
(\$1,100
B
(\$118,000
C
(\$48,000
D
(\$500
E
(\$12,400
\$800
\$100)
\$120,000 \$45,000
\$600
\$80)
\$12,500
rt(%) 25.00
\$800 \$15,000)
\$7,000)
\$120,000
\$45,000
22.22
\$12,500
3.33 11.20
\$600 \$1,500)
10.83
Risk preferences LG 1; Intermediate a.
The risk-neutral manager would accept Investments X and Y because these have higher returns than the 12% required return and the risk doesn’t matter. b. The risk-averse manager would accept Investment X because it provides the highest return and has the lowest amount of risk. Investment X offers an increase in return for taking on more risk than what the firm currently earns. c. The risk-seeking manager would accept Investments Y and Z because he or she is willing to take greater risk without an increase in return. d. Traditionally, financial managers are risk averse and would choose Investment X, since it provides the required increase in return for an increase in risk.
P8-4.
Risk analysis LG 2; Intermediate a. Expansion
Range
A
24%
16%
8%
B
30%
10%
20%
b. Project A is less risky, since the range of outcomes for A is smaller than the range for Project B. c. Since the most likely return for both projects is 20% and the initial investments are equal, the answer depends on your risk preference. d. The answer is no longer clear, since it now involves a risk-return tradeoff. Project B has a slightly higher return but more risk, while A has both lower return and lower risk. P8-5.
Risk and probability LG 2; Intermediate a. Camera
Range
R
30%
20%
10%
S
35%
15%
20%
b. Possible Outcomes Camera R
Camera S
c.
Probability Pri
Expected Return ri
Weighted Value (%)(ri Pri)
Pessimistic Most likely
0.25 0.50
20 25
5.00% 12.50%
Optimistic
0.25 1.00
30 Expected return
7.50% 25.00%
Pessimistic Most likely
0.20 0.55
15 25
3.00% 13.75%
Optimistic
0.25
35
8.75%
1.00
Expected return
25.50%
Camera S is considered more risky than Camera R because it has a much broader range of outcomes. The risk-return tradeoff is present because Camera S is more risky and also provides a higher return than Camera R.
P8-6.
Bar charts and risk LG 2; Intermediate a.
b. Market Acceptance
c.
Line J
Very Poor Poor Average Good Excellent
Line K
Very Poor Poor Average Good Excellent
Probability Pri
Expected Return ri
Weighted Value (ri Pri)
0.05 0.15 0.60 0.15 0.05 1.00 0.05 0.15 0.60 0.15 0.05 1.00
0.0075 0.0125 0.0850 0.1475 0.1625 Expected return 0.010 0.025 0.080 0.135 0.150 Expected return
0.000375 0.001875 0.051000 0.022125 0.008125 0.083500 0.000500 0.003750 0.048000 0.020250 0.007500 0.080000
Line K appears less risky due to a slightly tighter distribution than line J, indicating a lower range of outcomes.
P8-7.
Coefficient of variation: CV
r
r
LG 2; Basic
7% 0.3500 20% 9.5% B CVB 0.4318 22% 6% C CVC 0.3158 19% 5.5% D CVD 0.3438 16% b. Asset C has the lowest coefficient of variation and is the least risky relative to the other choices. a.
P8-8.
A CVA
Standard deviation versus coefficient of variation as measures of risk LG 2; Basic a. Project A is least risky based on range with a value of 0.04. b. The standard deviation measure fails to take into account both the volatility and the return of the investment. Investors would prefer higher return but less volatility, and the coefficient of variation provices a measure that takes into account both aspects of investors’ preferences. Project D has the lowest CV, so it is the least risky investment relative to the return provided. 0.029 c. A CVA 0.2417 0.12 0.032 B CVB 0.2560 0.125 0.035 C CVC 0.2692 0.13 0.030 D CVD 0.2344 0.128 In this case Project D is the best alternative since it provides the least amount of risk for each percent of return earned. Coefficient of variation is probably the best measure in this instance since it provides a standardized method of measuring the risk-return tradeoff for investments with differing returns.
P8-9.
Personal finance: Rate of return, standard deviation, coefficient of variation LG 2; Challenge a. Year 2009 2010 2011 2012 b. c.
Stock Price Variance Beginning End Returns (Return–Average Return)2 14.36 21.55 50.07% 0.0495 21.55 64.78 200.60% 1.6459 64.78 72.38 11.73% 0.3670 72.38 91.80 26.83% 0.2068 Average return 72.31% Sum of variances 2.2692 3 Sample divisor (n 0.7564 86.97%
d. e.
1.20
1)
Variance Standard deviation Coefficient of variation
The stock price of Hi-Tech, Inc. has definitely gone through some major price changes over this time period. It would have to be classified as a volatile security having an upward price trend over the past 4 years. Note how comparing securities on a CV basis allows the investor to put the stock in proper perspective. The stock is riskier than what Mike normally buys but if he believes that Hi-Tech, Inc. will continue to rise then he should include it. The coefficient of variation, however, is greater than the 0.90 target.
P8-10. Assessing return and risk LG 2; Challenge a.
Project 257 (1) Range: 1.00
(
.10)
(2) Expected return: r
1.10 n
ri Pri
i =1
Expected Return Rate of Return ri .10 0.10 0.20 0.30 0.40 0.45 0.50 0.60 0.70 0.80 1.00
Probability Pr i 0.01 0.04 0.05 0.10 0.15 0.30 0.15 0.10 0.05 0.04 0.01 1.00
Weighted Value r i Pr i
n
r i
ri 1
0.001 0.004 0.010 0.030 0.060 0.135 0.075 0.060 0.035 0.032 0.010 0.450
Pri
n
(3) Standard deviation:
(ri
r )2 Pri
ri
r
i 1
ri
r
0.10 0.10
0.450 0.450
0.20 0.30 0.40
0.450 0.450 0.450
0.45 0.50 0.60 0.70 0.80 1.00
0.450 0.450 0.450 0.450 0.450 0.450
0.550 0.350 0.250 0.150 0.050 0.000 0.050 0.150 0.250 0.350 0.550
( ri
r)2
( ri
Pr i
r)2
Pr i
0.3025 0.1225
0.01 0.04
0.003025 0.004900
0.0625 0.0225 0.0025
0.05 0.10 0.15
0.003125 0.002250 0.000375
0.0000 0.0025 0.0225 0.0625 0.1225 0.3025
0.30 0.15 0.10 0.05 0.04 0.01
0.000000 0.000375 0.002250 0.003125 0.004900 0.003025 0.027350
Project 257
(4) CV
0.027350
0.165378 0.450
Project 432 (1) Range: 0.50
0.165378
0.3675
0.10
(2) Expected return: r
0.40 n
ri Pri
i 1
Expected Return Rate of Return ri
Probability Pr i
Weighted Value ri Pri
0.10
0.05
0.0050
0.15 0.20
0.10 0.10
0.0150 0.0200
0.25
0.15
0.0375
0.30 0.35
0.20 0.15
0.0600 0.0525
0.40
0.10
0.0400
0.45
0.10
0.0450
0.50
0.05
0.0250
1.00
r
n
ri
i =1
0.300
Pri
n
(3) Standard deviation:
(ri
r )2 Pri
i 1
ri
r
0.10
0.300
0.20
0.0400
0.05
0.002000
0.15
0.300
0.15
0.0225
0.10
0.002250
0.20
0.300
0.10
0.0100
0.10
0.001000
0.25
0.300
0.05
0.0025
0.15
0.000375
0.30
0.300
0.00
0.0000
0.20
0.000000
0.35 0.40
0.300 0.300
0.05 0.10
0.0025 0.0100
0.15 0.10
0.000375 0.001000
0.45
0.300
0.15
0.0225
0.10
0.002250
0.50
0.300
0.20
0.0400
0.05
0.002000
ri
r
( ri
r )2
Pri
( ri
r )2
Pri
0.011250 Project 432
0.011250
0.106066 0.300 b. Bar Charts (4) CV
0.3536
0.106066
c.
Summary statistics Project 257
Project 432
Range
1.100
0.400
Expected return (r )
0.450
0.300
0.165
0.106
0.3675
0.3536
Standard deviation (
r
)
Coefficient of variation (CV)
Since Projects 257 and 432 have differing expected values, the coefficient of variation should be the criterion by which the risk of the asset is judged. Since Project 432 has a smaller CV, it is the opportunity with lower risk. P8-11. Integrative—expected return, standard deviation, and coefficient of variation LG 2; Challenge a.
Expected return: r
n
ri
Pri
i 1
Expected Return
Asset F
Rate of Return ri
Probability Pr i
Weighted Value ri Pri
0.40
0.10
0.04
0.10
0.20
0.02
0.00
0.40
0.00
0.05
0.20
0.01
0.10
0.10
0.01
r
n
ri Pri
i 1
0.04
Continued Asset G
0.35
0.40
0.14
0.10
0.30
0.03
0.20
0.30
0.06 0.11
Asset H
0.40
0.10
0.04
0.20
0.20
0.04
0.10 0.00
0.40 0.20
0.04 0.00
0.20
0.10
0.02 0.10
Asset G provides the largest expected return. n
b. Standard deviation:
(ri
r )2 xPri
i 1
ri
Asset F
( ri
r
r)2
Pr i
2
0.40
0.04
0.36
0.1296
0.10
0.01296
0.10
0.04
0.06
0.0036
0.20
0.00072
0.00
0.04
0.04
0.0016
0.40
0.00064
0.05
0.04
0.09
0.0081
0.20
0.00162
0.10
0.04
0.14
0.0196
0.10
0.00196 0.01790
Asset G
0.35
0.11
.24
0.0576
0.40
0.02304
0.10
0.11
0.01
0.0001
0.30
0.00003
0.20
0.11
0.31
0.0961
0.30
0.02883 0.05190
Asset H
0.40
0.10
.30
0.0900
0.10
0.009
0.20
0.10
.10
0.0100
0.20
0.002
0.10
0.10
0.00
0.0000
0.40
0.000
0.00
0.10
0.10
0.0100
0.20
0.002
0.20
0.10
0.30
0.0900
0.10
0.009 0.022
r
0.1338
0.2278
0.1483
Based on standard deviation, Asset G appears to have the greatest risk, but it must be measured against its expected return with the statistical measure coefficient of variation, since the three assets have differing expected values. An incorrect conclusion about the risk of the assets could be drawn using only the standard deviation.
c.
Coefficient of variation =
standard deviation ( ) expected value
0.1338 3.345 0.04 0.2278 Asset G: CV 2.071 0.11 0.1483 Asset H: CV 1.483 0.10 As measured by the coefficient of variation, Asset F has the largest relative risk. Asset F:
CV
P8-12. Normal probability distribution LG 2; Challenge a.
Coefficient of variation: CV
r
r
Solving for standard deviation: 0.75
0.189 0.75 0.189 0.14175 r b. (1) 68% of the outcomes will lie between 1 standard deviation from the expected value:
1 1
r
0.189 0.14175 0.33075 0.189 0.14175 0.04725
(2) 95% of the outcomes will lie between 2 2
0.189 (2 0.14175) 0.189 (2 0.14175)
2 standard deviations from the expected value:
0.4725 0.0945
(3) 99% of the outcomes will lie between 3 standard deviations from the expected value: 3 3
c.
0.189 (3 0.14175) 0.189 (3 0.14175)
0.61425 0.23625
P8-13. Personal finance: Portfolio return and standard deviation LG 3; Challenge a.
Expected portfolio return for each year: rp
Asset L (wL rL)
Year
rL)
(wM
Asset M (wM rM)
rM) Expected Portfolio Return rp
2013
(14%
0.40
5.6%)
(20%
0.60
12.0%)
17.6%
2014
(14%
0.40
5.6%)
(18%
0.60
10.8%)
16.4%
2015
(16%
0.40
6.4%)
(16%
0.60
9.6%)
16.0%
2016
(17%
0.40
6.8%)
(14%
0.60
8.4%)
15.2%
2017
(17%
0.40
6.8%)
(12%
0.60
7.2%)
14.0%
2018
(19%
0.40
7.6%)
(10%
0.60
6.0%)
13.6%
n
w j rj
j 1
b. Portfolio return: rp
rp c.
(wL
Standard deviation:
n 17.6 16.4 16.0 15.2 14.0 13.6 15.467 15.5% 6
i
(17.6% 15.5%)2 (15.2% 15.5%)2 rp
(2.1%)2
(0.9%)2
( 0.3%)2 rp
rp
(ri r )2 1 ( n 1)
n rp
(16.4% 15.5%)2 (14.0% 15.5%)2 6 1
(16.0% 15.5%)2 (13.6% 15.5%)2
(0.5%)2
( 1.5%)2 5
( 1.9%)2
(.000441 0.000081 0.000025 0.000009 0.000225 0.000361) 5
0.001142 0.000228% 0.0151 1.51% 5 d. The assets are negatively correlated. e. Combining these two negatively correlated assets reduces overall portfolio risk. rp
P8-14. Portfolio analysis LG 3; Challenge a.
Expected portfolio return: Alternative 1: 100% Asset F
rp
16% 17% 18% 19% 17.5% 4
Alternative 2: 50% Asset F
50% Asset G
Asset F (wF rF)
Year
Asset G (wG rG)
Portfolio Return rp
2013
(16%
0.50
8.0%)
(17%
0.50
8.5%)
16.5%
2014
(17%
0.50
8.5%)
(16%
0.50
8.0%)
16.5%
2015
(18%
0.50
9.0%)
(15%
0.50
7.5%)
16.5%
2016
(19%
0.50
9.5%)
(14%
0.50
7.0%)
16.5%
rp
16.5% 16.5% 16.5% 16.5% 16.5% 4
Alternative 3: 50% Asset F
50% Asset H
Asset F (wF rF)
Year
Asset H (wH rH)
Portfolio Return rp
2013
(16%
0.50
8.0%)
(14%
0.50
7.0%)
15.0%
2014
(17%
0.50
8.5%)
(15%
0.50
7.5%)
16.0%
2015
(18%
0.50
9.0%)
(16%
0.50
8.0%)
17.0%
2016
(19%
0.50
9.5%)
(17%
0.50
8.5%)
18.0%
rp
15.0% 16.0% 17.0% 18.0% 16.5% 4
b. Standard deviation:
(ri r )2 1 ( n 1)
n rp i
(1) [(16.0% 17.5%)2 F
[( 1.5%)2 F
F
F
(17.0% 17.5%)2 (18.0% 17.5%) 2 (19.0% 17.5%)2 ] 4 1
( 0.5%)2 (0.5%)2 3
(1.5%)2 ]
(0.000225 0.000025 0.000025 0.000225) 3 0.0005 3
.000167
0.01291 1.291%
(2) [(16.5% 16.5%)2
(16.5% 16.5%)2 (16.5% 16.5%)2 4 1
FG
[(0)2
(0)2
FG
FG
(0)2
(16.5% 16.5%)2 ]
(0)2 ]
3 0
(3) [(15.0% 16.5%)2
(16.0% 16.5%)2 (17.0% 16.5%)2 4 1
FH
[( 1.5%)2 FH
FH
FH
c.
( 0.5%)2 (0.5%)2 3
(18.0% 16.5%)2 ]
(1.5%)2 ]
[(0.000225 0.000025 0.000025 0.000225)] 3
0.0005 3
0.000167
Coefficient of variation: CV
r
0.012910 1.291% r
1.291% 0.0738 17.5% 0 CVFG 0 16.5% 1.291% CVFH 0.0782 16.5% d. Summary: CVF
rp: Expected Value of Portfolio
rp
Alternative 1 (F)
17.5%
1.291%
Alternative 2 (FG)
16.5%
0
Alternative 3 (FH)
16.5%
1.291%
CVp 0.0738 0.0 0.0782
Since the assets have different expected returns, the coefficient of variation should be used to determine the best portfolio. Alternative 3, with positively correlated assets, has the highest coefficient of variation and therefore is the riskiest. Alternative 2 is the best choice; it is perfectly negatively correlated and therefore has the lowest coefficient of variation.
P8-15. Correlation, risk, and return LG 4; Intermediate a.
(1) Range of expected return: between 8% and 13% (2) Range of the risk: between 5% and 10%
b. (1) Range of expected return: between 8% and 13% (2) Range of the risk: 0 risk 10% c.
(1) Range of expected return: between 8% and 13% (2) Range of the risk: 0 risk 10%
P8-16. Personal finance: International investment returns LG 1, 4; Intermediate 24,750 20,500 20,500
a.
Returnpesos
b.
Purchase price Sales price
c.
Returnpesos
4,250 20,500
Price in pesos Pesos per dollar
Price in pesos Pesos per dollar
20.50 9.21
0.20732 20.73%
\$2.22584 1,000 shares \$2,225.84
24.75 \$2.51269 1,000 shares \$2,512.69 9.85
2,512.69 2,225.84 2,225.84
286.85 2,225.84
0.12887 12.89%
d. The two returns differ due to the change in the exchange rate between the peso and the dollar. The peso had depreciation (and thus the dollar appreciated) between the purchase date and the sale date, causing a decrease in total return. The answer in part c is the more important of the two returns for Joe. An investor in foreign securities will carry exchange-rate risk. P8-17. Total, nondiversifiable, and diversifiable risk LG 5; Intermediate a. and b.
c.
Only nondiversifiable risk is relevant because, as shown by the graph, diversifiable risk can be virtually eliminated through holding a portfolio of at least 20 securities that are not positively correlated. David Talbot’s portfolio, assuming diversifiable risk could no longer be reduced by additions to the portfolio, has 6.47% relevant risk.
P8-18. Graphic derivation of beta LG 5; Intermediate a.
b. To estimate beta, the ―rise over run‖ method can be used: Beta
c.
Rise Run
Y X
Taking the points shown on the graph: Y 12 9 3 Beta A 0.75 X 8 4 4 Y 26 22 4 Beta B 1.33 X 13 10 3 A financial calculator with statistical functions can be used to perform linear regression analysis. The beta (slope) of line A is 0.79; of line B, 1.379. With a higher beta of 1.33, Asset B is more risky. Its return will move 1.33 times for each one point the market moves. Asset A’s return will move at a lower rate, as indicated by its beta coefficient of 0.75.
P8-19. Graphical derivation and interpretation of beta LG 5; Intermediate a. With a return range from 60% to 60%, Biotech Cures, exhibited in Panel B, is the more risky stock. Returns are widely dispersed in this return range regardless of market conditions. By comparison, the returns of Panel A’s Cyclical Industries Incorporated only range from about 40% to 40%. There is less dispersion of returns within this return range. b. The returns on Cyclical Industries Incorporated’s stock are more closely correlated with the market’s performance. Hence, most of Cyclical Industries’ returns fit around the upward sloping least-squares regression line. By comparison, Biotech Cures has earned returns approaching 60% during a period when the overall market experienced a loss. Even if the market is up, Biotech Cures has lost almost half of its value in some years. c. On a standalone basis, Biotech Cures Corporation is riskier. However, if an investor was seeking to diversify the risk of their current portfolio, the unique, nonsystematic performance of Biotech Cures Corporation makes it a good addition. Other considerations would be the mean return for both (here Cyclical Industries has a higher return when the overall market return is zero), expectations regarding the overall market performance, and level to which one can use historic returns to accurately forecast stock price behavior.
P8-20. Interpreting beta LG 5; Basic Effect of change in market return on asset with beta of 1.20: a. 1.20 (15%) 18.0% increase b. 1.20 ( 8%) 9.6% decrease c. 1.20 (0%) no change d. The asset is more risky than the market portfolio, which has a beta of 1. The higher beta makes the return move more than the market. P8-21. Betas LG 5; Basic a. and b. Asset
Beta
Increase in Market Return
A
0.50
0.10
B
1.60
C D
Expected Impact on Asset Return
Decrease in Market Return
Impact on Asset Return
0.05
0.10
0.05
0.10
0.16
0.10
0.16
0.20
0.10
0.02
0.10
0.02
0.90
0.10
0.09
0.10
0.09
c. Asset B should be chosen because it will have the highest increase in return. d. Asset C would be the appropriate choice because it is a defensive asset, moving in opposition to the market. In an economic downturn, Asset C’s return is increasing. P8-22. Personal finance: Betas and risk rankings LG 5; Intermediate a. Most risky Least risky
Stock
Beta
B
1.40
A
0.80
C
0.30
b. and c. Asset
Beta
Increase in Market Return
Expected Impact on Asset Return
Decrease in Market Return
Impact on Asset Return
A
0.80
0.12
0.096
0.05
0.04
B
1.40
0.12
0.168
0.05
0.07
C
0.30
0.12
0.036
0.05
0.015
d. In a declining market, an investor would choose the defensive stock, Stock C. While the market declines, the return on C increases. e. In a rising market, an investor would choose Stock B, the aggressive stock. As the market rises one point, Stock B rises 1.40 points.
n
P8-23. Personal finance: Portfolio betas: bp
w j bj
j 1
LG 5; Intermediate a. Portfolio A Asset
Beta
wA
1 2 3 4 5
1.30 0.70 1.25 1.10 0.90
0.10 0.30 0.10 0.10 0.40 bA
wA
Portfolio B wB
bA
0.130 0.210 0.125 0.110 0.360 0.935
0.30 0.10 0.20 0.20 0.20 bB
wB
bB
0.39 0.07 0.25 0.22 0.18 1.11
b. Portfolio A is slightly less risky than the market (average risk), while Portfolio B is more risky than the market. Portfolio B’s return will move more than Portfolio A’s for a given increase or decrease in market return. Portfolio B is the more risky. P8-24. Capital asset pricing model (CAPM): rj LG 6; Basic Case
rj
RF
RF
[bj
[bj
(rm
(rm
RF)]
RF)]
A
8.9%
5%
[1.30
(8%
B
12.5%
8%
[0.90
(13%
C
8.4%
9%
[ 0.20
D
15.0%
10%
[1.00
(15%
10%)]
E
8.4%
6%
[0.60
(10%
6%)]
(12%
5%)] 8%)] 9%)]
P8-25. Personal finance: Beta coefficients and the capital asset pricing model LG 5, 6; Intermediate To solve this problem you must take the CAPM and solve for beta. The resulting model is: r RF Beta rm RF a. b. c. d. e.
10% 5% 5% 0.4545 16% 5% 11% 15% 5% 10% Beta 0.9091 16% 5% 11% 18% 5% 13% Beta 1.1818 16% 5% 11% 20% 5% 15% Beta 1.3636 16% 5% 11% If Katherine is willing to take a maximum of average risk then she will be able to have an expected return of only 16%. (r 5% 1.0(16% 5%) 16%.) Beta
P8-26. Manipulating CAPM: rj LG 6; Intermediate a.
rj rj b. 15% RF c. 16% rm d. 15% bj
RF
[bj (rm
RF)]
8% [0.90 (12% 8%)] 11.6% RF [1.25 (14% RF)] 10% 9% [1.10 (rm 9%)] 15.36% 10% [bj (12.5% 10%) 2
P8-27. Personal finance: Portfolio return and beta LG 1, 3, 5, 6: Challenge a.
bp
(0.20)(0.80) (0.35)(0.95) (0.30)(1.50) 0.16 0.3325 0.45 0.1875 1.13
(0.15)(1.25)
b. rA
(\$20,000 \$20,000) \$1,600 \$20,000
\$1,600 \$20,000
8%
rB
(\$36,000 \$35,000) \$1,400 \$35,000
\$2,400 \$35,000
6.86%
rC
(\$34,500 \$30,000) 0 \$30,000
rD
(\$16,500 \$15,000) \$375 \$15,000
rP
(\$107,000 \$100,000) \$3,375 \$100,000
c.
\$4,500 15% \$30,000 \$1,875 \$15,000
12.5%
\$10,375 \$100,000
10.375%
d. rA 4% [0.80 (10% 4%)] 8.8% rB 4% [0.95 (10% 4%)] 9.7% rC 4% [1.50 (10% 4%)] 13.0% rD 4% [1.25 (10% 4%)] 11.5% e.
Of the four investments, only C (15% vs. 13%) and D (12.5% vs. 11.5%) had actual returns that exceeded the CAPM expected return (15% vs. 13%). The underperformance could be due to any unsystematic factor that would have caused the firm not do as well as expected. Another possibility is that the firm’s characteristics may have changed such that the beta at the time of the purchase overstated the true value of beta that existed during that year. A third explanation is that beta, as a single measure, may not capture all of the systematic factors that cause the expected return. In other words, there is error in the beta estimate.
P8-28. Security market line, SML LG 6; Intermediate a, b, and d.
c.
rj RF [bj (rm RF)] Asset A rj 0.09 [0.80 (0.13 0.09)] rj 0.122 Asset B rj 0.09 [1.30 (0.13 0.09)] rj 0.142 d. Asset A has a smaller required return than Asset B because it is less risky, based on the beta of 0.80 for Asset A versus 1.30 for Asset B. The market risk premium for Asset A is 3.2% (12.2% 9%), which is lower than Asset B’s market risk premium (14.2% 9% 5.2%). P8-29. Shifts in the security market line LG 6; Challenge a, b, c, d.
b. rj RF [bj (rm RF)] rA 8% [1.1 (12% 8%)] rA 8% 4.4% rA 12.4% c.
rA rA rA
6% [1.1 (10% 6% 4.4% 10.4%
6%)]
d. rA rA rA
8% [1.1 (13% 8% 5.5% 13.5%
8%)]
e.
(1) A decrease in inflationary expectations reduces the required return as shown in the parallel downward shift of the SML. (2) Increased risk aversion results in a steeper slope, since a higher return would be required for each level of risk as measured by beta.
P8-30. Integrative—risk, return, and CAPM LG 6; Challenge a. Project
rj
RF
[bj
A
rj
9%
[1.5
B
rj
9%
[0.75
C
rj
9%
[2.0
D
rj
9%
[0
E
rj
9%
[( 0.5)
(rm RF)] (14%
9%)]
(14% (14%
(14%
9%)] 9%)]
9%)]
(14%
9%)]
b. and d.
c.
Project A is 150% as responsive as the market.
16.5% 12.75% 19.0% 9.0% 6.5%
Project B is 75% as responsive as the market. Project C is twice as responsive as the market. Project D is unaffected by market movement. Project E is only half as responsive as the market, but moves in the opposite direction as the market. d. See graph for new SML. rA 9% [1.5 (12% 9%)] 13.50% rB 9% [0.75 (12% 9%)] 11.25% rC 9% [2.0 (12% 9%)] 15.00% rD 9% [0 (12% 9%)] 9.00% rE 9% [ 0.5 (12% 9%)] 7.50% e. The steeper slope of SMLb indicates a higher risk premium than SMLd for these market conditions. When investor risk aversion declines, investors require lower returns for any given risk level (beta). P8-31. Ethics problem LG 1; Intermediate Investors expect managers to take risks with their money, so it is clearly not unethical for managers to make risky investments with other people’s money. However, managers have a duty to communicate truthfully with investors about the risk that they are taking. Portfolio managers should not take risks that they do not expect to generate returns sufficient to compensate investors for the return variability. | 9,359 | 25,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.768658 |
https://answercult.com/question/eli5-probability-odds-help-im-not-math-genius-15-chance-3-times-but-if-i-hit-on-any-of-those-chances-i-get-the-whole-thing/ | 1,686,105,237,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00576.warc.gz | 123,091,311 | 20,913 | # ELi5 Probability odds help. Im not math genius 15% chance 3 times but if I hit on any of those chances I get the whole thing
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0
So I have a math probablity question.
Lets say there is a pile of money that I can win. I get 15 random numbers assigned to me (i get no say in the matter) and then a number out of 100 is drawn at random. If I hit on that 15% chance I win the money.
Now lets say I get three shots. Get 15 random numbers, number gets drawn out of 100 if I win I win game stops, If i lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose we do it again. Get 15 random numbers, number gets drawn out of 100 if I win I win if I lose I lose.
So basically I have a 15% chance of winning three times.. what is my overall chances of winning the money?
In: 2
[removed]
You need to look at the “complement”. The only way to *not* win is to lose all three times. There’s an 85% chance of losing on each attempt. 1-0.85^(3) ≈ 0.39 = 39% chance of winning.
Your odds of winning the first time are .15. Your odds of losing the first time and winning the second time are .85 * .15. Your odds of losing the first and second times but winning the third time are .85 * .85 * .15. Add those up and your odds of winning are about .3858, or 38.58%.
From your phrasing, it looks like the number you drew goes back in after each draw, so each trial is independent (whether you win on one trial doesn’t affect whether you win on another).
In general, the formula for the probability that BOTH of two events happen is:
P(X and Y) = P(X) * P(Y|X)
where P(Y|X) means “the chance that Y happened if you know that X happened”.
In this case, because the events are independent, knowing something about X tells you nothing about Y. That means the probability of Y doesn’t change if you know X happened, so P(Y|X) is just P(Y). In other words, **because these events are independent**:
P(X and Y) = P(X) * P(Y)
and the same goes for all three events together:
P(X and Y and Z) = P(X) * P(Y) * P(Z)
But in this case, we want to know the chance you win on **one** of your draws. There [are formulas](https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle) we could use for this, but it’s easier if we think about the problem a different way.
—-
*Winning* on *one* of your draws is the same thing as *not* losing on *all* of your draws. So instead of thinking about it as P(X or Y or Z) (which has a somewhat ugly formula), we can think of it as one minus P(not X and not Y and not Z) (since the chance something doesn’t happen is one minus the chance it does happen).
And that lets us use our formula from above:
P(not X and not Y and not Z) = P(not X) * P(not Y) * P(not Z)
P(not X) is the chance you lose on the first draw, or 0.85. Same for P(not Y) and P(not Z), so:
P(not X and not Y and not Z) = 0.85 * 0.85 * 0.85
which is 0.614125, or a bit more than 61%. This is the chance you **lose** all three draws, so the chance you *don’t* lose all three draws (that is, you win one of them) is 1 – 0.614125 = 0.385875 = about 39%.
Chance of losing in the first run is 85/100, in the second run it is 84/99, in the third run it is 83/98. So the chance of you losing is (85 * 84 * 83)/(100 * 99 * 98) = approx 0.61. The chance of winning is 1 – 0.61 = 0.39, or 39%. | 957 | 3,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-23 | latest | en | 0.92396 |
https://umassliving.wordpress.com/2009/07/31/jul-30-meeting-summary-wainwright-jordan-ch-7/ | 1,553,544,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00405.warc.gz | 667,580,129 | 18,430 | UMass LIVING
July 31, 2009
(Jul 30) Meeting Summary: Wainwright / Jordan, ch 7
Filed under: Uncategorized — umassliving @ 10:43 am
Tags: ,
Discussion from the meeting:
• Why is the Bethe approximation to the entropy $H_{Bethe}$ not generally concave?
• Since we know that $A^*$ is convex, and is the negative entropy, we know that the entropy is concave. However, $H_{Bethe}$ does not generally correspond to a true entropy. We assume that the edges form a tree and compute the entropy as if it were, subtracting on the mutual information on each edge, but this is not a true entropy unless the edges actually do form a tree, and hence is not generally concave.
• On page 167, in the discussion of the projection operation $\mapsto$, how do we know that $\mu(F) \in \mathcal{M}(F)$?
• To see this, we need to go back to the original definition of $\mathcal{M}$. Assume $\mu \in \mathcal{M}$. By the definition of $\mathcal{M}$, there exists some distribution $p$ such that $\mu_\alpha = \mathbb{E}_p[\phi_\alpha(x)]$ for all sufficient statistics indexed by $\alpha$, where the expectation is with respect to $p$. Clearly then, we can use the exact same $p$ to show that $\mu(F) \in \mathcal{M}(F)$ since the requirements are the same, but only for a subset of the sufficient statistics $\mathcal{I}(F)$.
• Understand 7.2.1. In particular, consider the distribution over spanning trees $\rho$ that places all its mass on one specific spanning tree. This seems very similar to structured mean field, where we just use the one spanning tree as the tractable subfamily we optimize over. Yet structured mean field gives us a lower bound on $A(\theta)$ whereas tree-reweighted Bethe gives us an upper bound. What accounts for this difference? For instance, why does structured mean field give a lower bound when we know that $A^*(\mu(F)) \leq A^*(\mu)$, and we are subtracting off $A^*$ in the objective function?
• In mean field, we have that $A^*(\tau) = A^*(\mu(F)) = A^*(\mu)$, as proven in last week’s meeting summary. Understanding this is the key to understanding this issue. In mean field, we assumed that the canonical parameters that were outside those defined for the tractable subgraph were 0. Therefore, $A(\theta)$ is the same, whether you include the inner product with the extra parameters, which are all zero, or not. The convex combination of spanning trees is different because of the projection operator $\mapsto$ above. We are still searching across the full space of possible mean parameters, but when we compute entropy, we use the projection operator to essentially ignore all the mean parameters that are not in our subgraph. This differs from mean field, where the mean parameters that are not in our subgraph are deterministic functions of the mean parameters that are in the subgraph, corresponding to the functions $g$ in Wainwright and $\Gamma$ in the structured mean field optimization paper. This difference means that you are searching over an outer bound as opposed to an inner bound, and are using an upper bound to the objective function, thus you are guaranteed to get an upper bound to $A(\theta)$.
• There should be a $\rho(T')$ in front of the second inner product in the second equation on page 176.
• This is for the $T'$ they mention that has $\Pi^{T'}(\lambda)$ different from zero. Additionally, if you are wondering why $A^*$ is strictly convex, it is a consequence of the fact that $A$ is strictly convex. The details are in Appendix B.3.
• Understand the general ideas in 7.2.2 and 7.2.3, the exact details are not as important. For instance, understand why $H_{ep}$ is generally not concave, but why using the re-weighting subject to $\sum_{\ell=1}^{d_I} \rho(\ell) = 1$ gives a concave function.
• Since entropy is concave, if we take a positive combination of entropy functions, that should also be concave. In $H_{ep}$, we have a negative amount $(1-d_I) H(\tau)$, so it is not concave. Re-weighting subject to $\sum_{\ell=1}^{d_I} \rho(\ell) = 1$, or actually, $\sum_{\ell=1}^{d_I} \rho(\ell) \leq 1$, will fix this problem, making the approximation concave.
• Understand the benefit of the convexified algorithms for algorithmic stability and what it means for an algorithm to be globally Lipschitz stable, why that is important, and how Figure 7.3 suggests that ordinary sum-product is not globally Lipschitz stable.
• One other important take-away from Figure 7.3 is why the approximations are better the lower the coupling strength (size of the parameter on the edges). This is because, the lower the coupling strength, the more independent the individual nodes, which in the extreme case of a coupling strength of zero becomes a simple product distribution. The higher the coupling strength, the more important the loops in the graph become, the more difficult inference becomes.
• Understand how the convexified algorithms can be used within learning in 7.5. We use the fact that when $B$ upper bounds the true cumulant function, then the surrogate likelihood lower bounds the true likelihood. Yet in chapter 6, we saw how using mean field within EM also gives us a lower bound on the true likelihood, despite the fact that mean field gives a lower bound on the cumulant function. What accounts for this discrepancy?
• This is a check on your understanding of mean field within EM. If you look at what mean field is approximating in EM, you should see why a lower bound on the cumulant gives a lower bound, whereas in 7.5 we need an upper bound on the cumulant to give a lower bound.
• Understand the intuitative interpretation of example 7.3.
• Another interesting point is the idea, in the final paragraph of the chapter, that it may be better to have an inconsistent estimator of the true parameter $\theta^*$, if your interest is in prediction (inference) and you must use an approximate inference algorithm.
Slightly more esoteric details:
• On page 184, why do $\mu_i$ show up along the diagonal of $N_1[\mu]$? Hint: each random variable $X_i$ is binary valued.
• The diagonal elements are $\mathbb{E}[X_i^2]$. Since $X_i$ is binary, we have $\mathbb{E}[X_i^2] = \sum_{0,1} X_i^2 p(X_i) = p(X_i = 1) = \mu_i$.
• Still unsure of whether you can go the other direction and show that $\text{cov}(X) \succeq 0$ implies $N_1[\mu] \succeq 0$. The wording in Wainwright seems to imply that you can, but it’s not clear to me how to do that using the Schur complement formula as they claim.
• Update: Kim, Kojima, Yamashtia, “Second Order Cone Programming Relaxation of Positive Semidefinite Constraint” has a simple proof of this equivalence. Let $A = \left( \begin{array}{cc} 1 & -\mu^T \\ 0 & \mathbf{I} \end{array} \right)$. Then $B = A^T N_1[\mu] A$ where $B = \left( \begin{array}{cc} 1 & 0 \\ 0 & \mathbb{E}[XX^T] - \mu \mu^T \end{array} \right)$. Clearly $B$ is positive semidefinite if and only if $\mathbb{E}[XX^T] - \mu \mu^T = \text{cov}(X) \succeq 0$, and since $A$ is nonsingular, $N_1[\mu]$ is psd if and only if $B$, and hence $\text{cov}(X)$, is psd. | 1,861 | 7,059 | {"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": 54, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-13 | latest | en | 0.8584 |
https://www.coursehero.com/file/5973304/MAT-109-Computer-Activity-5/ | 1,529,481,382,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00305.warc.gz | 768,510,718 | 67,403 | MAT 109 Computer Activity 5
# MAT 109 Computer Activity 5 - MAT 109 Computer Activity on...
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MAT 109 Computer Activity on the Central Limit Theorem Section 6-5 [40 total points] In this experiment, assume all dice have 6 sides. a. First run is the population. (5) One Die: Use STATDISK to simulate the rolling of a single die 1000 times. (Select Data , then Dice Generator .) Use Copy/Paste to transfer the results to the blank data window, Column 1. Once your data have been transferred, select Descriptive Statistics of Data to get the mean and standard deviation of this, the parent population. [Remember this first run requires the following entries in Dice Generator: 900 for sample size (which is really a population size, 1 for number of dice, and 6 for number of sides of each die. You cannot fill in random seed because you do not know it on this first (a) run. The computer will give you its pick for your random number. Record your random seed on this paper . You must use this random seed for each of the other three runs, b, c, and d. ] Use Histogram of Data to judge the shape of the distribution. If the bars are about the same size, write “approximately uniform.”
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# Question Video: Multiplying One-Digit Numbers by Four Using a Number Line Mathematics • Third Year of Primary School
Use the number line to find the product: 3 × 4 = ＿. Hint: How many jumps of 4 should you make?
01:06
### Video Transcript
Use the number line to find the product: Three times four equals what. Hint: How many jumps of four should you make?
In this question, we have to find the product of three times four. And we’re given a hint. How many jumps of four should we make on the number line? One jump has been made already, and we know that one times four is four. To find three times four, we need to make three jumps of four. The first one has been done. One four is four, two fours are eight, and three fours are 12. Four, eight, 12. We needed to skip count by fours to find three times four. Three times four equals 12.
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• Realistic Exam Questions | 283 | 1,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.907734 |
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DOI : 10.17577/IJERTCONV5IS04019
Text Only Version
#### Application of Differential Equations
Ms. K. brindha
Student, Kongunadu College of education, Trichy.
Abstract – In this paper we show that a method of embedding for a class of non-linear Volterra equations can be used in a novel fashion to obtain variation of parameters formulas for Volterra integral equations subjected to a general type of variation of the equation. The approach is of intrinsic interest. Our variation of parameters formulas generalize classical formulas for ordinary differential equations and for linear Volterra integral equations (based on resolvents). Illustrative examples are related to known results.
Keywords: Volterra equation, integral equation, variation of parameters, ordinary differential equations, linear equations, non linear equations.
INTRODUTION
Sir Isaac Newton (1642 – 1727) English scientist and mathematician famous for his discovary of the law of gravity and three laws of motion. Today these laws are known as Newtons laws of motion and descripe the motion of all objects on the scale we experience in our everyday lives. Solved his first differential equation, by the use of infinite series, eleven years after his discovery of calculus in 1665. Gottfried Wilhelm Leibniz (also known as von Leibniz) was a prominent German mathematician, philosopher, physicist and statesman. Noted for his independent invention of the differential and integral calculus, Gottfried Leibniz remains one of the greatest and most influential metaphysicians, thinkers and logicians in history. He also invented the Leibniz wheel and suggested important theories about force, energy and time.
DEFINITIONS
Differential equation:
Any relation involving the dependent variable, independent variable (or variables) and the differential coefficient (or coefficients) of the dependent variable with respect to the independent variable(or variables) is known as a differential equation. (Or)
A differential equation is a relationship between a function of time and its derivative.
For example:
1. = cot 5.2 + 4 = tan 2
2
2.2 + = 0 6. + =
2
3. = + ()3 7. 2
=
3
2 2
4.2
3
{1+() }
2 +
1 + ( )
= 0 8. =
2
2
Are all differential equations. VARIATION OF PARAMETER
equation
Variation of parameter is another method for finding a particular solution of the nth-order linear differential
() = () (1)
Once the solution of the associated homogeneous equation
() = 0 is known. If 1(), 2(), , () are n linearly independent solutions of () = 0,then the general solution of () = 0 is
= 11() + 22() + () (2)
The method:
A particular solution of () = () has the form
= 11 + 22 +
Where = ()( = 1,2,3, , )is given in (2) and ( = 1,2, , ) is an unknown function of x which still must be determined.
To find , first solve the following linear equations simultaneously for.
11 + 22 + + = 0
11 + 22 + + = 0
1(2) + 2(2) + + (2) = 0
1 2
1(1) + 2(1) + + (1) = ()
1 2
Then integrate each to obtain , disregarding all constraints of integration. This is permissible because we are seeking only one particular solution.
Example :
For the special case n=3, reduce to
Problems
1. + = sec .
Solution:
11 + 22 + 33 = 0
11 + 22 + 33 = 0
11 + 22 + 33 = ()
Given + = sec .
This is a third-order equation with
= 1 cos + 2 sin (1)
It follows from equation,
= 1 cos + 2 sin (2)
Using linear equation,
11 + 22 = 0
11 + 22 = ()
Here 1 = cos , 2 = sin & () = sec
1 cos + 2 sin = 0
* 1 cos = 2 sin
1 = 2 tan (3)
1( sin ) + 2 cos = sec
* 2(tan sin ) + 2 cos = sec
* 2
[sin2 +cos2 ] = sec
cos
2 = 1
= 1
2
cos
× cos = 1
(3) 1 = tan
Thus 1 = 1 = ( tan ) = log(cos )
2 = 2 = =
1 = log(cos x) & v2 = x
Substituting these values into (2) we obtain
= log(cos ) cos + sin
The general solution is
= +
= 1 cos + 2 sin + cos (log(cos )) + sin
2. 2 = 3
Solution:
This is second-order equation
Here n=2 & = 1 + 22
Hence = 1 + 22 (1)
Using linear equation is
11 + 22 = 0
11 + 22 = ()
1 = , 2 = 2 & () = 3,
1 + 22 = 0 (2)
1 = 22
1 = 23
& 1() + 222 = 0
22 + 222 = 3
322 = 3
2
=
3
(2) = 3
1 3
1
= 4
3
1
= 4
3
& 2
=
3
Thus = = 4
1 1 3
= 1 4 = 4
3 4 12
2
=
3
2 = 2 = 3
1
= 4
12
& 2
=
3
Substituting these values into (1) we obtain
= 4 + 2
12 3
= 3 + 3 = 3+43
The general solution is,
=
+
12 3 12
= 3
4
= + 2 + 3
1 2 4
CONCLUSION
I am very glad that I had an opportunity to do an independent paper in differential equation & their application. In the short span of time I have done best of my level. I scincerely hope that this small paper work will inspire the reader to do further reading in this field. This paper is mainly concerned with many interesting chapters such as basic concepts of differential equation , variation of parameter & etc.
REFERENCE:
1. Ahmad, Shair, Ambrosetti A textbook on Ordinary Differential Equations, Antonio 15th edition, 2014
2. Earl A. CoddingtonAn Introduction to Ordinary Differential Equations 1st Edition
3. Victor Henner, Tatyana Belozerova,Ordinary and Partial Differential Equations Mikhail Khenner January 29, 2013 by A K Peters/CRC Press
4. Dr. Richard Bronson , Differential Equations Third Edition.
5. Dr. Sudhir, , differential equation & convergence Sequence Publishing ,, 2013. | 1,553 | 5,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.919875 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/128/ES+(2,2)s4C4.html | 1,548,169,794,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583857913.57/warc/CC-MAIN-20190122140606-20190122162606-00576.warc.gz | 615,144,388 | 5,791 | Copied to
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## G = 2+ 1+4⋊4C4order 128 = 27
### 3rd semidirect product of 2+ 1+4 and C4 acting via C4/C2=C2
p-group, metabelian, nilpotent (class 3), monomial
Series: Derived Chief Lower central Upper central Jennings
Derived series C1 — C2×C4 — 2+ 1+4⋊4C4
Chief series C1 — C2 — C4 — C2×C4 — C22×C4 — C2×C4○D4 — C2.C25 — 2+ 1+4⋊4C4
Lower central C1 — C2 — C2×C4 — 2+ 1+4⋊4C4
Upper central C1 — C4 — C22×C4 — 2+ 1+4⋊4C4
Jennings C1 — C2 — C2 — C22×C4 — 2+ 1+4⋊4C4
Generators and relations for 2+ 1+44C4
G = < a,b,c,d,e | a4=b2=e4=1, c2=d2=a2, bab=a-1, ebe-1=ac=ca, ad=da, eae-1=a2d, bc=cb, bd=db, dcd-1=a2c, ece-1=a2bd, ede-1=a >
Subgroups: 556 in 274 conjugacy classes, 66 normal (18 characteristic)
C1, C2, C2 [×9], C4 [×2], C4 [×2], C4 [×8], C22, C22 [×2], C22 [×15], C8 [×2], C2×C4 [×2], C2×C4 [×6], C2×C4 [×28], D4 [×4], D4 [×28], Q8 [×4], Q8 [×8], C23, C23 [×2], C23 [×6], C42 [×2], C22⋊C4 [×2], C4⋊C4 [×2], C2×C8 [×4], M4(2) [×2], C22×C4, C22×C4 [×2], C22×C4 [×6], C2×D4, C2×D4 [×4], C2×D4 [×20], C2×Q8, C2×Q8 [×2], C2×Q8 [×6], C4○D4 [×8], C4○D4 [×36], C22⋊C8 [×2], C23⋊C4 [×2], D4⋊C4 [×2], Q8⋊C4 [×2], C4≀C2 [×4], C42⋊C2 [×2], C22×C8, C2×M4(2), C2×C4○D4, C2×C4○D4 [×2], C2×C4○D4 [×6], 2+ 1+4 [×2], 2+ 1+4 [×4], 2- 1+4 [×2], 2- 1+4 [×2], (C22×C8)⋊C2, C23.C23, C23.24D4 [×2], C42⋊C22 [×2], C2.C25, 2+ 1+44C4
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], C2×C4 [×6], D4 [×12], C23, C22⋊C4 [×12], C22×C4, C2×D4 [×6], C2×C22⋊C4 [×3], C22≀C2 [×4], C243C4, 2+ 1+44C4
Smallest permutation representation of 2+ 1+44C4
On 32 points
Generators in S32
```(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)
(1 2)(3 4)(5 6)(7 8)(9 12)(10 11)(13 28)(14 27)(15 26)(16 25)(17 18)(19 20)(21 32)(22 31)(23 30)(24 29)
(1 7 3 5)(2 8 4 6)(9 20 11 18)(10 17 12 19)(13 29 15 31)(14 30 16 32)(21 27 23 25)(22 28 24 26)
(1 12 3 10)(2 9 4 11)(5 19 7 17)(6 20 8 18)(13 30 15 32)(14 31 16 29)(21 28 23 26)(22 25 24 27)
(1 29 10 30)(2 14 9 13)(3 31 12 32)(4 16 11 15)(5 27 19 26)(6 22 18 23)(7 25 17 28)(8 24 20 21)```
`G:=sub<Sym(32)| (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), (1,2)(3,4)(5,6)(7,8)(9,12)(10,11)(13,28)(14,27)(15,26)(16,25)(17,18)(19,20)(21,32)(22,31)(23,30)(24,29), (1,7,3,5)(2,8,4,6)(9,20,11,18)(10,17,12,19)(13,29,15,31)(14,30,16,32)(21,27,23,25)(22,28,24,26), (1,12,3,10)(2,9,4,11)(5,19,7,17)(6,20,8,18)(13,30,15,32)(14,31,16,29)(21,28,23,26)(22,25,24,27), (1,29,10,30)(2,14,9,13)(3,31,12,32)(4,16,11,15)(5,27,19,26)(6,22,18,23)(7,25,17,28)(8,24,20,21)>;`
`G:=Group( (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), (1,2)(3,4)(5,6)(7,8)(9,12)(10,11)(13,28)(14,27)(15,26)(16,25)(17,18)(19,20)(21,32)(22,31)(23,30)(24,29), (1,7,3,5)(2,8,4,6)(9,20,11,18)(10,17,12,19)(13,29,15,31)(14,30,16,32)(21,27,23,25)(22,28,24,26), (1,12,3,10)(2,9,4,11)(5,19,7,17)(6,20,8,18)(13,30,15,32)(14,31,16,29)(21,28,23,26)(22,25,24,27), (1,29,10,30)(2,14,9,13)(3,31,12,32)(4,16,11,15)(5,27,19,26)(6,22,18,23)(7,25,17,28)(8,24,20,21) );`
`G=PermutationGroup([(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)], [(1,2),(3,4),(5,6),(7,8),(9,12),(10,11),(13,28),(14,27),(15,26),(16,25),(17,18),(19,20),(21,32),(22,31),(23,30),(24,29)], [(1,7,3,5),(2,8,4,6),(9,20,11,18),(10,17,12,19),(13,29,15,31),(14,30,16,32),(21,27,23,25),(22,28,24,26)], [(1,12,3,10),(2,9,4,11),(5,19,7,17),(6,20,8,18),(13,30,15,32),(14,31,16,29),(21,28,23,26),(22,25,24,27)], [(1,29,10,30),(2,14,9,13),(3,31,12,32),(4,16,11,15),(5,27,19,26),(6,22,18,23),(7,25,17,28),(8,24,20,21)])`
32 conjugacy classes
class 1 2A 2B 2C 2D 2E ··· 2J 4A 4B 4C 4D 4E 4F ··· 4K 4L 4M 4N 4O 8A 8B 8C 8D 8E 8F order 1 2 2 2 2 2 ··· 2 4 4 4 4 4 4 ··· 4 4 4 4 4 8 8 8 8 8 8 size 1 1 2 2 2 4 ··· 4 1 1 2 2 2 4 ··· 4 8 8 8 8 4 4 4 4 8 8
32 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 4 type + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C4 C4 D4 D4 D4 D4 2+ 1+4⋊4C4 kernel 2+ 1+4⋊4C4 (C22×C8)⋊C2 C23.C23 C23.24D4 C42⋊C22 C2.C25 2+ 1+4 2- 1+4 C22×C4 C2×D4 C2×Q8 C4○D4 C1 # reps 1 1 1 2 2 1 4 4 2 4 2 4 4
Matrix representation of 2+ 1+44C4 in GL4(𝔽17) generated by
16 0 0 8 4 0 13 1 0 13 0 4 4 0 0 1
,
1 0 0 9 0 0 13 16 0 4 0 13 0 0 0 16
,
1 0 15 0 13 0 4 16 1 0 16 0 0 1 4 0
,
13 0 8 0 0 0 1 13 0 0 4 0 0 13 1 0
,
8 2 15 2 0 14 5 12 5 2 15 0 5 14 5 14
`G:=sub<GL(4,GF(17))| [16,4,0,4,0,0,13,0,0,13,0,0,8,1,4,1],[1,0,0,0,0,0,4,0,0,13,0,0,9,16,13,16],[1,13,1,0,0,0,0,1,15,4,16,4,0,16,0,0],[13,0,0,0,0,0,0,13,8,1,4,1,0,13,0,0],[8,0,5,5,2,14,2,14,15,5,15,5,2,12,0,14] >;`
2+ 1+44C4 in GAP, Magma, Sage, TeX
`2_+^{1+4}\rtimes_4C_4`
`% in TeX`
`G:=Group("ES+(2,2):4C4");`
`// GroupNames label`
`G:=SmallGroup(128,526);`
`// by ID`
`G=gap.SmallGroup(128,526);`
`# by ID`
`G:=PCGroup([7,-2,2,2,-2,2,2,-2,224,141,422,2019,1018,521,248,1411,124]);`
`// Polycyclic`
`G:=Group<a,b,c,d,e|a^4=b^2=e^4=1,c^2=d^2=a^2,b*a*b=a^-1,e*b*e^-1=a*c=c*a,a*d=d*a,e*a*e^-1=a^2*d,b*c=c*b,b*d=d*b,d*c*d^-1=a^2*c,e*c*e^-1=a^2*b*d,e*d*e^-1=a>;`
`// generators/relations`
×
𝔽 | 3,346 | 5,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | longest | en | 0.378592 |
https://www.coursehero.com/file/6742553/MG375-Week-2-Homework/ | 1,485,169,248,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00079-ip-10-171-10-70.ec2.internal.warc.gz | 887,296,741 | 93,356 | MG375 Week 2 Homework
MG375 Week 2 Homework - Example 4.2 - Decision trees Value...
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Example 4.2 - Decision trees Value Revenue Build Cost Strong Growth \$765,000 \$975,000 \$210,000 0.55 Move \$585,000 Weak Growth \$365,000 \$575,000 \$210,000 0.45 Strong Growth \$863,000 \$950,000 \$87,000 0.55 \$703,750 Expand \$660,500 Weak Growth \$413,000 \$500,000 \$87,000 0.45 Expand \$843,000 \$930,000 \$87,000 Strong Growth \$850,000 0.55 Do Nothing \$703,750 Do Nothing \$850,000 \$850,000 \$0 Weak Growth \$525,000 \$525,000 \$0 0.45 Chapter 4, Problem 5 Value Revenue Building Cost High demand \$4,800,000 \$12,000,000 \$6,000,000 0.4 Small Facility \$4,800,000 Low demand \$6,000,000 \$10,000,000 \$6,000,000 0.6 High demand \$5,600,000 \$14,000,000 \$9,000,000 0.4 \$4,800,000 Large Facility \$2,600,000 Low demand \$6,000,000 \$10,000,000 \$9,000,000 0.6 High demand \$0 \$0 \$0 0 Do Nothing \$0 Low demand \$0 \$0 \$0 0 Hacker's Computer Store Expando Additional Factory
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Exhibit 4.4 - Decision Tree Analysis EV= \$585,000.00 \$765,000.00 Strong Growth 0.55 \$365,000.00 Weak Growth Move 0.45 EV= \$660,500.00 \$863,000.00 Strong Growth Expand
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Ask a homework question - tutors are online | 505 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.675336 |
https://imoneypage.com/9-4-worksheet-solving-right-triangles-with-trig/ | 1,566,111,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313715.51/warc/CC-MAIN-20190818062817-20190818084817-00402.warc.gz | 500,061,684 | 30,049 | # 9 4 Worksheet Solving Right Triangles With Trig
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Top | 1,517 | 7,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-35 | latest | en | 0.865739 |
http://www.dummies.com/education/science/physics/string-theory-strings-and-branes/ | 1,508,765,170,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826049.46/warc/CC-MAIN-20171023130351-20171023150351-00283.warc.gz | 417,918,686 | 10,637 | # String Theory: Strings and Branes
When string theory was originally developed in the 1970s, the filaments of energy in string theory were considered to be 1-dimensional objects: strings. (One-dimensional indicates that a string has only one dimension, length, as opposed to say a square, which has both length and height dimensions.)
These strings came in two forms — closed strings and open strings. An open string has ends that don’t touch each other, while a closed string is a loop with no open end. It was eventually found that these early strings, called Type I strings, could go through five basic types of interactions, as shown in this figure.
Type I strings can go through five fundamental interactions, based on different ways of joining and splitting. The interactions are based on a string’s ability to have ends join and split apart. Because the ends of open strings can join together to form closed strings, you can’t construct a string theory without closed strings.
This proved to be important, because the closed strings have properties that make physicists believe they might describe gravity! In other words, instead of just being a theory of matter particles, physicists began to realize that string theory may just be able to explain gravity and the behavior of particles.
Over the years, it was discovered that the theory required objects other than just strings. These objects can be seen as sheets, or branes. Strings can attach at one or both ends to these branes. A 2-dimensional brane (called a 2-brane) is shown in this figure. | 315 | 1,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.970888 |
https://www.maximintegrated.com/en/glossary/definitions.mvp/term/inverting-op-amp/gpk/1218 | 1,638,650,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00401.warc.gz | 962,856,141 | 19,667 | # Glossary Definition for inverting-op-amp
## Glossary Term: inverting-op-amp
Definition
An inverting op amp is an operational amplifier circuit with an output voltage that changes in the opposite direction as the input voltage. In other words, it is out of phase by 180o
## What is an inverting input?
An amplifier’s inverting input refers to the pin configuration. The inverting input is the terminal marked with a minus (-) sign, and the non-inverting input is marked with a plus (+) sign. These can also be referred to as negative and positive terminals. Circuit diagram symbol for an op amp with inverting and non-inverting inputs.
Circuit diagram symbol for an op amp with inverting (-) and non-inverting (+) inputs.
## How do inverting op amps work?
Inverting op amps work following the op amp golden rules:
1. The Current Rule: No current flows into the inputs of the op amp. (I+=I-=0)
2. The Voltage Rule: The output of the op amp attempts to ensure that the voltage difference between the two inputs is zero (V+=V-)
Inverting op amp circuit.
Consider the inverting op amp circuit shown above. Since the inverting input is tied to ground, by the Voltage Rule, the non-inverting input must also be at (virtual) ground.
The current flowing through R1is I=Vin/R1，and since the Current Rule states that the inputs draw no current, all of that current must then flow throughR2.
Since the inverting input is at virtual ground, the output of the inverting op amp is Vout=-IR2=-VinR2/R1.
This makes the gain of the inverting op amp circuit -R2/R1.The gain is negative, meaning the output is out of phase with the input.
## Op amp inverter
An op amp inverter is an inverting buffer constructed with an operational amplifier. An inverting buffer changes the direction of the signal without amplifying it, so the gain of the circuit is -1. We can see above that the inverting op amp circuit has a gain of -1 when the two resistors are equal, so an op amp inverter is an inverting op amp with R1=R2. | 474 | 2,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-49 | latest | en | 0.84921 |
http://jwilson.coe.uga.edu/EMAT6680Fa2012/Johnston/Asn6SwJ/Asn6SWJ.html | 1,542,817,436,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039749054.66/warc/CC-MAIN-20181121153320-20181121175320-00060.warc.gz | 157,459,647 | 2,097 | # Viewing Fine Art
### Sean Johnston
Problem: A 4x4 picture hangs on a wall such that its bottom edge is 2 feet above your eye level. How far back from the picture should you stand, directly in front of the picture, in order to view the picture under the maximum angle? Side view:
Solution: This problem is about optimal angles. We need to figure out where to move point B on the eye level line to make ABC as large as possible. We can see that if we move B very close to the picture, ABC gets very small, so we certainly should not stand there. If we move point B very far from the wall, ABC also gets very small. We can further explore the relationship of ABC to the position of B by looking at the circumcircle of ABC:
The measurement of ABC is exactly half of the measurement of AKC where K is the circumcenter of ABC. This fact is due to the Central Angle Theorem. Therefore, in order to maximize the measure of ABC, we need to maximize the measure of AKC. In order to do this, we need to minimize the size of the circumcircle of ABC. Since B is confined to the eye level line, we need to find the position of B where the circumcircle of ABC is tangent to the eye level line, like so:
Since the circumcircle of ABC is tangent to the eye level line at B in this case, B and K are the same distance from the wall in this case. By definition, K must lie along the perpendicular bisector of AC, which means that it is exactly 4 feet off of the ground. Therefore, |BK| = 4 ft. As the circumcenter, K must be equidistant from A, B, and C, so we know that |AK| = |CK| = |BK| = 4 ft. We can see now that we have two sets of equilateral triangles in this problem, like so:
We know that AKC is equilateral because we have calculated all of its side lengths to be 4 ft. Since, C is 2 ft above eye level, |CK| = 4ft, and |BK| = 4 ft, and BK is perpendicular to the eye level line, we have sufficient proof to say that KBC is equilateral, which means that |BC| = 4 ft. We can use this fact, along with the Pythagorean Theorem, to conclude that the distance from B to the wall is = 3.46 ft.
One other detail that we can notice about this problem is the measure of ABC. We know that AKC is 60 degrees because AKC is equilateral. The Central Angle Theorem tells us that ABC is half of AKC, making the measure of ABC 30 degrees in this optimum situation.
Return | 582 | 2,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2018-47 | latest | en | 0.95467 |
https://edulissy.com/shop/projects/lab-1-intro-to-logic-with-multimedia-logic-solution-2/ | 1,610,940,402,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00709.warc.gz | 321,451,343 | 27,364 | Lab 1: Intro to Logic with Multimedia Logic Solution
\$30.00
Category:
Description
Lab Objectives:
In this first lab, you will learn how to use the Multimedia Logic application in Windows to do schematic entry and simulation. MML is a free schematic entry and simulation tool. If you would like to get the tool for home (you still have to come to lab though for checkoff) here is a link to it http://www.softronix.com/logic.html. You will use this program to build some simple circuits.
You will also gain experience designing combinatorial logic circuits, and using combinatorial logic as part of more complex circuits.
Lab1 Preparation:
• Read through this Lab1 assignment.
• Review basic Logic Gates (Beginning of Chapter 3).
For every lab assignment in this course (write-ups, MML files, flowcharts, and code), you are required to include the following information in a header on the top of the first page:
• Lab number (in this case, Lab 1)
• Lab title
• Due Date
• The name of your TA
For MML files, each page should contain a header indicating the part of the lab that page contains.
• Each page should have a header describing which part of the lab the circuit is associated with, and a brief description of that circuit (eg: “Implementation of a truth table”)
• Each input should be labelled.
• If an input is several switches indicating a binary number, indicate which is the most and least significant bit.
• Each output should be labelled.
• If there are particularly complicated or confusing parts of your circuit, brief comments inside should describe what each section does (eg: “Compare bits 1a and 1b”)
Comments should also help people who are not familiar with your circuit to understand what it does, and how. In particular, your grader should be able to use your circuit, and even modify it slightly, without you there to explain it to them. Write your comments accordingly.
Part A (5pts): Warm Up
Do the tutorial for Multimedia Logic (Help->Tutorial) or look at the MML_Tutorial.pdf provided. This simple tutorial will walk you through building and simulating simple circuit, save the resulting file as Lab1.lgi.
Once you’ve built and tested the circuit, use the “Text” tool to put the required comments on your schematic.
Part B (5pts): Playing with Numbers
Now, build the schematic below. See if you can make it cleaner than the one given. Simulate the circuit and play around with the switches, getting a feel for inputting a binary number and seeing the result on the display.
The schematic below is not complete until you label your inputs and outputs!
Part C (10pts): Truth Table to Gates
Start a new schematic page and label it PART C. Design logic that implements the following truth table.
IN[2] IN[1] IN[0] Output 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 0
Things to note for Part C:
Connect the inputs to “Switch” tools and the output to an LED to verify your circuit works correctly. There should only be 3 inputs and one final output. (You may use additional debugging outputs if you like, but label them accordingly).
You may use any circuit topology you like: sum-of-products, product-of-sums, or a gate-reduced topology of your own. Whatever topology you choose, be sure you can reproduce your design process (guessing wildly or copying someone else will result in a poor grade!)
Part D (10pts): Guessing Game
Now you will create a fun guessing game in logic! Create a design on a new page labeled PART D that allows the user to play “guess the number.” In this game, the user presses a momentary pushbutton which generates a random 2-bit number. Then, the user can switch two toggle switches to attempt to guess that number. When the user is correct, an LED lights up.
Your circuit will need the following components:
• Use the random number generator circuit element. You only need to connect two of the outputs, which are the pins on the right side of the box (question: does it matter which two you use? Why or why not?).
• Use a momentary pushbutton switch to drive the random number “generation”. Connect the switch to “C+”. (To make a momentary pushbutton, start with a toggle switch and double-click to edit its properties).
• Use two toggle switches for user input, this will be the “guess”.
• Use combinational logic to test for equality; basically, is the output of the random number the same as that of the switches? If you’re having trouble figuring out how to make this circuit: one strategy is to write a truth table. Another strategy is to test for equality one bit at a time, and combine the results.
• Use an LED to indicate whether the user’s guess was correct or not.
• 2 LEDs to indicate the random secret number. This makes it a bad secret, but much easier to grade and debug! (You can cover up these with a sticky note while you play the guessing game, or if you’re ambitious, have another toggle switch disable them.)
In total, you should have 3 inputs (one momentary pushbutton and two toggles) and 3 outputs.
Lab Write-up (5 pts)
In every lab write-up for this course, we will be looking for the following things:
• Sensible lab writeup structure (eg: purpose, methods, results, analysis)
• Describe what you learned, what was surprising, what worked well and what did not
For this lab in particular, also address:
• How would you make your own 7-segment display from part B if Multimedia Logic didn’t provide one for you?
• How do you hypothesize the random number generator works? How can things be really random in a computer when it is made of logic gates, which are supposed to be deterministic?
Lab reports shouldn’t need to be long. One page is sufficient.
We do not break down the point values; instead, we will assess the lab report as a whole while looking for the following content in the report.
To alleviate file format issues we want lab reports in plain text. Feel free to use a word processor to type it up but please submit a plain text file and CHECK IT IN CANVAS to make sure it is readable.
Deliverables
You will submit two(2) files on canvas:
• lgi
• txt
Check-off
For this lab, as with all labs, you will need to demonstrate your lab when it is finished to the TA or tutor and get it signed off. The staff member will use your circuit, and ask you questions about its functionality and your design process. Checkoffs are due within 1 week of the submission date. GET CHECKED OFF IN YOUR OWN SECTION.
error: Content is protected !! | 1,481 | 6,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-04 | latest | en | 0.910533 |
https://www.javatpoint.com/postorder-traversal | 1,725,731,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00268.warc.gz | 789,217,591 | 27,707 | # Postorder Traversal
In this article, we will discuss the postorder traversal in data structure.
Linear data structures such as stack, array, queue, etc., only have one way to traverse the data. But in a hierarchical data structure such as tree, there are multiple ways to traverse the data. So, here we will discuss another way to traverse the tree data structure, i.e., postorder traversal. The postorder traversal is one of the traversing techniques used for visiting the node in the tree. It follows the principle LRN (Left-right-node). Postorder traversal is used to get the postfix expression of a tree.
The following steps are used to perform the postorder traversal:
• Traverse the left subtree by calling the postorder function recursively.
• Traverse the right subtree by calling the postorder function recursively.
• Access the data part of the current node.
The post order traversal technique follows the Left Right Root policy. Here, Left Right Root means the left subtree of the root node is traversed first, then the right subtree, and finally, the root node is traversed. Here, the Postorder name itself suggests that the tree's root node would be traversed at last.
### Algorithm
Now, let's see the algorithm of postorder traversal.
### Example of postorder traversal
Now, let's see an example of postorder traversal. It will be easier to understand the process of postorder traversal using an example.
The nodes with yellow color are not visited yet. Now, we will traverse the nodes of above tree using postorder traversal.
• Here, 40 is the root node. We first visit the left subtree of 40, i.e., 30. Node 30 will also traverse in post order. 25 is the left subtree of 30, so it is also traversed in post order. Then 15 is the left subtree of 25. But 15 has no subtree, so print 15 and move towards the right subtree of 25.
• 28 is the right subtree of 25, and it has no children. So, print 28.
• Now, print 25, and the postorder traversal for 25 is finished.
• Next, move towards the right subtree of 30. 35 is the right subtree of 30, and it has no children. So, print 35.
• After that, print 30, and the postorder traversal for 30 is finished. So, the left subtree of given binary tree is traversed.
• Now, move towards the right subtree of 40 that is 50, and it will also traverse in post order. 45 is the left subtree of 50, and it has no children. So, print 45 and move towards the right subtree of 50.
• 60 is the right subtree of 50, which will also be traversed in post order. 55 is the left subtree of 60 that has no children. So, print 55.
• Now, print 70, which is the right subtree of 60.
• Now, print 60, and the post order traversal for 60 is completed.
• Now, print 50, and the post order traversal for 50 is completed.
• At last, print 40, which is the root node of the given binary tree, and the post order traversal for node 40 is completed.
The final output that we will get after postorder traversal is -
{15, 28, 25, 35, 30, 45, 55, 70, 60, 50, 40}
## Complexity of Postorder traversal
The time complexity of postorder traversal is O(n), where 'n' is the size of binary tree.
Whereas, the space complexity of postorder traversal is O(1), if we do not consider the stack size for function calls. Otherwise, the space complexity of postorder traversal is O(h), where 'h' is the height of tree.
## Implementation of Postorder traversal
Now, let's see the implementation of postorder traversal in different programming languages.
Program: Write a program to implement postorder traversal in C language.
Output
Program: Write a program to implement postorder traversal in C++.
Output
Program: Write a program to implement postorder traversal in C#.
Output
After the execution of the above code, the output will be -
Program: Write a program to implement postorder traversal in Java.
Output
After the execution of the above code, the output will be -
So, that's all about the article. Hope the article will be helpful and informative to you.
Next TopicSparse Matrix | 955 | 4,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.831068 |
https://ask.sagemath.org/question/62464/listing-all-those-pairs-of-tuples-which-are-permutations-of-each-other/ | 1,721,524,787,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00701.warc.gz | 91,035,731 | 15,188 | listing all those pairs of tuples which are permutations of each other
Suppose I have a list of tuples like
import itertools
l=range(0,6)
m=itertools.combinations(l,2)
list(m)
How do I get those pairs of tuples (i,j);(x,y) such that (i,j)=(y,x)? Also, is there a way to generalize this, that is, I would to have pairs of tuples (a,b,c,...);(x,y,z...) such that (a,b,c...)=(z,y,x...). Is there any efficent way of doing this? Thanks beforehand.
edit retag close merge delete
I guess itertools.combinations(1,2) should be replaced by itertools.combinations(l,2).
( 2022-05-16 21:48:28 +0200 )edit
(I edited, changing 1 to l.)
( 2022-05-16 23:57:34 +0200 )edit
Please never use a single ell-letter for the name of a variable, even if it stays for a list. (In our case it is a range...)
( 2022-05-17 22:35:57 +0200 )edit
@dan_fulea: next you're probably going to suggest that we shouldn't use "O" for a variable, either...
( 2022-05-18 01:12:56 +0200 )edit
Sort by ยป oldest newest most voted
I am not sure to understand your question, but you can do:
sage: [((a,b),(b,a)) for (a,b) in m]
[((0, 1), (1, 0)),
((0, 2), (2, 0)),
((0, 3), (3, 0)),
((0, 4), (4, 0)),
((0, 5), (5, 0)),
((1, 2), (2, 1)),
((1, 3), (3, 1)),
((1, 4), (4, 1)),
((1, 5), (5, 1)),
((2, 3), (3, 2)),
((2, 4), (4, 2)),
((2, 5), (5, 2)),
((3, 4), (4, 3)),
((3, 5), (5, 3)),
((4, 5), (5, 4))]
If you have longer tuples, you can reverse them as follows:
sage: t = (1,2,5,6)
sage: t[::-1]
(6, 5, 2, 1)
So, you can do :
sage: m=itertools.combinations(l,4)
sage: [(t,t[::-1]) for t in m]
[((0, 1, 2, 3), (3, 2, 1, 0)),
((0, 1, 2, 4), (4, 2, 1, 0)),
((0, 1, 2, 5), (5, 2, 1, 0)),
((0, 1, 3, 4), (4, 3, 1, 0)),
((0, 1, 3, 5), (5, 3, 1, 0)),
((0, 1, 4, 5), (5, 4, 1, 0)),
((0, 2, 3, 4), (4, 3, 2, 0)),
((0, 2, 3, 5), (5, 3, 2, 0)),
((0, 2, 4, 5), (5, 4, 2, 0)),
((0, 3, 4, 5), (5, 4, 3, 0)),
((1, 2, 3, 4), (4, 3, 2, 1)),
((1, 2, 3, 5), (5, 3, 2, 1)),
((1, 2, 4, 5), (5, 4, 2, 1)),
((1, 3, 4, 5), (5, 4, 3, 1)),
((2, 3, 4, 5), (5, 4, 3, 2))]
more
If you want permutations, then you can use Permutations generator:
t = ['a','b','c','d']
for p in Permutations(t):
print(p)
more | 1,000 | 2,157 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.749817 |
https://www.convertunits.com/from/kilotonne/to/bag | 1,670,602,545,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00326.warc.gz | 768,053,076 | 24,085 | ››Convert kilotonne to bag [portland cement]
kilotonne bag
How many kilotonne in 1 bag? The answer is 4.263768278E-5.
We assume you are converting between kilotonne and bag [portland cement].
You can view more details on each measurement unit:
kilotonne or bag
The SI base unit for mass is the kilogram.
1 kilogram is equal to 1.0E-6 kilotonne, or 0.023453432147327 bag.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between kilotonnes and bag [portland cement].
Type in your own numbers in the form to convert the units!
››Quick conversion chart of kilotonne to bag
1 kilotonne to bag = 23453.43215 bag
2 kilotonne to bag = 46906.86429 bag
3 kilotonne to bag = 70360.29644 bag
4 kilotonne to bag = 93813.72859 bag
5 kilotonne to bag = 117267.16074 bag
6 kilotonne to bag = 140720.59288 bag
7 kilotonne to bag = 164174.02503 bag
8 kilotonne to bag = 187627.45718 bag
9 kilotonne to bag = 211080.88933 bag
10 kilotonne to bag = 234534.32147 bag
››Want other units?
You can do the reverse unit conversion from bag to kilotonne, or enter any two units below:
Enter two units to convert
From: To:
››Definition: Kilotonne
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilotonne = 103 tonnes.
The definition of a tonne is as follows:
A tonne (also called metric ton) is a non-SI unit of mass, accepted for use with SI, defined as: 1 tonne = 1000 kg (= 106 g).
››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! | 565 | 1,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.791165 |
http://www.mathplanet.com/education/algebra-1/how-to-solve-linear-equations/properties-of-equalities | 1,472,027,161,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982291592.14/warc/CC-MAIN-20160823195811-00251-ip-10-153-172-175.ec2.internal.warc.gz | 563,737,138 | 14,329 | # Properties of equalities
Two equations that have the same solution are called equivalent equations e.g. 5 +3 = 2 + 6. And this as we learned in a previous section is shown by the equality sign =. An inverse operation are two operations that undo each other e.g. addition and subtraction or multiplication and division. You can perform the same inverse operation on each side of an equivalent equation without changing the equality.
$5+3\, {\color{green} {-\, 2}}=6+2\, {\color{green}{ -\, 2}}$
This gives us a couple of properties that hold true for all equations.
The addition property of equality tells us that adding the same number to each side of an equation gives us an equivalent equation
$if\: a-b=c,then\: a-b\, {\color{green} {+\, b}}=c\, {\color{green} {+\, b}}, or \: a=c+b$
The same goes with the subtraction property of equality.
$if\: a+b=c,then\: a+b\, {\color{green} {-\, b}}=c\, {\color{green}{ -\, b}}, or \: a=c-b$
As well as it goes for the multiplication property of equality. If you multiply each side of an equation with the same nonzero number you produce an equivalent equation.
$if\: \frac{a}{b}=c, and\: b\neq 0,then\: \frac{a}{b}\, \cdot {\color{green} b}=c\cdot \, {\color{green} b}, or \: a=cb$
And naturally this goes for the division property of equality as well. You can divide each side of an equation with the same nonzero number to produce an equivalent equation
$if\: a\cdot b=c, and\: b\neq 0,then\: \frac{a\cdot b}{{\color{green} b}}\, =\frac{c}{{\color{green} b}}, or \: a=\frac{c}{b}$
This gives us a way to change an equation to our own liking. Anything is acceptable as long as you do the same thing on both sides.
There are a couple of other properties of equations that can be good to know as well
Example
George has cut down a 60ft high oak tree. He now wants to chop it into smaller pieces. He first cuts it into two pieces that are both 30ft. And then he continues on into making ten pieces that all are 6ft long before loading them onto his truck.
By looking at the different pieces of wood we can see that the following holds true.
$60=30+30=6+6+6+6+6+6+6+6+6+6$
This is called the reflexive property of equality and tells us that any quantity is equal to itself
$a=a$
We can also use this example with the pieces of wood to explain the symmetric property of equality. This property states that if quantity a equals quantity b, then b equals a.
$if\: a=b, \: then\: b=a$
Or if we use our example
$if\: 60=30+30, \: then\: 30+30=60$
Another property that can be explained by this is the transitive property of equality. It tells us that if a quantity a equals quantity b, and b equals the quantity, c, then a and c are equal as well.
$if\: a=b\: and\: b=c, \: then\: a=c$
Or in the numbers taken from the oak tree example
$if\: 60=30+30$
$and\: 30+30=6+6+6+6+6+6+6+6+6+6$
$then\: 60=6+6+6+6+6+6+6+6+6+6$
Since we know that 30 + 30 = 20 + 40 and that 30 + 30 = 60 we can substitute 30 + 30 for 20 + 40 and get 60 = 20 + 40. This is called the substitution property of equality.
If a = b, then a can be substituted for b in any expression.
## Video lesson
Solve these equations using inverse operations
$x + 8 = 10$
$x - 4 = 22$
$x\div 3 = 6$
$7x = 28$ | 954 | 3,239 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2016-36 | longest | en | 0.87394 |
https://gmatclub.com/forum/unlike-most-warbler-species-the-male-and-female-blue-winged-13444.html?fl=similar | 1,505,913,181,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687281.63/warc/CC-MAIN-20170920123428-20170920143428-00078.warc.gz | 650,065,798 | 41,599 | It is currently 20 Sep 2017, 06:13
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# Unlike most warbler species, the male and female blue-winged
Author Message
Director
Joined: 21 Sep 2004
Posts: 607
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Unlike most warbler species, the male and female blue-winged [#permalink]
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22 Jan 2005, 17:03
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22. Unlike most warbler species, the male and female blue-winged warbler are very difficult to tell apart.
(A) Unlike most warbler species, the male and female blue-winged warbler are very difficult to tell apart.
(B) Unlike most warbler species, the gender of the blue-winged warbler is very difficult to distinguish.
(C) Unlike those in most warbler species, the male and female blue-winged warblers are very difficult to distinguish.
(D) It is very difficult, unlike in most warbler species, to tell the male and female blue-winged warbler apart.
(E) Blue-winged warblers are unlike most species of warbler in that it is very difficult to tell the male and female apart.
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Director
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22 Jan 2005, 17:59
I chose C too. .But OA is E. can somebody please explain why is it E?
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Director
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### Show Tags
22 Jan 2005, 19:40
I got E and I shall explain:
You should compare most warbler species with blue-winged warbler, as in
'Unlike most warbler species, blue-winged warbler ....'
A, B, C do the comparison wrongly.
D is just ackward.
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Joined: 03 Jan 2005
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### Show Tags
22 Jan 2005, 23:49
Yes, C compares blue-winged warblers with "those in most warblers species". "those" here refers to genders, but comparing bird with genders is wrong.
Kudos [?]: 368 [0], given: 0
22 Jan 2005, 23:49
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Display posts from previous: Sort by | 886 | 3,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-39 | latest | en | 0.942387 |
http://mathhelpforum.com/pre-calculus/104468-quick-periodic-extension-question.html | 1,529,334,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00537.warc.gz | 208,137,279 | 9,990 | # Thread: Quick Periodic Extension Question
1. ## Quick Periodic Extension Question
If you were asked to do a periodic extension of a function and to increase the period, would you stretch the parameters out?
For example:
$\displaystyle f(x)=\left\{\begin{array}{cc}x^2,&\mbox{ if } [0,1)\\2-x, & \mbox{ if } [1,2]\end{array}\right.$
If were told to do a periodic extension, but change the period from 2 to 4, how would I go about doing this? Would I change the first interval from [0,1) to [0,2)?
2. Originally Posted by BlackBlaze
If you were asked to do a periodic extension of a function and to increase the period, would you stretch the parameters out?
For example:
$\displaystyle f(x)=\left\{\begin{array}{cc}x^2,&\mbox{ if } [0,1)\\2-x, & \mbox{ if } [1,2]\end{array}\right.$
If were told to do a periodic extension, but change the period from 2 to 4, how would I go about doing this? Would I change the first interval from [0,1) to [0,2)?
Look closely this picture.
I think that is understandable.
In the general case (on the real line) for your function $\displaystyle k \in \mathbb{Z}$
3. Wait.
Wait, wait, wait, wait.
...Nope, I lost it. Forgive me, I guess I'm not that bright. I see you were able to make a simple periodic extension of my function, but I do not understand how the k plays a part. Am I to extend function so that when k = 0, f(x) can go across 0->4? | 404 | 1,389 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.90797 |
https://datascience.stackexchange.com/questions/121568/does-a-classifier-based-on-optimal-bayes-classifier-equation-classify-every-new/121571 | 1,718,944,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00777.warc.gz | 163,350,567 | 40,209 | # Does a classifier based on optimal bayes classifier equation classify every new instance the same way?
I'm trying to understand how optimal bayes classifier works and I was wondering if, given that the function we try to maximize when making a new prediction does not depend on the instance we are trying to classify, is it correct to state that the optimal bayes classifier would always predict the most probable class no matter what the input is?
EDIT:
I've been studying the subject on "Machine Learning, Tom Mitchell, McGraw Hill, 1997" where is stated that the prediction for a new instance is the class $$v_j$$ for which the function $$\underset{v_j \in V}{\operatorname{arg max}}\sum_{h_i \in H}{P}(v_j|h_i){P}(h_i|D)$$
Where $$V$$ is the set of all possible classes, $$H$$ is the space of the hypothesis and $$D$$ is the train dataset.
• Hi Niccolò, welcome to DS stack exchange. Please include a reference for your statement, this would help clear the particular misunderstanding that it seems you're having. Anyway, the statement is wrong. The output does depend on the input. Commented May 16, 2023 at 10:00
• @KishKash thank you for the suggestion, I don't know if I'm misinterpreting the way the classification happens but to me it seems like all the values involved in the maximization problem depends on dataset, hypotesis space and class space Commented May 16, 2023 at 10:24
Therefore, the $$\arg \max$$ in the optimization problem above is over different functions mapping inputs to outputs, taking into account the entire training set.
• The books says that the sum in the argmax I wrote gives the probability ${P}(v_j|D)$ and that the optimal classification is the $v_j$ value which is a class value itself. So I don't agree with the fact that the problem is trying to obtain the best classification function. The book also differentiate the bayes optimal classifier from the MAP problem which is the one that gives the optimal hypotheses as output. Commented May 16, 2023 at 20:09 | 470 | 2,008 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.931008 |
http://www.topperlearning.com/forums/ask-experts-19/a-solid-sphere-and-a-disc-of-equal-radii-are-rolling-on-an-i-physics-system-of-particles-and-rotational-motion-62712/reply | 1,485,263,916,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284411.66/warc/CC-MAIN-20170116095124-00198-ip-10-171-10-70.ec2.internal.warc.gz | 721,848,500 | 37,078 |
Question
Tue February 05, 2013 By:
# a solid sphere and a disc of equal radii are rolling on an inclined plane without slipping. one reaches earlier than the other due to different radius of gyration?? why the answer cannot be moment of inertia??
Wed February 06, 2013
Either the answer is : one reaches earlier than the other due to different radius of gyration
OR, moment of inertia (not only this)and mass.MOment of inertia and mass both are into radius of gyration.
acceleration of the object=a=
k=radius of gyration and I=mom of inertia
Related Questions
Fri December 02, 2016
# A 6.5 metres long ladder rests along a vertical wall reaching a height of 6 metres .A 60kg man stands halfway up the ladder(a)find the torque of the force exerted by the man on the laddder about the upper end of the ladder(b)assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth,find the force exerted by the ground on the ladder.
Sun November 20, 2016
| 250 | 1,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-04 | longest | en | 0.914222 |
http://lastfiascorun.com/cuba/what-is-an-alternate-interior-angle.html | 1,670,460,404,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00191.warc.gz | 27,715,686 | 19,210 | # What Is An Alternate Interior Angle?
the angles which are inside the parallel lines and on alternate sides of the third line are called alternate interior angles. If two parallel lines are transected by a third line, the angles which are inside the parallel lines and on the same side of the third line are called opposite interior angles.
## What is alternative interior angle?
Alternate interior angles are the angles formed on the opposite sides of the transversal. In other words, when two parallel lines are intersected by a transversal, eight angles are formed.
## What is alternate interior angles example?
When two lines are crossed by another line (called the Transversal): Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal. In this example, these are two pairs of Alternate Interior Angles: c and f.
## Do alternate interior angles add up to 180?
Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles’. d and f are interior angles. These add up to 180 degrees (e and c are also interior). Any two angles that add up to 180 degrees are known as supplementary angles.
## What’s the meaning of alternate angles?
: one of a pair of angles with different vertices and on opposite sides of a transversal at its intersection with two other lines: a: one of a pair of angles inside the two intersected lines.
## What is alternate interior and exterior angles?
Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called exterior angles. Angles that are on the opposite sides of the transversal are called alternate angles e.g. 1 + 8.
You might be interested: Readers ask: How Long Does Aws Certificate Manager Take?
## What is an example of a vertical angle?
Vertical angles are pair angles formed when two lines intersect. Vertical angles are sometimes referred to as vertically opposite angles because the angles are opposite to each other. Real-life settings where vertical angles are used include; railroad crossing sign, letter “X”, open scissors pliers etc.
## What is the difference between alternate interior angles and alternate interior angles converse?
Converse of Alternate Interior Angles Theorem The Converse of the Alternate Interior Angles Theorem states that if two lines are cut by a transversal and the alternate interior angles are congruent, then the lines are parallel.
## Is alternate interior angles supplementary?
Yes alternate interior angles are supplementary.
## How do you solve alternate interior?
The alternate interior angles theorem states that, the alternate interior angles are congruent when the transversal intersects two parallel lines. Hence, it is proved. Alternate interior angles can be calculated by using properties of the parallel lines. Two consecutive interior angles are (2x + 10) ° and (x + 5) °.
## What are the rules of alternate interior angles?
The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate interior angles are congruent.
## Does supplementary mean 180?
Two angles are called supplementary when their measures add up to 180 degrees.
## Do opposite angles add up to 180?
The sum of the opposite angles of a quadrilateral in a circle is 180°, as long as the quadrilateral does not cross itself. | 714 | 3,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-49 | latest | en | 0.936245 |
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Topic: Test constantness of normality of residuals from linear regression
Replies: 9 Last Post: Jan 12, 2013 7:01 PM
Messages: [ Previous | Next ]
Ray Koopman Posts: 3,383 Registered: 12/7/04
Re: Test constantness of normality of residuals from linear regression
Posted: Jan 10, 2013 2:02 PM
On Jan 10, 6:59 am, Paul <paul.domaskis@gmail.com> wrote:
> On Jan 10, 12:48 am, Ray Koopman <koopman@sfu.ca> wrote:
>> On Jan 9, 7:35 pm, Paul <paul.domaskis@gmail.com> wrote:
>>> After much browsing of Wikipedia and the web, I used both normal
>>> probability plot and Anderson-Darling to test the normality of
>>> residuals from a simple linear regression (SLR) of 6 data points.
>>> Results were very good. However, SLR doesn't just assume that the
>>> residuals are normal. It assumes that the standard deviation of the
>>> PDF that gives rise to the residuals is constant along the horizontal
>>> axis. Is there a way to test for this if none of the data points
>>> have the same value for the independent variable? I want to be able
>>> to show that there is no gross curves or spreading/focusing of the
>>> scatter.
>>>
>>> In electrical engineering signal theory, the horizontal axis is time.
>>> Using Fourier Transform (FT), time-frequency domains can show trends.
>>> Intuitively, I would set up the data as a scatter graph of residuals
>>> plotted against the independent variable (which would be treated as
>>> time). Gross curves show up as low-frequency content. There should
>>> be none if residuals are truly iid. The spectrum should look like
>>> white noise. The usual way to get the power spectrum is the FT of
>>> the autocorrelation function, which itself should resemble an impulse
>>> at zero. This just shows indepedence of samples, not constant iid
>>> normal along the horizontal axis.
>>>
>>> As for spreading or narrowing of the scatter, I guess that can be
>>> modelled in time as a multiplication of a truly random signal by a
>>> linear (or exponential) attenuation function. The latter acts like
>>> a modulation envelope. Their power spectrums will then convolve in
>>> some weird way. I'm not sure if this is a fruitful direction for
>>> identifying trends in the residuals. It starts to get convoluted
>>> pretty quickly.
>>>
>>> Surely there must be a less klugy way from the world of statistics?
>>> I realize that my sample size will probably be too small for many
>>> conceptual approaches. For example, if I had a wealth of data points,
>>> I could segment the horizontal axis, then do a normality test on each
>>> segment. This would generate mu's and sigma's as well, which could
>>> then be compared across segments. So for the sake of conceptual
>>> gratification, I'm hoping for a more elegant test for the ideal case
>>> of many data points. If there is also a test for small sample sizes,
>>> so much the better (though I don't hold my breath).
>>
>> If y|x = a + b*x + e, where the errors are iid random variables with
>> zero means, and you do an ordinary least squares fit of that model to
>> (x1,y1), ..., (xn,yn), then the theoretical variance of the residual
>> for xi is 1 - 1/n - [(xi-m)^2 / sum{(xj-m)^2}], ...
That's incomplete. That whole expression needs to be multiplied
by the variance of the error distribution.
>> . . . . . . . . . . . . . . . . . . . . . . . . where m is the mean
>> of x1, ..., xn. In words, residuals whose x is far from the mean tend
>> to be smaller than those whose x is hear the mean. (This is known as
>> "leverage": points far from the mean have more "leverage" on the
>> regression line, pulling it closer to them.) Note that normality is
>> not required
>
> Ray,
>
> Thanks for the background. One of the 4 explicit assumptions of
> regression is that the PDF for the random errors are normal, according
> to Introductory Statistics by Prem S Man (3rd edition). Is this not
> correct? This is the reasons I am learngin about normality tests, and
> especially about the constantness of the PDF along the horizontal axis.
It all depends on what you want. Look up the Gauss-Markov theorem.
To justify the usual OLS estimates of the regression coefficients,
the errors need only to be unbiased, uncorrelated, and homoscedastic,
but to justify all the usual p-values and confidence regions, the
errors must be iid normal.
However, that's considering only the theoretical justification.
In practice, what matters is not whether the assumptions are right
or wrong, but how wrong they are -- they're never exactly right.
Normality is probably the least important assumption. The most
important things to worry about are the general form of the model
and whether it includes all the relevant predictor variables. Then
you ask how correlated and/or heteroscedastic the errors might be.
Finally, you might wonder about shapes of the error distributions.
Minor departures from normality are inconsequential. Nothing in the
real world is exactly normal, and any test of normality will reject
if the sample size is big enough.
Date Subject Author
1/9/13 Paul
1/9/13 Paul
1/10/13 Ray Koopman
1/10/13 Paul
1/10/13 Ray Koopman
1/10/13 Michael Press
1/10/13 Paul
1/12/13 Herman Rubin
1/10/13 David Jones
1/10/13 Paul | 1,336 | 5,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-22 | latest | en | 0.906292 |
https://www.physicsforums.com/threads/charge-distribution-question.733943/ | 1,521,937,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651465.90/warc/CC-MAIN-20180324225928-20180325005928-00710.warc.gz | 835,809,072 | 15,406 | # Charge distribution question.
1. Jan 21, 2014
### That Neuron
This post is referring to page 35-36.
I just find it odd that this book doesn't change it's mathematical description to align with the fact that a ring produces a component force up smaller than the force that it would produce if it weren't pointing straight up.
In other words, taking an infitesmal ring of small dipoles that interact with a single dipole of the same type above it by a force proportional to 1/(r^6) (dipole-dipole interaction), wouldn't the force be something involving the unit vectors of the vectors connecting the dipole being studied and the ring of dipoles?
More to the point, wouldn't it be the integral per portion of dipoles in the ring dq with some function based off of the distance from the dipoles(charge analogy), so the integral of (kdq/(x^2 + y^2)^(6/2)) cosθ = integral of (kxdq/(x^2 + y^2)^5/2)?
This might help illustrate what I'm talking about: http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html | 266 | 1,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.917121 |
https://la.mathworks.com/matlabcentral/cody/problems/730-how-many-trades-represent-all-the-profit/solutions/2187078 | 1,590,968,217,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00573.warc.gz | 413,095,618 | 15,735 | Cody
# Problem 730. How many trades represent all the profit?
Solution 2187078
Submitted on 1 Apr 2020 by Yin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
trades = [1 3 -4 2 -1 2 3] y_correct = 2; assert(isequal(trade_profit(trades),y_correct))
trades = 1 3 -4 2 -1 2 3
2 Pass
trades = [1 2 3 -5] y_correct = 1; assert(isequal(trade_profit(trades),y_correct))
trades = 1 2 3 -5
3 Pass
trades = [1 2 3 4 5 6] y_correct = 6; assert(isequal(trade_profit(trades),y_correct))
trades = 1 2 3 4 5 6
4 Pass
trades = [-2 3 -4 5 -6 1 2 3 4 5] y_correct = 3; assert(isequal(trade_profit(trades),y_correct))
trades = -2 3 -4 5 -6 1 2 3 4 5 | 299 | 766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-24 | latest | en | 0.592883 |
https://community.deeplearning.ai/t/measures-of-central-tendency-example-calculation-explanation/357138 | 1,713,798,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00032.warc.gz | 146,466,574 | 5,939 | # Measures of Central Tendency - example calculation explanation
Hey,
Can somebody explain the calculation behind the solution to these two examples, it makes sense visually but I’m getting confused during calculations.
I think for the 2nd example, the median might be wrong.
Given n =5 ( total 6 samples), therefore it would be the average of 2 and 3 = (2+3)/2 = 2.5
Median = 2.5, given 6 samples
I am not clear why median = 1.
Average is the weighted average (not just average) :
weighted average =
(0.1680 + 0.3601 + 0.380872 + 0.13233 + 0.028354 + 0,002435)/(0.168+0.360+0.38087+0.1323+0.02835)
weighted avg = 1.5
Mode is the biggest number: mode => 0.36, which is mode =1
1 Like
Hey thanks for the explanation, I understand the mean and mode part, but not sure how come median is 1.5, it’s supposed to be avg to middle two values in a sorted array right?
Can somebody explain the calculation of median in the 2nd example?
The median is wrong.
It should be median = 2.5. 6 Samples with x = (0,1,2,3,4,5) have to yield a median = 2.5.
There is no trick to it.
@Kevin_Shey The median in the picture is correct, for a binomial distribution the median is defined in terms of CDF, it’s the value of random variable where it’s CDF crosses 0.5.
source: My PhD mentor.
thanks for your help. | 378 | 1,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-18 | latest | en | 0.913936 |
http://math.stackexchange.com/questions/213515/a-wedge-sum-of-circles-without-the-gluing-point-is-not-path-connected | 1,469,529,814,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00187-ip-10-185-27-174.ec2.internal.warc.gz | 149,681,275 | 19,203 | A wedge sum of circles without the gluing point is not path connected
I am trying to prove that a wedge sum of two circles is not a topological manifold, to do so I am showing that the wedge sum without the gluing point is not path connected while the $\mathbb R^2$ without a point is path connected, this implies that the wedge sum is not a manifold since it is not homeomorphic to $\mathbb R^2$.
However, my question is how to prove formally that a wedge sum of circles without the gluing point is not path connected, I can see that intuitively, but it seems very difficult to prove formally. I've tried so hard, but I do not know even how to begin to solve the problem. I think if I can solve this question, I can use it as model to solve the similar ones, Anyone can help me please?
-
This doesn't show that the wedge sum is not a manifold. – Qiaochu Yuan Oct 14 '12 at 8:44
When you say circle, do you mean $S^1$ (which is a circle), or do you mean a closed ball in $\Bbb R^2$, which is actually a disk? – Brian M. Scott Oct 14 '12 at 8:44
@BrianM.Scott The $S^1$ – user42912 Oct 14 '12 at 9:03
The neighbourhood of $S^1\lor S^1$ around any point other than the glueing point looks like $\mathbb R^1$, so it is of no use to show that no neighbourhood of the glueing point loooks like $\mathbb R^2$. But you can still tke this path: Every neighbourhood of a point in $\mathbb R^1$ has a subneighbourhood such that it deleting one point produces exactly two connected components (not four). – Hagen von Eitzen Oct 14 '12 at 9:10
@HagenvonEitzen yes, I understood, thank you, your comment clarifies me a lot. However my problem remains unsolvable for me, how can I prove formally that the wedge sum of the two cicles has 4 components? – user42912 Oct 14 '12 at 9:18
The wedge sum of two copies of $S^1$ is not homeomorphic to $\Bbb R^2$ with or without the gluing point. With the gluing point it has a point, the gluing point, whose removal disconnects it, while $\Bbb R^2$ has no such point. Without it it’s disconnected, while $\Bbb R^2$ is connected. In any case, as Qiaochu said, this doesn’t show that the wedge sum of two circles is not a manifold.
HINT: Does the gluing point have a neighborhood homeomorphic to $R^n$ for any $n$? Note that if $p$ is the gluing point, $p$ has arbitrarily small neighborhoods that look like $+$; is this true of any point in any $\Bbb R^n$? Think about connectedness and numbers of connected components.
Added: We can realize $S^1\lor S^1$ as $$X=\Big\{\langle x,y\rangle\in\Bbb R^2:(x+1)^2+y^2=1\text{ or }(x-1)^2+y^2=1\Big\}\;,$$ with $p=\langle 0,0\rangle$ as the gluing point. (You shouldn’t have too much trouble writing down a homeomorphism between the formal description of $S^1\lor S^1$ and this $X$.) A typical small open neighborhood of $p$ in $X$ is $N_\epsilon=X\cap B(p,\epsilon)$, where $\epsilon>0$, and $B(p,\epsilon)=\{q\in\Bbb R^2:\|q\|<\epsilon\}$, the open $\epsilon$-ball centred at $p$. If $\epsilon\le2$, $N_\epsilon\setminus\{p\}$ has the following four components:
• $\{\langle x,y\rangle:(x-1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x,y>0\}$
• $\{\langle x,y\rangle:(x-1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x>0\text{ and }y<0\}$
• $\{\langle x,y\rangle:(x+1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x<0\text{ and }y>0\}$
• $\{\langle x,y\rangle:(x+1)^2+y^2=1\text{ and }x^2+y^2<\epsilon^2\text{ and }x,y<0\}$
Each is homeomorphic to $(0,1)$.
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@user42912: It could actually be $2,3$, or $4$, depending on what nbhd of $p$ you look at. $(S^1\lor S^1)\setminus\{p\}$ has two components; removing $p$ from a nbhd that includes all of one $S^1$ and an arc of the other leaves three components. Give me a few minutes, and I’ll add something about this to my answer. – Brian M. Scott Oct 14 '12 at 9:31
@user42912: You’re welcome; good luck! – Brian M. Scott Oct 14 '12 at 10:02
@user42912: Yes, that’s exactly what I intended. And those circles are tangent at the origin, so the origin acts as the gluing point. – Brian M. Scott Oct 14 '12 at 10:36
@user42912: It doesn’t fail: $t\mapsto\langle\cos 2\pi t,\sin 2\pi t\rangle$ maps $(0,1)$ homeomorphically onto all of the unit circle except the point $\langle 1,0\rangle$, so by choosing the right open subinterval of $(0,1)$, you can get any open arc of the unit circle. – Brian M. Scott Oct 15 '12 at 8:33
@user42912: That’s the hard way. Just show that its inverse is continuous; the inverse maps the point with polar coordinates $\langle 1,\theta\rangle$ to $\frac{\theta}{2\pi}\in(0,1)$, and that’s clearly a continuous map. (I’m about to go to bed, so it’ll be a few hours before I’m back.) – Brian M. Scott Oct 15 '12 at 12:47 | 1,500 | 4,688 | {"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} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.934735 |
https://www.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/1211811 | 1,579,454,207,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594662.6/warc/CC-MAIN-20200119151736-20200119175736-00384.warc.gz | 984,767,163 | 15,562 | Cody
# Problem 94. Target sorting
Solution 1211811
Submitted on 14 Jun 2017 by Georgios Stoumpis
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [1 2 3 4]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))
2 Pass
a = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))
3 Pass
a = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))
4 Pass
a = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct)) | 251 | 679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | latest | en | 0.637002 |
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A student rides a bicycle 2400 meters in four minutes to get to school. What is the student's speed?
a) 10 m/s
b) 10.1 m/s
c) 1.1 m/s
d) 0.1 m/s
What does a horizontal line on a velocity/time graph represent?
a) Zero Acceleration
b) Positive Acceleration
c) Negative Acceleration
d) Changing Acceleration
A rider finishes a 120 km bicycle trip in three hours. What is the average speed of the rider?
a) 3.6 km/hr
b) 4 km/hr
c) 40 km/hr
d) 360 km/hr
Which of the following best describes a situation in which the acceleration would be negative?
a) object speeds up
b) object slows down
c) object stops
d) object reverses
Where on Earth would a person weigh the most?
a) sea level
b) a mountain top
c) a valley below sea level
d) the top of a ten story building
What force acts on all objects on Earth?
a) friction
b) gravity
c) momentum
d) weight
A student drops a baseball and a feather from a third-floor window. Which force explains why the baseball hits the ground before the feather?
a) air resistance
b) gravity
c) mass
d) velocity
A student stops a bicycle by applying a brake. Which unbalance force causes the bicycle to stop?
a) energy
b) friction
c) gravity
d) inertia
Studen A and Student B are sitting on opposite sides of a see-saw. Student A's side is on the ground while student B's side is in the air. How does the mass of Student A compare to Student B?
a) Student A has more mass than Student B
b) Student A has less mass then Student B
c) Student A and B have the same mass
d) Student B has twice as much mass as Student A
What will happen to a moving object if an unbalanced force is applied to the object in the same direction the object is moving?
a) The object will speed up
b) The object will slow down
c) The object will change direction
d) The object will not be affected by the unbalanced force.
What happens to motion and direction when an unbalanced force acts on an object?
a) The motion stays the same and there is no direction
b) The motion stays the same and the direction stays the same
c) The motion changes and the direction is toward the larger force
d) The motion changes and the direction is opposite the larger force
Which of Newton's Laws states that an object will not move unless a force acts upon it?
a) Newton's 1st Law of Motion
b) Newton's 2nd Law of Motion
c) Newton's 3rd Law of Motion
d) Newton's 4th Law of Motion
Which of Newton's laws is also known as the Law of Inertia
a) Newton's 1st Law
b) Newton's 2nd Law
c) Newton's 3rd Law
d) Newton's 4th Law
Which property of matter causes an object to stay at rest unless acted upon a force?
a) friction
b) gravity
c) inertia
d) mass
A hockey puck is traveling on a frictionless surface. Which statement best describes the motion of the hockey puck?
a) the hockey puck will slow down
b) the hockey puck will travel in a curved pattern
c) the hockey puck will speed up
d) the hockey puck will continue to travel at constant speed until it hits something
Which of Newton's Laws of Motion best describes an object in constant motion?
a) 1st Law of Motion
b) 2nd Law of Motion
c) 3rd Law of Motion
d) 4th Law of Motion
An object is resting on a table. What would cause the object to move?
a) no force acts on it
b) a gravitational force acts on it
c) balanced forces act on it
d) unbalanced forces act on it
Which explains the motion of a rolling ball on a flat surface?
a) will stop quickly
b) will speed up quickly
c) will stay in motion until acted on by a force
d) will stay in motion for one hour
What is the tendency of a moving object to continue moving in a straight line?
a) inertia
b) gravity
c) mass
d) orbital speed
Which describes a car going 50 miles per hour east?
a) Acceleration
b) Momentum
c) Speed
d) Velocity
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- - time step and vortex shedding frequency (http://www.cfd-online.com/Forums/main/11601-time-step-vortex-shedding-frequency.html)
kevin June 8, 2006 10:00
time step and vortex shedding frequency
hi all, i found the vortex frequency are influenced by the time step i chose, what is your oppion about the time step if when want to get right vortex shedding frequency?
kevin
ganesh June 9, 2006 01:26
Re: time step and vortex shedding frequency
Dear Kevin,
Time accuracy is an important issue in unsteady flows. It is therefore imperative that you choose the right time step for a desired accuracy. You could compare your experience with the dependency of solution on grid and attempts to achieve a grid-independent solution. Therefore, you must use a time-step that gives you the acceptable temporal accuracy. In general, a physical time step of 1e-3 to 1e-4 should be good enough. For ant delt less than this, changes in shedding frequency should be minimal.
Hope this helps
Regards,
Ganesh
kevin June 9, 2006 03:47
Re: time step and vortex shedding frequency
thanks for your detailed reply,ganesh. i agree with you, and there is another problem, sometimes i found at T/4 and 3T/4, Cl achieve its max and min, and it is almost 0 at 0T/4 and 4T/4, but when i changed the time delt to check again, i got different results.
i'm wondering maybe the low pressure induced by the vortex account for the relationship between vortex shedding frequency and Cl(what i said above)
best regards.
kevin
Anton Lyaskin June 9, 2006 07:03
Re: time step and vortex shedding frequency
I guess you're modelling something like flow past the cylinder? Lift oscillations have frequency twice than vortex shedding. To get trustworthy results your time step should be at least 20 times smaller than lift oscillation period (so talking about absolute values of time step is bit meaningless for this case). And may be you'd better use some high order differencing in time (Cranck-Nikolson or Runge-Kutta), but in this case you'd probably need even smaller time step to satisfy stability criterion (CFL number)
Mani June 9, 2006 15:22
Re: time step and vortex shedding frequency
I get weary repeating myself, but I see this too often to hush up. You guys may have all the experience and expertise necessary to give good advice, but sometimes you're completely missing out on common sense. What does it mean, when I talk about a physical time step of 1e-3 to 1e-4? My physics teacher in high school would ask: What is that? Is it 1e-3 seconds, 1e-3 hours, 1e-3 potatoes, or some non-dimensionalized time?
Numbers are meaningless without unit, unless you can assume that everyone knows what you mean, and that's rarely the case. The only thing that makes sense in this context is a time step non-dimensionalized by an appropriate time scale, i.e. the (approximate) vortex shedding period. However, a non-dimensional time of 1e-3 times the vortex shedding period would mean to resolve each period by 1000 time steps. That may be necessary for an explicit time marching method but is very excessive if you're using an implicit method (which you really should).
With implicit time stepping the vortex shedding frequency can be (and has been) quite accurately obtained with 30-50 time steps per vortex shedding period on a reasonably fine mesh. For laminar 2-D vortex shedding over a cylinder (47<Reynolds number<180), a second order finite volume method will need about 300-400 grid cells around the cylinder circumference, according to a study by J. Alonso, and my own experience.
This case is quite simple because you know the result (shedding frequency, lift and drag amplitudes) from ample experimental data that's available to you. If you're not getting the right result, you may not have the necessary temporal or spatial resolution.
diaw June 9, 2006 22:38
Re: time step and vortex shedding frequency
A few physics questions:
What do the *physics* & the *form* of the N-S equations tell us about the selection of a 'suitable' time step?
Should this be a fixed, or variable time-step? Why?
Is the time-step linked to spatial coordinates? If so, why?
Do we *have to* scale everything, or could we elect to work in real physical coordinates - space & time?
What is the significance of the d()/dt term in the N-S? When does it 'become active'? Are there flow situations where it may be zero/non-zero?
Does the concept of numeric stability, numeric 'error' need to take precedence in all our computations - or, does physics dictate the end-game?
ganesh June 10, 2006 05:09
Re: time step and vortex shedding frequency
Dear Mani,
Your statements are true and I apologise for having left off the units. However, in case I was talking about a physical time step which was dimensional I would have actually put down the units, and therfore the time step had to be a non-dimensional one. The non-dimensionalisation is what I missed and this in my case is t* = t/(L/U_inf), where L and U_inf are the characteristic length adn reference velocity respectievely. With an implicit time stepping procedure(I use Crank-Nicholson), the physical time step roughly translates to 75-100 time steps per vortex shedding period.
Regards,
Ganesh
Mani June 12, 2006 08:38
Re: time step and vortex shedding frequency
Are 75+ steps really necessary, according to your experience? For which time-stepping method are you quoting these numbers (I'm assuming implicit)?
To add one question to diaw's list: In choosing an appropriate time step, should we not account for the (smallest) time scales that our spatial resolution is able to handle? Does it make sense to let delta_t go to zero and then speak of convergence, when the grid is fixed? Will there be stability issues, when the grid is too coarse to resolve the unsteady solution at a very small time step?
Mani June 12, 2006 09:04
Re: time step and vortex shedding frequency
fixed or variable time step:
I suppose this topic is equivalent to the question: fixed or adaptive grid. yes, you could go through the trouble to adapt your spatial and temporal resolution to the 'current' requirements throughout your computation. this may give you a performance benefit, as long as your adaptation is both efficient and effective.
linkage between spatial and temporal resolution:
under certain circumstances you have to take account of that. I posed a similar question in response to ganesh. I recall discussing a transonic flutter case, where a shock wave oscillates on an airfoil. if you keep decreasing the time step you will eventually get to a point where the solution from one time level to the next does not change, because the shock displacement is within the cell size. it makes no sense to further reduce the time step.
scaling or no scaling:
if you're smart you'll scale for two reasons: a) non-dimensional computation and data interpretation make your life easy for well known reasons, b) multiple scales lead to stiff systems of equations, and scaling may help to reduce that stiffness.
in response to the more fundamental questions on numerics versus physics:
you'll have to live with the physics you select to model. the challenge is to resolve the physical phenomena you are interested in, without exceeding the necessary resolution (time and cost). hence, your choice of equations, solution method, spatial and temporal resolution, etc., all depend on what you want to get out of it. as an example: if you want to test your unsteady method by evaluating the vortex shedding frequency for a cylinder, your time step should be chosen to resolve that frequency. will that give you all the physics, i.e. resolve all the vortices, get the correct wake dissipation rate, give you an accurate flow field at all points in space-time? not necessarily, but that's not what you were looking for.
ganesh June 12, 2006 09:38
Re: time step and vortex shedding frequency
Dear Mani,
I make use of a Dual time stepping method, which solves the pseudo-steady equations in pseudo-time, which makes things more easier (I can make use of convergence accelaeration devices here, icluding implicit stepping, residual smoothing etc.. , although I do not make use of all available ones). The physical time step which is important, is generally held constant. This time step is the one that decides the temporal accuracy. When I say 75+ steps, I mean 75 outer iterations or in other words 75 physical time steps (The DTS performs what I call "inner iterations" where the implicit time stepping can be employed). This choice of the physical time step has a bearing on both the temporal accuracy and stability and must therefore be chosen with care. This choice depends on the problem and the mesh resolution available. What possibly makes things easy for the standard problems like vortex shedding past cylinder is the knowledge if the shedding frequency, this should not be expected for all real-life problems.
As far as the spatial and temporal resolution are related, my thoughts are as follows. If we consider the very common linear convective equation (which is an unsteady case), and use a simple explicit scheme, we end up with the CFL criterion for stability. With the convective velocity and grid size known, the delt can be obtained. Any delt smaller than this quantity guarantees stability. So, there is nothing wrong when I speak of "temporal convergence" (the choice of time step that guarantees me a given accuracy at the least computational cost). On a given grid, for the transonic flutter problem, decreasing the time step does not any way guarantee "spatial accuracy"(the shock will be well within the grid size if the spactial resolution is good enough or you make use of an adaptive strategy in time), as you seem to point out, all that happens is that after sufficient number of iterations, the solution achieves a certain periodicity. The periodicity is what gets affected if there is a loss in temporal resolution. A grid too coarse to see no features is not useful, and a grid coarse enough to detect some significant activity with a small time step should not pose any stability problem, to my knowledge. In fact, on coarser grids, you can have a larger physical time step compared to the finer grid, much as you would expect for the linear convective problem, delt ~ delx. The time step for the inner iterations for the DTS is same as that while handling any steady-state problem.
Regards,
Ganesh
diaw June 12, 2006 10:13
Re: time step and vortex shedding frequency
Thanks Mani for the excellent & very complete reply.
diaw...
diaw June 12, 2006 10:22
Re: time step and vortex shedding frequency
Thanks Mani - excellent thought.
Mani wrote:
To add one question to diaw's list: In choosing an appropriate time step, should we not account for the (smallest) time scales that our spatial resolution is able to handle? Does it make sense to let delta_t go to zero and then speak of convergence, when the grid is fixed? Will there be stability issues, when the grid is too coarse to resolve the unsteady solution at a very small time step?
If I could add a few pre-questions to Mani's question.
Is there an appropriate time-step for a certain mesh/grid configuration? What happens if our selected time step is too large, or too small for this mesh?
Then afterwards: Is the grid spatial mesh the limiting factor in the simulation, or time-step, or both acting simultaneously?
If the mesh is refined during the simulation, does the time step have to be altered?
---------
I'm going to add in a small twist - "It depends"...
ganesh June 12, 2006 12:38
Re: time step and vortex shedding frequency
Dear Diaw,
The "appropriate" time step for a given grid and proble, should be the one that is stable, yet gives results to desired accuracy. The choice, in general could therefore be a difficult one. A too large time step would result in a lower temporal accuracy, reflected in the results or worse could cause stability issues. A very small time step is undesirable, purely because you spend more time than required in actually getting the solution. The choice for the "optimum" or "appropriate" time step is a tricky issue, though in some cases available information( such as frequency of vortex shedding etc..) could help us make a good decision.
In a unsteady flow simulation, both the spatial and temporal accuracies are important, spatial acuracy would resolve the flow features better and since the problem is time--dependent, the temporal accuracy would have its effect on the simulation.
For a dynamic adaptation, the spatial resolution gets changed during adaptation, I believe that the temporal resolution should also be altered accordingly, though I have not seen this actually being done. This is because the chosen physial time step for the initial mesh could prove unstable for the refined mesh after a few levels of refinement and the decision of optimally altring the time step I believe could be important, which takes us back to square one, the choice of "appropriate time step". For a general problem, the initial choice possibly comes out of prior experience and a general idea of the problem in hand.
Regards,
Ganesh
Praveen. C June 12, 2006 23:16
Re: time step and vortex shedding frequency
1. Convergence results are of the form:
<blockquote> ... the scheme converges to the exact solution as dt->0, dx->0 while the CFL number is held fixed ... </blockquote>
For an unsteady problem, the error depends both on dt and dx, and it does not make sense to talk of "spatial convergence" and "temporal convergence" separately.
2. You can talk of temporal resolution alone provided you have taken a sufficiently fine spatial grid. The CFL condition puts an upper bound on the allowable time-step Î"t<sub>s</sub>. Any time-step smaller than this is fine but as Ganesh wrote, taking a very small time-step is wasteful since then the spatial errors will dominate and you do not gain anything. The problem also has a physical time-scale Î"t<sub>p</sub>. Your time-step must be smaller than both
Î"t ≤ min(Î"t<sub>s</sub>, Î"t<sub>p</sub>)
3. Of course, Î"t<sub>p</sub> will vary from problem to problem. Just as we do grid convergence studies, we should also do Î"t convergence studies. What is an acceptable time step will depend on what quantities you are interested in (integral quantities like lift/drag/heat transfer or detailed spatio-temporal dynamics and structures).
4. It is better to use a higher order time-integration scheme than to reduce the time-step to a very small value.
diaw June 13, 2006 06:42
Re: time step and vortex shedding frequency
Excellent overviews, Mani, Ganesh & Praveen... good points.
I've tended to work in a sequence as follows:
1. Determine reasonable spatial scale; 2. Select Courant no. of choice; 3. Determine dt upper bound based on courant, dx,dy,dz;
If the dx etc change during run, then determine new appropriate dt.
----------
I agree entirely that all dx, dy, dz, dt must travel to zero - in the limit. I am convinced that along that path, they should also do so in *direct proportion* to each other, if true dynamic relationships are to be retained. If not, then term biase comes into play & we no longer solve the original pde, but a weighted version. In other words, a global scaling rule.
---------
Praveen, you refer to a dt,p. How would you pre-determine this for the problem at hand?
--------
With all the above in mind, & with all dx,dy,dz,dt moving in lockstep... *How* do we determine an appropriate dx, dy, dz for the model at hand?
I'd value debate around this last point particularly, as it is currently something I have been grappling with.
diaw...
Mani June 13, 2006 12:29
Re: time step and vortex shedding frequency
Good discussion.
Just one more comment:
Praveen (and everyone else, including me) said something like:
> What is an acceptable time step will depend on what quantities you are interested in (integral quantities like lift/drag/heat transfer or detailed spatio-temporal dynamics and structures)
:
... and that could be a textbook comment. However, to be fair, it's not as easy as we make it seem at first sight. It's hard to tell how much detail you need to predict in a flow in order to even get the integral quantities correctly. In the cylinder example, the drag is linked to the rate at which energy is pumped into the flow and then dissipated. How much detail (in both space and time) is required to get the vortex street right, and predict the right drag? In practice, you will find that not just the grid locally around the cylinder, but also the resolution in the whole wake region, has an impact on the integral quantities obtained on the cylinder surface.
Likewise, the drag will depend on the time step that you choose to resolve bigger or smaller vortex interaction events and dissipation at considerable distance downstream. In the turbulent case, all this becomes more complicated, as you need to model the turbulent dissipation rate accurately (not just in the boundary layers, but in the wake). In the extreme case, i.e. if you were to solve the steady-state equations for this problem, you would end up with a completely wrong result for drag (steady-state is not equal to time-average!), because you neglect all unsteady transport of energy through vortex shedding.
That's why, when I say 30-50 time steps per period are enough, it's not meant as universally applicable rule of thumb for unsteady problems, just specific experience on the laminar circular cylinder with a second-order implicit method.
alessios123 November 19, 2009 07:00
Hi, i would like that some holy man could solve this simple problem: i' m simulating the motion of a fluid post a cilinder of 13 mm diameter. my problem is that i can't create vortex shedding under reynold number of 60000, but we know the the vortex shedding phenomenon begins and it shows better at 80 untill 160 Re...
I put also an initial disturb like a initial velocity in direction perpendicular to the fluid motion, but nothing i can't create the vortex shedding..
i thank everybody who help me..
Alessio
All times are GMT -4. The time now is 19:52. | 4,081 | 18,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.898533 |
https://lifwynnfoundation.org/how-to-read-ounces-on-a-digital-scale/ | 1,660,384,041,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00413.warc.gz | 349,289,994 | 5,315 | Turn her digital range on and wait till the analysis is collection to 0.0.Place every little thing needs come be weighed on the scale, and keep yourself or your materials immobile till the weight reading is complete.Understand that many manufacturers measure ounces in 0.2-lb.
You are watching: How to read ounces on a digital scale
## How perform you usage a digital kitchen scale?
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## How execute you review a weighing scale?
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## How carry out you use a gram scale?
The many accurate method to measure grams is to usage a scale. Choose a digital or mechanical scale that uses the metric system, then push the tare button to zero that out. Ar your article on the facility of the scale, wait because that the scale’s digital display or needle to pertained to a stop, and record the grams.
## How do you calibrate a digital scale?
Steps
Place the scale on a sturdy, level surface. Ar one or two computer system mouse pads ~ above the table’s surface. Place your range on the mousepad and power top top the unit. Press the “Zero” or “Tare” button on your scale. Verify that your scale is set to “calibration” mode.
## How much is 4 ounces top top a scale?
A quarter weighs 113.4 grams, or 4 ounces. 112 grams is more common. How much does an ounce weigh on a scale?
## Can I use a kitchen scale to weigh packages?
Yes, girlfriend do. First off, a an excellent postage scale does not need to be expensive. And, girlfriend might currently have one – a great kitchen scale will be accurate sufficient to sweet packages, but typically don’t have very high weight limits. A great postage scale should be specific to a tenth the an oz at every ranges.
## What is Tare on kitchen scales?
Tare weight is accounted for in kitchen scales, analytical (scientific) and other weighing scales which encompass a button that resets the screen of the scale to zero as soon as an empty container is inserted on the weighing platform, in order consequently to display screen only the weight of the components of the container.
## What is the difference between tare and also zero ~ above a scale?
The ZERO function is used only as soon as the scale is empty and is no at gross ZERO due to material develop up. This is to be supplied to offset tiny amounts the product buildup on the scale. The Tare function is provided when you just wish to see a current change in weight, not the whole amount that weight that is ~ above the scale.
## How do you review a weighing scale in grams?
How to check out a range by Weighing Grams
Place an item or item on the platform of a digital scale.Observe the display screen screen ~ above a digital scale. Read the digital weight display in entirety grams to tenths of grams. Place an object on a mechanically scale’s platform.Read a mechanical scale by observing the guideline on the dial that reflects the weight of one item.
See more: How To Replace Wick On Zippo, Reflinting And Rewicking A Zippo Lighter
## How carry out you usage a balance load scale?
To weigh yourself using a digital or dial scale, ar the range on a level surface and also step top top it. Then, just read the number to uncover out exactly how much girlfriend weigh. Alternatively, if using a balance beam scale, action on the scale, readjust the weights, and include up the numbers.
## How perform you use a Detecto scale?
1:20
2:38
Suggested clip 63 seconds
Detecto physician Scales at QuickMedical – YouTube | 824 | 3,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.885449 |
http://www.math.wpi.edu/Course_Materials/MA1021A97/lab4/node3.html | 1,542,393,362,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743110.55/warc/CC-MAIN-20181116173611-20181116195611-00380.warc.gz | 477,105,552 | 2,164 | Next: Exercises Up: The Newton-Raphson Method Previous: Background
## Maple Usage
The CalcP package (which must be called using the with(CalcP); command) contains an implementation of the Newton-Raphson method called Newton. This Newton command takes three arguments. The first is the function that is set equal to zero, next is the initial guess x0, and third is the number of repetitions of Newton-Raphson.
For instance, to see the first 15 approximations to a root of x5-6x4+x3+2x2-5x+7=0 when x0 is taken as -9, the syntax would be as follows.
``` > with(CalcP);
```
``` > f := x -> x^5-6*x^4+x^3+2*x^2-5*x+7;
```
``` > Newton(f,x=-9,15);
```
This will display x0 and f(x0) and xn and f(xn) for n equal 1 to 15. By looking at the f(xn), one can see if f(xn) is tending to 0 as n increases.
If an expression is used as the first argument of Newton as in
``` > Newton(x^7-5*x^3-3*x+7,x=2,10);
```
then each line of output will be of the form
xi = . . . . . . , Expr(xi) = . . . . . . .
.
In the CalcP package there is a routine called NewtonPlot which can help you visualize the action of the Newton-Raphson algorithm. You may find it useful in dealing with some of the exercises below. Basically, it plots the sequence of approximations to a root generated by Newton-Raphson. The syntax is as follows where a is the initial guess for a root.
``` > NewtonPlot(f,x=a);
```
Next: Exercises Up: The Newton-Raphson Method Previous: Background
William W. Farr
9/30/1997 | 430 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.883958 |
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AmosWEB means Economics with a Touch of Whimsy!
FACTOR PRICE: The price paid for and received by the services of factor of productions (labor, capital, land, and entrepreneurship) when exchange through factor markets. Like prices in other markets, factor price adjusts to balance the forces of demand and supply. For factor demand and the factor demand curve, the factor price is negatively related to the quantity of factor services demanded. For factor supply and the factor supply curve, factor price is positively related to the quantity of factor services supplied. The key factor prices are wage rates, interest rates, rents, and profits. The rigidity or inflexibility of factor prices is an important aspect of the macroeconomic study of the short-run aggregate market.
GRAPHICAL ANALYSIS:
The process of investigating phenomena, especially economic phenomena, in a systematic manner using diagrams and graphs. Graphical analysis is commonly used to display abstract scientific relations, then to manipulate those relations to gain greater understanding of real world events. The market model is a primary example of graphical analysis.
Graphs are two-dimensional pictures used to represent economic relations between two (or more) variables. Graphical analysis is most interesting and useful when it combines two or more relations into a single diagram. The interaction among these relations is then analyzed for insight into the workings of the economic world.
### A Primer On Graphs
Consider three common types of graphs:
• Pie Chart: A graph commonly used to present the division of a total among parts is a pie chart. Click the [Pie Chart] button to illustrate. This particular pie chart represents the division of national income among different factor payments--wages, interest, rent, and profit. The pie is the total and each slice represents the portion distributed to each category. Pie charts are a handy way to present information, but are not well suited for more involved economic analysis.
• Bar Chart: A graph used to present data for discrete categories is a bar chart. Click the [Bar Chart] button to illustrate. This bar chart indicates the unemployment rate for each of five demographic groups--total population, males, females, whites, and non-whites. A bar chart provides a useful way to compare information about different groups or categories.
• Line Graph: A graph that tends to be most useful in the construction of graphical models and in doing economic analysis is a line graph. Click the [Line Graph] button to illustrate. This particular line graph shows the relation between two variables--price and quantity. Such line graphs are ideally suited for illustrating scientific principles and hypotheses. They can be used to show how one variable (quantity) is affected by changes in another variable (price).
### Two Relations
Two alternative relations are commonly illustrated with line graphs.
• Positive or Direct: One type of line graph illustrated by clicking the [Positive Relation] button represents a positive or direct relation between two variables, such as price and quantity. With this line, a higher price is related to a larger quantity. Another way of stating this is that the slope of the line is positive. A common positive relation in economics is the market supply curve.
• Negative or Indirect: A second type of line, one that represents a negative or indirect relation between two variables, such as price and quantity, can be seen by clicking the [Negative Relation] button. With this line, a lower price is related to a larger quantity. Another way of stating this is that the slope of the line is negative. A common negative relation in economics is the market demand curve.
<= GOVERNMENT SUBSIDIES LESS CURRENT SURPLUS OF GOVERNMENT ENTERPRISES GROSS DOMESTIC INCOME =>
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Maths Resources & WorksheetsYear 2 Maths Resources & WorksheetsSummer Block 4 (Mass, Capacity and Temperature)02 Measure Mass in Grams › Measure Mass in Grams Homework Extension Year 2 Mass, Capacity and Temperature
# Measure Mass in Grams Homework Extension Year 2 Mass, Capacity and Temperature
## Step 2: Measure Mass in Grams Homework Extension Year 2 Summer Block 4
Measure Mass in Grams Homework Extension provides additional questions which can be used as homework or an in-class extension for the Year 2 Measure Mass in Grams Resource Pack. These are differentiated for Developing, Expected and Greater Depth.
More resources for Summer Block 4 Step 2.
Free Home Learning Resource
It's just £4.83 for a month's subscription.
### What's included in the pack?
This pack includes:
• Measure Mass in Grams Homework Extension with answers for Year 2 Summer Block 4.
#### National Curriculum Objectives
Mathematics Year 2: (2M1) Compare and order lengths, mass, volume/capacity and record the results using >, < and =
Differentiation:
Questions 1, 4 and 7 (Varied Fluency)
Developing Match objects to the appropriate reading on a weighing scale. Scales in increments of 10 only.
Expected Match objects to the appropriate reading on a weighing scale. Scales in increments of 2, 5 and 10.
Greater Depth Match objects to the appropriate reading on a weighing scale. Scales in increments of 2, 5 and 10. Some measurements fall between increments on the scale.
Questions 2, 5 and 8 (Varied Fluency)
Developing Place the pointer in the correct position on a weighing scale. Scales in increments of 10 only.
Expected Place the pointer in the correct position on a weighing scale. Scales in increments of 2, 5 and 10.
Greater Depth Place the pointer in the correct position on a weighing scale. Scales in increments of 2, 5 and 10. Some measurements fall between increments on the scale.
Questions 3, 6 and 9 (Reasoning and Problem Solving)
Developing Find three ways to make a weight which has been doubled using 5g and 10g weights. Scales in increments of 10.
Expected Find three ways to make a weight which has been doubled using 2g, 5g and 10g weights. Scales in increments of 5.
Greater Depth Find three ways to make a weight which has been doubled using 2g, 5g and 10g weights when 10g has already been placed on the scale. Scales in increments of 10, measurement falls between increments on the scale. | 590 | 2,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.854536 |
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##### Home / [April-2016-Updated]200-355 Dump New Updated Version Free Download – Braindump2go
2016 April New — Cisco CCNA Wireless 200-355 WIFUND Exam Questions and Answers Updated Today!
Exam Introduction:
The Implementing Cisco Wireless Network Fundamentals (WIFUND) exam (200-355) is a 90-minute, 60–70 item assessment that is associated with the CCNA Wireless certification. This exam tests a candidate’s knowledge of Radio Frequency (RF) and 802.11 technology essentials along with installing, configuring, monitoring, and basic troubleshooting tasks needed to support Small Medium Business and Enterprise wireless networks.
Exam General Guidelines:
1. RF Fundamentals;
2. 802.11 Technology Fundamentals;
3. Implementing a Wireless Network;
4. Operating a Wireless Network;
5. Configuration of Client Connectivity;
6. Performing Client Connectivity Troubleshooting;
7. Site Survey Process;
QUESTION 377
What is the effect of increasing antenna gain on a radio?
A. focusing energy in a defined direction
B. adding energy creating a larger cell
C. aligning phase shifting
D. improving frequency specific diversity
QUESTION 378
Which signal strength reading indicates that the engineer is closest to the access point?
A. -43 dBm
B. -67 dBm
C. -87 dBm
D. -100 dBm
QUESTION 379
The new tablets for the sales department require an SNR of 18 or more to operate. Given a noise floor of – 88 dBm, what is the minimum RSSI that is needed?
A. -60 dBm
B. -70 dBm
C. -72 dBm
D. -80 dBm
E. -96 dBm
F. -106 dBm
QUESTION 380
Refer to the exhibit. A network engineer needs the far end of a wireless bridge to receive at -45dBM.
Based on the diagram, what value in dBm must the transmitter use to send to achieve the desired result?
A. -118
B. -28
C. -20
D. 20
E. 28
F. 118
QUESTION 381
Refer to the exhibit. A wireless engineer has an antenna with the radiation pattern shown in the exhibit. What type of antenna is it?
A. Patch
B. Yagi
C. Parabolic dish
D. Dipole
E. Internal omni
QUESTION 382
When reading a radiation pattern for an antenna, at which two drops in signal and power is the beamwidth measured? (Choose two.)
A. 3 dB
B. 6 dB
C. 9 dB
D. 10 dB
E. 1/2 power
F. 1/4 power
G. 1/8 power
H. 1/10 power
QUESTION 383
When calculating the link budget for a wireless point-to-point bridge, the engineer notices that one antenna has its gain marked as 2.85 dBd. With a 20-mW access point and 3-dBi loss for the cable, what is the approximate EIRP?
A. 15 dBm
B. 18 dBm
C. 22 dBm
D. 25 dBm
QUESTION 384
Due to the terrain, a deployment requires a point-to-point wireless bridge to allow for network connectivity to a remote building. What 5GHz band would permitted to use the highest power in the U.S.?
A. U-NII-3
B. U-NII-2 Extended
C. U-NII-2
D. U-NII-1
QUESTION 385
A wireless engineer is designing a network for the London branch of a company. Which 5-GHz band allows the branch to use the highest EIRP?
A. 2.4-GHz ISM
B. UNII-1
C. UNII-2
D. UNII-2 Extended | 930 | 3,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.720395 |
https://www.geeksforgeeks.org/algorithms-searching-question-6/ | 1,702,088,319,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00343.warc.gz | 837,545,380 | 47,099 | # Algorithms | Searching | Question 6
Which of the following is the correct recurrence for the worst case of Ternary Search?
(A)
T(n) = T(n/3) + 4, T(1) = 1
(B)
T(n) = T(n/2) + 2, T(1) = 1
(C)
T(n) = T(n + 2) + 2, T(1) = 1
(D)
T(n) = T(n – 2) + 2, T(1) = 1
Explanation:
Following is a typical implementation of Binary Search.
```// A recursive ternary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
int ternarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid1 = l + (r - l)/3;
int mid2 = mid1 + (r - l)/3;
// If x is present at the mid1
if (arr[mid1] == x) return mid1;
// If x is present at the mid2
if (arr[mid2] == x) return mid2;
// If x is present in left one-third
if (arr[mid1] > x) return ternarySearch(arr, l, mid1-1, x);
// If x is present in right one-third
if (arr[mid2] < x) return ternarySearch(arr, mid2+1, r, x);
// If x is present in middle one-third
return ternarySearch(arr, mid1+1, mid2-1, x);
}
// We reach here when element is not present in array
return -1;
}```
In ternary search, we divide the given array into three parts and determine which has the key (searched element). We can divide the array into three parts by taking mid1 and mid2 which can be calculated as shown below. Initially, l and r will be equal to 0 and n-1 respectively, where n is the length of the array. So the recurrence relation is T(n) = T(n/3) + 4, T(1) = 1
Hence Option (A) is the correct answer.
Quiz of this Question
Please comment below if you find anything wrong in the above post
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https://github.com/genomicsclass/labs/blob/master/linear/collinearity.Rmd | 1,537,897,130,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161902.89/warc/CC-MAIN-20180925163044-20180925183444-00341.warc.gz | 513,156,428 | 15,262 | # genomicsclass/labs
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title author date output layout
Collinearity
Rafa
January 31, 2015
html_document
page
library(knitr)
opts_chunk$set(fig.path=paste0("figure/", sub("(.*).Rmd","\\1",basename(knitr:::knit_concord$get('infile'))), "-"))
## Collinearity
If an experiment is designed incorrectly we may not be able to estimate the parameters of interest. Similarly, when analyzing data we may incorrectly decide to use a model that can't be fit. If we are using linear models then we can detect these problems mathematically by looking for collinearity in the design matrix.
#### System of equations example
The following system of equations:
\begin{align*} a+c &=1\ b-c &=1\ a+b &=2 \end{align*}
has more than one solution since there are an infinite number of triplets that satisfy $a=1-c, b=1+c$. Two examples are $a=1,b=1,c=0$ and $a=0,b=2,c=1$.
#### Matrix algebra approach
The system of equations above can be written like this:
# $$, \begin{pmatrix} 1&0&1\ 0&1&-1\ 1&1&0\ \end{pmatrix} \begin{pmatrix} a\ b\ c \end{pmatrix} \begin{pmatrix} 1\ 1\ 2 \end{pmatrix}$$
Note that the third column is a linear combination of the first two:
# $$, \begin{pmatrix} 1\ 0\ 1 \end{pmatrix} + -1 \begin{pmatrix} 0\ 1\ 1 \end{pmatrix} \begin{pmatrix} 1\ -1\ 0 \end{pmatrix}$$
We say that the third column is collinear with the first 2. This implies that the system of equations can be written like this:
# $$, \begin{pmatrix} 1&0&1\ 0&1&-1\ 1&1&0 \end{pmatrix} \begin{pmatrix} a\ b\ c \end{pmatrix} a \begin{pmatrix} 1\ 0\ 1 \end{pmatrix} + b \begin{pmatrix} 0\ 1\ 1 \end{pmatrix} + c \begin{pmatrix} 1-0\ 0-1\ 1-1 \end{pmatrix}$$
$$=(a+c) \begin{pmatrix} 1\ 0\ 1\ \end{pmatrix} + (b-c) \begin{pmatrix} 0\ 1\ 1\ \end{pmatrix}$$
The third column does not add a constraint and what we really have are three equations and two unknowns: $a+c$ and $b-c$. Once we have values for those two quantities, there are an infinity number of triplets that can be used.
#### Collinearity and least squares
Consider a design matrix $\mathbf{X}$ with two collinear columns. Here we create an extreme example in which one column is the opposite of another:
$$\mathbf{X} = \begin{pmatrix} \mathbf{1}&\mathbf{X}_1&\mathbf{X}_2&\mathbf{X}_3\ \end{pmatrix} \mbox{ with, say, } \mathbf{X}_3 = - \mathbf{X}_2$$
This means that we can rewrite the residuals like this:
$$\mathbf{Y}- \left{ \mathbf{1}\beta_0 + \mathbf{X}_1\beta_1 + \mathbf{X}_2\beta_2 + \mathbf{X}_3\beta_3\right}\ = \mathbf{Y}- \left{ \mathbf{1}\beta_0 + \mathbf{X}_1\beta_1 + \mathbf{X}_2\beta_2 - \mathbf{X}_2\beta_3\right}\ = \mathbf{Y}- \left{\mathbf{1}\beta_0 + \mathbf{X}_1 \beta_1 + \mathbf{X}_2(\beta_2 - \beta_3)\right}$$
and if $\hat{\beta}_1$, $\hat{\beta}_2$, $\hat{\beta}_3$ is a least squares solution, then, for example, $\hat{\beta}_1$, $\hat{\beta}_2+1$, $\hat{\beta}_3+1$ is also a solution.
#### Confounding as an example
Now we will demonstrate how collinearity helps us determine problems with our design using one of the most common errors made in current experimental design: confounding. To illustrate, let's use an imagined experiment in which we are interested in the effect of four treatments A, B, C and D. We assign two mice to each treatment. After starting the experiment by giving A and B to female mice, we realize there might be a sex effect. We decide to give C and D to males with hopes of estimating this effect. But can we estimate the sex effect? The described design implies the following design matrix:
$$, \begin{pmatrix} Sex & A & B & C & D\ 0 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 1 & 0 \ 1 & 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 & 1\ \end{pmatrix}$$
Here we can see that sex and treatment are confounded. Specifically, the sex column can be written as a linear combination of the C and D matrices.
# $$, \begin{pmatrix} Sex \ 0\ 0 \ 0 \ 0 \ 1\ 1\ 1 \ 1 \ \end{pmatrix} \begin{pmatrix} C \ 0\ 0\ 0\ 0\ 1\ 1\ 0\ 0\ \end{pmatrix} + \begin{pmatrix} D \ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 1\ \end{pmatrix}$$
This implies that a unique least squares estimate is not achievable.
## Rank
The rank of a matrix columns is the number of columns that are independent of all the others. If the rank is smaller than the number of columns, then the LSE are not unique. In R, we can obtain the rank of matrix with the function qr, which we will describe in more detail in a following section.
Sex <- c(0,0,0,0,1,1,1,1)
A <- c(1,1,0,0,0,0,0,0)
B <- c(0,0,1,1,0,0,0,0)
C <- c(0,0,0,0,1,1,0,0)
D <- c(0,0,0,0,0,0,1,1)
X <- model.matrix(~Sex+A+B+C+D-1)
cat("ncol=",ncol(X),"rank=", qr(X)$rank,"\n") Here we will not be able to estimate the effect of sex. ## Removing Confounding This particular experiment could have been designed better. Using the same number of male and female mice, we can easily design an experiment that allows us to compute the sex effect as well as all the treatment effects. Specifically, when we balance sex and treatments, the confounding is removed as demonstrated by the fact that the rank is now the same as the number of columns: Sex <- c(0,1,0,1,0,1,0,1) A <- c(1,1,0,0,0,0,0,0) B <- c(0,0,1,1,0,0,0,0) C <- c(0,0,0,0,1,1,0,0) D <- c(0,0,0,0,0,0,1,1) X <- model.matrix(~Sex+A+B+C+D-1) cat("ncol=",ncol(X),"rank=", qr(X)$rank,"\n") | 1,928 | 5,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-39 | latest | en | 0.617944 |
http://www.docstoc.com/docs/106920342/ideal_gas_equation | 1,427,415,783,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131293283.10/warc/CC-MAIN-20150323172133-00027-ip-10-168-14-71.ec2.internal.warc.gz | 457,710,220 | 31,783 | ; ideal_gas_equation
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# ideal_gas_equation
VIEWS: 13 PAGES: 2
• pg 1
``` Ideal gas equation
Combining the equations PV = constant, P/T = constant and V/T = constant gives:
PV /T = constant
If we use 1 mole of gas the constant is known as the molar gas constant (R).
So for one mole of the ideal gas equation is:
Ideal Gas equation (one mole): PV = RT
Example problem
3 o o
0.3 m of an ideal gas are heated at constant pressure from 27 C to 127 C. What is the new volume
of the gas?
3
V2 = [V1T2]/T1 = [0.3 x 400]/300 = 0.4 m
(Notice that the temperatures used are always in Kelvin!)
Now the volume of one mole of an ideal gas at Standard Temperature and Pressure (STP)
5 3
(1.014x10 Pa and 273.15 K) is 0.0224m and so
1.014x105 x 0.0224 = 1 x R x 273.15 and therefore R = 8.314 JK-1mol-1.
For n moles this equation becomes:
Ideal Gas equation (n moles): PV = nRT
For a change from P1, V1 and T1 to P2, V2 and T2 the equation can be written:
P1V1 = P2V2
T1 T2
Example problem
5 3
If 600g of argon have a pressure of 1.5 x 10 Pa and a volume of 0.3m what is the temperature of the
gas?
Molar mass of argon = 40g and so we have 15 moles.
Therefore:
5 o
T = PV = 1.5 x 10 x 0.3 = 361 K = 88 C
nR 15 x 8.31
We have of course assumed that argon behaves as an ideal gas.
1
Example problem
o
A petrol - air mixture in the piston of a car engine initially has volume of 50cc, a temperature of 27 C
5
and is at a pressure of 2 x 10 Pa. When it is ignited by the spark plug the volume increases to 450cc
4
and the pressure drops to 8 x 10 Pa. What is the final temperature of the mixture. (assume that it
behaves as an ideal gas!)
5 4
2 x 10 x 50 = 8 x 10 x 450
300 T2
4 o
Therefore final temperature (T2) = 8 x 10 x 450 x 300 = 1080 K = 807 C
5
2 x 10 x 50
The equation of state for 1 kg of the gas is:
PV = [R/M]T = rT
where M is the molar mass in kg (2 x 10-3 for hydrogen and 32 x 10-3 for oxygen, for
example) and r is a further constant depends on the gas under consideration. Therefore for
m kg of the gas we have:
PV= mRT/M = mrT
For a fixed mass of gas whose conditions are changed from P1, V1 and T, to P2, V2 and T2
the equation of state can be written:
P1V1/T1 = P2V2/T1
Note that the temperature must always be measured in kelvins.
PV (J)
20
10
200
Temperature (K)
2
```
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http://search.cpan.org/dist/Math-PlanePath/lib/Math/PlanePath/CincoCurve.pm | 1,529,867,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867050.73/warc/CC-MAIN-20180624180240-20180624200240-00409.warc.gz | 266,205,541 | 6,087 | search.cpan.org is shutting down
Kevin Ryde > Math-PlanePath > Math::PlanePath::CincoCurve
Math-PlanePath-126.tar.gz
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# CPAN RT
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Module Version: 126
# NAME
Math::PlanePath::CincoCurve -- 5x5 self-similar curve
# SYNOPSIS
``` use Math::PlanePath::CincoCurve;
my \$path = Math::PlanePath::CincoCurve->new;
my (\$x, \$y) = \$path->n_to_xy (123);```
# DESCRIPTION
This is the 5x5 self-similar Cinco curve
John Dennis, "Inverse Space-Filling Curve Partitioning of a Global Ocean Model", and source code from COSIM
http://www.cecs.uci.edu/~papers/ipdps07/pdfs/IPDPS-1569010963-paper-2.pdf
It makes a 5x5 self-similar traversal of the first quadrant X>0,Y>0.
``` |
4 | 10--11 14--15--16 35--36 39--40--41 74 71--70 67--66
| | | | | | | | | | | | | |
3 | 9 12--13 18--17 34 37--38 43--42 73--72 69--68 65
| | | | | |
2 | 8 5-- 4 19--20 33 30--29 44--45 52--53--54 63--64
| | | | | | | | | | | |
1 | 7-- 6 3 22--21 32--31 28 47--46 51 56--55 62--61
| | | | | | | |
Y=0 | 0-- 1-- 2 23--24--25--26--27 48--49--50 57--58--59--60
|
+--------------------------------------------------------------
X=0 1 2 3 4 5 6 7 8 9 10 11 12 13 14```
The base pattern is the N=0 to N=24 part. It repeats transposed and rotated to make the ends join. N=25 to N=49 is a repeat of the base, then N=50 to N=74 is a transpose to go upwards. The sub-part arrangements are as follows.
``` +------+------+------+------+------+
| 10 | 11 | 14 | 15 | 16 |
| | | | | |
|----->|----->|----->|----->|----->|
+------+------+------+------+------+
|^ 9 | 12 ||^ 13 | 18 ||<-----|
|| T | T ||| T | T || 17 |
|| | v|| | v| |
+------+------+------+------+------+
|^ 8 | 5 ||^ 4 | 19 || 20 |
|| T | T ||| T | T || |
|| | v|| | v|----->|
+------+------+------+------+------+
|<-----|<---- |^ 3 | 22 ||<-----|
| 7 | 6 || T | T || 21 |
| | || | v| |
+------+------+------+------+------+
| 0 | 1 |^ 2 | 23 || 24 |
| | || T | T || |
|----->|----->|| | v|----->|
+------+------+------+------+------+```
Parts such as 6 going left are the base rotated 180 degrees. The verticals like 2 are a transpose of the base, ie. swap X,Y, and downward vertical like 23 is transpose plus rotate 180 (which is equivalent to a mirror across the anti-diagonal). Notice the base shape fills its sub-part to the left side and the transpose instead fills on the right.
The N values along the X axis are increasing, as are the values along the Y axis. This occurs because the values along the sub-parts of the base are increasing along the X and Y axes, and the other two sides are increasing too when rotated or transposed for sub-parts such as 2 and 23, or 7, 8 and 9.
Dennis conceives this for use in combination with 2x2 Hilbert and 3x3 meander shapes so that sizes which are products of 2, 3 and 5 can be used for partitioning. Such mixed patterns can't be done with the code here, mainly since a mixture depends on having a top-level target size rather than the unlimited first quadrant here.
# FUNCTIONS
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
`\$path = Math::PlanePath::CincoCurve->new ()`
Create and return a new path object.
`(\$x,\$y) = \$path->n_to_xy (\$n)`
Return the X,Y coordinates of point number `\$n` on the path. Points begin at 0 and if `\$n < 0` then the return is an empty list.
## Level Methods
`(\$n_lo, \$n_hi) = \$path->level_to_n_range(\$level)`
Return `(0, 25**\$level - 1)`.
http://user42.tuxfamily.org/math-planepath/index.html | 1,343 | 3,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.59743 |
http://girdhargopalbansal.blogspot.com/2013/07/priority-queues-insertion-sort.html | 1,527,363,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00138.warc.gz | 121,164,476 | 27,459 | Tuesday, 2 July 2013
Priority Queues, Insertion Sort, Selection Sort
• A Priority Queue ranks its elements by key with a total order relation.
• Keys:
• Every element has its own key.
• Keys are not necessarily unique.
• A key is often a numerical value but the concept is general, i.e., not limited to numerics.
• Total Order Relation
• Denoted by £ .
• Reflexive:k£k.
• Antisymetric: if k1£k2 and k2£k1 then k1 = k2.
• Transitive: if k1£k2 and k2£k3 then k1£k3.
• A Priority Queue supports these fundamental methods:
• insertItem(Object k, Object e)
• Object removeMinElement()
Sorting with a Priority Queue
• A Priority Queue P can be used for sorting by inserting a Sequence S of n elements and calling removeMinElement until P is empty.
Algorithm PriorityQueueSort(S,P):
Input:
A sequence S storing n keys, on which a total order relation is defined, and a Priority Queue P that compares keys with the same relation. Output: The Sequence S sorted by the total order relation. while !S.isEmpty() do e ¬S.removeFirst()
P.insertItem(e,e)
while !P.isEmpty() do
e ¬ P.removeMinElement()
S.insertLast(e)
The Priority Queue ADT
• A Priority Queue ADT must support the following methods:
• size():
• Return the number of elements in P.Input: None; Output: integer
• isEmpty():
• Test whether P is empty.Input: None; Output: boolean
Two other methods of Interface Container.
• insertItem(k,e):
• Insert a new element e with key k into P.Input: Objects k,e; Output: none
• minElement():
• Return but don’t remove an element of P with smallest key; an error occurs if P is empty.Input: none: Output: Object e
The Priority Queue ADT (cont.)
• minKey():
• Return the smallest key in P; an error occurs if P is empty.Input: none; Output: Object k
• removeMinElement():
• Remove from P and return an element with the smallest key; an error condition occurs if P is empty.Input: none; Output: Object e
Item.java
public class Item {
private Object key, elem;
public Item (Object k, Object e) {
key = k;
elem = e;
}
public Object key() { return key; }
public Object element() { return elem; }
public void setKey(Object k) { key = k; }
public void setElement(Object e) { elem = e; }
}
Comparators
• The most general and reusable form of a priority queue makes use of comparator objects. We don’t want to have to implement a different priority queue for every key type and every way of comparing them.
• Comparator objects are external to the keys that are to be compared and compare two objects. If we required keys to compare themselves, ambiguities could result. For example, numerical vs. lexicographical comparison.
• When the priority queue needs to compare two keys, it uses the comparator it is given to do the comparison.
• Thus any priority queue can be general enough to store any object.
• The Comparator ADT includes the following methods, all of which return boolean values:
• isLessThan(Object a, Object b)
• isLessThanOrEqualTo(Object a, Object b)
• isEqualTo(Object a, Object b)
• isGreaterThan(Object a, Object b)
• isGreaterThanOrEqualTo(Object a, Object b)
• isComparable(Object a)
Implementation with an Unsorted Sequence
• Let’s try to implement a priority queue with an unsorted sequence S.
• The element-items of S are a composition of two objects, the key k and the element e.
• We can implement insertItem() by using insertFirst() of the Sequence ADT. This would take O(1) time.
• However, because we always insertFirst() regardless of the key value, out sequence is unordered.
Implementation with an Unsorted Sequence (cont.)
• Thus, for methods such as minElement(), minKey(), and removeMinElement(), we need to look at all elements of S. The worst-case time complexity for these methods is Q(n).
Implementation with a Sorted Sequence
• Another implementation uses a sequence S that is sorted by keys such that the first element of S has the smallest key.
• We can implement minElement(), minKey(), and removeMinElement() by accessing the first element of S. Thus, these methods are O(1), assuming our sequence has O(1) front-removal.
• However, these advantages come at a price. To implement insertItem(), we might now have to scan through the entire sequence. Thus inserItem() is O(n).
Cost of Operations
Sequence-Based Priority Queue
Method Unsorted Sequence Sorted Sequence size, isEmpty O(1) O(1) insertItem O(1) O(n) minElement, minKey, removeMinElement Q(n) O(1)
Class SequenceSimplePriorityQueue
• A sorted-sequence implementation of the SimplePriorityQueue ADT.
•
public class SequenceSimplePriorityQueue implements SimplePriorityQueue {
//# Implementation of a priority queue using a sorted sequence
protected Sequence seq = new NodeSequence();
protected Comparator comp;
// auxiliary methods
protected Object extractKey (Position pos) {
return ((Item)pos.element()).key();
}
protected Object extractElem (Position pos) {
return ((Item)pos.element()).element();
}
protected Object extractElem (Item kep) {
return kep.element();
}
Class SequenceSimplePriorityQueue (cont.)
// methods of the SimplePriorityQueue ADT
public SequenceSimplePriorityQueue (Comparator c) {
this.comp = c;
} public int size () {return seq.size(); }
public boolean isEmpty () { return seq.isEmpty(); }
public void insertItem (Object k, Object e) throws InvalidKeyException {
if (!comp.isComparable(k))
throw new InvalidKeyException("The key is not valid");
else
if (seq.isEmpty())
seq.insertFirst(new Item(k,e));
else
if (comp.isGreaterThan(k,extractKey(seq.last())))
seq.insertAfter(seq.last(),new Item(k,e));
else {
Position curr = seq.first();
while (comp.isGreaterThan(k,extractKey(curr)))
curr = seq.after(curr);
seq.insertBefore(curr,new Item(k,e));
}
}
Class SequenceSimplePriorityQueue (cont.)
public Object minElement () throws EmptyContainerException {
if (seq.isEmpty())
throw new EmptyContainerException("The priority queue is empty");
else
return extractElem(seq.first());
}
// methods minKey and removeMinElement are not
// shown.
}
Selection Sort
• Selection Sort is a variation of PriorityQueueSort that uses an unsorted sequence to implement the priority queue P.
• Phase 1, the insertion of an item into P takes O(1) time.
• Phase 2, removing an item from P takes time proportional to the number of elements in P.
Sequence S Priority Queue P Input {7,4,8,2,5,3,9} {} Phase 1: (a) (b) … (g) {4,8,2,5,3,9} {8,2,5,3,9} … {} {7} {7,4} … {7,4,8,2,5,3,9} Phase 2: (a) (b) (c) (d) (e) (f) (g) {2} {2,3} {2,3,4} {2,3,4,5} {2,3,4,5,7} {2,3,4,5,7,8} {2,3,4,5,7,8,9} {7,4,6,5,3,9} {7,4,8,5,9} {7,8,5,9} {7,8,9} {8,9} {9} {}
Selection Sort (cont.)
• Clearly, a bottleneck exists in phase 2. The first call to removeMinElement takes Q(n), the second Q(n-1), etc., until the last removal takes Q(1) time.
• The total time needed for phase 2 is then
.
• Recall that .
• The total time complexity of phase 2 is then Q(n2 ).
• Thus the total time complexity of the algorithm is
.
Insertion Sort
• Insertion sort is the sort that results when we perform a PriorityQueueSort implementing the priority queue with a sorted sequence.
• The cost of phase 2 is improved to O(n).
• However, phase 1 now becomes the bottleneck for the running time. The first insertItem takes O(1), the second O(2), until the last insertItem takes O(n).
• The run time of phase 1 is
.
• The overall run time of the algorithm is
O(n2 ) + O(n) = O(n2 ).
Insertion Sort (cont.)
Sequence S Priority Queue P Input {7,4,8,2,5,3,9} {} Phase 1: (a) (b) (c) (d) (e) (f) (g) {4,8,2,5,3,9} {8,2,5,3,9} {2,5,3,9} {5,3,9} {3,9} {9} {} {7} {4,7} {4,7,8} {2,4,7,8} {2,4,5,7,8} {2,3,4,5,7,8} {2,3,4,5,7,8,9} Phase 2: (a) (b) … (g) {2} {2,3} … {2,3,4,5,7,8,9} {3,4,5,7,8,9} {4,5,7,8,9} … {}
Comparison of Selection Sort and Insertion Sort
Phase 1 Phase 2 Total Selection Sort O(n) + Q(n2 ) = Q(n2 ) Insertion Sort O(n2) + O(n) = O(n2 )
• Selection sort and insertion sort both take O(n2 ).
• Selection sort will always take W(n2 ) time, no matter what the input sequence.
• The cost of insertion sort varies depending on the input sequence.
• If we remove from the rear of an already sorted input sequence S or remove from the front of a reverse-ordered sequennce, then the (best-case) cost of phase 1 of insertion sort is O(n). Thus, nearly sorted sequences will perform better for insertion sort. | 2,262 | 8,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-22 | longest | en | 0.663907 |
http://stackoverflow.com/questions/8598442/scaling-images-to-an-area | 1,432,523,921,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00218-ip-10-180-206-219.ec2.internal.warc.gz | 230,977,341 | 16,741 | # Scaling images to an area
I'm trying to scale a bunch of images so that they have the same area, but keep their aspect ratio, but am having trouble find a formula to do so.
Does anyone know a formula?
-
For a given area A,
``````newx * newy = A
newx / newy = oldx / oldy
``````
which gives you:
``````newy = A / newx
newy = newx / (oldx / oldy)
A / newx = newx / (oldx / oldy)
A * oldx / oldy = newx ^ 2
``````
which then solves to:
``````newx = sqrt(A * oldx / oldy)
newy = A / newx
``````
Then again, this is a maths question, not a programming one...
-
I need to program it into C#... – chris vdp Dec 22 '11 at 2:12
And what's the problem with that? The only thing that would change is `Math.sqrt`. And it's still a maths question. – Amadan Dec 22 '11 at 4:43
Given you have two images with sizes (w1, h1) and (w2, h2) and you want to scale the second image to the same area as the first while maintaining the aspect ratio, then
``````A = w1 * h1
new_w2 = sqrt(A * (w2 / h2))
new_h2 = A / new_w2
``````
-
I think it should be new_w2 = sqrt(A * (w2 / h2)) – chris vdp Dec 22 '11 at 2:52
Correct. Fixed. – Ivan Navarrete Dec 22 '11 at 10:46 | 398 | 1,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.939106 |
http://www.ehow.co.uk/how_4612507_build-octagon-deck.html | 1,532,354,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596463.91/warc/CC-MAIN-20180723125929-20180723145929-00290.warc.gz | 428,323,080 | 12,694 | DISCOVER
# How to build an octagon deck
Updated February 21, 2017
An octagon deck is often associated with outdoor gazebos. However, an eight-sided outdoor platform style wooden patio makes a nice addition to your backyard. That's true even without the overhead roof. The octagon space seems to be a congenial configuration. That gives everyone ample space to move around and enjoy the company of each other. In construction terms, an octagon is nothing more than a modified square. If you keep this geometric concept in mind during the construction process, then your deck should turn out just fine.
Acquire all materials and bring them to the job site. If the platform is twelve feet across or less then you can use 2 X 6's, otherwise you will need a 2 x 8 joist. Cedar or redwood is preferable to pressure treated wood. Overall, you will need eight posts, which will be placed in the ground or set on top of concrete piers. If you cannot find a cut post, then you will have to cut each post at an angle. These are rip cuts that need to run along the length of the post.
Lay out the deck, starting with the location of the upright posts first. The math for this step is really very simple, if you regard an octagon as a square shape with all four corners lopped off. For example if you want to build an octagon deck that is twelve feet across, then first you must lay out a twelve foot square area. Now divide each side into thirds and these marks will be the eight corners for your eight-sided deck. Each mark will be situated 4-feet along the outside perimeter from the corner of your original square. (This is not a perfect octagon, but it is very close in size to one. To get a real octagon you have to do some more advanced math that involves the hypotenuse of a triangle. The simpler version does not change the angle of your cuts and will look almost identical.)
Dig the holes for the post or concrete feet. You can do four piers at a time, to make things easier. Put in place two opposite pairs of posts at a time and things will go smoothly. That means that first you will lay out a rectangle that is 4 feet wide and 12 feet long. It is a good idea to put in temporary diagonal braces at each corner. Use 2 X 4's for this.
Make sure the top of all the posts are level. (Actually a slight slant in one direction is not a bad idea. It helps shed rain water). You can cut a post if you have to, but the idea is to level all posts with out using a cut to make a correction.
Measure and cut the eight-sided box frame that will go around the perimeter of the eight posts. 22 and a half degrees is the magic number here. It is the angle you will subtract from 90 degrees when you cut the end of each board. So set your saw at 67 and a half degrees and start cutting. Do one piece at a time.
Cut and fit the floor joists. Do the ones that are the longest first and gradually work your way out to the edge. All bevelled cuts should be done with your saw again set at 67 and a half degrees. Attach the joists with nails or building screws driven in from each end. The joists should be placed 16 inches on centre.
Finish the frame with another framing member attached to the outside of the box. The end result will be that a double frame that completely around the outside of the octagon.
Cut and fit the floor boards. Do the longer pieces first and cut each end flush with the edge of the frame. Leave a small gap of 1/4 or 1/2 inch between each board. Nail each board with #8 galvanised nails or special decking screws. Run the boards perpendicular to the floor joists.
Apply a sealer or paint finish if necessary.
#### Warning
Work on sawhorses and use safety glasses.
#### Things You'll Need
• Circular saw
• Post hole diggers
• Galvanised framing nails (or builder's screws)
• Tape measure
• Saw horses
• Hammer
• Framing square and carpenter's square
• Lumber | 889 | 3,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-30 | latest | en | 0.948624 |
https://en.doc.boardgamearena.com/SantoriniPowerSiren | 1,669,659,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00571.warc.gz | 260,108,320 | 9,283 | This is a documentation for Board Game Arena: play board games online !
# SantoriniPowerSiren
## Introduction
Sirens are famous for their ability to lure and charm people. And now, that can bewitch people to tumble down and undo their efforts! But without a bit of strategy, it cannot reach its maximum potential…
### God Power
At the start of the game, choose the direction of the Siren's song. Instead of moving, you can force one or more opponent workers one space in the direction of the Siren’s song to unoccupied spaces at any level.
Opponent forcing
### Explanation
In order to simplify the notation, every time we refer to an orthogonal wind, we mean direction South (↓) and whenever we refer to a diagonal wind, we mean direction South-East (↘).
The first thing that Siren needs to realize is that:
• If she sets the wind orthogonally, there are 5 columns of length 5 in which she can use her power. Moreover, there are 5 squares (the fifth row) over which Siren cannot use her power.
• If she sets the wind diagonally, there are more diagonals, but only one of length 5, which is the A5-E1 diagonal (we still have two diagonals of lengths 2, 3 and 4). Moreover, there are 9 squares (the fifth row and the A column) over which Siren cannot use her power.
So the first question to ask is “which wind direction is better, ↓ or ↘ ?”. Taking the previous observations into consideration, the natural choice would be the orthogonal direction, but there still could be some cases where Siren would prefer the diagonal direction. In the following analysis, the wind direction is assumed to be South (↓).
## Game Strategies
### Early Game
Siren should build in row 5 (north) and avoid at all cost to build in row 1. An opponent in row 5 can be forced four times while an opponent in row 1 can not even be forced. So you have to understand how important it is for Siren to play in row 5. Even if the opponent can create an attack in row 5, it will be very hard for them to win (due to several reasons).
However, if the opponent is playing in row 1, they can create some serious threats. A very deceiving (for the opponent) plan for Siren is to purposely let the opponent move to a good position in row 5 so that the opponent places more blocks near row 5 (benefiting Siren). In these cases, I would advise the opponent to just ignore Siren and don’t fall for her plan (play in row 1!).
### Mid Game
Siren should keep building near North and don’t be afraid to let the opponent move up to a level 2, since she can probably force the worker out of there in the next move. This strategy is especially effective if Siren builds a level 2 in B4 and there is a level 0 in B3. If the opponent moves to B4, it will fall in the next move. The opponent can still build in B2, but they can do this forever and eventually Siren will get a better position. For example,
1. S: a5-b5, build B4(2) [this would normally allow the opponent to control 8 winning squares…] O: a4-b4, build B3(1) [the opponent avoids falling two levels]
2. S: force b4-b3, build B4(3) O: b3-a4, build B4(X)
3. S: force a4-a3, build A5(2) [threatening to win in A5 in two turns] O: a3-a4, build A3(2) 4. S: force a4-a3, build C5(1) [and there is no way for the opponent to defend both A5 and C5]
I hope that from this example you have realized the importance of building in row 5. On the other hand, if there were more blocks near row 1, it would probably be possible for the opponent to come back into the game.
Furthermore, Siren should avoid domes, especially in row 1 or in the center. For example, a dome in B2, allows the opponent to stay in B3 without getting forced by Siren (and a worker in B5 can only be forced twice).
### Late Game
An endgame will usually be pretty good for Siren. If she can force the opponent, she can stay in the same square, which can be very good in such situations. She needs to be careful about where she creates winning threats (thus creating domes). For example, the opponent won’t be able to make use of a dome in row 5 (the opponent cannot be in row 6) so, if Siren creates a winning threat just to get a better position (not really intending to win), it is probably a good idea to do it in row 5.
## Specific Matchup
### Aphrodite
Whenever a power can control Aphrodite, it is usually very good (e.g. Charon, Dionysus, Eris) and this is what makes this matchup interesting. If Aphrodite isn’t careful enough, she will be dragged away from Siren’s workers and she will then not be able to defend. Still, it won’t be easy for Siren to create such situations and this is a balanced matchup in general.
### Apollo and Charon (and maybe Scylla)
Being two of the strongest powers of the game, it is amazing that Siren is one of the few powers which will give them a headache. It is definitely possible to win this matchup playing with Siren.
### Artemis
Siren can force Artemis one square, but Artemis can recover two squares. I would pick Artemis in this matchup but without underestimating Siren.
### Clio
This is a matchup in which Clio’s coins are very effective, especially if they are in the center. Not only a Clio’s worker standing on a coin cannot be forced, but also a worker cannot be forced into a coin. For example, if the wind is ↓, Clio is in C4 and there is a coin in C3, that worker cannot be forced south.
### Domers (Asteria, Atlas, Selene)
The more domes there are on the board, the weaker Siren becomes. For Atlas and Selene, they can place a dome in every move. For Asteria, she cannot do it, but when she places the dome, it is anywhere on the board (which can make up for the wait). All these gods are strong against Siren and I would not even recommend playing on Siren’s side (you will be smashed).
### Eris
It is interesting putting two controllers playing together. Although Eris should be stronger, there is always room for Siren to create deadly attacks.
### Europa & Talus
At each move, Europa can relocate the token and thus she has (almost) always the possibility to prevent that worker from being forced (since it can only be forced in one direction). This makes Europa very strong in this matchup and I would choose her without taking too much into consideration.
### Graeae
Forcing Graeae away from a defending zone is a nightmare and that’s why this will be very difficult for Graeae. I almost would say not to even try it, but if you do and are able to win with Graeae, give me a call.
### Harpies
Siren is very strong in this matchup. Being able not to move against Harpies is just great and so Harpies will have to play differently if they want to win.
### Hermes
The fact that Hermes has full control of his workers location, makes it easier for him to make Siren’s ability less effective. Anyhow, he will eventually have to move up, and this will be in a turn in which he can only move once. This is an interesting matchup.
### Pan
This is a very interesting matchup where there will be many situations where Siren cannot use her power since in the next turn Pan would move down two (or more) levels and win. This said, this matchup can go to any side and I would definitely recommend it.
### Pegasus
If, like me, you thought Pegasus would have no problem converting this game into a win (you are forced to the ground but then you can move up in the next turn), I’m afraid you’re wrong. It won’t be easy for Pegasus to play this matchup and this can actually turn into a very interesting combination.
### Poseidon
Another OP god against which Siren is actually able to put up a fight! Her ability is useful in three ways in this matchup: separate Poseidon’s workers (a separated Poseidon usually sucks), force Poseidon out of high levels and force Poseidon from a level 0 to a higher level (so that he loses his power). Putting this all together, it won’t be that easy for Poseidon to win and it is even possible for Siren to turn the tables around.
### Proteus
Proteus holds on very well and this is another interesting matchup for you to consider.
### Triton
Once again, Siren is actually able to face Triton. As stated before, I would advise a ↓Siren to build in row 5. Moreover, if Siren wants to have a chance in the endgame (most games involving Triton go to the endgame, if you don’t lose first), she should build domes in C5, C1, A3, E3 (especially C5) to reduce Triton’s mobility.
### Urania
If the song is set South (↓), Urania will just play in the first row and create threats Siren cannot defend. For example, a level 2 Urania in A1, can threaten to win in E1, A5 and E5 (and this worker cannot be forced out of there by Siren). Urania should have a very comfortable game. If the song is set South-East (↘), the strategy for Urania is similar as she can play in the first row and E-column, especially A1, E1 and E5 (and even create winning threats in A5).
## Conclusion
Siren is definitely not an easy power to master and it might take some games to get used to it. The wind can be set orthogonally or diagonally, which can lead to different possibilities just for the same matchup. Moreover, the first few moves are very important since they will dictate where the game will be played. Siren is not an excessively strong power, but if you are not careful enough, you might just be dragged away without being able to defend Siren’s threats.
Tier ranking: B | 2,276 | 9,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.951384 |
https://proofassistants.stackexchange.com/questions/1396/what-are-some-real-world-first-order-logical-theories-for-demos | 1,718,805,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00725.warc.gz | 418,276,227 | 27,623 | # What are some "real world" first order logical theories for demos?
I'm working on a tool for first order logical theories. I want to show the tool can work with real world logical theories. What are some good theories for demos?
I think I want demos that are:
• finitely axiomizable
• constructive
• familiar to ordinary mathematics
I have some ideas for theories but they are a bit ambiguous.
From easy to hard:
• theory of groups
• Peano arithmetic
• lambda calculus
• set theory
• simply typed lambda calculus
• calculus of constructions
• theory of complete ordered fields (Real numbers)
Some Problems Are:
I have heard there are multiple variants on the induction principle for Peano arithmetic.
I think the call by value lambda calculus with De Bruijn levels would be easiest to formalize. I only know this sort of detail because I know about computer stuff. I'm sure stuff like set theory has similar wrinkles.
There are many variants of set theory. Some are more constructive some are less. I'm not sure I want to get into proper classes but these make finitely axiomised presentations easier. I have heard of NBG (von Neumann–Bernays–Gödel) set theory but I don't know anything about it.
Type theory can be hacked into a first order logical theory with judgments as relation symbols but it's a little ugly.
• Set theory and higher-order theories require infinite "axiom schemas" (with syntactical rules being the same as second-order axioms, if you want to formalize them), like this one for taking subsets... Commented May 12, 2022 at 18:17
• @ZhanrongQiao IIRC NBG can be finitely axiomised and is a conservative extension. I would like to learn more about these sort of hairy details. I'm still not sure if I want to restrict to finitely axiomizable theories. Commented May 12, 2022 at 18:22
• Ah, that looks interesting! I'm also trying to write a toy prover for first-order logic, but was unaware of NBG and thought that I have to support second-order things for doing set theory... Commented May 12, 2022 at 18:40
• @ZhanrongQiao I think maybe you can really support any system in a first order finitely axiomised style but you have to do some ugly coding. Something like assume a sort P of formulas, assume function symbols for connectives and axiomize an entailment relation. Just it's pretty ugly. Sort of like writing a Lisp interpreter to write your programs in. Commented May 12, 2022 at 19:50
• I think in any way there must be a method for quantifying over functions/predicates if we want schemas. Even in Metamath, variables are really second-order (since they can be substituted for expressions that contain free variables), which may look ugly because MM allowed some "controlled naming clashes" among those introduced free variables. Metamath Zero is similar, but it gives a better control over the naming thing. (I personally feel that the "free-variable" approach is still less intuitive/natural than using lambda binders...) Commented May 12, 2022 at 21:22 | 687 | 3,000 | {"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} | 2.796875 | 3 | CC-MAIN-2024-26 | latest | en | 0.960041 |
https://cloud.tencent.com/developer/article/1010083 | 1,558,712,715,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257660.45/warc/CC-MAIN-20190524144504-20190524170504-00303.warc.gz | 441,089,578 | 14,783 | # LWC 58：724. Find Pivot Index
## LWC 58：724. Find Pivot Index
Problem:
Given an array of integers nums, write a method that returns the “pivot” index of this array. We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index. If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
Example 1:
Input: nums = [1, 7, 3, 6, 5, 6] Output: 3 Explanation: The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3. Also, 3 is the first index where this occurs.
Example 2:
Input: nums = [1, 2, 3] Output: -1 Explanation: There is no index that satisfies the conditions in the problem statement.
Note:
The length of nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].
``` public int pivotIndex(int[] nums) {
int n = nums.length;
int[] lf = new int[n];
int[] rt = new int[n];
int sum = 0;
for (int i = 0; i < n; ++i) {
lf[i] = sum;
sum += nums[i];
}
sum = 0;
for (int j = n - 1; j >= 0; --j) {
rt[j] = sum;
sum += nums[j];
}
for (int i = 0; i < n; ++i) {
if (lf[i] == rt[i]) {
return i;
}
}
return -1;
}```
``` public int pivotIndex(int[] nums) {
int n = nums.length;
int[] sums = new int[n + 1];
for (int i = 0; i < n; ++i){
sums[i + 1] = sums[i] + nums[i];
}
for (int i = 0; i < n; ++i){
if (sums[i + 1] == sums[n] - sums[i]){
return i;
}
}
return -1;
} ```
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3576 | 747 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.345363 |
https://www.paraisos-fiscales.info/answers/1995071-one-seventh-of-an-unknown-value | 1,708,886,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00553.warc.gz | 938,972,735 | 5,431 | # One seventh of an unknown value
Answer: one seventh of x = (1/7)x
where x is the unknown value
## Related Questions
Step-by-step explanation:
It is given that; () is the proportion by which distances are scaled down on a map. One can use this to make a proportion and solve to find how many () would represent (). () will represent the unknown or the number of () that one has to solve for.
Cross products,
Inverse operations,
Use a commutative property to complete the statement 2x + 16
2(x+8)
Step-by-step explanation:
Would give you brainliest whoever answers this correct first !!
6 is the answer to the question
Im still stuck.... help? :',)
This is because if you multiply, you’ll get the answer of 3/8 which is less than 1/2 (4/8) but not close to zero, thus leaving B as the best option.
Hope this helps!
PLZ HELP ME GEOMETRY
question is below
The missing justification is for the statement that three angles add to a particular angle. The appropriate choice is ...
Need Assistance With This
a =7.5
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2+ b^2 = c^2 where a and b are the legs and c is the hypotenuse
a^2 + 10 ^2 = 12.5^2
a^2 + 100 =156.25
Subtract 100 from each side
a^2 = 56.25
Take the square root of each side
sqrt(a^2) = sqrt( 56.25)
a =7.5 | 366 | 1,342 | {"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} | 4.09375 | 4 | CC-MAIN-2024-10 | latest | en | 0.899998 |
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basic_tools:variational_calculus:functional [2018/03/15 14:31]iiqof basic_tools:variational_calculus:functional [2020/04/12 14:41] (current)jakobadmin Both sides previous revision Previous revision 2020/04/12 14:41 jakobadmin 2018/03/15 14:31 iiqof 2018/03/14 16:00 iiqof 2018/03/14 15:58 iiqof 2018/03/10 17:28 iiqof Clarification2018/03/10 17:05 iiqof created Functional Page 2020/04/12 14:41 jakobadmin 2018/03/15 14:31 iiqof 2018/03/14 16:00 iiqof 2018/03/14 15:58 iiqof 2018/03/10 17:28 iiqof Clarification2018/03/10 17:05 iiqof created Functional Page Line 3: Line 3: Let $\Omega(\mathcal{Q})$ be the set of functions $q:\mathbb{R^n} \to \mathcal{Q}$, then a //functional// S is a map Let $\Omega(\mathcal{Q})$ be the set of functions $q:\mathbb{R^n} \to \mathcal{Q}$, then a //functional// S is a map - $$+ - S:\Omega \to \mathbb{R}; S[q] \mapsto \alpha \in\mathbb{R} +$$ S:\Omega \to \mathbb{R}; S[q] \mapsto \alpha \in\mathbb{R} .$$-$$ + So we can see how a functional is a //function of functions// as we said before, this is the reason why the notation $S[\cdot]$ instead of $S(\cdot)$, to remind that it is more that the eyes meet. So we can see how a functional is a //function of functions// as we said before, this is the reason why the notation $S[\cdot]$ instead of $S(\cdot)$, to remind that it is more that the eyes meet. | 530 | 1,486 | {"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} | 3.84375 | 4 | CC-MAIN-2020-24 | latest | en | 0.727787 |
http://forums.wolfram.com/mathgroup/archive/2006/Apr/msg00543.html | 1,722,851,178,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00555.warc.gz | 9,361,246 | 7,507 | error function
• To: mathgroup at smc.vnet.net
• Subject: [mg65985] error function
• From: "ann" <annjo.puphy at gmail.com>
• Date: Tue, 25 Apr 2006 05:19:24 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Hello everybody,
I would be greatful if someone can help me with the following
problem..
i have a task of integrating a complicated funtion which upon
execution gives the result with Erf[] how to get it solved.
To say the problem in detail , i have to do the following operation
Integrate[ (F^2 + C*G^2 ) * Real part of (F*complex conjugate of F *
exp(I*theta) ] with the limits [-infinity to infinity] where F and G
are of the form
F=x1 * exp{ (-(t-x2)^2/x3^2 + i/2 * x4(t-xc) + i x4* (t-xc) + i x5 and
g is of the same form with x replaced with y.
now this results in the erf[of another complicated function]
how to get it solved..or to how should i help mathematica to get it
completely solved??
expecting suggestions | 286 | 948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.892286 |
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