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https://learn.careers360.com/engineering/question-if-and-are-the-roots-of-the-equation-and-and-are-the-roots-of-the-equation-then-is-equal-tooption-1-option-2-option-3-option-4-112213/ | 1,621,157,510,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00279.warc.gz | 387,115,186 | 110,991 | # If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+px+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are the roots of the equation $2x^{2}+2px+1=0$, then $\left ( \alpha - \frac{1}{\alpha } \right )\left ( \beta -\frac{1}{\beta } \right )\left ( \alpha +\frac{1}{\beta } \right )\left ( \beta +\frac{1}{\alpha } \right )$ is equal to: Option: 1 Option: 2 Option: 3 Option: 4
$x^{2}+px+2=0$ $\alpha ,\beta$
$2x^{2}+2px+1=0$ $\frac{1}{\alpha}$ ,$\frac{1}{\beta}$
$\alpha +\beta =-p ,\alpha \beta =2$
$\frac{1}{\alpha }+\frac{1}{\beta }=-q,\frac{1}{\alpha \beta }=\frac{1}{2}$
$\left ( \alpha - \frac{1}{\alpha } \right )\left ( \beta -\frac{1}{\beta } \right )\left ( \alpha +\frac{1}{\beta } \right )\left ( \beta +\frac{1}{\alpha } \right )$
$=\left ( \alpha \beta -\frac{\alpha }{\beta } -\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }\right )\left ( \alpha \beta +2+\frac{1}{\alpha \beta } \right )$
$=\left ( \frac{5}{2}-\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }\right )\left ( \frac{9}{2} \right )$
$=\left ( \frac{5}{2}- \frac{p^{2}-4^{-1}}{2} \right )\left ( \frac{9}{2} \right )$
$=\frac{9}{4}(9-p^{2})$
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₹ 34999/- ₹ 14999/- | 755 | 2,166 | {"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": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.453405 |
https://www.physicsforums.com/threads/sequence-convergence-proof.388851/ | 1,695,311,965,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00460.warc.gz | 1,065,175,552 | 15,910 | Sequence Convergence Proof
Homework Statement
Let $$X=(x_n)$$ be a sequence of strictly positive numbers such that $$\lim(x_{n+1}/x_n)<1$$. Show for some $$0<r<1$$, and for some $$C>0$$, $$0<x_n<Cr^n$$
The Attempt at a Solution
Let $$\lim(x_{n+1}/x_n)=x<1$$
By definition of the limit, $$\lim(x_{n+1}/x_n)=x \Rightarrow \forall \epsilon>0$$ there exists $$\: K(\epsilon)$$ such that $$. \: \forall n>K(\epsilon)$$
$$|\frac{x_{n+1}}{x_n}-x|<\epsilon$$
Since i can pick any epsilon, let epsilon be such that $$\epsilon + x = r <1$$. Also, I know that since this is a positive sequence, $$\frac{x_{n+1}}{x_n}>0$$. Therefore, for large enough $$n$$,
$$0<\frac{x_{n+1}}{x_n}<r<1.$$
From here I am not sure where to go, any hints would be much appreciated! I cannot find out what this tells me about $x_n$
For large enough $n$ you also have the following:
$$\frac{x_{n+k}}{x_n} = \frac{x_{n+1}}{x_n} \frac{x_{n+2}}{x_{n+1}} \cdots \frac{x_{n+k}}{x_{n+k-1}} < r^k$$
Can I let $$x_n=C$$ and therefore say the sequence $$x_{n+k}<Cr^k$$? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!
Can I let $$x_n=C$$ and therefore say the sequence $$x_{n+k}<Cr^k$$? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!
You're not quite there yet.
First note that you need
$$x_{n} < C r^n$$
not
$$x_{n+k} < Cr^k$$
Also, it needs to be true for EVERY $n$, not just sufficiently large $n$.
First let's introduce some notation. It is easier to discuss if we give this "sufficiently large" $n$ a name, say $N$. For all $n \geq N$,
$$0 < \frac{x_{n+1}}{x_n} < r < 1$$
which yields
$$x_{N+k} < r^k x_N$$
for all $$k \geq 0$$.
I can get the index and exponent to agree by writing, equivalently,
$$x_{N+k} < r^{N+k} (x_N r^{-N})$$
Then if I set $$C_1 = x_N r^{-N}$$ I get
$$x_{N+k} < C_1 r^{N+k}$$
which looks pretty promising. However, this $C_1$ may not be large enough to work for ALL $n$, i.e. it may not be true that
$$x_n < C_1 r^n$$
for all $n < N$.
But note that there are only finitely many $x_n$ with $n < N$. Can you use that fact to find a $C$ that does work for all $n$?
Last edited:
So, for every $$j<N$$, Let $$M_j$$ be such that $$x_j<M_jr^j$$ (there exists such an M by the archimedean property). Now, if I take $$C=\sup\{M_j,C_1| j\in\mathbb{N},j<N\}$$, that should do the job since there are only finitely many M's, correct?
So, for every $$j<N$$, Let $$M_j$$ be such that $$x_j<M_jr^j$$ (there exists such an M by the archimedean property). Now, if I take $$C=\sup\{M_j,C_1| j\in\mathbb{N},j<N\}$$, that should do the job since there are only finitely many M's, correct?
Looks good to me. You can even call it "max" instead of "sup" since there are only N+1 numbers under consideration. | 968 | 2,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-40 | latest | en | 0.841261 |
https://math.stackexchange.com/questions/3001002/division-to-account-for-repetition-in-multiplication-principle | 1,568,920,110,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00475.warc.gz | 568,351,329 | 29,947 | # Division to account for repetition in multiplication principle
I have this very bad misconception regarding when to divide and when not to divide when counting. I have compiled some of these problems(some from the book of Principle of Techniques in Combinatorics and some from my school’s lecture slides). I usually distinguish very clearly between when to strictly use binomial coeffficients only and when to compute binomial coefficients with factorials(to arrange some items). However, after going through the problems, it seems one can apply the multiplication principle on binomial coefficients as well to mimic the behaviour of arrangement, based on the multiplication principle.
Here are the problems and my interpretation of how whether I need to consider any form of repetition given the context. Answers shown and my own perspective will only from the binomial coefficient perspective to highlight the difference.
Highlighted in green are the restrictions for the corresponding sub problems. Highlighted in yellow is the pool to make our subsets from(which is what the binomial coefficient is counting)
Prime factor:
Committee:
Pairing:
Spelling:
From all my previous experiences, I always believe it all bows down to common sense and phrasing in the question whether they want to consider any form of distinction after the restriction they impose, and what is the object we are actually "choosing" and I cannot accurately discern which is the right way to count quite frequently.
Is there a clearer way of explaining the anomaly of seeing division in the pairing problem or any alternate form of phrasing to make the other questions become a scenario that requires us to consider “division to account for lack of emphasis of order” to give me a wider perspective. | 330 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.929264 |
http://www2.cs.utah.edu/~zachary/computing/lessons/uces-8/uces-8/body-node6.html | 1,516,745,028,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892699.72/warc/CC-MAIN-20180123211127-20180123231127-00245.warc.gz | 556,555,554 | 1,504 | What Do We Want?
For now, the problem we want to solve is this:
Given the conditions described in the text above and in the diagrams, and given an angle for the squatter's ankle joint (call this angle ), we want to determine the X-Y position of the squatter's knee joint. Remember: the weight must always be directly over the feet.
The angle will be measured in degrees counterclockwise from the positive X axis. (Or more precisely, will be measured in degrees counterclockwise, from a ray in the positive X direction passing though the ankle joint, to the shin link.) So when the squatter is standing straight up, the ankle joint angle will be 90 degrees.
Eric N. Eide
Hamlet Project
Department of Computer Science
University of Utah | 164 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.889083 |
https://www.lmfdb.org/EllipticCurve/Q/100010d/ | 1,579,275,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589861.0/warc/CC-MAIN-20200117152059-20200117180059-00118.warc.gz | 954,982,106 | 5,418 | # Properties
Label 100010d Number of curves 1 Conductor 100010 CM no Rank 1
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("100010.d1")
sage: E.isogeny_class()
## Elliptic curves in class 100010d
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
100010.d1 100010d1 [1, 1, 0, 3, -169] [] 27136 $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curve 100010d1 has rank $$1$$.
## Modular form 100010.2.a.d
sage: E.q_eigenform(10)
$$q - q^{2} + 2q^{3} + q^{4} + q^{5} - 2q^{6} + q^{7} - q^{8} + q^{9} - q^{10} + 2q^{12} + 6q^{13} - q^{14} + 2q^{15} + q^{16} - q^{18} + 7q^{19} + O(q^{20})$$ | 288 | 709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | latest | en | 0.375728 |
https://physics.stackexchange.com/questions/734501/why-do-silicon-photodiodes-respond-to-a-wavelength-range-of-190-1100nm | 1,713,369,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00255.warc.gz | 407,975,966 | 40,965 | # Why do silicon photodiodes respond to a wavelength range of 190-1100nm?
Silicon photodiodes respond to a wavelength range of around 190-1100nm (source: wikipedia). I understand that photodiodes function by having a photon of sufficient energy create an electron-hole pair in a semiconductor via the photoelectric effect, which in turn increases the number of charge carriers, which in turn increases the photocurrent.
I know that the energy of a photon is proportional to its frequency as per the Planck-Einstein relation: $$E=hf$$. The minimum energy required to stimulate the creation of an electron-hole pair in the semiconductor material should therefore be a product of its bandgap. The bandgap for silicon at 302 Kelvin is approximately 1.14eV. The associated wavelength of the minimum energy should therefore be:
$$f = \frac E h = \frac {1.14eV} {6.62607015 \times 10^{−34}} = 2.75650774 \times 10^{14} Hz$$
$$\lambda = \frac c f = \frac {3 \times 10^8} {2.75650774 \times 10^{14}} = 1088 nm$$
The maximum energy is where I get a little confused. My intuition says that the ionisation energy would be an upper limit; I found 8.15168eV quoted for silicon. Repeating the equation finds the associated wavelength:
$$f = \frac E h = \frac {8.15168eV} {6.62607015 \times 10^{−34}} = 1.97106745 \times 10^{15} Hz$$
$$\lambda = \frac c f = \frac {3 \times 10^8} {1.97106745 \times 10^{15}} = 152nm$$
This is somewhat smaller than the 190nm figure claimed above.
My initial guess was that the maximum energy level is lowered as to avoid exceeding the free exciton binding energy, but the figure I found listed for that is 14.7eV (84nm) so that doesn't line up.
Where does the 190nm lower wavelength bound (upper energy bound) figure come from? Am I also right in thinking that not all wavelengths in the 190-1100nm range would elicit photocurrents, due to the requirement that photon energies match the quantum energy state transition levels?
• (+1) If the photodiode has a fused silica window, as mine do to protect the die surface, then the 190 nm may just be the typical fused silica short wavelength cutoff.
– Ed V
Oct 31, 2022 at 2:11
• @Polynomial It's complicated, not just a simple material property. Depends on the surface. $SiO_2$ is opaque below ~200 nm unless it's very thin. The penetration depth for UV in semiconductors is very short, so the photoionization may occur in undepleted semiconductor near the surface. The depleted bulk material starts at some depth below the surface, depending on doping. If the minority carriers from a shallow event can't make their way through the undepleted material, the diode won't collect them. Oct 31, 2022 at 13:26 | 697 | 2,681 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.886779 |
http://www.indiabix.com/placement-papers/dell/6609 | 1,638,495,414,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00369.warc.gz | 101,625,486 | 7,637 | # Placement Papers - Dell
## Why Dell Placement Papers?
Learn and practice the placement papers of Dell and find out how much you score before you appear for your next interview and written test.
## Where can I get Dell Placement Papers with Answers?
IndiaBIX provides you lots of fully solved Dell Placement Papers with answers. You can easily solve all kind of placement test papers by practicing the exercises given below.
## How to solve Dell Placement Papers?
You can easily solve all kind of questions by practicing the following exercises.
### Dell - Dell Technical Interview Questions
Posted By : Shana Rating : +37, -4
1. Write a program that asks for user input from 5 to 9 then calculates the average
2. Implement strcat {} function without using string function and optimize the function for speed and space
3. Write a function to reverse a linked list
4. Write a recursive function to get the factorial of a given number only if it is prime
5. Write a C++ programme without using any loop {if,for,while etc}to print number from 1 to 1-- and 100 to 1
6. Write a programme to Exchange two number without using a temporary variable
7 Write a function to find if the given number is a power of 2
8. Write a function in 2 different ways that will return f{7} = 4 and f (4) = 7
9. Write a function to remove duplicates in an array?
10. Convert integer number in binary without loops.
Section II -Oops:
1. What are the main features of OOPS? Explain them briefly
2. What is a virtual function? Explain with an example?
4. In inheritance explain the public-private protected modifiers behaviours on data members.
5. What is a factory method, give reference to an example class
Technical Interview Questions:
1. What is composition and aggregation?
2 What is the difference between pointers and references? why?
3. What is the order of calling of constructors and destructors in the class hierarchy? Explain.
4. What is RITI?
5. difference between an array, vector, list and dequeue?
6. Explain "Passing by value","passing by pointer" and "passing by reference"?
7. what is the difference between thread ad a processing are the used?
8. What is the difference between heap and stack memory? when are they used?
9. Implement an algorithm to sort an array?
10. Explain pre-order, post-order and in-order binary tree traversal?
11. What is the difference between char a [] ="string";and char .p ="star";?
12. What is the difference between union and struct?
13. Explain compiling and linking in brief?
14. What is a DLL? Explain a few advantages of using DLL's.
15. What is an interrupt?
16. What is the difference between UDP and TCP protocols?
17. Explain in brief Client Server technology?
18. How will you get the no.of instances created for a class from within the class?
19. What are RAM and ROM also explain the difference between them?
20. What is the difference between HTTP GET and HTTP Post?Explain
21. What is the difference between heap and stack memory? | 722 | 3,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.843751 |
https://www.smore.com/jwy6u-geometry?ref=my | 1,596,743,882,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00206.warc.gz | 743,015,019 | 12,779 | # Geometry
## Week at a Glance
This week will be focused on Volume. We will make a book and explore http://www.ixl.com/math/. Please work on memorizing your formulas as there are only 3 given on the EOC: V=Bh; V=1/3Bh; and V=4/3 pi r cubed. By the end of the week, you should be able to identify the 3-D shape by name (cylinder, prism, pyramid, cone and sphere) and identify and apply the appropriate formula that applies to the shape.
## Monday, February 02, 2015
***I am unavailable this morning as I am in a meeting!***
Today we will be learning how to identify and apply the formula for a pyramid. Time to enjoy my drawings! By the end of the period, you should know what a rectangular and square pyramid is and how apply the formula V=Bh to both of them.
Engaged Community - Support your fellow Rams
6:00PM – Baseball City Classic Tournament, RHS (RHS plays)
7:00PM – Boys BB Districts
## Tuesday,February 03, 2015
Today and tomorrow we will be creating our flip book to help us remember our formulas. You will receive minimal guidance as this is to help you and not a grade. You will be able to use this on your quizzes. You will not be able to use it on a test, as the EOC does not give you the formulas.
Periods 1 and 4: You also have a worksheet on Canvas to work on. I did include the answers as not all of the shapes have been covered. Give them a good try!
Engaged Community - Support your fellow Rams
Writing Workshop - 10th grade, Gym
6:00PM – Financial Aid Night, Media
6:00PM – Baseball City Classic Tournament, RHS
## Wednesday, February 04, 2015
Today and tomorrow we will be creating our flip book to help us remember our formulas. You will receive minimal guidance as this is to help you and not a grade. You will be able to use this on your quizzes. You will not be able to use it on a test, as the EOC does not give you the formulas.
Today, once you have completed your book, you are to work on http://www.ixl.com/math/ with the user name and password I give you.
Engaged Community - Support your fellow Rams
Writing Workshop - 10th grade, Gym
6:00PM & 7:30PM - Boys BB Districts - Ridgewood plays @ 7:30PM
6:00PM – Baseball City Classic Tournament, RHS
## Thursday, February 05, 2015
***I am unavailable before school today as I will be in a meeting***
Today we will be learning how to identify and apply the formula for a cone. Time to enjoy my drawings again! By the end of the period, you should know what a cone is and how apply the formula.
Engaged Community - Support your fellow Rams
Lunches – Spring Sports Meeting, RM 106
3:15PM – Girls Tennis vs. WRHS
6:00PM – Baseball City Classic Tournament, RHS
6:30PM – Volleyball Open Gym
7:00PM – Softball vs. JWMHS
## Friday, February 06, 2015
***I am unavailable before school today as I will be in a meeting***
Today we will be learning how to identify and apply the formula for a sphere. Time to enjoy my drawings again! By the end of the period, you should know what a sphere is and how apply the formula.
Engaged Community - Support your fellow Rams
6:00PM – Softball vs. SSHS
7:00PM - Boys BB Districts
Important future dates:
Friday 2/13 – Progress Reports distributed 1st period
Monday 2/16 – No school | 824 | 3,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.943391 |
https://www.jiskha.com/questions/1133389/Graph-the-piecewise-function-given-by-f-x-1-3x-2-x-lt-0-x-5-x-gt-0 | 1,537,379,075,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156252.62/warc/CC-MAIN-20180919161420-20180919181420-00301.warc.gz | 768,980,228 | 5,001 | # algebra
Graph the piecewise function given by
f(x)= -1/3x +2, x<0
x-5,x>0
1. if you can graph one line, surely you can graph two.
You can see the two lines at
http://www.wolframalpha.com/input/?i=plot+y+%3D+-1%2F3+x+%2B+2%2C+y%3Dx-5+for+x+%3D+-5+to+10
Just pick the blue line left of x=0 and the purple line to the right.
posted by Steve
## Similar Questions
1. ### Math
flickr.(dotcom)/photos/77304025@N07/6782789880/ Create a function that closely gives the graph given below. The function may be a piecewise function but it does not necessarily have to be a piecewise function. a. Give a brief
2. ### Calculus
flickr.(dotcom)/photos/77304025@N07/6782789880/ Create a function that closely gives the graph given below. The function may be a piecewise function but it does not necessarily have to be a piecewise function. a. Give a brief
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i am attempting to graph the piecewise function y=(-2/3)x-6 if -9<=x<-6 (1/3)x if -6<=x<0 (((x-3)^2)/9) + (((y-0)^2)/16)=1 if 0<=x<=6 on a ti 89 titanium. i believe you must use the when( feature, but i haven't a
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Consider the graph of f(x)=2[[x/3]] +4 with doman {x|0<_x<12}. Create the piecewise function that would produce the same graph over the indicated domain. [[ ]] stands for a step function
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Given the following piecewise function: Find the Given the following piecewise function: f(x)= {?(x+10 for -9 ?x<-2@3 for -2 ?x ?0@2x-6 for 0<x ?5 )? Find the domain. Find the range. Find the intercepts. Is f continuous on
7. ### Math (Semicircle)(Pre-calc)
goo.gl/photos/e6Ks3v4Qqj177Nun6 The graph below is made from two semicircles. The domain of the function in the graph is [−8,8] [-8,8] . Find a piecewise formula for the function f(x) f(x). Help please, I already on my last
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More Similar Questions | 793 | 2,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.780282 |
http://new.transum.com/Maths/Activity/Playing_Cards/ | 1,537,524,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157028.10/warc/CC-MAIN-20180921092215-20180921112615-00222.warc.gz | 177,906,803 | 6,739 | ##### ContentsLong LineMake 15Card TowerSquaresSnap!Times TablesGreaterMore ...
Imagine you are on a desert island with nothing but a pack of playing cards. Do you have to stop learning mathematics? Of course not, there are many fascinating mathematical activities you can enjoy and here are some ideas.
• Long Line Make a line of cards so that consectutive card totals are odd, even, prime or square;
• Make 15 Take turns to select a card. The first person to make a sum of 15 with their cards wins;
• Card Tower Find the rule for the number sequence created while building a tower or bridge of cards;
• Squares Use nine playing cards to form a magic, unmagic, prime or sum square;
• Snap! There are many ways to play snap and include numeracy facts in the rules;
• Times Tables The more ways you can practice your times tables the better. Here are some ideas;
• Greater A really interesting way to explore probability while having fun at the same time;
• More ... There is an ever-growing list of other activities for learning Mathematics with cards;
For Students:
For All:
## Description of Levels
Close
Level 1 - Contents
Level 2 - Long Line
Level 3 - Make 15
Level 4 - Card Tower
Level 5 - Squares
Level 6 - Snap!
Level 7 - Times Tables
Level 8 - Greater
Level 9 - More ...
Close | 299 | 1,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-39 | longest | en | 0.901253 |
http://www.cprogramming.com/tutorial/computersciencetheory/heap.html | 1,485,004,850,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00127-ip-10-171-10-70.ec2.internal.warc.gz | 410,185,247 | 8,535 | Tutorial Map
• Stacks
• Queues
• Heaps ← you are here
• Hash Tables
• Graphs
• Two-three trees
• # What is a heap?
A heap is a partially sorted binary tree. Although a heap is not completely in order, it conforms to a sorting principle: every node has a value less (for the sake of simplicity, we will assume that all orderings are from least to greatest) than either of its children. Additionally, a heap is a "complete tree" -- a complete tree is one in which there are no gaps between leaves. For instance, a tree with a root node that has only one child must have its child as the left node. More precisely, a complete tree is one that has every level filled in before adding a node to the next level, and one that has the nodes in a given level filled in from left to right, with no breaks.
Why use a heap?
A heap can be thought of as a priority queue; the most important node will always be at the top, and when removed, its replacement will be the most important. This can be useful when coding algorithms that require certain things to processed in a complete order, but when you don't want to perform a full sort or need to know anything about the rest of the nodes. For instance, a well-known algorithm for finding the shortest distance between nodes in a graph, Dijkstra's Algorithm, can be optimized by using a priority queue.
Heaps can also be used to sort data. A heap sort is O(nlogn) efficiency, though it is not the fastest possible sorting algorithm. Check out this tutorial heap sort for more information related to heap sort.
How do you implement a heap?
Although the concept of a heap is simple, the actual implementation can appear tricky. How do you remove the root node and still ensure that it is eventually replaced by the correct node? How do you add a new node to a heap and ensure that it is moved into the proper spot?
The answers to these questions are more straight forward than meets the eye, but to understand the process, let's first take a look at two operations that are used for adding and removing nodes from a heap: upheaping and downheaping.
Upheap: The upheap process is used to add a node to a heap. When you upheap a node, you compare its value to its parent node; if its value is less than its parent node, then you switch the two nodes and continue the process. Otherwise the condition is met that the parent node is less than the child node, and so you can stop the process. Once you find a parent node that is less than the node being upheaped, you know that the heap is correct--the node being upheaped is greater than its parent, and its parent is greater than its own parent, all the way up to the root.
Downheap: The downheap process is similar to the upheaping process. When you downheap a node, you compare its value with its two children. If the node is less than both of its children, it remains in place; otherwise, if it is greater than one or both of its children, then you switch it with the child of lowest value, thereby ensuring that of the three nodes being compared, the new parent node is lowest. Of course, you cannot be assured that the node being downheaped is in its proper position -- it may be greater than one or both of its new children; the downheap process must be repeated until the node is less than both of its children.
When you add a new node to a heap, you add it to the rightmost unoccupied leaf on the lowest level. Then you upheap that node until it has reached its proper position. In this way, the heap's order is maintained and the heap remains a complete tree.
Removing the root node from a heap is almost as simple: when you take the node out of the tree, you replace it with "last" node in the tree: the node on the last level and rightmost on that level.
Once the top node has been replaced, you downheap the node that was moved until it reaches its proper position. As usual, the result will be a proper heap, as it will be complete, and even if the node in the last position happens to be the greatest node in the entire heap, it will do no worse than end up back where it started.
Efficiency of a heap
Whenever you work with a heap, most of the time taken by the algorithm will be in upheaping and downheaping. As it happens, the maximum number of levels of a complete tree is log(n)+1, where n is the number of nodes in the tree. Because upheap or downheap moves an element from one level to another, the order of adding to or removing from a heap is O(logn), as you can make switches only log(n) times, or one less time than the number of levels in the tree (consider that a two level tree can have only one switch).
Want more detail? Check out an implementation of a heap | 1,044 | 4,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-04 | longest | en | 0.949422 |
https://sustainability.stackexchange.com/questions/8317/what-is-the-typical-embodied-energy-of-a-solar-photovoltaic-panel?noredirect=1 | 1,638,537,491,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00203.warc.gz | 637,533,593 | 36,177 | # What is the typical embodied energy of a solar photovoltaic panel?
What is the typical embodied energy of a solar photovoltaic panel, in terms of nominal power output or panel area? The ideal answer would be in kWh/WP, but kWh/m2 would also be acceptable.
I understand there is probably a range of answers to this question, so here are some simplifying assumptions:
• If the manufacturing process is more efficient for larger panels, then the answer for large panels (best case scenario) is preferred.
• If the manufacturing process has become more efficient over time, then an answer for the most modern and efficient process (best case scenario) is prefered.
• If different manufacture technologies (monocrystal, polycrystal, thin film) have different amounts of embodied energy, then the lowest answer (best case scenario) is preferred.
• – Nic
Mar 2 '20 at 15:54
• With such a significant positive return on energy investment, what's to stop solar panels taking over the world? Sure, there are not enough of them to meet our energy needs at the moment, but if a large proportion of energy output from solar panels is allocated to making new solar panels, the total number of panels can just keep growing exponentially with surplus electricity to spare. It's not like the world is going to run out of the raw materials (basically silicon & silicon dioxide) within any meaningful time frame. Am I missing something? Oct 14 at 12:58
According to the Wikipedia article on EROI, 585 kWh/m2 is a median value for the embodied energy of a photovoltaic panel, rated based on surface area.
The "energy invested" critically depends on technology, methodology, and system boundary assumptions, resulting in a range from a maximum of 2000 kWh/m² of module area down to a minimum of 300 kWh/m² with a median value of 585 kWh/m² according to a meta-study. [9]
Assuming a panel efficiency of 20% (typical of commercially available panels) and solar irradiance of 1000 W/m2, 1.0 m2 of panel would have a peak power output (WP) of about 200 W/m2. Or stated differently, it would take about 50 cm2 of panel to deliver 1 Watt, and manufacturing that panel would use about 2.9 kWh of energy.
So as a rough back-of-the-napkin calculation, the embodied energy of a solar panel, normalized for peak power output, is about 2.9 kWh/WP.
If the "embodied energy" of photo-voltaic tech is ~585kWh/m^2 and it generates ~0.2 kW/m^2 then; the answer to the question seems to be that the photo-voltaic tech will need to run for:
~(585kWh/m^2)/(0.2kW/m^2) hours = 2925 hours to generate it's embodied energy; yes ?
• Yes, 3000 hours at nameplate capacity (in full sun). Of course, the panel doesn't generate much power at dusk/dark/dawn so estimating the number of days/years for EROI depends on more variables.
– Nic
Feb 13 at 17:48
### Energy payback in 1-6 years
Taking manufacturing variables into account basically gives you a range of time from 1.25 - 6.5 years based on 5 sunny days out of seven and an average of 6-8 hours of sun per day.
But in reality none of this matters, here’s why...
### Fossil fuels never reach energy payback
Electricity as a commodity has been in use for over a century, it isn’t going away in our lifetime. A better question is, “What’s the best way to produce it?”
If you look at the embodied energy of a coal plant, or even a modern high efficiency natural gas generator, how long does it take to earn back their embodied energy? The answer is they never do. We forget, or more accurately ignore the fact that once you build a coal plant, or gas fired turbine, you then have to feed it fuel the rest of its life which it converts to electricity at a rate somewhere less than 100%. So they keep digging themselves a deeper and deeper embodied energy hole they can never crawl out of.
At least the solar panel, water wheel and wind turbine can one day get even. This is why renewable energy as a whole is better for us environmentally and economically than any form of fossil based fuel source.
• I disagree that fossil fuel plants wouldn't pay back the embodied energy. For example, the major oil and gas company Eni is planning that in 2050 the production is 90% natural gas, and emissions are net zero. Yes, that even includes emissions from the use of the produced fuels, not just emissions from extracting the gas. So they are planning to pump back the carbon dioxide back where the natural gas came from. One mole (/ cubic meter) of natural gas produces one mole (/ cubic meter) of carbon dioxide. So if there was room for natural gas under ground, you can fit the carbon dioxide there. Oct 14 at 15:25
• We will need either natural gas or hydrogen for generating electricity when intermittent renewables have zero or low production. There is not enough rainfall in the world to create all adjustable power as hydropower, and no battery is big enough to store energy for use during calm winter periods with no winds for example (and during winter in areas that have real winter, solar production is practically zero). The only competitor to natural gas is hydrogen. We won't know if green hydrogen will ever be cheap enough. Natural gas might be needed. Oct 14 at 15:27 | 1,194 | 5,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-49 | latest | en | 0.930783 |
http://openstudy.com/updates/4e44757e0b8b3609c71fdcfe | 1,448,418,755,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398444139.37/warc/CC-MAIN-20151124205404-00248-ip-10-71-132-137.ec2.internal.warc.gz | 180,538,111 | 9,590 | ## lalakyshani 4 years ago What kind of triangle has vertices (4,-4)(2,1) and (0,8) ? right isosceles right scalene obtuse isosceles obtuse scalene
1. abhisheksingh59
It's an obtuse triangle because it contains an angle over 90°. It's also a scalene, meaning no sides are equal.
2. abhisheksingh59
obtuse scalene | 106 | 316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2015-48 | longest | en | 0.717169 |
http://www.wikihow.com/Convert-Roman-Numerals | 1,506,256,512,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690016.68/warc/CC-MAIN-20170924115333-20170924135333-00307.warc.gz | 609,198,681 | 44,330 | # wikiHow to Convert Roman Numerals
Roman numerals is a system of numbers which uses the letters of the alphabet.[1] It is still occasionally used today, especially when representing years.
## Steps
1. 1
Know which letters stand for which numbers.
• I is 1, V is 5.
• X is 10, L is 50.
• C is 100, D is 500.
• M is 1000.
2. 2
Locate the largest symbol within the number. This should be either the first or the second symbol.
3. 3
If the largest symbol appears first, count how many times it appears in an uninterrupted row. This should be between one and three times. Multiply its value by the number of appearances and add this amount to the total.
4. 4
If the largest symbol appears second, subtract the value of the symbol before it from its value. Add this amount to the total.
5. 5
Repeat steps 2-4 with the numbers you have not yet accounted for.
6. 6
Add up the values of each set of numbers to find the total value of the entire numeral.
## Community Q&A
Search
• How do I convert MCMLXXVI?
Work from left to right: M is 1,000. CM is 900. LXX is 70. VI is 6. That's 1,976.
200 characters left
## Tips
• In modern Roman numerals, the following rules apply:
• Only I, X, C and M may be repeated, and each no more than three times in sequence.
• A large symbol can only require one smaller symbol to be subtracted from it at a time.
• Only I, X and C may be subtracted, and each only from values up to and including the next highest of the repeatable symbols.
• Older styles of writing Roman numerals may break the above rules.
## Article Info
Categories: Mathematics | Conversion Aids
Thanks to all authors for creating a page that has been read 7,853 times. | 434 | 1,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-39 | latest | en | 0.911153 |
https://www.physicsforums.com/threads/stoichiometry-help.66337/ | 1,508,667,807,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00357.warc.gz | 963,954,630 | 18,857 | # Stoichiometry help
1. Mar 7, 2005
### DB
A little stoichiometry...
Here's all that was given:
A container of HCl spilled on a marble table surface. The marble contains calcium carbonate, which reacts with the acid to produce calcium chloride, carbon dioxide and water.
a) If 500ml of 0.2 M acid spilled, what mass of calcium carbonate reacted?
So I wrote out the (combustion is it?) equation.
$$CaCO_3 + 2HCl\rightarrow CaCl_2 + CO_2 + H_2O$$
Then molar mass:
$$100g(CaCO_3) + 72g(2HCl)\rightarrow 110g(CaCl_2) + 44g(CO_2) +18g( H_2O)$$
So now I know that 72g of HCl is 2 mol (since 2HCl) so 0.2 mol = 7.2g
But know I'm stuck...a push in the right direction???
Also.
b)How many moles of $CO_2$ were formed during this reaction?
1 mole
c)How many molecules of water were formed during this reaction?
~6.022*10^23
I don't think there's any problems there.
Thanks in advance
2. Mar 7, 2005
### dextercioby
Yes,there are problems everywhere.So the first.U've computed 0.2 mols of HCl.How many mols of CaCO_{3} will react with it...?
Daniel.
3. Mar 7, 2005
### saltydog
According to the balanced equation, for every mole of HCl reacting, one-half of a mole of $CaCO_3$ reacts. So how many moles of HCl are in 500ml of a 0.2 molar solution? Suppose that's x number of moles. Now, from what I said above, one-half of that number of moles is the number of moles of $CaCO_3$ reacting. Same dif for $CO_2$ right (half as much) and water too for that matter.
Now, isn't there Avagadro's number of molecules in one mole? So using the infor above (the number of moles of water produced), figure out how many molecules of water.
4. Mar 7, 2005
### Gokul43201
Staff Emeritus
DB,
0.2 M does not mean 0.2 moles. It refers to a solution of strength 0.2 moles per liter. So, how many moles will there be in 500 mL of this solution ?
5. Mar 7, 2005
### dextercioby
Ooooooops,didn't see the big M,sorry.
Daniel.
6. Mar 7, 2005
### The Bob
Lets start here. You have the concentration and the volume. You can use these to see what the amount of moles is.
The equation is: $$Concentration (mol dm^{-3}) = \frac{Moles (moles)}{Volume (dm^3)}$$
Use that to find the number of moles of acid that have been spilled. You can then use this to find the mass: $$Moles (moles) = \frac{Mass (grams)}{Molar Mass (g mol^{-1})}$$
You then have the mass of the HCl. Use this you can find the mass of CaCl2 in the normal way, find ratios and molar masses and see what is produced.
Get this and then I will help with the rest.
The Bob (2004 ©)
7. Mar 7, 2005
### DB
Ok so, 0.1 moles reacts with $CaCO_3$ in a ratio of 2:1? So, half of 0.1 mol is 0.05 mole which is 5g of $CaCO_3$
for b and c, i think you told me that the answers are 0.5 mol and $N_A/2$, but i dont really understand why I'm dividing by 2...
8. Mar 7, 2005
### The Bob
I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl2 is 111 g mol-1.
The Bob (2004 ©)
9. Mar 7, 2005
### saltydog
Ok.
No. I think you mean 0.05 moles right? If 0.1 moles of HCl are reacting and the ratio is 2/1 then that means 0.05 moles of water and carbon dioxide are produced right? And if one mole has Avagadros number then one half of a mole would have one half of Avagadros number, one tenth of a mole would have one-tenth of Avagadros number and so if you have 0.05 moles of water then that means you have 0.05 of Avagadros's number of molecules right?
10. Mar 7, 2005
### DB
"I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl2 is 111 g mol-1.
The Bob (2004 ©)"
calium chloride? for a? the question is asking calcium carbonate
11. Mar 7, 2005
### The Bob
I apologise. In that case I get 50g of Calcium Carbonate.
The Bob (2004 ©)
12. Mar 7, 2005
### DB
Naww lol i meant 0.5. im confused. 0.1 moles of acid are reacting we know that.
So then 0.1 mole of both CO2 and water are...? Im sorry if im being anoying im just really confused.
13. Mar 7, 2005
### DB
no prob, just i got 5 g and saltydog said it was ok...
14. Mar 7, 2005
### The Bob
Here is what I did.
The amount of moles of the spilled Hydrochloric Acid is 0.2 mol dm-3 x 0.5 dm3 = 0.4 moles.
Molar Mas of HCl = 36.5 g mol-1
0.4 moles x 36.5 g mol-1 = 14.6g (now proven)
The equation you need to look at is the $$2HCl + CaCO_3$$
HCl = 35.6 g mol-1 x 2 moles = 73g
CaCO3 = 100 g mol-1 x 1 mol = 100g
Therefore HCl:CaCO3 = 73g:100g
So 1g of HCl : 1.37g of CaCO3
Therefore 14.6g of HCl : 1.37 x 14.6 = 20g of CaCO3
Ok. I have changed my mind because my original answer was wrong because I get the molar mass of HCl wrong (:rofl:) but I am now sure it is 20g.
The Bob (2004 ©)
15. Mar 7, 2005
### The Bob
The moles of Carbon Dioxide = 0.2 moles.
The moles of Water = 0.2 moles as well. I don't know about how many molecules but I personally think it might be the same.
The Bob (2004 ©)
Last edited: Mar 7, 2005
16. Mar 7, 2005
### DB
Thanks bob, i see the work u've done, i just cant understand how u got 0.4 moles of acid spilled, i always get 0.1....can u explain plz
17. Mar 7, 2005
### The Bob
$$Concentration (mol dm^{-3}) = \frac{Moles (mol)}{Volume (dm^3)}$$
M (or the molarity) is equal to the same number in mol dm-3 e.g. 0.2M = 0.2 mol dm-3.
Also the equation needs the amount in dm3. You have it in ml. This can be changed straight away into cm3 e.g. 300 ml = 300 cm3. Then you need it in dm3. As it is a cubic factor then you need to change the number by a cubic number. Let me explain this better.
1 dm = 10 cm. This is worked out by multiplying the dm by 10 or by dividing the cm by 10.
1 dm2 = 100 cm2. This time it is worked out by multiplying the dm by 102 (100) or by dividing the cm by 102 (100).
Therefore 1 dm3 = 1000 cm3 multiplying/dividing by 103 (1000). e.g. 300 ml = 300 cm3 = 0.3 dm3
The amount of moles of the spilled Hydrochloric Acid is 0.2M x 500 ml = 0.2 mol dm-3 x 500 cm3 = 0.2 mol dm-3 x 0.5 dm3 = 0.1 moles. (Oh man, I have done it again. I need to learn to check what I am doing. Sorry again ).
The Bob (2004 ©)
18. Mar 7, 2005
### The Bob
I get 5g as well . Sorry everyone. I has been good practise for me though and I can see my mistakes clearer and also you can see my method clearly and rather than jumping to conclusion it can be proven (although jumping seemed to work better).
The Bob (2004 ©)
19. Mar 7, 2005
### The Bob
It is now:
The moles of Carbon Dioxide = 0.05 moles.
The moles of Water = 0.05 moles.
The Bob (2004 ©)
P.S. Sorry again but I need to sleep now. I am really quite tired. Hope what I did helped in some way or another.
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## Critical Reasoning – The Assumption Question 1
The Assumption Question is a Critical Reasoning Question Type on the GMAT® that gives test-takers a certain amount of trouble (the most troublesome being the Boldfaced Question Type). Test-takers often say that they have trouble in attaining a certain level of consistency on this question type. In this post we shall look at a standard operating procedure that will help you increase your accuracy level and choose the right option when faced with two seemingly correct options.
## GMAT Quantitative – Approximation Problems
On the GMAT it is very likely that test-takers will encounter problems that involve pure approximation. The key to solving these problems is to be aware of two things: A. The answer need not be calculated precisely B. Eliminating incorrect answer options might be the best solution
## Sentence Correction – Usage 1
This is a common and seemingly easily resolvable grammatical conundrum — when does one use which and when does one use that? Most test-takers who have prepared for GMAT® Sentence Correction will have this answer at the tip of their tongues — essential/restrictive and non-essential/non-restrictive clause. What they mean is that which is used to state information that is not essential while that is used to state essential information. A easier way to remember this is by looking at the pair of sentences below: 1. These are the keys to the fourth car in the parking row, which is black. 2. These are the keys to the fourth car in the parking row that is black. From the first sentence you would get the keys to the fourth car in the parking row; the sentence gives you some additional information, namely that it is black in colour; even without this information you could have known which car you have the keys to — the fourth car in the row.
## Critical Reasoning – The Conclusion Question 2
In the previous Critical Reasoning post we discussed one specific kind of logic that is tested on the Conclusion Question Type. In this post we will take a look at the only other type of standard logic tested on the conclusion questions. If you are able to understand apply the technique to solve these two logical structures, most conclusion questions should be a breeze. Let us take a GMAT Critical Reasoning question to examine this further. Although aspirin has been proven to eliminate moderate fever associated with some illnesses, many doctors no longer routinely recommend its use for this purpose. A moderate fever stimulates the activity of the body’s disease-fighting white blood cells and also inhibits the growth of many strains of disease-causing bacteria.
## Will there be experimental questions on the GMAT?
All life is an experiment, the more experiments you make the better – Ralph Waldo Emerson The short answer to this question — yes. Now let’s get to the long answer. As we have discussed before in one of the earliest posts on this blog, the GMAT is an adaptive test. The Quant and Verbal sections will start with test-takers being posed a question of moderate difficulty and will proceed based on the test-taker’s response to that question. Depending upon whether the test-taker answers it correctly or incorrectly, the subsequent question will be easier or tougher. | 670 | 3,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.950601 |
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Correlation Coefficients of Extended Hesitant Fuzzy Sets and Their Applications to Decision Making
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Article
# Dual Hesitant Fuzzy Probability
by
Jianjian Chen
* and
Xianjiu Huang
*
Department of Mathematics, Nanchang University, Nanchang 330031, China
*
Authors to whom correspondence should be addressed.
Symmetry 2017, 9(4), 52; https://doi.org/10.3390/sym9040052
Submission received: 3 January 2017 / Revised: 30 March 2017 / Accepted: 1 April 2017 / Published: 7 April 2017
## Abstract
:
Intuitionistic fuzzy probabilities are an extension of the concept of probabilities with application in several practical problem solving tasks. The former are probabilities represented through intuitionistic fuzzy numbers, to indicate the uncertainty of the membership and nonmembership degrees in the value assigned to probabilities. Moreover, a dual hesitant fuzzy set (DHFS) is an extension of an intuitionistic fuzzy set, and its membership degrees and nonmembership degrees are represented by two sets of possible values; this new theory of fuzzy sets is known today as dual hesitant fuzzy set theory. This work will extend the notion of dual hesitant fuzzy probabilities by representing probabilities through the dual hesitant fuzzy numbers, in the sense of Zhu et al., instead of intuitionistic fuzzy numbers. We also give the concept of dual hesitant fuzzy probability, based on which we provide some main results including the properties of dual hesitant fuzzy probability, dual hesitant fuzzy conditional probability, and dual hesitant fuzzy total probability.
MSC:
03E72; 03E75
## 1. Introduction
After the introduction of the concept of fuzzy sets by Zadeh in [1], several surveys were conducted on possible extensions of the concept of fuzzy set. Among these extensions, one that has attracted the attention of much research in recent decades is dual hesitant fuzzy set theory, introduced by Zhu et al. [2]. In 2011, Xu and Xia [3] defined the concept of dual hesitant fuzzy element (DHFE), which can be considered as the basic unit of a DHFS, and is also a simple and effective tool used to express the decision makers’ hesitant preferences in the process of decision making. Since then, many scholars [4,5,6,7,8,9] have conducted much research work on aggregation, distance measures, correlation coefficient, and decision making with dual hesitant fuzzy information.
On the other hand, probability theory is an old uncertainty method which is appropriate to deal with another kind of uncertainty. However, since probabilities consider an absolute knowledge, and in many real situations this knowledge is partially known or uncertain, there were several ways to extend the notion of probabilities to deal with such situations. An original conception of connecting fuzzy set theory with probability theory was first introduced by Hirota and Wright [10]. For James Buckley [11], the probabilities of events, in practice, should be known exactly, however, these values are very often estimated or provided by experts, and therefore are of vague nature. He modeled this vagueness using fuzzy numbers. This approach has been applied in several subjects (see [11,12]), such as testing for HIV, and other blood type problems. Besides, according to fuzzy probability, James Buckley [11] also proposed fuzzy Markov chains, and joint fuzzy probability distributions, etc., then applied them to the fuzzy queuing decision problem, machine servicing problem, fuzzy decisions under risk and fuzzy reliability theory. In 2010, Costa et al. [13] introduced a generalization of the concept of fuzzy probabilities by using an original notion of intuitionistic fuzzy numbers instead of usual fuzzy numbers and gave some notions about intuitionistic fuzzy probabilities. In 2013, Costa et al. [14] investigated Atanassov’s intuitionistic fuzzy probability and Markov chains. Rashid et al. [15] introduced a convex hesitant fuzzy set and a quasi-convex hesitant fuzzy set with an example and investigated aggregation functions for convex hesitant fuzzy sets, which are available in the optimization problem.
As discussed above, a DHFS has its own desirable characteristics and advantages and appears to be a more flexible method—which is to be valued in multifold ways due to the practical demands—than the existing fuzzy sets, taking into account much more information given by decision makers, and at the same time, fuzzy probability and intuitionistic fuzzy probability play a crucial role in applications [11]; we urgently need to put forward dual hesitant fuzzy probability to satisfy the same problems. In this paper, we will introduce probabilities to a DHFS by using an original notion of dual hesitant fuzzy numbers instead of usual fuzzy numbers. Thus, we give an original generalization, in the context of a DHFS, to the intuitionistic fuzzy probability approach of Costa et al. in [13]; thus, we introduce a theory to deal with probabilities in a framework where it does not only model uncertainty in the probability of some events but also models the dual hesitation which is naturally present in the uncertainty.
Motivated by earlier research work, the remainder of the paper is organized as follows: In Section 2, some basic concepts of dual hesitant fuzzy sets are presented. In Section 3, we give some convexity for DHFSs, which contains a convex dual hesitant fuzzy set with respect to $( α , β ) − c u t s$ and a quasi-convex dual hesitant fuzzy set; we also introduce intuitionistic fuzzy numbers that are fundamental to the paper. In Section 4, we provide the main definitions and results of this work. Section 5 introduces a dual hesitant fuzzy extension of (fuzzy intuitionistic) conditional probability and it is proven that this extension satisfies more analogous properties than conditional probabilities. In Section 6, an example of the application of color blindness is given to show the actual need of dual hesitant fuzzy probability. A conclusion to the paper and further topics are given in Section 7.
## 2. Some Basic Concepts about DHFS
Zhu et al. [2] defined a DHFS—which is an extension of the hesitant fuzzy set, in terms of two functions that return two sets of membership values and nonmembership values, respectively, for each element in the domain—as follows.
Definition 1.
[2] Let X be a fixed set, then a DHFS D on X is defined as:
$D = { < x , h ( x ) , g ( x ) > | x ∈ X } ,$
where $h ( x )$ and $g ( x )$ are two sets of some values in [0, 1], denoting the possible membership degrees and nonmembership degrees of the element $x ∈ X$ to the set D, respectively, with the conditions: $0 ≤ γ , η ≤ 1$ and $0 ≤ γ + + η + ≤ 1$, where $γ ∈ h ( x )$, $η ∈ g ( x )$, $γ + ∈ h + ( x ) = ∪ γ ∈ h ( x ) m a x { γ }$, and $η + ∈ g + ( x ) = ∪ η ∈ g ( x ) m a x { η }$ for $x ∈ X$. For convenience, the pair $e ( x ) = { < h ( x ) , g ( x ) > }$ is called a DHFE denoted by all $e = { < h , g > }$.
The support of the DHFS D is the crisp set $S u p p ( D ) = { t ∈ X : m a x$ $h ( t ) ≠ 0$ and max $g ( t ) ≠ 1 } ⊆ X$, whereas the kernel of the DHFS D is the crisp set $K e r ( D ) = { t ∈ X : m a x$ $h ( t ) = 1$ and $m a x$ $g ( t ) = 0 } ⊆ X$. The DHFS D is said to be normal provided that it has a non-empty kernel.
From the above definition, we can see that it consists of two parts, that is, the membership hesitancy function and the nonmembership hesitancy function; this supports more exemplary and flexible access to assign values for each element in the domain, and two kinds of hesitancy can be handled in this situation. The existing sets, including fuzzy sets, intuitionistic fuzzy sets, hesitant fuzzy sets, and fuzzy multisets, can be regarded as special cases of DHFSs.
Example 1.
Let D be a dual hesitant fuzzy set in $X = { x 1 , x 2 , x 3 }$, and we give three dual hesitant fuzzy elements to a set D as follows:
• $h D ( x 1 ) = { { 0 . 7 , 0 . 5 , 0 . 3 } , { 0 . 2 , 0 . 1 } }$,
• $h D ( x 2 ) = { { 0 . 4 , 0 . 3 } , { 0 . 2 , 0 . 4 } }$,
• $h D ( x 3 ) = { { 0 . 6 , 0 . 3 } , { 0 . 3 , 0 . 1 , 0 . 2 } }$,
then $D = { < x 1 , { 0 . 7 , 0 . 5 , 0 . 3 } , { 0 . 2 , 0 . 1 } > ,$ $< x 2 , { 0 . 4 , 0 . 3 } , { 0 . 2 , 0 . 4 } > ,$ $< x 3 , { 0 . 6 , 0 . 3 } , { 0 . 3 , 0 . 1 , 0 . 2 } }$.
The following operations are also given by Zhu et al. [2], for any DHFEs, d, $d 1$ and $d 2$,
(1)
$⊕ − u n i o n$: $d 1 ⊕ d 2 =$ $∪ γ d 1 ∈ h d 1 , γ d 2 ∈ h d 2$ ${ { γ d 1 + γ d 2 − γ d 1 γ d 2 } , { η d 1 η d 2 } }$;
(2)
$⊗ − i n t e r s e c t i o n$: $d 1 ⊗ d 2 =$ $∪ γ d 1 ∈ h d 1 , γ d 2 ∈ h d 2$ ${ { γ d 1 γ d 2 } , { η d 1 + η d 2 − η d 1 η d 2 } }$;
(3)
$n d = ∪ γ d ∈ h d , η d ∈ g d { 1 − ( 1 − γ d ) n , ( η d ) n }$;
(4)
$d n = ∪ γ d ∈ h d , η d ∈ g d { ( γ d ) n , 1 − ( 1 − η d ) n }$, where n is a positive integral and all the results are also DHFEs.
Definition 2.
[2] Let $d i = { h d i , g d i }$ be any two DHFEs, $s ( d i ) = 1 ♯ h ∑ γ ∈ h γ − 1 ♯ g ∑ η ∈ g η ( i = 1 , 2 )$ the score function of $d i ( i = 1 , 2 )$, and $p ( d i ) = 1 ♯ h ∑ γ ∈ h γ + 1 ♯ g$ $∑ η ∈ g η ( i = 1 , 2 ) ( i = 1 , 2 )$ the accuracy function of $d i ( i = 1 , 2 )$, where $♯ h$ and $♯ g$ are the numbers of the elements in h and g, respectively, then
(i)
if $s ( d 1 ) > s ( d 2 )$, then $d 1$ is superior to $d 2$, denoted by $d 1 ≻ d 2$;
(ii)
if $s ( d 1 ) = s ( d 2 )$, then
(1)
if $p ( d 1 ) = p ( d 2 )$, then $d 1$ is equivalent to $d 2$, denoted by $d 1 ∼ d 2$;
(2)
if $p ( d 1 ) > p ( d 2 )$, then $d 2$ is superior to $d 1$, denoted by $d 2 ≻ d 1$.
In order to give the level set of a DHFS, we give the following notion: Let $d = { h , g }$ be a DHFE. Then, $s ( h ) = 1 ♯ h ∑ γ ∈ h γ$ and $s ( g ) = 1 ♯ g ∑ η ∈ g η$ are the score function of the membership degrees and nonmembership degrees of DHFEs, respectively, where $♯ h$ and $♯ g$ are the numbers of elements in h and g, respectively.
## 3. Dual Hesitant Fuzzy Numbers
In this section, we carry out a brief introduction to some kinds of convexity for dual hesitant fuzzy sets, based on which we give some concepts about dual hesitant fuzzy numbers for a better understanding of the main body of the paper. Next, let us start by recalling a unit triangular lattice.
Definition 3.
[16,17] Let $T = { ( x 1 , x 2 ) ∈ [ 0 , 1 ] × [ 0 , 1 ] : x 1 + x 2 ≤ 1 } .$ Define on T a binary relation $≤ T$ given by $( x 1 , x 2 ) ≤ T ( y 1 , y 2 )$ provided that $x 1 ≤ y 1$ and $y 2 ≤ x 2$. Then, $( T , ≤ T , 0 T , 1 T )$ is a bounded complete lattice, where $0 T$ stands for (0,1) and $1 T$ stands for (1, 0). This new lattice T is called the unit triangular lattice.
The unit triangular lattice T has been proven to be a helpful tool in the theory of representability of several ordered structures as total preorders, interval orders and semiorders (see [16,17] for further details).
Given a DHFS D, its supremum and infimum, with respect to $≤ T$, could be obtained as follows:
$s u p D = ( s u p { s ( h 1 ( x ) ) | x ∈ D } , i n f { s ( g 2 ( x ) ) | x ∈ X } ) ,$
and
$i n f D = ( i n f { s ( h 1 ( x ) ) | x ∈ D } , s u p { s ( g 2 ( x ) ) | x ∈ X } ) .$
Given $( α , β ) ∈ T$ and a DHFS D of universe X, the $( α , β ) − c u t$ of D is the set:
$D ( α , β ) = { x ∈ X | s ( h D ( x ) ) ≥ α a n d s ( g D ( x ) ) ≤ β } .$
Notice that $D ( α , β ) = ( D α , D β )$, where $D α = { x ∈ X | s ( h D ( x ) ) ≥ α }$ and $D β = { x ∈ X | s ( g D ( x ) ) ≤ β }$. On the other hand, every DHFS can be recovered from its $( α , β ) − c u t s$. In fact, $( h D ( x ) , g D ( x ) ) = s u p { ( α , β ) ∈ T | x ∈ A ( α , β ) }$, where the supremum is with respect to $≤ T$.
Definition 4.
Let D be a dual hesitant fuzzy set with universe R. D is convex if its cuts $D ( α , β )$ are convex subsets of X for all $α , β ∈ ( 0 , 1 ]$.
An equivalent statement is shown as follows:
Definition 5.
Let X be a vector space. A dual hesitant fuzzy set D of the universe X is said to be quasi-convex, if for all $x , y ∈ X$ and $λ ∈ [ 0 , 1 ]$ it holds that $( m i n { s ( h D ( x ) ) , s ( h D ( y ) ) } ,$ $m a x { s ( g D ( x ) ) , s ( g D ( y ) ) } )$ $≤ T$ $( s ( h D ( λ x + ( 1 − λ ) y ) ,$ $s ( g D ( λ x + ( 1 − λ ) y ) )$.
Lemma 1.
If $D = { < x , h D ( x ) , g D ( x ) > }$ is a DHFS and we denote by $D 1$ the hesitant fuzzy set defined by means of the membership function of D $( i . e . , D 1 = h D )$, then $D ( α , β ) = ( D 1 ) α$ for any $α ∈ ( 0 , 1 ]$. This means that $D ( α , β ) = { x ∈ X : h D ( x ) ≥ α }$.
Proof.
It is trivial that $D α , β ⊆ ( D 1 ) α$. Let $x ∈ ( D 1 ) α$, then $s ( h D ( x ) ) ≥ α$. Since D is a DHFS, we have that $m a x { h D ( x ) } + m a x { g D ( x ) } ≤ 1$, then $s ( g D ( x ) ) ≤ 1 − m a x$ $s ( h D ( x ) ) ≤ 1 − α$ and therefore $s ( g D ( x ) ) ≤ β ( β = 1 − α )$. Then, $x ∈ D ( α , β )$. ☐
Remark 1.
$( D 1 ) α$ and $α − c u t$ of the hesitant fuzzy set in Lemma 1 is given in [16].
Theorem 1.
Let X be a universe. The following statements are equivalent:
(1)
D is a quasi-convex DHFS;
(2)
any $( α , β ) − c u t s$ of DHFS D are convex crisp sets, for any $α , β ∈ ( 0 , 1 ]$;
(3)
$h D$ is the membership function of a convex hesitant fuzzy set with respect to $α − c u t s$ [16].
Proof.
$( 1 ) ⇔ ( 2 )$: To prove the direct implication, given $( α , β ) ∈ L$, for all $x , y ∈ X$ and $λ ∈ [ 0 , 1 ]$ such that $( α , β ) ≤ T$ $( m i n { s ( h D ( x ) ) , s ( h D ( y ) ) }$ $, m a x { s ( g D ( x ) ) , s ( g D ( y ) ) } )$, if D is a quasi-convex DHFS, we have that $( α , β ) ≤ T ( m i n { s ( h D ( x ) ) , s ( h D ( y ) ) }$ $, m a x { s ( g D ( x ) ) , s ( g D ( y ) ) } )$ $≤ T ( s ( h D ( λ x + ( 1 − λ ) y ) , s ( g D ( λ x + ( 1 − λ ) y )$. Then, any $( α , β ) − c u t s$ of a DHFS D are convex crisp sets, for any $α ∈ ( 0 , 1 ]$.
On the contrary, given $x , y ∈ X$ and $( α , β ) ∈ L$, if we call $α = m i n { s ( h D ( x ) ) , s ( h D ( y ) ) }$ and $β = m i n { s ( g D ( x ) ) , s ( g D ( y ) ) }$, it is clear that $x , y$ belong to the $α − c u t$ of the dual hesitant fuzzy set defined by $D ( α , β )$. By hypothesis, $λ x + ( 1 − λ ) y$ also lies in this $( α , β ) − c u t$, so that $( α , β ) ≤ T ( m i n { s ( h D ( x ) ) , s ( h D ( y ) ) }$ $, m i n { s ( g D ( x ) ) , s ( g D ( y ) ) } )$ $≤ T ( s ( h D ( λ x + ( 1 − λ ) y ) )$ $, s ( g D ( λ x + ( 1 − λ ) y ) ) )$. Thus, $( m i n { s ( h D ( x ) ) , s ( h D ( y ) ) }$ $, m a x { s ( g D ( x ) ) , s ( g D ( y ) ) } )$ $≤ T$ $( s ( h D ( λ x + ( 1 − λ ) y ) ,$ $s ( g D ( λ x + ( 1 − λ ) y ) )$.
$( 2 ) ⇔ ( 3 )$: Based on Lemma 1, it is easy to prove that it holds. ☐
Definition 6.
Let D be a DHFS with universe R. D is continuous if $s ( h D ( x ) )$ and $s ( g D ( x ) )$ are continuous; D is normalized if there exists $x ∈ R$ such that $( s ( h D ( x ) ) , s ( g D ( x ) ) ) = 1 T$.
Thus, for the convex DHFS, its $( α , β ) − c u t s$ are closed intervals of real numbers, and so it is possible to represent them via their end points, which can be obtained as follows:
$l ( D ( α , β ) ) = m a x ( m i n s ( h D ) ( α ) − 1 , m i n s ( g D ) ( β ) − 1 ) ,$
and
$r ( D ( α , β ) ) = m i n ( m a x s ( h D ) ( α ) − 1 , m a x s ( g D ) ( β ) − 1 ) ,$
that is, $D ( α , β ) = [ l ( D ( α , β ) ) , r ( D ( α , β ) ) ]$. Where $s ( h D ) ( α ) − 1 = { x | s ( h D ( x ) ) ≥ α }$ and $s ( g D ) ( α ) − 1 = { x | s ( g D ( x ) ) ≤ β }$.
Definition 7.
A DHFS D with the real universe is a dual hesitant fuzzy number (DHFN) if it is convex, normalized and continuous. Denote by ℑ the set of all DHFN.
Important subclasses of a DHFN are those in which $s ( h D )$, as well as its complement $s ( g D )$, have a triangular shape, that is, they are triangular fuzzy numbers in the usual sense; then, we call the corresponding dual hesitant fuzzy number a dual hesitant triangular fuzzy number (DHTFN). Thus, as can be seen in Figure 1, the DHTFN is completely determined by the score functions of $s ( h D )$ and $s ( g D )$ in Figure 1. We can denote it by $( S a , S b / S c / S d , S e )$; one advantage of the DHTFN is that its $( α , β ) − c u t$ can be expressed as follows:
$( S a , S b / S c / S d , S e ) ( α , β ) = [ m a x ( S a + ( S c − S a ) α , S b + ( S c − S b ) β ) , m i n ( S e + ( S e − S c ) α , S d + ( S d − S c ) β ) ] .$
Remark 2.
If the DHFS reduces to a fuzzy set or an intuitionistic fuzzy set, then the DHFN reduces to a fuzzy number or an intuitionistic fuzzy number.
Let A and B be two DHFNs. Then, define the addition, subtraction, multiplication and division of A with B from the corresponding interval arithmetic operations on their $( α , β ) − c u t s$. Let $( α , β ) ∈ L$,
$( A + B ) α = { x + y | x ∈ A ( α , β ) a n d y ∈ B ( α , β ) } = [ l ( A ( α , β ) ) + l ( B ( α , β ) ) , r ( A ( α , β ) ) + r ( B ( α , β ) ) ] ,$
$( A − B ) α = { x − y | x ∈ A ( α , β ) a n d y ∈ B ( α , β ) } = [ l ( A ( α , β ) ) − r ( B ( α , β ) ) , r ( A ( α , β ) ) − l ( B ( α , β ) ) ] ,$
$( A · B ) α = { x · y | x ∈ A ( α , β ) a n d y ∈ B ( α , β ) } = [ m i n S , m a x S ] ,$
$( A / B ) α = { x / y | x ∈ A ( α , β ) a n d y ∈ B ( α , β ) } = [ m i n T , m a x T ] ,$
where $S = { l ( A ( α , β ) ) · l ( B ( α , β ) ) , l ( A ( α , β ) ) · r ( B ( α , β ) ) ,$ $r ( A ( α , β ) ) · l ( B ( α , β ) ) , r ( A ( α , β ) ) · r ( B ( α , β ) ) }$ and $T = { l ( A ( α , β ) ) / l ( B ( α , β ) ) , l ( A ( α , β ) ) / r ( B ( α , β ) ) ,$ $r ( A ( α , β ) ) / l ( B ( α , β ) ) , r ( A ( α , β ) ) / r ( B ( α , β ) ) }$.
Notice that this method is equivalent to the extension principle method of fuzzy arithmetic [12] and intuitionistic fuzzy arithmetic [13]. Obviously, this procedure is more user friendly.
Theorem 2.
Let A and B be two DHFNs. Then, $A + B$, $A − B$, $A · B$ and $A / B ( 0 ∉ B )$ are DHFNs.
Proof. Straightforward.
It is possible to establish several orders on intuitionistic fuzzy numbers [11] which could be extended to DHFNs. Here, we give the following order on DHFNs:
$A ⪯ B ⇔ A ( α , β ) ⊴ B ( α , β ) f o r a l l ( α , β ) ∈ L ,$
where,
$A ( α , β ) ⊴ B ( α , β ) ⇔ r ( A ( α , β ) ) < r ( B ( α , β ) ) o r r ( A ( α , β ) ) = r ( B ( α , β ) ) a n d l ( B ( α , β ) ) ≤ l ( A ( α , β ) ) .$
A DHFN D is strictly positive if for each $( α , β ) − c u t$, $D ( α , β ) > 0$. Then, we have
Lemma 2.
D is strictly positive if $D ⪰ 0$.
Proof. Straightforward.
Lemma 3.
If A and B are strictly positive DHFNs, then
$( A · B ) ( α , β ) = [ l ( A ( α , β ) ) · l ( B ( α , β ) ) , r ( A ( α , β ) ) · r ( B ( α , β ) ) ] = A ( α , β ) · B ( α , β ) ;$
and
$( A / B ) ( α , β ) = [ l ( A ( α , β ) ) / r ( B ( α , β ) ) , r ( A ( α , β ) ) / l ( B ( α , β ) ) ] = A ( α , β ) / B ( α , β ) .$
Proof. Straightforward.
Theorem 3.
Let A, B and C be three strictly positive DHFNs. Then,
(i)
If $A ⪯ B$ then $A C ⪯ B C$;
(ii)
If $B ⊆ C$ then $A B ⊆ A C$;
(iii)
$B ⊆ A · B A$.
Proof.
Let $( α , β ) ∈ T$,
(i)
If $A ⪯ B$, we have $r ( A ( α , β ) ) < r ( B ( α , β ) ) o r r ( A ( α , β ) ) = r ( B ( α , β ) ) a n d l ( B ( α , β ) ) < l ( A ( α , β ) )$, thus $r ( A ( α , β ) ) l ( C ( α , β ) ) < r ( B ( α , β ) ) l ( C ( α , β ) )$ or $r ( A ( α , β ) ) l ( C ( α , β ) ) = r ( B ( α , β ) ) l ( C ( α , β ) )$ and $l ( A ( α , β ) ) r ( C ( α , β ) ) ≤ l ( B ( α , β ) ) r ( C ( α , β ) )$. Then $A C ⪯ B C$;
(ii)
If $B ⊆ C$, we have $[ l ( B ( α , β ) ) , r ( B ( α , β ) ) ]$ $⊆ [ l ( C ( α , β ) ) , r ( C ( α , β ) ) ]$, thus $[ l ( A ( α , β ) ) r ( B ( α , β ) ) , r ( A ( α , β ) ) l ( B ( α , β ) ) ]$ $⊆ [ l ( A ( α , β ) ) r ( C ( α , β ) ) , r ( A ( α , β ) ) l ( C ( α , β ) ) ]$, namely, $A B ⊆ A C$;
(iii)
As $[ l ( B ( α , β ) ) , r ( B ( α , β ) ) ]$ $⊆ [ l ( A ( α , β ) ) l ( B ( α , β ) ) r ( A ( α , β ) ) , r ( A ( α , β ) ) r ( B ( α , β ) ) l ( A ( α , β ) ) ]$, namely, $B ⊆ A · B A$.
☐
## 4. Dual Hesitant Fuzzy Probability
Let $X = { x 1 , . . . , x 2 }$ be a finite set. Let be the set of all strictly positive DHFN and let $d ¯ i ∈ ℑ$, $i = 1 , . . . , n ,$ with $d ¯ i ⪯ 1 ¯$, for all $i = 1 , . . . , n$. A family of fixed DHFNs such that there are $d ¯ i ∈ ( d ¯ i ) ( α , β )$ with $∑ i = 1 n d ¯ i = 1$. For each $( α , β ) ∈ T$, denote $( d ¯ 1 ) ( α , β ) × . . . × ( d ¯ n ) ( α , β )$ by $S α β$. Define also, for each $A ⊆ X$ and DHFS $( α , β ) ∈ L$, the following set:
$( P ¯ ( A ) ) ( α , β ) = { ∑ i ∈ I A d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ i = 1 n m a x i d i = 1 } ,$
where $I A$ is the set of indexes of A, that is, $I A = { i ∈ { 1 , . . . , n } | x i ∈ X }$. Conventionally, $∑ i ∈ ⌀ d ¯ i = 0$.
Theorem 4.
Let $A ⊆ X$, $( P ¯ ( A ) ) ( α , β )$ are the $( α , β ) − c u t s$ of a DHFN, $P ¯ ( A )$, that is, $P ¯ : ℘ ( X ) → ℑ$, where $℘ ( X )$ is the powerset of X.
Proof.
Let $S = { y 1 , y 2 , . . . , y n ) ∈ [ 0 , 1 ] n a n d ∑ i = 1 n m a x i y i = 1 } .$ For each $( α , β ) ∈ L$, define $D o m [ α , β ] = S ⋂ ∏ i = 1 n ( d ¯ i ) ( α , β )$ and $f : D o m [ α , β ] → [ 0 , 1 ]$ by:
$f ( d 1 , d 2 , . . . , d n ) = ∑ i ∈ I A d i .$
We have that f is continuous and $D o m$ is bounded and closed. Then, the images of f are some bounded and closed intervals of R. For each $( α , β ) ∈ L$, define $Γ [ α , β ] = f ( D o m [ α , β ] )$.
By Equation (13), we have that, for each $α , β ∈ ( 0 , 1 ]$, $( P ¯ ( A ) ) ( α , β ) = ⋃ j Γ [ α , β ]$. Therefore, $( P ¯ ( A ) )$ is a DHFN.
$P ¯$ is called a dual hesitant fuzzy probability function. ☐
Remark 3.
If the DHFS reduces to a fuzzy set or an intuitionistic fuzzy set, then the dual hesitant fuzzy probability reduces to a fuzzy probability or an intuitionistic fuzzy probability.
Theorem 5.
Let $X = { x 1 , x 2 , . . . , x n }$ and let $P ¯ : ℘ ( X ) → ℑ$ be a dual hesitant fuzzy probability function. Then, for each $A , B ⊆ X ,$ the following properties hold:
(i)
If $A ∩ B = ⌀$ then $P ¯ ( A ∪ B ) ⊆ P ¯ ( A ) + P ¯ ( B ) ;$
(ii)
If $A ⊆ B$ then $P ¯ ( A ) ⪯ P ¯ ( B ) ;$
(iii)
$P ¯ ( ⌀ ) = 0 ¯ ⪯ P ¯ ( A ) ⪯ P ¯ ( X ) = 1 ¯ ;$
(iv)
$1 ¯ ⪯ P ¯ ( A ) + P ¯ ( A c ) ;$
(v)
If $A ∩ B ≠ ⌀$ then $P ¯ ( A ∪ B ) ⊆ P ¯ ( A ) + P ¯ ( B ) − P ¯ ( A ∩ B ) .$
Proof.
(i)
Notice that $A ∩ B = ⌀$ if and only if $I A ∩ I B = ⌀$. In order to prove (i), we only need to prove that:
$P ¯ ( A ∪ B ) α ⊆ P ¯ ( A ) α + P ¯ ( B ) α .$
Namely, for each $( α , β ) ∈ L ,$
${ ∑ i ∈ I A d i +$ $∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β$ $a n d ∑ l = 1 n d l = 1 }$ lies in
${ ∑ i ∈ I A d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β$ $a n d ∑ l = 1 n d l = 1 }$ + ${ ∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n )$ $∈ S α β a n d ∑ l = 1 n d l = 1 }$, which is obvious;
(ii)
If $A ⊆ B$, we have ${ ∑ i ∈ I A d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } ⊆ { ∑ i ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$, thus $P ¯ ( A ) ⪯ P ¯ ( B ) ;$
(iii)
If $∅ ⊆ A ⊆ X$, $∅ ⊆ { ∑ i ∈ I A d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } ⊆ { ∑ i ∈ I X d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$, thus $P ¯ ( ⌀ ) = 0 ¯ ⪯ P ¯ ( A ) ⪯ P ¯ ( X ) = 1 ¯ .$
(iv)
As (i), we donate $B = A c$, then $P ¯ ( A ) + P ¯ ( A c )$ $⪰ P ¯ ( A ∪ A c ) ) = P ¯ ( X ) = 1 ¯$;
(v)
If $A ∩ B ≠ ⌀$, for each $( α , β ) ∈ L$, ${ ∑ u ∈ I A ∪ I B d u | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$ lies in ${ ∑ i ∈ I A d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$ + ${ ∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$${ ∑ k ∈ I A ∩ I B d k | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$, which proves that $P ¯ ( A ∪ B ) ⊆ P ¯ ( A ) + P ¯ ( B ) − P ¯ ( A ∩ B ) .$
☐
## 5. Dual Hesitant Fuzzy Conditional Probability
Let $A , B ⊆ X$ with $I A$ and $I B$ being their index sets, respectively. Define the dual hesitant fuzzy conditional probability of A with B as being the DHFS $P ¯ ( A | B )$ whose $( α , β ) − c u t s$ are:
$P ¯ ( A | B ) ( α , β ) = { ∑ i ∈ I A ∩ I B d i ∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } .$
Theorem 6.
$P ¯ ( A | B ) ⊆ P ¯ ( A ∩ B ) P ¯ ( B ) .$
Proof.
It is sufficient to prove that:
$P ¯ ( A | B ) ( α , β ) ⊆ P ¯ ( A ∩ B ) ( α , β ) P ¯ ( B ) ( α , β ) .$
Notice that $A ∩ B = ⌀$ if and only if $I A ∩ I B = ⌀$.
To prove Equation (17), it is sufficient to prove that for each $α ∈ [ 0 , 1 ] ,$
${ ∑ i ∈ I A ∩ I B d i ∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } ⊆ { ∑ i ∈ I A ∩ I B d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } { ∑ j ∈ I B d j | ( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 } ,$ which is obvious. ☐
Theorem 7.
$P ¯ ( A | B )$ is a DHFN.
Proof.
Analogous to the proof of Theorem 4 by replacing the function f in Equation (14) by $f ( d 1 , d 2 , . . . , d n ) = ∑ i ∈ I A ∩ I B d i ∑ j ∈ I B d j$. ☐
Theorem 8.
Let $X = { x 1 , x 2 , . . . , x n }$ and let $P ¯ : ℘ ( X ) → ℑ$ be a dual hesitant fuzzy probability function. Then, for each $A , B ⊆ X ,$ the following properties hold:
1.
If $A 1 ∩ A 2 = ⌀$ then $P ¯ ( A 1 ∪ A 2 | B ) ⊆ P ¯ ( A 1 | B ) + P ¯ ( A 2 | B )$;
2.
$0 ¯ ⪯ P ¯ ( A | B ) ⪯ 1 ¯$;
3.
$P ¯ ( A | A ) = 1 ¯$, $P ¯ ( A | A c ) = 0 ¯$;
4.
If $B ⊆ A$ then $P ¯ ( A | B ) = 1 ¯$;
5.
If $A ∩ B = ⌀$ then $P ¯ ( A | B ) = 0 ¯$.
Proof.
The proof of items 2–5, is trivial.
Notice that if $A 1 ∩ A 2 = ∅$, then $I A 1 ∩ I A 2 = ∅ .$ To prove item 1, it is sufficient to prove that for each $( α , β ) ∈ T$, $P ¯ ( A 1 ∪ A 2 | B ) ( α , β )$ $⊆ P ¯ ( A 1 | B ) ( α , β ) + P ¯ ( A 2 | B ) ( α , β )$.
However, this follows by the fact that $P ¯ ( A 1 ∪ A 2 | B ) α$ $= { ∑ i ∈ I A 1 ∩ I B d i + ∑ j ∈ I A 2 ∩ I B d j ∑ k ∈ I B d k |$ $( d 1 , d 2 , . . . , d n ) ∈ S α β a n d ∑ l = 1 n d l = 1 }$ lies in ${ ∑ i ∈ I A 1 ∩ I B d i ∑ k ∈ I B d k | ( d 1 , d 2 , . . . , d n )$ $∈ S α β a n d ∑ l = 1 n d l = 1 }$ + ${ ∑ i ∈ I A 2 ∩ I B d i ∑ k ∈ I B d k | ( d 1 , d 2 , . . . , d n )$ $∈ S α β a n d ∑ l = 1 n d l = 1 }$, which is obvious.
Notice that the last term in the above equality coincides with $P ¯ ( A 1 | B ) ( α , β ) + P ¯ ( A 2 | B ) ( α , β )$. Then, $P ¯ ( A 1 ∪ A 2 | B ) ⊆ P ¯ ( A 1 | B ) + P ¯ ( A 2 | B )$. ☐
Lemma 4.
Let $A , B , C ⊆ X$. Then, $P ¯ ( A ∩ B | C ) ⊆ P ¯ ( A | C ) · P ¯ ( B | A ∩ C ) .$
Proof.
We only need to prove that for each $( α , β ) ∈ T$, $P ¯ ( A ∩ B | C ) ( α , β ) ⊆ P ¯ ( A | C ) ( α , β ) · P ¯ ( B | A ∩ C ) ( α , β ) .$
Furthermore, by Equation (16), $P ¯ ( A ∩ B | C ) ( α , β ) = { ∑ i ∈ I A ∩ I B ∩ I C d i ∑ j ∈ I c d j | ( d 1 , d 2 , . . . , d n ) ∈ S ( α , β ) a n d ∑ l = 1 n d l = 1 }$ $⊆ { ∑ i ∈ I A ∩ I B ∩ I C d i ∑ j ∈ I A ∩ I C d j | ( d 1 , d 2 , . . . , d n ) ∈ S ( α , β ) a n d ∑ l = 1 n d l = 1 }$ $· { ∑ i ∈ I A ∩ I C d i ∑ j ∈ I C d j | ( d 1 , d 2 , . . . , d n ) ∈ S ( α , β ) a n d ∑ l = 1 n d l = 1 }$ $= P ¯ ( A | C ) · P ¯ ( B | A ∩ C ) ,$ which proves the Lemma. ☐
Theorem 9.
Let $A 1 , A 2 , . . . , A k$ be subsets of X, i.e., $A i ∩ A j = ∅$ when $i ≠ j$ and $X = ∪ i = 1 k A i$. Let $B , C ⊆ X$, then $P ¯ ( B | C ) ⊆ ∑ i = 1 k P ¯ ( A j | C ) · P ¯ ( B | A j ∩ C )$.
Proof.
Since $B = ∪ i = 1 k B ∩ A i$, by Theorem 8 item 1, we have that $P ¯ ( B | C ) ⊆ ∑ i = 1 k P ¯ ( B ∩ A i | C )$. So, by Lemma 4, we have $P ¯ ( B ∩ A j | C ) = P ¯ ( A j | C ) · P ¯ ( B | A j ∩ C )$, and therefore, $P ¯ ( B | C ) ⊆ ∑ i = 1 k P ¯ ( A j | C ) · P ¯ ( B | A j ∩ C )$. ☐
Lemma 5.
$P ¯ ( B ∩ C ) ⊆ P ¯ ( B | C ) · P ¯ ( B )$.
Proof.
Let $A = { x 1 , . . . , x k }$ and $B = { x l , . . . , x m }$ with $1 ≤ l ≤ k ≤ m ≤ n .$ Then, $A ∩ B = { x l , . . . , x k }$ and for any $( α , β ) ∈ T ,$ $P ¯ ( B ∩ C ) α = { ∑ i = l k d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β } ⊆$ ${ ∑ i = l k d i ∑ i = l m d i | ( d 1 j , d 2 , . . . , d n ) ∈ S α β } · { ∑ i = l m d i | ( d 1 , d 2 , . . . , d n ) ∈ S α β }$ $= P ¯ ( B / C ) ( α , β ) · P ¯ ( B ) ( α , β )$. ☐
Theorem 10.
Let $A 1 , A 2 , . . . , A k$ be subsets of $X = { x 1 , x 2 , . . . , x n }$ such that $A i ∩ A j = ∅$ when $i ≠ j$ and $X = ∪ i = 1 k A i$. Let D by any event, then $P ¯ ( D ) ⊆ P ¯ ( A i ) · P ¯ ( D | A i )$.
Proof.
Since $D = ∪ i = 1 k D ∩ A i$, then by Theorem 8 item 1, we have $P ¯ ( D ) ⊆ ∑ i = 1 k P ¯ ( A i | D ) .$ Therefore, by Lemma 5, $P ¯ ( D ) ⊆ P ¯ ( A i ) · P ¯ ( D | A i )$.
Next, we give the dual hesitant fuzzy Bayes’ theorem. ☐
Theorem 11.
Let $A 1 , A 2 , . . . , A k$ be subsets of $X = { x 1 , x 2 , . . . , x n }$ such that $A i ∩ A j = ∅$ when $i ≠ j$ and $X = ∪ i = 1 k A i$. Let D by any event, then $P ¯ ( A i | D ) ⊆ P ¯ ( A i ∩ D ) ∑ j = 1 k P ¯ ( A j ) P ¯ ( D | A j )$.
Proof.
From Theorems 6 and 7, we have that $P ¯ ( A i | D ) ⊆ P ¯ ( A i ∩ D ) P ¯ ( D ) ⊆ P ¯ ( A i ∩ D ) ∑ j = 1 k P ¯ ( A j ) P ¯ ( D | A j )$. ☐
## 6. Application of Color Blindness
Although fuzzy probabilities have been applied to the problem of color blindness by Buckley [11], sometimes they are not suitable for the practical environment. The following example is given to show the application of dual hesitant fuzzy probabilities.
Example 2.
Some people believe that red-green color blindness is more prevalent in males than in females. To test this hypothesis, we gather a random sample from the adult population. Let M be the event that a person is male, F is the event that a person is female, C is the event that a person has red-green color blindness and $C ′$ is the event that he/she does not have red-green color blindness.
Assume that researchers investigate three cities to reflect red-green color blindness in males and in females. As we cannot avoid information loss and some people do not participate in the survey, we can make it a DHFE. Thus, a dual hesitant fuzzy set of the point estimates of probabilities is given to indicate the results of the investigation:
• $p 1 = { { p ( M ∩ C } , { p ( M ∩ C ′ ) } }$ $= { < { 0 . 04 , 0 . 06 , 0 . 02 }$ $, { 0 . 4 , 0 . 35 , 0 . 45 } > }$;
• $p 2 = { { p ( F ∩ C } , { p ( F ∩ C ′ ) } }$ $= { < { 0 . 08 , 0 . 06 , 0 . 07 } ,$ ${ 0 . 3 , 0 . 4 , 0 . 2 } > }$.
Then, we obtain that
• $p 1 = { s ( p ( M ∩ C ) , s ( p ( M ∩ C ′ ) ) }$ $= { < 0 . 04 , 0 . 4 > }$;
• $p 2 = { s ( p ( F ∩ C ) , s ( p ( F ∩ C ′ ) }$ $= { < 0 . 07 , 0 . 3 > }$.
Assume that the uncertainty in these point estimates has been shown in their fuzzy values as follows:
We wish to calculate the dual hesitant fuzzy conditional probabilities $P ( M | C )$ and $P ( F | C )$. The $( α , β ) − c u t s$ of the first fuzzy probability are:
$P ( M | C ) ( α , β ) = { p 1 p 1 + p 2 | S } ,$
for $( α , β ) ∈ L$ and S donates "$p i ∈ ( p i ) ( α , β ) ( i = 1 , 2 , 3 )$ and $p 1 + p 2 + p 3 = 1$ ($p 3$ donates the information loss and some people not participating in the survey)". Let $H ( p 1 , p 2 ) = p 1 p 1 + p 2$, then H is an increasing function of $p 1$ but decreasing in $p 2$. According to Table 1, we obtain:
$P ( M | C ) ( α , β ) = [ l ( H ( p 11 , p 22 ) ) , r ( H ( p 12 , p 21 ) ) ] = [ m a x { 0 . 02 + 0 . 02 α 0 . 16 − 0 . 05 α , 0 . 3 + 0 . 1 β 0 . 9 − 0 . 2 β } , m i n { 0 . 06 − 0 . 02 α 0 . 06 + 0 . 05 α , 0 . 5 − 0 . 1 β 0 . 5 + 0 . 2 β } ] ≈ [ m a x { 1 8 + 11 40 α , 1 3 + 1 6 β } , m i n { 1 − 3 5 α , 1 − 1 2 β } ]$
for all $( α , β ) ∈ L$.
The $( α , β ) − c u t s$ of the second dual hesitant fuzzy probability are:
$P ( F | C ) ( α , β ) = { p 2 p 1 + p 2 | S } ,$
for $( α , β ) ∈ L$ and S donates "$p i ∈ ( p i ) ( α , β ) ( i = 1 , 2 , 3 )$ and $p 1 + p 2 + p 3 = 1$ ($p 3$ donates the information loss and some people not participating in the survey)". Let $G ( p 1 , p 2 ) = p 2 p 1 + p 2$, then G is a decreasing function of $p 1$ but increasing in $p 2$. According to Table 1, we obtain:
$P ( F | C ) ( α , β ) = [ l ( G ( p 12 , p 21 ) ) , r ( G ( p 11 , p 22 ) ) ] = [ m a x { 0 . 04 + 0 . 03 α 0 . 16 − 0 . 05 α , 0 . 2 + 0 . 1 β 0 . 9 − 0 . 2 β } , m i n { 0 . 1 − 0 . 03 α 0 . 06 + 0 . 05 α , 0 . 4 − 0 . 1 β 0 . 5 + 0 . 2 β } ] ≈ [ m a x { 1 4 + 7 20 α , 2 9 + 5 18 β } , m i n { 1 − 2 5 α , 4 5 − 3 10 β } ]$
for all $( α , β ) ∈ L$.
Based on Equation (3.11), we obtain that, for $( α , β ) ∈ L$, when $6 α − 3 β ≥ 2$, $P ( M | C ) ⪰ P ( F | C )$; when $6 α − 3 β < 2$, $P ( M | C ) ⪯ P ( F | C )$.
## 7. Conclusions and Further Study
This work extends the notion of dual hesitant fuzzy probabilities by representing probabilities through the dual hesitant fuzzy numbers, in the sense of Zhu et al., instead of intuitionistic fuzzy numbers. We also give the concept of dual hesitant fuzzy probability, based on which we provide some main results including the properties of dual hesitant fuzzy probability, dual hesitant fuzzy conditional probability, and dual hesitant fuzzy total probability. It is an extension of the approach of [13] to intuitionistic fuzzy probability, since all intuitionistic fuzzy sets (and thus intuitionistic fuzzy numbers and fuzzy probabilities) are a dual hesitant fuzzy set (with $h D = u , g D = v$).
This extension may be useful in some situations where some uncertain probabilities are hesitant, but where this uncertainty can be modeled using dual hesitant fuzzy numbers. As a future work, we intend, using this same approach, to generalize the notion of probability spaces as well as other concepts related to probability. We also expect to extend the notions of Markov chain, and the hidden Markov model, etc. Furthermore, DEHFSs are likely to play an important role in decision making with more studies on the theory and applications.
## Acknowledgments
This work has been supported by the National Natural Science Foundation of China (11461043, 11661053), the Provincial Natural Science Foundation of Jiangxi, China (20142BAB201005, 20161BAB201009) and the Science and Technology Project of Educational Commission of Jiangxi Province, China (150013).
## Author Contributions
Jianjian Chen and Xianjiu Huang contributed to the results; Jianjian Chen wrote the paper.
## Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1. Dual hesitant triangular fuzzy number.
Figure 1. Dual hesitant triangular fuzzy number.
Table 1. Dual hesitant triangular fuzzy number (DHTFN) of results.
Table 1. Dual hesitant triangular fuzzy number (DHTFN) of results.
Uncertainty in Membership Degree hUncertainty in Nonmembership Degree g
$p 1$(0.02/0.04/0.06)(0.3/0.4/0.5)
$p 2$(0.04/0.07/0.1)(0.2/0.3/0.4)
## Share and Cite
MDPI and ACS Style
Chen, J.; Huang, X. Dual Hesitant Fuzzy Probability. Symmetry 2017, 9, 52. https://doi.org/10.3390/sym9040052
AMA Style
Chen J, Huang X. Dual Hesitant Fuzzy Probability. Symmetry. 2017; 9(4):52. https://doi.org/10.3390/sym9040052
Chicago/Turabian Style
Chen, Jianjian, and Xianjiu Huang. 2017. "Dual Hesitant Fuzzy Probability" Symmetry 9, no. 4: 52. https://doi.org/10.3390/sym9040052
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https://www.zbmath.org/?q=an%3A0879.34006 | 1,618,561,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00142.warc.gz | 1,180,523,056 | 10,479 | zbMATH — the first resource for mathematics
Nonexistence of finite order solutions of certain second order linear differential equations. (English) Zbl 0879.34006
The author considers the second order differential equation $$f''+A(z)f'+ B(z)= 0$$. $$A(z)$$ and $$B(z)$$ are entire functions not identically zero. $$\rho$$ is the order of the entire function. This paper investigates the relationship of the order of the solution $$f$$ to the order of the coefficients of the differential equation. The author proves these results:
Theorem I: If (1) $$A(z)$$ is an entire function of finite nonintegral order $$\rho (A)>1$$, and all its zeros lie in the angular sector $$\theta_1\leq \arg z\leq \theta_2$$ satisfying $$\theta_2- \theta_1< {\pi \over q+1}$$, if $$q$$ is odd, and $$\theta_2- \theta_1< {\pi(2q-1) \over (q+1)2q}$$ if $$q$$ is even, $$q$$ is the genus of $$A(z)$$, (2) $$B(z)$$ an entire function with $$0<\rho (B)<1/2$$ then every nonconstant solution $$f$$ of the differential equation has an infinite order satisfying $$\varlimsup_{r\to \infty} {\log\log T(r,f) \over \log r} \geq\rho (B)$$.
The second result deals with the order of the solution of the differential equation when the coefficients are polynomials of equal degree over the complex numbers.
Reviewer: H.S.Nur (Fresno)
MSC:
34M05 Entire and meromorphic solutions to ordinary differential equations in the complex domain 30D20 Entire functions of one complex variable, general theory
Full Text:
References:
[1] P. O. BARRY, Some theorems related to the cos p theorem, Proc. London Math. Soc. (3), 21 (1970), 334-360. · Zbl 0204.42302 · doi:10.1112/plms/s3-21.2.334 [2] R. P. BOAS, Entire Functions, Academic Press, New York, 1954 · Zbl 0058.30201 [3] G. GUNDERSEN, Finite order solutions of second order linear differential equations, Trans. Amer. Math. Soc., 305 (1988), 415-429 Zentralblatt MATH: · Zbl 0669.34010 · doi:10.2307/2001061 · www.zentralblatt-math.org [4] G. GUNDERSEN, Estimates for the logarithmic derivative of a meromorphic func tion, plus similar estimates, J. London Math. Soc. (2), 37 (1988), 88-104. · Zbl 0638.30030 · doi:10.1112/jlms/s2-37.121.88 [5] W. HAYMAN, Meromorphic Functions, Clarendon Press, Oxford, 1964 · Zbl 0115.06203 [6] S. HELLERSTEIN, J. MILES AND J. Rossi, On the growth of solutions of /”+?/ + /=0, Trans. Amer. Math. Soc., 323 (1991), 693-706. · Zbl 0719.34011 · doi:10.2307/2001737 [7] K. KWON, On the growth of entire functions satisfying second order linear dif ferential equations, Bull. Korean Math. Soc. (2), 33 (1996), 487-496. · Zbl 0863.34007 [8] A. I. MARKUSHEVICH, Theory of Functions of a Complex Variable (Vol.11), Prentice-Hall, New Jersey, 1965 · Zbl 0142.32602 [9] M. OZAWA, On a solution of w^+e zw/+(az+b’)w=Q. Kodai Math. J., 3 (1980), 295-309 · Zbl 0463.34028 · doi:10.2996/kmj/1138036197 [10] G. VALIRON, Sur les functions entieres d’ordre fini et d’ordre nul, et en particu lier les functions a correspondance reguliere, Ann. Fac. Sci. Univ. Toulouse, 3 (1913), 117-257. · JFM 46.1462.03
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,071 | 3,408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.734884 |
https://www.enotes.com/homework-help/prove-following-tan-theta-cot-phi-cot-theta-tan-247371 | 1,695,620,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00779.warc.gz | 842,801,367 | 17,537 | # Prove the following: (tan A + cot B)(cot A - tan B) = cot A - tan A*tan B
I have replaced the theta with A and phi with B in the identity to make it easier to type.
We have to prove: (tan A + cot B)(cot A - tan B) = cot A - tan A * tan B
(tan A + cot B) (cot A - tan B)
Open the brackets and multiply the terms.
=> tan A * cot A - tan A * tan B + cot A * cot B - cot B * tan B
use the relation tan x = 1/cot x or tan x * cot x = 1
=> 1 - tan A * tan B + cot A * cot B - 1
=> cot A * cot B - tan A * tan B
The identity you have given has a missing term. You have mistyped cot A instead of cot A*cot B.
Therefore, the accurate identity (tan A + cot B)(cot A - tan B) = cot A *cot B - tan A * tan B is proved.
## See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Approved by eNotes Editorial Team | 283 | 930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.760526 |
https://support.nag.com/numeric/nl/nagdoc_27/flhtml/g01/g01haf.html | 1,702,071,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00214.warc.gz | 611,898,478 | 5,569 | # NAG FL Interfaceg01haf (prob_bivariate_normal)
## 1Purpose
g01haf returns the lower tail probability for the bivariate Normal distribution.
## 2Specification
Fortran Interface
Function g01haf ( x, y, rho,
Real (Kind=nag_wp) :: g01haf Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: x, y, rho
#include <nag.h>
double g01haf_ (const double *x, const double *y, const double *rho, Integer *ifail)
The routine may be called by the names g01haf or nagf_stat_prob_bivariate_normal.
## 3Description
For the two random variables $\left(X,Y\right)$ following a bivariate Normal distribution with
$EX=0, EY=0, EX2=1, EY2=1 and EXY=ρ,$
the lower tail probability is defined by:
$PX≤x,Y≤y:ρ=12π1-ρ2 ∫-∞y ∫-∞x exp- X2- 2ρ XY+Y2 21-ρ2 dXdY.$
For a more detailed description of the bivariate Normal distribution and its properties see Abramowitz and Stegun (1972) and Kendall and Stuart (1969). The method used is described by Genz (2004).
## 4References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
Genz A (2004) Numerical computation of rectangular bivariate and trivariate Normal and $t$ probabilities Statistics and Computing 14 151–160
Kendall M G and Stuart A (1969) The Advanced Theory of Statistics (Volume 1) (3rd Edition) Griffin
## 5Arguments
1: $\mathbf{x}$Real (Kind=nag_wp) Input
On entry: $x$, the first argument for which the bivariate Normal distribution function is to be evaluated.
2: $\mathbf{y}$Real (Kind=nag_wp) Input
On entry: $y$, the second argument for which the bivariate Normal distribution function is to be evaluated.
3: $\mathbf{rho}$Real (Kind=nag_wp) Input
On entry: $\rho$, the correlation coefficient.
Constraint: $-1.0\le {\mathbf{rho}}\le 1.0$.
4: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 4 in the Introduction to the NAG Library FL Interface for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{rho}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{rho}}\ge -1.0$.
On entry, ${\mathbf{rho}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{rho}}\le 1.0$.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
Accuracy of the hybrid algorithm implemented here is discussed in Genz (2004). This algorithm should give a maximum absolute error of less than $5×{10}^{-16}$.
## 8Parallelism and Performance
g01haf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The probabilities for the univariate Normal distribution can be computed using s15abf and s15acf.
## 10Example
This example reads values of $x$ and $y$ for a bivariate Normal distribution along with the value of $\rho$ and computes the lower tail probabilities.
### 10.1Program Text
Program Text (g01hafe.f90)
### 10.2Program Data
Program Data (g01hafe.d)
### 10.3Program Results
Program Results (g01hafe.r) | 1,211 | 4,435 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.641294 |
http://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator | 1,411,266,613,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657134511.2/warc/CC-MAIN-20140914011214-00322-ip-10-234-18-248.ec2.internal.warc.gz | 89,913,702 | 14,398 | Hilbert–Schmidt operator
In mathematics, a Hilbert–Schmidt operator, named for David Hilbert and Erhard Schmidt, is a bounded operator A on a Hilbert space H with finite Hilbert–Schmidt norm
$\|A\|^2_{HS}={\rm Tr} |(A^{{}^*}A)|:= \sum_{i \in I} \|Ae_i\|^2$
where $\|\ \|$ is the norm of H and $\{e_i : i\in I\}$ an orthonormal basis of H for an index set $I$.[1][2] Note that the index set need not be countable. This definition is independent of the choice of the basis, and therefore
$\|A\|^2_{HS}=\sum_{i,j} |A_{i,j}|^2 = \|A\|^2_2$
for $A_{i,j}=\langle e_i, Ae_j \rangle$ and $\|A\|_2$ the Schatten norm of $A$. In Euclidean space $\|\ \|_{HS}$ is also called Frobenius norm, named for Ferdinand Georg Frobenius.
The product of two Hilbert–Schmidt operators has finite trace class norm; therefore, if A and B are two Hilbert–Schmidt operators, the Hilbert–Schmidt inner product can be defined as
$\langle A,B \rangle_\mathrm{HS} = \operatorname{tr} (A^*B) = \sum_{i} \langle Ae_i, Be_i \rangle.$
The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on H. They also form a Hilbert space, which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces
$H^* \otimes H, \,$
where H* is the dual space of H.
The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, H is finite-dimensional.
An important class of examples is provided by Hilbert–Schmidt integral operators.
Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact.
Functionals
A mapping $\phi:H_1\times H_2\to C$ is a Hilbert-Schmidt functional if it is a bounded bilinear functional.[3]
A bounded linear mapping $L:H_1\times H_2\to K$ is weakly Hilbert-Schmidt if for all $v\in K$ the mapping
$\phi_v = (u_1,u_2)\mapsto\langle L(u_1,u_2), v\rangle$
is a Hilbert-Schmidt functional and $\|\phi_v\|\leq M\|v\|$ for some real number $M\geq 0$.[4]
References
1. ^ Moslehian, M.S. "Hilbert–Schmidt Operator (From MathWorld)".
2. ^ Voitsekhovskii, M.I. (2001), "H/h047350", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
3. ^ Kadison, Richard V.; Ringrose, John R. (1997), Fundamentals of the theory of operator algebras. Vol. I, Graduate Studies in Mathematics 15, Providence, R.I.: American Mathematical Society, ISBN 978-0-8218-0819-1, MR 1468229 (see p. 127)
4. ^ Kadison and Ringrose, (see p. 131) | 784 | 2,459 | {"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": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2014-41 | longest | en | 0.792125 |
https://justaaa.com/accounting/363729-income-statement-sales-15000-operating-expenses | 1,686,035,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00399.warc.gz | 381,232,376 | 12,021 | Question
# Income Statement Sales \$ 15,000 Operating expenses COGS \$ 11,000 Depreciation
Income Statement Sales \$ 15,000 Operating expenses COGS \$ 11,000 Depreciation 800 Admin expenses 1,500 Total operating expense 13,300 Operating income 1,700 Interest revenue 200 Gain on disposal of equipment 400 Income before taxes 2,300 Income tax 920 Net income \$ 1,380 Balance sheet Assets This year Last year Cash \$ 3,000 \$ 1,100 A/R 500 530 Inventory 850 820 Prepaid Ins 150 200 Equipment 1,800 2,400 Less: Acc Depr (900) (600) Total Assets \$ 5,400 \$ 4,450 Liabilities & SE A/P \$ 1,500 \$ 1,300 Unearned revenue \$ 500 \$ 400 L/T Note payable 300 500 Common Stock 800 900 Retained earnings 2,300 1,350 Total liabilties &SE \$ 5,400 \$ 4,450 Equipment was sold and dividends were paid. a. How much was paid out in dividends? b. How much was equipment sold for? c. What are total cash flows (Op+Inv+Fin)?
Solution a:
Amount of dividend paid = Beginning retained earnings + Net income - Ending retained earnings
= \$1,350 + \$1,380 - \$2,300 = \$430
Solution b:
Cost of equipment sold = \$2,400 - \$1,800 = \$600
Accumulated depreciation on equipment sold = Accumulated depreciation at begginning + Depreciation expense - Accumulated depreciation at year end
= \$600 + \$800 - \$900 = \$500
Carrying value of equipment sold = \$600 - \$500 = \$100
Gain on disposal of equipment = \$400
Sale value of equipment = Carrying value + Gain = \$100 + \$400 = \$500
Solution c:
Total cash flows = Change in cash = Ending cash - Beginning cash = \$3,000 - \$1,100 = \$1,900
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 471 | 1,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-23 | latest | en | 0.833628 |
https://dateandtime.info/distanceantarcticcircle.php?id=4876523 | 1,638,992,362,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00385.warc.gz | 279,539,100 | 9,630 | # Distance between Sioux City, Iowa, USA and the Antarctic Circle
12141 km = 7544 miles
During our calculation of the distance to the Antarctic Circle we make two assumptions:
1. We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
2. We calculate the distance between a point on the Earth’s surface and the Antarctic Circle as the length of the arc of the meridian passing through this point and crossing the Antarctic Circle.
Find out the distance between Sioux City and the North Pole, the South Pole, the Equator, the Tropic of Cancer, the Tropic of Capricorn, the Arctic Circle
Find out the distance between Sioux City and other cities
## Sioux City, USA
Country: USA
State: Iowa
Sioux City’s coordinates: 42°29′59″ N, 96°24′01″ W
Population: 82,684
Find out what time it is in Sioux City right now
Wikipedia article: Sioux City
## The Antarctic Circle
The Antarctic Circle is a circle of latitude or parallel on the Earth's surface. The polar night and the midnight sun phenomena can be observed to the south of the Antarctic Circle.
During the midnight sun there is no sunset and the Sun is over the horizon continuously for 24 hours or more.
During the polar night there is no sunrise and the Sun is below the horizon continuously for 24 hours or more.
For the points on the Earth's surface located at the Antarctic Circle the polar night and the midnight sun coincide with the winter solstice and the summer solstice correspondingly and their duration equals 24 hours.
For the points on the Earth's surface located to the south of the Antarctic Circle the duration of the polar night and the midnight sun grows with latitude.
The latitude of the Antarctic Circle depends on the tilt of the Earth's axis which changes with time.
The current latitude of the Antarctic Circle equals 66°33′43″ S.
Wikipedia article: the Antarctic Circle | 437 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.863761 |
http://mscroggs.co.uk/puzzles/start/8 | 1,596,729,307,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00310.warc.gz | 68,821,693 | 8,043 | mscroggs.co.uk
mscroggs.co.uk
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# Puzzles
## 17 December
Eve picks a three digit number then reverses its digits to make a second number. The second number is larger than her original number.
Eve adds her two numbers together; the result is 584. What was Eve's original number?
## 16 December
Arrange the digits 1-9 in a 3×3 square so that: the median number in the first row is 6; the median number in the second row is 3; the mean of the numbers in the third row is 4; the mean of the numbers in the second column is 7; the range of the numbers in the third column is 2, The 3-digit number in the first column is today's number.
median 6 median 3 mean 4 today's number mean 7 range 2
## 15 December
There are 5 ways to make 30 by multiplying positive integers (including the trivial way):
• 30
• 2×15
• 3×10
• 5×6
• 2×3×5
Today's number is the number of ways of making 30030 by multiplying.
## 14 December
During one day, a digital clock shows times from 00:00 to 23:59. How many times during the day do the four digits shown on the clock add up to 14?
## 13 December
Each clue in this crossnumber (except 5A) gives a property of that answer that is true of no other answer. For example: 7A is a multiple of 13; but 1A, 3A, 5A, 1D, 2D, 4D, and 6D are all not multiples of 13. No number starts with 0.
## 12 December
For a general election, the Advent isles are split into 650 constituencies. In each constituency, exactly 99 people vote: everyone votes for one of the two main parties: the Rum party or the Land party. The party that receives the most votes in each constituency gets an MAP (Member of Advent Parliament) elected to parliament to represent that constituency.
In this year's election, exactly half of the 64350 total voters voted for the Rum party. What is the largest number of MAPs that the Rum party could have?
## 11 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10. Today's number is the product of the red digits.
+ ÷ = 2 + ÷ ÷ ÷ ÷ = 3 ÷ - ÷ ÷ ÷ = 1 =2 =1 =1
Tags: grids, numbers
## 10 December
For all values of $$x$$, the function $$f(x)=ax+b$$ satisfies
$$8x-8-x^2\leqslant f(x)\leqslant x^2.$$
What is $$f(65)$$?
Edit: The left-hand quadratic originally said $$8-8x-x^2$$. This was a typo and has now been corrected.
## Archive
Show me a random puzzle
▼ show ▼ | 730 | 2,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-34 | latest | en | 0.935971 |
https://www.geeksforgeeks.org/find-number-diagonals-n-sided-convex-polygon/?ref=rp | 1,701,563,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100476.94/warc/CC-MAIN-20231202235258-20231203025258-00033.warc.gz | 874,921,368 | 53,203 | # Find number of diagonals in n sided convex polygon
Given n > 3, find number of diagonals in n sided convex polygon.
According to Wikipedia, In geometry, a diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge. Informally, any sloping line is called diagonal.
Examples :
```Input : 5
Output : 5```
Explanation: Five possible diagonals are : AC, AD, BD, BE, CE
Recommended Practice
Since for an n-sided convex polygon, from each vertex, we can draw n-3 diagonals leaving two adjacent vertices and itself. Following this way for n-vertices, there will be n*(n-3) diagonals but then we will be calculating each diagonal twice so total number of diagonals become n*(n-3)/2
Here is code for above formula.
## C++
`#include ` `using` `namespace` `std;` `// C++ function to find number of diagonals` `// in n sided convex polygon` `int` `numberOfDiagonals(``int` `n)` `{` ` ``return` `n * (n - 3) / 2;` `}` `// driver code to test above function` `int` `main()` `{` ` ``int` `n = 5;` ` ``cout << n << ``" sided convex polygon have "``;` ` ``cout << numberOfDiagonals(n) << ``" diagonals"``;` ` ``return` `0;` `}`
## Java
`// Java function to find number of diagonals` `// in n sided convex polygon` `public` `class` `Diagonals {` ` ``static` `int` `numberOfDiagonals(``int` `n)` ` ``{` ` ``return` `n * (n - ``3``) / ``2``;` ` ``}` ` ``// driver code to test above function` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int` `n = ``5``;` ` ``System.out.print(n + ``" sided convex polygon have "``);` ` ``System.out.println(numberOfDiagonals(n) + ``" diagonals"``);` ` ``}` `}` `// This code is contributed by Saket Kumar`
## Python3
`# Python3 program to find number of diagonals` `# in n sided convex polygon` `def` `numberOfDiagonals(n):` ` ``return` `n ``*` `(n ``-` `3``) ``/` `2` ` ` `# driver code to test above function` `def` `main():` ` ``n ``=` `5` ` ``print``(n , ``" sided convex polygon have "``)` ` ``print``(numberOfDiagonals(n) , ``" diagonals"``)` `if` `__name__ ``=``=` `'__main__'``:` ` ``main()` `#this code contributed by 29AjayKumar`
## C#
`// C# function to find number of diagonals` `// in n sided convex polygon` `using` `System;` `class` `GFG {` ` ` ` ``static` `int` `numberOfDiagonals(``int` `n)` ` ``{` ` ``return` `n * (n - 3) / 2;` ` ``}` ` ``// driver code to test above function` ` ``public` `static` `void` `Main()` ` ``{` ` ``int` `n = 5;` ` ``Console.Write(n + ``" sided convex polygon have "``);` ` ` ` ``Console.WriteLine(numberOfDiagonals(n) +` ` ``" diagonals"``);` ` ``}` `}` `// This code is contributed by Sam007`
## PHP
``
## Javascript
``
Output :
`5 sided convex polygon have 5 diagonals`
Time Complexity: O(1)
Auxiliary Space: O(1)
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 1,028 | 3,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.68233 |
https://djalil.chafai.net/blog/2011/11/ | 1,670,519,945,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00231.warc.gz | 232,536,221 | 29,743 | # Month: November 2011
Recently, I have spent some time with Krzysztof Burdzy writing a $\LaTeX$ document class for the periodicals Electronic Journal of Probability (EJP) and Electronic Communications in Probability (ECP). This document class is named ejpecp. The latest version is available on the Comprehensive TeX Archive Network (CTAN). It will serve starting from January 2012, together with a new editorial board and a new website. EJP and ECP are peer reviewed free access electronic periodicals publishing mathematical research articles in Probability Theory. They are sponsored by the Institute of Mathematical Statistics (IMS) and the Bernoulli Society. They share the same editorial board, with distinct Editor in Chief. They use the Open Journal System (OJS) free software developed by the non-profit organization Public Knowledge Project (PKP).
Last Updated on 2012-01-15
This post is devoted to central limit theorems for convex bodies.
The uniform law on the cube. Let ${X=(X_1,\ldots,X_n)}$ be a random variable distributed according to the uniform law on the cube
$B_{n,\infty}(\sqrt{3}) =\left\{x\in\mathbb{R}^n:\max_{1\leq i\leq n}|x_i|\leq\sqrt{3}\right\}$
The coordinates ${X_1,\ldots,X_n}$ are i.i.d. with uniform law on ${[-\sqrt{3},\sqrt{3}]}$. In particular,
$\mathbb{E}(X)=0 \quad\text{and}\quad \mathrm{Cov}(X)=I_n.$
The classical central limit theorem gives then that as ${n\rightarrow\infty}$,
$\frac{X_1+\cdots+X_n}{\sqrt{n}}\overset{d}{\longrightarrow}\mathcal{N}(0,1),$
where ${\mathcal{N}(0,1)}$ stands for the Gaussian law on ${\mathbb{R}}$ with mean ${0}$ and variance ${1}$. Actually, and more generally, by the classical Berry-Esseen theorem (the version for independent and possibly not identically distributed random variables), we know that for a well spread ${\theta}$ in the unit sphere (in the sense of a ratio of norms), as ${n\rightarrow\infty}$,
$\left<X,\theta\right>\overset{d}{\longrightarrow}\mathcal{N}(0,1).$
This does not hold for all ${\theta}$ in the unit sphere, and a simple counter example is given by ${\theta=e_1}$ for which ${\left<X,\theta\right>=X_1\sim\mathrm{Uniform}([-\sqrt{3},\sqrt{3}])}$ which is not Gaussian as ${n\rightarrow\infty}$.
The uniform law on the sphere. Let ${X=(X_1,\ldots,X_n)}$ be a random variable distributed according to the uniform law on the sphere
$S_{n,2}(\sqrt{n}) =\left\{x\in\mathbb{R}^n:\sqrt{x_1^2+\cdots+x_n^2}=\sqrt{n}\right\}.$
A way to simulate ${X}$ consists in using the identity in law
$X \overset{d}{=} \sqrt{n}\frac{G}{\sqrt{G_1^2+\cdots+G_n^2}}$
where ${G=(G_1,\ldots,G_n)\sim\mathcal{N}(0,I_n)}$. The classical law of large numbers gives ${G_1^2+\cdots+G_n^2=n(1+o(1))}$ almost surely. This gives that ${X_1\overset{d}{=}(1+o(1))G_1}$ converges in law to ${\mathcal{N}(0,1)}$ as ${n\rightarrow\infty}$. By rotational invariance, for any ${\theta}$ in the unit sphere, as ${n\rightarrow\infty}$,
$\left<X,\theta\right>\overset{d}{\longrightarrow}\mathcal{N}(0,1).$
In particular, for ${\theta=(1/\sqrt{n},\ldots,1/\sqrt{n})}$, we get, as ${n\rightarrow\infty}$,
$\frac{X_1+\cdots+X_n}{\sqrt{n}}\overset{d}{\longrightarrow}\mathcal{N}(0,1).$
From the sphere to the ball with the Archimedes principle. The Archimedes principle states that the projection of the uniform law of the unit sphere of ${\mathbb{R}^3}$, on a diameter, is exactly the uniform law on the diameter. More generally, if ${n\geq3}$ and ${X=(X_1,\ldots,X_n)}$ is a uniform random variable on the unit sphere ${S_{n,2}(1)}$ of ${\mathbb{R}^n}$ then ${(X_1,\ldots,X_{n-2})}$ is uniform on the unit ball ${B_{n-2,2}(1)}$ of ${\mathbb{R}^{n-2}}$. Letac has few pages on this nice antic observation. This does not work if one replaces ${n-2}$ by ${n-1}$. Note that the projection of the uniform law of the unit circle on a diameter gives an arc-sine law and not the uniform law.
The uniform law on the ball. Let ${X=(X_1,\ldots,X_n)}$ be a random variable following the uniform law on the ball
$B_{n,2}(\sqrt{n})=\left\{x\in\mathbb{R}^n:\sqrt{x_1^2+\cdots+x_n^2}\leq\sqrt{n}\right\}.$
A way to simulate ${X}$ is to use the Archimedes principle:
$X\overset{d}{=}\frac{\sqrt{n}}{\sqrt{n+2}}(Y_1,\ldots,Y_n)$
where ${Y=(Y_1,\ldots,Y_{n+2})}$ is uniform on ${S_{n+2,2}(\sqrt{n+2})}$. Now, from our preceding analysis on the sphere, we obtain that ${X_1\overset{d}{=}(1+o(1))Y_1}$ tends in law to ${\mathcal{N}(0,1)}$ as ${n\rightarrow\infty}$. By rotational invariance, for any ${\theta}$ in the unit sphere, as ${n\rightarrow\infty}$,
$\left<X,\theta\right>\overset{d}{\longrightarrow}\mathcal{N}(0,1).$
In particular, taking ${\theta=(1/\sqrt{n},\ldots,1/\sqrt{n})}$, we get, as ${n\rightarrow\infty}$,
$\frac{X_1+\cdots+X_n}{\sqrt{n}}\overset{d}{\longrightarrow}\mathcal{N}(0,1).$
Note that here again we have
$\mathbb{E}(X)=0 \quad\text{and}\quad \mathrm{Cov}(X)\approx I_n.$
It is worthwhile to mention an alternative simulation algorithm of the uniform law on ${B_{n,2}(1)}$. Namely, and following Barthe, Guédon, Mendelson, and Naor, if ${G=(G_1,\ldots,G_n)}$ and ${E}$ are independent with ${G\sim\mathcal{N}(0,I_n)}$ and ${E\sim\mathrm{Exponential}(1)}$ then
$X\overset{d}{=}\sqrt{n}\frac{G}{\sqrt{G_1^2+\cdots+G_n^2+E}}.$
Exercise. Show that the central limit theorem holds for the ${\ell^1}$ ball with suitable radius, using the probabilistic representation of the uniform law on the ${\ell^1}$ unit sphere
$\frac{E}{|E_1|+\cdots+|E_1|}$
where ${E=(E_1,\ldots,E_n)}$ has i.i.d. components following a symmetric exponential distribution.
The central limit theorem for convex bodies. So, well, the central limit theorem holds for the uniform law on the cube and on the ball, two remarkable convex bodies. We note also:
• normalization used: zero mean and identity covariance. In other words, the coordinates are centered, uncorrelated, with unit variance. This does not mean that they are independent. Their possible dependence is related to the geometry of the convex body (independent for the cube but dependent for the ball). For any ${\theta}$ in the unit sphere, since ${X}$ has zero mean and identity covariance,
$\mathbb{E}(\left<X,\theta\right>)=0 \quad\text{and}\quad \mathbb{E}(\left<X,\theta\right>^2) =\mathrm{Var}(\left<X,\theta\right>) =\left<\theta,\mathrm{Cov}(X)\theta\right> =\left\Vert\theta\right\Vert_2^2=1.$
Therefore ${\left<X,\theta\right>}$ and ${\mathcal{N}(0,1)}$ have identical first two moments.
• possible weights: the ${\theta}$ version may not work for all ${\theta}$, and the example of the cube shows that one may experience problems if the most of the mass of ${\theta}$ is supported by few coordinates. It is actually well known that even for independent random variables, the central limit theorem may fail in the presence of sparsity or localization.
Let ${K}$ be a convex body, in other words a convex and non empty compact sub-set of ${\mathbb{R}^n}$. Let us equip ${K}$ with the uniform law, which is the Lebesgue measure normalized to give a total mass ${1}$ to ${K}$. Let us assume that this law has zero mean and identity covariance (isotropy). All ${\ell^p}$ balls are of this kind. From the examples ${p=2}$ and ${p=\infty}$ considered above, it is quite natural to ask if a central limit theorem holds on a generic convex body such as ${K}$.
The celebrated Dvoretzky theorem is an encouraging result in this direction, dealing with sections instead of projections. Using ideas going back to Sudakov, it has been understood that a central limit theorem is implied by the thin shell property: the norm on ${K}$ is well concentrated around its mean and most of the mass of ${K}$ lies in a thin shell. This roughly reduces the problem to show that there exists a sequence ${\varepsilon_n\rightarrow0}$ which does not depend on ${K}$ such that if ${X}$ follows the uniform law of ${K}$ then
$\mathbb{P}\left(\left|\frac{\left\Vert X\right\Vert_2}{\sqrt{n}}-1\right| \geq\varepsilon_n\right) \leq\varepsilon_n.$
Note that ${\mathbb{E}(\left\Vert X\right\Vert_2^2)^{1/2}=\sqrt{n}}$. The details are in Anttila, Ball, and Perissinaki. In this kind of convex geometry, the study of uniform laws on convex bodies in naturally superseded by the study of ${\log}$-concave probability measures (the indicator of a convex set is ${\log}$-concave). Let ${f:\mathbb{R}^n\rightarrow\mathbb{R}_+}$ be a ${\log}$-concave probability density function with respect to the Lebesgue measure, with mean ${0}$ and covariance ${I_n}$. In other words, ${f=e^{-V}}$ with ${V}$ convex, and
$\int\!xf(x)\,dx=0\quad\text{and}\quad\int\!x_ix_jf(x)\,dx=\delta_{i,j}.$
Let ${X=(X_1,\ldots,X_n)}$ be a random vector of ${\mathbb{R}^n}$ with density ${f}$. The central limit theorem of Klartag states that if additionally ${f}$ is invariant by any signs flips on the coordinates (we say that it is unconditional, and this holds in particular for all ${\ell^p}$ balls), then, as ${n\rightarrow\infty}$,
$\frac{X_1+\cdots+X_n}{\sqrt{n}} \overset{d}{\longrightarrow} \mathcal{N}(0,1).$
Moreover, there exists a uniform quantitative version à la Berry-Esseen on cumulative distribution functions. If one drops the unconditional assumption, then a weaker version of the central limit theorem remains available. Moreover, the central limit theorem remains valid for ${\left<X,\theta\right>}$, for most ${\theta}$ in the unit sphere (some directions are however possibly bad, as shown for the cube). The work of Klartag is crucially based on a proof of the thin shell property. An alternative proof was also given by Fleury, Guédon, and Paouris (slightly after Klartag).
Further reading. The main source for this post is the expository paper by Franck Barthe entitled Le théorème central limite pour les corps convexes, d’après Klartag et Fleury-Guédon-Paouris, published in Séminaire Bourbaki (2010). One may also take a look at the survey High-dimensional distributions with convexity properties, by Klartag, and to the paper Variations on the Berry-Esseen theorem by Klartag and Sodin. On this blog, you may also be interested by the blog posts When the central limit theorem fails. Sparsity and localization and Independence and central limit theorems. It is worthwhile to mention that there exists other kinds of central limit theorems for convex bodies, such as the central limit theorem for the volume and the number of faces of random polytopes, obtained by Barany and Vu (it is another convex universe).
Last Updated on 2014-06-17
This post is devoted to few convex and compact sets of matrices that I like.
The set ${\mathcal{C}_n}$ of correlation matrices. A real ${n\times n}$ matrix ${C}$ is a correlation matrix when ${C}$ is symmetric, semidefinite positive, with unit diagonal. This means that
$C_{ii}=1, \quad C_{ji}=C_{ji},\quad \left<x,Cx\right>\geq0$
for every ${1\leq i,j\leq n}$ and for every ${x\in\mathbb{R}^n}$. In particular, we have ${C_{ij}=C_{ji}\in[-1,1]}$ for every ${1\leq i,j\leq n}$. The set ${\mathcal{C}_n}$ of all correlation matrices is convex and compact. It contains the identity matrix ${I_n}$ but is not stable by the matrix product. However, it is stable by the Schur-Hadamard entrywise product! It is the intersection of the cone of symmetric positive semidefinite matrices ${\mathcal{S}_n^+}$ with the affine subspace of equations ${S_{11}=\cdots=S_{nn}=1}$. A closer look at ${\mathcal{C}_3}$ gives, by simply computing the principal minors,
$\begin{pmatrix} 1 & a & c \\ a & 1 & b \\ c & b & 1 \end{pmatrix} \in\mathcal{C}_3 \quad\text{iff}\quad a,b,c\in[-1,1] \quad\text{and}\quad a^2+b^2+c^2\leq 1+2abc.$
This shows the ellipsoidal geometry of ${\mathcal{C}_n}$, coming from the spectral constraint, and related to the Schur complement. The set ${\mathcal{C}_n}$ is not a polytope of ${\mathbb{R}^{n\times n}}$ (i.e. is not the intersection of a finite collection of half spaces). Some authors use the name elliptope, an example of spectrahedron (set of matrices defined by linear matrix inequalities). The extremal points of ${\mathcal{C}_n}$ were studied by Ycart while the facial structure was explored by Laurent and Poljak. A useful parameterization of ${\mathcal{C}_n}$ using covariance graphs was developed by Kurowicka and Cooke.
The uniform probability measure on ${\mathcal{C}_n}$ is the normalized trace of the Lebesgue measure. Its simulation was studied by many people, and one can take a look at the works of Joe and of Lewandowski, Kurowicka, and Joe, where the volume of ${\mathcal{C}_n}$ is also computed in ${\mathbb{R}^{n(n-1)/2}}$:
$\mathrm{Volume}(\mathcal{C}_n) = \pi^{(n^2-1)/4}\frac{\prod_{k=1}^{(n-1)/2}\Gamma(2k)}{2^{(n-1)^2/4}\Gamma^{n-1}((n+1)/2)}$
if ${n}$ is odd, while if ${n}$ is even:
$\mathrm{Volume}(\mathcal{C}_n) = \pi^{n(n-2)/4}\frac{2^{(3n^2-4n)/4}\Gamma^n(n/2)\prod_{k=1}^{(n-2)/2}\Gamma(2k)}{\Gamma^{n-1}(n)}.$
If ${S}$ is a symmetric ${n\times n}$ positive semidefinite matrix then the matrix ${C}$ defined by
$C_{ij}=\frac{S_{ij}}{\sqrt{S_{ii}S_{jj}}}$
for every ${1\leq i,j\leq n}$ belongs to ${\mathcal{C}_n}$. If ${C}$ is a random element of ${\mathcal{C}_n}$ following the uniform law, then it is known that the off-diagonal entries of ${C}$, which are correlated, follow the same law, namely the Beta law of parameter ${(n/2,n/2)}$ on ${[-1,1]}$ (the density of the law of ${C}$ is also computable using a covariance graph representation, leading to the volumetric formulas above). This allows to show that if ${S}$ is the empirical covariance matrix of a standard Gaussian sample in ${\mathbb{R}^n}$ then the random correlation matrix constructed from ${S}$ (which is the empirical correlation matrix) does not follow the uniform law on ${\mathcal{C}_n}$. One may ask whether or not the asymptotic spectral properties of a uniform random element of ${\mathcal{C}_n}$, as ${n\rightarrow\infty}$, are identical to the ones of Gaussian empirical covariance matrices (Marchenko-Pastur quarter circular law). I have not tested this numerically by simulation.
The set ${\mathcal{M}_n}$ of stochastic matrices. A real ${n\times n}$ matrix ${M}$ is a stochastic matrix (we say also a Markov matrix) when all its entries are non-negative and all rows sum up to one. This means that ${M_{ij}\geq0}$ and ${M_{i1}+\cdots+M_{in}=1}$ for every ${1\leq i,j\leq n}$. In particular, ${M_{ij}\in[0,1]}$ for all ${1\leq i,j\leq n}$, and every row belongs to the non-negative portion of the ${\ell^1_n}$ unit sphere. The name Markov comes from the fact that such matrices are exactly the transition matrices of finite Markov chains on a state space of cardinal ${n}$. The set ${\mathcal{M}_n}$ of all ${n\times n}$ Markov matrices is convex and compact. It is a polytope of ${\mathbb{R}^{n\times n}}$ (i.e. intersection of half spaces) with ${n(n-1)}$ degrees of freedom. The set ${\mathcal{M}_n}$ contains the identity matrix ${I_n}$ and is stable by matrix product (it is thus a semigroup for this product).
If ${E}$ is a ${n\times n}$ matrix with non-negative entries then the ${n\times n}$ matrix ${M}$, defined by
$M_{ij}=\frac{E_{ij}}{E_{i1}+\cdots+E_{in}}$
for every ${1\leq i,j\leq n}$, belongs to ${\mathcal{M}_n}$. If the entries of ${E}$ are i.i.d. exponential random variables then the Markov matrix ${M}$ constructed from ${E}$ follows the uniform law on ${\mathcal{M}_n}$ which is the normalized trace of the Lebesgue measure. This kind of random matrices is called the Dirichlet Markov Ensemble since the rows of such random matrices are i.i.d. and follow the Dirichlet distribution with parameter ${(1,\ldots,1)}$. Note that the entries of this matrix are identically distributed and follow a Beta law, and are correlated on each row and independent across rows. The law of large numbers gives ${E_{i1}+\cdots+E_{in}=n(1+o(1))}$ and this suggests that the asymptotic spectral properties of such a random ${M}$ as ${n\rightarrow\infty}$ are identical to the ones of ${E}$ (Marchenko-Pastur quarter circular law and Girko circular law), see cb-pc-dc.
Stochastic group. Let us consider the set of all ${n\times n}$ real matrices such that ${M}$ is invertible and ${M_{i1}+\cdots+M_{in}=1}$ for all ${1\leq i\leq n}$ (note that the entries of ${M}$ are allowed to take negative values). This set is actually a group for the matrix product, sometimes called the stochastic group, and we refer to Poole for more details. This set is neither convex nor compact and I ignore what are its asymptotic spectral properties if equipped with a probability measure (notion of uniform law via constrained maximum entropy?).
The set ${\mathcal{B}_n}$ of doubly stochastic matrices. A real ${n\times n}$ matrix ${B}$ is doubly stochastic (we say also bistochastic) if ${B}$ and ${B^\top}$ are both stochastic matrices. This means that ${B_{i1}+\cdots+B_{in}=B_{i1}+\cdots+B_{in}=1}$ for any ${1\leq i\leq n}$. In particular, ${B_{ij}\in[0,1]}$ for every ${1\leq i,j\leq n}$. The set ${\mathcal{B}_n}$ is convex and compact, and forms a polytope of ${\mathbb{R}^{n\times n}}$ with ${(n-1)^2}$ degrees of freedom. It was shown by Canfield and McKay that in ${\mathbb{R}^{(n-1)\times(n-1)}}$,
$\mathrm{Volume}(\mathcal{B}_n) =\frac{\exp\left(\frac{1}{3}+n^2+o(1)\right)}{(2\pi)^{n-1/2}n^{(n-1)^2}n^{n-1}}.$
The Birkhoff polytope was remarkably studied by Barvinok, and a discrete version (contingency tables) by Barvinok and Hartigan.
The set ${\mathcal{B}_n}$ contains the identity matrix and is stable by the matrix product (semigroup!). The matrix with all entries equal to ${1/n}$ belongs to ${\mathcal{B}_n}$ and is called sometimes the Wedderburn matrix. The van der Waerden conjecture, solved by Falikman, states that the Wedderburn matrix achieves the minimum of the permanent on ${\mathcal{B}_n}$. The set ${\mathcal{B}_n}$ is referred to as the Birkhoff polytope (that is why we use the notation ${\mathcal{B}_n}$). A celebrated theorem of Birkhoff and von Neumann states that ${\mathcal{B}_n}$ is the convex envelope of permutation matrices. It can be shown, using the Minkowski-Carathéodory theorem, that each element of ${\mathcal{B}_n}$ is the convex combination of at most ${(n-1)^2+1}$ permutation matrices (which depend on the particular element). The set ${\mathcal{B}_n}$ is also known as the assignment polytope or as the transportation polytope, since each element corresponds to a transportation map between ${n}$ unit sources and ${n}$ unit targets. One should not confuse ${\mathcal{B}_n}$ with the permutahedron. The set ${\mathcal{B}_n}$ plays a role in diverse areas and can be used for instance for the proof of the Hoffman-Wielandt inequality in matrix analysis.
The uniform law ${\mathcal{L}_n}$ on ${\mathcal{B}_n}$ is the normalized trace of the Lebesgue measure. The law ${\mathcal{L}_n}$ has identical marginal laws, and is invariant by permutation of rows and of columns but is not exchangeable (i.e. invariant by any permutation of the entries). Let us consider the random walk on ${\mathcal{B}_n}$ obtained at each step by selecting randomly two rows and two columns and transferring a random amount of mass between the corresponding four entries. This is a Markov chain which admits ${\mathcal{L}_n}$ as an invariant law. Its speed of convergence was studied by Diaconis. More recently, Chatterjee, Diaconis, and Sly have shown that many asymptotic properties of random matrices drawn from ${\mathcal{L}_n}$ are identical to the ones of random matrices with i.i.d. exponential entries. In particular, the entries are close to the exponential distribution of mean ${1/n}$ and the Marchenko-Pastur quarter circular law holds true. It is conjectured that the Girko circular law is also true (we ignore how to adapt the Tao and Vu proof).
Unistochastic matrices. If ${U}$ is ${n\times n}$ unitary then the matrix ${(|U_{ij}|^2)_{1\leq i,j\leq n}}$ belongs to ${\mathcal{B}_n}$. Such matrices are sometimes called unistochastic by some physicists. However, it turns out that many elements of ${\mathcal{B}_n}$ are not unistochastic.
Tridiagonal doubly stochastic matrices. The subset of tridiagonal elements of ${\mathcal{B}_n}$ corresponds to birth and death processes on a finite state space of cardinal ${n}$ which are symmetric with respect to the uniform law. The uniform distribution on this remarkable subset was studied by Diaconis and Wood.
Permutation matrices. Recall that ${P}$ is an ${n\times n}$ permutation matrix when there exists a permutation ${\sigma}$ of ${\{1,\ldots,n\}}$ such that ${P_{ij}=\mathbf{1}_{i=\sigma(j)}}$ for every ${1\leq i,j\leq n}$. The set ${\mathcal{P}_n}$ of ${n\times n}$ permutation matrices is the set of extremal points of the convex polytope ${\mathcal{B}_n}$. The set ${\mathcal{P}_n}$ is a finite sub-group of ${\mathrm{GL}_n}$ homomorphic to the permutation group ${\Sigma_n}$. The spectrum of a permutation matrix ${P}$ associated to ${\sigma\in\Sigma_n}$ is actually a very nice object which is related to its cycles decomposition. Namely, if ${\sigma}$ has exactly ${1\leq r\leq n}$ cycles, with lengths ${\ell_1,\ldots,\ell_r}$, then ${P}$ has the spectrum of the bloc diagonal matrix
$B_1\oplus\cdots\oplus B_r$
where ${B_k}$ is ${\ell_k\times\ell_k}$ of the form (the cycle is a right shift on ${\mathbb{Z}/\ell_k\mathbb{Z}}$)
$\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 0 & 0 & \cdots & 0 \end{pmatrix}.$
The spectrum of ${B_k}$ is given by the solutions of the equation ${\lambda^{\ell(k)}=(-1)^{\ell(k)}}$ (roots of ${\pm}$ unity). The spectrum of ${P}$ is the union of the spectra of ${B_1,\ldots,B_r}$. In particular, the spectrum of ${P}$ is included in the unit circle ${\{z\in\mathbb{C};|z|=1\}}$ and depends only on ${c_1,\ldots,c_n}$ where ${c_k}$ is the number of cycles of length ${k}$ of ${\sigma}$. Note that ${c_1}$ is the number of fixed points of ${\sigma}$, and ${c_1+2c_2+\cdots+nc_n=n}$. Now if ${P}$ is uniformly distributed on ${\mathcal{P}_n}$ then ${\sigma}$ is uniformly distributed on ${\Sigma_n}$ and this allows to describe the spectrum of ${P}$ via an Ewens distribution with parameter ${\theta=1}$, which gives rise as ${n\rightarrow\infty}$ to a Poisson-Dirichlet distribution. The spectrum of uniform permutation matrices was studied for instance by Wieand, by Hambly, Keevash, O’Connell, and Stark, and more recently by Ben Arous and Dang. We ignore how to exploit these results in order to study the spectrum of the element of ${\mathcal{B}_n}$.
Note. I have learned the existence of the set of correlation matrices from Gérard Letac, an eclectic mind who was one of my teachers years ago.
Last Updated on 2014-06-17
I use like many others the free software GNU Octave for my numerical computations. Just like Scilab, Octave serves as a good free alternative to the commercial software Matlab. I prefer Octave over Scilab because it belongs to the GNU Project and because its syntax is more compatible with the one of Matlab (official objective of Octave). Scilab is quite popular in France for various reasons. Actually, one can also use the more recent Python packages matplotlib, SciPy, NumPy. You may browse the online matplotlib gallery. The installation of matplotlib/SciPy/NumPy is fearly easy on a Debian GNU/Linux system:
sudo apt-get install python-matplotlib ipython
If you do not know it yet, iphyton is “an interactive shell for the Python programming language that offers enhanced introspection, additional shell syntax, tab completion and rich history”.
Last Updated on 2014-06-17
Syntax · Style · . | 6,969 | 23,626 | {"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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.845812 |
http://newsgroups.derkeiler.com/Archive/Alt/alt.radio.digital/2009-08/msg00401.html | 1,369,071,938,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699138006/warc/CC-MAIN-20130516101218-00019-ip-10-60-113-184.ec2.internal.warc.gz | 184,238,743 | 4,237 | # Re: Roberts Gemini 49
"DAB sounds worse than FM" <dab.is@xxxxxxxxxx> wrote in message
news:7falvuF2k86fsU1@xxxxxxxxxxxxxxxxxxxxx
The context of the discussion is irrelevant. The claim you made was that
I'd made a mistake with my algebra, but in reality you didn't think about
the situation properly, or you simply didn't understand that it doesn't
matter whether you add or subtract a random signal.
The equations are as follows:
The signals received are: sum = L+R+n1, difference = L-R+n2
Left = sum + difference
Left = 0.5 * [(L + R + n1) + (L - R + n2)] = L + (n1 + n2)/2
Right = sum - difference
Right =0.5 * [(L + R + n1) - (L - R + n2)] = R + (n1 + n2)/2
You claimed that the + in front of n2 in the final equation was wrong, and
even after I'd explained teh maths to you you repeated that it was wrong.
You cannot now change the subject "because somethign else was being
discussed", because what was being discussed is irrelevant to whether the
equations are correct or not. This is maths, not the debating society.
Point out the error in my proof, or admit that you're wrong.
This *blatent* error in your "algebra" is *blisteringly* obvious to *anyone*
with even a basic understanding of all this.
The *error* is the final + in front of n2 in the final equation. It should
be a minus because, since L-R is subtracted to create R, then n2 - being the
noise present on L-R - is also subtracted to create R.
That's why n2 ends up being 180degrees out of phase between the two
channels, and that's why it totally cancels itself out when the L and R
channels are summed into mono.
I seem to recall saying this about a dozen times already, to a *total and
utter idiot* who still won't listen to reason, even when it hits him on the
head and then smacks him in the face.
You really are without clue, steven.
. | 498 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2013-20 | latest | en | 0.976972 |
https://www.indiabix.com/electronics-and-communication-engineering/exam-question-papers/discussion-10415 | 1,670,276,587,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00329.warc.gz | 850,091,213 | 5,380 | # Electronics and Communication Engineering - Exam Questions Papers - Discussion
17.
A 1000 kHz carrier wave modulated 40% at 4000 Hz is applied to a resonant circuit tuned to a carrier frequency and having Q = 140. What is the degree of modulation after passing the wave through this circuit?
[A]. 0.4 [B]. 0.2 [C]. 0.27 [D]. 0.554
Explanation:
Resulting depth of modulation is given by :
when δ =
fc = 1000 x 103 Hz
fm = 4 x 103 Hz
δ =
m0 =
= = 0.27 . | 139 | 465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-49 | latest | en | 0.880687 |
trigpapapodcasts.blogspot.com | 1,493,122,831,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120349.46/warc/CC-MAIN-20170423031200-00614-ip-10-145-167-34.ec2.internal.warc.gz | 398,700,568 | 14,732 | ## Friday, January 30, 2009
### Trig Ratios SOH CAH TOA - 20:58
KEY CONCEPTS:
ONLY for RIGHT ANGLED TRIANGLES
SOH-CAH-TOA
SOH - sinA=opp/hyp
Longest side of right angled triangles is called the HYPOTENUSE. Always found on the opposite side of the right angle of the triangle.
To find the angle use the inverse of the trig buttons on your calculator.
### Angle Relationships - 20:58
KEY CONCEPTS:
Complementary Angles = 90
Supplementary Angles = 180
Opposite Angles = one another
Parallel Lines with a Transverse through it:
Corresponding Angles = one another (F-Shape)
Co-Interior Angles add up to 180 | 177 | 612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-17 | longest | en | 0.723896 |
http://www.conversion-website.com/volume/ton-water-to-barrel-oil.html | 1,643,196,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00241.warc.gz | 88,275,946 | 4,581 | # Tons (water) to barrels (oil)
## Convert tons (water) to barrels (oil)
Tons (water) to barrels (oil) converter on this page calculates how many barrels (oil) are in 'X' tons (water) (where 'X' is the number of tons (water) to convert to barrels (oil)). In order to convert a value from tons (water) to barrels (oil) just type the number of tons (water) to be converted to bbl and then click on the 'convert' button.
## Tons (water) to barrels (oil) conversion factor
1 ton (water) is equal to 6.4050662693592 barrels (oil)
## Tons (water) to barrels (oil) conversion formula
Volume(bbl) = Volume (tons (water)) × 6.4050662693592
Example: Determine the number of barrels (oil) in 413 tons (water).
Volume(bbl) = 413 ( tons (water) ) × 6.4050662693592 ( bbl / ton (water) )
Volume(bbl) = 2645.2923692454 bbl or
413 tons (water) = 2645.2923692454 bbl
413 tons (water) equals 2645.2923692454 barrels (oil)
## Tons (water) to barrels (oil) conversion table
tons (water)barrels (oil) (bbl)
1276.860795232311
22140.9114579259
32204.9621206195
42269.01278331309
52333.06344600668
62397.11410870027
72461.16477139386
82525.21543408746
92589.26609678105
102653.31675947464
112717.36742216823
122781.41808486183
132845.46874755542
142909.51941024901
152973.5700729426
tons (water)barrels (oil) (bbl)
2001281.0132538718
2501601.2665673398
3001921.5198808078
3502241.7731942757
4002562.0265077437
4502882.2798212117
5003202.5331346796
5503522.7864481476
6003843.0397616155
6504163.2930750835
7004483.5463885515
7504803.7997020194
8005124.0530154874
8505444.3063289553
9005764.5596424233
Versions of the tons (water) to barrels (oil) conversion table. To create a tons (water) to barrels (oil) conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to tons (water) to barrels (oil) conversion
TableFormulaFactorConverterTop | 647 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-05 | latest | en | 0.551839 |
https://tw.knowledge.search.yahoo.com/search?p=get+a+move+on&context=%7Cgsmcontext%3A%3Acat%3A%3A%E5%85%B6%E4%BB%96+-+%E7%A7%91%E5%AD%B8%7C%7Cgsmcontext%3A%3Amarket%3A%3Ahk&ei=UTF-8&flt=age%3A1w&fr2=sortBy&b=31&pz=10&xargs=0 | 1,603,320,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00237.warc.gz | 590,773,007 | 18,120 | # Yahoo奇摩 網頁搜尋
1. ### Phy about work done 我唔要just copy and paste
...of work done: John moved a box from A to B in 3... question?? have does work on something?....funny english...please check.......I give you \$10, you get \$10. Total transfer is...
2. ### Phy about KE and PE
Q1: Potential energy of Mark, referrenced to B, when he is at O = 50g x (15 + 85) J = 50 000 J where g is the acceleration due to gravity, taken to be 10 m/s2 Hence, by conservation of energy, elastic potential energy in the string = 50 000 J...
3. ### Phy Qs. !!Urgent..pls help!
...positive. By v = u + at v = 50 - 10(1) v = 40 ms-1 The answer is A . 3. The answer is B. The net force on the apple when it is in free fall is its weight, which is...
4. ### free electron
...lattice of +ve metal ions in a sea of delocalized ... are free to move and thus got the name "free electrons... property lies on the fact that most metal atoms...
5. ### uniform circular motion
...will even topple when it is at rest on the banking) 2010-07-25 16:57:39...and the symmetry axis about the bus, α = a /h h = height of CG above the ...qestion don't give enough information to get the exact anumerical answer. This time...
6. ### 07年PHY CE MC 第31條
圖片參考：http://i.imagehost.org/0430/ScreenHunter_16_Apr_17_20_25.gif http://i.imagehost.org/0430/ScreenHunter_16_Apr_17_20_25.gif
7. ### phy gas
...predict its speed and moving direction, say, after 1 s. ...idela gas model is a simplified mathematical.... Based on the listed assumptions, you would be able to get simple formulae for...
8. ### 對流的問題 (20點)
... serves as a cooling mechanism, otherwise the temperature of the lamp may get too hot. 2)藏在棉花和羽毛...the hot pot is free to move and could carries away the heat. On the other hand, air insode the...
9. ### F4PHYSICS急!!!!!!!!!!!!!!!!!!!!
... the car is moving along the road and the car engine...only net force acting on the car is the frictional...of the car. By -f=ma, a =-f/m,thus the car ...,the car get hot as the work done...
10. ### proof of KE=1/2mv^2
...would overestimate the actual displacement that travelled. That's why you got twice the amount of the actual kinetic energy. The correct derivation is... | 638 | 2,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.834799 |
https://chemistry.stackexchange.com/questions/130809/how-to-compute-change-in-gibbs-energy-given-change-in-entropy-and-enthalpy-found | 1,718,993,945,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00009.warc.gz | 136,976,786 | 40,550 | # How to compute change in Gibbs energy given change in entropy and enthalpy found over a temperature range?
I was given the following question by my tutor:
The variation of enthalpy ($$\Delta H$$) and entropy ($$\Delta S$$) in a reaction carried out at constant pressure in the temperature range between $$\pu{300^\circ C}$$ and $$\pu{350^\circ C}$$, are $$\pu{1700 kcal}$$ and $$\pu{15 kcal/K}$$, respectively. Calculate the range of temperatures in which this reaction is spontaneous.
But my book on thermodynamics says that the change in Gibbs' energy for a process can only be calculated if we start and end at the same temperature (the temperature can vary during the process as heat is added or lost from the system to the surroundings, but the end points must have constant temperature and pressure.) In this question the changes in entropy and enthalpy are found over a range of temperatures where the final temperature is not the same as the initial temperature, but the pressure is always constant.
Is my tutor's question invalid, is my book wrong in saying that the temperature has to be the same at the initial and final values, or am I thinking about this wrong?
• $\ce{\Delta G}$ can be calculated by the expression ${\Delta G = \Delta H - T \Delta S}$. Why don't you calculate the change of H and S, then put ${\Delta G = 0}$. This an equation with one unknown T. A simple substraction plus a division gives you the minimum temperature for the reaction to become spontaneous. Commented Mar 30, 2020 at 17:04
• @Maurice Yes, this is the "solution" I was given, but my question is not regarding answering the question, it's whether the question is valid since the enthalpy and entropy were taken over a range of temperatures rather than at constant end points of temperature and pressure (Gibb's function is defined so that a spontaneous process with starting pressure $P_0$ and temperature $T_0$ and the same pressure $P_0$ and temperature $T_0$ at the end point yields a $\Delta G < 0$, according to my book!)
– SON1
Commented Mar 30, 2020 at 17:16
• @ Son1. You are right in stating that in order to calculate $\Delta G$, the temperature of the system must be the same in the beginning and at the end of the reaction. But it does not mean that this temperature must stay at a given value. The reaction may occur at any temperature, provided it is the same at the beginning and at the end of the reaction. $\Delta G$ can be calculated at any temperature, provided you know the change in enthalpy and entropy, and provided these changes in H and S are not changing with the temperature. Commented Mar 30, 2020 at 19:49
• I think what is (too loosely) being referred to in the problem statement is the standard changes in enthalpy and entropy for the reaction. I think the intent is for you to assume that they are nearly constant over the temperature range from 300 C to 350 C. And I think you are supposed to use the rule of thumb that a reaction is spontaneous at a given temperature if the change in Gibbs free energy is less than zero. Commented Mar 30, 2020 at 20:27
$$G=H-TS\Rightarrow\mathrm dG=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT$$
$$\Rightarrow \text{ (at constant T) } \mathrm dG=\mathrm dH-T\,\mathrm dS\Rightarrow\Delta G=\Delta H-T\,\Delta S$$
So, yes, you need $$\Delta T=0$$ for $$\Delta G=\Delta H-T\Delta S$$ to hold.
However, this holds at any $$T$$.
E.g., suppose you do the reaction at $$300\ \mathrm K$$. $$\Delta T=0$$, so $$\Delta G=\Delta H-T\,\Delta S$$ holds.
But suppose you repeat reaction at $$350\ \mathrm K$$. It's still the case that $$\Delta T=0$$, so $$\Delta G=\Delta H-T\,\Delta S$$ still holds.
As far as the question itself goes, I agree with Chet Miller that it appears to be badly worded. It seems what your tutor meant to say was:
"Between $$300\ \mathrm{^\circ C}$$ and $$350\ \mathrm{^\circ C}$$, assume that $$\Delta H$$ and $$\Delta S$$ are constant, with values of $$1700\ \mathrm{kcal}$$ and $$15\ \mathrm{kcal/K}$$, respectively. Calculate the range of temperatures over which $$\Delta G\le0$$, i.e., the range the textbook labels as 'spontaneous'". | 1,057 | 4,128 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.937138 |
https://math-webwork3.unl.edu/webwork2/html2xml?&answersSubmitted=0&sourceFilePath=UNL-Problems/100A-Problems/Workbook/systems_of_linear_equations/solve9JustHint.pg&problemSeed=123567890&displayMode=MathJax&courseID=OERProblemCourse&userID=anonymous&course_password=anonymous&outputformat=unl | 1,714,046,159,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297292879.97/warc/CC-MAIN-20240425094819-20240425124819-00172.warc.gz | 345,283,089 | 2,210 | Solve the system of linear equations if possible. If there is no solution, enter NONE for both $x$ and $y$. If there are infinitely many solutions, enter $x$ for $x$ and an expression in terms of $x$ for $y$ (i.e. solve for $y$).
\begin{cases} x-y&=9 \\ 10 x + 10 y &= 9 \end{cases} $x=$
$y=$ | 99 | 293 | {"found_math": true, "script_math_tex": 9, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-18 | latest | en | 0.696999 |
http://article-search-online.com/specifications-for-a-part-for-a-dvd-player-state/ | 1,679,338,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943555.25/warc/CC-MAIN-20230320175948-20230320205948-00006.warc.gz | 4,816,781 | 10,004 | # Specifications For A Part For A Dvd Player State
By | July 24, 2022
Specifications For A Part For A Dvd Player State. The process that produces the parts has a mean of 25 3 ounces and a standard deviation of 21 ounce. The distribution of output is normal.
12/11/2017 08:51 am due on: (round your z value and final answer to 2. Specifications for a critical part for a dvd player state that the part should weigh between 22 and 26 ounces.
### The Distribution Of Output Is Normal.a.what Percentage Of Parts Will Not Meet The Weight Specs?B.within What Values Will 95.44 Percent Of.
Specifications for a part for a dvd player state that the part should weigh between 24.7 the parts has. The distribution of output is normal. What percentage of parts will not meet the weight specs?
### Specifications For A Critical Part For A Dvd Player State That The Part Should Weigh Between 22 And 26 Ounces.
Specifications for a part for a dvd player state that the part should weigh between 24.2 and 25.2 ounces. The distribution of output is normal. A.what percentage of parts will not meet.
### What Percentage Of Parts Will Not Meet The Weight Specs?
Specifications for a part for a dvd player state that the part should weigh between 24.6 and 25.6 ounces. The distribution of the weights of the part is normal. Specifications for a part for a dvd player state that the part should weigh between 24 and 25 ounces.
### The Process That Produces The Parts Yields A Mean Of 24.5 Ounces And A Standard Deviation Of.2 Ounce.
The process that produces the parts has a mean of 24.7 ounces and a standard deviation of.22 ouces. A) what percentage of parts will meet the weight specifications? The distribution of output is normal.a.what percent pf parts will not meet the weigt specs?
### The Distribution Of Output Is Normal.
The distribution of output is normal. Specifications for a part for dvd player state that the part should weight between 24.9 and 25.9 ounces. The pricess that produces the parts has a mean of \$25.4 once and a standard deviation of 0.22 ounce. | 483 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.878947 |
https://www.aol.com/article/2011/09/15/learning-mathanese-how-to-calculate-the-p-e-ratio/20044007/?gen=1 | 1,495,648,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607849.21/warc/CC-MAIN-20170524173007-20170524193007-00126.warc.gz | 842,980,443 | 28,882 | # Learning Mathanese: How to Calculate the P/E Ratio
Math: the four-letter word you can say on TV yet so reviled that people go great lengths to avoid it, even when they know doing so puts their financial well-being in peril.
Wait! Don't click away. Today brings part two in a continuing series that spotlights the major computations found here at DailyFinance. Or, as we'll call it, a short course in Mathanese -- aka the numbers behind investing's (and life's) big equations.
Don't worry, there's nothing too complicated here, despite how difficult and intimidating professional money managers and Wall Street talking heads make it seem. We're talking about math most 10-year-olds have learned upon graduating fifth grade. And today, it's all about the P/E ratio.
The Two Things That Matter Most: Price and Earnings
As with all buyers of things, investors like shorthand. Like the P/E ratio: investing shorthand for the cost of buying a claim on a company's earnings.
See, investors are also owners. Buying a share of stock means buying a piece of a business. Each "share" has a claim on profits, so when public companies report quarterly, they report the amount earned per share -- the amount that owners would be entitled to if all profits were paid out rather than reinvested for growth.
Say you read that Google (GOOG) trades for 18.9 times earnings. The numbers represent the P/E ratio, and the ratio represents the premium per share you must pay in order to lay claim to a share of Google's \$28.12 in earnings reported over the last four quarters. Here's the math:
[Current price per share / four full quarters of earnings per share]
In this case, the numbers are: [\$532.07 / \$28.12] = 18.9
Now here's where the math gets tricky. The P/E ratio can cover any period where there are four quarters of reported or estimated earnings. So when you read that Google trades for less than 13 times "forward" earnings, the equation changes to dividing current price by the consensus estimate for earnings in the year ahead. The math stays the same even if the numbers change.
The P/E is a relative measure and as such can't tell you much without other ratios to compare with. Saying Google trades for 18.9 times earnings means little. However, saying Google trades for 18.9 times earnings while Apple's (AAPL) P/E ratio is 15.4 and Microsoft's (MSFT) is 9.9 says quite a bit. Google is more expensive, and needs to grow earnings faster in order to justify its share price.
Have questions or comments? Click here to send Tim an email.
Motley Fool contributor Tim Beyers is a member of the Motley Fool Rule Breakers stock-picking team. He owned shares of Apple and Google at the time of publication. Check out Tim's portfolio holdings and writings, or connect with him on Google+ or Twitter, where he goes by @milehighfool.
The Motley Fool owns shares of Google, Apple, and Microsoft. Motley Fool newsletter services have recommended buying shares of Google, Apple, and Microsoft, as well as creating a bull call spread positions in Apple and Microsoft. | 679 | 3,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-22 | latest | en | 0.953738 |
https://artofproblemsolving.com/wiki/index.php/2022_USAMO_Problems | 1,721,433,713,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00386.warc.gz | 99,344,803 | 13,851 | # 2022 USAMO Problems
## Day 1
### Problem 1
Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.
### Problem 2
Let $b\geq2$ and $w\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length 1.
We assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.
$[asy] size(10cm); real w = 2*Sin(18); real h = 0.10 * w; real d = 0.33 * h; picture wht; picture blk; draw(wht, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle); fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle, black); // draw(unitcircle, blue+dotted); // Original polygon add(shift(dir(108))*blk); add(shift(dir(72))*rotate(324)*blk); add(shift(dir(36))*rotate(288)*wht); add(shift(dir(0))*rotate(252)*blk); add(shift(dir(324))*rotate(216)*wht); add(shift(dir(288))*rotate(180)*blk); add(shift(dir(252))*rotate(144)*blk); add(shift(dir(216))*rotate(108)*wht); add(shift(dir(180))*rotate(72)*blk); add(shift(dir(144))*rotate(36)*wht); // White shifted real Wk = 1.2; pair W1 = (1.8,0.1); pair W2 = W1 + w*dir(36); pair W3 = W2 + w*dir(108); pair W4 = W3 + w*dir(216); path Wgon = W1--W2--W3--W4--cycle; draw(Wgon); pair WO = (W1+W3)/2; transform Wt = shift(WO)*scale(Wk)*shift(-WO); draw(Wt * Wgon); label("W", WO); /* draw(W1--Wt*W1); draw(W2--Wt*W2); draw(W3--Wt*W3); draw(W4--Wt*W4); */ // Black shifted real Bk = 1.10; pair B1 = (1.5,-0.1); pair B2 = B1 + w*dir(0); pair B3 = B2 + w*dir(324); pair B4 = B3 + w*dir(252); pair B5 = B4 + w*dir(180); pair B6 = B5 + w*dir(144); path Bgon = B1--B2--B3--B4--B5--B6--cycle; pair BO = (B1+B4)/2; transform Bt = shift(BO)*scale(Bk)*shift(-BO); fill(Bt * Bgon, black); fill(Bgon, white); label("B", BO); [/asy]$
Prove that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.
### Problem 3
Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x,y\in \mathbb{R}_{>0}$ we have $$f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y).$$
## Day 2
### Problem 4
Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.
### Problem 5
A function $f: \mathbb{R}\to \mathbb{R}$ is essentially increasing if $f(s)\leq f(t)$ holds whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.
Find the smallest integer $k$ such that for any 2022 real numbers $x_1,x_2,\ldots , x_{2022},$ there exist $k$ essentially increasing functions $f_1,\ldots, f_k$ such that$$f_1(n) + f_2(n) + \cdots + f_k(n) = x_n\qquad \text{for every } n= 1,2,\ldots 2022.$$
### Problem 6
There are 2022 users on a social network called Mathbook, and some of them are Mathbook-friends. (On Mathbook, friendship is always mutual and permanent.)
Starting now, Mathbook will only allow a new friendship to be formed between two users if they have at least two friends in common. What is the minimum number of friendships that must already exist so that every user could eventually become friends with every other user? | 1,283 | 3,637 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 45, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-30 | latest | en | 0.625779 |
https://www.chegg.com/homework-help/questions-and-answers/2-kg-mass-supported-spring-undergoes-shm-amplitude-10-cm-frequency-05-hz-t-0-velocity-mass-q1994603 | 1,529,578,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00629.warc.gz | 795,989,479 | 11,295 | A 2 kg mass is supported by a spring and undergoes SHM with an amplitude of 10 cm and a frequency of 0.5 Hz. At t=0 the velocity of the mass is zero.
a) Given a maximum amplitude of 10 cm, find the maximum velocity
b) Calculate the maximum acceleration
c) How would you graph the displacement, velocity, and acceleration of this motion for a period of 6 seconds? | 92 | 365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-26 | latest | en | 0.861818 |
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# Working with Receivables
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If a company had sales of \$180,000 during the month of October, which also included \$120,000 in credit sales. October collections were \$90,000.
September 30th we will debit balance in Accounts Receivable for \$28,000.
September 30th we will credit balance in Allowance for Uncollectible Accounts, \$1,000.
Uncollectible- Account expense, estimated as 2% of credit sales.
Write-offs of uncollectible receivables totaled \$1,200
Need help with:
1) Preparing journal to record sales, collections, uncollectible -account expense by the allowance method (percent of sales method), and write -offs of uncollectibles during October.
2) Show the ending balance in Accounts Receivable, Allowance for Uncollectible Accounts, and net Accounts receivable at October 31st. Question how much would the company expect to collect
#### Solution Preview
If a company had sales of \$180,000 during the month of October, which also included \$120,000 in credit sales. October collections were \$90,000.
September 30th we will debit balance in Accounts Receivable for \$28,000.
September 30th we will credit balance in Allowance for Uncollectible Accounts, \$1,000.
Uncollectible- Account expense, estimated as 2% of credit sales.
Write-offs of uncollectible receivables totaled \$1,200
Need help with:
1) Preparing journal to record sales, collections, uncollectible -account expense by the allowance method (percent of sales method), and write -offs of uncollectibles during October.
First, we will have to separate the amount between credit sales and cash sales. As the total sales in October is \$180,000 and that the credit sales in October is \$120,000, then the cash sales will be equaled to \$180,000 - \$120,000 = \$60,000.
Therefore, journal entries for sales will be: -
Cash 60,000
Accounts Receivable 120,000
Sales 180,000
If the ...
#### Solution Summary
This solution is comprised of a detailed explanation to prepare journal to record sales, collections, uncollectible -account expense by the allowance method (percent of sales method), and write-offs of uncollectibles during October and show the ending balance in Accounts Receivable,Allowance for Uncollectible Accounts, and net Accounts receivable at October 31st.
\$2.19
## Multiple Choice Questions on Working Capital: operating cycle, cash cycle, credit terms, accounts receivable period, accounts payable period, short-term financial policy, accounts receivable turnover, days in receivables, disbursements, net cash flow
Question 1
1. Which of the following statements is true?
b. Cash is decreased when new debt is issued to purchase holiday merchandise.
a. Accepting the credit offered by a supplier is a source of cash.
c. Increasing the use of trade credit offered by a supplier is a use of cash.
d. Collecting an accounts receivable is a use of cash.
Question 2
2. Which one of the following will increase the operating cycle?
e. increasing the inventory period
c. decreasing the cash cycle
b. decreasing the accounts payable period
a. increasing the accounts payable period
d. increasing the accounts receivable turnover rate
Question 3
3. Which one of the following actions should a manager take if he or she wants to decrease the operating cycle?
d. decrease the period of time for which credit is granted to customers
c. decrease the rate at which the average inventory is sold
a. delay payments to suppliers to decrease the cash cycle
b. increase the inventory level while maintaining constant sales
e. purchase all inventory with cash
Question 4
4. All else equal, which one of the following will decrease the cash cycle?
c. increasing the operating cycle
e. decreasing the accounts receivable turnover rate
d. decreasing the accounts payable period
b. increasing the inventory turnover rate
a. increasing the credit period granted to a customer
Question 5
5. Which one of the following credit terms is most apt to produce the shortest accounts receivable period?
b. net 10
c. 2/10, net 30
a. net 45
e. 2/20, net 45
d. 3/5, net 10
Question 6
6. Baker Industries offers credit terms of 2/20, net 60 to Charlie Co. Charlie Co. has an inventory period of 15 days and an operating cycle of 45 days. Given this, which of the following statements are correct? (I. The credit terms of Baker Industries are too restrictive; II. If Charlie Co. forgoes the discount on its purchases, it will have a negative cash cycle; III. Baker Industries is financing the accounts receivable of Charlie Co; IV. If Charlie Co. is delinquent in its payment, Baker Industries should be concerned)
b. III and IV only
a. I and II only
d. I, III, and IV only
c. II, III, and IV only
e. I, II, III, and IV
Question 7
7. Which one of the following statements is correct concerning the accounts payable period?
c. Managers generally prefer a shorter accounts payable period than a longer one.
a. The accounts payable period is equal to the cost of goods sold divided by the average accounts payable.
d. Extending the accounts payable period effectively decreases the cash needs of a firm.
e. Increasing the accounts payable turnover rate increases the accounts payable period.
b. An increase in the accounts payable period will increase the operating cycle, all else equal.
Question 8
8. A flexible short-term financial policy:
b. tends to cause more production interruptions than does a restrictive policy due to inventory shortages.
d. tends to lower the selling prices that can be charged versus the prices under a restrictive policy.
c. lowers the costs of maintaining current assets.
e. tends to indicate that the carrying costs of a firm are relatively high as compared to the shortage costs.
a. tends to increase the cash inflows of a firm in the future more so than a restrictive policy does.
Question 9
9. A firm which adopts a compromise short-term financial policy:
a. borrows sufficient long-term money so that short-term financing can be avoided.
c. relies primarily on short-term debt to meet all of its financing needs.
d. will sometimes have cash surpluses and sometimes have cash shortfalls.
b. finances its long-term assets with a combination of short-term and long-term debt.
e. will maintain a constant level of long-term debt as the firm increases in size.
Question 10
10. A negative net cash inflow on a cash budget indicates that a firm:
a. has cash outflows other than those related to accounts payable.
d. is facing bankruptcy.
e. has projected cash disbursements that exceed the projected cash collections.
c. has funds available for short-term investing.
b. utilizes both short and long-term debt.
Question 11
11. Nelson's Mulch has the following current account values for the year.
Account Beginning Balance Ending Balance
Accounts receivable \$1,300 \$1,450
Inventory 2,100 1,900
Accounts payable 1,500 1,250
These accounts represent a net _____ of cash for the year of:
e. use; \$250.
a. source; \$100.
b. source; \$150.
d. use; \$200.
c. use; \$100.
Question 12
12. Wayne's Wells has sales for the year of \$48,900 and an average inventory of \$8,800. The cost of goods sold is equal to 60 percent of sales and the profit margin is 5 percent. How many days on average does it take the firm to sell an inventory item?
a. 95 days
e. 109 days
d. 104 days
b. 99 days
c. 101 days
Question 13
13. The accounts receivable turnover rate for the Bedford Bedding Co. has gone from an average of 6.7 times to 7.2 times per year. The days in receivables has:
a. decreased by 7 days.
b. decreased by 4 days.
e. increased by 7 days.
d. increased by 5 days.
c. increased by 4 days.
Question 14
14. The Winters Co. has annual sales of \$918,700. Cost of goods sold is equal to 55 percent of sales. The firm has an average accounts payable balance of \$72,400. How many days on average does it take The Winters Co. to pay its suppliers?
c. 38 days
d. 46 days
e. 52 days
b. 34 days
a. 29 days
Question 15
15. The Sun Lee Co. has a receivables turnover rate of 11.5, a payables turnover rate of 9.8, and an inventory turnover rate of 13.6. What is the length of the firm's operating cycle?
c. 37 days
a. 15 days
b. 22 days
d. 59 days
e. 67 days
Question 16
16. Robert's International currently has an inventory turnover of 15, a receivables turnover of 18, and a payables turnover of 10. How many days are in the cash cycle?
a. 8 days
c. 45 days
e. 81 days
b. 22 days
d. 74 days
Question 17
17. Die Cast, Inc., has these projected sales estimates:
April May June July
Sales \$1,200 \$1,300 \$1,700 \$1,900
The company collects 15 percent of its sales in the month of sale, 70 percent in the following month, and another 12 percent in the second month following the month of sale. Die Cast never collects 3 percent of its sales. What is the amount of the June collections?
b. \$1,406
c. \$1,423
e. \$1,631
d. \$1,447
a. \$1,309
Question 18
18. The Thomas-Davis Co. has the following estimated sales:
Q1 Q2 Q3 Q4
Sales \$3,800 \$3,300 \$2,800 \$4,400
Purchases are equal to 67 percent of the following quarter's sales. The accounts receivable period is 45 days and the accounts payable period is 60 days. Assume that there are 30 days in each month. Thomas-Davis will purchase _____ of goods in quarter 3 and pay their suppliers _____ during quarter 3.
a. \$1,876; \$2,099
b. \$1,876; \$2,233
d. \$2,948; \$2099
c. \$2,948; \$1,876
e. \$2,948; \$2,233
Question 19
19. The Co-Co Co. purchases are equal to 55 percent of the following month's sales. The accounts payable period for the purchases is 60 days while all other expenditures are paid in the month during which they are incurred. Assume that each month has 30 days. The company has compiled this information:
April May June July
Sales \$4,500 \$5,200 \$5,700 \$6,100
Payroll expenses 400 500 550 575
Rent and other expenses 900 940 980 1,020
Taxes and insurance 2,500 100 2,500 0
What is the total amount of Co-Co's disbursements for the month of June?
d. \$7,165
c. \$6,890
b. \$5,352
a. \$4,360
e. \$8,048
Question 20
20. The Complete Co. has projected their first quarter sales at \$7,500, second quarter sales at \$8,000, and third quarter sales at \$8,400. The firm's cost of goods sold is equal to 55 percent of the next quarter's sales. The accounts receivable period is 45 days and the accounts payable period is 60 days. At the beginning of the first quarter, the firm has an accounts receivable balance of \$3,600 and an accounts payable balance of \$2,750. The firm pays \$1,200 a month in cash expenses and \$200 a month in taxes. At the beginning of the first quarter, the cash balance is \$300 and the short-term loan balance is zero. During the first quarter, the firm is planning on spending \$2,500 for some new equipment. The firm maintains a minimum cash balance of \$25. Assume that each month has 30 days. The net cash flow for the first quarter is _____ and the cumulative cash surplus (deficit) at the end of the first quarter, prior to any short-term borrowing, is:
b. -\$767; -\$492.
a. -\$767; -\$518.
c. -\$767; -\$467.
d. \$1,733; -\$518.
e. \$1,733; -\$492.
View Full Posting Details | 2,846 | 11,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-47 | longest | en | 0.95238 |
http://stackoverflow.com/questions/13347387/find-value-in-2d-array-and-return-value-in-adjacent-cell | 1,394,912,380,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678699096/warc/CC-MAIN-20140313024459-00017-ip-10-183-142-35.ec2.internal.warc.gz | 135,324,185 | 15,284 | # Find value in 2D array and return value in adjacent cell
``````**Sheet 1**
ColumnA B C D E F G H
------------------------------------------------------------
EURUSD 1.2765 1 ACCOUNT624 2 account125 1 account834
EURCAD 1.01 2 Account49 3 account45 2 account67
EURGBP 0.78 2 Account777 1 account45 2 account678
**Sheet 2**
ColumnA B C D
---------------------------------------
EURUSD 1.2765 Account 624 ?
EURUSD 1.2765 Account 125
EURUSD 1.2765 Account 834
``````
In Sheet 1 above each row shows a currency trade and what quantity goes to each account.In Sheet 2 each row shows 1 account only. I would like to populate columnd D in sheet 2 with the quantites from sheet 1.
Breaking it up into steps, i would like to:
1. Find the price in Sheet2!B1 in sheet1
2. On the same row in sheet1, find the cell containing the same account as Sheet2!C1
3. Return value in cell to the left of cell with matching account
I have used index/match before but I can't get it to work for 2 dimensional arrays. Can anyone help with a formula? Thanks in advance!
-
It's not pretty, but using what you requested - to find the match based upon price in column B (I would say your safer bet would be to use the Currency conversion "EURUSD", for example, since what if 2 currencies have the same ocnversion rate??), paste this formula in cell D1 on your second sheet:
``````=OFFSET(Sheet1!\$B\$1,MATCH(B1,Sheet1!\$B\$1:\$B\$3,0)-1,MATCH(C1,OFFSET(Sheet1!\$B\$1,MATCH(B1,Sheet1!\$B\$1:\$B\$3,0)-1,0,1,10),0))
``````
You can then drag it down / change ranges as needed.
(PS - I'm also assuming you made a mistake on sheet2 and that the account numbers will be typed the exact same in both sheets)
-
@ John Bustos Thanks. It's almost working,just needs some tweaks. firstly,this is returning the value in the cell to the right not left. Also sorry I wasnt very clear, I can have dozens of accounts so I would like it to read over to column EZ. Can you modify it please? – Brackers Nov 12 '12 at 17:05
`=OFFSET(Sheet1!\$B\$1,MATCH(B1,Sheet1!\$B\$1:\$B\$3,0)-1,MATCH(C1,OFFSET(Sheet1!\$B\$1,MATCH(B1,Sheet1!\$B\$1:\$B\$3,0)-1,0,1,150),0)-2)` - That should do it – John Bustos Nov 12 '12 at 17:16
looks good, thanks – Brackers Nov 13 '12 at 11:14 | 717 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2014-10 | latest | en | 0.847292 |
http://www.ck12.org/concept/Measuring-Earthquake-Magnitude/?difficulty=basic | 1,432,531,997,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928414.45/warc/CC-MAIN-20150521113208-00101-ip-10-180-206-219.ec2.internal.warc.gz | 389,194,256 | 15,949 | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Measuring Earthquake Magnitude
## Scientists use seismometers to measure earthquake magnitude.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Measuring Earthquake Magnitude
Covers the instruments scientists use to measure earthquake magnitude.
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• Lesson Plan
## A Whole Lot of Shakin' Goin' On! Lesson Plan
Students collect data to compare the 1906 and 1989 California earthquakes, and determine the factors that influence the amount of shaking that occurs in an area due to an earthquake.
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• Practice
0%
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## Measuring Earthquake Magnitude Quiz
Quiz on Measuring Earthquake Magnitude
0
• Critical Thinking
## Measuring Earthquake Magnitude Discussion Questions
A list of student-submitted discussion questions for Measuring Earthquake Magnitude.
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## Measuring Earthquake Magnitude Pre Read
To use prior knowledge of root words to learn new vocabulary words using Etymology.
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## Measuring Earthquake Magnitude Post Read
To reinforce and increase concept comprehension, and to analyze similarities and differences between topics using a Two Column Table.
0
• Real World Application
## Earth Polygraph
Students will look at actual seismograph data to identify S and P waves, determine magnitude, learn about finding epicenters, and calculating wave speeds.
1
• Real World Application
## Mexico City, 1985
Describes the plate tectonics setting of the 1985 earthquake and why Mexico City was so hard hit.
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• Study Guide
## Measuring Earthquake Magnitude Study Guide
This study guide summarizes the key points of Measuring Earthquake Magnitude. You can download and customize it to suit your needs and study habits.
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## Seismogram: Regional Earthquake
Illustrates a seismograph recording an earthquake 50 miles away.
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• Flashcards | 478 | 2,332 | {"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} | 3.03125 | 3 | CC-MAIN-2015-22 | latest | en | 0.821187 |
https://yetanothermathprogrammingconsultant.blogspot.com/2017/01/solving-takuzu-puzzles-as-mip-using-xor.html | 1,561,450,220,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00167.warc.gz | 971,945,165 | 18,087 | Sunday, January 1, 2017
Solving Takuzu puzzles as a MIP using xor
The Takuzu puzzle (1) can be solved using a MIP model. Interestingly we can use an xor condition in the model.
Assuming an $$n \times n$$ board (with $$n$$ even), we need to fill the cells $$(i,j)$$, with values 0 or 1, with the following restrictions:
1. The number of ones in each row and in each column is equal to $$\displaystyle\frac{n}{2}$$.
2. Then we have also: The number of zeroes in each row and in each column is equal to $$\displaystyle\frac{n}{2}$$.
3. There can be no duplicate rows and no duplicate columns.
4. In a row or column there cannot be more than two consecutive zeroes or ones.
A small example is given in (1):
Of course our central variable will be $$x_{i,j}\in\{0,1\}$$. Condition 1 can be easily modeled using two constraints:
\begin{align} &\sum_j x_{i,j} = \frac{n}{2}\>\> \forall i\\ &\sum_i x_{i,j} = \frac{n}{2}\>\> \forall j\end{align}
Condition 2 we don’t need to model explicitly: this follows from the above constraints.
Condition 3 (no duplicate rows or columns) is not very easy. One way to handle this, is to calculate a value for each row and column as follows:
\begin{align}&v_i = \sum_j 2^{j-1} x_{i,j}\\&w_j = \sum_i 2^{i-1} x_{i,j}\end{align}
and then make sure there are no duplicate $$v_i$$’s or duplicate $$w_j$$’s. This can be done using some all-different formulation (3). A big-M approach to implement $$v_i \le v_{i’}-1 \textbf{ or } v_i \ge v_{i’}+1, \forall i\ne i’$$ can look like:
\begin{align}&v_i \le v_{i’}-1 + M \delta_{i,i’}\\&v_i \ge v_{i’}+1 – M (1-\delta_{i,i’})\\&\delta_{i,i’} \in \{0,1\}\end{align}
However I want to attack this in a different way, using an xor condition (4). The constraints involved can be conceptualized as:
\begin{align}&\sum_j |x_{i,j} – x_{i’,j}| \ge 1 \>\> \forall i\ne i’\\&\sum_i |x_{i,j} – x_{i,j’}| \ge 1 \>\> \forall j\ne j’\end{align}
The value $$z_{i,j,i’,j’}=|x_{i,j}-x_{i’,j’}|$$ can be interpreted as $$z_{i,j,i’,j’}=x_{i,j} \textbf{ xor } x_{i’,j’}$$: zero if $$x_{i,j}$$ and $$x_{i’,j’}$$ are identical and one if they are different. This in turn can be written as a set of inequalities (4):
\begin{align}&z_{i,j,i’,j’}\le x_{i,j} + x_{i’,j’}\\&z_{i,j,i’,j’}\ge x_{i,j} - x_{i’,j’}\\&z_{i,j,i’,j’}\ge x_{i’,j’} - x_{i,j}\\&z_{i,j,i’,j’}\le 2 - x_{i,j} - x_{i’,j’}\\&z_{i,j,i’,j’} \in \{0,1\}\end{align}
After this we just need:
\begin{align}&\sum_j z_{i,j,i’,j} \ge 1 \>\> \forall i\ne i’\\&\sum_i z_{i,j,i,j’} \ge 1 \>\> \forall j\ne j’\end{align}
Due to the nature of these conditions on $$z$$ we can limit the xor constraints to
\begin{align}&z_{i,j,i’,j’}\le x_{i,j} + x_{i’,j’}\\&z_{i,j,i’,j’}\le 2 - x_{i,j} - x_{i’,j’}\\&z_{i,j,i’,j’} \in \{0,1\}\end{align}
Note that we do not need to compute all $$z$$ for all combinations $$(i,j)$$, $$(i’,j’)$$ (e.g. we never compare $$x_{2,3}$$ with $$x_{3,2}$$). Furthermore we can relax $$z$$ to be continuous between zero and one. Also note that if we compared row $$i$$ and $$i’$$ then we don’t need to compare row $$i’$$ and row $$i$$. Similarly for the columns. So we can do:
\begin{align}&\sum_j z_{i,j,i’,j} \ge 1 \>\> \forall i< i’\\&\sum_i z_{i,j,i,j’} \ge 1 \>\> \forall j< j’\end{align}
Finally we need to handle restriction 4. This can be implemented as:
\begin{align}&1\le x_{i,j}+x_{i,j+1}+x_{i,j+2}\le 2\\&1\le x_{i,j}+x_{i+1,j}+x_{i+2,j}\le 2\end{align}
Usually we need to write this as four inequalities (unless we can specify ranged equations). If the summation was longer I would have introduced extra variables with bounds one and two to reduce the number of constraints.
Note that the lower bound of one can also be inferred from $$(1-x_{i,j})+(1-x_{i,j+1})+(1-x_{i,j+2})\le 2 \Rightarrow x_{i,j}+x_{i+1,j}+x_{i+2,j} \ge 1$$.
Complete model
The variables are:
The dummy variable is related to a dummy objective: we look for a feasible integer solution only.
The different $$z$$ values we need to compare is encoded in a set compare:
Finally the model equations are:
The result looks like:
---- 77 VARIABLE x.L i1 i2 i3 i4 i1 1 1i2 1 1i3 1 1i4 1 1
A large problem
The site (2) contains some “very hard” 14 x 14 puzzles. An example is:
It is interesting to see how this model behaves on this problem. It turns out the problem can be solved completely in preprocessing. E.g. Cplex and Gurobi report that zero Simplex iterations were needed and solve this problem blazingly fast:
Tried aggregator 3 times.MIP Presolve eliminated 5491 rows and 2301 columns.MIP Presolve modified 18 coefficients.Aggregator did 396 substitutions.Reduced MIP has 92 rows, 48 columns, and 298 nonzeros.Reduced MIP has 48 binaries, 0 generals, 0 SOSs, and 0 indicators.Presolve time = 0.03 sec. (7.10 ticks)Found incumbent of value 0.000000 after 0.05 sec. (8.15 ticks) Root node processing (before b&c): Real time = 0.05 sec. (8.24 ticks)Sequential b&c: Real time = 0.00 sec. (0.00 ticks) ------------Total (root+branch&cut) = 0.05 sec. (8.24 ticks)MIP status(101): integer optimal solutionCplex Time: 0.05sec (det. 8.25 ticks)
Verifying Uniqueness
A well-formed puzzle has only one solution. To verify whether a solution is unique we can use a similar strategy as in (6):
1. First solve the model to obtain a solution $$x^*_{i,j}$$.
2. Formulate a constraint to forbid this solution.
3. Add this constraint to the model and resolve. The solver should not find a new solution but rather report that this model is infeasible.
Here the cut can be formulated as:
$\sum_{i,j} x^*_{i,j} x_{i,j} \le \frac{n^2}{2}-1$
Data Entry
Entering the problem data:
is an interesting issue. In GAMS “zero” and “does not exist” is identical, so we can not just use:
table init(i,j)
i1 i2 i3 i4
i1 1 0
i2 0
i3 0
i4 1 1 0
;
I used the following, which is not very clean:
table init(i,j)
i1 i2 i3 i4
i1 1 2
i2 2
i3 2
i4 1 1 2
;
x.fx(i,j)$(init(i,j)=1) = 1; x.fx(i,j)$(init(i,j)=2) = 0;
Not sure if this is better:
parameter init(i,j) / #i.#j NA /;
$onmulti table init(i,j) i1 i2 i3 i4 i1 1 0 i2 0 i3 0 i4 1 1 0 ; variable x(i,j); x.fx(i,j)$(init(i,j)<>
NA
) = init(i,j);
Note that AMPL has a "default" facility, so we can use in a data file:
param init default -1 : i1 i2 i3 i4 :=
i1 . 1 . 0
i2 . . 0 .
i3 . 0 . .
i4 1 1 . 0
;
It makes sense to use Excel for data entry. The initial configuration in Excel can look like:
To read this into GAMS and retain the zeros we can use the following code:
parameter init(i,j);
$call 'gdxxrw i=takuzu.xlsx sq=n par=init rng=b3 rdim=1 cdim=1'$oneps
$gdxin takuzu.gdx$loaddc init
display
init;
The Squeeze=no option in the call to GDXXRW will tell it to retain the zeros. The $OnEps command tells GAMS to read the zero as an EPS special value. EPS is very useful: it is physically stored in the sparse data structures, but as soon as you do a numerical operation on it, it becomes zero. This will show: ---- 12 PARAMETER init i1 i2 i3 i4 i1 1.000 EPSi2 EPSi3 EPSi4 1.000 1.000 EPS This will now allow us to fix variables as: binary variable x(i,j); x.fx(i,j)$init(i,j) = init(i,j);
This model turned out to have quite a few interesting angles.
References
1. Takuzu, https://en.wikipedia.org/wiki/Takuzu
2. Online Takuzu puzzles, http://www.binarypuzzle.com/
3. All-different and Mixed Integer Programming, http://yetanothermathprogrammingconsultant.blogspot.com/2016/05/all-different-and-mixed-integer.html
4. xor as Inequalities, http://yetanothermathprogrammingconsultant.blogspot.com/2016/02/xor-as-linear-inequalities.html
5. Similar puzzles are Sudoku and KenKen: http://yetanothermathprogrammingconsultant.blogspot.com/2016/10/mip-modeling-from-sudoku-to-kenken.html
6. Unique Solutions in KenKen, http://yetanothermathprogrammingconsultant.blogspot.com/2016/12/unique-solutions-in-kenken.html | 2,682 | 8,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-26 | latest | en | 0.731818 |
https://physics.stackexchange.com/questions/633569/a-question-on-proca-equation-and-real-massive-vector-fields | 1,656,610,559,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00330.warc.gz | 491,087,704 | 65,439 | # A question on Proca equation and real massive vector fields
The mode expansion of a real massive real field is given by:
$$A^{\mu} (x)= \sum_r \int dp [ c_r(p)\epsilon^\mu_r(p) e^{-ip\cdot x} + c_r^\dagger(p)\epsilon^\mu_r(p) e^{ip\cdot x} ]$$
where $$dp$$ is the Lorentz invariant integration measure, $$c^\dagger$$ and $$c$$ are the creation and annihilation operators, $$\epsilon^\mu_r$$ are the polarisation vectors.
Now I'm asked to show the following:
$$\langle 0 \lvert A^\mu(x) \lvert p,r\rangle = \epsilon^\mu_r(p) e^{-ip\cdot x}$$
where $$\vert p,r\rangle = c^\dagger_r(p) \vert 0 \rangle$$. $$A^\mu$$ satisifies the Proca equation.
My idea to do this was to compute
$$A^{\dagger \mu} \vert 0 \rangle = \left( \sum_r \int dp (c_r^\dagger(p)\epsilon^\mu_r(p) e^{ip\cdot x} + (c_r(p)\epsilon^\mu_r(p) e^{-ip\cdot x} \right) \vert 0 \rangle = \sum_r \int dp \ \epsilon^\mu_r(p) e^{ip \cdot x } \vert p,r \rangle$$
Since I thought that the annihilation operator would give zero when acting on the vacuum and we have $$c^\dagger_r |0 \rangle = |p,r\rangle$$.
Taking the conjuagate I got:
$$\langle 0 | A^\mu = \langle p,r | \sum_r \int dp \ \epsilon^\mu_r(p) e^{-ip \cdot x }$$
I'm not sure if this is correct but then I would "stack" $$c^\dagger_r(p) | 0 \rangle$$ to the right of this expression but that doesn't seem to give me the right answer.
• May 3, 2021 at 10:52
In order to compute your time contraction (because it is what it is), you just have to remember that $$\langle 0|a^\dagger=0$$. So you have the following: \begin{align*} \langle 0 | A^\mu |p,r\rangle & =\langle 0 | A^{-,\mu} c^\dagger_r (p)|0\rangle \\ &=\int dp' e^{-ip'_\mu x^\mu} \sum_{s=1}^3 \epsilon^\mu_s \langle 0| c_s (p') c^\dagger_r(p) |0 \rangle \\ &=\int dp' e^{-ip'_\mu x^\mu} \sum_{s=1}^3 \epsilon^\mu_s \langle 0| c^\dagger_r(p) c_r (p') + \delta_{rs}\delta^{(3)}(p-p')|0 \rangle \\ &=\epsilon^\mu_r e^{-ip_\mu x^\mu} \end{align*} | 690 | 1,938 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-27 | longest | en | 0.739969 |
https://www.chinascientificbooks.com/finite-automata-and-application-to-cryptography-p-6391/ | 1,713,091,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00045.warc.gz | 666,105,616 | 10,602 | # Finite Automata and Application to Cryptography
## Price: \$53.00
Qty.
Author: Tao Jiren
Language: English
ISBN/ISSN: 7302175308
Published on: 2008-01
Hardcover
Foreword by Arto Salomaa
Preface
1. Introduction
1.1 Preliminaries
1.1.1 Relations and Functions
1.1.2 Graphs
1.2 Definitions of Finite Automata
1.2.1 Finite Automata as Transducers
1.2.2 Special Finite Automata
1.2.3 Compound Finite Automata
1.2.4 Finite Automata as Recognizers
1.3 Linear Finite Automata
1.4 Concepts on Invertibility
1.5 Error Propagation and Feedforward Invertibility.
1.6 Labelled Trees as States of Finite Automata
2. Mutual Invertibility and Search
2.1 Minimal Output Weight and Input Set
2.2 Mutual Invertibility of Finite Automata
2.3 Find Input by Search
2.3.1 On Output Set and Input Tree
2.3.2 Exhausting Search
2.3.3 Stochastic Search
3. Ra Rb Transformation Method
3.1 Suificient Conditions and Inversion
3.2 Generation of Finite Automata with Invertibility
3.3 Invertibility of Quasi-Linear Finite Automata
3.3.1 Decision Criteria
3.3.2 Structure Problem
4. Relations Between Transformations
4.1 Relations Between Ra -Rb Transformations
4.2 Composition of Ra Rb Transformations
4.3 Reduced Echelon Matrix
4.4 Canonical Diagonal Matrix Polynomial
4.4.1 Ra Rb Transformations over Matrix Polynomial
4.4.2 Relations Between Ra Rb Transformation and Canonical Diagonal Form
4.4.3 Relations of Right-Parts
4.4.4 Existence of Terminating Re Rb Transformation Sequence
5. Structure of Feedforward Inverses
5.1 A Decision Criterion
5.2 Delay Free
5.3 One Step Delay
5.4 Two Step Delay
6. Some Topics on Structure Problem
6.1 Some Variants of Finite Automata
6.1.1 Partial Finite Automata
6.1.2 Nondeterministic Finite Automata
6.2 Inverses of a Finite Automaton
6.3 Original Inverses of a Finite Automaton
6.4 Weak Inverses of a Finite Automaton
6.5 Original Weak Inverses of a Finite Automaton
6.6 Weak Inverses with Bounded Error Propagation of a Finite
Automaton
7. Linear Autonomous Finite Automata
7.1 Binomial Coefficient
7.2 Root Representation
7.3 Translation and Period
7.3.1 Shift Registers
7.3.2 Finite Automata
7.4 Linearization
7.5 Decimation
8. One Key Cryptosystems and Latin Arrays
8.1 Canonical Form for Finite Automaton One Key Cryptosystems
8.2 Latin Arrays
8.2.1 Definitions
……
9 Finite Automaton Public Key Crytosystems
References
Index
You May Also Like | 696 | 2,367 | {"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} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.607732 |
https://www.toppr.com/guides/business-economics-cs/analysis-of-market/equilibrium-under-perfect-competition-ii/ | 1,713,483,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00860.warc.gz | 923,027,806 | 39,415 | # Equilibrium under Perfect Competition – II
Under Perfect Competition, we know that a firm is unable to affect the price of a product even if it modifies the quantity of its output. Also, in this market structure, the input and cost conditions are given. Therefore, a firm can change the quantity of the output of a product without affecting its price. The cost and revenue conditions of a firm determine its equilibrium state (maximum profits). In this article, we will talk about a firm’s long-run equilibrium under Perfect Competition.
## Equilibrium under Perfect Competition – II
A competitive firm is in equilibrium when it earns maximum profits. This invariably depends on the cost and revenue conditions of the firm. Further, the cost and revenue conditions vary on the short and long run. Therefore, a competitive firm has four equilibrium states based on its period of operation. These are:
1. Short-run equilibrium of a competitive firm
2. Long-run equilibrium of a competitive firm
3. Short-run equilibrium of a competitive industry
4. Long-run equilibrium of a competitive industry
Today, we will talk about a firm’s long-run equilibrium under Perfect Competition.
## A firm’s Long-run equilibrium under Perfect Competition
Long-term is the period in which the firm can vary all of its inputs. There are no fixed costs and therefore, the AFC or Average Fixed Cost curve vanishes. Also, the Average Cost (AC) curve represents the Average Total Cost (ATC) curve. Further, since the firm can vary all its inputs, it can close own and leave the industry.
We know that in the long-run, the AC curve which is formed by its short-run AC curves is also U-shaped. This means that up to a certain limit, the firm experiences increasing returns and the AC curve slopes downwards.
A phase of constant returns follows in which the AC curve neither rises nor falls. Subsequently, diminishing returns to scale phase starts in which the AC curve slopes upwards.
In the long-run, new firms can also enter the industry. This is the free entry and exit feature which has two implications:
1. There is no compulsion on the firm to operate under losses and it can leave the industry.
2. No firm can earn super-normal profits. This is because when a firm earns super-normal profits, it attracts new firms to the industry. This leads to an increase in the supply which results in lowering the prices and normalizing of profits.
The figure above describes the determination of long-run equilibrium under perfect competition. As you can see, the output is measured along the X-axis and the costs along the Y-axis. Also, the firm is a price-taker.
Further, its AR curve runs parallel to the X-axis and the MR curve coincides with it.
In order to determine the equilibrium of the firm, we will consider three alternative prices that the firm receives from the industry:
### Price #1
The price in the market is below the optimum cost of the firm (OP0). From this cost, we get a corresponding average revenue of AR0 and Marginal Revenue of MR0. As you can see in the figure, MR0 cuts the LMC curve at two points – E and E0.
However, none of these points is the long-run equilibrium of the firm. At point ‘E’, the LMC curve cuts the MR0 curve from above while at point E0, it cuts the curve from below. But, since AR0 < LAC, the firm incurs losses.
### Price #2
The price of the firm’s product is more than the optimum cost or the least possible average cost of the firm. In such cases, the firm is not in a state of stable equilibrium. If this price is OP2 with the average revenue curve AR2 and the marginal revenue curve MR2, then we can see that
• The LMC curve intersects the MR2 curve from below at point E2
• AR2 > LAC
This means that the firm is enjoying super-normal profits. However, this attracts new firms to the industry which increases the supply and the price falls until no firm can earn super-normal profits.
### Price #3
The price of the firm’s product is equal to its optimum cost of production. If this price is OP1 with the average revenue curve AR1 and the marginal revenue curve MR1, then we can see that
• The MR1 curve cuts the LMC curve from below at the lowest point E1
• AR1 = LAC
Therefore, the firm neither incurs a loss nor earn a super-normal profit. Therefore, there is no incentive for the existing firms to leave the market or new ones to join it. Also, the corresponding equilibrium output is OM1.
Hence, we can note that in long-run equilibrium, the firm produces an optimum output at the lowest possible average cost. Therefore, the firm operates under constant returns to scale. Also, we have
• MC = AC
• MC = MR
• AC = AR
Therefore, we have AC = AR = MC = MR.
## Solved Question on Perfect Competition
Q1. What are the four equilibrium states of a competitive firm?
Answer: The four equilibrium states are:
1. Short-Run equilibrium of a Competitive Firm
2. Long-Run equilibrium of a Competitive Firm
3. Short-Run equilibrium of a Competitive Industry
4. Long-Run equilibrium of a Competitive Industry
Q4. What is the long-run equilibrium of a competitive firm?
Answer: In long-run equilibrium, a firm produces an optimum output at the lowest possible average costs. Also, AR = AC = MC = MR.
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Get Started | 1,268 | 5,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.936597 |
https://www.doubtnut.com/question-answer/the-cost-of-30-mangoes-is-rs-150-what-is-the-cost-of-60-mangoes-in-rs-40379801 | 1,653,315,572,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00465.warc.gz | 849,334,304 | 72,018 | # The cost of 30 mangoes is Rs. 150. What is the cost of 60 mangoes (in Rs)?
Updated On: 17-04-2022
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Cost of 30 mangoes = Rs. 150 <br> Cost of one mango = (150)/(30) = Rs. 5 <br> therefore " Cost of 60 mangoes "= 60 xx 5 = Rs. 300 <br> Hence, the correct option is (d). | 124 | 375 | {"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} | 3.15625 | 3 | CC-MAIN-2022-21 | latest | en | 0.912565 |
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# In a bag there are 44 identical cards with figure of circle or square on them. There are 24 circles, of which 9 are blue and rest are green and 20 squares of which 11 are blue and rest are green. One card is drawn from the bag at random. Find the probability that it has the figure of green square.
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## Text solutionVerified
There are 44 identical cards with figure of circle or square in the bag.
∴ Total number of outcomes = 44
Number of cards with green square = 20 − 11 = 9
So, the favourable number of outcomes is 9.
∴ P(card drawn has the figure of green square) = ]
37
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Question Text In a bag there are 44 identical cards with figure of circle or square on them. There are 24 circles, of which 9 are blue and rest are green and 20 squares of which 11 are blue and rest are green. One card is drawn from the bag at random. Find the probability that it has the figure of green square. Topic Probability Subject Mathematics Class Class 10 Answer Type Text solution:1 Upvotes 37 | 382 | 1,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-22 | latest | en | 0.919532 |
http://www.pedagonet.com/puzzles/tins.html | 1,438,518,019,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042989043.35/warc/CC-MAIN-20150728002309-00202-ip-10-236-191-2.ec2.internal.warc.gz | 642,973,276 | 3,414 | Videos Test Yourself Books MathFacts Math Diagnostics Math Tricks Daily PhysEd Class Photos Worksheets Insectclopedia Musiclopedia Techlopedia Teacher Timesavers Study Guides
The Five Tea Tins
Sometimes people will speak of mere counting as one of the simplest operations in the world; but on occasions, as I shall show, it is far from easy. Sometimes the labour can be diminished by the use of little artifices; sometimes it is practically impossible to make the required enumeration without having a very clear head indeed. An ordinary child, buying twelve postage stamps, will almost instinctively say, when he sees there are four along one side and three along the other, "Four times three are twelve;" while his tiny brother will count them all in rows, "1, 2, 3, 4," etc. If the child's mother has occasion to add up the numbers 1, 2, 3, up to 50, she will most probably make a long addition sum of the fifty numbers; while her husband, more used to arithmetical operations, will see at a glance that by joining the numbers at the extremes there are 25 pairs of 51; therefore, 25×51=1,275. But his smart son of twenty may go one better and say, "Why multiply by 25? Just add two 0's to the 51 and divide by 4, and there you are!" A tea merchant has five tin tea boxes of cubical shape, which he keeps on his counter in a row, as shown in our illustration. Every box has a picture on each of its six sides, so there are thirty pictures in all; but one picture on No. 1 is repeated on No. 4, and two other pictures on No. 4 are repeated on No. 3. There are, therefore, only twenty-seven different pictures. The owner always keeps No. 1 at one end of the row, and never allows Nos. 3 and 5 to be put side by side. The tradesman's customer, having obtained this information, thinks it a good puzzle to work out in how many ways the boxes may be arranged on the counter so that the order of the five pictures in front shall never be twice alike. He found the making of the count a tough little nut. Can you work out the answer without getting your brain into a tangle? Of course, two similar pictures may be in a row, as it is all a question of their order. See answer
Brain Teaser Of The Week | 531 | 2,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.938878 |
https://podcast.mathisfigureoutable.com/1062400/12280028-ep-144-fraction-division-pt-3 | 1,718,335,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00359.warc.gz | 427,179,846 | 28,007 | # Ep 144: Fraction Division Pt 3
March 21, 2023 Pam Harris Episode 144
Ep 144: Fraction Division Pt 3
Math is Figure-Out-Able with Pam Harris
Math is Figure-Out-Able with Pam Harris
Ep 144: Fraction Division Pt 3
Mar 21, 2023 Episode 144
Pam Harris
Sure, simple fraction division is figure-out-able, but what about gnarly fractions? In this episode Pam and Kim uncover simple relationships that make division of any fractions figure-out-able!
Talking points:
• What happens when the fractions aren't as nice, especially fractions where their denominators are not factors of each other?
• Thinking quotitively
• The connection between division and fractions
• Find common denominators to divide
• Use relative size of the fractions to estimate the quotient
• We don't teach rules for students to mimic, we develop relationships
See Episodes 142, 143 and 155 for more about fraction division.
Check out our social media
Instagram: Pam Harris_math
Facebook: Pam Harris, author, mathematics education
Sure, simple fraction division is figure-out-able, but what about gnarly fractions? In this episode Pam and Kim uncover simple relationships that make division of any fractions figure-out-able!
Talking points:
• What happens when the fractions aren't as nice, especially fractions where their denominators are not factors of each other?
• Thinking quotitively
• The connection between division and fractions
• Find common denominators to divide
• Use relative size of the fractions to estimate the quotient
• We don't teach rules for students to mimic, we develop relationships
See Episodes 142, 143 and 155 for more about fraction division.
Check out our social media
Instagram: Pam Harris_math
Facebook: Pam Harris, author, mathematics education
Pam:
Hey, fellow mathematicians! Welcome to the podcast where Math is Figure-Out-Able. I'm Pam.
Kim:
And I'm Kim.
Pam:
And you found a place where math is not about memorizing and mimicking, waiting to be told or shown what to do. But it's about making sense of problems, noticing patterns, and reasoning using mathematical relationships. We can mentor students to think and reason like mathematicians. Not only are algorithms really not helpful in teaching mathematics, but rotely repeating steps actually keep students from being the mathematicians they can be.
Kim:
So, in the last two episodes, we've been answering your question from a follower of our Math is Figure-Out-Able, teacher Facebook group, and we've been diving into fraction division. So, today, we're going to hopefully finish and continue to answer Beth's question.
Pam:
So, I'm going to be honest. It might be our last episode that we do on fraction division for a minute because we have some other things we're going to do. I don't know that we're going to finish fraction division today, but I think we'll give you some really good things to think about and continue your journey, at least about thinking quotitively about fraction division.
Kim:
Yeah.
Pam:
So, what about fractions divided by fractions, especially fractions where their denominators are not factors of each other? I say that because in last week's episode, we did a bunch of fraction problems, fraction division problems, things like thirds divided by sixths, and thirds divided by twelfths. And so thirds and sixths. 3 and 6, and 3 and 12, 4 and 12. Those are all related.
Kim:
Right.
Pam:
Their factors of each other. And so, there's maybe some different thinking that happens. And so today, we kind of wanted to dive in. What if the fractions are not quite as nice? And let's do it. Are you ready, Kim?
Kim:
Yep.
Pam:
Alright. So, your first problem today... I'm getting my pen out? Do you have your pencil out?
Kim:
I do, of course. (unclear) eraser?
Pam:
Yeah, I have no eraser because I have a pen. I have a gel pen today. Let's see if it smears on my... Actually, I have a really nice gel pen. I don't think this one smears. Anyway, moving on. What is 3 divided by 2? And I know we said fraction division, but bear with me. What's 3 divided by 2?
Kim:
One and a 1/2.
Pam:
Okay, and I was going to write 1.5 until you said 1 and a 1/2, and then, I wrote 1 and the fraction 1/2. But it could be either, right? 1.5, 1 and a 1/2. How do you know?
Kim:
Because I can think about it like I can fit 2 into 3, one time. Like, there's room for 2. But then, I can only fit 2 into 3, one-half again. So, there's already one and a half 2s in 3. Yeah,
Pam:
One and a half 2s in 3. So, you're really thinking quotitively, which I kind of asked you to.
Kim:
Yeah.
Pam:
How many 2s are in 3? And you're like, "Well, I can fit a whole 2 in 3."
Kim:
Yep.
Pam:
And I've only got 1 leftover. And how many 2s are in that 1 leftover? Just a half. Cool. Okay, so you can think about 3 divided by 2 is 1 and a 1/2. Listeners, if you've never thought about something like that, 3 divided by 2, quotitively. If you've never thought about, "How many 2s can I fit into 3." You might want to pause the podcast and think about that maybe a little bit more before we go on. Maybe. Maybe think about how many. Like, if I said, "5 divided by 2," you might think "how many 2s are in 5?" Kind of see if you can kind of make some generalizations about that. Okay, so next problem. What about three-sixths divided by one-sixth. Three-sixth divided by one-sixth? What do you got?
Kim:
Three.
Pam:
Because?
Kim:
Because there's one-sixth, three times in three-sixths.
Pam:
Cool. And so, I might could draw. In fact, I should have maybe talked about 3 divided by 2. I might have had 3 candy bars. And in this case, I'm going to say, "If I wanted to share that with 2 people..." No. That's not how I want to say that. Nope,
Kim:
Yeah.
Pam:
Nope, nope. I want to say if, "I'm going to give..." No. How would I say that? I hadn't thought about a model for this one Kim ahead of time, and that's not... That wasn't smart on my part.
Kim:
Well, I was just thinking about like of a 1 size candy bar is 3 pieces long. and another size candy bars 2 pieces long. How many of the 2 piece candy bars would fit in the 3 piece candy bar? You know like, 3 bars long. If it's an extra large candy bar.
Pam:
The pieces would have to be the same size, right?
Kim:
Yeah.
Pam:
So, how many 2 piece candy bars can fit in a 3 piece candy bar?
Kim:
Yeah, (unclear).
Pam:
That's perfect. Perfect. Nicely done. Cool. Alright, so we've got 3 divided by 2 is 1 and a 1/2. 3/6 divided by 1/6 is 3. What's three-sixths divided by two-sixths?
Kim:
Oh, you just asked me three-sixths divided by one-sixth, and that was three. But now the pieces are twice as big, so there's going to be half as many of them. So, it's going to be 1 and a 1/2.
Pam:
Because half of 3 is 1 and a 1/2. So, you kind of used the relationship with the problem before. So, if I say 3/6 divided by 1/6 is 3, then three-sixths divided by something twice as big has to be half as... The answer's half as much?
Kim:
Yeah.
Pam:
So, 3/6 divided by 2/6 is 1 and a 1/2?
Kim:
Mmhmm.
Pam:
Can I say that in a slightly different way?
Kim:
Sure.
Pam:
If 3/6 divided by 1/6. How many 1/6 are in 3/6? Is 3. Then, when I asked 3/6 divided by 2/6, how many 2/6, how many something twice as big are in the same total? That would have to be half as many. 1 and a 1/2.
Kim:
Yeah. So, I actually was just thinking while you were talking about how I thought about the first problem. Like, how many 2s are in 3?
Pam:
Mmhmm.
Kim:
And so, for this problem, you could think about how many two-sixths fit inside three-sixth? So, there's going to be one 2/6 inside that 3/6. But I can only fit half of the two-sixths the second time.
Pam:
Like, in the leftover ones?
Kim:
In the leftover. Mmhmm.
Pam:
So, that's interesting. What I'm hearing you say is. I've given you three problems. To do the third problem, first you thought about the second problem to help you.
Kim:
Yeah.
Pam:
But then you also said,"But I can also think about the first problem to help me."
Kim:
Yep.
Pam:
So, listeners, this might be. If you've never actually pulled out a pen and pencil while you're listening to the podcast, this one might be one that you want to. Yeah.
Kim:
(unclear). Pull over on the side of the road.
Pam:
So, if we can just focus on how you use the first problem. It's almost like I hear you saying...the first problem you thought about 3 divided by 2....you thought about how many 2s are in 3. And the third problem, three-sixths divided by two-sixths, you thought about how many two-sixths are in three-sixths. It's almost like you were thinking about how many two-somethings could fit into three-somethings.
Kim:
Mmhmm.
Pam:
And in the first problem, you were like 2 wholes. How many 2 wholes could fit into 3 holes? And then, the third problem you thought, how many two 1/6 could fit into three 1/6? And both times the answer was 1 and a 1/2.
Kim:
Yep.
Pam:
1 and a 1/2 of those 2 whatevers could fit into those 3 whatever's. Interesting. Okay, next problem. Three-fifths divided by one-fifth?
Kim:
Three 1/5s.
Pam:
Fit into 3/5s.
Kim:
Fit into three-fifths.
Pam:
Okay, (unclear).
Kim:
Did I say that wrong?
Pam:
No, you said it, right. It's just depending on how you're... Yeah, okay. So, 3/5 divided by 1/5 is 3. Cool. Next problem. Three-fifths divided by two-fifths?
Kim:
So two-fifths can fit inside of three-fifths, one time, and then half again. So, it's going to be 1 and a 1/2.
Pam:
Again.
Kim:
Oh, and that was like the first problem, Pam.
Pam:
Say more.
Kim:
You said 3 of something, divided by 2 of something. It has been 1 and a 1/2. So, three 1/5s divided by two 1/5s is still going to be 1 and a 1/2.
Pam:
Still 1 and a 1/2. So, I wonder... Like, I'm adding a problem late in the game here. So, if I said something like three-sevenths divided by two-sevenths?
Kim:
Yeah.
Pam:
You think it still might be?
Kim:
One and a 1/2.
Pam:
One and a 1/2?
Kim:
Yep.
Pam:
What if it was 3/29 divided by 2/29?
Kim:
Yep.
Pam:
Still 1 and a half?
Kim:
Still 1 and a 1/2. Yep.
Pam:
Because 2 whatevers can fit 1 and a 1/2 times into 3 whatever's.
Kim:
Yeah.
Pam:
Whatever they are. Interesting.
Kim:
Yep.
Pam:
Alright, next next problem of the string. 10 divided by 9.
Kim:
Well, you're asking me to think quotitively, so 9 can fit into 10, one time. And then...
Pam:
But there's some stuff leftover.
Kim:
There's some stuff leftover. And 9 can fit into the stuff leftover only 1/9. So, 1 and 1/9.
Pam:
You're saying 10 divided by 9 is 1 and 1/9?
Kim:
Yeah.
Pam:
Say that... Why again?
Kim:
Because 9 can fit...
Pam:
How many 9s are in 10? You're thinking about how many 9s are in 10. Sorry to interrupt.
Kim:
Yeah. That's okay. So, 9 can fit into 10, one full time.
Pam:
And then there's what leftover?
Kim:
There's a tenth leftover.
Pam:
I think there's one leftover, yeah? A tenth of 10 is 1.
Kim:
Sorry. So, it can fit into 10, one time, and then there's 1 leftover. And then, 9 can only fit into that 1 leftover, 1/9(unclear).
Pam:
Yeah.
Kim:
It can't fit the whole time. It's just one-ninth.
Pam:
We can only get one-ninth. Yeah.
Kim:
Yeah.
Pam:
There's only... Yeah. That's crazy kind of reasoning, right? That's interesting. So, I have on my paper written 10"division sign" 9, "equals" 1 and 1/9.
Kim:
Yes.
Pam:
I also want to remind us that sometimes we can write 10 divided by 9 as 10 "fraction bar" 9.
Kim:
Yeah.
Pam:
10/9.
Kim:
Yep.
Pam:
And is 10/9 equivalent to 1 and 1/9? Like, you just reason quotitively.
Kim:
Yes.
Pam:
Oh, sure enough, sure enough. So, a couple different connections that we can make. Now, I kind of forced you to think quotitively, and you were like, "Alright, I can think about how many 9s fit into 10." And well done doing that. But we can also sort of... That connection between division and fractions, we can think about 10 divided by 9 as 10/9. Cool. Nice thinking. Alright, ready? Next problem. Ten-twelfths. So, the fraction ten-twelfths divided by nine-twelfths. Ten-twelfths divided by nine-twelfths.
Kim:
Nine-twelfths fits inside 10/12, one time.
Pam:
With some stuff leftover.
Kim:
With some stuff over. And there's a twelfth leftover. And nine-twelfths can fit into one-twelfth, a ninth.
Pam:
Like, it really doesn't fit, right. (unclear).
Kim:
It doesn't fit. It can only fit (unclear).
Pam:
(unclear).
Kim:
But you know what would have been like a little bit easier for me?
Pam:
Kim:
Pam:
Mmhmm?
Kim:
And what we kind of just talked about.
Pam:
Yeah?
Kim:
And 10/12. The thing is the twelfths. So, 10 something, 10/12, divided by 9 something, 9/12 is going to be the same. It's an equivalent answer to 10 divided by 9. So, it's still 1 and 1/9.
Pam:
So, you're saying 10/12 divided by 9/12 is equivalent to 10 divided by 9?
Kim:
Mmhmm.
Pam:
Because it's almost like 10 things divided by 9 things?
Kim:
Yep.
Pam:
It's 10 divided by 9, which is 1 and 1/9 or 10/9.
Kim:
Yep.
Pam:
Fascinating. Okay. Okay. So, similar to what we were doing before. If we can kind of think about 10/9, 10 divided by 9, we can think about 10 anythings divided by 9 anythings. There's still going to be equivalent to 10/9.
Kim:
Yep.
Pam:
Cool. Alright. Next problem. How about five-sixths divided by three-fourths? Now, this isn't related to anything we've done, so just random problem. Five-sixths divided by.
Kim:
Well, I'm glad I've been writing the problems down because.
Pam:
So, listeners, you might want to actually take the advice we have, (unclear) writing stuff down.
Kim:
So, I'm comparing the problem you just gave me and the problem prior to that. So, five-sixths is equivalent to ten-twelfths from the previous problem.
Pam:
Okay.
Kim:
And three-fourths is equivalent to the nine-twelfths in the previous problem.
Pam:
Okay.
Kim:
So, the answer is going to be the same. 1 and 1/9.
Pam:
Or, 10/9. So, Kim, what did you think, though? Like, when you saw five- sixths divided by three-fourths, is there a first thought that you thought, "Okay, how many three-fourths are in..." Or did you just go, "It's a Problem String. I'm going to see if they're equivalent."
Kim:
I totally did. I'm not going to lie. I totally did. I was like, "There's got to be something here because that problem is funky."
Pam:
Well, so when you said that problem is funky, then I actually want to pause that brief second because I think you really quickly were like, "That problem is funky. It's a Problem String. I'm going to look for a pattern. I'm going to look for something."
Kim:
Yeah.
Pam:
And I think we could potentially kind of skip over the fact that you actually considered five-sixth divided by three-fourths first. Like, mathematicians consider that. You don't just look to(unclear).
Kim:
Oh, yeah, yeah. I looked at it. And I thought about it. And I was like, "Ugh."
Pam:
And then, you said to yourself, "What else do I know?" Oh, great mathematical thinking. And you're like, "Hey, sweet! It's equivalent to the one before, and so the answer is going to be equivalent."
Kim:
Yeah.
Pam:
Let's maybe parse out a little bit. Five-sixths divided by three-fourths? You could think about how many three-fourths are in five-sixths. And then, that's like. Wow, like fourths and sixths. And those are not delightfully related.
Kim:
Right.
Pam:
(unclear) in a super, you know like, "If I just cut them in half, I get the other one," or whatever. So, then, you kind of said to yourself, "Since they're not, I wonder what else I know." And yeah, the problem before was just sitting there. But in a way, the problem before when you said, "What else do I know?" you kind of found a problem that had the same sort of pieces.
Kim:
Mmhmm.
Pam:
Like, often when we add and subtract fractions, we say to ourselves, "Oh, I can't really add sixths and fourths, so let me find equivalent fractions that have the same kind of pieces." Maybe twelfths. And then, "Hey, I could add them together." It's almost like this time you said to yourself, "I wonder if I could find the same number of pieces, so that I could divide?" I don't know that we typically have done that, teachers. I wonder, you know like, when you said,"Five-sixths divided three-fourths... Hey, but they both can turn into twelfths. And if I can turn them both into twelfths, then I could just think about the twelfths."
Kim:
Yeah. So, I think there's two things that happened too that I'm aware of that I want for our students. Like, in this moment is, One, what I was going to say to you is, I think this is the funkiest problem that we've done in the last couple of episodes.
Pam:
Yeah.
Kim:
And I want for kids to think about the fact that there are kind of different... I don't want to say "levels of problems", but like, some are a little funkier than others. And so like, I want them to be a little bit more discerning about, "I have this variety of strategies, and this problem is one that I might have to think about a little bit differently." But also, I know, you just gave me the ten-twelves divided by nine-twelfths, but I looked at it and was like, "Oh, those are equivalent," so I had to know that those were equivalent to be able to make use of it.
Pam:
Absolutely. Hey, one other thing that we should probably mention is, in this Problem String if I was doing it with students and not Kim, I think that I... Well, I guess I could have done it with you too. But I might have said, "Hey, let's actually talk about five-sixths and three-fourths. How do they relate? If I'm asking how many three-fourths fit into five-sixths, am I going to have a lot of three-fourths and five-sixths? Am I going to have... Is the answer a fraction? Three-fourths doesn't fit? Is three-fourths less than five-sixths? Like, how do those relate?" And, Kim, how would you answer that? How do those relate?
Kim:
You know, the first thing I'm thinking about...
Pam:
(unclear). I was just going to say just relative size. But go ahead. What was the first thing you were thinking?
Kim:
Well, so I know that I know a lot about three-fourths, and so I was actually thinking about what's 3/4 of 6 to see if 5/6 is more than 3/4 or less than 3/4.
Pam:
[Pam laughs] Okay. Alright. Let's go there in a minute. Can we slow that down a little bit?
Kim:
[Kim laughs] Sorry.
Pam:
I want to slow you down just a handful. So, if we were just to compare five-sixths and three-fourths, which one's bigger?
Kim:
Five-sixths.
Pam:
Five-sixths.
Kim:
Pam:
Sure, sure. And in that way, we would want to do some work with students. Five-sixths is just one unit fraction away from six-sixths. It's just one-sixth away from six-sixths. And three-fourths is just one unit fraction, one-fourth away from four-fourths. And so, I can think about what's bigger? One-fourth or one-sixth? And that can help me think about what's closer to 1. Since one-sixth is smaller than one-fourth, five-sixths is closer to 1 than three-fourths. And so, now, I have this comparison. Three-fourths is smaller than five-sixths. So, when I ask "How many three-fourths are in five-sixths?" I should get an answer of "Well, at least one. One and some extra." Right?
Kim:
Mmhmm.
Pam:
Did I say that right? Okay, cool.
Kim:
Yeah.
Pam:
Then, I could go to what you were talking about that(unclear).
Kim:
Well, I think I compared as well. I just compared it in a different way than you did. I think it's the same. Like, we were both seeking to compare.
Pam:
Oh, interesting. What is 3/4 of 6? I think your way is more multiplicative. I'll just say that. Mine was less multiplicative. Mine was more based on the definition of fractions, and I think yours was a bit more multiplicative. So, you said three-fourths of six. Why would you ask three-fourths of six?
Kim:
Because I know I have five, 1/6s, and I want to know if that's more than three-fourths or less than three-fourths. So, I thought about what is 3/4 of 6, and I know that it's 4 and a 1/2. So, if four and a half of 6 is 3/4 than 5/6 is going to be more than 3/4.
Pam:
I love it. That's brilliant thinking. Well done. Alright, cool.
Kim:
Moving on.
Pam:
So, the upshot of this problem is, we can look at a cranky problem like five-sixths divided by three-fourths, and we can say to ourselves, "I wonder if we can find the same kind of pieces, and then deal with that problem." So, you said, "Yeah, I know ten-twelfths divided by nine-twelfths. Bam! I can think about that like 10 divided by 9." Woah!
Kim:
Yep.
Pam:
Cool. Last problem set.
Kim:
Okay.
Pam:
9 divided by 4. What's 9 divided by 4?
Kim:
I just took a deep breath(unclear) a lot of talking. Okay, so I know...
Pam:
A lot of thinking.
Kim:
Yeah. So, I'm going to say that that is... I can fit two 4s in 9. Two 4s in 9. So, then I'm going to say 2 and a 1/4.
Pam:
2 and 1/4. Because two 4s is 8, so you've got 1 leftover.
Kim:
Yep.
Pam:
And then, you're thinking about how many 4s are in 1?
Kim:
Yep.
Pam:
And you're like, "That's a fourth, one-fourth."
Kim:
Yep.
Pam:
Okay, cool. So, 9 divided by 4 is 2 and a 1/4. Could I also remind you that we can think about that as 9/4?
Kim:
Yep.
Pam:
So, 9 divided by 4 is 9/4, 2 and a 1/4. Those are all equivalent. We want all that going on.
Kim:
Yep.
Pam:
Cool. Next question. What's nine-twelfths divided by four-twelfths.
Kim:
That's 9. It's similar to 9 divided by 4. So, it's still going to be 2 and a 1/4.
Pam:
Because?
Kim:
Because I have 9 somethings divided by 4 somethings, so that's the same as 9 divided by 4. I could fit four 1/12s into nine-twelfths, twice. And a fourth again.
Pam:
Another fourth leftover. Cool. So, similarly, if I know something about 9 divided by 4, then I can reason about 9 anythings divided by 4 anythings, would also have that equivalent 9/4 solution. Nice. Last problem. Three-fourths divided by one-third. How many one-thirds are in three-fourths?
Kim:
That's nice.
Pam:
[Pam laughs].
Kim:
No, it really is because I'm thinking about how I could...I can scale both those up, and I could use twelfths. I could use a couple of different things, but I'm going to go with twelfths. And if I scale the three-fourths, that would be equivalent to nine-twelfths. And if I scale the one-third to twelfth, that would be four-twelfths. And then, I land back in the same place where it's 9/12 divided by 4/12, or 9 divided by 4, which is 2 and a 1/4.
Pam:
And so, you're saying the answer to 3/4 divided by 1/3 is 2 and a 1/4 because it's equivalent to the two problems that we did before?
Kim:
Yeah.
Pam:
Or 9/4. Either one.
Kim:
I'm going to go looking for problems like these.
Pam: | 6,400 | 22,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.887989 |
https://www.milieu-nieuws.nl/MILgrinding/2018/1014.html | 1,627,272,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00357.warc.gz | 929,275,388 | 8,167 | how to find out ball mill rpm from motor and gear box output rpm
## how to find out ball mill rpm from motor and gear box output rpm
#### What it is the optimun speed for a ball mill ...
Motor Wiring Diagram D.C. Motor Connections Your motor will be internally connected according to one of the diagrams shown below. These connections are in accordance with NEMA MG-1 and American Standards Publication 06. 1 - 1956. Use figure 1 if your motor has a single voltage shunt field. Use figure 2 if your motor has a dual voltage shunt field.
#### RPM Calculator | Engine Rpm Calculator
Jan 28, 2014· Setting the clock speed too low and your machine speed too high can have a huge negative effect on machine performance (missed steps, limited speed). This clock speed is governed by the number of steps the motor must turn to move the machine an inch, and the maximum top speed the machine will obtain in inches per minute.
#### Speed Reducers - Grainger Industrial Supply
Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input? Formula: GPM x PSI x 36.77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.
#### Pulley and RPM Calculator - Culver Armature and Motor
May 07, 2012· Find the amount of Torque that the 1.6 Hp motor delivers at 3600 RPM. Taking the torque formula one step at a time, divide the HP of 1.6 by the RPM of 3600. The answer will be 0.00044. This is a very small number. Next multiply 0.00044 times the constant of 63,025. The final Torque value of the motor is 28.16 ft-lbs of Torque.
#### Pulley Calculator. RPM, Belt Length, Speed, Animated Diagrams
Oct 19, 2006· What it is the optimun speed for a ball mill ? - posted in Pyrotechnics: I have done a ball mill, recenly finished, but the motor has too rpms, is too fast for use in a ball mill (the pvc cylinder that i use, left of the shafts). With the motor i will use a 40 mm pulley, because i have a 50 mm driven pulley, in one of my two shafts.
#### Pump Power Calculation Formula | Specific speed of a ...
For example, combining a Fisher Price and CIM motor would require an extra 3:1 gear reduction for the Fisher Price motor because its output speed is approximately 3 times faster than a CIM's. If the output speeds are not matched, it will cause added resistance in the gearbox and negate any benefits of having multiple motors.
#### Bearing Handbook for Electric Motors
Optimize speed and performance while increasing equipment longevity with speed reducers from Grainger. The motor transmits power to the input shaft of the reducer during equipment operation. The speed reducer coverts the power to a lower output speed while transmitting to …
#### How to Calculate Electric Motor Torque - RC Groups
Jan 05, 2015· Revolutions per minute, or RPM, are a measure of how fast a rotating object turns. Knowing how fast an object turns is important in determining wind speed, gear ratios, how powerful a motor is, and how well bullets fly and penetrate.There are a number of ways to calculate RPM, depending on what the value is needed for; we'll stick with some of the simplest.
#### Power, Torque and RPM Calculations - RACELOGIC Support Centre
By input we understand the source of power, in our case could be an electric motor. By output we mean where the power is delivered (e.g. hydraulic pump). The gear ratio i can be calculated in two ways: as a ratio between the number of teeth of the output gear and the number of teeth of the input gear [i=frac{z_{OUT}}{z_{IN}}]
#### How To Find Out Ball Mill Rpm From Motor And Gear
Output RPM formula = Input RPM times number of driving gear teeth divided by number of driven gear teeth. A small transmission in a mixer has a 12-tooth gear on the motor shaft driving a 72-tooth gear on the mixer blade. The motor is turning at 1,800 RPM. What is the mixer blade speed?
#### Calculating Output Speed Using Pulley Diameters and Input ...
Oct 16, 2019· Using the idea of gear ratios, it's easy to figure out how quickly a driven gear is rotating based on the "input" speed of the drive gear. To start, find the rotational speed of your drive gear. In most gear calculations, this is given in rotations per minute (rpm), though other units of …
#### How to calculate the shaft diameter from the torque ...
Use the RPM calculator to determine what RPM your engine will run based on gear ratio, tire diameter, transmission and MPH. You can try various combinations to come up with an appropriate combination of ring and pinion gear ratio and tire diameter for your vehicle. The RPM calculator should be used as a guideline for comparative purposes only.
#### Hydraulic Motor Calculations - Womack Machine Supply Company
Sep 01, 2008· At 12 pole speed (50Hz here, so only about 360 rpm) it 'only' puts out 2 HP, but to put out that much power at that speed, the torque has to be the same as a 12HP 2 pole motor. Consequently it's a massive frame: two strong men couldn't lift it.
#### Working out Revolutions Per Minute
Use the RPM calculator to determine what RPM your engine will run based on gear ratio, tire diameter, transmission and MPH. You can try various combinations to come up with an appropriate combination of ring and pinion gear ratio and tire diameter for your vehicle. The RPM calculator should be used as a guideline for comparative purposes only.
#### 4 Easy Ways to Determine Gear Ratio (with Pictures)
May 14, 2018· Inputs that we need to calculate shaft diameter. As we know Torsional equation (Read more about the torsional equation here). Where. τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).. r = Radius of the shaft.. T = Twisting Moment or Torque.. J = Polar moment of inertia.. C = Modulus of rigidity for the shaft material.. l = Length of the shaft.
#### Gear Reducing Formulas - Engineering ToolBox
In the example below the DRIVER gear is large than the DRIVEN gear. The general rule is - large to small gear means 'multiply' the velocity ratio by the rpm of the first gear. Divide 60 teeth by 30 teeth to find the velocity ratio. Multiply this number (2) by the rpm (120). This gives an answer of 240rpm.
#### Spur Gear Calculator
Performing a conversion from RPM to speed in a linear direction involves two steps: first convert the RPM to a standard angular velocity, and then use the formula v = ωr to convert to linear velocity. You divide the figure in RPM by 60, multiply by 2π and then multiply by the radius of the circle.
#### Pump Power Calculation Formula | Specific speed of a ...
May 28, 2020· RPM Output/RPM Input = Multiplication Factor On a belt-driven fan the two primary speeds required are the RPM of the Motor and the RPM of the fan. For example, if you know the motor speed is 1778 rpm and the fan speed is 944 rpm the multiplication factor would be: 944/1778 = .5309
#### Spur Gear Calculator
Speed And Feed Calculators. Ball Mill Finish Calculator. G & M Code Characters. Standard End Mill Sizes. Standard Drill Sizes. Spur Gear Calculator. Number of Teeth. Diametric Pitch. Pitch Diameter. Circular Pitch. Outside Diameter. Addendum. Dedendum.
#### Differential Gear Ratio Calculator | West Coast Differentials
PREMIUM EFFICIENCY MOTOR SELECTION AND APPLICATION GUIDE | i ACKNOWLEDGMENTS The Premium Efficiency Motor Selection and Application Guide and its companion publication, Continuous Energy Improvement in Motor-Driven Systems, have been developed by the U.S. Department of Energy (DOE) Office of Energy Efficiency and Renewable Energy (EERE) with support …
#### Motor Power Calculator | Magtrol
Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input? Formula: GPM x PSI x 36.77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.
#### How to Convert RPM to Linear Speed | Sciencing
Speed ratings (RPM) Deep groove ball bearings 6200 & 6300 series Warning: The new reference and limiting speeds are not to be usedas a direct substitution for the previous oil and grease speed ratings. See page 4/5 or contact SKF Applications Engineering.
#### What it is the optimun speed for a ball mill ...
Motor Wiring Diagram D.C. Motor Connections Your motor will be internally connected according to one of the diagrams shown below. These connections are in accordance with NEMA MG-1 and American Standards Publication 06. 1 - 1956. Use figure 1 if your motor has a single voltage shunt field. Use figure 2 if your motor has a dual voltage shunt field.
#### RPM Calculator | Engine Rpm Calculator
Jan 28, 2014· Setting the clock speed too low and your machine speed too high can have a huge negative effect on machine performance (missed steps, limited speed). This clock speed is governed by the number of steps the motor must turn to move the machine an inch, and the maximum top speed the machine will obtain in inches per minute.
#### Speed Reducers - Grainger Industrial Supply
Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input? Formula: GPM x PSI x 36.77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.
#### Pulley and RPM Calculator - Culver Armature and Motor
May 07, 2012· Find the amount of Torque that the 1.6 Hp motor delivers at 3600 RPM. Taking the torque formula one step at a time, divide the HP of 1.6 by the RPM of 3600. The answer will be 0.00044. This is a very small number. Next multiply 0.00044 times the constant of 63,025. The final Torque value of the motor is 28.16 ft-lbs of Torque.
#### Pulley Calculator. RPM, Belt Length, Speed, Animated Diagrams
Oct 19, 2006· What it is the optimun speed for a ball mill ? - posted in Pyrotechnics: I have done a ball mill, recenly finished, but the motor has too rpms, is too fast for use in a ball mill (the pvc cylinder that i use, left of the shafts). With the motor i will use a 40 mm pulley, because i have a 50 mm driven pulley, in one of my two shafts.
#### Pump Power Calculation Formula | Specific speed of a ...
For example, combining a Fisher Price and CIM motor would require an extra 3:1 gear reduction for the Fisher Price motor because its output speed is approximately 3 times faster than a CIM's. If the output speeds are not matched, it will cause added resistance in the gearbox and negate any benefits of having multiple motors.
#### Bearing Handbook for Electric Motors
Optimize speed and performance while increasing equipment longevity with speed reducers from Grainger. The motor transmits power to the input shaft of the reducer during equipment operation. The speed reducer coverts the power to a lower output speed while transmitting to …
#### How to Calculate Electric Motor Torque - RC Groups
Jan 05, 2015· Revolutions per minute, or RPM, are a measure of how fast a rotating object turns. Knowing how fast an object turns is important in determining wind speed, gear ratios, how powerful a motor is, and how well bullets fly and penetrate.There are a number of ways to calculate RPM, depending on what the value is needed for; we'll stick with some of the simplest.
#### Power, Torque and RPM Calculations - RACELOGIC Support Centre
By input we understand the source of power, in our case could be an electric motor. By output we mean where the power is delivered (e.g. hydraulic pump). The gear ratio i can be calculated in two ways: as a ratio between the number of teeth of the output gear and the number of teeth of the input gear [i=frac{z_{OUT}}{z_{IN}}]
#### How To Find Out Ball Mill Rpm From Motor And Gear
Output RPM formula = Input RPM times number of driving gear teeth divided by number of driven gear teeth. A small transmission in a mixer has a 12-tooth gear on the motor shaft driving a 72-tooth gear on the mixer blade. The motor is turning at 1,800 RPM. What is the mixer blade speed?
#### Calculating Output Speed Using Pulley Diameters and Input ...
Oct 16, 2019· Using the idea of gear ratios, it's easy to figure out how quickly a driven gear is rotating based on the "input" speed of the drive gear. To start, find the rotational speed of your drive gear. In most gear calculations, this is given in rotations per minute (rpm), though other units of …
#### How to calculate the shaft diameter from the torque ...
Use the RPM calculator to determine what RPM your engine will run based on gear ratio, tire diameter, transmission and MPH. You can try various combinations to come up with an appropriate combination of ring and pinion gear ratio and tire diameter for your vehicle. The RPM calculator should be used as a guideline for comparative purposes only.
#### Hydraulic Motor Calculations - Womack Machine Supply Company
Sep 01, 2008· At 12 pole speed (50Hz here, so only about 360 rpm) it 'only' puts out 2 HP, but to put out that much power at that speed, the torque has to be the same as a 12HP 2 pole motor. Consequently it's a massive frame: two strong men couldn't lift it.
#### Working out Revolutions Per Minute
Use the RPM calculator to determine what RPM your engine will run based on gear ratio, tire diameter, transmission and MPH. You can try various combinations to come up with an appropriate combination of ring and pinion gear ratio and tire diameter for your vehicle. The RPM calculator should be used as a guideline for comparative purposes only.
#### 4 Easy Ways to Determine Gear Ratio (with Pictures)
May 14, 2018· Inputs that we need to calculate shaft diameter. As we know Torsional equation (Read more about the torsional equation here). Where. τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).. r = Radius of the shaft.. T = Twisting Moment or Torque.. J = Polar moment of inertia.. C = Modulus of rigidity for the shaft material.. l = Length of the shaft.
#### Gear Reducing Formulas - Engineering ToolBox
In the example below the DRIVER gear is large than the DRIVEN gear. The general rule is - large to small gear means 'multiply' the velocity ratio by the rpm of the first gear. Divide 60 teeth by 30 teeth to find the velocity ratio. Multiply this number (2) by the rpm (120). This gives an answer of 240rpm.
#### Spur Gear Calculator
Performing a conversion from RPM to speed in a linear direction involves two steps: first convert the RPM to a standard angular velocity, and then use the formula v = ωr to convert to linear velocity. You divide the figure in RPM by 60, multiply by 2π and then multiply by the radius of the circle.
#### Pump Power Calculation Formula | Specific speed of a ...
May 28, 2020· RPM Output/RPM Input = Multiplication Factor On a belt-driven fan the two primary speeds required are the RPM of the Motor and the RPM of the fan. For example, if you know the motor speed is 1778 rpm and the fan speed is 944 rpm the multiplication factor would be: 944/1778 = .5309
#### Spur Gear Calculator
Speed And Feed Calculators. Ball Mill Finish Calculator. G & M Code Characters. Standard End Mill Sizes. Standard Drill Sizes. Spur Gear Calculator. Number of Teeth. Diametric Pitch. Pitch Diameter. Circular Pitch. Outside Diameter. Addendum. Dedendum.
#### Differential Gear Ratio Calculator | West Coast Differentials
PREMIUM EFFICIENCY MOTOR SELECTION AND APPLICATION GUIDE | i ACKNOWLEDGMENTS The Premium Efficiency Motor Selection and Application Guide and its companion publication, Continuous Energy Improvement in Motor-Driven Systems, have been developed by the U.S. Department of Energy (DOE) Office of Energy Efficiency and Renewable Energy (EERE) with support …
#### Motor Power Calculator | Magtrol
Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input? Formula: GPM x PSI x 36.77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information.
#### How to Convert RPM to Linear Speed | Sciencing
Speed ratings (RPM) Deep groove ball bearings 6200 & 6300 series Warning: The new reference and limiting speeds are not to be usedas a direct substitution for the previous oil and grease speed ratings. See page 4/5 or contact SKF Applications Engineering. | 3,775 | 16,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.863442 |
https://www.univerkov.com/how-much-wood-must-be-burned-in-order-to-heat-50-liters-of-water-in-a-10-kg/ | 1,721,434,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00431.warc.gz | 898,380,104 | 6,535 | # How much wood must be burned in order to heat 50 liters of water in a 10 kg
How much wood must be burned in order to heat 50 liters of water in a 10 kg iron boiler from 15 degrees Celsius to 65 degrees Celsius?
Vv = 50 l = 50 * 10 ^ -3 m ^ 3.
ρw = 1000 kg / m ^ 3.
Sv = 4200 J / kg * ° C.
Сж = 460 J / kg * ° C.
mk = 10 kg.
t1 = 15 ° C.
t2 = 65 ° C.
λd = 1 * 10 ^ 7 J / kg.
md -?
When burning firewood Q, the amount of heat is released, which is determined by the formula: Q = λd * md, where λd is the specific heat of combustion of firewood, md is the mass of firewood.
The amount of heat Q, which is necessary for heating the pot and water in it, will be determined by the formula: Q = Sv * mw * (t2 -t1) + Szh * mk * (t2 -t1), where Sv, Szh is the specific heat of water and iron , mw, mk – mass of water and pot.
We find the mass of water by the formula: mw = Vw * ρw, where Vw is the volume of water, ρw is the density of water.
Let’s write down the heat balance equation: λd * md = Sv * Vw * ρw * (t2 -t1) + Szh * mk * (t2 -t1).
md = (Sv * Vw * ρw * (t2 -t1) + Szh * mk * (t2 -t1)) / λd.
md = (4200 J / kg * ° C * 50 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * (65 ° C – 15 ° C) + 460 J / kg * ° C * 10 kg * ( 65 ° C – 15 ° C)) / * 10 ^ 7 J / kg = 1.07 kg.
Answer: it is necessary to burn md = 1.07 kg of firewood.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 554 | 1,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.869206 |
http://www.ask.com/question/are-all-parallelograms-rectangles | 1,408,659,101,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500821666.77/warc/CC-MAIN-20140820021341-00274-ip-10-180-136-8.ec2.internal.warc.gz | 245,359,558 | 15,751 | # Are All Parallelograms Rectangles?
All parallelograms are not considered rectangles. A rectangle has four corners that represent ninety degree angles. A parallelogram has four corners, but the angles can be any measurement.
Reference:
Q&A Related to "Are All Parallelograms Rectangles?"
because that is how rectangles are defined. that's like asking why are all circles round. http://wiki.answers.com/Q/Why_are_all_rectangles_p...
A rectangle is a four-sided http://www.chacha.com/question/why-are-all-rectang...
All rectangles are, but a trapezoid is not a parallelogram because only one pair of sides is parallel. http://www.kgbanswers.com/why-arent-all-rectangles...
http://www.math.com/school/subject3/less…. Move the cursor and it will give you the special characteristics. isosceles is only a triangle. http://www.mathsteacher.com.au/year7 http://answers.yahoo.com/question/index?qid=201203...
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A parallelogram is a four sided shape like a square, diamond, and a rectangle. A rectangle is always a parallelogram. ...
The other parallelograms that are not rectangles are the square and the rhombus. The rhombus has four sides of equal length, equal opposite angles and diagonals, ...
Yes, a trapezoid is a parallelogram. It has four sides and the opposite sides are of equal length. A square is also a parallelogram, as is a rectangle. ... | 330 | 1,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.866739 |
https://quant.stackexchange.com/questions/74687/delta-hedging-frequency-directly-affects-pnl-and-not-just-pnl-smoothness-and-va | 1,716,561,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00740.warc.gz | 410,016,130 | 42,182 | # Delta-hedging frequency directly affects PnL, and not just PnL smoothness and variance?
The information I have found about delta hedging frequency and (gamma) PnL on this site and numerous others all reiterate the same thing: that the frequency at which you delta-hedge only has an effect on the smoothness and variance of your PnL.
Nivel Egres: From the perspective of gamma pnl, the only thing that matters is the change in your asset price. Frequency is irrelevant - you can rebalance at different time periods or when delta exceeds a threshold or many other things - it is still an approximation of continuous integral and your expected P&L would be the same. For example, in real life, you are usually trying to optimize your hedging for a balance of (a) smoothness of P&L, (b) transaction costs, (c) vol dampening or amplifying (d) mean reversion found in the asset you're trading and (e) your risk limits.
How is this true though? Delta-hedging frequency has a direct effect on your PnL, and not just the smoothness of it.
So the thought here is that a trader who delta-hedges every minute, and a trader who hedges every end of day at market close, will both have the same expected profit at option expiry and only their PnL smoothness/variance will differ. Let's put this to the test.
We'll use this hypothetical scenario. And for the sake of simplicity we ignore transaction costs:
Two traders have bought a 100 strike ATM straddle (long gamma) that expires in a week on stock XYZ. The stock price is 100. They are both initially delta neutral. Throughout expiry, Trader A delta-hedges every minute, and trader B hedges every end of day at market close.
At 2:00 PM, stock XYZ drops -2% to 98. Trader A is delta-hedging the straddle every minute anyways so he buys shares to hedge and become delta-neutral again. The stock doesn't move at all until at 2:05 PM stock XYZ shoots up 4% from 98 to 101.92. Trader A sells the shares he previously bought, and then short sells some more shares to become delta-neutral. He has effectively locked in his gamma scalping PnL. The stock doesn't move at all again until 3:59 PM where the stock drops -1.88% from 101.92 to exactly 100. Trader A buys back his short shares, locks in his gamma PnL again, and is completely delta neutral.
Meanwhile it's the end of the day and time for Trader B to hedge, but he has nothing to delta-hedge because the stock is 100 at the end of the trading day, the same price at which he bought the ATM straddle and his delta of the position is 0.
Trader A has made some hefty PnL, meanwhile Trader B comes out with nothing at all and his missed out on volatility during the trading day which he could've profited off of had he been continuously hedging instead of just once a day.
So how does delta-hedging frequency just affect the smoothness and variance of PnL if we can clearly see it affects PnL itself in this example?
• Your example measures P&L over a specific price path. And this depends on the rebalancing frequency. But "expected P&L" refers to an average over all possible price paths. So there is not necessarily a contradiction here. Feb 18, 2023 at 20:08
• @nbbo2 I'm using the specific price path in the example for a reason, it disproves the basis of delta-hedging frequency not directly affecting PnL. And I mean "expected P&L" as the option premium (PnL) replicated by delta-hedging a position which can be calculated by subtracting realized volatility from implied volatility. So, is it correct to say then delta-hedging rebalancing frequency directly affects the amount of P&L then?
– user46424
Feb 18, 2023 at 20:19
• What do you mean by ‘directly affects p/l’? If you mean it affects the expected p/l then you are wrong. If you mean it affects p/l on a particular path then you are right.
– dm63
Feb 19, 2023 at 3:41
• Can you be a bit more specific about the source of the quote in your post, i.e. name of book or article, date, page no, etc.? Thx. Feb 19, 2023 at 9:18
• The quote is a comment by Mr. Nivel Egres in this Stackexchange and this post quant.stackexchange.com/questions/32503/… Feb 19, 2023 at 10:21
If you look at just a single example, it may seem like the frequency of hedging directly effects the EV/Avg(Pnl), like in the situation you described where hedging every minute proved to be more profitable. But you need to think about the question in a bigger picture sense. How would hedging frequency affect the results over thousands of simulations?
If you hedge every minute, you wouldn't realize the full pnl of the larger SD moves but you do capture the full pnl of the smaller intraday moves. Conversely, if you only hedge once per day, you won't realize the full pnl from the smaller intraday moves (like in your example) but you would in return realize the full pnl from the larger SD moves. Infrequent larger gains from larger SD moves balance out against the more frequent missed gains from smaller intraday moves in EV/Avg(PnL). [Sinclair does a great job explaining this in his books, pictures below provided].
At the end of the day, the EV/Avg(PNL) boils down to iv vs rv of stock. If those two are equal, then the EV/PNL will be the same for both traders regardless of hedging frequency. The only difference will be the variance of their PNL as described above.
Food for thought, the below example is just in theory and doesn't exist:
Now, in the above explanation, we assumed the stock was performing on some constant vol at all moments in time. What if the intraday vol diverges significantly from the daily vol? Ie: As an EXAGGERATION, say you look at some stock and you calculate from the past 10 day closing prices that the stock is performing on a 1 vol. Pretty much closes where it opened each day. You then decide to look closer and measure vol in 30 minute increments rather than by daily closing prices. When you look intraday/30 min increments, you see the stock moves a lot, but based on closing prices performs still on a 1 vol. In this case, when we measure vol in smaller 30 min increments, we can see it is significantly different than vol measured on close to close prices. Both traders buy the straddle on a 1 vol let's say, who do you think would be better off? The person who hedges several times a day or the person who hedges once at the end of the day? In this case, the stock is not performing at some constant vol at all moments in time over the duration of the life of the option and throughout each day, instead we can see the intraday vol is significantly different that the daily close to close vol.
In theory, if the above example existed, I WOULD THINK this would imply that hedging more frequently would be more profitable. HOWEVER, THIS ALSO WOULD IMPLY THAT THERE IS SOME PROFITABLE OUTRIGHT STOCK TRADING STRATEGY, WHICH WOULD IMPLY WE KNOW WHICH DIRECTION THE STOCK IS GOING TO GO NEXT, WHICH IN REALITY WE DON'T KNOW. Ie: If we know the stock is going to close near the opening price because it always performs on a 1 vol, and its noon and the stock is down -10%, we know that it has to go higher in the last few hours of the day and we could just outright buy stock to make money.
Under the assumptions of GBM - namely that periodic returns are independent of one another - then hedging frequency will have 0 impact on the expected P/L over time. If there is autocorrelation in the intraday return process that you choose to hedge at (which will in turn affect daily annualised volatility), then your P/L is definitely affected by your choice of hedging interval.
However, the existence of significant autocorrelation in the return process would hint that we are able to trade using futures/linear products on a intraday horizon which would probably (after accounting for liquidity and theta) prove more profitable to trade than the delta hedging strategy.
Over any longer period of time, there is not often a statistically significant autocorrelation in high frequency returns. If there was, then the above would be applicable which would dampen the effect.
The net effect of all that is that increased delta hedging frequency does just have the smoothing effect on P/L over long enough time horizons. But like you indicate you are exposed to one-off or rare mean reversion (or trend) effects, but these dissipate over large samples. | 1,931 | 8,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.941081 |
https://www.numbersaplenty.com/35655553 | 1,638,363,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.0/warc/CC-MAIN-20211201113241-20211201143241-00595.warc.gz | 931,969,476 | 3,179 | Search a number
35655553 is a prime number
BaseRepresentation
bin1000100000000…
…0111110000001
32111002111020001
42020000332001
533111434203
63312120001
7612032023
oct210007601
974074201
1035655553
11191435aa
12bb36001
13750527b
144a42013
1531e491d
hex2200f81
35655553 has 2 divisors, whose sum is σ = 35655554. Its totient is φ = 35655552.
The previous prime is 35655541. The next prime is 35655577. The reversal of 35655553 is 35555653.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 33558849 + 2096704 = 5793^2 + 1448^2 .
It is a cyclic number.
It is not a de Polignac number, because 35655553 - 213 = 35647361 is a prime.
It is not a weakly prime, because it can be changed into another prime (35655593) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 17827776 + 17827777.
It is an arithmetic number, because the mean of its divisors is an integer number (17827777).
Almost surely, 235655553 is an apocalyptic number.
It is an amenable number.
35655553 is a deficient number, since it is larger than the sum of its proper divisors (1).
35655553 is an equidigital number, since it uses as much as digits as its factorization.
35655553 is an evil number, because the sum of its binary digits is even.
The product of its digits is 168750, while the sum is 37.
The square root of 35655553 is about 5971.2270933201. The cubic root of 35655553 is about 329.1362566648.
The spelling of 35655553 in words is "thirty-five million, six hundred fifty-five thousand, five hundred fifty-three". | 487 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-49 | latest | en | 0.874669 |
http://modelingcommons.org/browse/one_model/5269 | 1,539,917,882,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00409.warc.gz | 236,916,184 | 8,620 | # tax
Al Pugh (Author)
### Tags
inequality & taxation
Tagged by Al Pugh 11 months ago
Visible to everyone | Changeable by the author
Model was written in NetLogo 5.3.1 • Viewed 173 times • Downloaded 16 times • Run 0 times
# Economic inequality & tax redistribution
## Background
Adapted from a model popularized by Bryan Hayes in American Scientist. Agents meet randomly to wager a small amount on the outcome of a coin flip. With this simple model the distribution of agent wealth will become increasingly skewed and asymptotically leads to all the wealth being held by one agent.
Should one be tempted to at least consider the possibility that the above model represents reality (however tenuously), it begs the question if anything can be done to stabilize it.
Two methods of taxation (for wealth redistribution) are examined. One is to set aside a percentage of each transaction between agents – a sort of an excise or sales tax. The second is to tax those with a wealth above some threshold value, somewhat like a property tax. Wealth redistribution can also be done is two ways: equally divided among all agents, or a larger share going to poor agents.
## Design
Agents are homogeneous and unsophisticated with the single property of wealth. They move randomly about the environment, follow rules for wealth exchange upon encountering another and paying any taxes assessed. The environment includes user-adjustable parameters for sales and wealth tax rates, tax redistribution frequency, and the method by which taxes are redistributed. Those parameters serve as inputs. The primary output is the level of inequality as represented by the Gini coefficient.
Initially, each agent is endowed with a small amount of wealth drawn from a constant plus a uniform random deviate (specifically, \$90 + URD \$0-21 which yields an expected value of \$100 per agent). They are then set free to randomly roam about the environment.
Should two agents meet, an amount equal to a fifth (1/5) of the lesser of the agent's wealth is awarded the winner (determined by coin flip). Sales tax and a wealth tax applied to those with greater than 10 times the average wealth may be applied.
Tax receipts are evenly distributed to all agents at a chosen interval, or a greater share is allocated to poor agents if one toggles the switch on the interface. Here the specific method of enhanced redistribution is to grant all agents an equal share of half the collected amount, with the other half granted to agents with wealth below average.
Inequality is shown by displaying the percentage of total wealth owned by the wealthiest agent, a histogram of agent wealth and by the Gini coefficient change over time. This metric varies between 0 (indicating uniform wealth distribution) and 1 (indicating one agent has acquired all the wealth). A plot the total amount traded during each period is shown to indicate economic activity. Note it contains essentially the same information as the Gini plot, except in noisier form. For clarity, agents are color-scaled in proportion to their relative wealth.
## Results & analysis
The results are not completely unexpected. Taxation appears to forestall the collapse of the economy. Both forms of taxation will prevent, or profoundly delay, the onset of collapse. If the option to give a greater portion of tax receipts to the poorer agents is selected, the Gini metric is reduced significantly. However, once the economy has decayed to the point that any semblance of a middle class has evaporated, taxation of wealth seems to be slightly more effective in restoring vitality. In any scenario, redistribution favoring poorer agents is effective. The redistribution frequency appears to have minimal effect other than adding noise to the metric.
What does this simple model imply about the nature of economies? The first is that the really simple model exhibits characteristics in agreement with actual data, the most important of these is the distribution of wealth. The second is that an economic system based upon unencumbered trade is likely to collapse. Lastly, collapse might be avoided through the use of taxation for the redistribution of wealth, especially if it is biased toward the poor.
One might complain this model is based upon agents that are so simple-minded that it could not possibly represent reality. But richly complex systems can emerge from the interaction of simple elements. There is also empirical evidence – the similarity of wealth distributions for real economies and the simulated economy is remarkable.
## Things to observe
* When no or minimal taxes are imposed, and tax redistribution is equally allocated, the Gini coefficient (the metric for inequality) soars.
* If the above-mentioned conditions are left to run for some time, it is difficult to recover.
* The Gini coefficient and economic activity appear inversely correlated.
* Even with heavy taxation and redistribution favoring the poor, the histogram of agent wealth remains weighted toward the poorer side.
## References & credits
* Gini-coefficient calculation algorithm due to: Wilensky, U. (1998). [NetLogo Wealth Distribution model.] (http://ccl.northwestern.edu/netlogo/models/WealthDistribution) Center for Connected Learning and Computer-Based Modeling, Northwestern University, Evanston, IL.
* NetLogo software: Wilensky, U. (1999). [NetLogo.] (http://ccl.northwestern.edu/netlogo/) Center for Connected Learning and Computer-Based Modeling, Northwestern University, Evanston, IL.
* Hayes, Bryan. Follow the Money, **American Scientist**, v90, September-October 2002, pp 400–405.
Click to Run Model
```turtles-own [ wealth ] ; A economic inequality ABM
globals [ exc tax totaltax gini counter sumexc ] ; © Al Pugh, 2016
to setup
ca set gini 0 set counter 0 set totaltax 0
ask patches [ set pcolor green ] ; set green background for contrast
ask n-of 400 patches [ ; ask some patches to create agents
sprout 1 [set shape "person" set wealth (90 + random 21)
set heading random 360 ] ]
ask turtles [ set color yellow ] ; initialize agent color
update-gini reset-ticks
end
to go
set sumexc 0
set heading heading + random 10 fd 1 ; move agent randomly
wealthtax ; tax wealthy
encounter ] ; resolve agent encounters
redistribute ; redistribute taxes
update-gini ; update Gini coefficient
; checkwealth ; (for debugging only)
tick
end ; (plot set up commands in plots)
to wealthtax
if ( wealth > 1000 ) [ ; tax those with 10x mean starting wealth
set tax ( wealthtaxrate / 100. * wealth )
set wealth wealth - tax
set totaltax totaltax + tax ]
end
to encounter
if ( count turtles-here = 2 ) [ ; for every interaction of 2 agents:
let n2 one-of other turtles-here
set exc ( min [ wealth ] of turtles-here / 5.0 ) ; determine exchange amount
set sumexc sumexc + exc ; add to sum of exchanges
set tax ( exc * salestaxrate / 100. ) ; extract sales tax
set totaltax totaltax + 2 * tax
ifelse ( random 2 ) = 0 ; determine winner
[ set wealth ( wealth - exc - tax )
ask n2 [set wealth ( [wealth] of n2 + exc - tax ) ] ]
[ set wealth ( wealth + exc - tax )
ask n2 [set wealth ( [wealth] of n2 - exc - tax ) ] ]
set color scale-color yellow ; color agents by wealth
wealth 0 int max [ wealth ] of turtles ]
end
to redistribute
set counter counter + 1
if ( totaltax > 0 ) [ ; distribute tax receipts
ifelse ( poordist? )
[ let share ( totaltax / count turtles ) / 2.0
ask turtles [ set wealth wealth + share ] ; all agents get reduced share
let npoor count turtles with [ wealth < 100.0 ] ; define poor as < mean starting wealth
let poorshare share * count turtles / npoor ; poor get additional share
ask turtles with [ wealth < 100.0 ] [
set wealth wealth + poorshare ]
set totaltax 0 ]
[ let share ( totaltax / count turtles ) ; agents share tax redistribution equally
ask turtles [ set wealth wealth + share ]
set totaltax 0 ] ]
end
to update-gini ; find Gini coefficient
let sorted-wealths sort [wealth] of turtles
let total-wealth sum sorted-wealths let wealth-sum-so-far 0
let index 0 let gini-index-reserve 0
repeat count turtles [
set wealth-sum-so-far (wealth-sum-so-far + item index sorted-wealths)
set index (index + 1)
set gini-index-reserve gini-index-reserve + (index / count turtles)
- (wealth-sum-so-far / total-wealth) ]
set gini (gini-index-reserve / count turtles) / 0.5
end
to checkwealth ; check total wealth remains unchanged
print sum [ wealth ] of turtles + totaltax
end
```
There is only one version of this model, created 11 months ago by Al Pugh.
## Attached files
File Type Description Last updated
tax.png preview Preview for 'tax' 11 months ago, by Al Pugh Download
This model does not have any ancestors.
This model does not have any descendants. | 2,024 | 9,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-43 | longest | en | 0.953009 |
https://www.explorelinux.com/operators-in-python/ | 1,620,872,667,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00093.warc.gz | 784,710,665 | 19,395 | # Operators in Python
An operator is a symbol that tells the Python interpreter to perform specific arithmetic or logical operations on variables and values.
For example –
```# Here + and = are examples of operators in Python
sum = 5+6
print(sum)```
Python has the following type of built-in operators.
• Arithmetic operators
• Relational operators
• Logical operators
• Bitwise operators
• Assignment operators
• Python special operators
## Arithmetic operators
These operators are used to perform mathematical operations such as addition, subtraction, multiplication, division, modulus, etc. The list of arithmetic operators used in Python is given below.
OperatorDescriptionUsage
`+`Addition`a+b`
`-`Subtraction`a-b`
`*`Multiplication`a*b`
`/`Division`a/b`
`%`Modulus`a%b`
`**`Exponentiation`a**b`
`//`Floor division`a//b`
#### Floor division –
This can be new to many of you when you use `//` operator it performs floor division. The division which results into the whole number is known as floor division. You can differentiate between normal and floor by understanding the given example.
```x=18
y=5
# Normal division
print(f"x/y is {x/y}")
# Floor division
print(f"x//y is {x//y}")```
You can see the output of this code in the given image –
### Example of using arithmetic operators –
```a=10
b=4
print(f"Addition of a and b is {a+b}")
print(f"Subtraction of a and b is {a-b}")
print(f"Multiplication of a and b is {a*b}")
print(f"Division of a and b is {a/b}")
print(f"The remainder when a is divided by b is {a%b}")
print(f"a to the power b is {a**b}")
print(f"Floor division of a and b is {a//b}")```
Execute this code on your own system to see the output.
## Relational operators
A relational or comparison operator is used to compare two values. The following table shows the list of relational operators used in Python.
OperatorDescriptionUsage
`>`Greater than`a>b `
`<`Less than`a<b`
`==`Equal (Check if two values equal or not)`a==b`
`>=`Greater than or equal to`a>=b`
`<=`Less than or equal to`a<=b`
`!=`Not equal`a!=b`
It compares two values if the condition goes true it returns `True` otherwise, it returns `False`.
For example –
```a=4
b=5
#Since a is not greater than b this will return False
print("a >b is", a>b)
```
See the output in the image below –
Similarly, you can use other comparison operators in Python.
## Logical operators
Logical operators in Python are used to combine two conditional statements. The list of logical operators is given in the table below.
OperatorDescriptionExample
`and`Returns True if both the conditional statements are true`a>10 and b<7`
`or`Returns True if both the conditional statements are true`a>5 or b>2`
`not`It returns False if condition is True and vice versa`not(a>5 or b<2)`
For example –
```a=4
b=5
# Here both the statements are true the and operation will return True
print("a<10 and b>4 is",a<10 and b>4 )```
You can see the output in the given image –
## Bitwise operators
The bitwise operators work on bits these perform operation bit by bit. The following table shows the list of bitwise operators that are used in Python.
OperatorDescriptionUsage
`&`AND`a&b`
`|`OR`a|b`
`~`NOT`~a`
`^`XOR`a^b`
`>>`Right Shift`a>>2`
`<<`Left Shift`a<<2`
For example-
Lets a=4 and b=5 that means in 4-bit binary a will be 0100 and b will be 0101
```a b a&b
0 0 0
1 1 1
0 0 0
0 1 0```
After performing bitwise and operation i.e. a&b will be 4 you can see the output in the given image-
## Assignment operators
Assignment operators are used to assigning a value to a variable. Equal to i.e. `=` is a simple assignment operator we can combine this with other types of operators to calculate and assign a value to a variable. For example, `+ =` will add and assign the calculated value to the variable.
The given table shows the list of assignment operators used in Python.
OperatorsExampleExample is same asDescription
`=``x=10``x=10`Assign 10 to variable x
`+=``x+=10``x=x+10`Add and assign to variable x
`-=``x-=10``x=x-10`Subtract and assign to variable x
`*=``x*=10``x=x*10`Multiply and assign to variable x
`/=``x/=10``x=x/10`Divide and assign to variable x
`%=``x%=10``x=x%10`Find remainder and assign to x
`//=``x//=10``x=x//10`Divide and assign a whole number to x (remove fractional part)
`**=``x**=10``x=x**10`Find x to the power 10 and assign to the variable x
`&=``x&=10``x=x&10`Perform bitwise AND operation and assign the value to variable x
`|=``x|=10``x=x|10`Perform bitwise OR operation and assign the value to variable x
`^=``x^=10``x=x^10`Perform bitwise XOR operation and assign the value to variable x
`>>=``x>>=10``x=x>>10`Shift value 10 bit right and assign it to variable x
`<<=``x<<10``x=x<<10`Shift value 10 bit left and assign it to variable x
## Python special operators
Python has some special operators i.e. identity and membership operators.
### Identity operators
Identity operators check if the two values or variables are equal and at the same location in memory. There are two identity operators, you can see them in the given table.
OperatorDescriptionUsage
`is`Return True if both the values are identicala is b
`is not`Return True if both the values are not identicala is not b
For example –
```a=5
b=5
c=[6,7,8]
d=c
# b is identical to a so the given statement will be True
print("b is a is",b is a)
# c is identical to d so the given statement will be False
print("c is not d", c is not d)```
You can see the output in the image below –
### Membership operators
You can use membership operators to test whether a value or variable is in the sequence or not. You can see membership operators in the given table.
OperatorDescriptionUsage
`in`Return True if value found in the sequencea in b
`in not`Returns True if value not found in the sequencea not in b
```a="Hello World!"
b="World"
c=[6,7,8]
d=[10]
# b is in a so the given statement will be True
print("b in a is",b in a)
# d is not in c so the given statement will also return True
print("d not in c", d not in c)```
You can see the output in the given image –
## Operator precedence and associativity
An expression in programming is the combination of operators and operands. The 5+2*3 is an example of a valid Python expression. Python interpreter performs the calculation on the basis of operator precedence and associativity.
For example in the expression 5+2*3 Python interpreter will first perform multiplication and later it will perform addition. That means * operator has higher precedence over +.
The following table shows the operator precedence and associativity. It is in descending order that means upper groups have higher precedence than lower ones.
OperatorsDescription
`()`Parentheses
`**`Exponentiation(raise to the power)
`+`, `-`, `~`Unary plus, Unary minus, Bitwise NOT
`*`, `/`,`//`,`%`Multiplication, Division, Floor division, Modulus
`+`, `-`Addition, Subtraction
`<<`,`>>`Left and right shift
`&`Bitwise AND
`^`Bitwise XOR
`|`Bitwise OR
`==`,` !=`, `>`,` >=`, `<`,` <=`, `is`,` is not`, `in`,` not in`Comparisons, Identity, Membership operators
`not`Logical NOT
`and`Logical AND
`or`Logical OR
Now which operator will be calculated first if they fall under the same group? Here associativity helps to determine the order of operations. So almost all the groups have left to right associativity that means the operator on the left side will be calculated first.
## Conclusion
For better understanding try to evaluate Python expressions on your own and then try to find them with code. Now If you find any difficulty then write us in the comments below.
Previous Keywords In Python
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#### Python if else and nested if else statements
In programming conditional statements provides a way in which the computer takes a decision based on whether a condition is true or false. The if and else statements are common | 2,187 | 8,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-21 | latest | en | 0.837514 |
https://crypto.stackexchange.com/questions/35569/is-this-hash-function-almost-xor-universal | 1,576,160,782,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543850.90/warc/CC-MAIN-20191212130009-20191212154009-00197.warc.gz | 321,623,658 | 29,683 | # Is this hash function almost XOR universal?
Consider this hash function:
The key $K$ is $2^n$ bits. The message $M$ is divided into $2^n$ bit blocks. Arithmetic is performed modulo $2^{n+1}$. The hash is defined by:
• $H_0 = 1$.
• $H_{i+1} = (2M_i + H_i)\times (2K + 1)$ where K is a (secret) hash key.
• Output $H' = \frac{H_{|M|/2^n}}2$
Is this hash (the iterative one) almost-XOR-universal?
Is this hash almost-XOR-universal?
No.
Consider the two 1 block messages $M = \{0\}$ and $M' = \{2^{n-1}\}$.
We have:
$$H(M) = ((2\cdot 0 + 1)(2K+1) \bmod 2^{n+1}) /2 = K$$
and
$$H(M') = ((2\cdot 2^{n-1} + 1)(2K+1) \bmod 2^{n+1})/ 2 = K + 2^{n-1} \bmod 2^n$$
As $H(M) \oplus H(M') = 2^{n-1}$ has a nontrivial probability of being true, $H$ is not almost-XOR-universal.
Similar logic (using the two messages $\{0, 2^{n-1}\}$ and $\{2^{n-1}, 0\}$) will show that $H$ isn't even almost-universal. | 362 | 902 | {"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} | 3.5625 | 4 | CC-MAIN-2019-51 | longest | en | 0.798047 |
https://www.clutchprep.com/chemistry/practice-problems/143689/what-size-tank-would-be-needed-to-contain-this-same-amount-of-helium-at-atmosphe | 1,643,272,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00149.warc.gz | 747,080,980 | 19,591 | # Problem: What size tank would be needed to contain this same amount of helium at atmospheric pressure (1 atm)?Express the size in liters to three significant figures.Application of Boyle's lawA 12-liter tank contains helium gas pressurized to 160 atm.
94% (58 ratings)
###### Problem Details
What size tank would be needed to contain this same amount of helium at atmospheric pressure (1 atm)?
Express the size in liters to three significant figures.
Application of Boyle's law
A 12-liter tank contains helium gas pressurized to 160 atm. | 116 | 544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.862669 |
http://mathhelpforum.com/advanced-algebra/142985-operator-print.html | 1,495,781,599,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608642.30/warc/CC-MAIN-20170526051657-20170526071657-00236.warc.gz | 289,071,260 | 3,854 | # Operator
• May 4th 2010, 07:43 AM
gralla55
Operator
Need some help with this... doesn't seem to be a decent example in my book. Appreciate any help!
• May 4th 2010, 08:23 AM
tonio
Quote:
Originally Posted by gralla55
Need some help with this... doesn't seem to be a decent example in my book. Appreciate any help!
1) Apply the transformation T to each and every element of E and write the result as a linear combination of E itself;
2) Take the transpose , if you apply transformations from the left and vectors from the right), or directly (otherwise) the coefficients matrix of the above: this is $[T]_E$.
For example:
$T(1)=1+1 = 2=2\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3$
$T(x)=x+x-2=-2+2x=(-2)\cdot 1+2\cdot x +0\cdot x^2+0\cdot x^3$ ...
So the first two columns of $[T]_E$ are $\begin{pmatrix}2\\0\\0\\0\end{pmatrix}\,,\,\,\begi n{pmatrix}\!\!-2\\2\\0\\0\end{pmatrix}$ (or the first two rows if you write the map on the right of the vector).
Tonio
• May 6th 2010, 01:03 AM
gralla55
T(x) = -2 + 2x ? Shouldn't that just equal 2?
Thanks again.
• May 6th 2010, 02:26 AM
HallsofIvy
Quote:
Originally Posted by gralla55
T(x) = -2 + 2x ? Shouldn't that just equal 2?
Thanks again.
Yes, that must have been a typo. Since T(f)= f(x)+ f(2- x), T(x)= x+ (2- x)= 2.
• May 6th 2010, 03:08 AM
gralla55
In that case 2 is the answer for all four. And a linear combination would just be every element of E times 1/2 times each column vector? And won't the transpose of this matrix is the same as the matrix itself?
• May 6th 2010, 04:09 AM
tonio
Quote:
Originally Posted by gralla55
In that case 2 is the answer for all four.
Why? $T(x^2)=x^2+(2-x)^2=2x^2-4x+4$ ...
Tonio
And a linear combination would just be every element of E times 1/2 times each column vector? And won't the transpose of this matrix is the same as the matrix itself?
.
• May 6th 2010, 04:18 AM
gralla55
Don't you just substitute the "x" for "x^2" ? I don't see why the ^2 suddenly goes outside the parantheses. But of course, I might be wrong here...
• May 6th 2010, 05:53 AM
gralla55
What confuses me is the notation T(f(x))... In case you're right:
T(1) = 1 + (2-1) = 2
T(x) = x + (2x-2) = 2
T(x^2) = 4 - 4x + 2x^2
T(x^3) = 8 - 12x + 8x^2 - 2x^3
How to proceed?
• May 6th 2010, 06:48 AM
tonio
Quote:
Originally Posted by gralla55
What confuses me is the notation T(f(x))... In case you're right:
T(1) = 1 + (2-1) = 2
T(x) = x + (2x-2) = 2
T(x^2) = 4 - 4x + 2x^2
T(x^3) = 8 - 12x + 8x^2 - 2x^3
How to proceed?
Write each outcome as a linear combination of the given basis and thus the coefficients in each case become the wanted matrix's columns...
Tonio
• May 6th 2010, 06:58 AM
gralla55
So:
2 2 4 8
0 0 -4 -12
0 0 2 8
0 0 0 -2
and then I take the transpose:
2 0 0 0
2 0 0 0
4 -4 2 0
8 -12 8 -2
And that's it?
• May 6th 2010, 08:39 AM
gralla55
Nevermind, I figured it out! And I also did a mistake, the correct answer is:
2 2 4 8
0 0 -4 -12
0 0 2 6
0 0 0 0
Now, how can I find a solution to T(f) = (x-1)^2 ? | 1,151 | 3,005 | {"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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.880908 |
https://certifiedcalculator.com/neg-am-mortgage-calculator/ | 1,709,401,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00889.warc.gz | 156,096,672 | 44,722 | # Neg Am Mortgage Calculator
Managing finances can be a daunting task, especially when it comes to mortgages. For those considering or currently holding a negative amortization (Neg Am) mortgage, understanding the financial implications is crucial. To simplify this process, a Neg Am Mortgage Calculator proves to be an invaluable tool.
Formula:
The formula used to calculate the monthly payment for a negative amortization mortgage is derived from the standard mortgage formula. It incorporates the loan amount, interest rate, and loan term to determine the monthly payment. The formula takes into account the negative amortization feature, which allows for a minimum payment that may not cover the full interest due, resulting in the remaining interest being added to the loan balance.
How to Use:
1. Enter the loan amount, interest rate, and loan term in the respective fields.
2. Click on the “Calculate” button.
3. The calculator will compute the monthly payment based on the provided information.
Example:
Let’s consider a scenario where the loan amount is \$200,000, the interest rate is 5% per annum, and the loan term is 30 years. Upon calculation, the monthly payment would be approximately \$1,073.64.
FAQs:
1. What is a Neg Am mortgage?
• A Neg Am mortgage is a type of mortgage where the minimum payment may not cover the full interest due, resulting in the unpaid interest being added to the loan balance.
2. How does negative amortization affect my loan balance?
• Negative amortization causes your loan balance to increase over time, as the unpaid interest is added to the principal amount.
3. Is negative amortization common?
• Negative amortization mortgages were more prevalent before the 2008 financial crisis but have become less common since then.
4. Are Neg Am mortgages risky?
• Yes, Neg Am mortgages can be considered risky due to the potential for increasing loan balances and higher future payments.
5. Can I make larger payments to avoid negative amortization?
• Making larger payments can help reduce negative amortization and mitigate its effects on your loan balance.
6. Are Neg Am mortgages suitable for everyone?
• Neg Am mortgages are typically more suitable for borrowers who have irregular income streams or those who anticipate higher future earnings.
7. How does the interest rate affect negative amortization?
• A higher interest rate can accelerate negative amortization, as more interest accrues on the outstanding balance.
8. What happens if I reach the negative amortization limit?
• Once you reach the negative amortization limit, your loan may be recast, resulting in higher monthly payments to amortize the outstanding balance.
9. Can I refinance a Neg Am mortgage?
• Refinancing a Neg Am mortgage may be an option to obtain a more favorable loan structure or to mitigate the effects of negative amortization.
10. How do I determine if a Neg Am mortgage is right for me?
• Consult with a financial advisor or mortgage specialist to assess your financial situation and determine if a Neg Am mortgage aligns with your long-term goals.
Conclusion:
A Neg Am Mortgage Calculator serves as a valuable resource for individuals seeking clarity on their mortgage obligations. By understanding the intricacies of negative amortization and its impact on loan repayment, borrowers can make informed decisions regarding their financial future. Utilizing this calculator empowers individuals to plan effectively and navigate the complexities of mortgage financing with confidence. | 677 | 3,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.942559 |
https://www.texasgateway.org/search-standards?tid=&external_1=All&external_2=All&external_3=All&page=1 | 1,531,704,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589029.26/warc/CC-MAIN-20180716002413-20180716022413-00438.warc.gz | 1,011,158,491 | 16,139 | # +Find resources aligned to Texas Essential Knowledge and Skills (TEKS)
Standards explanation. Given a situation that can be modeled by a quadratic function or the graph of a quadratic function the student will determine the domain.
TEKS Number
Student Expectation
1(3)(B)
make predictions based on observable patterns; and
1(3)(B)
describe and measure calendar time by days, weeks, months, and years; and
1(3)(B)
use objects and pictorial models to solve word problems involving joining, separating, and comparing sets within 20 and unknowns as any one of the terms in the problem such as 2 + 4 = [ ]; 3 + [ ] = 7; and 5 = [ ] - 3
1(3)(C)
use common syllabication patterns to decode words, including:
1(3)(C)
describe what scientists do.
1(3)(C)
create a calendar and simple timeline.
1(3)(C)
compose 10 with two or more addends with and without concrete objects
### Making Ten is Easy as Pie!
• Lesson Study ID: TXLS021
• Subject: Math
Students will practice composing 10 by interacting with a counting story, playing a dice game with ten frames and response sheets, and participating in a small group to extend the learning with three addends.
### K–2 Diagnostic Assessments
• Resource ID: MATH_RA_K2
• Subject: Math
This resource provides two types of diagnostic tools available to K–2 math teachers: a rapid assessment tool and a flexible interview assessment.
1(3)(C)(i)
use common syllabication patterns to decode words, including: closed syllable (CVC) (e.g., mat, rab-bit);
1(3)(C)(ii)
use common syllabication patterns to decode words, including: open syllable (CV) (e.g., he, ba-by);
1(3)(C)(iii)
use common syllabication patterns to decode words, including: final stable syllable (e.g., ap-ple, a-ble);
1(3)(C)(iv)
use common syllabication patterns to decode words, including: vowel-consonant-silent 'e' words (VCe) (e.g., kite, hide);
1(3)(C)(v)
use common syllabication patterns to decode words, including: vowel digraphs and diphthongs (e.g., boy-hood, oat-meal); and
1(3)(C)(vi)
use common syllabication patterns to decode words, including: r-controlled vowel sounds (e.g., tar); including er, ir, ur, ar, and or;
1(3)(D)
decode words with common spelling patterns (e.g., -ink, -onk, -ick);
1(3)(D)
apply basic fact strategies to add and subtract within 20, including making 10 and decomposing a number leading to a 10
• Resource ID: T2T02
• Subject: Math
1(3)(E)
read base words with inflectional endings (e.g., plurals, past tenses);
1(3)(E)
explain strategies used to solve addition and subtraction problems up to 20 using spoken words, objects, pictorial models, and number sentences
• Resource ID: T2T02
• Subject: Math
1(3)(F)
use knowledge of the meaning of base words to identify and read common compound words (e.g., football, popcorn, daydream);
1(3)(F)
generate and solve problem situations when given a number sentence involving addition or subtraction of numbers within 20
1(3)(G)
identify and read contractions (e.g., isn't, can't);
1(3)(H)
identify and read at least 100 high-frequency words from a commonly used list; and
1(3)(I)
monitor accuracy of decoding.
1(4)(A)
confirm predictions about what will happen next in text by 'reading the part that tells';
1(4)(A)
collect, record, and compare information using tools, including computers, hand lenses, primary balances, cups, bowls, magnets, collecting nets, notebooks, and safety goggles; timing devices, including clocks and timers; non-standard measuring items such as paper clips and clothespins; weather instruments such as classroom demonstration thermometers and wind socks; and materials to support observations of habitats of organisms such as aquariums and terrariums; and
1(4)(A)
locate places using the four cardinal directions; and
1(4)(A)
identify U.S. coins, including pennies, nickels, dimes, and quarters, by value and describe the relationships among them
1(4)(B)
ask relevant questions, seek clarification, and locate facts and details about stories and other texts; and
1(4)(B)
measure and compare organisms and objects using non-standard units.
1(4)(B)
describe the location of self and objects relative to other locations in the classroom and school.
1(4)(B)
write a number with the cent symbol to describe the value of a coin
1(4)(C)
establish purpose for reading selected texts and monitor comprehension, making corrections and adjustments when that understanding breaks down (e.g., identifying clues, using background knowledge, generating questions, re-reading a portion aloud).
1(4)(C)
use relationships to count by twos, fives, and tens to determine the value of a collection of pennies, nickels, and/or dimes
### K–2 Diagnostic Assessments
• Resource ID: MATH_RA_K2
• Subject: Math
This resource provides two types of diagnostic tools available to K–2 math teachers: a rapid assessment tool and a flexible interview assessment.
1(5)
Students read grade-level text with fluency and comprehension. Students are expected to read aloud grade-level appropriate text with fluency (rate, accuracy, expression, appropriate phrasing) and comprehension.
1(5)(A)
classify objects by observable properties of the materials from which they are made such as larger and smaller, heavier and lighter, shape, color, and texture; and
1(5)(A)
create and use simple maps such as maps of the home, classroom, school, and community; and
1(5)(A)
recite numbers forward and backward from any given number between 1 and 120
1(5)(B)
predict and identify changes in materials caused by heating and cooling such as ice melting, water freezing, and water evaporating.
1(5)(B)
locate the community, Texas, and the United States on maps and globes.
1(5)(B)
skip count by twos, fives, and tens to determine the total number of objects up to 120 in a set
### K–2 Diagnostic Assessments
• Resource ID: MATH_RA_K2
• Subject: Math
This resource provides two types of diagnostic tools available to K–2 math teachers: a rapid assessment tool and a flexible interview assessment.
1(5)(C)
use relationships to determine the number that is 10 more and 10 less than a given number up to 120
1(5)(D)
represent word problems involving addition and subtraction of whole numbers up to 20 using concrete and pictorial models and number sentences
• Resource ID: T2T02
• Subject: Math
1(5)(E)
understand that the equal sign represents a relationship where expressions on each side of the equal sign represent the same value(s)
1(5)(F)
determine the unknown whole number in an addition or subtraction equation when the unknown may be any one of the three or four terms in the equation
1(5)(G)
apply properties of operations to add and subtract two or three numbers
### Making Ten and Applying Properties of Operations
• Resource ID: TEKS12_MATH_01_001
• Subject: Math
In these activities students will explore and identify patterns in related addition and subtractions number sentences.
1(6)(A)
identify words that name actions (verbs) and words that name persons, places, or things (nouns);
1(6)(A)
identify and discuss how different forms of energy such as light, heat, and sound are important to everyday life;
### Energy in Everyday Life
• Resource ID: K4SCI001
• Subject: Science
This resource provides sample activities for teachers to use in helping students to identify and discuss how different forms of energy such as light, heat, and sound are important to everyday life.
1(6)(A)
identify and describe the physical characteristics of places such as landforms, bodies of water, natural resources, and weather;
1(6)(A)
classify and sort regular and irregular two-dimensional shapes based on attributes using informal geometric language
### K–2 Diagnostic Assessments
• Resource ID: MATH_RA_K2
• Subject: Math
This resource provides two types of diagnostic tools available to K–2 math teachers: a rapid assessment tool and a flexible interview assessment.
1(6)(B)
determine the meaning of compound words using knowledge of the meaning of their individual component words (e.g., lunchtime);
1(6)(B)
predict and describe how a magnet can be used to push or pull an object;
### Magnets Push and Pull
• Resource ID: R4SCI0049 | 1,957 | 8,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-30 | latest | en | 0.865479 |
https://s2.smu.edu/~olinick/riot/detail_calc.html | 1,542,286,275,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742685.33/warc/CC-MAIN-20181115120507-20181115142507-00246.warc.gz | 744,420,804 | 5,696 | RIOT -- Major League Baseball --
### Introduction
As the end of the season approaches, baseball fans like to engage in speculation about which teams will advance to the post season and whether their favorite team is still in contention for a playoff berth. Newspapers frequently run headlines in their sports sections declaring that a particular team has been eliminated from contention or that the first place team has clinched a playoff berth. These determinations can often be made by looking at the standings and making a few simple calculations. The RIOT baseball standings can actually tell you when a team has locked up a playoff spot or fallen out of contention days before it is reported anywhere else!
The key is taking into account the actual matchups that remain in the season and enumerating all possible outcomes. Until the final weeks of the season, however, the number of possible outcomes can be astronomically large, which makes such an enumeration impractical even on a very fast computer. By using a powerful and pervasive analytics technique called Network Optimization, it is possible to accurately perform the necessary calculations without explicitly enumerating all possibilities. Consequently, it is possible for RIOT to determine the playoff prospects of a team a few days, or even weeks, before it is reported by the popular media. The following example, using the American League standings on the morning of Wednesday, September 5, 2012 shows how this works.
AL East
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL Magic Number 1st Postseason 1st Postseason Baltimore 76 59 - 0.563 27 - 26 22 5 2 New York 76 59 - 0.563 27 - 26 22 5 2 Tampa Bay 75 61 1.5 0.551 26 - 26 23 6 3 Boston 63 74 14 0.460 25 - * * 18 15 Toronto 60 75 16 0.444 27 - * * 21 18
Since Baltimore and New York are tied for first place in the AL East with identical records, neither team has a traditionally calculated magic number. However, if either team wins 26 of their remaining 27 games, they will clinch first place. This is because they still have a four-game series left on the schedule. For example, suppose Baltimore wins 26 more games. That would have to include at least three wins against New York. The final win totals in this scenario would be 102 for Baltimore, and at most 100 for New York. Therefore, RIOT reports that Baltimore has a first place clinch number of 26. Using the same logic, New York's first place clinch number is also 26.
Tampa Bay does not have a traditional magic number because they are not in first place. However, they have a four-game series left with New York, and a six-game series left with Baltimore. If they win all 26 of their remaining games, then they will win the division with 101 wins while New York has at most 99 wins and Baltimore has at most 97. Therefore, RIOT reports that Tampa Bay's first place clinch number is 26.
AL Central
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL Magic Number 1st Postseason 1st Postseason Chicago 73 62 - 0.541 27 27 25 25 2 2 Detroit 72 63 1 0.533 27 - 26 26 3 3 Kansas City 61 74 12 0.452 27 - * * 14 14 Cleveland 58 78 15.5 0.426 26 - * * 17 17 Minnesota 56 80 17.5 0.412 26 - * * 19 19
Chicago can clinch first place in the AL Central with 100 wins because Detroit's best possible win total at end of the season is 72+27 = 99.
Since Chicago already has 73 wins, their traditionally calculated magic number is 100 - 73 = 27.
It's easy to see that Chicago can clinch first place by winning all 27 of their remaining games, but they really only need to win 25 to clinch because they still have a four-game series with Detroit left on their schedule. If they split that series, and both teams win all of their other remaining games, then Chicago will finish the season with 98 wins to Detroit's 97. Therefore, RIOT reports that Chicago's first place clinch number is 25.
If Detroit wins 26 of their remaining 27 games, that would have to include at least three wins against Chicago. In this scenario, Detroit would win the division with 98 wins while Chicago would finish the season with at most 87. Therefore, Detroit's first place clinch number is 26.
AL West
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL Magic Number 1st Postseason 1st Postseason Texas 80 55 - 0.593 27 24 22 18 3 0 Oakland 76 59 4 0.563 27 - 26 22 7 2 Los Angeles 73 63 7.5 0.537 26 - * 25 10 5 Seattle 66 71 15 0.482 25 - * * 17 12
Texas can clinch first place in the AL West with 104 wins because Oakland's best possible win total at the end of the season is 76+27 = 103.
Since Texas already has 80 wins, their traditionally calculated magic number is 104 - 80 = 24.
But, Texas really only needs to win 22 more games to clinch the division because they still have a seven-game series against Oakland left on their schedule. If they win 22 out of their remaining 27 games, that would have to include at least two wins against Oakland. The final win totals in this scenario would be 102 for Texas, and at most 101 for Oakland. Therefore, RIOT reports that Texas has a first place clinch number of 22. Using similar logic, RIOT reports that Oakland's first place clinch number is 26.
In another example, RIOT's standings page for the National League on the morning of August 30, 2012 showed that Houston had already been eliminated from the playoffs even though it wasn't obvious from their won-lost record.
NL East
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL 1st Postseason 1st Postseason Washington 78 51 - 0.605 33 28 24 0 0 Atlanta 74 57 5 0.565 31 * 28 4 0 New York 61 69 17.5 0.469 32 * * 17 12 Philadelphia 61 69 17.5 0.469 32 * * 17 12 Miami 59 72 20 0.450 31 * * 19 14
NL Central
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL 1st Postseason 1st Postseason Cincinnati 80 52 - 0.606 30 24 22 0 0 St. Louis 71 59 8 0.546 32 * 31 9 2 Pittsburgh 70 60 9 0.538 32 * 32 10 3 Milwaukee 62 67 16.5 0.481 33 * * 18 11 Chicago 49 80 29.5 0.380 33 * * 31 24 Houston 40 90 39 0.308 32 * * Elim Elim
NL West
RIOT Numbers Clinch Avoid Elim Team W L GB PCT GL 1st Postseason 1st Postseason San Francisco 73 57 - 0.562 32 28 28 4 1 Los Angeles 70 61 3.5 0.534 31 31 31 7 3 Arizona 64 67 9.5 0.489 31 * * 13 9 San Diego 61 71 13 0.462 30 * * 16 12 Colorado 53 76 19.5 0.411 33 * * 24 20
It is easy to see from the standings above that Houston is eliminated from first place in the NL Central. If Houston were to win all 32 of their remaining games, they would finish with 72 wins. But, Cincinnati has already won 80 games; so no matter what else happens in the season, Cincinnati will finish ahead of Houston in the NL Central. However, the fact that Houston has also been eliminated from reaching the postseason as a wild card team is not readily apparent from the standings.
Since Washington, Atlanta, Cincinnati, and San Francisco already have better records than Houston, their only chance for the playoffs is to clinch the second wild card spot. St. Louis is currently in the lead for that spot with 71 wins. Since Houston still has a chance, however remote, of finishing the season with 72 wins, it looks as though they could still make the postseason as a wild card team. A closer look at the schedule, however, reveals that even a miracle winning streak would not allow Houston to make the playoffs.
Los Angeles, who have 70 wins so far, and St. Louis still have a four-game series to play before the end of season. There are no ties in baseball, which means that at least one of these teams will finish with at least 73 wins. If St. Louis wins two or more games in the series, then they will finish the season with at least 73 wins; and if they win one or zero of those games, then Los Angeles will finish with at least 73 wins. Either way, Houston is eliminated from the playoffs because there will be at least five teams in the National League with better records at the end of the season.
Determining that a particular team has been eliminated can often be much more difficult than it is in this example. Follow this link for a more complicated example involving all the teams in the American League East.
### Optimization Details
Calculating the clinching and elimination numbers for the RIOT baseball standings involves systematically searching for scenarios in which particular teams finish with or without gaining playoff berths. For example, we determined that San Francisco was eliminated from first place in the National League West on September 8th by proving that no feasible scenario exists in which the Giants win the division. The problem of determining whether a team can advance to playoffs given the current league standings and schedule of remaining games can be solved by a single maximum flow calculation (see Hoffman and Rivlin [1] and Schwartz [2]). By introducing additional constraints, we extend this maximum flow formulation to derive integer linear programming problems which find the minimum number of games a given team must win to clinch a playoff spot or avoid elimination from post season play. Robinson [3] takes a similar approach to finding a scenario which maximizes a given team's lead in the final standings. Interested readers should also consult Gusfield and Martel [4], who show how to find the minimum number of games a team must win to avoid elimination from first place by solving a parametric minimum cut problem.
For more information on the integer programming formulations and implementation of the RIOT baseball standings, send mail to the RIOT Baseball Project and see our paper "Baseball, Optimization and the World Wide Web", which appeared in the March/April 2002 issue Interfaces.
### References
[1] A. Hoffman and J. Rivlin. "When is a team 'mathematically' eliminated ?" In H.W. Kuhn, editor, Princeton Symposium on Math Programming (1967), Princeton University Press, Princeton, NJ, 1970, pp. 391 - 401.
[2] B. Schwartz. "Possible winners in partially completed tournaments." SIAM Review. 8:302-308. 1966
[3] L. W. Robinson "Baseball playoff eliminations: An application of linear programming." Operations Research Letters. 10:67 -74. 1991
[4] D. Gusfield and C. Martel. "A Fast Algorithm for the Generalized Parametric Minimum Cut Problem and Applications." Algorithmica. 7:499 -519. 1992
[ Home | American League | National League | Numbers Explanation | Problem Explanation | Frequently Asked Questions ] | 2,627 | 10,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.937415 |
https://www.kodytools.com/units/angularacc/from/gradpns2/to/milpmin2 | 1,713,147,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00738.warc.gz | 777,539,333 | 17,610 | # Gradian/Square Nanosecond to Mil/Square Minute Converter
1 Gradian/Square Nanosecond = 5.76e+22 Mil/Square Minute
## One Gradian/Square Nanosecond is Equal to How Many Mil/Square Minute?
The answer is one Gradian/Square Nanosecond is equal to 5.76e+22 Mil/Square Minute and that means we can also write it as 1 Gradian/Square Nanosecond = 5.76e+22 Mil/Square Minute. Feel free to use our online unit conversion calculator to convert the unit from Gradian/Square Nanosecond to Mil/Square Minute. Just simply enter value 1 in Gradian/Square Nanosecond and see the result in Mil/Square Minute.
Manually converting Gradian/Square Nanosecond to Mil/Square Minute can be time-consuming,especially when you don’t have enough knowledge about Angular Acceleration units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Gradian/Square Nanosecond to Mil/Square Minute converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gradian/Square Nanosecond to Mil/Square Minute, but not every online tool gives an accurate result and that is why we have created this online Gradian/Square Nanosecond to Mil/Square Minute converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gradian/Square Nanosecond to Mil/Square Minute (grad/ns2 to mil/min2)
By using our Gradian/Square Nanosecond to Mil/Square Minute conversion tool, you know that one Gradian/Square Nanosecond is equivalent to 5.76e+22 Mil/Square Minute. Hence, to convert Gradian/Square Nanosecond to Mil/Square Minute, we just need to multiply the number by 5.76e+22. We are going to use very simple Gradian/Square Nanosecond to Mil/Square Minute conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gradian/Square Nanosecond} = 1 \times 5.76e+22 = \text{5.76e+22 Mil/Square Minute}$$
## What Unit of Measure is Gradian/Square Nanosecond?
Gradian per square nanosecond is a unit of measurement for angular acceleration. By definition, if an object accelerates at one gradian per square nanosecond, its angular velocity is increasing by one gradian per nanosecond every nanosecond.
## What Unit of Measure is Mil/Square Minute?
Mil per square minute is a unit of measurement for angular acceleration. By definition, if an object accelerates at one mil per square minute, its angular velocity is increasing by one mil per minute every minute.
## What is the Symbol of Mil/Square Minute?
The symbol of Mil/Square Minute is mil/min2. This means you can also write one Mil/Square Minute as 1 mil/min2.
## How to Use Gradian/Square Nanosecond to Mil/Square Minute Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gradian/Square Nanosecond and in the first input field, enter a value.
• From the second dropdown, select Mil/Square Minute.
• Instantly, the tool will convert the value from Gradian/Square Nanosecond to Mil/Square Minute and display the result in the second input field.
## Example of Gradian/Square Nanosecond to Mil/Square Minute Converter Tool
1
Mil/Square Minute
5.76e+22
# Gradian/Square Nanosecond to Other Units Conversion Table
ConversionDescription
1 Gradian/Square Nanosecond = 900000000000000000 Degree/Square Second1 Gradian/Square Nanosecond in Degree/Square Second is equal to 900000000000000000
1 Gradian/Square Nanosecond = 900000000000 Degree/Square Millisecond1 Gradian/Square Nanosecond in Degree/Square Millisecond is equal to 900000000000
1 Gradian/Square Nanosecond = 900000 Degree/Square Microsecond1 Gradian/Square Nanosecond in Degree/Square Microsecond is equal to 900000
1 Gradian/Square Nanosecond = 0.9 Degree/Square Nanosecond1 Gradian/Square Nanosecond in Degree/Square Nanosecond is equal to 0.9
1 Gradian/Square Nanosecond = 3.24e+21 Degree/Square Minute1 Gradian/Square Nanosecond in Degree/Square Minute is equal to 3.24e+21
1 Gradian/Square Nanosecond = 1.1664e+25 Degree/Square Hour1 Gradian/Square Nanosecond in Degree/Square Hour is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 6.718464e+27 Degree/Square Day1 Gradian/Square Nanosecond in Degree/Square Day is equal to 6.718464e+27
1 Gradian/Square Nanosecond = 3.29204736e+29 Degree/Square Week1 Gradian/Square Nanosecond in Degree/Square Week is equal to 3.29204736e+29
1 Gradian/Square Nanosecond = 6.224263236e+30 Degree/Square Month1 Gradian/Square Nanosecond in Degree/Square Month is equal to 6.224263236e+30
1 Gradian/Square Nanosecond = 8.96293905984e+32 Degree/Square Year1 Gradian/Square Nanosecond in Degree/Square Year is equal to 8.96293905984e+32
1 Gradian/Square Nanosecond = 54000000000000000000 Arcmin/Square Second1 Gradian/Square Nanosecond in Arcmin/Square Second is equal to 54000000000000000000
1 Gradian/Square Nanosecond = 54000000000000 Arcmin/Square Millisecond1 Gradian/Square Nanosecond in Arcmin/Square Millisecond is equal to 54000000000000
1 Gradian/Square Nanosecond = 54000000 Arcmin/Square Microsecond1 Gradian/Square Nanosecond in Arcmin/Square Microsecond is equal to 54000000
1 Gradian/Square Nanosecond = 54 Arcmin/Square Nanosecond1 Gradian/Square Nanosecond in Arcmin/Square Nanosecond is equal to 54
1 Gradian/Square Nanosecond = 1.944e+23 Arcmin/Square Minute1 Gradian/Square Nanosecond in Arcmin/Square Minute is equal to 1.944e+23
1 Gradian/Square Nanosecond = 6.9984e+26 Arcmin/Square Hour1 Gradian/Square Nanosecond in Arcmin/Square Hour is equal to 6.9984e+26
1 Gradian/Square Nanosecond = 4.0310784e+29 Arcmin/Square Day1 Gradian/Square Nanosecond in Arcmin/Square Day is equal to 4.0310784e+29
1 Gradian/Square Nanosecond = 1.975228416e+31 Arcmin/Square Week1 Gradian/Square Nanosecond in Arcmin/Square Week is equal to 1.975228416e+31
1 Gradian/Square Nanosecond = 3.7345579416e+32 Arcmin/Square Month1 Gradian/Square Nanosecond in Arcmin/Square Month is equal to 3.7345579416e+32
1 Gradian/Square Nanosecond = 5.377763435904e+34 Arcmin/Square Year1 Gradian/Square Nanosecond in Arcmin/Square Year is equal to 5.377763435904e+34
1 Gradian/Square Nanosecond = 3.24e+21 Arcsec/Square Second1 Gradian/Square Nanosecond in Arcsec/Square Second is equal to 3.24e+21
1 Gradian/Square Nanosecond = 3240000000000000 Arcsec/Square Millisecond1 Gradian/Square Nanosecond in Arcsec/Square Millisecond is equal to 3240000000000000
1 Gradian/Square Nanosecond = 3240000000 Arcsec/Square Microsecond1 Gradian/Square Nanosecond in Arcsec/Square Microsecond is equal to 3240000000
1 Gradian/Square Nanosecond = 3240 Arcsec/Square Nanosecond1 Gradian/Square Nanosecond in Arcsec/Square Nanosecond is equal to 3240
1 Gradian/Square Nanosecond = 1.1664e+25 Arcsec/Square Minute1 Gradian/Square Nanosecond in Arcsec/Square Minute is equal to 1.1664e+25
1 Gradian/Square Nanosecond = 4.19904e+28 Arcsec/Square Hour1 Gradian/Square Nanosecond in Arcsec/Square Hour is equal to 4.19904e+28
1 Gradian/Square Nanosecond = 2.41864704e+31 Arcsec/Square Day1 Gradian/Square Nanosecond in Arcsec/Square Day is equal to 2.41864704e+31
1 Gradian/Square Nanosecond = 1.1851370496e+33 Arcsec/Square Week1 Gradian/Square Nanosecond in Arcsec/Square Week is equal to 1.1851370496e+33
1 Gradian/Square Nanosecond = 2.24073476496e+34 Arcsec/Square Month1 Gradian/Square Nanosecond in Arcsec/Square Month is equal to 2.24073476496e+34
1 Gradian/Square Nanosecond = 3.2266580615424e+36 Arcsec/Square Year1 Gradian/Square Nanosecond in Arcsec/Square Year is equal to 3.2266580615424e+36
1 Gradian/Square Nanosecond = 30000000000000000 Sign/Square Second1 Gradian/Square Nanosecond in Sign/Square Second is equal to 30000000000000000
1 Gradian/Square Nanosecond = 30000000000 Sign/Square Millisecond1 Gradian/Square Nanosecond in Sign/Square Millisecond is equal to 30000000000
1 Gradian/Square Nanosecond = 30000 Sign/Square Microsecond1 Gradian/Square Nanosecond in Sign/Square Microsecond is equal to 30000
1 Gradian/Square Nanosecond = 0.03 Sign/Square Nanosecond1 Gradian/Square Nanosecond in Sign/Square Nanosecond is equal to 0.03
1 Gradian/Square Nanosecond = 108000000000000000000 Sign/Square Minute1 Gradian/Square Nanosecond in Sign/Square Minute is equal to 108000000000000000000
1 Gradian/Square Nanosecond = 3.888e+23 Sign/Square Hour1 Gradian/Square Nanosecond in Sign/Square Hour is equal to 3.888e+23
1 Gradian/Square Nanosecond = 2.239488e+26 Sign/Square Day1 Gradian/Square Nanosecond in Sign/Square Day is equal to 2.239488e+26
1 Gradian/Square Nanosecond = 1.09734912e+28 Sign/Square Week1 Gradian/Square Nanosecond in Sign/Square Week is equal to 1.09734912e+28
1 Gradian/Square Nanosecond = 2.074754412e+29 Sign/Square Month1 Gradian/Square Nanosecond in Sign/Square Month is equal to 2.074754412e+29
1 Gradian/Square Nanosecond = 2.98764635328e+31 Sign/Square Year1 Gradian/Square Nanosecond in Sign/Square Year is equal to 2.98764635328e+31
1 Gradian/Square Nanosecond = 2500000000000000 Turn/Square Second1 Gradian/Square Nanosecond in Turn/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Turn/Square Millisecond1 Gradian/Square Nanosecond in Turn/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Turn/Square Microsecond1 Gradian/Square Nanosecond in Turn/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Turn/Square Nanosecond1 Gradian/Square Nanosecond in Turn/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Turn/Square Minute1 Gradian/Square Nanosecond in Turn/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Turn/Square Hour1 Gradian/Square Nanosecond in Turn/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Turn/Square Day1 Gradian/Square Nanosecond in Turn/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Turn/Square Week1 Gradian/Square Nanosecond in Turn/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Turn/Square Month1 Gradian/Square Nanosecond in Turn/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Turn/Square Year1 Gradian/Square Nanosecond in Turn/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 2500000000000000 Circle/Square Second1 Gradian/Square Nanosecond in Circle/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Circle/Square Millisecond1 Gradian/Square Nanosecond in Circle/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Circle/Square Microsecond1 Gradian/Square Nanosecond in Circle/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Circle/Square Nanosecond1 Gradian/Square Nanosecond in Circle/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Circle/Square Minute1 Gradian/Square Nanosecond in Circle/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Circle/Square Hour1 Gradian/Square Nanosecond in Circle/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Circle/Square Day1 Gradian/Square Nanosecond in Circle/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Circle/Square Week1 Gradian/Square Nanosecond in Circle/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Circle/Square Month1 Gradian/Square Nanosecond in Circle/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Circle/Square Year1 Gradian/Square Nanosecond in Circle/Square Year is equal to 2.4897052944e+30
1 Gradian/Square Nanosecond = 16000000000000000000 Mil/Square Second1 Gradian/Square Nanosecond in Mil/Square Second is equal to 16000000000000000000
1 Gradian/Square Nanosecond = 16000000000000 Mil/Square Millisecond1 Gradian/Square Nanosecond in Mil/Square Millisecond is equal to 16000000000000
1 Gradian/Square Nanosecond = 16000000 Mil/Square Microsecond1 Gradian/Square Nanosecond in Mil/Square Microsecond is equal to 16000000
1 Gradian/Square Nanosecond = 16 Mil/Square Nanosecond1 Gradian/Square Nanosecond in Mil/Square Nanosecond is equal to 16
1 Gradian/Square Nanosecond = 5.76e+22 Mil/Square Minute1 Gradian/Square Nanosecond in Mil/Square Minute is equal to 5.76e+22
1 Gradian/Square Nanosecond = 2.0736e+26 Mil/Square Hour1 Gradian/Square Nanosecond in Mil/Square Hour is equal to 2.0736e+26
1 Gradian/Square Nanosecond = 1.1943936e+29 Mil/Square Day1 Gradian/Square Nanosecond in Mil/Square Day is equal to 1.1943936e+29
1 Gradian/Square Nanosecond = 5.85252864e+30 Mil/Square Week1 Gradian/Square Nanosecond in Mil/Square Week is equal to 5.85252864e+30
1 Gradian/Square Nanosecond = 1.1065356864e+32 Mil/Square Month1 Gradian/Square Nanosecond in Mil/Square Month is equal to 1.1065356864e+32
1 Gradian/Square Nanosecond = 1.593411388416e+34 Mil/Square Year1 Gradian/Square Nanosecond in Mil/Square Year is equal to 1.593411388416e+34
1 Gradian/Square Nanosecond = 2500000000000000 Revolution/Square Second1 Gradian/Square Nanosecond in Revolution/Square Second is equal to 2500000000000000
1 Gradian/Square Nanosecond = 2500000000 Revolution/Square Millisecond1 Gradian/Square Nanosecond in Revolution/Square Millisecond is equal to 2500000000
1 Gradian/Square Nanosecond = 2500 Revolution/Square Microsecond1 Gradian/Square Nanosecond in Revolution/Square Microsecond is equal to 2500
1 Gradian/Square Nanosecond = 0.0025 Revolution/Square Nanosecond1 Gradian/Square Nanosecond in Revolution/Square Nanosecond is equal to 0.0025
1 Gradian/Square Nanosecond = 9000000000000000000 Revolution/Square Minute1 Gradian/Square Nanosecond in Revolution/Square Minute is equal to 9000000000000000000
1 Gradian/Square Nanosecond = 3.24e+22 Revolution/Square Hour1 Gradian/Square Nanosecond in Revolution/Square Hour is equal to 3.24e+22
1 Gradian/Square Nanosecond = 1.86624e+25 Revolution/Square Day1 Gradian/Square Nanosecond in Revolution/Square Day is equal to 1.86624e+25
1 Gradian/Square Nanosecond = 9.144576e+26 Revolution/Square Week1 Gradian/Square Nanosecond in Revolution/Square Week is equal to 9.144576e+26
1 Gradian/Square Nanosecond = 1.72896201e+28 Revolution/Square Month1 Gradian/Square Nanosecond in Revolution/Square Month is equal to 1.72896201e+28
1 Gradian/Square Nanosecond = 2.4897052944e+30 Revolution/Square Year1 Gradian/Square Nanosecond in Revolution/Square Year is equal to 2.4897052944e+30 | 4,466 | 14,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | longest | en | 0.88906 |
https://www.gamedev.net/blogs/blog/468-ramblings-of-a-partialy-sane-programmer/?page=1&sortby=entry_name&sortdirection=desc | 1,571,078,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00129.warc.gz | 934,330,559 | 53,612 | • entries
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Ramblings of programmyness, tech, and other crap.
## Yes I be a contemplater
Ok after much thought about what I want to do game development wise my final decision is to use Java and JOGL. I made this decision for a few reasons.
1. Ease of cross platform distrobution via Web start.
2. I am extreamily more fluent in Java then any other language basically because it was one of my first languages and I studied it in college.
3. I already have books for reference on Java and OpenGL incase I need to look something up.
It is amazing what contemplation can do especially with a goal in mind.
## Wow Long Time
Holy crap has it been a long time since I posted here. I have been so tied up with school and work that I kind of just fell of the face of the earth
being totally swamped with no real time to do much of anything.
I just recently due to school got back into doing some programming. Partially because of the nature of the class and me being as lazy as I could possibly be just not wanting to go through all the repeditive steps.
Right now I am taking a statistics class and calculating all of the probability stuff can get very very long and repedative to find out the various different answers. For instance when finding the binomial probability of a range of numbers in a set you might have to calculated 12 different binomial probabilities and then add them together so you can then caluculate the complement of that probability to find the other side of the range of numbers. It is just way too repedative in my liking.
The advantage of this is it really re-kindled my love of the Python language. I just wish the language was a bit more useful for game development sadly. The performance hits are just way too high when you progress onto 3D.
After I finished my homework I decided to do a comparison of the Python and C++ code required for calculating the binomial probability of a number in a set. This is the overall gist of the post because it is really amazing to see the difference in the code of two examples of the same program and it is simple enough to really demonstrate both in a reasonable amount of time. The interesting thing here is from a outside perspective runing both they appear to be run instantaniously with no performance difference at all. So here is the code btw it is indeed a night and day difference in readability and understandability.
Python (2.7.3)
def factorial(n):
if n n = 1
return 1 if n == 1 else n * factorial(n - 1)
def computeBinomialProb(n, p, x):
nCx = (factorial(n) / (factorial(n-x) * factorial(x)))
px = p ** x
q = float(1-p)
qnMinx = q ** (n-x)
return nCx * px * qnMinx
if __name__ == '__main__':
n = float(raw_input("Value of n?:"))
p = float(raw_input("Value of p?:"))
x = float(raw_input("Value of x?:"))
print "result = ", computeBinomialProb(n, p, x)
C++
#include
#include
int factorial(int n)
{
if (n n = 1;
return (n == 1 ? 1 : n * factorial(n - 1));
}
float computeBinomialProb(float n, float p, float x)
{
float nCx = (factorial(n) / (factorial(n - x) * factorial(x)));
float px = pow(p, x);
float q = (1 - p);
float qnMinx = pow(q, (n - x));
return nCx * px * qnMinx;
}
int main()
{
float n = 0.0;
float p = 0.0;
float x = 0.0;
float result = 0.0;
std::cout std::cin >> n;
std::cout std::cin >> p;
std::cout std::cin >> x;
result = computeBinomialProb(float(n), float(p), float(x));
std::cout return 0;
}
Sorry for no syntax highlighting I forget how to do this.
The bigest thing you can notice is that in Python you don't need all the type information which allows for really easy and quick variable declarations which actually slims the code down quite a bit. Another thing to notice is you can prompt and gather information in one go with the Python where in C++ you need to use two different streams to do so. I think the Python is much more readible but the C++ is quite crisp as well.
## Wow it has been a while
Hello everyone it has been another long while since I posted been so busy. Work and School seems to occupy my life and what little spare time I have had I use to play COD Black Ops on my PS3.
School is going great I am acing my classes so far. I must say I never expected going to school online to be so much more difficult then going right to a campus and classroom. The main reason I am going online is because I really have no choice with my job and what schools offer around here. It really takes a lot of time and dedication to stay on track when going to school online. There are no reminders you need to make sure everything is planned and organized when it comes to your time and that helps avoid the procrastination bug that people tend to have.
On another note as of late I have been learning the workings of the android framework. I have never done any sort of embedded development before so this is new territory for me when it comes to game development. So not only have I had to learn the layout of the phones architecture but I also have to brush up a bit on my Java. While doing this I have been watching some of Googles I/O videos about best practices for embedded android development to avoid some common architecture performances issues for when I start my project. From the looks of it to build a solid Android app you really need to watch what you do because so many things we take for granted doing desktop development are really expensive in performance cost when you toss it on a 1 ghz ARM processor.
That is all for now I will keep posted on the project process as I move further into development.
## Woot Woot D3 is announced and other stuff!!!!!!!
A little late on the jump had other stuff running through my mind as per my last journal entry. I am so Psyched up about this game I have been waiting for it for 8 years. The ideas that Blizzard can put into game is just pure genious. From the looks of it this will be another genra statement in the action rpg relm. Once again blizz will set the standard. For one the new boss multiplayer loot system is simple but very new. The fact that each person taking place in the kill gets his own loot generated is great. Also, new new health loot orbs allowing the front line to heal the back line is a great concept as well. Lets hope for the best to see it soon in action. Hurry Blizzard I need this game.
-----------------
On another note my new book on xna is much better then my last one. The previous books was Professional XNA Game Programming for XNA2.0 published by wrox. Horrible book just plain horrid. The fact that the auther changes a key part of the code and fails to tell you hey you got to modify this too is just horrible style of writing. Now with my new book Beginning XNA 2.0 Game Programming (from novice to professional) is just great. It really takes the time to explain everything before making your first 2d game and even 3d game. It starts with game dev concepts that all game developers should know. Then it explains a bit about the way xna works. Next it walks you through the steps of rendering a sprite and some simple collision detection, input and sound. Once that is done you make your first 2d game Rock Rain. I am on the part where I get to make the 2d game and I already understand alot more about game dev than I have ever did. Once I get through that section I think I am going to take a break from the book and make some 2d games to make sure I understand the concepts better. Then I will move to the 3d sections of the book.
Why could their not be a book written this good for C++ and OpenGL. If it was the case I would have been making games a long time ago instead of stupid little tech demos of rendering triangles and cubes.
If you are looking to learn XNA get this book it will really help out.
## Web site mock up complete.
So I am in the process of starting up a small indy game company. It will start as a one man operation pritty much and the plan is to release the games for free. Just a little something to occupy some time really. Who knows maybe it will take off and get bigger. But not worried about that. So through some research I realized that now a days a lot more goes into a website then XHTML/CSS and PHP. So I broke out the rusty Gimp skills and put together a entire web site mock up. This is the result of many hours of work I would estimate about 7 hours of work with research time included. I wanted to get this layout just right so I was tweaking stuff constantly. The next step will be to splice up the site into small images and then use those with XHTML/CSS via image replacement to build the functional site. I will also be using PHP to deliver database driven news.
I would like to know your opinions on the mock up. I am a big fan of constructive criticism. Click the thumbnail to make the mock up bigger.
## Visual C++ 2010 Express Beta 2
I just installed Express 2010 beta 2. One thing I can say it is very fast. In the coming days we will see how it works out. So far I was able to build SFML and Boost with it. I did stumble into some issues with SFML but resolved them. Turns out the /INCREMENTAL option was turned off in one of the Build Configs. Once I flipped the switch it seems to build alright. We will see if it actually works at a later date.
## Version Control.
First I must open with the simple fact that a lot of people just don't use version control. The main reason for this is because in all honesty a lot of people just don't understand it. Not to mention the major version control war going on at the current moment just tends to confuse people all together even more. I am going to do my best to give some informative information on the different version control systems (VCS) out there to try and help make sense of the decision I need to make for my next project as well as hopefully help others make the decision for their projects.
First there are currently two types of VCS's out there at the current moment first is the CVCS and next is the DVCS. CVCS systems like subversion and cvs have a central server that every client must connect to. The client basically pulls revision information from these system to a working copy on your hard drive. These working copies tend to be very small because it pulls only the information it needs to be considered up to date basically the latest revision. The major gripe people seem to have with these types of systems at the current moment is the lack of enough information to do a proper merge of a branch back into the main code base. DVCS systems are what people call distributed. I hate that term because I think it makes things harder to understand. Examples of these are git, mercurial, and bazaar. The reason I hate the term distributed is because currently a lot of people use DVCS systems just like a CVCS system but with benefits. Typically the main code base is stored in a central location so that people can stay properly up to date with the code base. DVCS pulls the entire repository to your system including all of the revision history not just the latest. This allows for you to not be connected to the server and do everything right from your machine without a internet connection. What I like about this is in its own right everyone has a complete copy so if something happens to the central repository numerous people have a backup of the code and revision history. The nice thing about DVCS and why I think so many people are fanatic about it is the easy branching and merging. Because you have the total revision history it allows you to easily branch merge and cherry pick changes at will without a lot of risk of "pain". So when looking at the two different types of systems think CVCS (most recent revision only in working copy), DVCS (total history in working copy).
*Warning opinionated*
My main gripes with the current arguments out there have to do with the fact of branching. The DVCS group of people seem to like the idea of branching every time they make a code change or add a new feature. They argue that this is insane and painful to do in SVN because of horrible revision/merge tracking. Ok I agree branching and merging can be painful in subversion but at the same time it is not as bad as people say because they are taking it to a extreme and not using the system properly. I am not the kind of person that likes to make a branch for every feature I add into a application. I feel branching should only be used for major changes, re-factors, and other changes that have a large chance to BREAK the current working development tree. Maybe this mentality is why I never really had to many issues with subversion to begin with when it came to branches and merges. Maybe it is because I was using subversion as the developers INTENDED according to the svn red-book. I don't know maybe I am just a hard sell.
So which one should you use. It is hard to say each system has its pros and cons as you have seen. The one feature I love about the DVCS camp is the speed at which you can get up and running and the fact that sites like GitHub and Bitbucket are amazing places to host your code. I also like the speed of DVCS systems, because you are doing everything from your local machine and not over a network DVCS is blazing fast compared to CVCS. Lastly the thing I like the most about DVCS is the fact that they are very flexible and you can use them with whatever workflow you desire. The main cons I have for DVCS are the lack of repository permissions, large file support is really not that good because of full repo history pulls, no locking for binary files and other code files to prevent major issues if both are modified at the same time. For example if you only want to branch when you feel something has a good chance to break the entire main line of development you can do so. If you want to be considered the central repo instead of a host like GitHub you can do so. What I like about CVCS is the fact that it can handle very large files quite well. Another thing I like is the fact you can have repository permissions making sure you know who can write to the repository. I also like the fact that you can lock binary files and other files if you wish to prevent other people from making changes to the file as you make your changes. This alone can save tons of time if you are working as a team and your art assets are in version control with your code. The major issues I see with CVCS are the required network connection and can cause speed issues *coffee anyone*, not having the full revision history present on the machine making it difficult to cherry pick or inject merges into different parts of the development line.
Keep in mind the pros and cons are as I see them other people may have different pros and cons. Yes I listed less cons for subversion however, there are times when these cons can definitely outweigh the pros. The cons of DVCS are hard to ignore and can outweigh the pros at times as well. So with these things in mind I am sure you can better make the decision you need to make. As for me I have used both and I like both a lot which make the decision for me extra hard. The one big pull factor for me is hosting and DVCS has the huge win there. So for my next project I will be using DVCS because I feel the pros outweigh the cons more so under most circumstances. Not to mention I really like the speed and the fact of having a whole repository backup on my machine. Ultimately the decision is yours but with this information hopefully you can weed through the google fluff that is out there.
In the future I just might have to return with my experiences and if anyone wants more detail into the inner workings of Version control systems let me know in the comments and I will go out and find the video that goes into the details of the internal differences of how revisions are stored.
## Update on where things are heading
Hey GDNet,
I know I don't post often enough a lot of this has to due with me being boged down with school + a full time job. The other reason is that I don't really tinker around with game programming that much anymore either. I still want to learn opengl at some point or another but this has been put on the back burner. Hopefully I can return to this goal at a later date when there are some better resources available aka if the new red book turns out to be written right this time.
On another note one thing I have wanted to get into for a long time is embedded development through microcontrollers (MCU). The reasoning behind this is it overall can make you a better developer. You have very small ammount of resources available that you need to use sparingly. Not to mention more often then not you get to use Assembly. I have always wanted to learn Assembly not to use for a project but to make myself a better developer. The reason this holds true is that in order to utilize Assembly you need to understand the bare metal architecture of the chip you are using. x86 and x86_64 are very complex architectures with huge ammounts of instructions and it make it difficult to learn. So one way is to instead use a MCU and then gradually work your way up.
My end goal project for this would be to make an 8-bit game I write like say asteroids run on a MCU. I asked for advice on a forum on what hardware I should look at to get to this goal and I was told I should look into Atmel Mega chips. Initially I was looking at the 8-bit PIC chips made by microchip. On the microchip forums I was told I am in for a big learning curve and PIC is probably a bad choice for an 8-bit game because the call stack is small and the ram/flash space is tiny. They also said the C compilers are bloated unless you buy a professional one. UH this is the point. The original gameboy ran a modified Z80 chip made by sharp. The actual specs of the chip are easily matched by the PIC 8-bit MCU's. So I decided to go with PIC anyway because from what I have read they have the better dev tools and are more then capable to compete with a Atmel Mega and are cheaper to get started with and have tons of documentation.
So despite this advice I made my order. This is what I bought there is a link to the store page if you are interested on this description page.
http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=1406&dDocName=en559587 In the side bar there is a link to buy/sample options if you want to look at buying one yourself.
I think this will be a great chip to start with as it has 12 tutorials in assembly & c the IDE as well as the programmer demo board and 2 MCU chips a PIC16F and PIC18F. The PIC16 is the mid range PIC 8-bit MCU and the PIC18 is the High End PIC 8-bit MCU. The tutorials cover both chips.
Wish me luck this is going to be FUN!!!!! I will try to post my progress here if you are interested. I still may end up making an outside blog instead not sure yet but if I do I will for sure kick a linkback here.
That is all for now have fun and code well.
## Update on learning XNA
In my last journal entry I mentioned I was going to give XNA a shot. So I did and I started learning XNA. As I am going through the book XNA feels kind of clunky and overly simplified. Not sure if it is the authors doing or the frameworks doing atm. But right now all I can say is I was more comfortable in C++ and OpenGL a few years ago. C++ and OpenGL made more sence to me I guess. This comes as a shocker because I do like C# as a whole. Maybe it is because I have used C++ longer then C# as it was one of my first languages I learned but have not yet totally mastered even after 7+ years of using it. Not sure what to do atm.
## Turns out it was the book
The Xna Programming Professional book by wrox is just horrible. The author fails to mention to the reader when important changes to the source are nessesary. I spent 20 minutes today hacking my way through his sample code trying to find the piece of code that was missing to get the damn pong menu to show up. Worst book I have bought in years. So down to the books store for me in a few hours to try and find a different book to learn from.
## Tool chain's for the stubborn
Hello GDNet. It has bee a while since I last posted partially because I have been really busy and just really did not have too much to post about. Today I am going to talk about toolchain's for the stubborn people among us. Every so often I see a thread come by on for beginners asking about command line compilation and various out of the ordinary editors. This entry is for those people. Like them I am stubborn and often prefer to do things the "more difficult way". So in order to do this properly I am going to have to make a few assumptions and here they are...
1. You like me are stubborn. This is defined as an individual who refuses to "see the light" and don't understand why everyone uses IDE's. You fail to see the benefits to your productivity and feel the feature they provide often get in the way and also feel that they are bloated, cumbersome, and restrictive to your workflow.
2. You want to develop cross platform and are completely clueless to the whole GCC/Make world and don't want to restrict yourself to a pure Microsoft compiler so that you can carry your new found knowledge over to Linux, BSD, and Mac OSX.
Note: As of this posting I don't have a particular editor of choice to put in the tool chain list I am still trying to find one I am 100% comfortable with.
Attention: At the end of this post I will also provide a self contained zip file with the appropriate files for you to try this out yourself. I will be using an example from the OpenGL Superbible 5 for this as this is GDNet. In order to compile and run this example your video card must support a minimum of OpenGL 3.0.
What we will be working with:
1. MinGW GCC + Msys
2. Makefiles
3. Editor of choice
Ok that is really all you need.
Installation:
First we need to install the MinGW GCC + Msys collection onto our Windows System.
2. Run this exe file to start the installation. Ensure you install this to C:\MinGW and make sure you also click the check box for C++ and Msys base.
3. Next we need to set a environment variable so that we can use these tools from the windows command line for simplicity of use.
- Open My Computer a.k.a. Computer (Windows 7) from the start menu and in the top bar select System Properties.
- Next select advanced system properties from the side bar. (Windows 7 not sure about XP it has been a while).
- Then click the Environment Variables button.
- Under user variables press new and call it PATH then under value type C:\MinGW\bin;C:\MinGW\msys\1.0\bin; and then press ok.
- Now press ok on the rest of the windows to close them out.
- We are now ready to go.
The Infamous Makefile:
Ok now in order to make larger projects and projects with dependencies easier to compile this thing called the Makefile was invented. The Makefile is a small script that the make tool reads so it knows what steps it must take to compile the code.
A Makefile consists of a few properties usually what are known as Variables and Targets. The Variables are designed so you can save yourself a lot of typing in lengthy Makefiles and Targets are basically labels of what you want to compile. When you use the make command the syntax is make the target tells make where to enter in the Makefile so it knows what to compile.
Here is the Makefile for the Triangle project don't worry I will break it down...
[source lang="make"]MAIN = Triangle
SRCPATH = ./src/
LIBDIRS = -L../libs/freeglut-2.6.0/lib -L../libs/GLTools/lib
INCDIRS = -I../libs/freeglut-2.6.0/include -I../libs/GLTools/include -I../libs/GLTools/include/GL
CXX = g++
CXXFLAGS = $(INCDIRS) LIBS = -lfreeglut32_static -lgltools -lglew -lwinmm -lgdi32 -lopengl32 -lglu32 all :$(MAIN)
$(MAIN).o :$(SRCPATH)$(MAIN).cpp$(MAIN) : $(MAIN).o$(CXX) $(CXXFLAGS) -o$(MAIN) $(LIBDIRS)$(SRCPATH)$(MAIN).cpp$(LIBS)
clean :
rm -f *.o
[/source]
Ok so what does all this mean. Well at first glance it may seem overwhelming due to variables but in all honesty it is quite simple.
First line 1-8 are variables. The syntax for variables is as follows... =
Very simple. Here in this Makefile I am using variables to handle the Main application name, the source code path, the library path, the include path, the compiler to use, the flags for the compiler, and the libraries that need to be linked to the binary file. In GCC -L is used for library path, -I is use for include path and -l is used for linking.
The rest of the file consists of targets, dependencies and rules(commands).
The general syntax for the structure of a target is as follows.
:
It is important to note that for the Rules section you must use a hard tab not a tab that was converted to spaces.
The first target is "all" This is executed when you type make or more explicitly make all at the command prompt. The dependency to this target is $(MAIN) which is the variable from above that actually points to our target later in the file known as$(MAIN).
The second target is $(MAIN).o with a dependency of$(MAIN).cpp. In this case remember our variable resolves to Target so this is the equivalent of dictating what the dependencies for our programs object file are.
The next target is $(MAIN) with a dependency of$(MAIN).o This target is what actually compiles and links our application. It has a dependency on our Triangle.cpp's object file. This target is slightly different in that it actually contains a rule. This rule is pretty straight forward if you resolve the variables. It is the actual call to the g++ compiler to compile and then link our application.
The last target is clean with no dependency. This has one rule which deletes all of the object files in our program path so we can get a clean compile from running make. To execute this target just like any target you need to type make clean.
That concludes our look at Makefiles so how do I actually use this?
Basically open your command prompt navigate to the projects root directory (The one with the Makefile in it. Makefiles are simply named Makefile or something with Makefile in the name) and then simply type make and hit enter and it will compile the application and the directory will now contain the exe file you can run.
Conclusion:
Now go and be proud of your stubborn behavior because now you have no excuse to not be stubborn.
Remember: You can try this out yourself just install the tools as stated in the article and extract the archive I provide below to your hard disk.
## Time to take a small switch
Well I have been learning DirectX as my last entry stated. After 8 chapters of the Luna shader approach book and a few days of self experimentation I can say I am comfortable with what I learned.
Now I want to take a look back at OpenGL. Quite a few years back I started messing about with OpenGL. I think at this point it is time to revisit for a few reasons. First a foremost I am a much better developer then I was back when I first started experimenting with it. Secondly I have a much better math background. When I first started dabbling with graphics programming in 2D/3D I found a passion for math. I use to hate it. But now I have a practical reason for using it making me find it much more interesting.
So I will start to screw around with OpenGL. I don't have a OpenGL book the quality of the luna book but I understand how to set up Win32 and OpenGL Devices already. It will just more or less be lets convert one of the luna book examples to OpenGL and see which API I like better.
I will make sure I post a update with my findings.
## The Woes
A little off topic today. Turns out I really miss my Linux install. I have been using Linux for a very long time it turns out. I learned a real lot about computers in my time using Linux. I officially abandoned Linux a few months ago because all my hopes of seeing commercial quality games come to it were very far fetched. So I switched to vista just to play games. Then I discovered XNA and I kinda stuck to Vista for those very reasons. I am passionate about games and come to think of it since I started learning XNA I haven't really been playing games at all. My passion for gaming escalated when I was learning how to make them myself. I have always been a techie guy since I used my first computer in second grade. Turns out I really developed a strong passion not only for games but for computers in general. So I have 2 techie loves in my life Games and Linux sadly the two don't meld together very well. I would really love the day I can see them both intertwine.
Yea I possibly could change the future by starting my own game company and putting games out there for Linux however, It seems a lot harder then that. I know by experience the user base is there but it would take more then one small dev company to change the face of gaming. So here I am yet again at a cross roads use the OS I love to make games or use a decent OS to make games and play games. Oh the woes of choices.
--------------------
There are a few reasons why Linux is not really a target platform for game companies. One simply is the massive amount of distros out there. Another one is no common package management system for all distros. Not to mention some support earlier versions of required libs for game dev then others. These are things that need to be solved to bring mainstream gaming to Linux.
I have thought of a possible solution not only to help the developers and save them money when wanting to cross release a game but also a way to help linux appeal to the game devs.
However, would it be worth trying to implement such a solution and hope it takes hold?
Would all the time and effort be worth it just to see it flop?
Would it be worth stopping my progress in XNA or game dev all together to try and solve a long standing problem?
In all honesty I really do not know.
Maybe some of you have the answer.
## The Results are out.
I am not going to bother with posting the code I used. I am going to try to keep this as short and sweet as possible because we all know benchmarks are to be taken with a grain of salt. The main conclusion I came up with is beware of for loops they can really bog down code so use them sparingly. So here are the results which are not surprising that java is faster. So here are the results.
First results are with a for loop that reruns the code 1000 times. By resetting positional values. This shows that Java with a for loop is about 2.3x as fast as Python mathematically.
Python average after 5 runs 17.7692 Seconds
Java average after 5 runs 7.6814 Seconds
These second results are without the for loop and just running the math through one iteration showing that Java is 5.1x as fast as python without a for loop.
Python average without for loop after 5 runs 0.041
Java average without for loop after 5 runs 0.008
Now these numbers seem insignificant, however, from my experience once you add in collision detection, with the physics and the rendering these numbers can become exponential and cause some bottle necks in your game.
Based off experience and the fact I have many languages under my belt and knowledge of core programming and game programming data structures and practices I am going to be looking into either going back to C++ or Java for my game coding. Because I know somewhere in my newest game with python I will have to optimize a lot of code by moving it to C++ anyway.
## The project from insanity
Wow has it really been this long since I posted a journal entry. Man time really flies right by it is just insane. Over the last few months I have been going through the motions of designing a project. The project is rather over ambitious for sure and
99% of the worlds population would probably call me insane. Even as I was going through and laying out the design I realized how insane
I really was but it does not matter I want to work on something long term a huge almost impossible endeavor just because I can. I know I have the
capability to complete said project and at this point it is more figuring out how to approach the project effectively. So lets get into some of the
decisions I have to make to do so after I give a brief layout of what it is I want to do. First and foremost my game is a RPG but not the typical RPG. I don't want to create just another RPG or ARPG to add to the meat grinder.
I want to create a RPG that can evolve and hold longevity without costing a player 1 penny. This project is not about making money or creating a
business it is about creating a community. The key goals of this project all combine around this fact of community and having friends be able to gather together and adventure.
A modular scenario based system (the ability to mod in your own custom adventures in a easy way)
A Turn based Action system
The ability to customize the rule set
The ability to customize various actions in the game (spells, attacks, etc...)
The ability to use premade or custom assets for the scenario's
The ability to play solo or with friends
Open Source/Cross Platform (This project is very ambitious and 100% free + I love open source)
As far as technologies to use I have no clue at the moment. I ruled out Unity/UE4 simply because the do not fit the open source motto even if
they would be great to use they just do not fit the project. I also need something very flexible that will allow me to create the necessary
tools needed to create a good environment for building the custom scenario modules. Since I have a wide variety of applicable programming skills I began evaluation of some potential target technologies. Currently I am evaluating
JME3 which just so happens to be very nice to work with despite some of the quirks and lack of direction in its tooling the core engine itself is
really well done and easy to pick up on. +1 for great documentation. The only thing I really do not like here is the Netbeans based SDK as I find
it very off putting for some reason or another, however, it may be possible to work outside of the sdk and develop some custom tooling to replace
some of the features. The goal is to abstract creators away from needing to actually touch the programming language behind the game and from
having to install the whole engine + sdk to create scenarios. I have also looked at SDL/SFML way back in the past but the new versions for sure are very slick, however, I am not sure I want to go the route of
a 2D game. It for sure would work and it would solve the issue quite quickly of having to work around the JME3 SDK system. This approach could
however remove some people from wanting to help contribute to the project due to the use of C/C++ . Sure there are other bindings but they tend to be quirky and awkward to use because they rarely follow the structure the other languages are known for. Any input on other tech that I did not mention would be much appreciated just leave it in the comments and if you want you can even just
comment to call me insane. Can't think of anything else to type so see you again soon.
## The Mosin Nagant is here
As I promised the Mosin Nagant has arrived. The Mosin I have received is 1942 Izhevsk 91/30. I think it would be best to give some background before the pictures.
The Mosin Nagant was originally designed by the Russians in 1891. The approximate pronunciation of Mosin Nagant is (Moseen Nahgahn) due to the Russians emphasizing vowels over consonants. Over the years they made some modifications to the rifle the most obvious modification was the switch from a Hex to a Round receiver to produce more accuracy. My particular year is a very interesting year for the Russians. In 1942 the Russians were in some very heated and significant battles to protect their homeland from the unstoppable German war machine. One such example was Stalingrad which everyone here should even know about. This meant the Russians were in a tight bind and really needed to get more weaponry out to the soviet soldiers so often the refurb process in the arsenals was quick and half assed so to speak in order to the the rifle out on the field. In 1942 the Mosin Nagant was still a mainstay weapon for the Russians due to their lack of a efficient Assault Rifle. This meant they suffered in medium range combat as their only other weapons were really the PPSH sub machine gun and some shovels and grenades.
The Mosin Nagant was a top notch rifle and very rugged. Accuracy was a key point in designing the Model 91/30 and other models as the sport a whopping 28 3/4" barrel or larger in some early models. They were designed and sited in to use the Bayonet all the time as it was Soviet doctrine to never remove the Bayonet. Hand picked the most accurate 91/30's were retrofitted with a bent bolt and often a PU scope or some other model scope for the snipers. The 91/30 was used as the Russian sniper rifle all the way up to the cold war when they designed the Dragonov sniper rifle based off the AK-47. Even during the post war time up to and including the cold war Mosin Nagant's were still in use and manufactured but in a Carbine form known as the M44. Numerous other countries also used the Mosin as many of them were part of the Soviet Block at some point or another including Poland, Hungaria, Finnland, and Bulgaria. Many other countries outside the Block used them as well including China, and the North Vietcong. Even today there have been reports of terrorist forces in Iraq, and Afghanistan are using Mosin Nagant rifles.
As stated above the rifle was designed for accuracy. The 7.62x54R was designed as a high velocity cartridge. To give some perspective with some Russian Surplus ammunition ballistic test using 148gr LPS ammunition which is a Light Ball ammo with a steel core instead of led. The muzzle velocity (this is as the bullet leaves the barrel aka 0 yards) sits around 2800 feet per second+. The impact force under 50 yards sits around 2800 foot lbs per sq inch. With the right configuration of load this rifle and push over 3000 feet per second. For those who do not know velocity and twist ratio really decide the accuracy of the rifle from a ballistic perspective. These rifles can easily hit out to 1000 meters if needed.
Ok now more about my rifle. My rifle was manufactured in 1942 by the Izhevsk arsenal in Soviet Russia. This is a wartime rifle in a wartime stock meaning the stock was not replaced post war. The rifle has been refinished by a Soviet Arsenal even though it appears that the refinishing stamps are missing, however, this is normal they forgot this stuff all the time. The rifle is also known as all matching numbers. This means the serial numbers on all the parts match which is good. I am 99% sure the rifle was force matched which is well known for military surplus as the fonts look slightly different on the stamps. There are no lineouts on the old serial numbers they were probably totally ground off and then re-stamped. There is lots of black paint on the rifle as well which was common to hide the rushed bluing jobs and light pitting. One thing you will also notice is a amazing stock repair job done by the Russians on the front of the stock. When it was done I do not know but it really adds to the unique character and history of the rifle.
The best part of this rifle is the fact that it is one heck of a good shooter. Had her down the range and it still functions great. The trigger does take some getting use to I estimate the trigger pull is around 8 - 9 lbs possibly 10 lbs. I would estimate the rifle weighs in at about 12 - 13 lbs or so.
As promised here are some pictures. Due to there being some 18 pictures or so I will just post the link to the album and you can check out a piece of history. http://s752.photobucket.com/user/blewisjr86/media/DSC_0001_zpsfbd2b09e.jpg.html?sort=9&o=0
## The hunt for an alternative
I am on the hunt at http://partsaneprog.wordpress.com/
## The beginnings of PIC (Hello World)
Hello GDNet
First keep in mind this is a rather long post. I also have images in a entry album for you.
So my PIC Micro Controller starter kit arrive a few days ago and I started to tinker around with it. I really like this piece of hardware.
The circuit build on the development board is very clean. It contains a 6 pin power connector, a 14 pin expansion header, a potentometer (dial), push button, and 4 LED's. There is also 7 resistors and 2 capacitors on the board. By the looks of it there is 1 resistor for each LED so you don't overload them, 2 for the push button, 1 for the expansion header, 1 capacitor for the potentometer and 1 capacitor for the MCU socket. This is just by looking at the board not quite sure if this is acurate would have to review the schematic which I am not quite good at yet.
The programmer (PICkit 3) has a button designed fast wipe the micro controller with a specified hex file. It also has 3 LED's to indicate what is happening.
First before I get into HelloWorld I would like to the pain in the ass features I found with the MPLABX IDE.
First I spent hours trying to figure out why the hell the ide could not find the chip on my development board to program it. Turns out by default the IDE assumes you are using a variable range power supply to power the board so I needed to change the options in the project to power the development board through the PICkit 3 programmer.
The dreaded device ID not found error. Next the IDE could not find the device ID of my MCU wtf!!!!. 2 hours later I stumbled apon an answer. THE MPLABX IDE MUST BE RUN IN ADMINISTRATOR MODE!!!!! WTF!!!!!!! The users manual stated nothing of the sort. So to get it working I needed to start the IDE in admin mode and after it is started I need to plug the programmer into the usb port. If it is not done in that order you will get errors when trying to connect to the programmer and the chip.
Ok now onto HelloWorld WARNING ASSEMBLY CODE INCLUDED!!!!!
Here is a little quick overview of the specific chip I used for this into project I find typing this stuff out helps me remember anyway.
There are 3 types of memory on the PIC16 enhanced mid range. Program memory (Flash), Data memory, and EEPROM memory.
Program memory stores the program, data memory handles all the components, EEPROM is persistant memory.
Data memory is separated into 32 banks on the PIC16 enhanced mid range.
Banks: You deal with these the most. It contains your registers and other cool stuff.
Every bank contains the core registers, the special function registers are spread out amongst all the banks, every bank has general purpose ram for variables, and every bank has a section for shared ram which is accessible from all banks.
The HelloWorld project uses 4 instructions, and 4 directives. Instructions instruct the MCU and directives instruct the assembler.
Directives:
banksel: Tells the assembler to select a specific memory bank. This is better to use then the raw instruction because it allows you to select by register name instead of by memory bank number.
errorlevel: Used to suppress errors and warnings the assembler spits out.
org: Used to set where in program memory the following instructions will reside
Labels: used to modularize code it is not a directive per se but a useful thing to use.
end: tells the assembler to stop assembling.
Instructions:
bsf: bit sets a register (turns it on) sets value to a 1.
bcf: bit clear a register (turns it off) sets value to a 0.
clrf: initializes a registers bits to 0 so if you have 0001110 it will be come 0000000
goto: move to a labeled spot in memory not as efficient as alternative methods
Registers:
LATC: Is a data LATCH. This one is a LATCH for PORTC allows read-modify-write. We use this to write to the appropriate I/O pin for the LED. You allways write with LATCHES it is better to read from PORT
PORTC: Reads the pin values for PORTC always write to LATCHES never to PORTS
TRISC: Determins if the pin is a input(1) or an output(0)
Explanation of Project:
So generally speaking assembler is very verbose especially on the PIC16 enhanced because you need to ensure you are in the proper bank before trying to manipulate the appropriate register. So in order to light the LED we need to ensure the I/O pin for the LED we want to light is set to an output. We should then initialize the data LATCH which is an array so that all bits are 0. Then we need to turn on (high)(1) the appropriate I/O port that our LED sits on in this case it is RC0 which is wired to LED 1 on DS1.
The code to do this follows forgive the formatting assembler is very strict in that labels can only be in column 1 and include directives can only be in column 1. Everything else must be indented. Also there are some configuration settings for the MCU in the beginning of the file. I am not sure what each one does yet has I did not get a chance to read the specific details yet in the data sheet. These may mess up formatting a bit because it seems they need to be on the same line unwrapped etc... which makes it extend out very far. I will need to look into how to wrap these for readability.
Lastly the code is heavily commented to go with the above explanation.
; --Lesson 1 Hello World; LED's on the demo board are connected to I/O pins RC0 - RC3.; We must configure the I/O pin to be an output.; When the pin is driven high (RC0 = 1) the LED will turn on.; These two logic levels are derived from the PIC MCU power pins.; The PIC MCU's power pin is VDD which is connected to 5V and the; source VSS is ground 0V a 1 is equivalent to 5V and 0 is equivalent to 0V.; -----------------LATC------------------; Bit#: -7---6---5---4---3---2---1---0---; LED: ---------------|DS4|DS3|DS2|DS1|-; ---------------------------------------#include ; for PIC specific registers. This links registers to their respective addresses and banks. ; configuration flags for the PIC MCU __CONFIG _CONFIG1, (_FOSC_INTOSC & _WDTE_OFF & _PWRTE_OFF & _MCLRE_OFF & _CP_OFF & _CPD_OFF & _BOREN_ON & _CLKOUTEN_OFF & _IESO_OFF & _FCMEN_OFF); __CONFIG _CONFIG2, (_WRT_OFF & _PLLEN_OFF & _STVREN_OFF & _LVP_OFF); errorlevel -302 ; supress the 'not in bank0' warning ORG 0 ; sets the program origin for all subsequent codeStart: banksel TRISC ; select bank1 which contains TRISC bcf TRISC,0 ; make IO Pin RC0 an output banksel LATC ; select bank2 which contains LATC clrf LATC ; init the data LATCH by turning off all bits bsf LATC,0 ; turn on LED RC0 (DS1) goto \$ ; sit here forever! end
## Taking High Level Programming for Granted
Recently there have been some posts around about people considering using C over C++ for god knows what reasons they have. As per usual the forum crowd advises them to stay away from C and just learn C++. This is great in theory because over all despite the insane complexity of C++ it is a much safer language to use then C. C is a very elegant language because of it's simplicity. It is a very tiny language that is extremely cross platform (more then C++) and has a very straight forward and tiny standard library. This makes C very easy to learn but at the same time makes C very difficult to master. This is because you have to do a lot of things by hand in C because there is no standard library equivalent or particular language features that cover all bases. C++ is safer in a lot of situations because of type safety. C++ keeps track of the type of everything where C actually discards all type information at compile time.
With that little blurb aside I personally feel a lot of programmers should learn C. Not as a first language but at some point I think they should learn it. This is because it allows you to understand how the high level features of modern day high level languages work and people take these things for granted now a days. Today there are not that many programmers that actually understand what templates are doing for them and what disadvantages/advantages they have. The same goes for objects. A lot of programmers fail to understand how objects work internally. This information can make you a much better programmer over all.
Over the last few years I have spent a lot of time in C compared to C++ or other high level languages. This is not only to understand the internals of high level features I have used in the past but because I am preparing for a up coming project I am designing. This project almost has to be done in C for portability, performance, and interoperability reasons. This project not only will be targeted towards the Linux desktop but possibly embedded devices as well. So today I am going to show something that C++ gives you that C does not and how to get the same functionality in C anyway. Then I will explain why the C version is more efficient then the C++ version but at the same time not as safe because of programmer error potential. We will keep the example simple instead of making a Generic Stack we are going to make a Generic Swap function and for simplicities sake I am going to keep the two examples as close as possible.
C++ gives us a feature known as templates. Templates are a powerful meta programming feature that will actually generate code for us based of the the type of data it receives. They can do more then just this but this is a very common use. The main downfall of this particular method of creating swap is that if you pass in over 50 different types to swap during the course of the application you are actually generating over 50 different functions that have to be added by the compiler. So with that said here is a generic swap function in C++ using templates.
[source lang="cpp"]
template
void swap(T &v1, T &v2)
{
T temp;
temp = v1;
v1 = v2;
v2 = temp;
}
[/source]
There are 2 specific features to C++ we are using here. First we are using templates to generalize the type we are swapping and lastly we are using references so that we are actually swapping the variables passed in. This basically means we take in any type of variable and then we swap the address those variables hold. When this is compiled for each different version of swap we call C++ will generate a type specific function for us.
Now we need to make the equivalent of this function in C. The first thing to note is C does not have templates and C does not have the concept of references and it does not retain type information after compile time. So with some C trickery and clever assumptions based of the specs we can achieve the same result. There are other ways to do this but I am going to do it the 100% portable way this is both ANSI and POSIX compliant standards wise. Here is the code and the explanation of why I can do what I am doing will be explained afterwards.
[source lang="cpp"]
void swap(void *vp1, void *vp2, int size)
{
char *buffer = (char *)malloc(size * sizeof(char));
assert(buffer != NULL);
memcpy(buffer, vp1, size);
memcpy(vp1, vp2, size);
memcpy(vp2, buffer, size);
free(buffer);
}
[/source]
Ok so there is a lot there. First a void ptr is a generic pointer we can do this because pointers are always 4 bytes in size so the compiler does not care what is in them because we are just pointing to the storage location. Since we also don't know how big the data being pointed to is we also need to pass in the size. Now we need to find a replacement for our temp variable we used in C++. We don't know what type is stored in our void pointer we need to figure out what to store and calculate how big of a space in storage we need. We don't care what is stored we just want to hold that bit pattern. Because we know in C that a char is only 1 byte we can use that to our advantage to do the pointer arithmetic necessary to calculate the size of our storage. So we will dynamically allocate an array of char types to store our bit pattern. We will also do an assertion to make sure that we are not null before we attempt to copy data into this location. The assertion will bail if we have no space allocated. Next we need to use memcpy this will copy our bit patterns around for us. Lastly we need to make sure we free our temp storage location.
The main advantage of this is the application does not generate a new function for each different type we call through it. This uses the same assembly no matter what we pass into it. This efficiency does come with a price. If swap is not called properly we don't know what we will get back. Also because we are using void pointers the compiler will not complain it actually suppresses what compiler checking we do have. Also you must keep in mind that if the 2 types being swapped are actually different say a double and an int or int and a char* we enter the realm of undefined behavior and have no idea what will happen.
When calling swap with 2 ints you would call it as
swap(&val1, &val2, sizeof(int));
If you are swapping 2 character strings you need to call it as
swap(&val1, &val2, sizeof(char *));
With the character strings you still need to pass the address and you need to pass the size of the pointer to that address range.
This is important because a character string or char * is actually a pointer to an array of characters so you need to make sure you are pointing to the address of that
array of characters.
With all that said you can see how the C++ makes things like this very easy at a price of generating duplicate instructions. With the C you can see of a very efficient way to do the same thing with its own set of drawbacks on the caller side. It is very similar to what the C++ would do internally behind the scenes the only difference is they are passing through hidden type information so that they can generate exact type casting so you retain your type safety. This is a great and simple demonstration of what we take for granted when we use the various high level features of different programming languages. So next time you use these features stop and say thank you to the designers because without their efforts your features would not exist and you would have to do pointer arithmetic on a daily basis.
Last note if you read this and you still are thinking of using C over C++ the decision is ultimately up to you. Personally I love C it is a very elegant and clean language and I really enjoy using it, however, ask yourself if it is the right tool for the job because in C you have to reinvent the wheel constantly to achieve the functionality that newer languages give you almost for free.
## SVN=Slow
Not really journal worthy or at least typical for my Journal.
So I am sitting here right now on my Linux Dual boot getting ready to setup some stuff for the OpenGL SuperBible 5th Edition that I am working my way through. I wanted to get everything set up on my Linux development side of things because unless I am playing Eve or doing school work I am using Linux anyway.
My first step is to pull down the code and I am ready to fall asleep. Basically all of the latest code for the book with bug fixes is located in a google code subversion repository. As of late I am really hammering home on git because in reality it is a really nice VCS once you get use to it. So I decided to pull down the code with git-svn. *SLEEP* So this has been pulling the code down now for the last 10 min or so. Still not done. Before you say anything it is not git-svn. Beleive it or not git-svn is going a lot faster then my first attempt at pulling it down with the svn client.
This is just rediculous and all this is just so I can compile GLTools :S
## Still Alive
Hello everyone just popping in to say I am still alive. Been pretty busy the last few with prepping for final exams and such. Now that exams are out of the way I can get back to relaxing and working on school work plus learn some more about Objective-C before my next final exam in 8 weeks. One thing about going to school Online with a compressed program structure is you really hammer through content fast in a class. I mean reading 2 - 3 chapters a week, plus discussions and assignments and a quiz here and there. I must say if anything can prepare you for tight deadlines it is a compressed school course. For instance it is now Week 1 of the spring session I read 2 chapters got my 2 discussions to take care of and a 2 page paper that is due at the end of next week + next weeks work on top of that paper. Can get kind of rough and really taxes the organizational skills.
On another note I ended up getting another Cocoa book to get a second perspective on Objective-C + Cocoa and I must say I like this book a lot better then the other one. The other one was good but this one ups it in every way shape and form. This book is what people call the Hillegass book also known as Cocoa Programming for Mac OS X done by the world famous Cocoa teacher. I must say it really shows his explanations and style Put Cocoa and Objective-C Up and Running to shame. My favorite feature of the book is at the end of each chapter the author presents challenges for the reader to go out and write code on your own which is important when learning something new. It is not a book that is all about copy down the example and see what happens he actually gives challenge assignments that make use of the concepts you have learned thus far in the book. Very good stuff. As I move through this book everything is starting to make sense. Cocoa + Objective-C is a whole new perspective on development that goes against the particular trend of modern languages today and right now as things click I can see why so many developers are in love with the system and why Mac OSX applications are so robust and solid compared to the applications on other operating systems. Not to say windows and Linux don't have good software they do but Mac seems to have more of it and part of the reason for that is Cocoa.
I will definitely make sure I keep you updated as things mould together and I really can't wait to start my own first Cocoa Application. I am hoping to actively talk about my project I will be starting as well in this journal. Even tho the application is not a game I still find it important to talk about it because a lot of the issues and concepts I will be dealing with from both a design and programming standpoint are great for everyone to learn from. This is a developer journal after all and I don't see anything stating it has to be game related in nature so all gloves are off and heck it beats having to go out and find another place for a blog and cross link like I tried previously. The new journal system is so much nicer then the old one.
## Some Updates on whats going on
Hey Everyone,
Not many people read my journal as much as they did in the past when I was heavily into game dev but it really does not matter. For one I very seldom even post anymore. The reasoning behind this is I have greatly drifted from game development and focus more on embedded stuff.
Right now I have been very busy actually finishing up my degree WOOOOO!!!!!. The stress is building up as the work load ever increases but I know all this hard work I have been doing will pay off. For those who did not know I am getting a BS in Information Systems Security and so far while working 48 hrs/week at my dead end job, working on hobby electronics, enjoying my firearm shooting hobby, and school I have been able to stay on the dean's list. *pat pat* I am really getting excited as this is a huge step for me.
I really do not blog much like I have stated. I have tried to do my own external blog but I always seem to not have the time hopefully one day I can get one going regularly again as I really do like writing.
In the name of my firearm hobby I am adding a new weapon to my collection. Currently I have a Springfield 1911 Range Officer edition .45 cal. I will be adding within the next few days (Can't wait for it to arrive longest 7 days of my life) a WW2 Russian Mosin Nagant bolt action rifle. This rifle shoots a 7.62x54R cartridge. Very powerful round which can easily punch right through cinder block. The bullet itself is a .30 cal right with the 308 and 30-06. The R means it is a rimmed cartridge the 54 basically dictates the cartridge size if I remember correctly. One of these rounds packs more of a wallop then a AK-47 round which is a 7.62x39. The 7.62x54R was designed as a long range round optimized for velocity which increases the accuracy and distance. This round is very accurate from 300 - 500 meters and can easily hit a human body sized target out towards 1000 meters. The Mosin was not just a infantry rifle for the longest time it was also the Russian sniper rifle of choice until the Dragonov was developed. So excited can't wait. I will be sure to post some pics when I get it.
As for hobby electronics I will be posting some more info on this project here as well hopefully. I am currently building what I call an audio trigger system. Essentially the micro controller waits for an audio pulse and uses this pulse to trigger an action. In my case the first project using this small subproject will be a audio triggered stopwatch. Then after this the trigger subsystem will also move to an audio visualizer project.
This project really stretches my electronics knowledge as there were some interesting hiccups I have had to design around. The code is simple the circuits are the hard part for a guy like me. Because of this project I am learning and actually understanding what is going on. The design of this project needs some preparation for a post so it may be a little while and quite long. Hope to get that together soonish.
Feels good to write again cya guys around.
## Some Orbis Work
Well I did some work on orbis today. Not much a bit short on time. I got the initial backdrop of the title screen done. To do this I used the dreaded GameComponent feature of XNA. GameComponent derived objects can be quite handy and abstract the code nicely for reuse if they are coded properly. I however can see these GameComponent objects being quite a sloppy mess and hard to follow in larger projects. In my book I am learning XNA from they overuse game components I think. They litterally make everything a GameComponent. Not bad in principle but as I said as the projects get large I think they can be a burden. But anyway onto the code..... (note main game code missing just the components are shown.)
Scene.cs (this is the main scene component all scenes are derived from this)
using System;
using System.Collections.Generic;
using Microsoft.Xna.Framework;
using Microsoft.Xna.Framework.Audio;
using Microsoft.Xna.Framework.GamerServices;
using Microsoft.Xna.Framework.Graphics;
using Microsoft.Xna.Framework.Input;
using Microsoft.Xna.Framework.Storage;
using Microsoft.Xna.Framework.Content;
namespace Orbis
{
///
/// This is a game component that implements IUpdateable.
///
public class Scene : Microsoft.Xna.Framework.DrawableGameComponent
{
// components belonging to the scene
public Scene(Game game)
: base(game)
{
components = new List(); // initialize the component list
// set the state of the scene
Visible = false;
Enabled = false;
}
///
/// Allows the game component to perform any initialization it needs to before starting
/// to run. This is where it can query for any required services and load content.
///
public override void Initialize()
{
base.Initialize();
}
///
/// Allows the game component to update itself.
///
/// Provides a snapshot of timing values.
public override void Update(GameTime gameTime)
{
// makes sure the child components of this scene are updated when they are enabled(active)
for(int i = 0; i {
if(components.Enabled)
{
components.Update(gameTime);
}
}
base.Update(gameTime);
}
public override void Draw(GameTime gameTime)
{
// draw the drawable child components of this scene if they are visible
for (int i = 0; i {
GameComponent component = components;
if ((component is DrawableGameComponent) && ((DrawableGameComponent)component).Visible)
{
((DrawableGameComponent)component).Draw(gameTime);
}
}
base.Draw(gameTime);
}
///
/// returns the values in the components list belonging to this scene
///
public List Components
{
get { return components; }
}
///
/// Set the state of the scene to display it
///
public virtual void Show()
{
// set state of the scene
Visible = true;
Enabled = false;
}
///
/// Set the state of the scene to not display it
///
public virtual void Hide()
{
// set state of the scene
Visible = false;
Enabled = false;
}
}
}
StartupScene.cs (this is the main title scene of the game not fully complete)
using System;
using System.Collections.Generic;
using Microsoft.Xna.Framework;
using Microsoft.Xna.Framework.Audio;
using Microsoft.Xna.Framework.GamerServices;
using Microsoft.Xna.Framework.Graphics;
using Microsoft.Xna.Framework.Input;
using Microsoft.Xna.Framework.Storage;
using Microsoft.Xna.Framework.Content;
namespace Orbis
{
///
/// This is a game component that implements IUpdateable.
///
public class StartupScene : Scene
{
// sprite batch for drawing
private SpriteBatch spriteBatch = null;
// background texture
private Texture2D texture;
// background texture rect
private Rectangle backRect;
public StartupScene(Game game, Texture2D texture)
: base(game)
{
// initialize spriteBatch from a SpriteBatch in the game services
spriteBatch = (SpriteBatch)Game.Services.GetService(typeof(SpriteBatch));
// initialize background texture
this.texture = texture;
// initialize backRect
backRect = new Rectangle(0, 0, Game.Window.ClientBounds.Width, Game.Window.ClientBounds.Height);
}
///
/// Allows the game component to perform any initialization it needs to before starting
/// to run. This is where it can query for any required services and load content.
///
public override void Initialize()
{
base.Initialize();
}
///
/// Allows the game component to update itself.
///
/// Provides a snapshot of timing values.
public override void Update(GameTime gameTime)
{
base.Update(gameTime);
}
public override void Draw(GameTime gameTime)
{
// draw the background
spriteBatch.Begin();
spriteBatch.Draw(texture, backRect, Color.White);
spriteBatch.End();
base.Draw(gameTime);
}
}
}
now for the great and all mighty picture (note thumbnail click for full)
## Some More Progress.
As I said in my last entry it has been a long time since I touched C#. So as I was doing the port of the D3DApp framework for the DirectX 10 Luna Book, I realized quite quickly that I was trying to write C++ in C#. Not going to work. So I took a dive into the SlimDX sample MiniTri to get a general quick refresher on how a C# programmer does things. This sample + the MSDN helped a lot on giving me a nice refresher.
The code is not fully finished yet. I still need to get the font display coded and the GameTimer coded, but none the less I got a fully functional Direct3D 10 window running. The window is cleared to blue and I tacked in the functionality to turn on and off VSync. I hate when the video card transistors scream at high frame rates.
There are two things I learned from sitting down to do this. One it is really a excellent way to reinforce the learning curve of the massive DirectX API. You can't just sit there and copy the code or look at the code and port it. The best way I found to go about things is the read the chapter then implement it on my own in a C# way. The advantage is you actually learn the API not how to copy code and you can get a nice clean framework to use compared to what the author provides.
The second thing I learned is probably the biggest one. Why was a torturing my self with C++ all these years. C# is a very clean and powerful language and just looking at the code you can tell what it does. I will say the code I put together is not perfect yet or finalized in anyway but it is indeed a lot cleaner then the authors C++ code. The main reason I can see that this is so would be the removal of Macros and Preprocessor directives. Not to mention the exclusion of header files.
Now again keep in mind this is not finalized yet still needs some tweaking. Plus some more implementation but here is what I got so far.
using System;
using System.Drawing;
using System.Windows.Forms;
using SlimDX;
using D3D10 = SlimDX.Direct3D10;
using DXGI = SlimDX.DXGI;
using SlimDX.Windows;
namespace InitDirect3D
{
class D3DApp
{
private RenderForm m_Window;
private D3D10.Device m_D3DDevice;
private DXGI.SwapChain m_SwapChain;
private D3D10.RenderTargetView m_RenderTargetView;
private D3D10.DepthStencilView m_DepthStencilView;
private D3D10.Texture2D m_DepthStencilBuffer;
private int m_VSync;
public D3DApp()
{
m_Window = null;
m_D3DDevice = null;
m_SwapChain = null;
m_RenderTargetView = null;
m_DepthStencilView = null;
m_DepthStencilBuffer = null;
m_VSync = 0;
}
~D3DApp()
{
m_RenderTargetView.Dispose();
m_DepthStencilView.Dispose();
m_DepthStencilBuffer.Dispose();
m_D3DDevice.Dispose();
m_SwapChain.Dispose();
}
public void Initialize(string windowCaption)
{
m_Window = new RenderForm(windowCaption);
// Setup the SwapChain and Device
DXGI.SwapChainDescription swapDesc = new DXGI.SwapChainDescription()
{
BufferCount = 1,
ModeDescription = new DXGI.ModeDescription(m_Window.ClientSize.Width, m_Window.ClientSize.Height,
new Rational(60, 1), DXGI.Format.R8G8B8A8_UNorm),
IsWindowed = true,
OutputHandle = m_Window.Handle,
SampleDescription = new DXGI.SampleDescription(1, 0),
Usage = DXGI.Usage.RenderTargetOutput,
Flags = DXGI.SwapChainFlags.None,
};
D3D10.Device.CreateWithSwapChain(null, D3D10.DriverType.Hardware, D3D10.DeviceCreationFlags.Debug,
swapDesc, out m_D3DDevice, out m_SwapChain);
this.OnResize();
// Disable Alt+Enter FullScreen
DXGI.Factory factory = m_SwapChain.GetParent();
factory.SetWindowAssociation(m_Window.Handle, DXGI.WindowAssociationFlags.IgnoreAll);
m_Window.Resize += new EventHandler(m_Window_Resize);
m_Window.KeyDown += new KeyEventHandler(m_Window_KeyDown);
}
void m_Window_KeyDown(object sender, KeyEventArgs e)
{
if (e.KeyCode == Keys.V)
{
if (m_VSync == 0)
{
m_VSync = 1;
}
else
{
m_VSync = 0;
}
}
}
void m_Window_Resize(object sender, EventArgs e)
{
this.OnResize();
}
public void Run()
{
MessagePump.Run(m_Window, () =>
{
DrawScene();
m_SwapChain.Present(m_VSync, DXGI.PresentFlags.None);
});
}
private void OnResize()
{
if (m_RenderTargetView != null)
{
m_RenderTargetView.Dispose();
}
if (m_DepthStencilView != null)
{
m_DepthStencilView.Dispose();
}
if (m_DepthStencilBuffer != null)
{
m_DepthStencilBuffer.Dispose();
}
// Setup the RenderTargetView and DepthStencilBuffer/View
D3D10.Texture2D backbuffer = D3D10.Texture2D.FromSwapChain3D10.Texture2D>(m_SwapChain, 0);
m_RenderTargetView = new D3D10.RenderTargetView(m_D3DDevice, backbuffer);
backbuffer.Dispose();
D3D10.Texture2DDescription depthStencilDesc = new D3D10.Texture2DDescription()
{
Width = m_Window.ClientSize.Width,
Height = m_Window.ClientSize.Height,
MipLevels = 1,
ArraySize = 1,
Format = DXGI.Format.D24_UNorm_S8_UInt,
SampleDescription = new DXGI.SampleDescription(1, 0),
Usage = D3D10.ResourceUsage.Default,
BindFlags = D3D10.BindFlags.DepthStencil,
CpuAccessFlags = D3D10.CpuAccessFlags.None,
OptionFlags = D3D10.ResourceOptionFlags.None
};
m_DepthStencilBuffer = new D3D10.Texture2D(m_D3DDevice, depthStencilDesc);
m_DepthStencilView = new D3D10.DepthStencilView(m_D3DDevice, m_DepthStencilBuffer);
// Bind views to pipeline
m_D3DDevice.OutputMerger.SetTargets(m_DepthStencilView, m_RenderTargetView);
// Set the viewport transform.
D3D10.Viewport vp = new D3D10.Viewport(0, 0, m_Window.ClientSize.Width, m_Window.ClientSize.Height, 0.0f, 1.0f);
m_D3DDevice.Rasterizer.SetViewports(vp);
}
{
}
public virtual void DrawScene()
{
m_D3DDevice.ClearRenderTargetView(m_RenderTargetView, new Color4(1.0f, 0.0f, 0.0f, 1.0f));
m_D3DDevice.ClearDepthStencilView(m_DepthStencilView, D3D10.DepthStencilClearFlags.Depth | D3D10.DepthStencilClearFlags.Stencil, 1.0f, 0);
}
}
}
## Some Minor Progress
I recently just picked up the newest luna book on DirectX 10. Just when I thought DX9 was very clean DX10 takes it to a much higher level. The logical flow of the API is much better.
It is going to take me some time to get into this however. I have made the choice to step back from C++ somewhat. I love C++ don't get me wrong but I can't help but be intrigued by SlimDX. If I did not at least check it out I would not be content with myself. I am always enthusiastic about new ways to do things. For this I must shift to C#.
I have not used C# in quite a few years so it is going to be a decent shift in mindset. Going from a native way of thinking to a managed way of thinking so far has been awkward. I am going to be porting the code to C# as I go so I will make sure to post up lots of code as I move along.
This will not be a port purse because I don't think a direct port is going to be pretty. After all the code should look and feel like C#. So it will be more of a redesign. Sharp Develop is up and running SlimDX installed. Luna book is in hand lets see what I can do. | 16,381 | 72,932 | {"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} | 2.875 | 3 | CC-MAIN-2019-43 | latest | en | 0.964982 |
http://openstudy.com/updates/56671762e4b08c245addae08 | 1,516,603,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00057.warc.gz | 258,745,927 | 8,431 | John Anderson bought a home with a 10.5% adjustable rate mortgage for 30 years. He paid $9.99 monthly per thousand on his original loan. At the end of 5 years, he owes the bank$55,000. Now that interest rates have gone up to 12.5%, the bank will renew the mortgage at this rate or John can pay $55,000. John decides to renew and will now pay$10.68 monthly per thousand on his loan. (You can ignore the small amount of principal that has been paid.) What is the amount of the old monthly payment? What is the amount of the new monthly payment? | 136 | 542 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-05 | latest | en | 0.971107 |
https://brainly.my/tugasan/140949 | 1,484,779,962,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00141-ip-10-171-10-70.ec2.internal.warc.gz | 802,289,027 | 10,096 | # Andrea bought 12 bagels and 10 muffins at the bakery of these items 2/3of the bagels were Multigrain and3/5of the muffins were bran muffins how many Multigrain bagels did Andrea buy
1
dari ey1cyru5lnurullaSy
## Jawapan
2016-03-17T17:45:07+08:00
Two-third of the bagels were multigrain bagels. to find the number of multigrain bagels, you have to multiply the fraction of multigrain bagels by the total number of bagels brought.
2/3×12=8 | 141 | 442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-04 | latest | en | 0.881787 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-8-polynomials-and-factoring-8-7-factoring-special-cases-practice-and-problem-solving-exercises-page-514/16 | 1,537,946,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163704.93/warc/CC-MAIN-20180926061824-20180926082224-00303.warc.gz | 764,952,725 | 14,075 | ## Algebra 1
$(2r+9)^{2}$
Given the polynomial $(2r)^{2}$ + 36r + $(9)^{2}$ We see that the polynomial has the first and last term squared and the middle term is +2 times the first and last term. Thus it follows the rule of $a^{2}$+2ab+$b^{2}$=$(a+b)^{2}$ In this polynomial a= 2r and b=9 $(2r)^{2}$ + 2(2r)(9) + $(9)^{2}$ = $(2r+9)^{2}$ | 134 | 338 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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-2018-39 | longest | en | 0.783199 |
https://www.moveitmaththesource.com/equals/mathbalancesolve.html | 1,600,473,278,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00697.warc.gz | 1,002,989,104 | 10,695 | The main cause of confusion about the equal sign in grade school math is the way it is presented in every textbook-based elementary school math curriculum in America today. Overwhelmingly, it is used to separate a problem from its answer with the answer appearing to the right of the problem. This does not mean that we should do away with math textbooks and math worksheets. What is needed is teacher or parental intervention to counter the incorrect mindset that is currently being promoted, however inadvertently. The following are strategies for doing so.
Model “equals” as “balanced” on a teeter-totter and/or a math balance.
Supplement written work with problems that ask for answers to be placed on either side of the equal sign, like 9 + 3 = ___, ___ = 5 + 8, ___ = 12 – 9, 17 – 9 = ___, 8 x 6 = ___, ___ = 4 x 7, ___ = 42 ÷ 6, 56 ÷ 8 = ___.
Explain why 1/2=2/4, 3/5=6/10, 5/8=10/16 and the like are “true.” (How to do this easily and effectively is explained in the fractions key.)
Balancing activities like playing on a teeter-totter, walking along a narrow board, or carrying books on one’s head develop a literal feel for “equals” as “balanced.” If no teeter-totter is available,
A is a desktop teeter-totter. Its main feature is a beam that is suspended in the middle that displays numbered pegs on either side of its balance point. By hanging weights on the pegs, children can explore and discover huge amounts of mathematics years in advance of its usual place in the curriculum, like algebra in kindergarten to squelch rumors spread by older kids that it is hard. When the beam is perfectly horizontal, whatever is shown on one side “equals” or “is the same as” that on the other side.
A major educational plus for the math balance is that it is self-correcting, thus giving students ownership of the math they discover. When teachers have all the answers, it shows how smart they are about somebody else’s math. When students work out the answers on their own, it shows how smart they are about their math.
Watch the videos below to see how to use a math balance to teach “equals” (as meaning “balanced” or “is the same as”), less than, greater than, addition, subtraction, multiplication, division, time, money, measurement, algebra, and even arithmetic word problems. The math balance in the videos—an EquaBeam™—can be purchased from this website. can be made with a pencil, paper cup, and card stock using bent paperclips for weights. | 585 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-40 | longest | en | 0.951164 |
https://www.geeksforgeeks.org/perpendicular-distance-between-a-point-and-a-line-in-2-d/?ref=rp | 1,590,619,735,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00256.warc.gz | 733,315,532 | 29,955 | # Perpendicular distance between a point and a Line in 2 D
Given a point (x1, y1) and a line (ax + by + c = 0). The task is to find the perpendicular distance between the given point and the line.
Examples :
Input: x1 = 5, y1 = 6, a = -2, b = 3, c = 4
Output:3.32820117735
Input: x1 = -1, y1 = 3, a = 4, b = -3, c = – 5
Output:3.6
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: The distance (i.e shortest distance) from a given point to a line is the perpendicular distance from that point to the given line. The equation of a line in the plane is given by the equation ax + by + c = 0, where a, b and c are real constants. the co-ordinate of the point is (x1, y1)
The formula for distance between a point and a line in 2-D is given by:
`Distance = (| a*x1 + b*y1 + c |) / (sqrt( a*a + b*b))`
Below is the implementation of the above formulae:
Program 1:
## C
`// C program to find the distance ` `// between a given point and a ` `// given line in 2 D. ` `#include ` `#include ` ` ` `// Function to find distance ` `void` `shortest_distance(``float` `x1, ``float` `y1, ` ` ``float` `a, ``float` `b, ` ` ``float` `c) ` `{ ` ` ``float` `d = ``fabs``((a * x1 + b * y1 + c)) / ` ` ``(``sqrt``(a * a + b * b)); ` ` ``printf``(``"Perpendicular distance is %f\n"``, d); ` ` ``return``; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``float` `x1 = 5; ` ` ``float` `y1 = 6; ` ` ``float` `a = -2; ` ` ``float` `b = 3; ` ` ``float` `c = 4; ` ` ``shortest_distance(x1, y1, a, b, c); ` ` ``return` `0; ` `} ` ` ` `// This code is contributed ` `// by Amber_Saxena. `
## Java
`// Java program to find ` `// the distance between ` `// a given point and a ` `// given line in 2 D. ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Function to find distance ` ` ``static` `void` `shortest_distance(``float` `x1, ``float` `y1, ` ` ``float` `a, ``float` `b, ` ` ``float` `c) ` ` ``{ ` ` ``double` `d = Math.abs(((a * x1 + b * y1 + c)) / ` ` ``(Math.sqrt(a * a + b * b))); ` ` ``System.out.println(``"Perpendicular "` `+ ` ` ``"distance is "` `+ d); ` ` ``return``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main (String[] args) ` ` ``{ ` ` ``float` `x1 = ``5``; ` ` ``float` `y1 = ``6``; ` ` ``float` `a = -``2``; ` ` ``float` `b = ``3``; ` ` ``float` `c = ``4``; ` ` ``shortest_distance(x1, y1, a, b, c); ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by Mahadev. `
## Python
`# Python program to find the distance between ` `# a given point and a given line in 2 D. ` ` ` `import` `math ` ` ` `# Function to find distance ` `def` `shortest_distance(x1, y1, a, b, c): ` ` ` ` ``d ``=` `abs``((a ``*` `x1 ``+` `b ``*` `y1 ``+` `c)) ``/` `(math.sqrt(a ``*` `a ``+` `b ``*` `b)) ` ` ``print``(``"Perpendicular distance is"``),d ` ` ` ` ` `# Driver Code ` `x1 ``=` `5` `y1 ``=` `6` `a ``=` `-``2` `b ``=` `3` `c ``=` `4` `shortest_distance(x1, y1, a, b, c) `
## C#
`// C# program to find ` `// the distance between ` `// a given point and a ` `// given line in 2 D. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ``// Function to find distance ` ` ``static` `void` `shortest_distance(``float` `x1, ``float` `y1, ` ` ``float` `a, ``float` `b, ` ` ``float` `c) ` ` ``{ ` ` ``double` `d = Math.Abs(((a * x1 + b * y1 + c)) / ` ` ``(Math.Sqrt(a * a + b * b))); ` ` ``Console.WriteLine(``"Perpendicular "` `+ ` ` ``"distance is "` `+ d); ` ` ``return``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main () ` ` ``{ ` ` ``float` `x1 = 5; ` ` ``float` `y1 = 6; ` ` ``float` `a = -2; ` ` ``float` `b = 3; ` ` ``float` `c = 4; ` ` ``shortest_distance(x1, y1, a, b, c); ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by inder_verma.. `
## PHP
` `
Output:
```Perpendicular distance is 3.32820117735
```
Program 2:
## C
`// C program to find the distance ` `// between a given point and a ` `// given line in 2 D. ` `#include ` `#include ` ` ` `// Function to find distance ` `void` `shortest_distance(``float` `x1, ``float` `y1, ` ` ``float` `a, ``float` `b, ` ` ``float` `c) ` `{ ` ` ``float` `d = ``fabs``((a * x1 + b * y1 + c)) / ` ` ``(``sqrt``(a * a + b * b)); ` ` ``printf``(``"Perpendicular distance is %f\n"``, d); ` ` ``return``; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``float` `x1 = -1; ` ` ``float` `y1 = 3; ` ` ``float` `a = 4; ` ` ``float` `b = -3; ` ` ``float` `c = - 5; ` ` ``shortest_distance(x1, y1, a, b, c); ` ` ``return` `0; ` `} ` ` ` `// This code is contributed ` `// by Amber_Saxena. `
## Java
`// Java program to find the distance ` `// between a given point and a ` `// given line in 2 D. ` `class` `GFG ` `{ ` `// Function to find distance ` `static` `void` `shortest_distance(``double` `x1, ``double` `y1, ` ` ``double` `a, ``double` `b, ` ` ``double` `c) ` `{ ` ` ``double` `d = Math.abs((a * x1 + b * y1 + c)) / ` ` ``(Math.sqrt(a * a + b * b)); ` ` ``System.out.println(``"Perpendicular distance is "` `+ d); ` ` ``return``; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``double` `x1 = -``1``; ` ` ``double` `y1 = ``3``; ` ` ``double` `a = ``4``; ` ` ``double` `b = -``3``; ` ` ``double` `c = - ``5``; ` ` ``shortest_distance(x1, y1, a, b, c); ` `} ` `} ` ` ` `// This code is contributed ` `// by mits `
## Python
`# Python program to find the distance between ` `# a given point and a given line in 2 D. ` ` ` `import` `math ` ` ` `# Function to find distance ` `def` `shortest_distance(x1, y1, a, b, c): ` ` ` ` ``d ``=` `abs``((a ``*` `x1 ``+` `b ``*` `y1 ``+` `c)) ``/` `(math.sqrt(a ``*` `a ``+` `b ``*` `b)) ` ` ``print``(``"Perpendicular distance is"``),d ` ` ` ` ` `# Driver Code ` `x1 ``=` `-``1` `y1 ``=` `3` `a ``=` `4` `b ``=` `-``3` `c ``=` `-` `5` `shortest_distance(x1, y1, a, b, c) `
## C#
`// C# program to find the distance ` `// between a given point and a ` `// given line in 2 D. ` `using` `System; ` ` ` `class` `GFG ` `{ ` `// Function to find distance ` `static` `void` `shortest_distance(``double` `x1, ``double` `y1, ` ` ``double` `a, ``double` `b, ` ` ``double` `c) ` `{ ` ` ``double` `d = Math.Abs((a * x1 + b * y1 + c)) / ` ` ``(Math.Sqrt(a * a + b * b)); ` ` ``Console.WriteLine(``"Perpendicular distance is "` `+ d); ` ` ``return``; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main() ` `{ ` ` ``double` `x1 = -1; ` ` ``double` `y1 = 3; ` ` ``double` `a = 4; ` ` ``double` `b = -3; ` ` ``double` `c = - 5; ` ` ``shortest_distance(x1, y1, a, b, c); ` `} ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai `
## PHP
` `
Output:
```Perpendicular distance is 3.6
```
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Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. | 3,000 | 8,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-24 | longest | en | 0.696107 |
http://zerohome.ml/top4698-formula-of-ratio-of-stock-price-to-book-value-per-share.html | 1,527,065,923,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00137.warc.gz | 513,258,183 | 7,838 | formula of ratio of stock price to book value per share
# formula of ratio of stock price to book value per share
Valuation Ratios. Earnings per share Cash earnings per share Dividend per share Book value per share Price/Earning.It expresses the relationship between the cost of revenue from operations and average inventory. The formula for its calculation is as follows The market-to-book ratio is calculated by dividing the current closing price of the stock by the most current quarters book value per share.The Market to Book formula is: Market Capitalization / Net Book Value. or. Market Value Ratios relate an observable market value, the stock price, to book values obtained from the firms financial statements.(Earnings per share are calculated by dividing net income by the number of shares outstanding.) Price to Earnings Ratio Price (High, Low, or Close)/Earnings per Share 12 Months Moving.For example, compound or least-squares growth rates of various index data such as earnings per share, price, dividends per share, annualized dividend rate, and book value per share can be calculated Earnings per Share, Price Earnings Ratios, Book Value per Share, andFor a corporation with only common stock, book value per share is easy to calculate: total stockholders equity divided by common shares outstanding at the end of the accounting period. The price-to-book ratio, or P/B ratio, is a financial ratio used to compare a companys current market price to its book value. It is also sometimes known as a Market-to-Book ratio. The calculation can be performed in two ways, but the result should be the same each way. Book Value Book Value Per Share And PB Ratio Explained In Hindi - Duration: 10:04.Warren Buffett Value Formula separates weak industries from strong ones. - Duration: 3:49.
Price/Book Ratio: Use it to Find Outperforming Stocks - Duration: 12:13. Earnings Per Share is adjusted for stock splits and stock dividends. If data is not available for 12 months or more, "N/A" is displayed for EPS.Formula: Current Market Price / Book Value Per Share A valuation ratio calculated by dividing the current market price by the most recent quarters (mrq) What is the price earnings ratio? What is premium on common stock?Related Business Forms. Book Value per Share of Common Stock. Balance Sheet: Retail/Wholesale - Corporation. The formulas and examples for calculating book value per share with and without preferred stock are given below: (1). If company has issued only common stock and no preferred stock Calculate the Book value per Share of the international corporation. Given, Stock holders equity 2000000 Preferred Stock 500000 Total outstanding shares 300000.
Cash Certificates Issue Prices. Price meaning the stock price and the book value specifies the value obtained when the liabilities are taken away from the tangible assets.P/B ratio is price to book value ratio, the formula for which is: Market price per share/ Book Value per share. If the market value of equity refers to the market value of equity of common stock outstanding, the book value of common equity should be used in the denominator.Price Book Value Ratio for a Stable Growth Firm: Example. price/book value ratio — Compares a stock s market value to the value of total assets less total liabilities ( liability) ( book value). Determined by dividing current stock price by common stockholder equity per share> ( book value), adjusted for stock splits. Book value per share: About Price to Book Ratio Calculator.The price to book ratio (P/B ratio) is a financial ratio used to compare a companys book value to its current market price. Book Value Per Share (BVPS) ( Total Equity Preferred Stock) /. Shares Outstanding. Lets break each variable a little bit to give us a better idea of what they are so weAnother benefit of calculating this formula is that it helps calculate another ratio which is the price to book value, or the P/B. Ratio: Book Value Per Share. Formula: Common Stockholders Equity. Common Shares Outstanding.Current share price. Trailing 12 months earnings per share Value/Use: A method of assessing or estimating the value embodied in shares of a particular stock. Key Highlights of Price to book value ratio of Automobile Companies.source: ycharts. Why Price to Book Value ratio can be used to value Banking Stocks.EPS/Book value per share formula is ROE (remember, ROE Net Income / Shareholders Equity or Book Value). Thus, causing the stock price to increase quarter over quarter. The book value of the company hasnt changed though. The business still has no assets.The price-to-book ratio formula is calculated by dividing the market price per share by book value per share. 1. Market value of equity MV Market price per share P X Number of issued Ordinary share (Common Stock). P current stock price in the market and S number of outstanding share.The formula is given by Calculations (formula). Price/Book Value Ratio Stock Price Per Share / Shareholders Equity Per Share.The book value considers original purchase price of an item not the current market price which leads to measurement inaccuracies. 1. Difference Between Market Value and Intrinsic Value. 2. How to Calculate the Value of Stock With the Price-to-Earnings Ratio.Book Value Per Share Definition. The book value of a company represents how much a company is worth based strictly on its balance sheet. Formula. Price to Book Value Share Price / Book Value Per Share. YCharts uses Total Shareholders Equity and the most recent quarters common shares outstanding to calculate Book Value Per Share. The price-to-book ratio is another ratio used in investing, mostly by value investors. It is one of those indicators they use to determine the value of a stock and how much more of it they can benefit from.It is easy to get the share price, but what does book value of equity mean? This is a thorough guide on how to calculate Price to Book Value Ratio (P/B) with detailed interpretation, analysis, and example.The formula to measure the Price to Book value is as follows: Price to Book (P/B) Stock Price Per Share / Book Value Per Share. Price Earning Ratio or P/E Ratio (Earnings Yield Ratio). Market value to Book Value Ratio.The formula to find out dividend payout ratio is as follows. Dividend Payout Ratio Dividend per Equity Share / Earnings Per Share. Formula: Book value per Share (Stock holders equity - Preferred Stock) / Total outstanding shares.Price To Sales Ratio. Formula The Price to Book ratio or P/B ratio is a multiple that compares the current market price of a company to its book value (shareholders equity) .The ratioThe Formula for the P/Book ratio is fairly straight forward and can be calculated as follows: Price/Book Ratio Stock Price Per Share / Shareholders The book value of a companys stock is simply the stockholders equity per common share of stock, equal to the net asset value, equal to total assets minus intangibleFor an individual company, the Q ratio is equal to the market price of the firm divided by its replacement cost. Tobins Q Ratio Formula. Price/Book Ratio. Ratio comparing market value of stock with total value of assets minus liabilities. Assets and liabilities are taken at book value. Formula: current stock price dividend by common shareholding equity per share. Stock splits are adjusted against this equity value. Investors who had an eye on the Price to Book Value ratio found that even if the company wound up its operations at its book value, they would still be left with more book value per share than the then prevailing market price per share. The stock price per share can be found as the amount listed as such through the secondary stock market.Issues with the Price to Book Value Formula. One may argue that a ratio under one implies that the company is perceived as being a worse investment than if it were above one. Price to Book Ratio . Current Share Price. Book Value per Share.Book Value per Share . Total Shareholders Equity. Total Number of Shares Outstanding. P/B ratio can also be calculated using the following formula Book value per share is used in relative valuation of companies as part of price to book value ratio inFormula. Book value per share is determined by dividing common shareholders equity by total number ofTotal Outstanding Shares Total Number of Shares Issued Shares as Treasury Stock. Book Value Per Share (total shareholders equity) / (shares outstanding).Is a price to earning ratio the price vs. the stocks earnings or the companies sales earnings? Formula: PEG Price to Earnings Ratio / Annual EPS Growth. And for a fairly priced company, P/E G. In other words, this implies that a fairly priced company will have a PEG ratio of 1.It is calculated by dividing the current price of the stock by the latest quarters book value per share. Arithmetically, it is the ratio of market to book value.First formula needs per share information whereas the second one needs the total values of the elements. Example. Assume there is a company X whose publicly traded stock price is 20 and it has 100,000 outstanding equity shares. Shares 128. Price To Book Ratio, often simply referred to as P/B Ratio, can be used to make a comparison between the current market price of a stock and the total book value of all the assets that company has on the balance sheet.
The lesser price to book value ratio normally less than 1 may indicate that the company undervalued its stocks. It is also known as Price Equity Ratio. Formula P/B (Price to Book) Value Ratio Stock Price per Share/Book Value of Equity. Book Value per Share is an easy formula to calculate, and it can tell us whether our bank is valued correctly compared to the price in the market.Book Value Per Share (BVPS) ( Total Equity Preferred Stock) /.A great way to find undervalued companies is to look at the price to book ratio Relating book value per share to market priceWhat a low P/B ratio is telling us?metric of stock selection. Look for companies with low P/B and high and enduring ROE. The book value per share of preferred stock represents the amount of shareholders equity that is clearly assignable to preferred stock on a per share basis."Business Ratios and Formulas: A Comprehensive Guide", Hoboken, NJ: John Wiley Sons. The book value per share formula is used to calculate the per share value of a company basedFor instance, if the market value per share is lower than the book value per share, then the stock price may be undervalued.Do You Know How to Calculate the Solvency Ratios of Your Company? Market-to-Book Ratio. Price-to-Pre-Tax Profit PS.Market-to-Book Ratio, is the ratio of the current share price to the book value per share.Register now to create your own custom streaming stock watchlist. By accessing the services available at ADVFN you are agreeing to be bound by ADVFNs Therefore, Book Value per Share Book Value / Shares Outstanding. Book value per share formula above assumes common stock only.This is commonly expressed as the ratio of Price to Book. Formula: The market value per share is a companys current stock price, and it reflects a value Price-to-book (P/B) ratio as a valuation multiple is useful for value.Formulas related to Book Value per Share Return on Equity (ROE) P/B Ratio. Price-To-Book Ratio - P/B Ratio. Share.The price-to-book ratio (P/B Ratio) is a ratio used to compare a stocks market value to its book value. Calculate price per share by dividing the market value per share by the earnings per share. This is also known as the price-earnings ratio or P/EThere are a number of price per share formulas used for stocks, depending on the type and time of investment. The formula for calculating the price-earnings ratio for any stock is simple: the market value per share divided by the earnings per share (EPS). Price to Book Current Market Price/Book Value per share. Although price to book ratio still has some utility today, the world has changed since Ben Grahams day. When the market was dominated by capital-intensive firms that owned factories, land, rail track, and inventory | 2,488 | 12,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.897399 |
http://bonsaicode.wordpress.com/2011/04/12/programming-praxis-house-of-representatives/ | 1,386,257,463,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163046759/warc/CC-MAIN-20131204131726-00027-ip-10-33-133-15.ec2.internal.warc.gz | 22,482,464 | 15,491 | ## Programming Praxis – House Of Representatives
In today’s Programming Praxis exercise, our goal is to calculate the amount of seats each state gets in the United States House of Representatives. Let’s get started, shall we?
Some imports:
```import Control.Arrow
import qualified Data.List.Key as K
import qualified Data.Map as M```
To calculate the seat distribution we need the population data for each state.
```popData :: M.Map String Integer
("Arizona",6392017), ("Arkansas",2915918), ("California",37253956),
("Florida",18801310), ("Georgia",9687653), ("Hawaii",1360301),
("Idaho",1567582), ("Illinois",12830632), ("Indiana",6483802),
("Iowa",3046355), ("Kansas",2853118), ("Kentucky",4339367),
("Louisiana",4533372), ("Maine",1328361), ("Maryland",5773552),
("Massachusetts",6547629), ("Michigan",9883640), ("Minnesota",5303925),
("Mississippi",2967297), ("Missouri",5988927), ("Montana",989415),
("New Jersey",8791894), ("New Mexico",2059179), ("New York",19378102),
("North Carolina",9535483), ("North Dakota",672591), ("Ohio",11536504),
("Oklahoma",3751351), ("Oregon",3831074), ("Pennsylvania",12702379),
("Rhode Island",1052567), ("South Carolina",4625364), ("South Dakota",814180),
("Tennessee",6346105), ("Texas",25145561), ("Utah",2763885),
("Vermont",625741), ("Virginia",8001024), ("Washington",6724540),
("West Virginia",1852994), ("Wisconsin",5686986), ("Wyoming",563626)]```
The algorithm itself is fairly simple. Start with one seat per state and then keep assigning one seat to the state with the highest geometric mean until the desired number of seats is reached.
```house :: Int -> M.Map String Integer
house seats = M.map fst \$ iterate add (M.map ((,) 1) popData) !! k where
```main :: IO () | 494 | 1,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-48 | longest | en | 0.63786 |
https://www.adamsmiddle.org/apps/classes/863191/assignments/ | 1,534,708,412,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215284.54/warc/CC-MAIN-20180819184710-20180819204710-00693.warc.gz | 825,182,933 | 29,193 | # Assignments
### Period 1, Math 7 (Period 1)
Instructor
Leslie Wendorf
Term
2017-2018 School Year
Department
Mathematics
Description
# Upcoming Assignments
No upcoming assignments.
# Past Assignments
### Due: Sunday, June 10
Home Practice 5/22/18
1) Work on project; Video Due June 10 by 8pm (you may submit earlier, but not later). Google Classroom has directions. Project Worksheet due June 11 at beginning of class.
Make sure the day you chose to present isn't the day you go to the Getty. If it is, write down the day you can present and tell me in class on Friday 5/25--we will figure out a time then.
2) Practice for Final Benchmark Exam (get on IXL, Khan Academy, HRW Interactive Guide)
Math 7 Q4 Benchmark Exam June 7-8
NO MULTIPLE CHOICE; ALL OPEN-ENDED
2B.3: Multiply and divide fractions
1A.2: Determine whether two quantities are in a proportional relationship by testing ratios or observing whether the graph is a straight line through the origin.
1A.3: Identify the constant of proportionality (unit rate) in tables, graphs, equations, etc.
1A.4: Represent proportional relationships by equations.
3A.2: Use the distributive property, and I can use properties of operations to combine like terms within an expression, including rational coefficients to show equivalency.
3B.1: Write and solve multi-step equations from word problems using rational numbers and variables to represent unknown quantities.
4B.1: Know and use the formulas for the area and circumference of a circle to solve problems.
### Due: Friday, June 8
Home Practice 5/25/18
1) Work on video project--Due 6/10 -8pm shared to me via Google Drive
(Project worksheet due 6/11/18)
2) Practice and study for Quarter 4 Benchmark Exam --6/7--6/8/18
2B.3: Multiply and divide fractions
1A.2: Determine whether two quantities are in a proportional relationship by testing ratios or
observing whether the graph is a straight line through the origin.
1A.3: Identify the constant of proportionality (unit rate) in tables, graphs, equations, etc.
1A.4: Represent proportional relationships by equations.
3A.2: Use the distributive property, and I can use properties of operations to combine like terms
within an expression, including rational coefficients to show equivalency.
3B.1: Write and solve multi-step equations from word problems using rational numbers and variables to represent unknown quantities.
4B.1: Know and use the formulas for the area and circumference of a circle to solve problems.
### Due: Friday, June 8
Practice for Final Benchmark
Suggestions 1) complete and correct Spriral Reviews #1--#6; KEYS ATTACHED
2) Study error analysis from quick checks
3 to videos, click on "Math On The Spot" then practice online
4) Online practice (HRW, IXL, Khan Academy, do or re-do QUIZIZZ home practices); min score should be 85.
### Due: Thursday, May 31
Home Practice 5/29--5/30
1) Parent Signature -Quick Check 5B & 5C
2) Continue to work on video project (due June 10; sheet due 6/11)
3) Continue to PRACTICE and study for Q4 Benchmark Exam (6/7--6/8)
4) CHECK VIDEO PRESENTATION UPDATE (attached)
5) SUGGESTED: Quizizz LT3A.2 and LT3B.1 for Spiral Review #2:
CODE: 307465 by Friday 8am
### Due: Tuesday, May 22
Home Practice 5/21/18
1) Q4 BM Spiral Review #1; Show ALL STEPS-no work, no credit
2) Suggestion: Review notes/AP/HP/Quick Check corrections for SBAC Testing
### Due: Monday, May 21
Home Practice 5/18/18
1) Finish solving equations worksheet (show all steps; shade with 1 colored pencil)
2) Spend about 25 minutes on practice test website for SBAC test (show some evidence on paper, about 4 problems);
3Select Grade 7 Math Practice Test (12 minutes)--show some evidence on paper
### Due: Tuesday, May 15
Home Practice 5/14/18
1) Complete 5B & 5C Review (show all steps)--Key attached
2) Practice and study for QUICK CHECK Units 5B and 5C on Tuesday 5/15
SUGGESTIONS:
Use notes, active and home practices, explore sheetspractice on IXL (see LT Log), practice on Khan Academy, and practice on TenMarks:
1) Understanding the Probability of an Event
2) Identifying the Probability of Compound Events
3) Identifying a Probability Model
4) Approximating the Probability of an Event
5) Understanding Overlap of Data Distributions
6) Understanding Compound Events -
7) Using Uniform Models to Calculate Probability
### Due: Monday, May 14
Home Practice 5/10-5/11:
1) 5C.3 Home Practice Worksheet (passed out in class)--show all work-use lined paper or back if needed. Use key (ATTACHED) to check answers. Write score at the top of your page as a ratio.
2) Practice and study for QUICK CHECK UNITS 5B & 5C on Tuesday 5/15 (all 5B & 5C targets)
SUGGESTIONS:
1) review explore, notes, active and home practices; re-do active/home practices
2) IXL
5B: IXL Grade 7 (CC.1--CC.5; CC.7;) on LT Log
5C IXL Grade 7 (DD.1, DD.3, DD.4, DD.5) on LT Log
3) TenMarks (any/all assigned 5/10 due 5/15) on LT Log
### Due: Friday, May 11
Home Practice 5/8--5/9 DUE 5/10--5/11
1) 5C.2 & 5C.4 Home Practice Worksheet (passed out in class)-Show all work. Check answers with key (attached) after completion; write score at top of page as a ratio
2) Parent Signature on Quick Check 3B.1, 4B.3, 4B.4
### Due: Wednesday, May 9
Home Practice MONDAY 5/7/18: Title each section of work with assignment letter/number/label
1) IXL Grade 7:J.12 (15-20 minutes)--Show evidence on paper; Use 5A.1 notes (using proportions to make predictions)
2) IXL Grade 7: DD.1 (15-20 minutes)--Show evidence on paper; use 5C.1 Notes
TOTAL 30 minutes minimum (40 minutes maximum)--you can do more if you choose.
### Due: Monday, May 7
Home Practice 5/3--5/4
1) 5C.1 Home Practice worksheet (passed out in class and posted here); choose 3 from each section (9 total); SHOW ALL WORK. Make corrections after completion and write score (as a ratio) on the top of you paper.
2) Practice for Quick Check MONDAY 5/7(SUGGESTION: Get on IXL; Get on Khan Academy; Study corrections on QUICK CHECKS; USE NOTES to practice & re-do homework since answer keys are posted) REMEMBER TO SHADE THE BASE FIRST (for surface area and volume)
4B.3 (replacement; except if you have a 4 already) Surface Area
(IXL Gr. 7 AA.8)
4B.4 (replacement; except if you have a 4 already) Volume
(IXL Gr. 7 AA.7)
3B.1 Write and solve multi-step equations
(IXL Gr. 7 S.5, S.6, S.7, S.8, S.9)
### Due: Friday, May 4
Home Practice 5/1--5/2
5B.2 Home Practice Worksheet
CHOOSE 2: 1 box plot and 1 dot plot
show all steps
### Due: Wednesday, May 2
Home Practice 4/30: HP 5B.2 Worksheet #2
Read directions to CHOOSE appropriate PROBLEMS
SHOW ALL STEPS on lined paper Title: "HP 5B.1 #2"
### Due: Monday, April 30
1)5B.1 Home Practice worksheet (Passed out in class and attached with key)
Use notes to complete
Score after completion. Write score on homework
2) Quick Check 4B.3 & 4B.4 Signature
3) Practice/Study for MONDAY'S Quick Check 5A.1, 2B.4, 1A.2 (SUGGESTION: IXL & Khan Academy):
5A.1Samples and Populations
2B.4Order of Operations (with fractions, decimals, integers and exponents)
1A.2 Identifying Proportional Relationships in tables, graphs and verbal descriptions,and equations
### Due: Friday, April 27
Home Practice 4/25
5A.1 Home Practice Worksheet and Key attached.
Use notes. Show work. Score after completion.
Suggestions:
1)Identify Samples and Populations 2) Valid Claims 3) Make Inferences from Random Samples
2) Finish IXL (minimum score 80)
### Due: Wednesday, April 25
1) Finish mind map for all Unit 4B learning targets (use Unit 4 Learning Target Log, your notes for steps and hints; use active practice & home practice worksheets, quick checks, and online textbook for examples). DIRECTIONS and EXAMPLE attached.
2) PRACTICE for Quick Check 4B.3 & 4B.4 (use active practices posted in Google Classroom; IXL, HRW Interactive Guide etc)--Review Notes.
### Due: Monday, April 23
Home Practice 4/20/18
4B.4 Worksheet (double-sided).
Read directions . Show all work .
Quick Check 4B.3 surface area and 4B.4 volume Wed 4/25
### Due: Friday, April 20
Home Practice 4/18/18
Choose 3 problems (#1-7) on first page; must do 8 & 9 on first page; choose 3 problems from 2nd page (pyramids)
TOTAL 8 problems
Show all work -use nets and formulas (but not both for the same problem)
#8 3380ft2
#9 448in2
### Due: Wednesday, April 18
Khan Academy "Surface Area Using Nets"
-show some evidence on paper ( picture of object, formulas used & steps)
### Due: Friday, April 6
NO HOME PRACTICE 4/3--4/4
### Due: Wednesday, April 4
Practice for Quarter 3 Benchmark (4/3--4/4/18)
-use HRW interactive guide / IXL/ Khan Academy;
CODE:
-study quick checks and assessments / error analysis / notes / active practice / home practice
The Quarter 3 Benchmark Exam includes:
Multiple Choice
Unit 1A (4 questions)
Unit 2B (4 questions)
LT 3B.2
LT 4A.1
LT 4A.3
Open Ended (free- response)
LT 3A.2
LT 3B.1
LT 4B.1
LT 4B.2
### Due: Wednesday, April 4
Home Practice 4/2/18
1) Practice for benchmark exam (HRW Interactive Guide/ IXL/ Khan Academy)
2) Review old quick check corrections and error analysis
3) Study/Finish Quarter 3 Benchmark Prep Active Practice (key attached)
Q3 Benchmark includes:
Multiple Choice
Unit 1A (4 questions)
Unit 2B (4 questions)
LT 3B.2
LT 4A.1
LT 4A.3
Open Ended (free- response)
LT 3A.2
LT 3B.1
LT 4B.1
### Due: Monday, April 2
Home Practice 3/30/18
1) Finish Quarter 3 Benchmark Prep (in Google Classroom) from class today.
2) Practice for benchmark exam (HRW Interactive Guide/ IXL/ Khan Academy)
Q3 Benchmark includes:
Multiple Choice
Unit 1A (4 questions)
Unit 2B (4 questions)
LT 3B.2
LT 4A.1
LT 4A.3
Open Ended (free- response)
LT 3A.2
LT 3B.1
LT 4B.1
LT 4B.2
### Due: Friday, March 30
Home Practice 3/28/18
TenMarks (3 assignments; 5 problems each)--started in class; Must show evidence (using and solving formulas) for each problem (15)
1) Area and Circumference of a Circle : Use LT4B.1 notes
2) Finding Area; Use LT4B.2 notes
3) Finding Area, Volume, Surface Area of Composite Figures: Use LT4B.2 notes
### Due: Wednesday, March 28
Home Practice 3/26/18
25 minutes minimum / 30 minutes max
OR
minimum score 80.
Show some evidence on paper
### Due: Monday, March 26
Home Practice 3/21/18
LT 4B.2 Worksheet--show work (all steps for each formula) on lined paper--Follow the directions-you have a choice.
Videos and notes in Google Classroom
Practice for QUICK CHECK MONDAY 3/26
LT 4B.1 Suggested Practice: IXL Grade 7 AA.5 & AA.6
LT 4B.2 Suggested Practice: IXL Grade 7 AA.12
LT 4A.2 (replacement) Suggested Practice: IXL Grade 7 Z.1; Z.3; Z.4
LT 1A.4 (replacement) Suggested Practice: IXL Grade 7 K.5
### Due: Wednesday, March 21
Home Practice 3/19--3/20
"HP 4B.1 Worksheet" --double-sided --follow directions--you have choices. ATTACHED
*Key also attached
Monday 3/26/18 : Quick Check LT4B.1, LT4B.2, LT4A.2 replacement, and 1 previous LT (from Unit 1, 2, or 3)
### Due: Monday, March 19
Home Practice 3/14-3/15/18
1) HP 4A.1 & 4A.3 Pt2 Worksheet (score after using notes to complete; key attached); Use lined paper (don't squeeze work like I did for the bottom)
2) Practice for Quick Check 4A.1 & 4A.3 3/19-3/20 (and 1 surprise target so flip through your notes Unit 1 and 2 notes)
### Due: Wednesday, March 14
Home Practice 3/13/18
1) 4A.3 Worksheet
2) IXL Grade 6 : CC.12 (no evidence on paper); 10 minutes max or minimum score of 75
3) IXL Grade 7: W.12 (no evidence on paper); 20 minutes max or minimum score of 75
4) Turn in "LT4A.1 Home Practice" and Parent Signature for Quick Check 3B.2 (due 3/8-3/9)
### Due: Tuesday, March 13
Home Practice 3/12/18
4A.1 & 4A.3 Home Practice worksheet
--use notes and videos in Google Classroom
--score after completion using key.
**Quick Check 4A.1 & 4A.3 Next Week: 3/19--3/20/18**
### Due: Monday, March 12
Home Practice 3/8--3/9
1) LT4A.1IXL Grade 7 J.7: 25 minutes max or minimum score of 75; show evidence on paper
2) Practice for Unit 3 Exam (including 4A.2) MONDAY 3/12/18
--Refer to your learning target log to see what to practice in IXL, Khan Academy and HRW
--Study from your quick checks and error analysis
### Due: Friday, March 9
Home Practice 3/7/18
"LT 4A.1 Home Practice" worksheet
--Use explore/notes; show all steps
Score AFTER completion with notes.
### Due: Wednesday, March 7
HOME PRACTICE 3/5/18
Code: 453312
use notes and graphic organizer
### Due: Monday, March 5
HOME PRACTICE 3/2/18
1) IXL Grade 7: Z.4 (20mins or minimum score of 80)
2) 4A.2 worksheet --USE NOTES/Graphic organizer/videos in Google Classroom (key attached)
3) Practice for 3B.2 Quick Check Monday 3/5
**UNIT 3 EXAM 3/12/18**
### Due: Friday, March 2
"Two-step Inequalities"
Show evidence on paper
2) Practice for QUICK CHECK 3B.2 (and surprise target ) THUR 3/1 and FRI 3/2
-review notes, active practice, home practice
-practice on Khan Academy, IXL and HRW Interactive Guide Lesson 7.2 & 7.3
### Due: Wednesday, February 28
IXL
T.7
30 minutes max or minimum score of 80
Show some evidence on paper
## 1) "HP 3B.2 pt2" worksheet
SHOW ALL STEPS (use line paper if necessary); USE YOUR NOTES
**no steps, no score, no credit***
## 2) Practice for 3B.1 write and solve multi-step equations (and surprise target) Quick Check Mon 2/26
### Due: Friday, February 23
1) Home Practice 3B.2 Worksheet
USE YOUR NOTES TO COMPLETE--SHOW ALL STEPS
Use answer key to score AFTER completion
2) PRACTICE/STUDY for 3A.2 Quick Check Thur 2/22-Fri 2/23
### Due: Wednesday, February 21
2) Parent Signature on QC 3A.1 & 3A.2
3) PRACTICE for Quick Check 3A.2 replacement on 2/20 Per.2 and 2/21 Per.1
### Due: Wednesday, February 14
1) TenMarks: Solve multi-step equations with fractions and decimals (use 3B.1 notes)
Show evidence on paper
### Due: Monday, February 12
1) "HP 3B.1 Multi-step Equations" Worksheet
Use notes to complete
Score using answer key AFTER completion
2) Practice for Quick Check 3B.1 2/13--2/14/18
UPDATE: Quick Check 3A.2 and 1 other "free-for-all target" Feb 20--Feb21
Quick Check 3B.1 Feb26
### Due: Friday, February 9
1) "LT3B.1 Home Practice" worksheet--SHOW ALL STEPS and use answer key to score AFTER completion
2) Practice for Quick Check 3A.1& 3A.2 Thur 2/8 and Friday 2/9
### Due: Wednesday, February 7
HP 3A.2 Simplifying Expressions and Perimeter Worksheet
Key attached
Complete using notes/AP...SHOW ALL STEPS. Score using answer key after completion.
### Due: Friday, February 2
"Simplifying Expressions Worksheet" --passed out in class (key attached)
Use 3A.2 Notes and your Exit Ticket Error Analysis for the correct steps. SHOW ALL YOUR WORK.
Score after completion on your worksheet and stamp sheet.
### Due: Monday, January 29
1) 3A.2 Home Practice
Do the rest (9 problems) using previous night's home practice
Use notes to complete, all steps, THEN score with answer key on your paper and stamp sheet.
IF YOU DID all 16 the first night, do IXL Grade 7 R.15 & R.16 for a total of 20 minutes.
2) PRACTICE FOR Q2 BENCHMARK EXAM USING YOUR GOAL SHEET, LT LOGS, & ONLINE RESOURCES, (REVIEW ERROR ANALYSIS)
Multiple Choice (4 Questions Each)
Unit 2A
Unit 2B
Unit 1 (LT 1A.1, 1A.4, 1B.2, 1B.3)
Open Ended
LT 1A.2 (graph from table, determine proportional)
2B.4
### Due: Friday, January 26
16 problems on page (choose 2 from #1-6, choose 2 from #7-12, choose 3 from #13-16) Check your answers online after using your notes to complete the assignment -SHOW ALL STEPS. Write score on paper and stamp sheet.
### Due: Monday, January 22
Home Practice 1/19
"Factoring 3A.1 HP Worksheet"
Choose 10 to solve
Watch video below for help
Use all 3 -steps from the notes
1) Find GCF
2) Divide each term by GCF
3) Write quotients inside ( ) and GCF outside touching the ( ).
### Due: Friday, January 19
Home Practice 1/7--1/18
TenMarks: show work on paper for credit
• Identify Equivalent Expressions: Distributive Property (10 problems)
Use 3A.1 explore/notes, and active practice for assistance
### Due: Wednesday, January 17
"LT 3A.1 Home Practice" Worksheet--CHOOSE 10 to complete-show work
Use Explore sheet and VIDEO below (stop at 1:35) for help
PRACTICE and study for QUICK CHECK 1A --It is tomorrow Wed 1/17 NOT Friday because Friday is minimum day (22 minutes-not enough time).
1A; Unit Rate, Proportional Realtionships (tables, graphs, equations, y=kx; k=y/x)
Use TenMarks, IXL, and Khan Academy to practice and be prepared--also study notes from last week's notes (1/10--1/11) when we reviewed Unit 1A with Q1 Benchmark questions
SUGGESTED PRACTICE:
IXL (K.1, K.2, K.4, K.5, K.6, M.3, J.5)
TenMarks
### Due: Tuesday, January 16
TenMarks: Show work on paper (do not copy problems or multiple choice answer-just the work/ steps or rule used to prove answer.) USE all 1A notes and active practices to help you.
-Identify Proportional Relationship (5 problems: Review of LT1A.2--1A.5
Use Proportional Relationships foldable (yellow, date 9/7--9/8)
-Calculate Unit Rate (5 problems: Review of LT1A.1)
Use Rate Unit Rate foldable (green, date 9/5--9/6)
1) Go through CLEVER (EASIEST way to login -see ACCESSING WEBSITES attachment)
2) IF that doesn't work, look at PERIOD 1 or PERIOD 2 attachment and follow those directions to login
## Suggested Practice for Quick Check 2B Friday 1/12/18
TenMarks
1) Fractions with Number Lines 0-4
2) Rational Numbers on Number Lines and Coordinate Planes
3) Applying Addition and Subtraction to Rational Numbers
4) Applying Multiplication and Division to Rational Numbers
IXL
E.1
E.2
E.2
E.3
E.5
E.6
HRW
Interactive Guides
3.1
3.2
3.3
3.4
### Due: Friday, January 12
Home Practice
Per.2 1/11 Due 1/12
Per.1 1/10 Due 1/12
1)Textbook pg. 101 (1-8) and pg.107 (18-21)--SHOW ALL STEPS (Use all 2B notes) & score after completion
2) Study notes and PRACTICE FOR UNIT 2B Assessment Friday 1/12/18
3) Parent Signature on Quick Check 2A
### Due: Wednesday, January 10
Passed out in class. Use all 2B notes and 2B.4 Active Practice to complete it with all the correct steps. Score it when you are done. Key attached
### Due: Monday, January 8
Home Practice 1/4--1/5
2B Rational Numbers Review (all odd or all even); 12 problems only
SHOW ALL STEPS
### Due: Friday, January 5
Home Practice 1/2-1/3
1) 2B Riddle (passed out in class): show ALL steps (key attached)
2) Suggested PRACTICE for QUICK CHECK 2A (Per.1 on 1/5 Per.2 on 1/4)
TenMarks
1) Understanding and Identifying Opposite Quantities | 5,630 | 18,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.863493 |
https://www.adventuresinmachinelearning.com/finding-the-best-fit-introducing-the-akaike-information-criterion-in-python/ | 1,726,063,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00425.warc.gz | 593,609,520 | 16,084 | # Finding the Best Fit: Introducing the Akaike Information Criterion in Python
## Introducing the Akaike Information Criterion
When it comes to analyzing data, one of the most common tasks is fitting a regression model. This process involves selecting the best model from a set of candidate models.
However, with so many possible models to choose from, how can we know which one is the best fit? That’s where the Akaike Information Criterion (AIC) comes into play.
In this article, we’ll explore what the AIC is, how it’s calculated, and how to use it to compare models. We’ll also show you how to apply the AIC in Python, using a simple example.
## Definition of AIC
The Akaike Information Criterion (AIC) is a metric commonly used to compare statistical models. It provides a way of comparing different models that have been fit to the same data, in order to select the best one.
The AIC takes into account the number of model parameters and the goodness of fit, penalizing models that have more parameters but not a corresponding improvement in fit.
## Calculation of AIC
The AIC is calculated as follows: AIC = -2log(L) + 2k, where L is the likelihood function of the model and k is the number of model parameters.
The likelihood function quantifies how well the model fits the data, while the second term in the equation penalizes models with more parameters.
This balance between goodness of fit and model complexity allows the AIC to identify models that are neither too simple nor too complex, but rather strike a balance between the two.
## Use of AIC for Model Comparison
Once the AIC values have been calculated for each model, they can be compared to identify the best fit. The model with the lowest AIC value is generally considered to be the best fit.
While the AIC value itself has no intrinsic meaning, the difference between AIC values for different models (ΔAIC) can be used to determine the relative support for each model.
A general rule of thumb is that models with a difference in AIC values of less than 2 are considered equally supported, while models with a ΔAIC greater than 10 are strongly disfavored.
## Applying AIC in Python
Now that we understand what AIC is and how it’s used, let’s see how to apply it in Python. We’ll use a simple example to illustrate the process.
The goal is to fit a linear regression model to predict the price of a house based on its size and number of bedrooms.
First, we need to load the data into Python and select our predictor variables. We’ll be using the Housing Price dataset from the Statsmodels package.
``````import statsmodels.api as sm
from statsmodels.formula.api import ols
import pandas as pd
data = sm.datasets.get_rdataset("Housing", "Ecdat").data
X = data[["lotsize", "bedrooms"]]
y = data["price"]
``````
### Fitting and AIC Calculation for Model 1
Next, we’ll fit our first model using the “ols” function from the statsmodels package.
``````model1 = ols("price ~ lotsize + bedrooms", data=data).fit()
a1 = model1.aic
print("Model 1 AIC: ", a1)
``````
This code specifies the model formula and fits the model to our data.
The AIC value for this model is displayed using the ‘aic’ method.
### Fitting and AIC Calculation for Model 2
Next, we’ll fit our second model using a different combination of predictor variables.
``````model2 = ols("price ~ lotsize", data=data).fit()
a2 = model2.aic
print("Model 2 AIC: ", a2)
``````
This code fits our second model, which uses only the “lotsize” variable as a predictor.
Again, we calculate the AIC value using the ‘aic’ method.
### Model Comparison and Selection
Finally, we can compare the AIC values for our two models to determine which one is the best fit.
``````if a1 < a2:
print("Model 1 is the better fit.")
else:
print("Model 2 is the better fit.")
``````
In this case, the AIC value for Model 1 is lower than the AIC value for Model 2.
Therefore, we select Model 1 as the better fitting model.
## Conclusion
In this article, we’ve introduced the Akaike Information Criterion and shown how it can be used to compare regression models. By taking into account both model complexity and goodness of fit, the AIC provides a useful metric for identifying the best fitting model.
We’ve also shown how to apply the AIC in Python, using a simple example. Next time you’re analyzing data, consider using the AIC to help you select the best fitting model.
The Akaike Information Criterion (AIC) is a critical metric used to compare statistical models and identify the best fit among them. It achieves this by balancing the model complexity against the goodness of fit.
We learned how to calculate AIC values using a simple example and presented how AIC helps in selecting the right model to work with. Stats models recognize AIC scores as important in choosing the best model, and Python programming language, which has become a popular tool for statisticians, scientists, and data analysts, provides an easy way of computing these scores.
By understanding AIC, we have a powerful tool in our data analysis toolkit. | 1,159 | 5,073 | {"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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.895763 |
https://number.academy/44386 | 1,660,448,675,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00729.warc.gz | 393,293,384 | 12,026 | # Number 44386
Number 44,386 spell 🔊, write in words: forty-four thousand, three hundred and eighty-six . Ordinal number 44386th is said 🔊 and write: forty-four thousand, three hundred and eighty-sixth. The meaning of number 44386 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 44386. What is 44386 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 44386.
## What is 44,386 in other units
The decimal (Arabic) number 44386 converted to a Roman number is (XL)(IV)CCCLXXXVI. Roman and decimal number conversions.
#### Weight conversion
44386 kilograms (kg) = 97853.4 pounds (lbs)
44386 pounds (lbs) = 20133.4 kilograms (kg)
#### Length conversion
44386 kilometers (km) equals to 27581 miles (mi).
44386 miles (mi) equals to 71433 kilometers (km).
44386 meters (m) equals to 145622 feet (ft).
44386 feet (ft) equals 13530 meters (m).
44386 centimeters (cm) equals to 17474.8 inches (in).
44386 inches (in) equals to 112740.4 centimeters (cm).
#### Temperature conversion
44386° Fahrenheit (°F) equals to 24641.1° Celsius (°C)
44386° Celsius (°C) equals to 79926.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
44386 seconds equals to 12 hours, 19 minutes, 46 seconds
44386 minutes equals to 1 month, 2 days, 19 hours, 46 minutes
### Codes and images of the number 44386
Number 44386 morse code: ....- ....- ...-- ---.. -....
Sign language for number 44386:
Number 44386 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 44386
### Multiplications
#### Multiplication table of 44386
44386 multiplied by two equals 88772 (44386 x 2 = 88772).
44386 multiplied by three equals 133158 (44386 x 3 = 133158).
44386 multiplied by four equals 177544 (44386 x 4 = 177544).
44386 multiplied by five equals 221930 (44386 x 5 = 221930).
44386 multiplied by six equals 266316 (44386 x 6 = 266316).
44386 multiplied by seven equals 310702 (44386 x 7 = 310702).
44386 multiplied by eight equals 355088 (44386 x 8 = 355088).
44386 multiplied by nine equals 399474 (44386 x 9 = 399474).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 44386
Half of 44386 is 22193 (44386 / 2 = 22193).
One third of 44386 is 14795,3333 (44386 / 3 = 14795,3333 = 14795 1/3).
One quarter of 44386 is 11096,5 (44386 / 4 = 11096,5 = 11096 1/2).
One fifth of 44386 is 8877,2 (44386 / 5 = 8877,2 = 8877 1/5).
One sixth of 44386 is 7397,6667 (44386 / 6 = 7397,6667 = 7397 2/3).
One seventh of 44386 is 6340,8571 (44386 / 7 = 6340,8571 = 6340 6/7).
One eighth of 44386 is 5548,25 (44386 / 8 = 5548,25 = 5548 1/4).
One ninth of 44386 is 4931,7778 (44386 / 9 = 4931,7778 = 4931 7/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
44386
#### Is Prime?
The number 44386 is not a prime number. The closest prime numbers are 44383, 44389.
#### Factorization and factors (dividers)
The prime factors of 44386 are 2 * 22193
The factors of 44386 are 1 , 2 , 22193 , 44386
Total factors 4.
Sum of factors 66582 (22196).
#### Powers
The second power of 443862 is 1.970.116.996.
The third power of 443863 is 87.445.612.984.456.
#### Roots
The square root √44386 is 210,679852.
The cube root of 344386 is 35,406419.
#### Logarithms
The natural logarithm of No. ln 44386 = loge 44386 = 10,700679.
The logarithm to base 10 of No. log10 44386 = 4,647246.
The Napierian logarithm of No. log1/e 44386 = -10,700679.
### Trigonometric functions
The cosine of 44386 is -0,008194.
The sine of 44386 is 0,999966.
The tangent of 44386 is -122,041411.
### Properties of the number 44386
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 44386 in Computer Science
Code typeCode value
44386 Number of bytes43.3KB
Unix timeUnix time 44386 is equal to Thursday Jan. 1, 1970, 12:19:46 p.m. GMT
IPv4, IPv6Number 44386 internet address in dotted format v4 0.0.173.98, v6 ::ad62
44386 Decimal = 1010110101100010 Binary
44386 Decimal = 2020212221 Ternary
44386 Decimal = 126542 Octal
44386 BASE64NDQzODY=
44386 MD5f1f904947b93e2b75be65a0930d1b695
44386 SHA1a1c4c1f5459e621e66929cc66c7813d3505958a1
44386 SHA224841e7b8b2dc911825e3b3fb4dd36a18c8777832f7d585744510e739a
44386 SHA256d77f3fd782aeb89ee45af7e5fbc0b896172d926419f6a04a98a24aafb8fcc8ba
44386 SHA384e7c1a8dcf5bff4611c9907b54d1d0e6a6540ebe5d5acafbc376251559827c5e6e9f70021b23483ee060486a3f5b64be0
More SHA codes related to the number 44386 ...
If you know something interesting about the 44386 number that you did not find on this page, do not hesitate to write us here.
## Numerology 44386
### Character frequency in number 44386
Character (importance) frequency for numerology.
Character: Frequency: 4 2 3 1 8 1 6 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 44386, the numbers 4+4+3+8+6 = 2+5 = 7 are added and the meaning of the number 7 is sought.
## Interesting facts about the number 44386
### Asteroids
• (44386) 1998 SV61 is asteroid number 44386. It was discovered by LONEOS from Anderson Mesa on 9/17/1998.
## № 44,386 in other languages
How to say or write the number forty-four thousand, three hundred and eighty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 44.386) cuarenta y cuatro mil trescientos ochenta y seis German: 🔊 (Anzahl 44.386) vierundvierzigtausenddreihundertsechsundachtzig French: 🔊 (nombre 44 386) quarante-quatre mille trois cent quatre-vingt-six Portuguese: 🔊 (número 44 386) quarenta e quatro mil, trezentos e oitenta e seis Chinese: 🔊 (数 44 386) 四万四千三百八十六 Arabian: 🔊 (عدد 44,386) أربعة و أربعون ألفاً و ثلاثمائةستة و ثمانون Czech: 🔊 (číslo 44 386) čtyřicet čtyři tisíce třista osmdesát šest Korean: 🔊 (번호 44,386) 사만 사천삼백팔십육 Danish: 🔊 (nummer 44 386) fireogfyrretusinde og trehundrede og seksogfirs Dutch: 🔊 (nummer 44 386) vierenveertigduizenddriehonderdzesentachtig Japanese: 🔊 (数 44,386) 四万四千三百八十六 Indonesian: 🔊 (jumlah 44.386) empat puluh empat ribu tiga ratus delapan puluh enam Italian: 🔊 (numero 44 386) quarantaquattromilatrecentottantasei Norwegian: 🔊 (nummer 44 386) førti-fire tusen, tre hundre og åtti-seks Polish: 🔊 (liczba 44 386) czterdzieści cztery tysiące trzysta osiemdziesiąt sześć Russian: 🔊 (номер 44 386) сорок четыре тысячи триста восемьдесят шесть Turkish: 🔊 (numara 44,386) kırkdörtbinüçyüzseksenaltı Thai: 🔊 (จำนวน 44 386) สี่หมื่นสี่พันสามร้อยแปดสิบหก Ukrainian: 🔊 (номер 44 386) сорок чотири тисячi триста вiсiмдесят шiсть Vietnamese: 🔊 (con số 44.386) bốn mươi bốn nghìn ba trăm tám mươi sáu Other languages ...
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If you know something interesting about the number 44386 or any natural number (positive integer) please write us here or on facebook. | 2,473 | 7,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.712363 |
https://na-inet.jp/na/gslsample/linear_system.html | 1,685,395,885,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00600.warc.gz | 465,171,401 | 2,317 | # GSL sample: Linear System of Equations
Last Update: Oct. 4, 2007
### Sample code
linear_system.c
```/***********************************************/
/* T.Kouya's GSL sample program collection */
/* Solving Linear System of Equations */
/* LU decomposition */
/* Written by Tomonori Kouya */
/* */
/* Version 0.01: 2007-10-02 */
/***********************************************/
#include <stdio.h>
#include <gsl/gsl_math.h>
#include <gsl/gsl_blas.h>
#include <gsl/gsl_vector.h>
#include <gsl/gsl_matrix.h>
#include <gsl/gsl_linalg.h>
/* Dimension of Matrix and Vectors */
#define DIM 5
int main(void)
{
int i, j, lotkin_signum, frank_signum;
gsl_matrix *lotkin_a, *frank_a;
gsl_vector *x, *lotkin_b, *frank_b, *lotkin_x, *frank_x;
gsl_permutation *lotkin_perm, *frank_perm;
/* allocate a, x, b */
lotkin_a = gsl_matrix_alloc(DIM, DIM);
frank_a = gsl_matrix_alloc(DIM, DIM);
x = gsl_vector_alloc(DIM);
lotkin_b = gsl_vector_alloc(DIM);
frank_b = gsl_vector_alloc(DIM);
lotkin_x = gsl_vector_alloc(DIM);
frank_x = gsl_vector_alloc(DIM);
/* set x = [1 2 ... DIM] */
for(i = 0; i < DIM; i++)
gsl_vector_set(x, i, (double)i);
/* set Lotkin matrix */
/* a_ij = 1 (i = 1) or 1/(i+j-1) (i != 1) */
for(i = 0; i < DIM; i++)
gsl_matrix_set(lotkin_a, 0, i, 1.0);
for(i = 1; i < DIM; i++)
for(j = 0; j < DIM; j++)
gsl_matrix_set(lotkin_a, i, j, 1.0 / (double)(i + j + 1));
/* set Frank matrix */
/* a_ij = DIM - min(i,j) + 1 */
for(i = 0; i < DIM; i++)
for(j = 0; j < DIM; j++)
gsl_matrix_set(frank_a, i, j, (double)DIM - (double)GSL_MAX(i, j) );
/* Print matrix */
printf("Lotkin Matrix(DIM = %d)\n", DIM);
for(i = 0; i < DIM; i++)
{
printf("%3d: ", i);
for(j = 0; j < DIM; j++)
printf("%g ", gsl_matrix_get(lotkin_a, i, j));
printf("\n");
}
printf("\n");
printf("Frank Matrix(DIM = %d)\n", DIM);
for(i = 0; i < DIM; i++)
{
printf("%3d: ", i);
for(j = 0; j < DIM; j++)
printf("%g ", gsl_matrix_get(frank_a, i, j));
printf("\n");
}
printf("\n");
/* b = A * x */
gsl_blas_dgemv(CblasNoTrans, 1.0, lotkin_a, x, 0.0, lotkin_b);
gsl_blas_dgemv(CblasNoTrans, 1.0, frank_a, x, 0.0, frank_b);
/* LU decomposition and forward&backward substition */
lotkin_perm = gsl_permutation_alloc(DIM);
gsl_linalg_LU_decomp(lotkin_a, lotkin_perm, &lotkin_signum);
gsl_linalg_LU_solve(lotkin_a, lotkin_perm, lotkin_b, lotkin_x);
gsl_permutation_free(lotkin_perm);
frank_perm = gsl_permutation_alloc(DIM);
gsl_linalg_LU_decomp(frank_a, frank_perm, &frank_signum);
gsl_linalg_LU_solve(frank_a, frank_perm, frank_b, frank_x);
gsl_permutation_free(frank_perm);
/* print */
printf("index Lotkin Frank True\n");
for(i = 0; i < DIM; i++)
printf("%3d %25.17e %25.17e %25.17e\n", i, gsl_vector_get(lotkin_x, i), gsl_vector_get(frank_x, i), gsl_vector_get(x, i));
/* free a, x, b */
gsl_matrix_free(lotkin_a);
gsl_matrix_free(frank_a);
gsl_vector_free(x);
gsl_vector_free(lotkin_b);
gsl_vector_free(frank_b);
gsl_vector_free(lotkin_x);
gsl_vector_free(frank_x);
return 0;
}```
### Example execution
```\$ ./linear_system
Lotkin Matrix(DIM = 5)
0: 1 1 1 1 1
1: 0.5 0.333333 0.25 0.2 0.166667
2: 0.333333 0.25 0.2 0.166667 0.142857
3: 0.25 0.2 0.166667 0.142857 0.125
4: 0.2 0.166667 0.142857 0.125 0.111111
Frank Matrix(DIM = 5)
0: 5 4 3 2 1
1: 4 4 3 2 1
2: 3 3 3 2 1
3: 2 2 2 2 1
4: 1 1 1 1 1
index Lotkin Frank True
0 -2.82440737464639824e-13 -1.42108547152020045e-15 0.00000000000000000e+00
1 1.00000000000282596e+00 1.00000000000000355e+00 1.00000000000000000e+00
2 1.99999999999155187e+00 1.99999999999999645e+00 2.00000000000000000e+00
3 3.00000000000982148e+00 3.00000000000000178e+00 3.00000000000000000e+00
4 3.99999999999608313e+00 4.00000000000000000e+00 4.00000000000000000e+00```
<- Go back | 1,465 | 3,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.250097 |
https://msp.ucsd.edu/syllabi/170.13f/course-notes/node2.html | 1,709,314,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475422.71/warc/CC-MAIN-20240301161412-20240301191412-00601.warc.gz | 399,298,444 | 11,208 | Next: 3. Spectra Up: course-notes Previous: 1. Sounds, Signals, and Index
Subsections
# 2. Sinusoids
For several different reasons, sinusoids pop up ubiquitously in both theoretical and practical situations having to do with sound. For one thing, sinusoids occur naturally in a variety of ways, and if one happens to couple physically with the air and is of audible frequency and amplitude, we'll hear it. Second, sinusoids behave in simple and predictable ways when the elementary operations (amplification. mixing, delay; section 1.5) are applied to them. Third, one can add up sinusoids to make arbitrary signals or digital recordings (with some provisos having to do with convergence); this ability is extraordinarily useful for analyzing and synthesizing sounds.
# 2.1 Elementary Operations on Sinusoids
Here is a picture that might help visualize the mathematics of sinusoids. Imagine a point on the rim of a spinning bicycle wheel:
The progress in space of the point has horizontal () and vertical () components. If we forget the vertical component and graph just the horizontal component over time we get a sinusoid. If the point is initially at an angle from the axis, we get the familiar formula:
where , the frequency, is the number of revolutions per unit time, and the amplitude is the radius of the wheel.
Now for the three elementary operations. First, amplification, say by a linear gain , replaces above with
If the gain is specified in decibels (say, ), then we convert from decibels to a linear gain by applying the definition of decibels backward:
Applying a delay to a sinusoid equal to (or, if a recording, a time shift forward or backward by a positive or negative number equivalent to a time ) has the effect of replacing with in the formula:
This leaves the amplitude and the frequency unchanged, but subtracts an offset from the initial phase.
The effect of mixing two sinusoids (the third elementary operation) is more complicated. We'll start by supposing the two have equal frequencies (but not necessarily the same amplitudes or initial phases). Here is a picture:
The parallelogram represents the initial situation at time zero; the entire thing rotates about the origin as indicated by the arrows, without changing size or shape. If the initial phases of the two are and , the angle between them is either plus or minus and, by the law of cosines, we get
(it doesn't matter which order and appear in the formula, since the cosine of the difference is the same either way). Depending on the phase difference, may lie anywhere between (if so that the two sinusoids are exactly out of phase) and (if so that they are perfectly in phase.)
The resulting initial phase depends in a complicated way on all of , , , and --the easiest way to compute it would be to convert everything to rectangular coordinates and back, but we will put that off for another day.
If the two frequencies are not equal--call them and --we can still apply the same reasoning, at least qualitatively. At time we still get a parallelogram, but now the two summands are rotating about the origin at different rates, so that the difference between the two phases, initially , is itself increasing or decreasing by a rate equal to the difference of the two component frequencies, that is, . As a result, exactly times every unit of time, the parallelogram runs through its entire range of shapes and the resultant amplitude runs back and forth between its minimum and maximum possible values, and .
If and differ by less than about 30 Hz., you can hear these changes in amplitude. This effect is called beating. At greater frequency separations you are likely to hear two separate tones, unless indeed they act as we'll describe in the next section:
# 2.2 Periodic and aperiodic tones
So far we have tacitly assumed that our ears can actually hear sinusoids as separate sounds, and that, presented with two or more sinusoids, we would be likely to perceive them as separate sounds. The truth is somewhat stranger: under the right conditions, our ears appear to have evolved to be able to distinguish periodic signals from each other, even if several of them with different periodicities are mixed together. (This is a good adaptation because it allows us to perceive the voices of other humans, which are approximately periodic most of the time, but rarely if ever sinusoidal.)
A signal is called periodic when, for some nonzero time duration , we have
for all . We can apply this equation repeatedly to get:
In other words, the signal repeats forever. Knowing the value of the function for one period, for example from to , determines the function for all other values of .
If a function repeats after time units, it also repeats after , , ..., time units.. The smallest value of at which the signal repeats is called the signal's period.
A sinusoid whose frequency is has period . But an infinitude of other sinusoids repeat after time units. A sinusoid of frequency has period , and so repeats twice in a time interval lasting . In general a sinusoid whose frequency is any integer multiple of repeats (perhaps for the th time) after an elapsed time of . More generally, any signal obtained by amplifying and mixing sinusoids of frequencies that are all multiples of will repeat after units of time, and therefore have a period of (if not some smaller submultiple of ).
Under reasonable conditions ( at least about 30; sinusoids at lower multiples of having enough relative amplitude compared to the whole; no signal frequency other than having an amplitude greater than the sum of all the others; at least some energy in odd-numbered multiples of ; etc.) we would hear such a mixture as a single tone whose pitch corresponds to , which is then called the fundamental frequency of the mixture. The mixture will have the general form:
only stopping, for a digital recording, at the Nyquist frequency, and possibly continuing forever for an analog signal.
Such a sum of harmonically sinusoids is known as a Fourier series, and although we won't prove it here, it's known that any reasonable" periodic signal, (having a certain continuity property in time that any real signal should have) can be expressed as a Fourier series. Its digital recording can as well. This means that, in principle at least, you can synthesize any periodic signal if you can synthesize sinusoids.
The whole mixture is sometimes called a complex periodic tone, and the individual sinusoids that make it up are called harmonics. If all goes well, the perceived pitch of a complex periodic tone is that of its first harmonic, corresponding to the frequency , which you can compute as in section 1.3.
It sometimes happens that a mixture of sinusoids that aren't collectively periodic somehow are perceived by the ear as a single entity (a tone) anyhow. Such a mixture could be written as:
and is called a complex inharmonic tone. The individual sinusoids that make it up are then called partials or components, and not harmonics - that term is reserved for the periodic case described earlier.
# 2.3 Special case: combining two equal-amplitude sinusoids
Suppose two sinusoids have the same amplitude and frequency , but different initial phases, and . Our formula for the amplitude of the sum (from section 2.1) reduces to:
We can apply a standard trigonometric identity to get:
This outcome is clear even if we don't remember that sort of identity; we can look at what the previous figure becomes when the two amplitudes are equal:
So not only is the amplitude increased (or decreased) by twice the cosine of half the phase difference; we also see that the initial phase of the resulting sinusoid (which would have been complicated to calculate in general) is the average of the two initial phases and .
As long as the amplitudes of the two sinusoids are the same, we can use the same picture to find the result of adding (mixing) two sinusoids of different frequencies and . To reduce clutter we'll leave out the initial phases to get the following formula:
This formula will recur often. I call it the Fundamental Law of Electronic Music, although perhaps that's overstating things a bit.
# 2.4 Power
Although the nominal (peak) amplitude of a sinusoid is a perfectly good measure of its overall strength, most signals in real life aren't sinusoids, and their peak amplitudes don't necessarily give a realistic measure of their strength. Also, you could wish to have a measure of strength that was additive, in the sense that, at least in good conditions, when you add two signals their measured strengths are added as well. The nearest thing we have to such a measure is the average power, which we will first motivate from physical considerations, then define, then show that it (at least sometimes) works the way we would wish.
The simplest way to motivate the definition of power is by considering a real-world analog, electrical signal. The amplitude (a function of time) is in this instance the time-varying voltage, customarily given the variable name . We now suppose the signal is connected to a load of some sort, which has an electrical resistance , measured in ohms. Power is voltage times current. To find the current we apply Ohm's law to get:
and finally
We conclude that power, like amplitude, is a function of time; it is proportional to the square of amplitude. It is always either zero or positive.
Although we aren't ready to discuss real sounds in the air yet (we will be able to put that off until chapter 5 or perhaps even 6), the same reasoning will apply. The amplitude is the (space-dependent) pressure. One can measure the power flowing through a specified area as follows: the pressure exerts a force on the area; as a result some air flows through the area, and the force times the velocity gives energy per second, which is the physical definition of power. The speed at which the air flows is proportional to the pressure (it's pressure divided by impedance)--a concept that generalizes resistance to describe reluctance to move" in whatever medium we're talking about.) Power is then amplitude squared divided by impedance.
For digital recordings, we don't have a notion of physical impedance and so we just arbitrarily set it to one, giving
where denotes the amplitude of the recorded signal. (Note that we're abusing notation here; recordings aren't functions of time, so really stands for the time at which we mean to play the sample, or else the time at which we recorded it. The only true way to describe the variation of a recording is by talking about the memory addresses, or indices, of the sample points.)
So far we've only described instantaneous power, which is a time-varying function. The measure we're interested in is a signal's or recording's average power, which is simply the average, over some suitable period of time or range of samples, of the instantaneous power.
What is the average power of a sinusoid? Well, its square is
(we're dropping the initial phase which won't affect out calculation). Now use my Fundamental Law of Electronic Music with (and omitting its own, arbitrary value of ):
(We omitted the term because and the cosine of zero is one.) If we multiplied the right hand size by we would get the desired instantaneous power, so we multiply through by and swap sides, giving
We want to know the average power. When we average the right-hand side of the equation, the cosine term averages out to zero, and so the average power of the original sinusoid is given by:
Here's what it looks like:
What happens when we add two sinusoids? Well, case one, they have the same frequency, and their amplitudes are and . Let denote the amplitude of the resulting sinusoid (which will also have the same frequency). As we saw above, the three are related by the law of cosines:
The three sinusoids have average power
so
About this we can at least say that, if we don't know what the relative phases of the two are, on average" we expect the power to be additive because the cosine term is just as likely to be negative as positive.
Once again, we can deal with sinusoids of differing frequencies and by just letting the phase difference precess in time at a frequency . In this case the cosine term really does average out to zero no matter what the initial phases were. The power of the sum of the two sinusoids is the sum of the powers of the two summands.
In fact, the cosine term can be considered as the two sinusoids beating. If we want to measure the power accurately we must wait at least a few beats--the closer the two sinusoids are in frequency, the longer it will take our measurement to converge on the correct answer.
To calculate the average power of uniform white noise of amplitude we have to do a quick integral; we get
Noise also has the property that it contributes power additively to a signal (as long as you don't add it to itself; see the next paragraph.)
It might seem that it is almost always true that adding two signals, with average power and , respectively, gives a signal of average power ; but beware the following counterexample: if you add a signal to itself you will double all its values and so the average power will be multiplied by 4, not 2. If you add a signal to its additive inverse (which has the same power as the original), the power of the sum will be zero. Also, if two sinusoids have the same frequency the average power of their sum will depend on the phase difference. There is a term for the situation in which you can simply add the average power of two signals to get the average power of the sum: such signals are said to be uncorrelated.
In general, scaling a signal (that is, multiplying all its values) by a factor of scales the average power by a factor , whereas accumulating unrelated signals should be expected only to multiply the power by on average.
## 2.4.1 Expressing Power In Decibels
In the previous chapter, we developed the notion of decibels for comparing the amplitudes of sinusoids. At that point we had no precise way to describe the amplitudes of signals in general, but now we do: by measuring their average power. If two signals have average power and , their level difference in decibels is:
We can quickly check that this is compatible with our earlier formula in terms of amplitude: two sinusoids of amplitude and would have average power , and the above formula reduces to:
so this new definition agrees with th eearlier one in section 1.3.
# Exercises and Project
1. Two sinusoids with the the same frequency (440 Hz., say), and with peak amplitudes 2 and 3 are added (or mixed, in other words). What are the minimum and maximum possible peak amplitude of the resulting sinusoid?
2. Two sinusoids with different frequencies, whose average powers are 3 and 4 respectively, are added. What is the average power of the resulting signal?
3. Two sinusoids, of period 4 and 6 milliseconds, respectively, are added. What is the period of the resulting waveform?
4. Two sinusoids are added (once again)... One has a frequency of 1 kHz . The resulting signal beats" 5 times per second. What are the possible frequencies of the other sinusoid?
5. A signal - any signal - is amplified, multiplying it by three. By how many decibels is the level raised?
6. What is the pitch, in octaves, of the second harmonic of a complex harmonic tone, relative to the first harmonic?
Project: comb filtering. In this project you will use the phase-dependent effect of combining two sinusoids to build the simplest type of digital filter, called a comb filter.
To start with, make a single sinusoid of frequency 100 Hz (using the sinusoid object in the course library for Pd). You can check the level of its output using the meter" object; it should be about 97 dB.
Now put the sinusoid into a vdelay" (variable delay) object, and connect the delay output as well as the original sinusoid output to the meter. When the delay is zero you should see something 6 decibels higher, about 103.
Now measure and graph the amplitudes you measure, changing the delay in ten steps from 0 to 0.005 seconds. (Hint: to make the graph readable, don't make the vertical axis linear in decibels; instead, perhaps make equal spaces for 0, 94, 97, 100, and 103). But if you really want a nice-looking graph and don't mind 5 extra minutes of effort, convert from decibels to power.
Now do the same thing (on the same graph with a different color or line style) with the sinusoid at 200 Hz. instead of 100 Hz. Do you see a relationship between the two?
Now put six sinusoids at 100, 200, 300, 400, 500, 600 Hz. into a switch" object (that's primarily for convenience; connecting the six to the switch will add them.) Connect the switch output to both the delay and directly to the output as before. As you change the delay between 0 and 10 milliseconds, what do you hear? What special thing happens when you choose a 5 millisecond delay?
Next: 3. Spectra Up: course-notes Previous: 1. Sounds, Signals, and Index
msp 2014-11-24 | 3,661 | 17,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.901863 |
https://www.proprofs.com/discuss/q/886553/method-involved-charges-uncharged-metal-object-rearranged-wi | 1,591,454,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00103.warc.gz | 847,010,884 | 83,660 | What method is involved when charges in an uncharged metal object are - ProProfs Discuss
# What method is involved when charges in an uncharged metal object are rearranged without direct contact with the charged object?
A. Friction
B. Induction
C. Convection
D. Conduction
This question is part of Static Electricity
Asked by Misrael, Last updated: Jun 04, 2020
#### C. Perez
Just getting better day by day
C. Perez, Writer, Writer, Cleveland
The correct answer to this question is B, Induction. This occurs when movement is used to create electricity. This is different from when electricity creates movement. For induction to occur, the circuit must have alternating current.
When it does, it flows through and creates another current in another circuit. This can occur by just placing them with each other. Induction is a principle part of many things around us every day.
For example, if one has a rice cooker in their kitchen, they are cooking with induction. Another example that most people already have in their homes are electric toothbrushes that are rechargeable.
#### A. Cook
Find happiness in writing new things.
A. Cook, English Professor, M.A, Ph.D, Kentucky
In science, friction is known the motion when two things are rubbed together. It creates heat usually. Induction is when the electrical charge occurs due to an electrical conductor. Convection occurs with heat. The heat during convection moves due to the liquid or gas that has the heated areas. Conduction also deals with the movement of heat. However, these parts are usually solid because they don’t move.
All of these terms have something in common and this is that they all have to do with heat. They either create heat or move heat in some form or fashion. The method that is involved when charges in an uncharged metal object are rearranged without direct contact with the charged object would be through induction.
Misrael | 403 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.967959 |
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May 5, 2016
# Homework Help: maths
Posted by claire philips on Saturday, May 12, 2007 at 11:07am.
which three statements are true?
a) if x= -10^4 then log 10 = -4
b)if x= 2^8 then log 2x = 8
c) log2 2= 4
d) if x= 3 then log10 3=x
e) log 10 256-2log 10 a/log 10 b
f)log 10 (a-b)= log 10 a/log 10 b
g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e
h) the gradient of the graph of y= e^x at x= 2is 2e | 190 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2016-18 | longest | en | 0.758663 |
http://computer-programming-forum.com/47-c-language/2b120028a01fa222.htm | 1,695,537,008,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00822.warc.gz | 9,518,260 | 7,965 | Convert four chars to a long and a long to four chars
Author Message
Convert four chars to a long and a long to four chars
if i have an array of four chars, how can i make it a long?
the reason i want to do it (transfering numbers on the net) is not c
related BUT MY QUESTION IS SO DONT "FLAME" ME
(btw, long = four bytes; char = 1 byte; in case thats platform specific)
what i want to do can be understood from this:
unsigned long thelong = 1002;
char thechars[4];
//how do i do this?
thechars = thelong
//and this
thelong = thechars
//replace the equals signs with an explanation please! :-)
thanks for all your help everyone
-joe
--
Posted via CNET Help.com
http://www.*-*-*.com/
Mon, 26 Jan 2004 17:30:17 GMT
Convert four chars to a long and a long to four chars
Quote:
> if i have an array of four chars, how can i make it a long?
> the reason i want to do it (transfering numbers on the net) is not c
> related BUT MY QUESTION IS SO DONT "FLAME" ME
It's a C question, worry not.
Quote:
> (btw, long = four bytes; char = 1 byte; in case thats platform specific)
It is indeed. As is "chars have 8 bits", and "the lowest addressed char
is the most significant part of the long".
Quote:
> what i want to do can be understood from this:
> unsigned long thelong = 1002;
> char thechars[4];
> //how do i do this?
> thechars = thelong
> //and this
> thelong = thechars
> //replace the equals signs with an explanation please! :-)
Bit by bit; or rather, byte by byte.
#include <limits.h>
unsigned long charsToLong( unsigned char *theChars )
{
return
((unsigned long) theChars[0] << 24)
| ((unsigned long) theChars[1] << 16)
| ((unsigned long) theChars[2] << 8)
| ((unsigned long) theChars[3])
;
}
void longToChars( unsigned long it, unsigned char *theChars )
{
theChars[0] = (it >> 24);
theChars[1] = (it >> 16);
theChars[2] = (it >> 8);
theChars[3] = it;
}
EXERCISES.
(a) bytes are CHAR_BITS wide [from limits.h]; revise the code so that
the shifts take this into account.
(b) longs need not be 4 bytes wide; revise the code to take this
into account. Is a loop needed? How likely is it we can do without
one? How good is your compiler at unrolling, and does it matter?
(c) the author of this code makes no warrenty as to its effectiveness.
Is his paranoia justified?
--
Chris "electric hedgehog" Dollin
C FAQs at: http://www.faqs.org/faqs/by-newsgroup/comp/comp.lang.c.html
Mon, 26 Jan 2004 18:04:04 GMT
Convert four chars to a long and a long to four chars
Quote:
> if i have an array of four chars, how can i make it a long?
> the reason i want to do it (transfering numbers on the net) is not c
> related BUT MY QUESTION IS SO DONT "FLAME" ME
> does anyone have an answer?
> (btw, long = four bytes; char = 1 byte; in case thats platform specific)
sizeof (char) is always 1, but CHAR_BIT may be greater than 8. The
long type must be at least 32 bits wide.
Quote:
> what i want to do can be understood from this:
> unsigned long thelong = 1002;
> char thechars[4];
Better make those unsigned chars.
Quote:
> //how do i do this?
> thechars = thelong
> //and this
> thelong = thechars
You could try mempcy() or something similar, but you might run into
endianness issues.
You could build on this code:
#include <stdio.h>
#include <limits.h>
int main(void)
{
unsigned long l, temp;
unsigned char c[sizeof l];
size_t i;
l = 0xabcdef01UL;
printf("l: %lx\n", l);
printf("c: ");
for (i = 0; i < sizeof l; i++)
{
temp = l & ((unsigned long) UCHAR_MAX << CHAR_BIT * i);
c[i] = temp >> (CHAR_BIT * i);
printf("%02x ", (unsigned int) c[i]);
}
putchar('\n');
l = 0;
for (i = 0; i < sizeof l; i++)
{
l |= c[i] << CHAR_BIT * i;
}
printf("l: %lx\n", l);
return 0;
Quote:
}
HTH,
Gergo
--
QOTD:
How can I miss you if you won't go away?
Mon, 26 Jan 2004 18:13:01 GMT
Convert four chars to a long and a long to four chars
Quote:
>if i have an array of four chars, how can i make it a long?
>the reason i want to do it (transfering numbers on the net) is not c
>related BUT MY QUESTION IS SO DONT "FLAME" ME
>(btw, long = four bytes; char = 1 byte; in case thats platform specific)
>what i want to do can be understood from this:
>unsigned long thelong = 1002;
>char thechars[4];
#include <limits.h>
unsigned long strlong(char *s) {
unsigned long r = 0; int i;
unsigned chars *S = (unsigned char*)s;
for (i=0; i<sizeof r && *S; i++) {
r = (r<<CHAR_BIT) | *S++;
}
return r;
}
char *longstr(unsigned long l) {
static char r[(sizeof l)+1];
char *R = r + (sizeof l)+1;
*R-- = 0;
while (R>r && l!=0) {
*R-- = l; l >>= CHAR_BIT;
}
return R;
}
on irix and gnu compilers strlong("xyzq")=='xyzq'
some compilers may reject multicharacter literals
or store the bytes in a different order
Mon, 26 Jan 2004 18:11:38 GMT
Convert four chars to a long and a long to four chars
Quote:
> if i have an array of four chars, how can i make it a long?
> the reason i want to do it (transfering numbers on the net) is not c
> related BUT MY QUESTION IS SO DONT "FLAME" ME
> does anyone have an answer?
> (btw, long = four bytes; char = 1 byte;
> in case thats platform specific)
It is platform specific.
Quote:
> what i want to do can be understood from this:
> unsigned long thelong = 1002;
> char thechars[4];
For starters, use unsigned char instead of char,
whenever you're manipulating bytes and or bits.
Quote:
> //how do i do this?
> thechars = thelong
> //and this
> thelong = thechars
Is the high byte of thelong, thechars[0] or thechars[3]?
--
pete
Mon, 26 Jan 2004 18:09:02 GMT
Convert four chars to a long and a long to four chars
Quote:
> unsigned long thelong = 1002;
> char thechars[4];
Make it an array of unsigned char.
Quote:
> //how do i do this?
> thechars = thelong
thechars[0] = (thelong >> 24) & 0xFF;
thechars[1] = (thelong >> 16) & 0xFF;
thechars[2] = (thelong >> 8) & 0xFF;
thechars[3] = (thelong >> 0) & 0xFF;
Quote:
> //and this
> thelong = thechars
thelong = ((unsigned long) thechars[0] << 24)
| ((unsigned long) thechars[1] << 16)
| ((unsigned long) thechars[2] << 8)
| ((unsigned long) thechars[3] << 0);
These are portable ways to encode an at most 32-bit unsigned
integer to big-endian format and decode it back. If the unsigned
long type of some platform has more than 32 bits, these
statements ignore the extras. Likewise, if unsigned char has
more than 8 bits, these statements leave the extra bits zero.
If you need little-endian format, swap the array indexes around.
Mon, 26 Jan 2004 18:04:17 GMT
Convert four chars to a long and a long to four chars
Quote:
> if i have an array of four chars, how can i make it a long?
> the reason i want to do it (transfering numbers on the net) is not c
> related BUT MY QUESTION IS SO DONT "FLAME" ME
> does anyone have an answer?
> (btw, long = four bytes; char = 1 byte; in case thats platform specific)
It is indeed platform-specific. The only thing you are guaranteed is
that chars are at least 8 bits but maybe more, and longs are at least 32
bits but maybe more.
Furthermore, you should be aware that not all platforms use the same
byte ordering:
16909060 (decimal) is represented by some platforms as 0x01020304 and by
others as 0x04030201, and those aren't the only two possibilities!
Quote:
> what i want to do can be understood from this:
> unsigned long thelong = 1002;
> char thechars[4];
> //how do i do this?
> thechars = thelong
> //and this
> thelong = thechars
> //replace the equals signs with an explanation please! :-)
The easiest way is to cheat. That you are using unsigned long makes
things easier. :-) But do make thechars unsigned as well:
unsigned long thelong = 1002;
unsigned char thechars[sizeof(long)] = {0}; /* = {0} zeroes out the
array */
unsigned char *p, *q;
size_t i;
p = (unsigned char *)&thelong;
q = thechars;
for(i = 0; i < sizeof thelong; i++)
{
*q++ = *p++;
Quote:
}
but this solution does depend upon byte ordering. You can get
byte-ordering-independent code fairly easily (or rather, you can
normalise the byte ordering), simply by exploring each bit:
#include <limits.h>
int main(void)
{
unsigned long thelong = 1002;
unsigned char thechars[sizeof(long)] = {0};
int bit = 0;
while(thelong > 0)
{
thechars[bit / CHAR_BIT] |= ((thelong & 1) << (bit % CHAR_BIT));
thelong >>= 1;
++bit;
}
return 0;
Quote:
}
Not tested, or even compiled, so be wary of loop boundary condition
bugs.
If this gives you the bits in the opposite order to the one you prefer,
simply change int bit = 0; to int bit = sizeof thelong, move ++bit to be
the first statement in the loop, and change it to --bit.
--
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
Mon, 26 Jan 2004 18:36:20 GMT
Convert four chars to a long and a long to four chars
I'm piggybacking, rather than responding to Richard. A bit of experimenting
sugested this:
static unsigned long charsToLong( unsigned char *them )
{
register unsigned long result = 0;
if (sizeof (unsigned long) > 0) result = (result << CHAR_BIT) | them[0];
if (sizeof (unsigned long) > 1) result = (result << CHAR_BIT) | them[1];
if (sizeof (unsigned long) > 2) result = (result << CHAR_BIT) | them[2];
if (sizeof (unsigned long) > 3) result = (result << CHAR_BIT) | them[3];
if (sizeof (unsigned long) > 4) result = (result << CHAR_BIT) | them[4];
if (sizeof (unsigned long) > 5) result = (result << CHAR_BIT) | them[5];
if (sizeof (unsigned long) > 6) result = (result << CHAR_BIT) | them[6];
if (sizeof (unsigned long) > 7) result = (result << CHAR_BIT) | them[7];
return result;
}
So long as sizeof(unsigned long) <= 8, which covers all the machines I
dare think about, this does it portably and with no loops. If your compiler
won't optimise away the sizeof tests, either turn up the optimisation
level a tad or get a better compiler.
Critiques requested.
--
Chris "now, where's that mask ... and the cloak ... and the ring ..." Dollin
C FAQs at: http://www.faqs.org/faqs/by-newsgroup/comp/comp.lang.c.html
Mon, 26 Jan 2004 19:19:34 GMT
Convert four chars to a long and a long to four chars
Quote:
> I'm piggybacking, rather than responding to Richard. A bit of experimenting
> sugested this:
> static unsigned long charsToLong( unsigned char *them )
> {
> register unsigned long result = 0;
> if (sizeof (unsigned long) > 0) result = (result << CHAR_BIT) | them[0];
> if (sizeof (unsigned long) > 1) result = (result << CHAR_BIT) | them[1];
> if (sizeof (unsigned long) > 2) result = (result << CHAR_BIT) | them[2];
> if (sizeof (unsigned long) > 3) result = (result << CHAR_BIT) | them[3];
> if (sizeof (unsigned long) > 4) result = (result << CHAR_BIT) | them[4];
> if (sizeof (unsigned long) > 5) result = (result << CHAR_BIT) | them[5];
> if (sizeof (unsigned long) > 6) result = (result << CHAR_BIT) | them[6];
> if (sizeof (unsigned long) > 7) result = (result << CHAR_BIT) | them[7];
> return result;
> }
> So long as sizeof(unsigned long) <= 8, which covers all the machines I
> dare think about, this does it portably and with no loops. If your compiler
> won't optimise away the sizeof tests, either turn up the optimisation
> level a tad or get a better compiler.
> Critiques requested.
Adding a while loop after the last if could catch really humungous ul's,
at the expense of one redundant test on the more 'normal' architectures.
--
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
Mon, 26 Jan 2004 19:50:55 GMT
Convert four chars to a long and a long to four chars
Quote:
>> I'm piggybacking, rather than responding to Richard. A bit of experimenting
>> sugested this:
>> static unsigned long charsToLong( unsigned char *them )
>> {
>> register unsigned long result = 0;
>> if (sizeof (unsigned long) > 0) result = (result << CHAR_BIT) | them[0];
>> if (sizeof (unsigned long) > 1) result = (result << CHAR_BIT) | them[1];
>> if (sizeof (unsigned long) > 2) result = (result << CHAR_BIT) | them[2];
>> if (sizeof (unsigned long) > 3) result = (result << CHAR_BIT) | them[3];
>> if (sizeof (unsigned long) > 4) result = (result << CHAR_BIT) | them[4];
>> if (sizeof (unsigned long) > 5) result = (result << CHAR_BIT) | them[5];
>> if (sizeof (unsigned long) > 6) result = (result << CHAR_BIT) | them[6];
>> if (sizeof (unsigned long) > 7) result = (result << CHAR_BIT) | them[7];
>> return result;
>> }
>> So long as sizeof(unsigned long) <= 8, which covers all the machines I
>> dare think about, this does it portably and with no loops. If your compiler
>> won't optimise away the sizeof tests, either turn up the optimisation
>> level a tad or get a better compiler.
>> Critiques requested.
> Adding a while loop after the last if could catch really humungous ul's,
> at the expense of one redundant test on the more 'normal' architectures.
Like:
if (sizeof (unsigned long) > 8)
{
size_t count = sizeof (unsigned long) - 8;
int index = 8;
while (count--) result = (result << CHAR_BIT) | them[index++];
}
??
--
Chris "electric hedgehog" Dollin
C FAQs at: http://www.faqs.org/faqs/by-newsgroup/comp/comp.lang.c.html
Mon, 26 Jan 2004 19:58:53 GMT
Convert four chars to a long and a long to four chars
Quote:
> Is the high byte of thelong, thechars[0] or thechars[3]?
If it doesn't matter (though I suspect it does)
then this would be the fatsest way:
/* BEGIN new.txt */
thelong = 0XAABBCCDD;
The intial value of thelong is AABBCCDD
*(long*)thechars = thelong;
thechars[0] DD
thechars[1] CC
thechars[2] BB
thechars[3] AA
theotherlong = *(long*)thechars;
The value of the other long is AABBCCDD
/* END new.txt */
/* BEGIN new.c */
#include <stdio.h>
#include <stdlib.h>
#define thechars (tchars.tchar)
int main(void)
{
long thelong, theotherlong;
union{
long lin;
unsigned char tchar[sizeof(long)];
}tchars;
size_t index;
char *fn = "new.txt";
FILE *fp = fopen(fn, "wt");
if(!fp){
fprintf(stderr, "fopen(%s) error\n", fn);
exit(EXIT_FAILURE);
}
fprintf(fp, "/* BEGIN %s */\n\n", fn);
thelong = 0XAABBCCDD;
fprintf(fp, "thelong = 0XAABBCCDD;\n");
fprintf(fp, "The intial value of thelong is %X\n\n", thelong);
*(long*)thechars = thelong;
fprintf(fp, "*(long*)thechars = thelong;\n");
for (index = 0; index != sizeof(long); ++index)
fprintf(fp, "thechars[%lu] %X\n",
(unsigned long)index, thechars[index]);
theotherlong = *(long*)thechars;
fprintf(fp, "\ntheotherlong = *(long*)thechars;\n");
fprintf(fp, "The value of the other long is %X\n", theotherlong);
fprintf(fp, "\n/* END %s */\n", fn);
fclose(fp);
return 0;
Quote:
}
/*
btw, long = four bytes; char = 1 byte; in case thats platform specific)
what i want to do can be understood from this:
unsigned long thelong = 1002;
char thechars[4];
//how do i do this?
thechars = thelong
//and this
thelong = thechars
//replace the equals signs with an explanation please! :-)
thanks for all your help everyone
-joe
*/
/* END new.c */
--
pete
Mon, 26 Jan 2004 20:22:58 GMT
Convert four chars to a long and a long to four chars
: static unsigned long charsToLong( unsigned char *them )
: {
: register unsigned long result = 0;
: if (sizeof (unsigned long) > 0) result = (result << CHAR_BIT) | them[0];
: if (sizeof (unsigned long) > 1) result = (result << CHAR_BIT) | them[1];
: if (sizeof (unsigned long) > 2) result = (result << CHAR_BIT) | them[2];
: if (sizeof (unsigned long) > 3) result = (result << CHAR_BIT) | them[3];
: if (sizeof (unsigned long) > 4) result = (result << CHAR_BIT) | them[4];
: if (sizeof (unsigned long) > 5) result = (result << CHAR_BIT) | them[5];
: if (sizeof (unsigned long) > 6) result = (result << CHAR_BIT) | them[6];
: if (sizeof (unsigned long) > 7) result = (result << CHAR_BIT) | them[7];
: return result;
: }
:
: So long as sizeof(unsigned long) <= 8, which covers all the machines I
: dare think about, this does it portably and with no loops. If your compiler
: won't optimise away the sizeof tests, either turn up the optimisation
: level a tad or get a better compiler.
Well, if your compiler won't automatically unroll a loop with a small
constant number of iterations, either turn up the optimisation level a
tad or get a better compiler. :)
D:\>type catol.c
#include <limits.h>
unsigned long catol(const unsigned char *bp) {
int sz = sizeof(unsigned long);
unsigned long ul = 0;
do {
ul = (ul << CHAR_BIT) | *bp++;
} while (--sz);
return ul;
Quote:
}
D:\>uname -a
CYGWIN_NT-5.0 AVALON 1.3.2(0.39/3/2) 2001-05-20 23:28 i686 unknown
D:\>gcc --version
3.0
D:\>gcc -O3 -fomit-frame-pointer -funroll-loops catol.c -S
D:\>type catol.s
.file "catol.c"
.text
.align 16
.globl _catol
.def _catol; .scl 2; .type 32; .endef
_catol:
movl 4(%esp), %ecx
movzbl (%ecx), %eax
movzbl 1(%ecx), %edx
sall \$8, %eax
orl %edx, %eax
movzbl 2(%ecx), %edx
sall \$8, %eax
orl %edx, %eax
movzbl 3(%ecx), %edx
sall \$8, %eax
orl %edx, %eax
ret
-- Mat.
Mon, 26 Jan 2004 21:04:55 GMT
Convert four chars to a long and a long to four chars
Quote:
> >> Critiques requested.
> > Adding a while loop after the last if could catch really humungous ul's,
> > at the expense of one redundant test on the more 'normal' architectures.
> Like:
> if (sizeof (unsigned long) > 8)
> {
> size_t count = sizeof (unsigned long) - 8;
> int index = 8;
> while (count--) result = (result << CHAR_BIT) | them[index++];
> }
> ??
Yes, exactly like that. :-)
--
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
Mon, 26 Jan 2004 21:01:08 GMT
Convert four chars to a long and a long to four chars
Quote:
> long thelong, theotherlong;
> fprintf(fp, "The intial value of thelong is %X\n\n", thelong);
> fprintf(fp, "thechars[%lu] %X\n",
> (unsigned long)index, thechars[index]);
> fprintf(fp, "The value of the other long is %X\n", theotherlong);
For the benefit of fprintf, those lines should be changed to:
long unsigned thelong, theotherlong;
fprintf(fp, "The intial value of thelong is %lX\n\n", thelong);
fprintf(fp, "thechars[%lu] %X\n",
(long unsigned)index, (unsigned)thechars[index]);
fprintf(fp, "The value of the other long is %lX\n", theotherlong);
--
pete
Mon, 26 Jan 2004 21:02:39 GMT
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Relevant Pages | 5,638 | 18,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-40 | latest | en | 0.806643 |
http://www.mathwarehouse.com/calculus/derivatives/how-to-find-equations-of-tangent-lines.php | 1,537,776,653,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160233.82/warc/CC-MAIN-20180924070508-20180924090908-00559.warc.gz | 362,149,985 | 8,668 | How to Find Equations of Tangent Lines and Normal Lines
# How to Find Equations of Tangent Lines and Normal Lines
### Quick Overview
• To find the equation of a line you need a point and a slope.
• The slope of the tangent line is the value of the derivative at the point of tangency.
• The normal line is a line that is perpendicular to the tangent line and passes through the point of tangency.
### Examples
##### Example 1
Suppose $$f(x) = x^3$$. Find the equation of the tangent line at the point where $$x = 2$$.
Step 1
Find the point of tangency.
Since $$x=2$$, we evaluate $$f(2)$$.
$$f(2) = 2^3 = 8$$
The point is $$(2,8)$$.
Step 2
Find the value of the derivative at $$x = 2$$.
$$f'(x) = 3x^2\longrightarrow f'(2) = 3(2^2) = 12$$
The the slope of the tangent line is $$m = 12$$.
Step 3
Find the point-slope form of the line with slope $$m = 12$$ through the point $$(2,8)$$.
\begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 8 & = 12(x-2) \end{align*}
$$y - 8 = 12(x-2)$$
For reference, here is the graph of the function and the tangent line we just found.
##### Example 2
Suppose $$f(x) = x^2 - x$$. Find the equation of the tangent line with slope $$m = -3$$.
Step 1
Find the derivative.
$$f'(x) = 2x -1$$
Step 2
Find the $$x$$-value where $$f'(x)$$ equals the slope.
\begin{align*} f'(x) & = 2x -1\\[6pt] -3 & = 2x -1\\[6pt] -2 & = 2x\\[6pt] x & = -1 \end{align*}
Step 3
Find the point on the function where $$x = -1$$.
$$f(-1) = (-1)^2 - (-1) = 1 + 1 = 2$$
The point is $$(-1, 2)$$.
Step 4
Find the equation of the line through the point $$(-1,2)$$ with slope $$m=-3$$.
\begin{align*} y -y_1 & = m(x-x_1)\\[6pt] y - 2 & = -3(x - (-1))\\[6pt] y - 2 & = -3(x+1) \end{align*}
$$y - 2 = -3(x+1)$$
For reference, here's the graph of the function and the tangent line we just found.
### Tangent Lines to Implicit Curves
The procedure doesn't change when working with implicitly defined curves.
##### Example 3
Suppose $$x^2 + y^2 = 16$$. Find the equation of the tangent line at $$x = 2$$ for $$y>0$$.
Step 1
Find the $$y$$-value of the point of tangency.
\begin{align*} \blue{x^2} + y^2 & = 16\\[6pt] \blue{2^2} + y^2 & = 16\\[6pt] \blue{4} + y^2 & = 16\\[6pt] y^2 & = 12\\[6pt] y & = \pm\sqrt{12}\\[6pt] y & = \pm\sqrt{4\cdot 3}\\[6pt] y & = \pm2\sqrt 3 \end{align*}
Since the problem states we are interested in $$y>0$$, we use $$y = 2\sqrt 3$$.
The point of tangency is $$(2, 2\sqrt 3)$$.
Step 2
Find the equation for $$\frac{dy}{dx}$$.
Since the equation is implicitly defined, we use implicit differentiation.
\begin{align*} 2x + 2y\,\frac{dy}{dx} & = 0\\[6pt] 2y\,\frac{dy}{dx} & = -2x\\[6pt] \frac{dy}{dx} & = -\frac{2x}{2y}\\[6pt] \frac{dy}{dx} & = -\frac x y \end{align*}
Step 3
Find the slope of the tangent line at the point of tangency.
At the point $$(2,2\sqrt 3)$$, the slope of the tangent line is
\begin{align*} \frac{dy}{dx}\bigg|_{(\blue{2},\red{2\sqrt 3})} & = -\frac {\blue 2} {\red{2\sqrt 3}}\\[6pt] & = -\frac 1 {\sqrt 3}\\[6pt] & = -\frac 1 {\sqrt 3}\cdot \blue{\frac{\sqrt 3}{\sqrt 3}}\\[6pt] & = -\frac{\sqrt 3} 3 \end{align*}
The slope of the tangent line is $$m = -\frac{\sqrt 3} 3$$.
Step 4
Find the equation of the tangent line through $$(2,2\sqrt 3)$$ with a slope of $$m=-\frac{\sqrt 3} 3$$.
At the point $$(2,2\sqrt 3)$$, the slope of the tangent line is
\begin{align*} y - y_1 & = m(x-x_1)\\[6pt] y - 2\sqrt 3 & = -\frac{\sqrt 3} 3(x-2) \end{align*}
The equation of the tangent line is $$y - 2\sqrt 3 = -\frac{\sqrt 3} 3(x-2)$$
For reference, the graph of the curve and the tangent line we found is shown below.
### Normal Lines
Suppose we have a a tangent line to a function. The function and the tangent line intersect at the point of tangency. The line through that same point that is perpendicular to the tangent line is called a normal line.
Recall that when two lines are perpendicular, their slopes are negative reciprocals. Since the slope of the tangent line is $$m = f'(x)$$, the slope of the normal line is $$m = -\frac 1 {f'(x)}$$.
##### Example 4
Suppose $$f(x) = \cos x$$. Find the equation of the line that is normal to the function at $$x = \frac \pi 6$$.
Step 1
Find the point on the function.
$$f\left(\frac \pi 6\right) = \cos \frac \pi 6 = \frac{\sqrt 3} 2$$
The point is $$\left(\frac \pi 6, \frac{\sqrt 3} 2\right)$$.
Step 2
Find the value of the derivative at $$x = \frac \pi 6$$.
$$f'(x) = -\sin x\longrightarrow f'\left(\frac \pi 6\right) = -\sin\frac\pi 6 = -\frac 1 2$$
The slope of the tangent line is $$m = -\frac 1 2$$. Since we are looking for the line that is perpendicular to the tangent line, we want to use $$m = 2$$.
Step 3
Find the equation of the line through the point $$\left(\frac \pi 6, \frac{\sqrt 3} 2\right)$$ with a slope of $$m =2$$.
\begin{align*} y -y_1 & = m(x-x_1)\\[6pt] y - \frac{\sqrt 3} 2 & = 2\left(x - \frac \pi 6\right) \end{align*}
The line normal to the function at $$x = \frac \pi 6$$ is $$y - \frac{\sqrt 3} 2 = 2\left(x - \frac \pi 6\right)$$. | 1,871 | 5,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2018-39 | latest | en | 0.781311 |
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FratBro23
Category:
Other
Price: \$10 USD
Question description
Evaluation of Correlation
Data are gathered regarding the length of tenure top
executives have at a major corporation and whether those executives have been
divorced. The HR department is
evaluating this data to drive decision-making in regard to their hiring
process. The data for 8 executives is:
Tenure Divorce
9.0 No
9.5 No
11.0 Yes
11.5 Yes
10.0 Yes
9.75 No
10.0 No
10.25 Yes
In a 3-5 page paper, answer the following questions to
analyze the data. Include clearly
labeled calculations, if applicable.
Calculations conducted in Excel must be copied and pasted into the Word
document.
What’s the most appropriate procedure for evaluating the
relationship between tenure and divorce?
What is the correlation and how can it be interpreted in
terms of magnitude, direction and practical importance?
How much of whether executives have been divorced can be
accounted for by their length of tenure with the organization?
How much of tenure can be explained by whether there has
been a divorce?
Make a logical argument for why lengthy tenure may be
causing divorce.
Make another logical argument for why divorce may be causing
lengthy tenure.
Calculations conducted in Excel must be copied and pasted
into the Word doc. Clearly label all calculations.
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# Ch4_2 - 1 Section 4.2 Elementary Statistics Elementary...
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Next Previous Section 4.2: Elementary Statistics 11 ® Elementary Statistics ® Sample Distributions ® Mean ® Median ® Standard Deviation ® Sample size effects Section 4.2: Elementary Statistics
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Next Previous Section 4.2: Elementary Statistics 22 ® Usually, only a sample of a population is measured ® The results are then taken to represent the entire population ® Example: Determine the amount of lumber in a wooded area ® Method: ® Measure a few trees in a sample area to obtain an approximate average diameter and height ® Count the number of trees in sample area ® Use results to estimate the total wood in the sample area ® Extrapolate results to entire forest The usefulness of statistics
Next Previous Section 4.2: Elementary Statistics 33 ® Magnitude: Size of a variable ® Observation: Result of a single measurement ® Parent Population: Collection of all possible measurements ® Sample: A finite group of measurements ® Error: Deviation from the true value ® Accuracy: Closeness to the true value ® Precision: Closeness of a grouping to each other ® Precision k Accuracy ® Maybe you can measure precisely to D 0.1 mm ® But you could be quite wrong if ruler is 25 mm too short Some useful statistical terms
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Next Previous Section 4.2: Elementary Statistics 44 Histogram: Frequency of occurrence vs. Value Example: Grades on a quiz grade # of grades value 4 2 5 4 6 7 7 8 8 6 9 3 10 2 32 total grades Frequency Value The distribution of values grade values # of grades
Next Previous Section 4.2: Elementary Statistics 55 Value Frequency
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Ch4_2 - 1 Section 4.2 Elementary Statistics Elementary...
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Ask a homework question - tutors are online | 529 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.806765 |
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a man travelled a distance of 22km in one hour partly in a car at a speed of 30 km/hour and partly on a motorcycle at 18km/hour. find the distance travelled by him in the car
1. time in car ---- t hrs
time on motorcycle --- 1 - t hrs
30t + 18(1 - t) = 22
solve for t, sub into 30t to get the distance in km
posted by Reiny
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https://www.gurufocus.com/term/tax/GLNG/Tax+Expense/Golar+LNG%252C+Ltd | 1,516,448,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889567.48/warc/CC-MAIN-20180120102905-20180120122905-00440.warc.gz | 898,053,883 | 40,591 | Switch to:
Golar LNG Ltd (NAS:GLNG) Tax Expense: \$1.5 Mil (TTM As of Sep. 2017)
Golar LNG Ltd's tax expense for the months ended in Sep. 2017 was \$0.4 Mil. Its tax expense for the trailing twelve months (TTM) ended in Sep. 2017 was \$1.5 Mil.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Golar LNG Ltd Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Tax Expense 2.77 -3.40 -1.11 -3.05 -0.59
Golar LNG Ltd Quarterly Data
Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Sep17 Tax Expense 0.25 0.45 0.19 0.46 0.42
Calculation
Tax paid by the company. It is computed in by multiplying the income before tax number, as reported to shareholders, by the appropriate tax rate. In reality, the computation is typically considerably more complex due to things such as expenses considered not deductible by taxing authorities ("add backs"), the range of tax rates applicable to various levels of income, different tax rates in different jurisdictions, multiple layers of tax on income, and other issues.
Tax Expense for the trailing twelve months (TTM) ended in Sep. 2017 was 0.45 (Dec. 2016 ) + 0.189 (Mar. 2017 ) + 0.458 (Jun. 2017 ) + 0.423 (Sep. 2017 ) = \$1.5 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the long run, income before tax and taxable income will likely be more similar than they are in any given period. If the one is less in earlier years, then it will be greater in later years. Deferred taxes will reverse themselves in the long run and in total will zero out, unless there is something like a change in tax rates in the intervening period. A deferred tax payable results from a tax break in the early years and will reverse itself in later years; a deferred tax receivable results from more taxes being paid in early years than the tax expense reported to shareholders and will again reverse itself in later years. The deferred tax amount is computed by estimating the amount and the timing of the reversal and multiplying that by the appropriate tax rates.
Related Terms | 571 | 2,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | latest | en | 0.939632 |
https://math.stackexchange.com/questions/3977394/proof-verification-if-f-omega-to-omega-has-two-fixed-points-is-conformal-a | 1,618,808,573,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00395.warc.gz | 467,377,561 | 36,877 | # Proof verification: if $f:\Omega\to \Omega$ has two fixed points, is conformal and $f^n=\text{Id}$
What I want to prove is exactly as stated it the title:
Let $$f:\Omega\to \Omega$$ be a holomorphic function and $$\Omega\subsetneq \mathbb{C}$$ a connected open subset of $$\mathbb{C}$$ such that its complement contains at least two point. Assume also that $$\exists a,b\in \Omega:f(a)=a\neq b=f(b)$$. Prove that $$f$$ is conformal and there exists a natural number $$n:f^{[n]}=f\circ \dots\circ f=\text{Id}$$
I proved it as follows: since $$\Omega$$ avoids at least two points, it is hyperbolic (i.e. its universal cover is the unit disc $$\mathbb{D}$$). Thus we have the following diagram ($$\pi$$ is the covering map):
$$\require{AMScd} \begin{CD} \mathbb{D} @>{\tilde f}>> \mathbb{D}\\ @VVV @VVV \\ \Omega @>{f}>> \Omega; \end{CD}$$ where $$\tilde f$$ is the lifting of $$f\circ \pi$$ such that $$\tilde f(\tilde b)=\tilde b$$ with $$\pi(\tilde b)=b$$. Let $$\phi$$ be an automorphism of $$\mathbb{D}$$ sending $$\tilde b$$ to $$0$$ and viceversa. We define $$g:=\phi \tilde f\phi$$. By Schwarz's lemma, $$|g'(0)|\le 1$$ and it is $$1$$ iff $$g$$ is a rotation. Suppose $$|g'(0)|<1$$. By Montel's theorem we can extract a convergent sequence of $$g^{[n_k]}$$, which is equal to $$\phi \tilde{f}^{[n_k]}\phi$$. Since $$|g'(0)|<1$$, $$0$$ is an attracting fixed point for a neighbourhood of itsel, which implies that $$\lim g^{[n_k]}$$ is constant in a neighbourhood of $$0$$ and so it is constant on $$\mathbb{D}$$. Thus, $$\lim \tilde{f}^{[n_k]}=\lim \phi g^{[n_k]} \phi=\phi \left(\lim g^{[n_k]}\right)\phi\equiv \phi(0)=\tilde b$$. However, by definition $$\tilde{f}(\pi^{-1}(a))\subseteq \pi^{-1}(a)$$, and since this is a discrete set not containing $$\tilde b$$, we get a contradiction. So $$|g'(0)|=1$$, and thus $$g$$ is a rotation. Suppose its angle is not a rational multiple of $$\pi$$. Then $$\overline{\{g^{n}(z_0)\}}=\{z:|z|=|z_0|\}$$. This is true in particular for every element of $$\phi(\pi^{-1}(a))$$. This implies that $$\overline{\pi^{-1}(a)}$$ contains at least one circle, contradicting the fact that $$\pi^{-1}(a)$$ is discrete. Now the result follows, since $$\pi=\pi \tilde f^{n}= f^n\pi$$, and inverting locally we get $$f^{n}=Id$$, which implies by the identity principle $$f^{n}=Id$$
Is my proof correct? Does this theorem have a name? I have seen something similar (a holomorphic map $$f:\Omega\to \Omega$$ on a bounded region, with two fixed points is conformal) being referred to as Cartan's theorem, but I am not sure whethere the attribution is correct. | 864 | 2,597 | {"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": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-17 | latest | en | 0.827014 |
https://www.instructables.com/id/Learn-to-talk-Binary-The-efficienthard-way/ | 1,569,090,117,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574588.96/warc/CC-MAIN-20190921170434-20190921192434-00513.warc.gz | 877,819,114 | 19,314 | # Learn to Talk Binary (The Efficient/hard Way)
3,998
64
6
In this tutorial, I will be teaching you how to talk... IN BINARY! Hope you Enj01!
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Take a Look at the Alphabet.
So, There are uppercase and lowercase letters, but for now we will only look at uppercase.
A = 0100 0001
B = 0100 0010
C = 0100 0011
At first glance, we can already recognize a pattern. The first 4 digits we can ignore for now. The last 4 digits go 0001, 0010, 0011, 0100, etc.
I would recommend trying to memorize a few benchmarks, for example:
A = 0001
B = 0010
D = 0100
H = 1000
Example: "A Bird Died Here".
Then, at P the first 4 digits change from 0100 to 0101! Now some other benchmarks would be:
P = 0000
Q = 0001
R = 0010
T = 0100
X = 1000
## Step 2: Now, Just Practice!
All you have to do now is PRACTICE. If you truly want to get good at this (or anything for that matter), you just have to keep on practicing. Talk to your friends in binary! Teach them how to talk in binary! Pass notes around in class in binary!
Also, try decoding this (Actually try, I know you have that "binary to text" tab ready).
01000011 01001111 01001110 01000111 01010010 01000001 01010100 01010011 00100001 00100000 01011001 01001111 01010101 00100000 01000001 01010010 01000101 00100000 01001110 01001111 01010111 00100000 01001111 01000110 01000110 01001001 01000011 01001001 01000001 01001100 01001100 01011001 00100000 01000001 00100000 01000010 01001001 01001110 01000001 01010010 01011001 00100000 01000111 01010101 01011001 00100001 00100000 01001000 01001111 01010000 01000101 00100000 01011001 01001111 01010101 00100000 01001100 01001001 01001011 01000101 00100000 01010100 01001000 01001001 01010011 00100000 01010100 01010101 01010100 01001111 01010010 01001001 01000001 01001100 00100001
## Recommendations
• ### Internet of Things Class
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## 6 Discussions
011011000110111101101100001000000110100100100000011000110110100001100101011000010111010001100101011001000010000001100001011011100110010000100000011101010111001101100101011001000010000001110100011010000110010100100000011000110110111101101110011101100110010101110010011101000110010101110010001000000110001001110101011101000010000001101001001000000111011101101001011011000110110000100000011101000111001001111001
If any of you think I can improve this instructable, tell me! Like that you can help me help you. Also, if you have any questions feel free to ask!
This is cool! I'm going to a Binary Ball in a couple weeks and I should totally learn to do this! Thank you for sharing. | 893 | 2,703 | {"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} | 4.28125 | 4 | CC-MAIN-2019-39 | latest | en | 0.643979 |
https://www.physicsoverflow.org/38212/determinant-of-the-dirac-operator-and-discrete-eigenvalues | 1,716,004,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00413.warc.gz | 841,023,609 | 26,333 | # Determinant of the Dirac operator and discrete eigenvalues
+ 2 like - 0 dislike
2020 views
Consider the determinant of the euclidean form of the Dirac operator:
$$\text{det}(iD), \quad iD = i\gamma_{\mu}(\partial_{\mu}+A_{\mu})$$
It is an elliptic operator, so has a discrete spectrum on compact manifolds. The euclidean manifold $R^{4}$ is not a compact manifold. However, people typically writes
$$\text{det}(iD) = \prod_{i = 1}^{\infty}\lambda_{i},$$
where $\lambda_{i}$ defines the $i$th eigenvalue:
$$iD\psi_{i} = \lambda_{i}\psi_{i}$$
Why can they do this?
asked Jan 12, 2017
+ 2 like - 0 dislike
If it is done in the framework of QFT, it is usually quite a formal expressions which then must be regularized in some way. The same story as for other determinants, traces, etc.
answered Jan 12, 2017 by (904 points)
@NAME_XXX It may be a kind of philosophic question. How do we know that the Path Integral is correct? Basically, it is because it works, and in all the experimentally verified situations it works correctly with enormous precision.
One not only replaces the manifold by a compact one but afterwards takes the infrared limit (infinite radius) that restores the original manifold. Under this limit the spectrum becomes continuous. of course this is not fully rigorous, and for massless fields unexpected things may happen....
@AndreyFeldman : but is there some reason which allows us to do that? All of the theory quantities must be projected on $S^{4}$ coordinates, and corresponding action must be invariant.
@ArnoldNeumaier : but is there some projection rule? I.e., if I want to construct the theory on $S^{4}$, I need to construct corresponding quantities defined on $S^{4}$.
@NAME_XXX In general, compactification of $\mathbb{R}^4$ to get $S^4$ is subtle -- it may give masses to massless fields, etc. If you are looking for some universal recipe, I'm afraid that it doesn't exist.
@NAME_XXX Usually one replaces a non-compact space by a compact one, which gives the former in the limit of infinite radius (say, $\mathbb{R}$ may be replaced by $S^1$), evaluates the index on this manifold, and then tends the radius to infinity.
Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor) Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification | 793 | 3,221 | {"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} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.895478 |
https://chat.stackoverflow.com/transcript/10?m=53592294 | 1,656,125,325,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00524.warc.gz | 215,560,759 | 7,183 | 1:57 AM
@Lapys how the f is ' '*13 160 i thought it to be 416 cuz it's like 32*13
7 hours later…
9:02 AM
@Agent_A (32 * 13) % 256
9:18 AM
Mod 256? But why
Oh ascii upper bound
@Lapys I think I know C better than whoever wrote that crap. Let's start with question 1. The correct answer is "at least 2", but we can't really say any more than that. The size of each object is always at least 1. But we could have a system with 16-bit bytes, in which case both `char` and `int` can be 1 byte apiece, giving a size of 2. Bytes are more often 8 bits, in which case the minimum size is 3, as `int` is only required to have a range from -32767 to +32767.
Skipping forward to your question (3, I guess), again his answer is almost complete nonsense. First of all, `char` is unique among integer types in one respect: other integer types are always signed by default. `char` may be either signed or unsigned by default. If it's unsigned, then the overflow produces defined results, but we don't know for sure what they are without knowing how many bits there are in a char--which has to be at least 8, but could be more.
If it's signed, the result is simply undefined. The answer he gives is what you'd expect on with a signed, 8-bit, 2's complement char. But in general, it's just plain wrong.
Oh, for question 2, the correct answer also depends. His assumes that sizeof(int) == 4, and `sizeof(short) == 2`, so the result will be 0. That's fairly common, but not required. It's also possible to have sizeof(int) == sizeof(short) == 2, in which case the result would be true (1). And this isn't purely an abstract possibility either--at one time it was quite common.
9:35 AM
> His assumes that `sizeof(int) == 4`, and `sizeof(short) == 2`
Have you read the actual anwers? I think that point is addressed in the explanations
Or are you just going from the proposed answers?
@Morwenn Guilty. But the fact that he's asking stupid questions doesn't strike me as promising. Looking through the rest, literally none of his questions has a really good answer given.
Well, having people answer "I don't know" to get shit right is bad
Because we clearly know but there isn't the answer we expect
@Morwenn But even looking through his explanations, they're still semi-poor. He gives one explanation for why we don't know the result in each case, but in some of them there are two or three different problems, of which he really explains only one.
I only skimmed through tbh
Didn't care enough to get angry at it
@Morwenn Nothing is so frustrating as people who think they know everything. Especially to those of us who really do. :-)
9:48 AM
I know that I know nothing =w=
@Morwenn I'm not smart enough to realize I'm an ignorant putz.
It's not about being smart or not, it's about preserving yourself :')
@Morwenn Hmm...not sure what preservation would have to do with it. Most people who know me seem to think I'm actually pretty well preserved for my age, but I doubt that really means much.
2 hours later…
11:48 AM
`Top soil, free to good home!`
I can understand if you are giving out your pet and want the pet to be given to a good household, but do the soil or used timber (with nails still in it) need to go to a good home as well?
If I was in the mood, I could troll the website with 'Cockroaches, free to good homes! '
12:29 PM
Didn't know raspbian have mathematica installed.
1:19 PM
@JerryCoffin I could actually see it getting more common again too. AI stuff seems to thrive off of 16bit values. So I could absolutely see an environment where everything is 16bit at the low end. Or someone could pull an Alpha AXP and just make `int` 64bit because why have hardware to do math at partial registers sizes.
3 hours later…
3:51 PM
Oh no, I finished "Mother of Learning". I think I'll have to actually work now :(
I should also give the author some money.
4 hours later…
7:40 PM
hello
I've been patching random CMakefiles for 6 hours now. What is my life
also, why do all the ML folks stack frameworks on top of frameworks and name them so confusingly. Why would I need to know that onnxruntime 1.9.0 is not compilable with onnx 1.9.0
8:23 PM
Doing CMakeFiles into ML vicinity? I remember that as a kind of torture
I was just being dumb, OBVIOUSLY I needed to compile onnxruntime 1.8.2 with onnx 1.9.0
lol
I could've spend 10 minutes looking for readmes, but why would I do that when I can waste a whole work-day
We need a word like ornithologist, but not for people who watch birds as a hobby, for people who watch compiler output for a hobby
9:05 PM
@PeterT Patching CMake files? Your life is hell.
3 hours later…
11:52 PM
would anybody like to discuss the one true god
Is it Bjarne? | 1,229 | 4,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-27 | longest | en | 0.973308 |
http://www.jiskha.com/display.cgi?id=1332301475 | 1,498,511,287,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320865.14/warc/CC-MAIN-20170626203042-20170626223042-00599.warc.gz | 580,306,840 | 3,698 | # math
posted by .
pei fen walked a distance of 526 m.then she walked for another 5minutes at a speed of 70m/min.she took a total of 12 min to walk the entire distance.
A.find the total distance that pei fen walked
b.what was pei fen's average speed for the whole journey?
• math -
total distance = 526 + 5 * 70
avg speed = total distance / 12 min
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http://www.ta-teachers.eu/teacher-diaries-science/aquiring-an-understanding-of-proportion-and-its-applications-part-1.html | 1,701,727,681,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00688.warc.gz | 81,108,604 | 11,865 | PART 1
This task was developed because I realised students could not progress in their 'scientific method strategies' without independent analysis skills. Both myself and the maths teacher want the students to acquire a working understanding of proportion and how it's used in maths, sciences and in general. To develop the language and visuals (diagrams etc) for describing proportional relationships. To extend/connect this understanding to fractions, ratios and percentages. To extend/connect and extend this understanding to straight line graphs which go through the origin.
In this task at the final stage, looking at any one variable (parameter) will not work no matter how many variables you check out. They will ultimately need to look at the relationship between 2 variables.
STEP 1: Increasing room for learning...Recognising proportion (3 class periods of 45 mins)
Starting point: SORTING GAME: MATERIALS: Lots of pictures of young/adult animals (dogs, cats, elephants, rhino, birds etc), young/adult people, young/adult human and animal cartoons and young/adult human skeletons. Also: Pencils, rubbers, rulers, plain and graph paper.
• Students (13) initially working on one large table asked to sort all pictures into groups (emphasis on plural). No other instructions. Sorting/no. of groups is entirely up to them. Each time a set of groups was completed the students would have to explain their choices and legitimacy of their groupings.
Initially they chose 8 groups which appeared to work well. They spoke freely of variables and values and said how there were hundreds of variables if you included all the pictures. However, upon close examination certain members of one group could also be included in other groups, due to the group descriptions. e.g. a cartoon puppy could belong to both 'cartoons' and 'dogs'. Therefore I argued with them about their descriptions. Sometimes the groups could work if they adjusted their descriptions. Idea here is to focus students on making sure their chosen description really does apply.
• Students allowed to reorganise groupings and refine definitions. Examples are here.
Went through several rotations of this until eventually they were able to confidently recognise/reorganise and define groups that worked. This is an important step so that they understand the point where the task becomes 'impossible' for them.
• Then I asked for all pictures to be sorted into 2 groups only! (Not impossible yet - I am looking for students to reach 2 groups of 'young' vs 'adult').
This caused some protest, so at this point I divided the class into 2 groups (boys 5 and girls 8 for a change) to allow fresh working realtionships.
Students pretty quickly came up with many groups of 2 e.g. 'celebrities' & 'non-celebrities', 'male' & 'female', 'clothed' (in any way at all) & 'naked', 'birds' & 'mammals' and so on. I kept going around between the groups and checking if they worked (though they were very good at this by now). Then I would say ok now find another criteria that gives only 2 groups (e.g. with the birds v mammals I simply removed the 2 pictures of birds). The students treated this part as a game until they couldn't come up with any more groups.
Just as I thought I'd have to direct the finding of the groups 'adult' & 'young' both groups of students came up with it themselves as in two versions - A. 'kids' and 'adults' and B. 'cute' and 'not cute' which was redefined and 'grown-ups' and 'babies'.
STEP 1 - 2: NOW Take away the fish... (now make the task impossible)
Now they had the idea for 2 groups BUT for the first time they could NOT express/define the criteria by which the pictures were grouped. They could not legitimise/explain their choices in any way or form. I asked HOW DO YOU KNOW this pic (holding pic of baby rhino/puppy in my hand) belongs in this group ('young').
The usual reponse was 'you can see it's a baby or a kid' and variations on that theme. As I said they had become very good at justifying their choices by now so not being able to EXPRESS something which to them was very OBVIOUS was quite uncomfortable. They knew absolutely what their brains were telling them but didn't know either how to communicate it or evaluate it. I chose this quite deliberately because young animals of any species are very obvious indeed but we are totally unaware of how we make this evaluation or even that we are making the evaluation, it's automatic.
The argument progressed to 'well it's instinct' (i.e. recognising it's a subconscious evaluation) and 'we just know because we know from a young age'.
STEP 2: BUILD with the students.
Then some students started to revise the description like this 'the head is bigger for the puppy compared to his body' and many other variations. This idea caught on like wildfire between the groups. Suddenly they could all 'see' it.
STEP 3: TEST their thinking and REFLECT.
So now they were 'finding' many other comparisons of variables that also seemed to work. The size of the paws compared to the body, the width of the head compared to the length of the head.
I asked them to go through all the pictures again one by one and view them all 'through this lens' of comparing one variable with another. They decided that in all cases it allowed identification of the younger ones.
We looked at how simply looking at the size of the eyes didn't work, it had to be the size of the eyes COMPARED to the size of the head.
STEP 3 back to STEP 1: (Take away the fish again).
Ok they have definitely accepted this discovery that the comparison of variables allows them to clearly identify juveniles vs adults. Using the language "X compared to Y shows this one belongs in this group" works and they can complete the task :)
Now I asked them to prove it! In other words they now have a testable statement/a hypothesis. To test it means to make actual measurements and find a way to compare the numbers that allows them to identify the 'juveniles vs adults' again. They can no longer use the language above.
STEP 1 - STEP 2:
Now they have to find a new way to solve the same grouping. We need to learn a new way. This is where we stopped the lesson although some were starting to have the idea of dividing the numbers... We will resume this task tomorrow.
FIND PART 2 HERE
# Alexander Sokol 2011-02-04 08:34
Deirdre, thanks a lot. It's always a pleasure to read your detailed descriptions.
A few things I wanted to specify, though.
1. At Step 2, building the model / hypothesis together with the students, did you ask them to 'fix' their hypothesis in any way? If yes, how? And how many different 'ways of fixing / representation' were there as a result?
2. At Step 3, reflection, there are normally at least two different types of reflection: the one dealing with the actual model / hypothesis and how it worked and the other dealing with the way of developing / testing the hypothesis (a kind of metacognitive level). I more or less imagine how you dealt with the first type but I wonder if you also touched upon the second one? And how did it look like if you did?
Many thanks.
# Deirdre Jennings 2011-02-07 14:59
First for number 1. At step 2 it was very hard to encourage the students to explore possibilities during this activity without giving away the answer i.e. without using words like compare. The idea was for them to develop the language of comparison between 2 variables. They are used to spotting variables and measuring them and so on. They focused quickly on the fact that "children's head are bigger than adults" (but of course they are not). But a compound variable x/y or x*y is an unfamiliar and abstract concept (thought they may have used them in the past without appreciating them as proportions!). They could not at first come up with the language SO I just kept asking questions laced with encouragement e.g. "yes you're on the right track and I know you know what it is, BUT what you are saying/writing doesn't mean that" because it's the concept/language we're after...or just point out "kids heads are not bigger than adults". So after some provocation they came to the point that discussing isolated variables didn't work. Then they started comparing them. They all kind of went for a comparison eyes/head or feet/body or legs/body or head/body. The 'fixing' therefore was making sure the language they used communicated what they wanted to say/show.
Step 3. Students themselves suggested going to the pre-K and measuring them to test the hypothesis - brilliant idea :) we did that today and their reaction was interesting as they realised kids' heads are similar in size to an adult! This made it more concrete too and I don't think some believed it until they measured the small kids.
# Alexander Sokol 2011-02-07 22:41
Did I get you right that:
1. They didn't actually 'put down' the model they came up with as it felt that using it in communication was enough. Right? If so, are you planning to come back to it in some time and see if it was?
I am asking as I am rather sceptical about students remembering things. In my experience, they forget pretty fast and, unless it's 'fixed' somewhere in the written form, it takes a long time to remind them what they already understood at some point.
However, this may be just my 'bad' experience...
Re Step 3. In my view, going to Pre-K and measuring them would be the first kind of reflection, wouldn't it? I still think it'd be important to get them to reflect at the more metacognitive level as well, don't you think?
Joomla SEF URLs by Artio | 2,127 | 9,563 | {"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} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.952085 |
http://forum.enjoysudoku.com/fun-puzzle-so-many-methods-to-solve-no-chains-required-t4905-45.html | 1,542,241,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742322.51/warc/CC-MAIN-20181114232605-20181115014605-00521.warc.gz | 115,185,945 | 8,494 | Fun puzzle. So many methods to solve. No CHAINS required.
Post puzzles for others to solve here.
Four biggish steps, for me.
Code: Select all
`.45...23.8...4...76..2.8..9..34.69...8.....7...97.53..5..8.9..32...3...5.98...41.`
wapati
2010 Supporter
Posts: 527
Joined: 13 September 2006
wapati wrote:'Nother one that takes me several big steps.
Code: Select all
`. 9 . | 2 . . | 1 4 .. 8 . | . 3 4 | . 2 5. 4 3 | . . 1 | . . 8---------------------1 . . | . . 3 | 8 6 .. . . | . 6 . | . 5 .6 . . | 1 . . | . . 7---------------------. . . | . . . | 4 . .. 1 . | . . . | 5 8 2. . . | 3 . 2 | . . .`
Code: Select all
` b2 - 6 Locked Candidate (1) c169 - 9 Swordfishr3c4 <> 69 Unique Rectangle Type 1r7c4 <> 7 XYZ-Wing on [r8c4]`
daj95376
2014 Supporter
Posts: 2624
Joined: 15 May 2006
Code: Select all
`# Crazy 8s*-----------------------*| 4 . 2 | . 3 . | 6 . . || 1 7 . | . . . | . 5 . || 3 . . | . . 6 | . 1 . ||-------+-------+-------|| . . . | . 9 . | 3 4 7 || . . . | 5 6 4 | . . . || 2 4 9 | . 7 . | . . . ||-------+-------+-------|| . 1 . | 2 . . | . . 4 || . 2 . | . . . | . 3 5 || . . 4 | . 1 . | 2 . 6 |*-----------------------*`
daj95376
2014 Supporter
Posts: 2624
Joined: 15 May 2006
Code: Select all
`# What Fun*-----------------------*| 4 3 . | . . 6 | . 9 2 || 1 . 8 | . . 9 | . 7 . || 6 . . | . . . | . . . ||-------+-------+-------|| . . . | . 6 . | 4 . . || 9 . . | 7 . 8 | . . 6 || . . 4 | . 9 . | . . . ||-------+-------+-------|| . . . | . . . | . . 8 || . 4 . | 2 . . | 9 . 3 || 8 9 . | 6 . . | . 1 7 |*-----------------------*`
daj95376
2014 Supporter
Posts: 2624
Joined: 15 May 2006
What Fun
Code: Select all
` *--------------------------------------------------------------* | 4 3 57 | 158 157 6 | 18 9 2 | | 1 25 8 | 345 235 9 | 6 7 45 | | 6 257 9 | 1458 1257 157 | 138 35 145 | |---------------------+--------------------+-------------------| | 2357 8 12357 | 135 6 1235 | 4 235 9 | | 9 15 1235 | 7 4 8 | 13 235 6 | | 235 6 4 | 135 9 1235 | 7 8 15 | |---------------------+--------------------+-------------------| | 23 157 6 | 9 1357 1357 | 25 4 8 | | 57 4 157 | 2 8 157 | 9 6 3 | | 8 9 23 | 6 35 4 | 25 1 7 | *--------------------------------------------------------------*`
1. [r2c4]=3=[r2c5](=2=[r3c5])-3-[r79c5]=3=[r7c6]-{TILA(7): r1c3|r1c5|r7c5|r7c2|r3c2}.
So, r2c5<>3.
2. {ATILA(7): r1c3|r1c5|r3c5|r7c5|r7c2|r3c2},
and so r3c5=7.
Carcul
Carcul
Posts: 724
Joined: 04 November 2005
daj95376 wrote:
Code: Select all
`# What Fun`
Great puzzle! (Added to my collection, thanks! )
wapati
2010 Supporter
Posts: 527
Joined: 13 September 2006
I see lots to do without chains, cool puzzle!
Code: Select all
`Soaring Bird. . . | . . . | . 4 2. 5 . | . 2 3 | 1 . .. . . | 9 . 7 | . . 6---------------------. . 5 | . 4 . | . 2 .. 2 . | 1 7 . | . . .. 3 1 | . . 8 | . . .---------------------. 7 . | . . . | 6 8 .9 . . | 7 . . | 2 . 55 . 2 | . . . | . 3 .`
wapati
2010 Supporter
Posts: 527
Joined: 13 September 2006
Hairy, below chins.
Code: Select all
`..13...5......7..95.4.61...6....2.9...9....3..731....4......8..3..72..4..2...8...`
wapati
2010 Supporter
Posts: 527
Joined: 13 September 2006
So little left, so much pain.
Code: Select all
`*-----------------------*| . . . | . 4 5 | 2 3 . || 8 3 . | . . . | . 6 . || . . 4 | . . . | 5 . . ||-------+-------+-------|| 3 . 2 | 4 . . | . . 9 || . 4 . | . . . | . 7 . || 1 . . | . . 6 | 8 . 5 ||-------+-------+-------|| . . 7 | . . . | 9 . . || . 8 . | . . . | . 2 1 || . 2 6 | 7 1 . | . . . |*-----------------------*`
daj95376
2014 Supporter
Posts: 2624
Joined: 15 May 2006
Daj, I so much dislike colours and chains that I just don't read threads that deal with them.
Carcul does brilliance, finding a single step to solve what takes me many, I like knowing that there is a better way, or simpler.
Daj, I am not sure why you are posting chain puzzles here?
I started this thread for puzzles that never need chains, yet need many patterns. I like that. Chains just do not do it, for me.
<sigh>
wapati
2010 Supporter
Posts: 527
Joined: 13 September 2006
Soaring Bird
Code: Select all
` *-------------------------------------------------------* | 13 189 3789 | 68 5 16 | 79 4 2 | | 6 5 79 | 4 2 3 | 1 79 8 | | 2 148 48 | 9 18 7 | 3 5 6 | |------------------+------------------+-----------------| | 78 69 5 | 3 4 69 | 78 2 1 | | 48 2 69 | 1 7 5 | 48 69 3 | | 47 3 1 | 2 69 8 | 5 679 479 | |------------------+------------------+-----------------| | 13 7 34 | 5 139 2 | 6 8 49 | | 9 468 3468 | 7 368 46 | 2 1 5 | | 5 1468 2 | 68 1689 1469 | 479 3 479 | *-------------------------------------------------------*`
R7c5=3 or r9c2=4. But:
[r7c5]-3-[r8c5]=3=[r8c3]=6=[r5c3]=9=[r4c2]=6=[r4c6]-6-[r1c6]-1-
-[r3c5]-8-[r3c3]-4-[r7c3]-3-[r7c5],
and so r7c5<>3 => r9c2=4 and the puzzle is solved.
Alternatively:
[r7c5]=9=[r7c9]=4=[r7c3]-4-[r3c3]-8-[r3c5](-1-[r7c5])-1-[r1c6]-6-[r4c6]
=6=[r4c2]=9=[r5c3]=6=[r8c3]=3=[r8c5]-3-[r7c5],
which implies r7c5<>1,3 => r7c5=9 and the puzzle is solved.
Carcul
Carcul
Posts: 724
Joined: 04 November 2005
Wapati wrote:Hairy, below chins
This puzzle is too easy:
Code: Select all
` *-----------------------------------------------------* | 7 6 1 | 3 8 9 | 4 5 2 | | 8 3 2 | 45 45 7 | 16 16 9 | | 5 9 4 | 2 6 1 | 3 7 8 | |-----------------+----------------+------------------| | 6 4 58 | 58 3 2 | 17 9 17 | | 1 58 9 | 568 7 4 | 2 3 56 | | 2 7 3 | 1 9 56 | 56 8 4 | |-----------------+----------------+------------------| | 49 15 567 | 4569 145 3 | 8 2 1567 | | 3 158 568 | 7 2 56 | 9 4 156 | | 49 2 567 | 4569 145 8 | 1567 16 3 | *-----------------------------------------------------*`
{ATILA(5,6): r8c9|r5c9|r6c7|r6c6|r8c6}
and so r8c9 must be "1", solving the puzzle.
Carcul
Carcul
Posts: 724
Joined: 04 November 2005
Carcul wrote:
Wapati wrote:Hairy, below chins
This puzzle is too easy:
Already forgot the remote pairs Wapati ?
ravel
Posts: 998
Joined: 21 February 2006
Daj95376 wrote:So little left, so much pain.
Code: Select all
` *--------------------------------------------------* | 7 9 1 | 6 4 5 | 2 3 8 | | 8 3 5 | 29 279 279 | 1 6 4 | | 2 6 4 | 138 38 138 | 5 9 7 | |---------------+------------------+---------------| | 3 5 2 | 4 78 78 | 6 1 9 | | 6 4 8 | 159 59 19 | 3 7 2 | | 1 7 9 | 23 23 6 | 8 4 5 | |---------------+------------------+---------------| | 45 1 7 | 2358 2358 2348 | 9 58 6 | | 459 8 3 | 59 6 49 | 7 2 1 | | 59 2 6 | 7 1 89 | 4 58 3 | *--------------------------------------------------*`
[r3c4]=1=[r3c6]=3=[r7c6]-3-[r7c45]=3|5=[r7c45]-5-[r8c4]-9-[r2c4]-2-
-[r249c6]-9-[r5c6]-1-[r5c4]=1=[r3c4],
=> r3c4=1 and the puzzle is solved.
Carcul
Carcul
Posts: 724
Joined: 04 November 2005
Carcul wrote:{ATILA(5,6): r8c9|r5c9|r6c7|r6c6|r8c6}
and so r8c9 must be "1", solving the puzzle.
Looks a lot like overlapping turbot fish.
ronk
2012 Supporter
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA
PreviousNext | 3,505 | 7,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-47 | latest | en | 0.261988 |
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## How do you find the density of a liquid?
To measure the density of a liquid you do the same thing you would for a solid. Mass the fluid, find its volume, and divide mass by volume. To mass the fluid, weigh it in a container, pour it out, weigh the empty container, and subtract the mass of the empty container from the full container.
## What is the unit of measurement for liquid density?
• grams per milliliter or g/ml is the derived unit that is usually used to express the density of liquids. In measuring the mass of a liquid, they use grams or kilograms, and uses liter, milliliters and cubic centimeters to measure the volume.
## How to find the density of a liquid?
• We can calculate density of a each liquid using the formula: Density= Mass/Volume where mass is that for just the liquid (you must subtract out the mass of the graduated cylinder).
## What are facts about liquid density?
• Density of a liquid determines how it will layer (heaviest to lightest).
• If the liquid is the most dense it will sink to the bottom
• If the liquid us least dense it float to the bottom.
• Layers will remain separated because each liquid is actually floating on top of the more dense liquid beneath it. | 267 | 1,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.869951 |
https://number.academy/102040 | 1,660,929,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00081.warc.gz | 396,621,328 | 12,052 | # Number 102040
Number 102,040 spell 🔊, write in words: one hundred and two thousand and forty . Ordinal number 102040th is said 🔊 and write: one hundred and two thousand and fortieth. Color #102040. The meaning of number 102040 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 102040. What is 102040 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 102040.
## What is 102,040 in other units
The decimal (Arabic) number 102040 converted to a Roman number is (C)MMXL. Roman and decimal number conversions.
#### Weight conversion
102040 kilograms (kg) = 224957.4 pounds (lbs)
102040 pounds (lbs) = 46285.0 kilograms (kg)
#### Length conversion
102040 kilometers (km) equals to 63405 miles (mi).
102040 miles (mi) equals to 164218 kilometers (km).
102040 meters (m) equals to 334773 feet (ft).
102040 feet (ft) equals 31103 meters (m).
102040 centimeters (cm) equals to 40173.2 inches (in).
102040 inches (in) equals to 259181.6 centimeters (cm).
#### Temperature conversion
102040° Fahrenheit (°F) equals to 56671.1° Celsius (°C)
102040° Celsius (°C) equals to 183704° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
102040 seconds equals to 1 day, 4 hours, 20 minutes, 40 seconds
102040 minutes equals to 2 months, 2 weeks, 20 hours, 40 minutes
### Codes and images of the number 102040
Number 102040 morse code: .---- ----- ..--- ----- ....- -----
Sign language for number 102040:
Number 102040 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 102040
### Multiplications
#### Multiplication table of 102040
102040 multiplied by two equals 204080 (102040 x 2 = 204080).
102040 multiplied by three equals 306120 (102040 x 3 = 306120).
102040 multiplied by four equals 408160 (102040 x 4 = 408160).
102040 multiplied by five equals 510200 (102040 x 5 = 510200).
102040 multiplied by six equals 612240 (102040 x 6 = 612240).
102040 multiplied by seven equals 714280 (102040 x 7 = 714280).
102040 multiplied by eight equals 816320 (102040 x 8 = 816320).
102040 multiplied by nine equals 918360 (102040 x 9 = 918360).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 102040
Half of 102040 is 51020 (102040 / 2 = 51020).
One third of 102040 is 34013,3333 (102040 / 3 = 34013,3333 = 34013 1/3).
One quarter of 102040 is 25510 (102040 / 4 = 25510).
One fifth of 102040 is 20408 (102040 / 5 = 20408).
One sixth of 102040 is 17006,6667 (102040 / 6 = 17006,6667 = 17006 2/3).
One seventh of 102040 is 14577,1429 (102040 / 7 = 14577,1429 = 14577 1/7).
One eighth of 102040 is 12755 (102040 / 8 = 12755).
One ninth of 102040 is 11337,7778 (102040 / 9 = 11337,7778 = 11337 7/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
102040
#### Is Prime?
The number 102040 is not a prime number. The closest prime numbers are 102031, 102043.
#### Factorization and factors (dividers)
The prime factors of 102040 are 2 * 2 * 2 * 5 * 2551
The factors of 102040 are
1 , 2 , 4 , 5 , 8 , 10 , 20 , 40 , 2551 , 5102 , 10204 , 12755 , 20408 , 25510 , 51020 , 102040
Total factors 16.
Sum of factors 229680 (127640).
#### Powers
The second power of 1020402 is 10.412.161.600.
The third power of 1020403 is 1.062.456.969.664.000.
#### Roots
The square root √102040 is 319,437005.
The cube root of 3102040 is 46,729394.
#### Logarithms
The natural logarithm of No. ln 102040 = loge 102040 = 11,53312.
The logarithm to base 10 of No. log10 102040 = 5,00877.
The Napierian logarithm of No. log1/e 102040 = -11,53312.
### Trigonometric functions
The cosine of 102040 is 0,479588.
The sine of 102040 is 0,877494.
The tangent of 102040 is 1,829683.
### Properties of the number 102040
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 102040 in Computer Science
Code typeCode value
PIN 102040 It's recommendable to use 102040 as a password or PIN.
102040 Number of bytes99.6KB
CSS Color
#102040 hexadecimal to red, green and blue (RGB) (16, 32, 64)
Unix timeUnix time 102040 is equal to Friday Jan. 2, 1970, 4:20:40 a.m. GMT
IPv4, IPv6Number 102040 internet address in dotted format v4 0.1.142.152, v6 ::1:8e98
102040 Decimal = 11000111010011000 Binary
102040 Decimal = 12011222021 Ternary
102040 Decimal = 307230 Octal
102040 Decimal = 18E98 Hexadecimal (0x18e98 hex)
102040 BASE64MTAyMDQw
102040 SHA1fc15f203955eeacd33dc02f7c626aa2c8b7e6c2d
102040 SHA224e6a78b69f79a4c0917f63ab5dc9fb96e0827440fa8994aafba2a44e9
102040 SHA2567fea288781d8217619442688c7aa582de55ba67ce8172bcd01b9bb9ebf119af2
102040 SHA384c1a735324836f546c2c472fc498b7af425ceda82d663421d78195b391ed476f6ff726060ba3bb375452fbc2187d40985
More SHA codes related to the number 102040 ...
If you know something interesting about the 102040 number that you did not find on this page, do not hesitate to write us here.
## Numerology 102040
### Character frequency in number 102040
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 3 2 1 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 102040, the numbers 1+0+2+0+4+0 = 7 are added and the meaning of the number 7 is sought.
## Interesting facts about the number 102040
### Asteroids
• (102040) 1999 RG112 is asteroid number 102040. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 9/9/1999.
## № 102,040 in other languages
How to say or write the number one hundred and two thousand and forty in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 102.040) ciento dos mil cuarenta German: 🔊 (Anzahl 102.040) einhundertzweitausendvierzig French: 🔊 (nombre 102 040) cent deux mille quarante Portuguese: 🔊 (número 102 040) cento e dois mil e quarenta Chinese: 🔊 (数 102 040) 十万二千零四十 Arabian: 🔊 (عدد 102,040) مائة و اثنان ألفاً و أربعون Czech: 🔊 (číslo 102 040) sto dva tisíce čtyřicet Korean: 🔊 (번호 102,040) 십만 이천사십 Danish: 🔊 (nummer 102 040) ethundrede og totusindfyrre Dutch: 🔊 (nummer 102 040) honderdtweeduizendveertig Japanese: 🔊 (数 102,040) 十万二千四十 Indonesian: 🔊 (jumlah 102.040) seratus dua ribu empat puluh Italian: 🔊 (numero 102 040) centoduemilaquaranta Norwegian: 🔊 (nummer 102 040) en hundre og to tusen og førti Polish: 🔊 (liczba 102 040) sto dwa tysiące czterdzieści Russian: 🔊 (номер 102 040) сто две тысячи сорок Turkish: 🔊 (numara 102,040) yüzikibinkırk Thai: 🔊 (จำนวน 102 040) หนึ่งแสนสองพันสี่สิบ Ukrainian: 🔊 (номер 102 040) сто двi тисячi сорок Vietnamese: 🔊 (con số 102.040) một trăm lẻ hai nghìn lẻ bốn mươi Other languages ...
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## Comment
If you know something interesting about the number 102040 or any natural number (positive integer) please write us here or on facebook. | 2,409 | 7,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-33 | latest | en | 0.651801 |
https://studylib.net/doc/18855295/problem-solving---information-for-parents-problem-solving... | 1,726,494,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00015.warc.gz | 513,356,950 | 11,278 | Problem Solving ~ Information for Parents Problem Solving, What
```Problem Solving ~ Information for Parents
Problem Solving, What & Why?
One of the main reasons for studying mathematics is to develop the ability to solve problems. Problem
solving is the process of applying what we already know to new and unfamiliar situations. This ability is
not only critical to our children’s future needs, but also to a productive society and even human progress
itself. In these early years, children develop attitudes and beliefs as to their ability to learn mathematics.
By learning mathematics through problem solving, children can make sense of why they need to know
their basic facts.
How Can Parents Help?
• Be enthusiastic. Let your child see how excited you are about solving a problem.
• Provide time and talk about problem solving. Be patient with your child. Let them work at
their own pace. Talk, talk, talk! Talk about options, strategies and ideas for problem solving.
• Reinforce risk taking. Children need a great deal of security to risk being wrong. When they
begin to realize that they can learn from their mistakes, they will try harder to complete the
problem.
• Reward perseverance. Instant success is not always possible in learning mathematics.
Encourage children to keep trying by asking them questions that will lead them in the right
direction.
• Use children’s experiences. As often as possible, base problems on children’s everyday
experiences at school and at home.
The best way for your children to become good problem solvers is for them to solve problems, lots of
problems! Also, it benefits children to think about how they solved the problem afterwards. In this way
they may use their particular strategy to solve similar problems in the future. There are no best ways of
solving a problem. We are interested in what makes sense to each individual. Here are some strategies to
~ act it out
~ use objects or models
~ make a drawing
~ make a graph or chart
~ make a list
~ guess and check
~ sort and order items
~ look for a pattern
~ look for all possibilities
~ solve a simpler problem
~ choose an operation
~ think logically, use what you know
Here is how we tackle problem solving in the classroom:
1. Think. Make sure that you know what the problem is all about. Do you understand all the
vocabulary? Do you know what information is given? Is anything missing? Can you restate the
problem in your own words? Do you understand exactly what you are asked to do or to find out? | 534 | 2,508 | {"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} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.965306 |
https://itecnotes.com/electrical/electrical-from-transfer-function-to-frequency-response/ | 1,726,157,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00435.warc.gz | 291,368,342 | 6,662 | # Electrical – From transfer function to frequency response
frequency responsetransfer function
In this question I asked about the difference between transfer function and frequency response. One user replied that "the frequency response is the transfer function where the transients are assumed to be completely dissipated". He showed an example, to prove his statement. It was like this:
Take, as an example, a sinusoid, \$\small \sin(\omega t) \rightarrow \dfrac{\omega}{s^2+\omega^2}\$, applied to a simple first order lag,
\$\small G(s)=\dfrac{1}{1+s}\$. The response is: \$\small > R(s)=\dfrac{\omega}{(s^2+\omega^2)(1+s)}\$, and this can be expressed
in partial fractions:
$$\small \frac{\omega}{(s^2+\omega^2)(1+s)}=\frac{A+Bs}{(s^2+\omega^2)}+\frac{C}{(1+s)}$$
Inverse LT gives:$$\small r(t)=\frac{A}{\omega}\sin(\omega t)+ B\cos(\omega t)+Ce^{-t/\tau}$$
The exponential term decays to zero, leaving the steady-state response
as:
$$\small \frac{A}{\omega}\sin(\omega t)+B\cos(\omega t)= X\sin(\omega t+\phi)$$
Solving for \$\small X\$ and \$\small\phi\$ gives \$\frac{1}{\sqrt{1+\omega^2}}\$, and \$\small \arctan{(-\omega)}\$,
respectively, as is obtained using \$\small s\rightarrow j\omega\$ in
the Laplace TF.
I don't quite understand the last part.
How does he calculate \$\small X\$ and \$\small\phi\$ and what does he deduce by plugging \$\small s\rightarrow j\omega\$ into the transfer function? How is his original statement verified?
I don't quite understand the last part. How does he calculate X and ϕ
This is just applying the trigonometric identity $$\sin\left(\alpha + \beta\right)=\sin\alpha\cos\beta + \cos\alpha\sin\beta$$ with \$\alpha=\omega{}t\$ and \$\beta=\phi\$.
Using this identity on the r.h.s. gives $$X\sin\left(\omega{}t+\phi\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$ so our equation becomes $$\frac{A}{\omega}\sin\left(\omega{}t\right)+B\cos\left(\omega{}t\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$
Which we can break into two parts, $$\frac{A}{\omega}\sin\left(\omega{}t\right)=X\cos\phi\sin{}\omega{}t$$ and $$B\cos\left(\omega{}t\right)=X\sin\phi\cos\omega{}t$$
So, $$\frac{A}{\omega} = X\cos\phi$$ and $$B=X\sin\phi$$
From there you should be able to get the conclusions from your source. | 735 | 2,300 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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-2024-38 | latest | en | 0.715552 |
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# Test: Simple Harmonic Motion
Test Description
## 10 Questions MCQ Test Physics Class 11 | Test: Simple Harmonic Motion
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Test: Simple Harmonic Motion - Question 1
### The velocity of a particle moving with simple harmonic motion is____ at the mean position.
Detailed Solution for Test: Simple Harmonic Motion - Question 1
Equation of SHM particle:
Y=a sinωt
V=aω sinωt
Vmax = aω
So the velocity is maximum at mean position
Test: Simple Harmonic Motion - Question 2
### The periodic time (tp) is given by:
Detailed Solution for Test: Simple Harmonic Motion - Question 2
Periodic time is the time taken for one complete revolution of the particle.
∴ Periodic time (tp) = 2 π/ω seconds.
Test: Simple Harmonic Motion - Question 3
### A frequency of 1Hz corresponds to:
Detailed Solution for Test: Simple Harmonic Motion - Question 3
Frequency is defined as time taken to perform one oscillation by the object.
Hence, 1Hz corresponds to 1 vibration per sec.
Test: Simple Harmonic Motion - Question 4
A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:
Detailed Solution for Test: Simple Harmonic Motion - Question 4
• Time Period, T = 2π √(l/g')where,
l = Length of seconds pendulum
g’ = Apparent Gravity
• For the period of oscillations of Seconds Pendulum to decrease, the Apparent gravity (g’) has to increase because:
• Hence, Time Period of oscillations of Seconds Pendulum will decrease when the rocket is ascending up with uniform acceleration.
Test: Simple Harmonic Motion - Question 5
A particle of mass 10 gm lies in a potential field v = (50x2+100) J/kg. The value of frequency of oscillations in Hz is
Detailed Solution for Test: Simple Harmonic Motion - Question 5
Test: Simple Harmonic Motion - Question 6
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is:
Detailed Solution for Test: Simple Harmonic Motion - Question 6
We know that in a simple harmonic motion the maximum velocity,
Vmax = A⍵
Here A = 50 mm
And ⍵ = 2π / T
= 2π / 2
= π
Hence Vmax = 50 x 10-3
= 0.15 m/s
Test: Simple Harmonic Motion - Question 7
In simple harmonic motion the displacement of a particle from its equilibrium position is given by . Here the phase of motion is
Test: Simple Harmonic Motion - Question 8
Find the amplitude of the S.H.M whose displacement y in cm is given by equation y= 3 sin157t + 4 cos157t, where t is time in seconds.
Detailed Solution for Test: Simple Harmonic Motion - Question 8
When the displacement of a SHM is:
y=a sin wt+ b cos wt
• Amplitude of the SHM will be:
A=√a2+b2
Here, a = 3, b = 4
Amplitude, A= √(32+42) = 5 cm
Hence option B is correct.
Test: Simple Harmonic Motion - Question 9
What will be the phase difference between bigger pendulum (with time period 5T/4 ) and smaller pendulum (with time period T) after one oscillation of bigger pendulum?
Detailed Solution for Test: Simple Harmonic Motion - Question 9
After one oscillation of a bigger pendulum i.e. 5T/4, ¼ of the total phase is travelled by the smaller pendulum while the bigger is still at initial position.
Thus, the phase difference between two is: ¼ (2π) - 0 = π/2
Test: Simple Harmonic Motion - Question 10
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
Detailed Solution for Test: Simple Harmonic Motion - Question 10
∴ We get, ω = √3 s-1
T = 2π / √3
## Physics Class 11
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## Physics Class 11
124 videos|464 docs|210 tests
### How to Prepare for NEET
Read our guide to prepare for NEET which is created by Toppers & the best Teachers | 1,298 | 5,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-23 | latest | en | 0.860021 |
http://www.jiskha.com/display.cgi?id=1173545338 | 1,493,551,928,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125074.20/warc/CC-MAIN-20170423031205-00309-ip-10-145-167-34.ec2.internal.warc.gz | 574,296,248 | 3,742 | # Word Problem
posted by on .
A wire from the top of a TV tower to the ground makes an angle of 52 degrees with the ground and touches the ground 180 feet from the base of the tower. How high is the tower?
a)290 feet
b)230 feet
c)110 feet
d)140 feet
I'm pretty sure we can eliminate c and d because they are less than 180 feet and weknow the tower has to be at least that right? Or is my thinking wrong, Please help!!!!
Why are you guessing? Why not solve the problem and pick the answer?
Draw a diagram so you can see the problem.
tan 52o = height/180.
solve for height. | 149 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.963447 |
http://mathhelpforum.com/number-theory/111690-remainder-summation-factorials.html | 1,481,026,863,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541905.26/warc/CC-MAIN-20161202170901-00339-ip-10-31-129-80.ec2.internal.warc.gz | 168,142,332 | 10,806 | # Thread: Remainder of Summation and Factorials
1. ## Remainder of Summation and Factorials
My problem is
What is the remainder when 1!+2!+...+100! is divided by 13?
How do you simplify a factorial to find its remainder?
Also, is the remainder of the summation of factorials the same as the remainder of the sum of the remainders of the factorials?
Thanks.
2. Yes, you can find the remainder of each, add them up, and reduce again mod 13 to get the remainder of the sum.
Note that 13!, 14!, ..., 100! all leave a remainder of 0 when divided by 13. That should simplify your calculations a lot.
3. Originally Posted by Bruno J.
Note that 13!, 14!, ..., 100! all leave a remainder of 0 when divided by 13. That should simplify your calculations a lot.
So any factorial greater than or equal to the divisor is 0, is this right?
This leads into another question that I have
find the remainder when 40!/((2^20) * 20!) is divided by 8.
Can I find the remainder of each separately, or do I have to simply the factorials first?
I'm guessing that I have to simplify the factorial first, because 40!/20! does not contain 8. So, 40!/20! = 40*39*38*...*22*21. Now since 40!/20! is a multiple of 32, and 8 divides 32, then 8 must divide 40!/20! with remainder 0.
But here's where I'm unsure. I think that 2^4 == 0 (mod 8) implies 2^20 == 0 (mod 8). But now I'm left with 0/0. Does this mean the remainder is 0, indeterminate, or did I make a mistake?
,
,
,
,
,
,
,
,
,
,
,
,
,
,
# remainder when factorial divided by a number
Click on a term to search for related topics. | 448 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2016-50 | longest | en | 0.933636 |
https://brainmass.com/business/strategy-and-business-analysis/sales-strategy-selecting-the-better-strategy-462073 | 1,712,930,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00668.warc.gz | 132,325,024 | 7,761 | Purchase Solution
# Sales Strategy: Selecting the Better Strategy
Not what you're looking for?
Gobi Inc. has sales of \$40,000,000. The contribution margin is 40% and the fixed costs are \$3,000,000. The variable cost per unit is \$12. The company is considering two different strategies for increasing their profits:
1) Spend \$2,000,000 in advertising: the results is expected to increase the company's sales by 25%
2) Reduce the price by 20%; the price-demand elasticity is -3.0
Which of the two strategies will generate the highest overall profits?
##### Solution Summary
This solution determines which strategy gives higher profits. Formula and calculations are provided and answers are explained.
##### Solution Preview
Current Sales=40,000,000
Contribution margin ratio=CMR=40%
Contribution margin (\$)=CM=Sales*CMR=40000000*40%=\$16,000,000
Fixed Costs= 3,000,000
Operating profit=CM- fixed Cost=16,000,000-3,000,000=\$13,000,000
1. Spend \$2,000,000 in advertising; results expected to increase sales by 25%
Expected Sales=40000000*(1+25%)=\$50,000,000
Contribution margin ratio=CMR=40%
Contribution margin ...
Solution provided by:
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This quiz intended to help students understand change and resistance in organizations | 570 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.904444 |
https://picos-api.gitlab.io/picos/api/picos.constraints.con_quadratic.html | 1,685,693,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00293.warc.gz | 504,677,241 | 7,262 | Classes
Bound on a nearly convex quadratic expression.
Nearly convex means that the bilinear form representing the quadratic part of the expression that the constraint poses to be at most zero has exactly one negative eigenvalue. More precisely, if the constraint is of the form , then Q - R needs to have exactly one negative eigenvalue for the constraint to be nearly convex. Such constraints can be posed as conic constraints under an additional assumption that some affine term is nonnegative.
At this point, this class may only be used for nonconvex constraints of the form with representable as a squared norm. In this case, the additional assumptions required for a conic reformulation are and .
Whether a constraint of this type is strengthened to a conic constraint or relabeled as a NonconvexQuadraticConstraint depends on the assume_conic option.
Example
>>> from picos import Options, RealVariable
>>> x, y, z = RealVariable("x"), RealVariable("y"), RealVariable("z")
>>> C = x**2 + 1 <= y*z; C
<Conic Quadratic Constraint: x² + 1 ≤ y·z>
>>> P = C.__class__.Conversion.convert(C, Options(assume_conic=True))
>>> list(P.constraints.values())[0]
<4×1 RSOC Constraint: ‖fullroot(x² + 1)‖² ≤ y·z ∧ y, z ≥ 0>
>>> Q = C.__class__.Conversion.convert(C, Options(assume_conic=False))
>>> list(Q.constraints.values())[0]
<Nonconvex Quadratic Constraint: x² + 1 ≤ y·z>
Note
Solver implementations must not support this constraint type so that the user’s choice for the assume_conic option is respected.
class Conversion[source]
Nearly convex quadratic to (rotated) second order cone conversion.
classmethod convert(con, options)[source]
Implement convert.
classmethod predict(subtype, options)[source]
Implement predict.
__init__(lhs, relation, rhs)[source]
Construct a ConicQuadraticConstraint.
See NonconvexQuadraticConstraint.__init__ for more.
Bound on a convex quadratic expression.
class ConicConversion[source]
Convex quadratic to (rotated) second order cone conversion.
classmethod convert(con, options)[source]
Implement convert.
classmethod predict(subtype, options)[source]
Implement predict.
__init__(lhs, relation, rhs)[source]
Construct a ConvexQuadraticConstraint.
See NonconvexQuadraticConstraint.__init__ for more.
Bases: Constraint
Bound on a nonconvex quadratic expression.
__init__(lhs, relation, rhs)[source]
Construct a NonconvexQuadraticConstraint.
Parameters
property greater
Greater-or-equal side of the constraint.
property le0[source]
Quadratic expression constrained to be at most zero.
property smaller
Smaller-or-equal side of the constraint. | 592 | 2,611 | {"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} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.767769 |
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Lectures TuTh 5:00-6:30pm, 155 Dwinelle Hall . 1. Es bleibt aber natürlich dauerhaft online – auch für die, die nicht auf Instagram unterwegs sind. Klasse Lehrplans. Kostenlose Übungen & Aufgaben mit Lösungen für das Fach Mathe Klasse 1 in der Grundschule Arbeitsblätter Übungsblätter Unbegrenzt herunterladen. All Chapters. Diese ersten, ganz einfachen Übungen machen Kindern Spaß machen und führen spielerisch ans Rechnen heran. Die Kinder lernen neben klassischen Plus- und Minus-Aufgaben auch neue Aufgabenformen wie Zahlenmauern und … Klasse. Klasse - 3. Math 1A Calculus. Math 1A, Exam #1. Homework Help. Log In. report. Die nachfolgenden Aufgabenblätter dienen dem Auswendiglernen kleiner Additionen. math 1b berkeley, *Chemistry 1B is required for BMB track 2 only. Mathe 2. **For transfer students, MCB will accept any two-semester combination of Math 1B, Math 53, Math 54, Math 55, and/or Stats 2 or 20. Announcements. Jetzt Material & Übungen gratis downloaden! Study Resources. (b). Klasse kostenlos als PDF-Datei. best. 13. Back to Department. Platzhalter im ZR 10 üben. 100% Upvoted. Prerequisites: Three years of high school math, including trigonometry, plus a satisfactory grade in one of the following: CEEB MAT test, an AP test, the UC/CSU math diagnostic exam, or 32. The most common math sequences that students use to fulfill requirements are below: Math 16A, 16B OR Math 1A, 1B OR Math 10A, 10B OR Math 1A, 16B OR Math 1B, 53, OR Math 53, 54 ; View Worksheet 1 Answers 10AF20 (1).pdf from MATH 10A at University of California, Berkeley. A deficient grade in Math 55 may be removed by completing Math N55. Addition, subtraction, place value, and logic games that boost first grade math skills. Vielfältige, kostenlose Übungen und Aufgaben für einen leichten Einstieg in die Mathematik, die alle im Zahlenraum 20 bleiben. Get Exam Ready Get Exam Ready. report. Mathe, 1. Adeamus 1C; Adeamus 1B ; Agite plus 1; Proben für alle Schulbücher; lernwolf.de. Mathe, 1. 2. (tan! Klasse 4. 7. 282 Klassenarbeiten und Übunsgblättter zu Mathematik 1. Log in or sign up to leave a comment Log In Sign Up. The official prerequisite for taking this course is Math 104: Introduction to Analysis. Klasse. 7 comments. 89% Upvoted . math 185 berkeley, A bibliography of online papers in 17th/18th Century Philosophy. Math 185. Summary: Philosophy in the 17th- and 18th-centuries can be characterized by an increased concern with questions relating to epistemology, human subjectivity, and the foundations of natural science. Fall and/or spring: 15 weeks - 3 hours of lecture and 2 hours of discussion per week. How to Sign Up for the Spring 2021 Math 1A Adjunct Course. hide. Please fill out this Google form to receive the Zoom ID for the Math 1B (Reshetikhin) adjunct.. 12/08/09: Office hours 1–2PM, Wed, Dec 09 and 1… Econ 101B after Math 1B? Complex Analysis Department of Mathematics University of California, Berkeley Fall 2009 This is an introductory course on complex analysis. MATH 1B. 36 pages. Find (a) !"!(1+!"#! 1. Gratisübungsblätter; einfaches Lernen; sicherer Übertritt; bequeme Downloads; Sie sind hier: lernwolf.de; Mathe; Übungen Mathe Kla Please fill out this Google form to receive the Zoom ID for the Math 1A adjunct. no comments yet. level 1. Be the first to share what you think! Mathe, 1. Klasse als gratis Download und zum Ausdrucken. While not required for admission to the university or the major, it is strongly recommended that all students take an introductory statistics course prior to the start of the upper division cou Klasse - 4. Lerne online oder mit der iOS-App oder Android-App alle Inhalte des 1. Paulin! Klasse / Mathe. Mathe. Access 6; On Track 2; Grammatik; Sachthemen; Französisch. Log in or sign up to leave a comment Log In Sign Up. Get ready with premium notes and study guides! Klasse 2. Sort by. hide. Math 1A on Bcourses. thinking of taking chem 1a, math 1b, eecs 16a, and an AC class. I’m in his 1B class right now and he’s an absolute gem. Mathe 1. Klasse ... Umso besser die Schüler das Zerlegen der Zahlen von 1 bis 10 beherrschen, desto weniger werden sie mit zukünftigen Rechenschwierigkeiten konfrontiert werden. Dieses Material ist das 19. Prev 1 2 Next. Exams are coming! If you are interested in enrolling in the Spring 2021 Math 1A adjunct course you must enroll in or be on the waiting list for Math 1A. Klasse Mathematik in der Grundschule an, die in der Nachhilfe, zu Hause, in der Schule oder der Mittagesbetreuung eingesetzt werden können. 7 pages. Prerequisites: Mathematical maturity appropriate to a sophomore math class. Klasse. Sie sollten aufsteigend verwendet werden. Klasse 3. MATH 1B - Calculus free online testbank with past exams and old test at UC Berkeley (UCB) Office hours W 2:00-4:00pm, Student Learning Center, and by appt. Would really appreciate some more tips on preparing for this math series; thanks guys and gals. share. Klasse zur Verfügung. Header search input. Download the best MATH 1B textbook notes at University of California - Berkeley to get exam ready in less time! Please fill out this Google form to receive the Zoom ID for the Math 1B adjunct.. Table of Contents. math 10a berkeley, Math 16A - Analytic Geometry and Calculus -- [3 units] Course Format: Three hours of lecture, one and one-half hours of discussion per week. Lernjahr 1; Latein. Credit Restrictions: Students will receive no credit for Math 55 after completion of Math N55 or Computer Science 70. Klasse Der Zahlenraum 10. I'm a rising sophomore trying to do grad school in Economics after graduation, and was told that Econ 101B would give me an advantage over Econ 100B. best. Related Courses. If you are interested in enrolling in the Spring 2021 Math 1B adjunct courses you must enroll in or be on the waiting list for the appropriate section of Math 1B. 1. Posted by 19 days ago. Türchen des Materialadventskalenders auf Instagram. math 10a berkeley, The downside to that is that I heard Math 1B is really hard, so there really is no winning with the Berkeley math "weeder" courses. Prove that (sin(!!)+cos(!!))!"≤2!. Math 1A on Piazza. save. Klasse 2. 1st grade math games for free. Syllabus; Sections and GSIs; Resources (books, notes, more) Academic honesty; Assignments; Exams and assessment. Die gehen einfach immer – geeignet für 1. und 2. Klasse - 2. How to Sign Up for the Spring 2021 Math 1B Adjunct Courses. Kostenlose Arbeitsblätter und Unterrichtsmaterial zum Thema Addition für Lehrer in der Grundschule. share. However, I've also been told I should take Math 53 and 54 before Econ 101B, and I've only got Math 1B under my belt so far. Klasse - 10. Klasse 1; Klasse 2; Klasse 3; Klasse 4; Brüche; römische Zahlen; Sachkunde (HSU) Englisch. Sort by. Access study documents, get answers to your study questions, and connect with real tutors for MATH 1a-1b at University Of California, Berkeley. Dezember 2020. save. File 3.docx. I've started going through khan academy and picked up the 1A textbook but thats about it. Kostenlose Arbeitsblätter und Unterrichtsmaterial für das Fach Mathe in der 2. ANTON ist die neue Lern-App für die Schule. Diese ersten, ganz einfachen Übungen sollen Kindern Spaß machen und führen spielerisch ans Rechnen heran. University of California - Berkeley. 452. | 2,835 | 9,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.472759 |
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