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New Member
## can't calculate DATEDIFF using Max function
Hi all,
I'm pretty new to power bi so I would like some help on this
I am trying to calculate the month difference between the max date and the selected date
I tried 2 different methods:
Method 1 using today() and it worked well
datediff = DATEDIFF(SELECTEDVALUE(DATA[date].[date]),today(),MONTH)
however when I tried to use the max function replacing the today() function- my result returned to 0
datediff2 = DATEDIFF(SELECTEDVALUE(DATA[date].[date]),max(DATA[date].[Date]),MONTH)
does anyone know why is my max() not returning the max date of the table?
2 ACCEPTED SOLUTIONS
Community Champion
You could try this:-
``````datediff2 =
var max_date = calculate(max(DATA[date]),ALL(DATA[date]))
return DATEDIFF(SELECTEDVALUE(DATA[date]),max_date,MONTH)``````
BR,
Samarth
Best Regards,
Samarth
If this post helps, please consider accepting it as the solution to help the other members find it more quickly.
New Member
okay I already figured out myself - in order to use the max function
i need to use maxdate= calculate(max(data[date]),all(data)); so the max date will be fixed and therefore i can use it to calculate the datediff
3 REPLIES 3
New Member
okay I already figured out myself - in order to use the max function
i need to use maxdate= calculate(max(data[date]),all(data)); so the max date will be fixed and therefore i can use it to calculate the datediff
Community Champion
You could try this:-
``````datediff2 =
var max_date = calculate(max(DATA[date]),ALL(DATA[date]))
return DATEDIFF(SELECTEDVALUE(DATA[date]),max_date,MONTH)``````
BR,
Samarth
Best Regards,
Samarth
If this post helps, please consider accepting it as the solution to help the other members find it more quickly.
New Member
thanks so much!! | 488 | 1,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.813711 |
http://de.metamath.org/mpeuni/df-lmi.html | 1,721,346,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00099.warc.gz | 7,032,319 | 4,273 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > df-lmi Structured version Visualization version GIF version
Definition df-lmi 25467
Description: Define the line mirroring function. Definition 10.3 of [Schwabhauser] p. 89. See islmib 25479. (Contributed by Thierry Arnoux, 1-Dec-2019.)
Assertion
Ref Expression
df-lmi lInvG = (𝑔 ∈ V ↦ (𝑚 ∈ ran (LineG‘𝑔) ↦ (𝑎 ∈ (Base‘𝑔) ↦ (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏))))))
Distinct variable group: 𝑎,𝑏,𝑔,𝑚
Detailed syntax breakdown of Definition df-lmi
StepHypRef Expression
1 clmi 25465 . 2 class lInvG
2 vg . . 3 setvar 𝑔
3 cvv 3173 . . 3 class V
4 vm . . . 4 setvar 𝑚
52cv 1474 . . . . . 6 class 𝑔
6 clng 25136 . . . . . 6 class LineG
75, 6cfv 5804 . . . . 5 class (LineG‘𝑔)
87crn 5039 . . . 4 class ran (LineG‘𝑔)
9 va . . . . 5 setvar 𝑎
10 cbs 15695 . . . . . 6 class Base
115, 10cfv 5804 . . . . 5 class (Base‘𝑔)
129cv 1474 . . . . . . . . 9 class 𝑎
13 vb . . . . . . . . . 10 setvar 𝑏
1413cv 1474 . . . . . . . . 9 class 𝑏
15 cmid 25464 . . . . . . . . . 10 class midG
165, 15cfv 5804 . . . . . . . . 9 class (midG‘𝑔)
1712, 14, 16co 6549 . . . . . . . 8 class (𝑎(midG‘𝑔)𝑏)
184cv 1474 . . . . . . . 8 class 𝑚
1917, 18wcel 1977 . . . . . . 7 wff (𝑎(midG‘𝑔)𝑏) ∈ 𝑚
2012, 14, 7co 6549 . . . . . . . . 9 class (𝑎(LineG‘𝑔)𝑏)
21 cperpg 25390 . . . . . . . . . 10 class ⟂G
225, 21cfv 5804 . . . . . . . . 9 class (⟂G‘𝑔)
2318, 20, 22wbr 4583 . . . . . . . 8 wff 𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏)
249, 13weq 1861 . . . . . . . 8 wff 𝑎 = 𝑏
2523, 24wo 382 . . . . . . 7 wff (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏)
2619, 25wa 383 . . . . . 6 wff ((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏))
2726, 13, 11crio 6510 . . . . 5 class (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏)))
289, 11, 27cmpt 4643 . . . 4 class (𝑎 ∈ (Base‘𝑔) ↦ (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏))))
294, 8, 28cmpt 4643 . . 3 class (𝑚 ∈ ran (LineG‘𝑔) ↦ (𝑎 ∈ (Base‘𝑔) ↦ (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏)))))
302, 3, 29cmpt 4643 . 2 class (𝑔 ∈ V ↦ (𝑚 ∈ ran (LineG‘𝑔) ↦ (𝑎 ∈ (Base‘𝑔) ↦ (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏))))))
311, 30wceq 1475 1 wff lInvG = (𝑔 ∈ V ↦ (𝑚 ∈ ran (LineG‘𝑔) ↦ (𝑎 ∈ (Base‘𝑔) ↦ (𝑏 ∈ (Base‘𝑔)((𝑎(midG‘𝑔)𝑏) ∈ 𝑚 ∧ (𝑚(⟂G‘𝑔)(𝑎(LineG‘𝑔)𝑏) ∨ 𝑎 = 𝑏))))))
Colors of variables: wff setvar class This definition is referenced by: lmif 25477 islmib 25479
Copyright terms: Public domain W3C validator | 1,545 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.194505 |
https://math.stackexchange.com/questions/1408729/f-mathbbn-rightarrow-mathbbn-is-a-one-to-one-function-such-that-fmn-f | 1,571,177,694,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00114.warc.gz | 617,807,573 | 29,727 | # $f:\mathbb{N}\rightarrow \mathbb{N}$ is a one-to-one function such that $f(mn)=f(m)f(n).$ Find the lowest possible value of $f(999)$. [duplicate]
$f:\mathbb{N}\rightarrow \mathbb{N}$ is a one-to-one function such that $f(mn)=f(m)f(n).$ Find the lowest possible value of $f(999)$.
The answer is given as $24$ but I never get that.
## marked as duplicate by Chris Culter, user91500, Jeremy Rickard, Claude Leibovici, EmpiricistAug 25 '15 at 8:13
• Wow, a one-to-one function which takes the same value at $mn$ and $n$... :). – Erick Wong Aug 25 '15 at 4:30
• @Erik Wong Sorry just edited it. – tatan Aug 25 '15 at 4:32
• Hint: Start by showing that $f(n)$ is determined by the values of $f(p)$ at all primes $p$. Which primes determine $f(999)$? – Erick Wong Aug 25 '15 at 4:33
• $3,37$ are the primes. – tatan Aug 25 '15 at 4:37
• Now $f(3), f(37)$ cannot be $1,$ so we might take $\begin{cases}f(3)=2\\f(37)=3\end{cases}.$ And the lowest value is $2^3\times3.$ – awllower Aug 25 '15 at 4:40
$$f(999)=f(37)\cdot f(3)\cdot f(3)\cdot f(3)$$
Now, $$f(3)$$ or $$f(37)$$ cannot be $$1.$$ Because, if, for example, $$f(3)=1,$$ then $$f(999)=f(37)$$ which implies $$999=37$$ (because the function is one-to-one).
Therefore, $$f(3)$$ can have the lowest value $$2.$$ And hence $$f(37)$$ can have the lowest value $$3.$$ Thus, $$f(999)=3\cdot 2\cdot 2\cdot 2=24.$$
• Please use MathJax formatting on this site--enclose math in dollar signs, and use \cdot for times. See the MathJax basic tutorial and quick reference. – 6005 Aug 25 '15 at 4:59
• Re: "$f(3)$ can have lowest value $2$. And hence $f(37)$ can have lowest value $3$": It could be the other way around. However, that would make $f(999) = 2\cdot 3\cdot 3\cdot 3 > 3 \cdot 2 \cdot 2 \cdot 2$. – Christopher Carl Heckman Aug 25 '15 at 5:02
$$m=n=1 \Rightarrow f(1)=1$$ so we have $f(3)\ge 2,f(37)\ge 2$ since $f(3)\neq f(37)$
$$f(999)=f(27)\cdot f(37)=(f(3))^3 \cdot f(37)\ge 2^3\cdot 3=24$$
Let us define $f(2)=37,f(3)=2,f(37)=3$ and $f(p)=p$ for all primes $p\neq 2,p\neq 3,p\neq 37$
Let us show the function is injective.
The condition implies $f(\prod_{i=1}^k p_i^{e_i})=\prod_{i=1}^k f(p_i)^{e_i}$
let $m=2^{a}\cdot 3^{b}\cdot 37^{c}\cdot\prod_{i=1}^k p_i^{e_i},n=2^{p}\cdot 3^{q}\cdot 37^{r}\cdot\prod_{i=1}^l p_i^{e_i},$ be prime factorizations of $m$ and $n$ and $f(m)=f(n)$ then
$$f(m)=f(n) \Leftrightarrow 37^{a}\cdot2^{b}\cdot 3^{c}\cdot \prod_{i=1}^k p_i^{e_i}=37^{p}\cdot2^{q}\cdot 3^{r}\cdot \prod_{i=1}^l p_i^{e_i} \Rightarrow$$
$$a=p,b=q,c=r,\land \prod_{i=1}^k p_i^{e_i}=\prod_{i=1}^l p_i^{e_i} \Rightarrow m=n$$
so the funciton is injective | 1,052 | 2,626 | {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-43 | latest | en | 0.673408 |
https://www.jiskha.com/users?name=Kelly+Brians | 1,597,264,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00391.warc.gz | 722,372,310 | 3,326 | # Kelly Brians
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https://www.physicsforums.com/threads/combinatorics-intense-question.571121/ | 1,542,561,154,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744513.64/warc/CC-MAIN-20181118155835-20181118181835-00553.warc.gz | 943,771,242 | 13,641 | # Homework Help: Combinatorics intense question!
1. Jan 26, 2012
### newchie
1. The problem statement, all variables and given/known data
How many non- empty subsets of {1,2,3...,15} have the following two properties?
1) No two consecutive integers belong to S.
2)If S contains K elements, then S contains no number less than K .
2. Relevant equations
choosing identities, not sure which one
3. The attempt at a solution
Tried casework, but I dont understand how the solution gets
n-k+1 choose k
2. Jan 26, 2012
### genericusrnme
you said n-k+1 choose k
what is n?
3. Jan 26, 2012
### newchie
If we have ways are there to choose k elements from an ordered n element set without choosing two consecutive members
4. Jan 26, 2012
### genericusrnme
Okay, so what does it mean when you take a choose b?
You're taking the number of ways to choose b items out of a, right?
So why would you suppose the answer to your question is n-k+1 choose k?
What is the significance of n-k+1?
5. Jan 26, 2012
### newchie
http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_12A_Problems/Problem_25 [Broken]
I dont understand the solution
Last edited by a moderator: May 5, 2017
6. Jan 26, 2012
### genericusrnme
What about it do you not understand?
7. Jan 26, 2012
### newchie
Defining the boxes into n-2k+1 elements, then the general choosing statement, i get the first part, but the second part is a bit opaque
8. Jan 27, 2012
### obafgkmrns
Interesting problem! Fiddling with pencil & paper and looking for patterns suggests that for n integers, the answer involves sums of triangle numbers, T(n) = n(n+1)/2, but I don't see a good way to get there from simple combinetrics.
Last edited: Jan 27, 2012 | 490 | 1,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-47 | latest | en | 0.859204 |
https://physics.stackexchange.com/questions/159929/what-does-the-equation-of-a-refracted-ray-trace-mean | 1,563,880,377,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529276.65/warc/CC-MAIN-20190723105707-20190723131707-00077.warc.gz | 495,794,902 | 36,997 | # What does the equation of a refracted ray trace mean?
I am doing a ray trace to a refracted vector. I read in some PDF files that: vector($\mathbf{t}$) is the refracted ray, vector($\mathbf{i}$) is the incident ray, the angle $\theta_t$ is the refraction angle and angle $\theta_i$ is the incident angle. To calculate the refracted ray we use this equation:
$$\mathbf{t}=\frac{\eta_1}{\eta_2}\mathbf{i}+\left(\frac{\eta_1}{\eta_2}\cos \theta_i-\sqrt{1-\sin^2 \theta_t} \right)$$
But I really didn't get what this equation represents. What information does it give about vector $\mathbf{t}$?
• Right off the bat I can see that something is wrong here, the first part of your equation is a vector (scalar multiplied by $\hat{\imath}$), but it is added to a scalar. Maybe the second part is supposed to be multiplied by the surface normal, $\hat{n}$? – 2012rcampion Jan 17 '15 at 20:24
Since it looks to me like there's a mistake in the equation, I'll try to re-derive it. Hopefully that will make it a little clearer what's going on.
We'll start with Snell's law: $$\frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1}$$ Now there are several vectors we're going to deal with, I'll go through them one by one.
• $\hat\imath$: The vector in the direction of the incoming ray.
• $\hat{t}$: The vector in the direction of the transmitted ray.
• $\hat{n}$: The vector perpendicular to the surface, the normal vector.
• $\hat{n}_\perp$: A vector perpendicular to the normal vector, in the plane of $\hat\imath$ and $\hat{t}$.
Let's write out some relations between these vectors. We can express $\hat\imath$ and $\hat{t}$ in terms of the direction of the surface and their angles to the surface.
$$\hat\imath = -\cos\theta_1\hat{n}+\sin\theta_1\hat{n}_\perp \\ \hat{t} = -\cos\theta_2\hat{n}+\sin\theta_2\hat{n}_\perp$$ (the minus sign comes from the fact that the ray is headed towards the surface.)
Since we don't know $\hat{n}_\perp$, let's eliminate it from those equations. $$\hat{n}_\perp = \csc\theta_1\hat\imath+\cot\theta_1\hat{n} \\ \hat{n}_\perp = \csc\theta_2\hat{t}+\cot\theta_2\hat{n} \\ \csc\theta_1\hat\imath+\cot\theta_1\hat{n} = \csc\theta_2\hat{t}+\cot\theta_2\hat{n}$$ Let's try to solve for $\hat{t}$. $$\csc\theta_2\hat{t}=\csc\theta_1\hat\imath+\cot\theta_1\hat{n}-\cot\theta_2\hat{n} \\ \hat{t}=\frac{\sin\theta_2}{\sin\theta_1}\hat\imath+\cos\theta_1\frac{\sin\theta_2}{\sin\theta_1}\hat{n}-\cos\theta_2\hat{n}$$ Those ratios of $\sin$'s are in Snell's law, so let's replace them and consolidate terms. $$\hat{t}=\frac{n_1}{n_2}\hat\imath+\left(\cos\theta_1\frac{n_1}{n_2}-\cos\theta_2\right)\hat{n}$$ Finally let's replace the $\cos$ with a $\sin$. $$\hat{t}=\frac{n_1}{n_2}\hat\imath+\left(\cos\theta_1\frac{n_1}{n_2}-\sqrt{1-\sin^2\theta_2}\right)\hat{n}$$ So we've recovered (almost) your original expression. It gives you the direction of the transmitted ray given the direction of the surface normal and incoming ray.
Note that $\cos\theta_1=\hat\imath\cdot\hat{n}$ and $\sin\theta_2=\frac{n_1}{n_2}\sqrt{1-\cos^2\theta_1}$.
In order to implement this equation, you'll need to add and multiply vectors, so let's talk vector arithmetic. For a more complete understanding I recommend you do some reading up on the subject.
A vector $\vec{v}$ has 3 components (in 3d space), I'll denote them $v_x$, $v_y$, and $v_z$.
The product of a vector $\vec{v}$ and a scalar (regular number) $a$ is a vector: $$a\times(v_x,v_y,v_z) = (av_x,av_y,av_z)$$
The sum of two vectors $\vec{v}$ and $\vec{u}$ is a vector: $$(v_x,v_y,v_z)+(u_x,u_y,u_z) = (v_x+u_x,v_y+u_y,v_z+u_z)$$
The dot product of two vectors $\vec{v}$ and $\vec{u}$ is a scalar: $$(v_x,v_y,v_z)\cdot(u_x,u_y,u_z) = v_x u_x+v_y u_y+v_z u_z$$
The magnitude (length) of a vector $\vec{v}$ is $|\vec{v}|=\sqrt{\vec{v}\cdot\vec{v}}$. Directions, like the directions of the light rays in raytracing, are represented by unit vectors: vectors with unit magnitude. We typically denote unit vectors with a hat instead of an arrow (e.g. $\hat{n}$). For a unit vector $\hat{n}$, it's always true that $\hat{n}\cdot\hat{n}=1$. To make a unit vector (normalize) any vector, simply divide that vector by its magnitude (equivalent to multiplying by one over the magnitude).
Note that you'll sometimes see the three coordinate axes labeled by $i$, $j$, and $k$ or $1$, $2$, and $3$. Also, you'll see vectors written as sums of the unit vectors along the coordinate axes, e.g: $$\vec{v}=(v_x,v_y,v_z) \\ =v_x\hat{x}+v_y\hat{y}+v_z\hat{z} \\ =v_x\hat{i}+v_y\hat{j}+v_z\hat{k} \\ =v_x\hat{e}_1+v_y\hat{e}_2+v_z\hat{e}_3$$
• thank you for your post. but actually I didn't get the main point I need, if I need to deal with the refracted vector (I have it as 3D) as its (x,y,z) components.Can I know them from the last equation you wrote?@2012rcampion – user3159060 Jan 17 '15 at 21:49
• @user3159060 I'll modify my post to include a little primer on vector arithmetic. – 2012rcampion Jan 18 '15 at 1:48
• @user3159060 If this answered your question, please mark this as the correct answer by clicking on the check mark to the left of the answer. – 2012rcampion Jan 25 '15 at 19:04 | 1,701 | 5,148 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-30 | latest | en | 0.89355 |
https://www.electronicspoint.com/threads/calculating-the-half-power-frequency.232161/ | 1,487,983,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171632.91/warc/CC-MAIN-20170219104611-00236-ip-10-171-10-108.ec2.internal.warc.gz | 819,649,359 | 14,007 | # Calculating the half power frequency
Discussion in 'Circuit Help' started by kdoug, Mar 1, 2011.
1. ### kdoug
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I would like to calculate the half power frequency of the circuit attached.
I have combined the impedances of R2 and C1 with the 1/Z = 1/Z_1 + 1/Z_2 ... forumla. To get R2 / (R2*j*w*C + 1).
Then use the voltage divider on this combination and R1 to find that,
Vin / Vout = 1 / ( R1jwC + (R1/R2 +1) )
I believe the gain now is |Vin / Vout| = 1 / sqrt( (R1wC)^2 + (R1/R2 + 1)^2 )
At this point, I'm unsure how to proceed. The Internet has told me that the half power frequency occurs when | Vin / Vout | = 1/sqrt(2). If I set what's contained the the radical above equal to 2 and solve for w, I can find the frequency f. I turns out to 141.5 Hz which doesn't agree with a computer generated Bode plot.
Can anyone suggest what I'm doing wrong?
Sorry for the poor Maths formatting.
Thanks for your time,
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kdoug, Mar 1, 2011
2. ### LaplaceVIP Member
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The gain of a network is Vo/Vi. The gain of this network is a function of frequency. Since this is a low-pass filter, the highest gain will be found at f=0 so the gain at the cutoff frequency will be half the power of the gain at zero frequency.
An alternative method is to realize that the cutoff frequency of a simple low-pass filter is Fc=1/(2piRC) where C=C1 and R=R1| R2. My calculator says Fc=175 kHz. What does the computer generated Bode plot say?
Laplace, Mar 2, 2011
3. ### kdoug
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Thank you for the response, Laplace.
I've attached the generated Bode plot. I've highlight the cursor representing your answer in green, and what I would have though the right answer is in red (70.7% of 5V, makes for a frequency of approx 325 kHz), though, what I think to be the right answer, appears to have occured after the main "break down" in the curve.
It's likely I've misunderstood something again, or I'm not using my tools properly. I used PSpice to generate this Bode plot from the circuit I attached previously.
Thanks for your time & help.
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kdoug, Mar 2, 2011
4. ### LaplaceVIP Member
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Here are my calculations doing it the long way.
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Laplace, Mar 3, 2011
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# Classs Test (Class ) Subject-Mathematics M.M: Q1) Multiple choice questions. (I) A tent of conical shape of base radius and slant height is th made from a certain material. What is the total cost of clothing materi the rate of material is (Take ) a) ₹ 140800 b) ₹ 150000 c) d) ₹ 140000 (II) The volume of two spheres are in the ratio of ' . The ratio of the surface area is : a) b) c) d) (III) The ratio of height and the diameter of a right circular cone is a its volume is . Then, its height is (Take ) a) b) c) d) (IV )The circumference of the circle i.e. base of a cone is . If the s height of cone is , then its curved surface area will be: a) b) c) d)
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http://andrewcannon.net/graphing-stories-worksheet/ | 1,623,737,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00015.warc.gz | 4,985,860 | 17,755 | # Graphing Stories Worksheet
Posted in Worksheet, by Kimberly R. Foreman
Worksheet focuses on reading a situation or story and comparing the changes over time to other variables like speed, distance, or temperature and matching to a graph. students choose from one of three graphs making comparisons to match the correct situation to the correct.
subjects math, algebra, word problems. Graphing stories worksheet. . date every graph tells a story interpreting graphs. ms. teaches an advanced placement class during period. at the end of the year, these students take an advanced placement test.
## List of Graphing Stories Worksheet
The graph shows the number of students who story displaying top worksheets found for this concept. some of the worksheets for this concept are graphing stories, graphing lines, stories from graphs, concept writing graphing inequalities, story of a graph work, baseball bar graph, chemistry i summer assignment graphing and analyzing, match it.
This activity will help students make the transition from representations e. g. , number lines to the representation of the coordinate plane. students will watch second videos and translate them into graphs with your help. special thanks to for his videos.
### 1. 9 Reading Straight Line Graphs Worksheet Line Graph
Microsoft word scary graphing stories. docx author, created date pm. Graphing stories short video stories that help students learn to graph on a plane. Graphing worksheets for practice. here is a graphic preview for all of the graphing worksheets. you can select different variables to customize these graphing worksheets for your needs.
the graphing worksheets are randomly created and will never repeat so you have an endless supply of quality graphing worksheets to use in the classroom or at home. Chunking graphing stories into act math tasks. and worked collaboratively with a group of math teachers to create some really impressive resources for and many other relationships comparing a dependent variable over time.
Answers. ,. ,. massacre,. redcoats soldiers,. die. the colonists in that were upset with parliament were called the sons of. the colonists felt that the sugar act, act, massacre and now the tea act were unfair. the colonists dressed as, onto ships, and threw tea into the harbor as A printable worksheet on, the first person to die for the revolutionary cause he died in the massacre.
the printout has information on life, questions about him, and a picture to color. answers. ,. The tea party was over. the colonists in had ruined a lot of tea. would react very strongly. they passed a new set of laws. the new laws would punish. the laws were known as the intolerable acts.
### 3. Learning Parts Plot Worksheet Line Plot Worksheets
Draw arrows to show which bits of the story connect to each paragraph. how can you tell when stopped just by looking at the graph how can you work out the speed travelled for different parts of his with a story, but as a group. you will be given ten graph cards and ten story cards.
in your group take a graph and find a story that matches it. alternatively, you may want to take a story and find a graph that matches it. take turns at matching pairs of cards. each time you do this, explain your thinking clearly and carefully. if in this math lesson students will interpret graphs to find the story that the graph is telling and then make their own graph and story using an event in their own life.
Students will love a lesson that allows them to see how math can tell a story and then relate it to stories in their own life. These graph worksheets will produce a chart of data for two lines and a single coordinate grid to graph the data on. you may select the difficulty of the graphing task.
double line graph comprehensions worksheets these graph worksheets will produce a single coordinate double line graph and questions based off the graph to answer. May, the graph in the video is about two runners who run the final race. their distance time graphs are told as a moving and exciting story.
### 5. Making Graphs Stories Practice Sheet Worksheets
The is the graph i used in the story and i have included the script, so you can do it with your classes. This matching stories with graphs worksheet is suitable for grade. in this five page worksheet, students are provided with five sets of graphs from which they must answer questions.
students need to describe in words and with appropriate units what information the graphs are displaying. Intro graph functions and stories displaying top worksheets found for this concept. some of the worksheets for this concept are stories from graphs, math, functions graphs and graphing tasks learning and teaching, function table and graph work, definition of a function and evaluating a function domain, algebra name lesson graphing absolute value, algebra i notes relations and First grade graphing data worksheets and learning how to collect, organize, and plot data on a graph is one of the many important components of early education math.
### 6. Matching Activity Helps Students Understand
After explaining the differences between bar, tally, and picture graphs, our first grade graphing worksheets turn kids loose to sharpen their counting, addition, and deductive reasoning skills. Exploring basic behavior of graphs, including, , increasing, decreasing, constant, maximum, and minimum.
Bar graph worksheets. represent data as rectangular bars of different heights in this collection of bar graph worksheets. included here are topics like graphing by coloring, comparing tally marks, reading and drawing bar graphs and double bar graphs. Pictograph worksheets.
### 7. Math Magical Swipe Stories
The worksheets on this page contain picture graphs with symbols. bar graph worksheets. printable bar graph worksheets with questions. pie graph worksheets. a collection of circle graph pie graph worksheets. line plot worksheets. print these line plot worksheets to teach frequency distribution of numbers.
Graphing stories name. time seconds time seconds. L ho. x e rd n. x c mo. worksheet by software software infinite algebra name graphing lines date worksheets for preschool and kindergarten including reading bar charts, grouping, sorting and counting items to complete a bar chart, and analyzing a bar chart.
### 8. Plot Diagram Worksheet Short Story Teaching Plot Plot
See document summaries page for details. doc addams,Argument analysis text what is the title who is the author what type of text is it where and when was it published speaker who is making the argument what is their background what biases might they have audience who is this piece for how do you know context what was happening in the world at the time this piece was analysis worksheet the moons of based on a worksheet prepared by the national archives and records administration, dc.
### 9. Procedural Writing Images
When you graph the story, you do not need to use exact data, but you do need to label each axis. Distance vs time graph worksheet. the accompanying graph shows distance from home a to work f at various times during her drive. a left her briefcase at home and had to return to get it.
state which point represents when graphing stories collection is a collaborative project run by and the team. each story comprises a video that shows a practical situation where a measure, such as height or weight, varies with time. the situation is then replayed at a slower rate to help enable the viewer to assess more clearly how the.
### 10. Sea Life Math Addition Subtraction Roll Cover
E la x f. r pezdh. t g y. h worksheet by software software infinite algebra name solving systems of equations by graphing date, graphing is one of many keystone mathematical skills for which early exposure makes all the difference. schools today teach their pupils to graph and interpret data and charts as soon as possible, and this leads to more success later more Sep, explore wideners board qualitative graphs, followed by people on.
see more ideas about middle school math, graphing, high school math. May, this fun math activity is a great twist on traditional graphing worksheets this post contains affiliate links. its week five in the k math activities series with to and this reading mama this week our theme is data.
### 11. Additional Graphing Worksheet Titles
Today sharing a second grade activity. i have a second grader at my house, but my. The second recurring structure from yesterdays lesson is graphing stories video. please take a look at the graphing stories section of yesterdays lesson for a detailed description of how i enact this part of the lesson.
here are the my key points for using one of these videos keep it simple and stay quiet. , editorialize, or say anything that might prevent. If an object is not moving, a horizontal line is shown on a graph. time is always plotted on the bottom of the graph.
### 12. Sequencing Events Story Worksheets
The further to the right on the axis, the longer the time from the start. distance is plotted on the side of the graph. the higher up the graph, the further from the start. These worksheets focus on the basic elements of stories including characters, plot and setting.
choose your grade topic grade story elements worksheets. identify the characters, setting and main plot of short texts. grade character, setting and plot, graphing stories videos. the graphing stories website has short videos that are examples of these types of stories that show functions in real life.
### 13. Story Elements Worksheet Images Story Elements
First, students watch a video showing a story or situation. students follow along and draw an accompanying graph. then, in the video they show a graph overlaying the video and how the graph. In this graph stories lesson, students also create their own stories to go along with graphs given.
get free access see review. in this graphing movements worksheet, students create graphs of either position vs. time or velocity vs. time given written stories. get free access see review. We also offer free math worksheets for offline use kids can use these tools to practice understanding diagrams and interpreting picture graphs, tally charts and tables interpreting and creating bar graphs, pictographs, and line plots graphing images and mapping decimal and fractional, wanting to stress contexts from day, i used the site graphing stories by and, as an opener a few times within the first weeks of class.
### 14. Story Sequencing Worksheets Kindergarten Planting
The site provides seconds video of a scenario such as time vs height on a rope swing, and students use graph paper to develop a graph which matches the scenario. Line graph worksheets have ample practice skills to analyze, interpret and compare the data from the graphs.
exercises to draw line graphs and double line graphs with a suitable scale labeling the axes giving a title for the graph and more are included in these printable worksheets May, now, seven months later has created a graphing stories website and very generously offered to create beautiful graphing stories videos from stories we all send in i am excited about this extension of their graphing stories for my students.
### 15. Systems Graphing Stories Worksheet Arithmetic
This is perfect timing for them as we have just begun reviewing material in preparation for their final. Standards literature. ccss. literacy. rl. describe the overall structure of a story, including describing how the beginning introduces the story and the ending concludes the action.
ccss. literacy. rl. refer to parts of stories, dramas, and poems when writing or speaking about a text, using terms such as chapter, scene, and stanza describe how each successive part. These graphing worksheets are perfect for any child that is just trying to learn the concept of bar type charts and they are simple enough that kids will be able to understand and complete.
### 16. Thanksgiving Math Worksheets Grade Thanksgiving
We have several different themes to choose from as well including shapes, food, sports, animals and more. Lets algebra teachers make perfectly customized linear functions worksheets, activities, and assessments in seconds. start by browsing the selection below to get word problems, projects, and more.
puts the kind of material you find in software, math aids,, and illustrative mathematics all in one place. This page contains all our printable worksheets in section data, graphs, and probability of first grade math. as you scroll down, you will see many worksheets for tally tables, picture graphs, interpreting data, bar graphs, diagrams, possible or impossible, more likely or less likely, and more.
### 17. Trip Mall Graph Story Graphing Common Core Math
A brief description of the worksheets is on each of the worksheet widgets. Kinematics graph worksheet or google doc kinematics graph worksheet or google doc kinematics graph worksheet or google doc kinematics graph worksheet or google doc rubric here must be logged in to school account kinematics graph worksheet or google doc rubric here must be logged in to school is a crucial skill for many subjects, and its important for understanding the news, jobs, and beyond.
our graphing worksheets help students of all levels learn to use this tool effectively. introduce younger students to the basics of collecting and organizing data. Using the, try to create the graph of each of the following lines. each person on your team must try to walk each story below.
### 18. Twist Graphing Kindergarten Crayons Graphing
As each person tries, sketch the line they created and write a suggestion for what should be done to make their graph more closely match the story given. a line that rises at a steady rate. number story displaying top worksheets found for graph number story.
some of the worksheets for this concept are safari adventure, baseball bar graph, distance vs time graph work, concept writing graphing inequalities, graphing lines, lesson interpreting graphs, favorite girl scout cookie, inequalities. Simple story with at least different motions.
### 19. Kindergarten Math Winter Worksheets Educational
Three motion graphs displaying the movement of the character in the story. these may be displayed or on graphs demonstrating the entire story. o position graph o velocity graph o acceleration graph scoring rubric total story plot motions position worksheets.
we cover each form of charting and graphing data. we really just have graph paper here. we have a huge curriculum of printable worksheets and lessons for this skill. we cover all types of graphs. how to create graphs and how to interpret them. Graphing a system of equations algebra.
### 20. Worksheet Ideas Elements Story Worksheet Worksheets
More holiday worksheets. puzzles brain teasers. brain teasers. logic addition squares. mystery graph pictures. number detective. lost in the. more thinking puzzles. teacher helpers. teaching tools. award certificates. more teacher helpers. An engaging worksheet for your students to practise the skill of setting out data in a stem and leaf plot.
### 21. Image Result Story Graph Narrative Story Book
Unit writing and graphing quadratics worksheet practice packet learning targets unit. i can use the discriminant to determine the number and type of. modeling with quadratic functions. i can identify a function as quadratic given a table, equation, or is the best collection of free sequencing worksheets you will find on the internet and they are free it is an important building block for children to be able to put together events in the appropriate sequence.
### 22. Free Printable Worksheets Kids Great
Each of the worksheets below shows parts of a specific story Find worksheets about coordinate picture graphing. coordinate picture graphing. coordinate pictures are a way of helping to reinforce plotting skills with a game of. each series of points connects to form a line.
### 23. Blank Plot Diagram Luxury Story Climax
Begin with the plot of your story and build up from there up to the critical moments and the character development, creating an impressive outline for your future success. Jan, its okay if it relates exactly the same or somewhat differently, its just important that you understand the themes of both separately before combining them together in an outline and short story.
### 24. Character Analysis Story Graph Close Reading
The larger velocity-time graph shows the motion of some hypothetical object over time. Worksheet constant velocity position-time graphs match the description provided about the behavior of a cart along a linear track to its best graphical representation.
negative slopes represent motion in a negative direction - zero slopes represent an object remaining in one position, that is, at rest. at rest traveling slowly in. This graphical analysis worksheet students will work on in their groups to extend their practice with graphing and analysis of graphs.
### 25. Coordinate Worksheets Printable Plot Coordinates
Interpreting. Reading charts and graphs worksheet mode and range worksheets though little readers should enjoy filling in the blanks the attached worksheet, lesson and key is for the more advanced students. reading a bar graph worksheet you not only learn how to read these graphs, but the ultimate goal is to be able to make.
### 26. Desert Sage Gifted Talented Line Plot
Bar graph worksheets from www. superteacherworksheets. com information is labeled and points and identifies different parts. visit the reading worksheet section on our in our reading comprehension worksheet section you find varying grade levels of reading worksheets that also come with multiple choice, free.
### 27. Hungry Caterpillar Theme Free Graph Worksheets
Some of the worksheets displayed are. Free printable graphic organizer worksheets and blank charts for k teachers and students browse our selection and click on your choice free to print. selection includes items such as brainstorming, concept circle maps, diagrams, family tree charts, circular flow charts, graph paper, story elements, weekly planner, compare and contrast, and many more.
### 28. Fall Worksheets Images Fall Kids
Line graph worksheets from graphing lines worksheet, sourcemathworksheetskids. com. over free stories followed by comprehension exercises, as well as. dimensional analysis worksheet answer key. great reading comprehension worksheets for Graph stories worksheet.
### 29. Gingerbread Graphing Lesson Plans Mailbox
We found some images about graph stories, these worksheets are perfect as revision after you have created a picture graph with the kids from the story using the free food picture printable that i have created. i created two worksheets because i let the children work in a pair, each doing a different worksheet.
### 30. Gingerbread Writing Freebie Math
Help kids learn about graphing with this fruit themed graphic worksheet. children are asked to look at the bar graph which shows kids favorite fruit. the first page shows is a bar graph for apples, oranges, strawberries and bananas. Sep, thoughts on mountain climber problem a nice follow up to graphing stories , at pm.
### 31. Graphing Stories Situations Worksheet Activity
Love the moment at the end. a sense of surprise in math class is a rare and pretty wonderful thing. As the name suggests, graph it is a math worksheet that requires kids to draw a simple chart. giving kids just the basics to start off with, this graphs worksheet requires them to count the number of various objects and mark it correctly on the chart.
### 32. Graphing Story Lesson Worksheet Real Life Math
Once mastered the basics, they will soon be able to analyze graphs the worksheet thing, i love to give the students the big picture application, before giving them the practice. but having both is so important to most students feeling successful in math.
### 33. Hearts Lesson Plans Mailbox Graphing
I will be adding a link to this from my website. thanks again for sharing your work. worksheets. unit rates and graphs worksheet integers this problem worksheet features graphs that represent everyday situations. some of the unit rates are obvious, but on some problems students will have to analyze the graph scale to identify the correct unit rate.
### 34. Genre Worksheets Grade Graph Worksheets Line Plot
Dec, graphing stories is a new website that was developed by and buzz math. graphing stories features short videos that tell a story that students can graph to tell the mathematical story happening in the video. i recommend reading this post by to get a full sense of how this works and where the idea came from.
### 35. Distance Time Graphs Distance Time Graphs Worksheets
Use the graph below to answer the questions. number of runs players name number of runs mark patty. how many runs did have. Created for a grade introduction to motion graphs acceleration, constant, and at rest. print and paste on index cards.
### 36. Step Worksheet Steps Teaching Plot
, made a searching and fearless moral inventory of ourselves. step four of program of recovery is infamously the scary one, probably because its a crucial step towards effective and lasting recovery. since the overall philosophy of alcoholics anonymous is that alcoholism is just a symptom of a spiritual disease, the real problem is in character flaws that need.
Sep, , celebrate recovery step inventory worksheets. saved from housview. com. celebrate recovery step inventory worksheets. saved by. steps anon therapy worksheets free worksheets alcoholics anonymous narcotics. Id language school subject as a second language age main content reading comprehension other contents add to my workbooks download file embed in my website or blog add to google worksheet printable step worksheets, source image i.
### 37. Coordinate Math Worksheet Practice Math Images
This will involve the basic operations, ratios, fractions, rounding off and percentages. of course related to the grade year the student is in. Nov, enzyme graphing worksheet answer key or this video shows how graphs can be used to tell stories the graph worksheet, we tried to locate some good of enzyme graphing worksheet answer key or this video shows how graphs can be used to tell stories the graph image to suit your needs.
### 38. Hidden Pictures Printable Kindergarten
Year title description dot grid dots spaced at apart. mainly for picks formula and transformations reflections, rotations translations. straight line geometry angles on straight lines and in triangles. transformations this worksheet explores the relationship between reflection and the other two transformations translation and rotation.
### 39. Kindergarten Math Math Grade Math
Draw your own graph for this graphing story. use straight line segments in your graph to model the elevation of the man over different time intervals. label your axis and Matching graphs to situations worksheet. problem , and are studying words for a spelling bee.
### 40. Blank Plot Diagram Blank Plot Diagram Worksheet
Started by learning how to spell many words each day but then learned fewer and fewer words each day. learned how to spell the same number of words each day. started by learning how to spell the same number of words each day but then learned Free worksheets and more since.
Used under license. Demonstrate your problem solving skills by interpreting and sketching graphs. this interactive exercise focuses on how line graphs can be used to represent mathematical data and provides an opportunity to translate actions from a story into graph form, then to write your own story to coincide with a line graph.
Graphing your story lesson and worksheet in this math lesson students will interpret graphs to find the story that the graph is telling and then make their own graph and story using an event in their own life. students will love a lesson that allows them to see how math can tell a story and then relate it to stories in their own. | 4,644 | 24,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-25 | latest | en | 0.913099 |
https://water-fire-damage-restoration-urbandale-ia.com/qa/question-what-is-inductance-used-for.html | 1,618,434,963,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00622.warc.gz | 722,310,411 | 6,999 | # Question: What Is Inductance Used For?
## How is inductance value calculated?
Calculate the inductance using the following formula: Inductance = µ (N squared) A / length, where N is the number of turns in the coil, A is the cross-sectional area of the coil, and length is the length of the coil..
## Does inductance depend on frequency?
The inductance of an inductor depends upon its construction and is a constant for a given inductor. In other terms, the inductance of the inductor is a constant and does not depend on the current or on the frequency.
## What is an inductor used for?
Inductors are used as the energy storage device in many switched-mode power supplies to produce DC current. The inductor supplies energy to the circuit to keep current flowing during the “off” switching periods and enables topographies where the output voltage is higher than the input voltage.
## What is inductance in a circuit?
In electromagnetism and electronics, inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. … An electronic component designed to add inductance to a circuit is called an inductor. It typically consists of a coil or helix of wire.
## What causes inductance?
Inductance is caused by the magnetic field generated by electric currents flowing within an electrical circuit. Typically coils of wire are used as a coil increases the coupling of the magnetic field and increases the effect.
## What is inductor in simple words?
An inductor, also called a coil, choke, or reactor, is a passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it. An inductor typically consists of an insulated wire wound into a coil.
## Is inductance a function of frequency?
The inductive reactance of an inductor increases as the frequency across it increases therefore inductive reactance is proportional to frequency ( XL α ƒ ) as the back emf generated in the inductor is equal to its inductance multiplied by the rate of change of current in the inductor.
## How does an inductor work?
An inductor is a passive electronic component which is capable of storing electrical energy in the form of magnetic energy. Basically, it uses a conductor that is wound into a coil, and when electricity flows into the coil from the left to the right, this will generate a magnetic field in the clockwise direction.
## What is the difference between inductance and impedance?
The opposition offered to the flow of current in an AC circuit because of resistance, capacitance and inductance is known as Impedance. Resistance occurs in both AC and DC circuit, whereas Impedance takes place only in an AC circuit. … Resistance is denoted by (R) whereas impedance by (Z).
## What is difference between inductance and capacitance?
Capacitance, as we now know, is the ability to store energy in the form of an electric field. Inductance, which is measured in henries and denoted by the letter L, is the ability to store energy in the form of a magnetic field.
## Where is self inductance used?
The applications of self-inductance include the following.Tuning circuits.Inductors used as relays.Sensors.Ferrite beads.Store energy in a device.Chokes.Induction motors.Filters.More items… | 706 | 3,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.939876 |
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1. Jul 16, 2012
### LindaNguyen
1. The problem statement, all variables and given/known data
Suppose three masses are arranged as shown, connected by a rodless mass.
(2 kg)-------(6 kg)--------(4 kg)
2m 3m
a) If this object is free to rotate in space, about what point will it spin?
Cm = 2(0) + 6(2) + 4(5)
_______________ = 2.7m
12
b) A small mass (m=0.1 kg) drops vertically onto the2 kg mass traveling 100 m/s and buries intself into the mass. What is the period of rotation of the system immediately after impact?
I'm not sure how I would begin setting this problem up.
c) What is the rotational kinectic energy of the system after the impact?
2. Jul 16, 2012
### cepheid
Staff Emeritus
Welcome to PF, LindaNguyen!
Looks good, except, are you sure that there wasn't some rounding error in your final answer? It's just a bit off from what I got. Oh, and you wrote "rodless mass", which I got a bit of a kick out of. (Not making fun of you, just thought it was a neat transposition).
The key physics principle here is that angular momentum around the centre of rotation has to be conserved. Even the small mass, when it is just moving through space, has a certain amount of angular momentum. The amount depends on the point around which you measure the angular momentum, since L = r x p = r x mv. At the moment of impact, the vector r between the centre of rotation and the little mass is perpendicular to the vector v, so this cross product is simple to evaluate. Since angular momentum is conserved, you can equate this to the angular momentum of the whole system (rod + embedded mass) after the impact. You should be able to compute this easily enough (you'll need to find the moment of inertia of the system around the rotation point). EDIT: once you know the angular momentum, the angular speed should follow form that.
This is just a matter of applying the right equation.
3. Jul 16, 2012
### LindaNguyen
Thank you, Cepheid! I'm not very good at posting on forums so bare with me. I'll learn how to quote and etc. soon enough!
Part A) Sorry for the typo, hehe. I have a million things going on right now. As for the rounding error, I'm still getting 2.666666. I'm trying to find the center of mass here, correct? The 2m was supposed to be the distance between the first 2 masses and the 3m was supposed ot be the distance between the second 2 masses. I'm not sure why it came out like that, but...
[ 2(0) + 6(2) + 4(5) ] / 12 ----> 32/12 = 2.7m.
I'm working out the other parts right now, will get back to you!
4. Jul 16, 2012
### cepheid
Staff Emeritus
The problem was that I can't do arithmetic.
Carry on.
5. Jul 16, 2012
### LindaNguyen
I'm still a tad bit lost on part B. Should I be finding the angular momentum about its center of mass?
6. Jul 16, 2012
### cepheid
Staff Emeritus
Yes, you want to find the angular momentum of the little mass around the centre of mass of the rod system at the moment of impact. Then equate this to the angular momentum of the entire system around this point after the collision. This is because angular momentum is conserved, so it should be the same before and after the collision.
I gave you the equation for finding the angular momentum of a single point mass around a point in space. The position vector r is a vector giving the position of the point mass relative to the location around which you're computing the angular momentum (which is the centre of mass of the rod in this case).
7. Jul 16, 2012
### LindaNguyen
Ah, I get it now (kind of)
L=IW (true for any object spinning around it's cm)
L=RxP (around an exterior point)
-----------------------------------------------------------
so, after I find the angular momentum about the small mass, I can set it to the whole system using L=IW, correct?
also.. if I'm finding the angular momentum of the small ***, would it look something like: (2.7i, 0j, 0k) x (0i, -10j, 0k) ---> -10 coming from mv (velocity is going down?)
8. Jul 16, 2012
### LindaNguyen
Here's what I've come up with:
r x p = i'm getting 27.
if I do Ii Wi = If Wf
--->using I=mr^2 ; w=v/r ; r=CM=2.7m
I end up with 27 also.
now If Wf = 2(2.7)^2 + 6(0.7)^2 + 4(2.3)^2 * w
I know I solve for W after this but I'm not sure if I have the numbers in the ( ) correct. | 1,164 | 4,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-43 | longest | en | 0.94077 |
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Bollinger Bands: A Powerful Trading Tool
Bollinger Bands: A Powerful Trading Tool
Bollinger Bands are a versatile technical analysis tool used by traders to measure market volatility, identify trends, and spot potential trade opportunities. Developed by John Bollinger in the 1980s, Bollinger Bands are comprised of three lines that are plotted on a price chart. These lines indicate the upper and lower bounds of a security’s price range and are used to identify potential buy and sell signals.
In this article, we’ll explore the basics of Bollinger Bands, how to calculate them, how to interpret readings, and outline some common trading strategies. We’ll also discuss the advantages and disadvantages of using Bollinger Bands and highlight some common mistakes that traders make when using this tool.
Calculating Bollinger Bands: The Basics
Bollinger Bands are calculated using three lines plotted on a price chart. The middle line is a moving average, typically a 20-period simple moving average (SMA), and represents the security’s average price over the specified time period. The upper and lower bands are plotted two standard deviations away from the middle line. The standard deviation is a measure of volatility, so the farther the bands are from the middle line, the higher the volatility.
The formula for calculating the upper and lower bands is:
Upper band = Middle band + (2 x standard deviation of price)
Lower band = Middle band – (2 x standard deviation of price)
The Three Lines of Bollinger Bands
As mentioned earlier, Bollinger Bands consist of three lines – the middle line, the upper band, and the lower band. The middle line represents the security’s moving average, and the upper and lower bands represent the price’s standard deviation from the moving average.
When the security’s price is trading within the upper and lower bands, it is considered to be in a “normal” range of volatility. If the price moves above the upper band, it may be considered overbought, and if it moves below the lower band, it may be considered oversold.
How to Interpret Bollinger Band Readings
Bollinger Bands are used to identify potential buy and sell signals. When the price moves above the upper band, it may signal a potential sell signal, and when it moves below the lower band, it may signal a potential buy signal. Traders can also look for price action within the bands, such as the price bouncing off the upper or lower bands, as potential trading opportunities.
Bollinger Bands can also be used to identify trends. When the bands are moving in a parallel direction, it indicates that the security is trending. When the bands are contracting, it suggests that the security is moving sideways.
Bollinger Bands can be used in a variety of trading strategies. One popular approach is to buy when the price touches the lower band and sell when it touches the upper band. Another strategy involves waiting for the price to break out of the bands and then taking a position in the direction of the breakout.
Traders can also look for price action within the bands as potential trading opportunities. For example, if the price bounces off the upper band, it may signal a potential sell opportunity, and if it bounces off the lower band, it may signal a potential buy opportunity.
One advantage of using Bollinger Bands is that they are relatively easy to use and require no additional technical indicators. They are also versatile, as they can be used in any time frame and on any security.
One potential disadvantage of using it is that they are a lagging indicator, meaning that they do not predict future price movements. They also may not work well in certain market conditions, such as during periods of low volatility.
Common Mistakes When Using Bollinger Bands
One common mistake when using Bollinger Bands is relying solely on them to make trading decisions. It’s important to use other technical indicators and to consider market fundamentals before taking a position. Another mistake is using Bollinger Bands in isolation, without considering other factors such as volume and price patterns.
Traders may also make the mistake of using it on too short of a time frame, leading to false signals and a potential loss of capital.
Conclusion: BB as a Valuable Resource
Bollinger Bands are a powerful tool that can help traders identify potential trade opportunities, measure market volatility, and spot trends. However, it’s important to use Bollinger Bands in conjunction with other technical indicators and to consider market fundamentals before taking a position. By using Bollinger Bands in combination with other tools, traders can increase their chances of success and make more informed trading decisions.
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what must be the concentration of Cl- to just start precipitation of CuCl from a solution which is 4.896x10^-6M in CuNO3? Ksp = 1.9x10^-7
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# Did we just miss an opportunity to guaranteed make \$240M?
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posted on May, 19 2013 @ 08:58 AM
The Powerball jackpot was at \$590M.
The odds of winning the jackpot: 1 in 175Mio.
Ticket costs \$2, to buy 175M tickets/combinations cost \$350M.
Net winnings: \$240M, guaranteed.
Of course, this works only when there is a massive jackpot that pays more than what it would cost to buy 175M tickets.
(Statistically you would not even need the entire 175Mio tickets since you will likely hit the right numbers already with less). I found that calculation interesting....
posted on May, 19 2013 @ 09:01 AM
If you bought 175m tickets you would have the same numbers as many other people so would share the jackpot and lose out big time.
Common sense can't be bought.
posted on May, 19 2013 @ 09:04 AM
I am aware of the technical problems, but I am implying that the machines would only generate each set of numbers once and you had a way to somehow purchase 175M combinations - and preferrably a way no one else could buy the same numbers.
That the odds are AGAINST the lottery and in the ticket buyer's favor (at least on paper) is interesting, nevertheless.
posted on May, 19 2013 @ 09:13 AM
Whos gonna pick though 175m tickets to find the winner?
You may just spend the rest of your life looking for it...
I think the odds just dropped a bit
edit on 19-5-2013 by Akragon because: (no reason given)
posted on May, 19 2013 @ 09:13 AM
It is posts like yours that proves WHY statistics CANNOT be trusted, and had been the downfall of mankind through the latest financial crisis. You failed to look at OTHER VARIABLES, just like the economists and statisticians.
Just because you buy all the combinations, it does not mean you alone own those combinations. There may be 300 million citizens with that same amount of numbers - some buying like you, all 175M combinations, or some just got lucky with 10s of tickets of the correct combination, some with just one ticket, and another slew of flawed mortal human errors with the purchase.
When that happens - say goodbye to your \$350M.
And also, do not forget if one wins the entire sum, it is still subject to taxes, with leaves you probably only slightly more than half, no matter how big the pot will be, such as the 'fat one' in Spain - \$900M.
Stick to the tried and tested sure ways - honest labour and savings, you are better off than gambling. When you win in gambling, you are happy, but when you lose, you can lose everything, even your life. You tend to lose more than you win in gambling, as it is often structured in the way the house wins it all, not you. You are only the sucker who provided the pool in the pot.
posted on May, 19 2013 @ 09:15 AM
I'm missing your premise entirely here.... \$350 million spent to cover the ticket combinations...just one time through. After all, you'd have to get them all. The way luck and Mr Murphy play, you could get 99.99999999% of the winners and it will always be the ONE you didn't figure you needed that hits the win.
Same problem someone else said though. So you spend well over a 1/4 of a billion to absolutely have one of the winners. I follow so far.....but then an 18 year old playing her first ticket gets the other one a 50% split. The genius of buying the whole game just became the humiliation of going broke to a first timer with an impulse ticket.
I've thought of the same thing on the pick 3's and other games at the state level. There, the jackpot is usually just under what the cost to buy all combinations comes to ..so it's never a winning idea, or like this, one other winner on a split isn't just a bad outcome...it's a financial calamity of a mistake.
So much for cornering the Lottery. lol..... It wouldn't be gambling if it were that easy, and every rich person would corner the games just for the % in return it would bring, IMO.
edit on 19-5-2013 by Wrabbit2000 because: (no reason given)
posted on May, 19 2013 @ 09:27 AM
If I had \$175 million for tickets, why would I be playing the lottery?
posted on May, 19 2013 @ 09:39 AM
Ah like the Family Guy episode where Peter buys several thousand lottery tickets.
They count them all. Then he tells the family that was a test, and then throws more boxes down and tells them these are the real tickets.
Yeah forget that.
I'm smart enough not to gamble on a game with infinitesimal odds.
posted on May, 19 2013 @ 09:40 AM
There are people like me that, if picking their own numbers for a lottery, will pick two of the exact same numbers. That way if myself and one other person wins, I get 2/3 and they get a third.
posted on May, 19 2013 @ 09:50 AM
Lottery's are taxes for the gullible lol
In the UK you dont pay tax on winnings, you pay tax on the purchase price, the ticket itself. Its just a stealth tax, millions of pounds/dollars taken from the poor every month because people think...it could be me.
And no, if one other person bought the same winning numbers you would be screwed. If it were that easy, Google would have bought all the combinations, or some other rich person or company. Im pretty sure the rich dont waste their time on the lottery, there are quicker more reliable ways to make money than a 1 in 175 million chance or whatever it is.
posted on May, 19 2013 @ 09:51 AM
Originally posted by Benevolent Heretic
If I had \$175 million for tickets, why would I be playing the lottery?
Because greed is the American dream now. Sad isn't it?
posted on May, 19 2013 @ 09:52 AM
i'd hate to be in line behind the guy who is buying 175 million tickets!
posted on May, 19 2013 @ 09:59 AM
I think they would have to call that a not-for-profit homicide. You didn't do it to rob the shmuck. You took one for the team and all those waiting behind you, who weren't close enough to do it themselves.
A grateful store would salute your courageous sacrifice, I have no doubt. (You get the impression, I've waiting behind a few inconsiderate "winners"? lol)
posted on May, 19 2013 @ 07:26 PM
wow did i ?
posted on May, 19 2013 @ 07:30 PM
There was a russian research group a few years back that realized there was a profit margin if they simply bought up every single number combination once the jackpots reached certain levels. So they did it.
The fatal flaw? They did not account for shared jackpots. Multiple people won the jackpot that they played for, and they ended up taking a huge loss.
You simply cannot beat the odds on something where the odds are so greatly stacked against you.
edit on 19-5-2013 by captaintyinknots because: (no reason given)
posted on May, 19 2013 @ 07:32 PM
Originally posted by AmberLeaf
Lottery's are taxes for the gullible lol
In the UK you dont pay tax on winnings, you pay tax on the purchase price, the ticket itself. Its just a stealth tax, millions of pounds/dollars taken from the poor every month because people think...it could be me.
And no, if one other person bought the same winning numbers you would be screwed. If it were that easy, Google would have bought all the combinations, or some other rich person or company. Im pretty sure the rich dont waste their time on the lottery, there are quicker more reliable ways to make money than a 1 in 175 million chance or whatever it is.
I call it the idiot tax.
posted on May, 19 2013 @ 07:39 PM
Originally posted by tinhattribunal
i'd hate to be in line behind the guy who is buying 175 million tickets!
175 million tickets...say the machine can print one ticket per second (which is probably too quick anyway).
175 million seconds = 2916666 minutes = 48611 hours = 2025 days....better start early.
This, above everything else is why this strategy could never possibly work.
posted on May, 20 2013 @ 08:59 AM
Originally posted by buster2010
Originally posted by Benevolent Heretic
If I had \$175 million for tickets, why would I be playing the lottery?
Because greed is the American dream now. Sad isn't it?
Nope. Greed exists in pretty much every human being. Doesn't have to be American thing
posted on May, 20 2013 @ 10:31 AM
\$2 a week is a pretty small price to pay for me maintaining the dream of being a multi-millionaire. So, for \$104 a year, I have multiple chances that my family and I will be set for life. Someone has to win, no reason it can't be me.
By the way, EVERY winning ticket of the Powerball has been a quick-pick, not someone's numbers. Every single one. I do a quick pick each time (so I never see any KNOWN numbers come up and kick myself for not playing). There are two drawings every week, but I just play the Saturday one, as I'm not willing to spend \$208 per year for such entertainment.
posted on May, 20 2013 @ 11:52 AM
Originally posted by flobot
Originally posted by tinhattribunal
i'd hate to be in line behind the guy who is buying 175 million tickets!
175 million tickets...say the machine can print one ticket per second (which is probably too quick anyway).
175 million seconds = 2916666 minutes = 48611 hours = 2025 days....better start early.
This, above everything else is why this strategy could never possibly work.
You had based the purchase of buying 175M tickets from one outlet, and thus it would take 2025 days.
The lottery was sold by many outlet across many states. All it takes is for an Iphone with an app to divide up the 175M numbers into lots and purchase them at 2025 outlets, and hey presto- within one day you would have bought all the numbers.
The technique is there. It's only maths. But there is the issue of trust - who else can you trust with all that money to buy, from the other 2024 outlets, as it is unlikely for you to travel across states and hunt down those outlets alone.
Thus, do not ever presume liner thinking will work when it comes to economics. It is more than just maths and stats. There are other variables involved.
top topics
0 | 2,436 | 9,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.953711 |
https://www.coursehero.com/file/6634449/mock-final/ | 1,512,987,678,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00082.warc.gz | 696,588,138 | 23,599 | mock final
# mock final - ENGR 232 Dynamic Engineering Systems:Mock...
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Unformatted text preview: ENGR 232 Dynamic Engineering Systems:Mock Final 1 1. Solve the following first-order differential equa- tions: (a) (1- x 3 ) dy dx = 3 x 2 y (b) t 3 y + 4 t 2 y = e- t , y (- 1) = 0 (c) dy dx = xy 2- x- y 2 +1 xy- 3 y +2 x- 6 (d) ( x 3- y 3 ) dx- 2 x 2 ydy = 0 (e) (1 + e x ) dy dx + e x y = 0 (f) cosx dy dx + ysinx = 1 2. For the following system of two first-order differ- ential equations, determine the • Eigen Values • Corresponding eigen vectors • General Solutions • Constants subject to the initial conditions (a) dx dt =- 1 1- 4 x , x(0) = 2 3 (b) dx dt = 1 1 4 1 x , x(0) = 4 6 (c) dx dt = 3- 2 4- 1 x 3. Transform the following second order differen- tial equations to a system of first order differen- tial equations. Find the solution by determining the Eigen values and corresponding eigen vec- tors, the general solution and the value of the constants based on the initial conditions (a) y 00- 2 y- 3 y = 0, y (0) = 1, y (0) = 0 (b) y 00 + 4 y = 0, y (0) = 1, y (0) =- 2 4. A large tank is filled with 500 gallons of pure4....
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Ask a homework question - tutors are online | 535 | 1,643 | {"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} | 3.28125 | 3 | CC-MAIN-2017-51 | latest | en | 0.690747 |
https://brainly.com/question/281085 | 1,485,207,288,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00128-ip-10-171-10-70.ec2.internal.warc.gz | 779,246,480 | 8,867 | 2015-01-30T08:43:34-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
When dividing exponents with like bases, keep the base and subtract the exponents.
12^6 / 12^2 = 12^(6-2) = 12^4 | 114 | 447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-04 | latest | en | 0.931295 |
http://oldwiki.tcl-lang.org/9087 | 1,720,839,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00225.warc.gz | 23,683,230 | 5,165 | Updated 2012-05-14 02:11:48 by RLE
Richard Suchenwirth 2003-06-10 - To teach programming to children, we need simple examples (simple both in contents and code) which are easily understood, and yet convey some of the fascination we old guys find in Tcl. Here's what I came up with, after a few evenings of |_Ps (drinks):
Imagine the computer couldn't compute, i.e. do arithmetics with numbers, but could only do string operations (everything is a string). Now we want to teach it arithmetics on natural numbers (0 included) by expressing them as strings in a simple way:
``` 1 <-> x
5 <-> xxxxx```
and so forth: a natural number is expressed by so many x's. (Introduce "" as notation for the empty string, length 0, here.) At least we can type in decimal numbers on the keyboard, so the first step is to write converters between natural numbers and x-strings:
``` proc s<-num {n} {string repeat x \$n}
proc s->num {s} {string length \$s}```
Obviously, addition of such "string-numbers" is easily achieved by just concatenation:
` xx + xxx = xxxxx`
which can be had without any function calls:
``` set a xx
set b xxx
puts \$a\$b```
but we want to preserve this first solution in a proc, so:
` proc s+ {a b} {return \$a\$b}`
Multiplication is also easy, we just have to repeat the string repeat of s<-num, but this time with one argument which might be longer (or shorter) than an "x":
` proc s* {a b} {string repeat \$a [s->num \$b]}`
Heavy testing should of course happen after each of these steps - I'll save them for the end of this page. For subtraction we introduce regsub which does the job very simply, but leaves the minuend unchanged if the subtrahend exceeds it in length (which case is undefined for natural numbers):
` proc s- {a b} {regsub \$b \$a ""}`
As this doesn't work in 8.4a2 on PocketPC, I had to rewrite it slightly wordier:
` proc s- {a b} {regsub \$b \$a "" a; return \$a}`
Division is again done with regsub, but this time using the -all switch to subtract b as often as possible, and give a target variable; so regsub returns the number of substitutions, which we have to convert into a "string-number" again:
` proc s/ {a b} {s<-num [regsub -all \$b \$a "" a]}`
Now, with only the string and regsub commands (plus a little help from proc and return), we have recreated four-species math on natural numbers (as long as virtual memory suffices to store possibly very long "string-numbers" - at least 32 bits is not the limit). I wonder how kids react... I at least am again amazed by the simplicity and power of Tcl.
Though we at first should not expect that extending these games to reals, rationals, or even integers, is another child's play, we can of course think of more uses. For example, modulo is even simpler than division:
` proc s% {a b} {regsub -all \$b \$a ""}`
But again, I had to work around 8.4a2, so make that:
``` proc s% {a b} {
regsub -all \$b \$a "" a
return \$a
}```
...and squaring is just a special case of multiplication:
` proc s2 {a} {string repeat \$a [s->num \$a]}`
Taking a to the b-th power introduces the concept of looping, which can go over the x's of the counter:
``` proc s^ {a b} {
set res x ;# neutral element
foreach i [split \$b ""] {
set res [s* \$res \$a]
}
set res
}```
...but let's go testing, which is long overdue. Here's a simple but strict "test harness", which only succeeds if no test fails:
``` proc test cases {
foreach {cmd expected} \$cases {
catch \$cmd res
if {\$res != \$expected} {
error "\$cmd=\$res, expected \$expected"
}
}
puts "Passed all tests."
}
test {
{set x [s<-num 7]} xxxxxxx
{set y [s<-num 4]} xxxx
{s<-num 0} ""
{s->num xxxxx} 5
{s->num ""} 0
{s->num [s+ \$x \$y]} 11
{s->num [s+ \$x ""]} 7
{s->num [s- \$x \$y]} 3
{s->num [s- \$x ""]} 7
{s->num [s- "" ""]} 0
{s->num [s* \$x \$y]} 28
{s->num [s* \$x ""]} 0
{s->num [s/ \$x \$y]} 1
{s->num [s/ "" \$y]} 0
{s->num [s% \$x \$y]} 3
{s->num [s2 \$y]} 16
{s->num [s2 ""]} 0
{s->num [s^ xx xxx]} 8
{s->num [s^ xxx ""]} 1
{s->num [s^ "" ""]} 1
}```
This test suite passes on my little box - note that it's longer than the few lines of source code that are tested, but that's normal and even good (but might bore kids). The earlier a bug is found, the better, so best spend at least as much time on test design as on coding :-
Just remembering the old computer folk joke, "back then, all we had was 0 and 1, and at some times we even didn't have 1": replace "x" with "0" in the above examples, and you get five(6?)-species math with 0 only :)
` s+ 00 000 -> 00000`
FW: Or, replace x with 1 to make it correct base 1 notation, which is what it really is.
Another application: determining whether a number is a prime, with regexp:
``` proc isprime x {
expr {\$x>1 && ![regexp {^(xx+?)\1+\$} [string repeat x \$x]]}
} ``` | 1,422 | 4,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.9203 |
https://math.answers.com/other-math/The_commutative_property_of_addition_allows_one_to_say_that_3_plus_6_is_the_same_as_what | 1,721,091,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00322.warc.gz | 349,687,090 | 48,835 | 0
# The commutative property of addition allows one to say that 3 plus 6 is the same as what?
Updated: 4/28/2022
Wiki User
13y ago
3+6=6+3
3 plus 6 equals 6 plus 3
Commutative is just the two numbers in an addition equation reversed.
Wiki User
13y ago
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### What is the property x plus y plus z equals x plus z plus y?
x + y + z = x + z + y is the commutative property of addition.
### Which property is in this equation 2 plus 8 equals 8 plus 2?
2 plus 8 equals 8 plus 2 demonstrates the commutative property of addition | 271 | 956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.858032 |
https://studysoup.com/tsg/math/52/contemporary-abstract-algebra/chapter/813/13 | 1,603,347,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00459.warc.gz | 530,990,090 | 12,043 | ×
×
# Solutions for Chapter 13: Contemporary Abstract Algebra 8th Edition
## Full solutions for Contemporary Abstract Algebra | 8th Edition
ISBN: 9781133599708
Solutions for Chapter 13
Solutions for Chapter 13
4 5 0 326 Reviews
12
5
##### ISBN: 9781133599708
This expansive textbook survival guide covers the following chapters and their solutions. Chapter 13 includes 76 full step-by-step solutions. This textbook survival guide was created for the textbook: Contemporary Abstract Algebra , edition: 8. Contemporary Abstract Algebra was written by and is associated to the ISBN: 9781133599708. Since 76 problems in chapter 13 have been answered, more than 45675 students have viewed full step-by-step solutions from this chapter.
Key Math Terms and definitions covered in this textbook
• Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.
• Cayley-Hamilton Theorem.
peA) = det(A - AI) has peA) = zero matrix.
• Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).
• Determinant IAI = det(A).
Defined by det I = 1, sign reversal for row exchange, and linearity in each row. Then IAI = 0 when A is singular. Also IABI = IAIIBI and
• Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
• Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.
• Factorization
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.
• Graph G.
Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.
• Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
• Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.
• Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.
• Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
• Length II x II.
Square root of x T x (Pythagoras in n dimensions).
• Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.
• Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
• Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
• Schur complement S, D - C A -} B.
Appears in block elimination on [~ g ].
• Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
• Stiffness matrix
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.
• Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn- 1 with P(Xi) = bi. Vij = (Xi)j-I and det V = product of (Xk - Xi) for k > i.
× | 910 | 3,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-45 | latest | en | 0.866123 |
http://weblogs.sqlteam.com/PETERL/archive/2010/01/23/Celko-Stumper---The-Class-Scheduling-Problem.aspx | 1,524,646,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947759.42/warc/CC-MAIN-20180425080837-20180425100837-00326.warc.gz | 321,892,602 | 8,384 | # Thinking outside the box
Patron Saint of Lost Yaks
## Celko Stumper - The Class Scheduling Problem
Joe Celko has posted a new Stumper - The Class Scheduling Problem
here http://www.simple-talk.com/sql/t-sql-programming/celkos-sql-stumper-the-class-scheduling-problem/
Here is one suggestion to solve the problem. It's linear in time so it should be very fast. It's based on the "Descending order" approach, and the "paths" columns are used to store valid room and classes.
-- Initialize and find the valid combinations
DECLARE @HowManySeatsFree INT = 0 -- Set to zero for maximum seating, and to 1 for letting in late pupils.
DECLARE
@Source TABLE
(
room_nbr CHAR(2),
class_nbr CHAR(2),
recID INT IDENTITY(1, 1) PRIMARY KEY CLUSTERED
)
INSERT @Source
(
room_nbr,
class_nbr
)
SELECT r.room_nbr,
c.class_nbr
FROM dbo.Rooms AS r
INNER JOIN dbo.Classes AS c ON c.class_size <= r.room_size - @HowManySeatsFree
ORDER BY r.room_size DESC,
c.class_size DESC
-- Iterate the possibilities and return the unique answers
;WITH cteYak(recID, room_nbr, roomPath, class_nbr, classPath, isPresent)
AS (
SELECT recID,
room_nbr,
'/' + CAST(room_nbr AS VARCHAR(MAX)) + '/' AS roomPath, -- List of taken rooms
class_nbr,
'/' + CAST(class_nbr AS VARCHAR(MAX)) + '/' AS classPath, -- List of taken classes
CAST(0 AS BIGINT)
FROM @Source
WHERE recID = 1
UNION ALL
SELECT recID,
room_nbr,
CASE isPresent -- If room never encountered before (isPresent=0), take it!
WHEN 0 THEN roompath + CAST(room_nbr AS VARCHAR(MAX)) + '/'
ELSE roompath
END AS roompath,
class_nbr,
CASE isPresent -- If class never encountered before (isPresent=0), take it!
WHEN 0 THEN classpath + CAST(class_nbr AS VARCHAR(MAX)) + '/'
ELSE classpath
END AS classpath,
isPresent
FROM (
SELECT s.recID,
s.room_nbr,
y.roomPath,
s.class_nbr,
y.classpath,
CHARINDEX('/' + CAST(s.room_nbr AS VARCHAR(MAX)) + '/', y.roompath)
+ CHARINDEX('/' + CAST(s.class_nbr AS VARCHAR(MAX)) + '/', y.classpath) AS isPresent -- See if room or class is already taken. If so, isPresent is greater than 0, otherwise it will be 0.
FROM @Source AS s
INNER JOIN cteYak AS y ON y.recID + 1 = s.recID
) AS d
)
SELECT room_nbr,
class_nbr
FROM cteYak
WHERE isPresent = 0 -- Only present the combination never taken/found before
OPTION (MAXRECURSION 0) -- Allow up to 32767 possible combinations.
Print | posted on Saturday, January 23, 2010 1:31 PM | Filed Under [ Optimization SQL Server 2008 Algorithms SQL Server 2005 SQL Server 2000 ]
## #re: Celko Stumper - The Class Scheduling Problem
I see you went for the (class_size <= room_size) rather than the (class_size < room_size) T-JOIN. Can you put in some commentary? Why did you put an ORDER BY on an INSERT INTO Statement? I thinhk you are the first answer, by the way.
1/23/2010 11:02 PM | Joe Celko
## #re: Celko Stumper - The Class Scheduling Problem
I did put an ORDER BY to ensure matches between class size and room size are done in "proper" order because of the nature of CTE.
The recID value holds a "ticket" for which order to check combinations.
1/24/2010 8:01 AM |
## #re: Celko Stumper - The Class Scheduling Problem
Oh, and the REASON for using ORDER BY is to simulate a MERGE JOIN becuase I believe this task will be faster when items are pre-sorted, much like a MERGE JOIN need to.
1/24/2010 8:05 AM |
## #re: Celko Stumper - The Class Scheduling Problem
Oh, and the REASON for using ORDER BY is to simulate a MERGE JOIN becuase I believe this task will be faster when items are pre-sorted, much like a MERGE JOIN need to. It also makes sure I start with the largest class and end with the smallest class.
1/24/2010 8:10 AM |
## #re: Celko Stumper - The Class Scheduling Problem
Hmmmm. looks eerily familiar.
Always enjoy checking this blog.
All the best, Charlie.
1/25/2010 4:39 PM | Transact Charlie
## #re: Celko Stumper - The Class Scheduling Problem
Just spotted this one... here is mine...
[code]
--------------------------------------------------------------------------
-- build a results table
--------------------------------------------------------------------------
CREATE TABLE results
(
id INT IDENTITY(0,1) PRIMARY KEY,
class_nbr CHAR(2),
class_size INT,
room_nbr CHAR(2),
room_size INT
)
INSERT results
SELECT
class_nbr,
class_size,
room_nbr,
room_size
FROM
Classes c
JOIN Rooms r ON r.room_size >= c.class_size
--------------------------------------------------------------------------
--build a function to find the min unused room size
--------------------------------------------------------------------------
GO
CREATE FUNCTION dbo.udfGetMinUnusedRoomsize (@id INT, @class_nbr CHAR(2))
RETURNS INT
AS
BEGIN
DECLARE @room_size int;
SELECT @room_size = MIN(room_size)
FROM results
WHERE class_nbr = @class_nbr AND room_nbr NOT IN (SELECT room_nbr WHERE id < @id)
RETURN @room_size
END
GO
--------------------------------------------------------------------------
--result
--------------------------------------------------------------------------
SELECT
c.class_nbr,
c.class_size,
r.room_nbr,
r.room_size
FROM
Classes c
JOIN
(
SELECT
c.class_nbr,
MIN(dbo.udfGetMinUnusedRoomsize(c.id, c.class_nbr)) AS min_room_size
FROM
results c
GROUP BY
c.class_nbr
) d ON d.class_nbr = c.class_nbr
JOIN Rooms r ON r.room_size = d.min_room_size
ORDER BY c.class_size DESC
--------------------------------------------------------------------------
--clean up
--------------------------------------------------------------------------
DROP TABLE results
DROP FUNCTION dbo.udfGetMinUnusedRoomsize
[/code]
2/24/2010 6:19 AM | ehorn
Comments have been closed on this topic. | 1,434 | 5,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.521303 |
https://www.biostars.org/p/474259/#9500113 | 1,643,195,374,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00534.warc.gz | 730,238,320 | 5,803 | How to understand Structural Variation in bedpe format?
1
2
Entering edit mode
14 months ago
Shicheng Guo ★ 8.9k
Hi All,
How to understand the structural variation with bedpe format?
For example, the first one is duplication, how to know which region is duplicated ? how many duplicates happened?
what does start1 and end1 mean? why they are always having 1b difference?
chrom1 start1 end1 chrom2 start2 end2 sv_id pe_support strand1 strand2 svclass
1 10428600 10428601 1 10582238 10582239 SVMERGE93 80 - + DUP
1 26939174 26939175 1 27518317 27518318 SVMERGE102 89 + - DEL
1 29834388 29834389 1 33723647 33723648 SVMERGE100 107 - - t2tINV
1 29834417 29834418 1 33723675 33723676 SVMERGE101 87 + + h2hINV
1 32728334 32728335 1 32845969 32845970 SVMERGE99 33 - + DUP
1 78236345 78236346 1 78427706 78427707 SVMERGE98 71 + - DEL
1 151858704 151858705 13 44927717 44927718 SVMERGE151 75 - + TRA
1 151873175 151873176 1 183511505 183511506 SVMERGE152 41 + - DEL
1 151887225 151887226 13 44703672 44703673 SVMERGE163 65 - - TRA
Thanks.
SV bedpe • 532 views
0
Entering edit mode
7 weeks ago
jxcao9812 • 0
start1/end1 are the endpoints of confidence intervals of left breakpoints, and start2/end2 are the same for the right breakpoints | 525 | 1,439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.572327 |
http://northstar-www.dartmouth.edu/doc/idl/html_6.2/Azimuthal_Projections.html | 1,550,301,149,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479967.16/warc/CC-MAIN-20190216065107-20190216091107-00532.warc.gz | 193,095,333 | 4,514 | Using IDL: Map Projections
## Azimuthal Projections
With azimuthal projections, the UV plane is tangent to the globe. The point of tangency is projected onto the center of the plane and its latitude and longitude are the points at the center of the map projection, respectively. Rotation is the angle between North and the v-axis.
Important characteristics of azimuthal maps include the fact that directions or azimuths are correct from the center of the projection to any other point, and great circles through the center are projected to straight lines on the plane.
The IDL mapping package includes the following azimuthal projections:
### Orthographic Projection
The orthographic projection was known by the Egyptians and Greeks 2000 years ago. This projection looks like a globe because it is a perspective projection from infinite distance. As such, it maps one hemisphere of the globe into the UV plane. Distortions are greatest along the rim of the hemisphere where distances and land masses are compressed.
The following figure shows an orthographic projection centered over Eastern Spain at a scale of 70 million to 1.
Figure 9-1: Orthographic Projection
### Stereographic Projection
The stereographic projection is a true perspective projection with the globe being projected onto the UV plane from the point P on the globe diametrically opposite to the point of tangency. The whole globe except P is mapped onto the UV plane. There is great distortion for regions close to P, since P maps to infinity.
The stereographic projection is the only known perspective projection that is also conformal. It is frequently used for polar maps. For example, a stereographic view of the north pole has the south pole as its point of perspective.
The following figure shows an equatorial stereographic projection with the hemisphere centered on the equator at longitude -105 degrees.
Figure 9-2: An Azimuthal Projection
### Gnomonic Projection
The gnomonic projection (also called Central or Gnomic) projects all great circles to straight lines. The gnomonic projection is the perspective, azimuthal projection with point of perspective at the center of the globe. Hence, with the gnomonic projection, the interior of a hemispherical region of the globe is projected to the UV plane with the rim of the hemisphere going to infinity. Except at the center, there is great distortion of shape, area, and scale. The default clipping region for the gnomonic projection is a circle with a radius of 60 degrees at the center of projection.
The projection in the following figure is centered around the point at latitude 40 degrees and longitude -105 degrees. The region on the globe that is mapped lies between 20 degrees and 70 degrees of latitude and -130 degrees and -70 degrees of longitude.
Figure 9-3: A Gnomonic Projection
### Azimuthal Equidistant Projection
The azimuthal equidistant projection is also not a true perspective projection, because it preserves correctly the distances between the tangent point and all other points on the globe. Any line drawn through the tangent point reports distance correctly. Therefore, this projection type is useful for determining flight distances. The point P opposite the tangent point is mapped to a circle on the UV plane, and hence, the whole globe is mapped to the plane. There is infinite distortion close to the outer rim of the map, which is the circular image of P.
The following Azimuthal projection is centered at the South Pole and shows the entire globe.
Figure 9-4: An Azimuthal Equidistant Projection
### Aitoff Projection
The Aitoff projection modifies the equatorial aspect of one hemisphere of the azimuthal equidistant projection, described above. Lines parallel to the equator are stretched horizontally and meridian values are doubled, thereby displaying the world as an ellipse with axes in a 2:1 ratio. Both the equator and the central meridian are represented at true scale; however, distances measured between the point of tangency and any other point on the map are no longer true to scale.
An Aitoff projection centered on the international dateline is shown in the following figure.
Figure 9-5: An Aitoff Projection
### Lambert's Equal Area Projection
Lambert's equal area projection adjusts projected distances in order to preserve area. Hence, it is not a true perspective projection. Like the stereographic projection, it maps to infinity the point P diametrically opposite the point of tangency. Note also that to preserve area, distances between points become more contracted as the points become closer to P. Lambert's equal area projection has less overall scale variation than the other azimuthal projections.
The following figure shows the Northern Hemisphere rotated counterclockwise 105 degrees, and filled continents.
Figure 9-6: A Lambert's Equal Area Projection
### Hammer-Aitoff Projection
Although the Hammer-Aitoff projection is not truly azimuthal, it is included in this section because it is derived from the equatorial aspect of Lambert's equal area projection limited to a hemisphere (in the same way Aitoff's projection is derived from the equatorial aspect of the azimuthal equidistant projection). In this derivation, the hemisphere is represented inside an ellipse with the rest of the world in the lunes of the ellipse.
Because the Hammer-Aitoff projection produces an equal area map of the entire globe, it is useful for visual representations of geographically related statistical data and distributions. Astronomers use this projection to show the entire celestial sphere on one map in a way that accurately depicts the relative distribution of the stars in different regions of the sky.
A Hammer-Aitoff projection centered on the international dateline is shown in the following figure:
Figure 9-7: The Hammer-Aitoff Projection
### Satellite Projection
The satellite projection, also called the General Perspective projection, simulates a view of the globe as seen from a camera in space. If the camera faces the center of the globe, the projection is called a Vertical Perspective projection (note that the orthographic, stereographic, and gnomonic projections are special cases of this projection), otherwise the projection is called a Tilted Perspective projection.
The globe is viewed from a point in space, with the viewing plane touching the surface of the globe at the point directly beneath the satellite (the sub-satellite point). If the projection plane is perpendicular to the line connecting the point of projection and the center of the globe, a Vertical Perspective projection results. Otherwise, the projection plane is horizontally turned G degrees clockwise from the north, then tilted w degrees downward from horizontal.
The map in the accompanying figure shows the eastern seaboard of the United States from an altitude of about 160km, above Newburgh, NY.
Figure 9-8: Satellite Projection | 1,383 | 6,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2019-09 | latest | en | 0.917458 |
https://qanda.ai/en/solver/popular-problems/1011047 | 1,642,995,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00423.warc.gz | 506,233,643 | 17,113 | # Calculator search results
Formula
Solve the quadratic equation
Answer
$$2 x ^ { 2 } - 5 x - 3 = 0$$
$\begin{array} {l} x = 3 \\ x = - \dfrac { 1 } { 2 } \end{array}$
Solve quadratic equations using the square root
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0 }$
Divide both sides by the coefficient of the leading highest term
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ 0 }$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ 0 }$
Convert the quadratic expression on the left side to a perfect square format
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x - \dfrac { 5 } { 4 } \right ) ^ { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = 0$
Move the constant to the right side and change the sign
$\left ( x - \dfrac { 5 } { 4 } \right ) ^ { 2 } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
$\left ( x - \dfrac { 5 } { 4 } \right ) ^ { 2 } = \dfrac { 3 } { 2 } + \left ( \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } }$
When raising a fraction to the power, raise the numerator and denominator each to the power
$\left ( x - \dfrac { 5 } { 4 } \right ) ^ { 2 } = \dfrac { 3 } { 2 } + \dfrac { \color{#FF6800}{ 5 } ^ { \color{#FF6800}{ 2 } } } { \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 ^ { 2 } } { 4 ^ { 2 } } }$
Organize the expression
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 49 } { 16 } }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ \dfrac { 49 } { 16 } }$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 49 } { 16 } } }$
$\color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 4 } } = \pm \sqrt{ \color{#FF6800}{ \dfrac { 49 } { 16 } } }$
Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 7 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ \dfrac { 7 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
Separate the answer
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 4 } } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 4 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 4 } } \end{array}$
Organize the expression
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{array}$
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Question C please. What is the diffrence between the totla travel time of system optimal vs UE?
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Problem 1 (30%) Consider the following static route choice problem where 110 vehicles travel from point A to point B. The corresponding travel time (in minutes) of each link is as follows: t1 =x1,t2 =x2 +20: t3 =X3 + 10, t4, =31" where x: denotes the number of vehicles that choose link i. Find the number of vehicles that travel on each link when a. The user equilibrium condition (UE) is satisfied, where vehicles select the route with the minimum travel time; and b. The system optimum condition (50) is satisfied, where the total travel time is minimised. c. Report the total delay savings when satisfying 50 instead of UE.
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Our Expert Tutors provide step by step solutions to help you excel in your courses | 361 | 1,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-21 | latest | en | 0.848293 |
http://slideplayer.com/slide/5792979/ | 1,725,870,917,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00032.warc.gz | 30,710,244 | 19,356 | # 1 of 84 S HAPE AND S PACE Circles. 2 of 84 L ET US DEFINE C IRCLE A ___________is a simple shape that is the set of all points in a plane that are at.
## Presentation on theme: "1 of 84 S HAPE AND S PACE Circles. 2 of 84 L ET US DEFINE C IRCLE A ___________is a simple shape that is the set of all points in a plane that are at."— Presentation transcript:
1 of 84 S HAPE AND S PACE Circles
2 of 84 L ET US DEFINE C IRCLE A ___________is a simple shape that is the set of all points in a plane that are at a given distance from a given point, the center. A circle is a simple shape that is the set of all points in a plane that are at a given distance from a given point, the center.
3 of 84 Radius is the distance from the center to the edge of a circle. Radius is half of the diameter
4 of 84 Diameter is a segment that passes through the center and has its endpoints on the circle. The diameter is twice the length of the radius
5 of 84 T HE VALUE OF We use the symbol π because the number cannot be written exactly. π = 3.141592653589793238462643383279502884197169 39937510582097494459230781640628620899862803482 53421170679821480865132823066470938446095505822 31725359408128481117450284102701938521105559644 62294895493038196 (to 200 decimal places)!
6 of 84 In circles the AREA is equal to 3.14 ( ) times the radius (r) to the power of 2. Thus the formula looks like: A= r 2 In circles the circumference is formula looks like: 2 r The circumference of a circle is the actual length around the circle which is equal to 360°. π is equal to 3.14.
7 of 84 T HE CIRCUMFERENCE OF A CIRCLE Use π = 3.14 to find the circumference of this circle. C = 2 πr 8 cm = 2 × 4 = 8 π R = 4
8 of 84 T HE CIRCUMFERENCE OF A CIRCLE Use π = 3.14 to find the circumference of the following circles: C = 2 πr 4 cm = 2 × 2 = 4 π cm C = 2 πr 9 m = 2 × π × 9 = 18 π m C = 2 πr = 2 × 12 = 12 π mm C = 2 πr 58 cm = 2 × π × 58 = 116 π cm 24mm
9 of 84 F ORMULA FOR THE AREA OF A CIRCLE We can find the area of a circle using the formula radius Area of a circle = πr 2 Area of a circle = π × r × r or
10 of 84 A REA OF A CIRCLE
11 of 84 T HE CIRCUMFERENCE OF A CIRCLE Use π = 3.14 to find the area of this circle. A = πr 2 4 cm = π × 4 × 4 = 16 π cm 2
12 of 84 T HE AREA OF A CIRCLE Use π = 3.14 to find the area of the following circles: A = πr 2 2 cm = π × 2 2 = 4 π cm 2 A = πr 2 10 m = π × 5 2 = 25 π m 2 A = πr 2 23 mm = π × 23 2 = 529 π mm 2 A = πr 2 78 cm = π × 39 2 = 1521 π cm 2
13 of 84 F IND THE AREA OF THIS SHAPE Use π = 3.14 to find area of this shape. The area of this shape is made up of the area of a circle of diameter 12cm and the area of a rectangle of width 6cm and length 12cm. 6 cm 12 cm Area of circle = π × 6 2 = 36 π cm 2 Area of rectangle = 6 × 12 = 78 cm 2 Total area = 36 π + 78
14 of 84 ? F INDING THE RADIUS GIVEN THE CIRCUMFERENCE Use π = 3.14 to find the radius of this circle. C = 2 πr 12 cm How can we rearrange this to make r the subject of the formula? r = C 2π2π 12 2 × π = 6 π =
15 of 84 F IND THE PERIMETER OF THIS SHAPE Use π = 3.14 to find perimeter of this shape. The perimeter of this shape is made up of the circumference of a circle of diameter 13 cm and two lines of length 6 cm. 6 cm 14 cm Perimeter = 14 x 6 Circumference = 2 πr
16 of 84 C IRCUMFERENCE PROBLEM The diameter of a bicycle wheel is 50 cm. How many complete rotations does it make over a distance of 1 km? 50 cm The circumference of the wheel = π × 50 Using C = 2 πr and π = 3.14, = 157 cm
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Similar presentations | 1,233 | 3,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.852383 |
https://slopeinterceptform.net/slope-intercept-form-worksheet-key/ | 1,653,169,949,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00118.warc.gz | 599,730,913 | 14,601 | # Slope Intercept Form Worksheet Key
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope Intercept Form Worksheet Key – One of the many forms that are used to represent a linear equation one that is frequently seen is the slope intercept form. The formula of the slope-intercept identify a line equation when you have the slope of the straight line and the y-intercept. It is the point’s y-coordinate where the y-axis meets the line. Find out more information about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations, namely the standard, slope-intercept, and point-slope. Although they may not yield the same results when utilized, you can extract the information line that is produced more quickly through an equation that uses the slope-intercept form. As the name implies, this form utilizes an inclined line, in which you can determine the “steepness” of the line determines its significance.
This formula can be utilized to discover the slope of a straight line. It is also known as the y-intercept (also known as the x-intercept), where you can utilize a variety available formulas. The line equation of this specific formula is y = mx + b. The straight line’s slope is symbolized through “m”, while its y-intercept is indicated via “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world in the real world, the slope intercept form is commonly used to show how an item or problem changes in it’s course. The value that is provided by the vertical axis demonstrates how the equation tackles the magnitude of changes in the amount of time indicated by the horizontal axis (typically in the form of time).
An easy example of this formula’s utilization is to find out how much population growth occurs in a specific area in the course of time. Based on the assumption that the area’s population increases yearly by a certain amount, the values of the horizontal axis will increase one point at a moment each year and the point values of the vertical axis is increased to show the rising population by the fixed amount.
Also, you can note the starting value of a question. The beginning value is at the y’s value within the y’intercept. The Y-intercept marks the point where x is zero. In the case of the problem mentioned above the beginning value will be at the point when the population reading begins or when the time tracking begins along with the associated changes.
So, the y-intercept is the location at which the population begins to be recorded in the research. Let’s assume that the researcher starts to perform the calculation or take measurements in the year 1995. This year will be”the “base” year, and the x = 0 point will be observed in 1995. So, it is possible to say that the 1995 population corresponds to the y-intercept.
Linear equation problems that use straight-line equations are typically solved this way. The beginning value is represented by the yintercept and the rate of change is represented by the slope. The principal issue with this form typically lies in the horizontal interpretation of the variable especially if the variable is attributed to an exact year (or any other type of unit). The most important thing to do is to ensure that you know the definitions of variables clearly. | 727 | 3,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-21 | longest | en | 0.932313 |
https://sodocumentation.net/cplusplus/topic/5693/recursion-in-cplusplus | 1,652,799,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00356.warc.gz | 596,575,463 | 14,247 | C++ Recursion in C++
Using tail recursion and Fibonnaci-style recursion to solve the Fibonnaci sequence
The simple and most obvious way to use recursion to get the Nth term of the Fibonnaci sequence is this
``````int get_term_fib(int n)
{
if (n == 0)
return 0;
if (n == 1)
return 1;
return get_term_fib(n - 1) + get_term_fib(n - 2);
}
``````
However, this algorithm does not scale for higher terms: for bigger and bigger `n`, the number of function calls that you need to make grows exponentially. This can be replaced with a simple tail recursion.
``````int get_term_fib(int n, int prev = 0, int curr = 1)
{
if (n == 0)
return prev;
if (n == 1)
return curr;
return get_term_fib(n - 1, curr, prev + curr);
}
``````
Each call to the function now immediately calculates the next term in the Fibonnaci sequence, so the number of function calls scales linearly with `n`.
Recursion with memoization
Recursive functions can get quite expensive. If they are pure functions (functions that always return the same value when called with the same arguments, and that neither depend on nor modify external state), they can be made considerably faster at the expense of memory by storing the values already calculated.
The following is an implementation of the Fibonacci sequence with memoization:
``````#include <map>
int fibonacci(int n)
{
static std::map<int, int> values;
if (n==0 || n==1)
return n;
std::map<int,int>::iterator iter = values.find(n);
if (iter == values.end())
{
return values[n] = fibonacci(n-1) + fibonacci(n-2);
}
else
{
return iter->second;
}
}
``````
Note that despite using the simple recursion formula, on first call this function is \$O(n)\$. On subsequent calls with the same value, it is of course \$O(1)\$.
Note however that this implementation is not reentrant. Also, it doesn't allow to get rid of stored values. An alternative implementation would be to allow the map to be passed as additional argument:
``````#include <map>
int fibonacci(int n, std::map<int, int> values)
{
if (n==0 || n==1)
return n;
std::map<int,int>::iterator iter = values.find(n);
if (iter == values.end())
{
return values[n] = fibonacci(n-1) + fibonacci(n-2);
}
else
{
return iter->second;
}
}
``````
For this version, the caller is required to maintain the map with the stored values. This has the advantage that the function is now reentrant, and that the caller can remove values that are no longer needed, saving memory. It has the disadvantage that it breaks encapsulation; the caller can change the output by populating the map with incorrect values. | 644 | 2,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.766134 |
https://www.12000.org/my_notes/Murphy_ODE/reports/maple_2021_1_and_mma_12_3_1/KERNELsubsubsection920.htm | 1,716,509,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00445.warc.gz | 537,236,471 | 3,233 | ##### 4.19.20 $$\left (a^2+x^2\right ) y'(x)^2=b^2$$
ODE
$\left (a^2+x^2\right ) y'(x)^2=b^2$ ODE Classification
[_quadrature]
Book solution method
Missing Variables ODE, Dependent variable missing, Solve for $$y'$$
Mathematica
cpu = 0.164934 (sec), leaf count = 101
$\left \{\left \{y(x)\to \frac {1}{2} b \log \left (1-\frac {x}{\sqrt {a^2+x^2}}\right )-\frac {1}{2} b \log \left (\frac {x}{\sqrt {a^2+x^2}}+1\right )+c_1\right \},\left \{y(x)\to -\frac {1}{2} b \log \left (1-\frac {x}{\sqrt {a^2+x^2}}\right )+\frac {1}{2} b \log \left (\frac {x}{\sqrt {a^2+x^2}}+1\right )+c_1\right \}\right \}$
Maple
cpu = 0.071 (sec), leaf count = 40
$\left [y \left (x \right ) = b \ln \left (x +\sqrt {a^{2}+x^{2}}\right )+\textit {\_C1}, y \left (x \right ) = -b \ln \left (x +\sqrt {a^{2}+x^{2}}\right )+\textit {\_C1}\right ]$ Mathematica raw input
DSolve[(a^2 + x^2)*y'[x]^2 == b^2,y[x],x]
Mathematica raw output
{{y[x] -> C[1] + (b*Log[1 - x/Sqrt[a^2 + x^2]])/2 - (b*Log[1 + x/Sqrt[a^2 + x^2]
])/2}, {y[x] -> C[1] - (b*Log[1 - x/Sqrt[a^2 + x^2]])/2 + (b*Log[1 + x/Sqrt[a^2
+ x^2]])/2}}
Maple raw input
dsolve((a^2+x^2)*diff(y(x),x)^2 = b^2, y(x))
Maple raw output
[y(x) = b*ln(x+(a^2+x^2)^(1/2))+_C1, y(x) = -b*ln(x+(a^2+x^2)^(1/2))+_C1] | 638 | 1,248 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-22 | latest | en | 0.256632 |
http://aux.planetmath.org/proofoftheoremonequivalentvaluations | 1,521,546,497,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00602.warc.gz | 29,978,523 | 4,110 | # proof of theorem on equivalent valuations
It is easy to see that $|\cdot|$ and $|\cdot|^{c}$ are equivalent valuations for any constant $c>0$ — it follows from the fact that $0\leq x^{c}<1$ if and only if $0.
Assume that the valuations $|\cdot|_{1}$ and $|\cdot|_{2}$ are equivalent. Let $b$ be an element of $K$ such that $0<|b|_{1}<1$. Because the valuations are assumed to be equivalent, it is also the case that $0<|b|_{2}<1$. Hence, there must exist positive constants $c_{1}$ and $c_{2}$ such that $|b|_{1}^{c_{1}}={1\over 2}$ and $|b|_{2}^{c_{2}}={1\over 2}$.
We will show that show that $|x|_{1}^{c_{1}}=|x|_{2}^{c_{2}}$ for all $a\in K$ by contradiction.
Let $a$ be any element of $k$ such that $0<|a|_{1}<1$. Assume that $|a|_{1}^{c_{1}}\neq|a|_{2}^{c_{2}}$. Then either $|a|_{1}^{c_{1}}<|a|_{2}^{c_{2}}$ or $|a|_{1}^{c_{1}}>|a|_{2}^{c_{2}}$. We may assume that $|a|_{1}^{c_{1}}<|a|_{2}^{c_{2}}$ without loss of generality.
Since $|a|_{2}^{c_{2}}/|a|_{1}^{c_{1}}>1$, there exists an integer $m>0$ such that $(|a|_{2}^{c_{2}}/|a|_{1}^{c_{1}})^{m}>2$. Let $n$ be the least integer such that $2^{n}|a|_{2}^{mc_{2}}>1$. Then we have
$2^{n}|a|_{1}^{mc_{1}}<2^{n-1}|a|_{2}^{mc_{2}}<1<2^{n}|a|_{2}^{mc_{2}}.$
Since $2=|b^{-1}|_{1}^{c_{1}}=|b^{-1}|_{2}^{c_{2}}$, this implies that
$\left|{a^{m}\over b^{n}}\right|_{1}^{c_{1}}<1<\left|{a^{m}\over b^{n}}\right|_% {2}^{c_{2}},$
but then
$\left|{a^{m}\over b^{n}}\right|_{1}<1$
and
$\left|{a^{m}\over b^{n}}\right|_{2}>1,$
which is impossible because the two valuations are assumed to be equivalent.
Q.E.D
Title proof of theorem on equivalent valuations ProofOfTheoremOnEquivalentValuations 2013-03-22 14:55:40 2013-03-22 14:55:40 rspuzio (6075) rspuzio (6075) 12 rspuzio (6075) Proof msc 13A18 | 757 | 1,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 34, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-13 | latest | en | 0.724182 |
https://byjus.com/question-answer/how-to-place-fractions-on-number-line/ | 1,718,217,589,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00399.warc.gz | 133,302,327 | 21,947 | 1
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# How to place fractions on number line
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Solution
## As you know 0.7 lies between 0 and 1 it lies 1 to 10 also but 0 to 1 is smallest intercellular so we can divide into small parts like 0.1 parts so by adding 7 parts of 0.7 we will get our number on line.but when we consider 0 to 10 as a interval then we need to divide this interval to large number of parts for example to show .7 on number line we have to make 100 subparts of 0.1 so the number line beacame to long.thats why we choose 0 t0 1 interval. hope this helps:)
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Join BYJU'S Learning Program | 213 | 760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-26 | latest | en | 0.882994 |
https://guillaumeboivin.com/what-is-the-relationship-between-the-mass-and-volume.html | 1,660,592,807,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00266.warc.gz | 278,434,209 | 12,661 | # What is the relationship between the mass and volume?
For any pure substance or homogenous mixture, density is an intensive quantity, which means that its value is the same no matter how much of the substance is present. Density expresses the proportional relationship between mass and volume of the substance.
Keeping this in consideration, what happens to volume if mass increases?
This happens because the mass of the rod stays the same, but its volume increases. The increase in the volume of matter with increasing temperature is called expansion. When cooled down, most matter decreases in volume and increases in density. This decrease in volume is called contraction.
What is the relationship between the mass and volume of water?
Just like a solid, the density of a liquid equals the mass of the liquid divided by its volume; D = m/v. The density of water is 1 gram per cubic centimeter. The density of a substance is the same regardless of the size of the sample.
How does mass affect volume?
The density of an object or quantity of matter is its mass divided by its volume. This is usually measured under standard conditions for temperature and pressure: 0°C and 1 atmosphere of pressure. One factor affecting the density of a material is how concentrated the atoms are in a given volume.
## Can you use mass and volume to predict?
Can mass alone be used to accurately predict whether an object will sink or float? -No, you need both mass and volume to determine its desity to see if it can float. Can volume alone be used to predict whether an object will sink or float? -No, you need both mass and volume to find its density to see if it can float.
## What is the relationship between the mass and volume of an object?
Density is the amount of matter filling the object’s space. Adding mass to an object without changing its volume, increases the object’s density. Objects that have a large mass and small volume have a high density.
## Are mass and volume inversely or directly proportional?
Here Volume is constant and mass is proportional to density. Now if all the cooking gas in the same cylinder is transferred in a lager cylinder, the mass of gas will be the same, but its volume will increase and density will decrease, this is because density is inversely proportional to volume.
## What is the relationship between pressure and volume?
The law itself can be stated as follows: Or Boyle’s law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant.
## How is density related to mass and volume?
Density is defined as the mass per unit volume. The SI unit for density is kilograms per cubic meter. However, often grams per cubic centimeter or grams per milliliter, which are equal (1 cubic centimeter = i milliliter), is used. In order to find the density of a substance, you divide its mass by its volume.
## How are volume and displacement related?
Displacement (fluid) The volume of the fluid displaced can then be measured, and from this, the volume of the immersed object can be deduced (the volume of the immersed object will be exactly equal to the volume of the displaced fluid). An object that sinks displaces an amount of fluid equal to the object’s volume.
## What is the definition of mass and volume?
Mass and volume are two units used to measure objects. Mass is the amount of matter an object contains, while volume is how much space it takes.
## Is density proportional to mass and volume?
Here Volume is constant and mass is proportional to density. Now if all the cooking gas in the same cylinder is transferred in a lager cylinder, the mass of gas will be the same, but its volume will increase and density will decrease, this is because density is inversely proportional to volume.
## What is the relationship between mass and weight?
Mass is a measure of how much matter an object contains, while weight is a measure of the force of gravity on the object. An object has the same composition, and therefore mass, regardless of its location. For example, a person with a mass of 70 kg on Earth has a mass of 70 kg in space as well as on the moon.
## Why do you think it would be difficult to prove the law of conservation of mass when gas is produced?
The law of conservation of mass indicates that mass cannot be created nor destroyed. If a gas is produced during a reaction, which mass is often forgotten when calculating the final mass because the students are unable to see the gas.
## What is the volume of the object?
Subtract the new water level from the starting water level. The amount of water the object displaces is equal to the volume of the object itself measured in cubic centimeters. Liquids are generally measured in milliliters, however, one milliliter is equal to one cubic centimeter.
## How does an object’s density affect the buoyant force acting on it?
For a floating object, the buoyancy force is equal to the gravity force on the object. Hence, the buoyancy force doesn’t change with a denser fluid. Instead the displaced volume decreases to cancel out the effect of the increased fluid density.
## What is the relationship between force and pressure?
Well Pressure is defined as ‘Force per unit area’—- Pressure=force/area. So, obviously , Force and Pressure are related , i.e. , Force is directly proportional to Pressure , which means , the more force you apply upon a fixed area , the more pressure you create .
## What is the relationship between density and buoyancy?
Buoyancy force is the weight of displaced fluid. So the buoyant force given by a liquid is directly proportional to the density of that liquid. Buoyant force = density of the liquid * volume of the object inside the liquid * gravitational constant g. Buoyancy and density are definitely related.
## How does density change with temperature?
Cooling a substance causes molecules to slow down and get slightly closer together, occupying a smaller volume that results in an increase in density. Hot water is less dense and will float on room-temperature water. Cold water is more dense and will sink in room-temperature water.
## What is the formula for measuring density of all matter?
Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a volume (the same volume as 1 milliliter). In contrast, iron is dense.
## What is the formula for volume in science?
The formula that relates density, mass, and volume looks like this: Here, m represents the mass of an object or material, V represents the volume, and the curly p (which is the Greek letter rho) represents the density. Sometimes, people can use non-standard or non-scientific units.
## How can density be used to determine the identity of a substance?
Density as a Characteristic Property. The density of a substance is a characteristic of that substance. Therefore, density is a property that can be used to help identify a substance. Properties used to help identify substances are called characteristic properties.
## Is the relationship between mass and volume?
To apply density relationships in problem solving. For any pure substance or homogenous mixture, density is an intensive quantity, which means that its value is the same no matter how much of the substance is present. Density expresses the proportional relationship between mass and volume of the substance.
## When volume increases does mass increase?
This happens because the mass of the rod stays the same, but its volume increases. The increase in the volume of matter with increasing temperature is called expansion. When cooled down, most matter decreases in volume and increases in density. This decrease in volume is called contraction.
Originally posted 2022-03-31 06:03:27.
Categories FAQ | 1,620 | 7,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-33 | latest | en | 0.935189 |
https://medium.com/@vicompany/vertical-rhythm-the-holy-grail-of-front-end-development-8f86eb152f18?ref=webdesignernews.com | 1,534,743,462,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215843.55/warc/CC-MAIN-20180820042723-20180820062723-00095.warc.gz | 718,323,344 | 24,275 | # Vertical rhythm: the Holy Grail of front-end development
By Olaf Muller, designer at VI Company
Just like a horizontal grid, the vertical grid is used for structure and balance in design. Simply put: a vertical grid is the basis for good typography and harmonious alignment.
View the demo or the Github page.
Grids have been around for quite some time in print. Applying the same technique in the browser is a complicated and time-consuming task. There are a lot of factors and formula’s to which all elements should comply. Our solution is SASS-based and does all the calculations automatically after defining the font size and modular scale for the body text.
Versatile and fully supported cross-browser, our vertical rhythm is a long-awaited step in the direction of better in-browser typography.
“Vertical rhythm is the Holy Grail of front-end development.”
- Someone at VI Company
For the best reading experience, a basic knowledge of the following subjects is required:
• Typography
• (Vertical) grids
• HTML and SASS
• Basic mathematics
So how did we get around creating a stable solution? Like any other design problem, we started with a question: how can we implement a vertical rhythm without too much hassle?
The first step was reading up on typography and all its wonders. We soon realized that starting with the most simple settings was the best way to get the formulas correct. After lots of trial and error, the project was beginning to materialize into a simple solution that focuses on the core of typography: typefaces, font size, line height, leading and modular scales.
The input needed to make the vertical rhythm work is encapsulated in a list of variables that can be easily set by the designer. These variables are run through a mixin that outputs the correct values to align an element to the baseline.
### Base typography
The goal of typography is to create balanced, readable and legible text. Good typography depends on many factors, luckily there are calculators that give a good indication of what combinations work best. Because we know designers are going to use the code, we wanted to start things off easy by letting them define the base font size and line height in pixels. These values are automatically converted to rems.
`\$px-font-size: 16; \$px-line-height: 24;`
### Modular scale
The next step is selecting a modular scale. A modular scale is nothing more than a list of values that share the same relationship. For example: if we apply the golden section (1.618) to a base font size of 16px, the outcome would be:
• Font size xs: 16 * 1.618^-1 = 9.89px
• Font size s (starting point): 16px
• Font size m: 16 * 1.618 = 25.89px
• Font size l: 16 * 1.618² = 41.89px
• Font size xl: 16 * 1.618³ = 67.77px
There are 17 predefined modular scales:
`\$modular-scale: "golden section"; `
`\$modular-scales: ( "minor second": 1.067, "major second": 1.125, "minor third": 1.2, "major third": 1.25, "perfect fourth": 1.333, "augmented fourth": 1.414, "perfect fifth": 1.5, "minor sixth": 1.6, "golden section": 1.618, "major sixth": 1.667, "minor seventh": 1.778, "major seventh": 1.875, "octave": 2, "major tenth": 2.5, "major eleventh": 2.667, "major twelfth": 3, "double octave": 4 );`
### Applying the vertical rhythm
Applying the vertical rhythm should be as easy as possible. For this, we’ve predefined 5 font sizes. To apply one of the font sizes to an element, you’ll only need the following code (the mixin does the calculations):
`small { @include vertical-rhythm(\$font-size-xs); } p { @include vertical-rhythm(\$font-size-s); } h3 { @include vertical-rhythm(\$font-size-m); } h2 { @include vertical-rhythm(\$font-size-l); } h1 { @include vertical-rhythm(\$font-size-xl); }`
### Alignment
To round it all up, aligning other elements can be done using predefined distance and size variables that are based on the vertical rhythm itself:
`div { margin: \$distance-s; } img { height: \$size-m; }`
### Conclusion
We think our solution makes the process of creating harmonious typography easier but there’s still a long way to go. When using this vertical rhythm, it is essential that all other elements on the site are aligned as well. We’ve done this by using predefined sizes and distances based on the rhythm of the vertical grid. We have the basics working and are eager to continue improving the vertical rhythm.
Originally published at www.vicompany.nl.
Like what you read? Give VI Company a round of applause.
From a quick cheer to a standing ovation, clap to show how much you enjoyed this story. | 1,116 | 4,607 | {"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} | 2.765625 | 3 | CC-MAIN-2018-34 | longest | en | 0.923984 |
https://www.stata.com/statalist/archive/2012-04/msg00457.html | 1,582,759,287,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00448.warc.gz | 893,560,422 | 4,084 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# Re: st: Multicollinearity and Time Trends
From "Justina Fischer" To statalist@hsphsun2.harvard.edu, statalist@hsphsun2.harvard.edu Subject Re: st: Multicollinearity and Time Trends Date Wed, 11 Apr 2012 14:47:45 +0200
```Hi Marten,
I thought similarly....
So the best way of indeitfying these three different trends in the same model would be to start the time trend with a minus sign, letting it cross the zero-line.
I'll keep this in kind for my own research....
Thanks
Justina
-------- Original-Nachricht --------
> Datum: Wed, 11 Apr 2012 14:35:44 +0200
> Von: Maarten Buis <maartenlbuis@gmail.com>
> An: statalist@hsphsun2.harvard.edu
> Betreff: Re: st: Multicollinearity and Time Trends
> On Wed, Apr 11, 2012 at 1:31 PM, Stefan Pichler wrote:
> > I want to detrend time series data and allow not only for linear trends,
> but also for quadratic and cubic trends. Here is a minimalistic example of
> the data:
> >
> > Y...outcome variable (some randomly typed numbers)
> > year....the year of the observation
> > gen year2=year^2
> > gen year3=year^3
> >
> > gen year1=year-1950 (so that year starts from 1)
> > gen year12=year1^2
> > gen year13=year1^3
>
> > If I tell Stata: "reg Y year year2 year3", Stata omits
> year because of collinearity, however if I regress "Y year1
> year12 year13" no variable is omitted.
>
> Think of this this way: Stata can distinguish between year12 and year1
> because they are non-linearly related. This non-linearity is greatest
> near 0 (the minimum of the parabola). However, if you get further and
> further away from 0, it becomes almost linear.
>
> Consider these three graphs:
>
> twoway function y = x^2, range(1951 1960)
> twoway function y = x^2, range(1 10)
> twoway function y = x^2, range(-4.5 4.5)
>
> In the last graph it is easiest to distinguish x from its square. This
> is true for us when we look at the graph, but also for computers.
>
> Hope this helps,
> Maarten
>
> --------------------------
> Maarten L. Buis
> Institut fuer Soziologie
> Universitaet Tuebingen
> Wilhelmstrasse 36
> 72074 Tuebingen
> Germany
>
>
> http://www.maartenbuis.nl
> --------------------------
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
--
Justina AV Fischer, PhD
COFIT Fellow
University of Bern
homepage: http://www.justinaavfischer.de/
e-mail: javfischer@gmx.de. justina.fischer@wti.org
papers: http://ideas.repec.org/e/pfi55.html
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 854 | 2,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.853561 |
http://worksheets.mathsbuilder.com.au/games/selector/Addition_and_Subtraction/3/ | 1,527,175,215,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00424.warc.gz | 331,067,087 | 9,485 | Hosted by Maths Builder - the ground-breaking primary maths teaching and learning program. Learn more >
Click on a level then select a game.Maths Builder Champion consists of over 1500 interactive games that help students master important concepts. Whole Numbers, Addition and Subtraction, and the first game in each of the other 14 topics are free – no logon is required.Each game is linked to a Maths Builder worksheet. Maths Builder is designed to master every skill introduced in the Primary School Maths Syllabus. [x] Close 0 1 2 3 4 5 6
3:01 Addition and subtraction...
Interactive Game
Interactive Game
Interactive Game
3:02A Addition facts to 20
Interactive Game
3:02E Addition and subtraction...
Interactive Game
3:03 The jump strategy
Interactive Game
3:04 Addition using Base 10...
Interactive Game
3:04C Addition using Base 10...
Interactive Game
3:05 Addition algorithm, no...
Interactive Game
Interactive Game
3:07 More addition strategies
Interactive Game
3:08 Subtraction (blocks)
Interactive Game
3:09 Subtraction, no trading
Interactive Game
3:09A Relating addition and...
Interactive Game
3:09C The jump strategy
Interactive Game
3:10 Subtraction, no trading
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
3:14 Shopping
Interactive Game
3:14C Giving change
Interactive Game
3:15 Subtraction with trading
Interactive Game
3:15C Subtraction with trading
Interactive Game
3:16 Subtraction with trading
Interactive Game
3:17 Subtraction with trading
Interactive Game
3:17C Subtraction with trading
Interactive Game
Interactive Game
3:19 Subtraction with trading
Interactive Game
3:19E Addition and subtraction...
Interactive Game
3:20 Estimation with addition
Interactive Game
3:21 Estimation with...
Interactive Game
3:22 Decimal addition and...
Interactive Game
3:23 Addition and subtraction
Interactive Game
3:23E Addition and subtraction
Interactive Game
3:24 Addition using place...
Interactive Game
3:24A Adding large numbers...
Interactive Game
3:25 Subtraction using place...
Interactive Game
3:25A Subtraction using place...
Interactive Game
3:NA01 Addition and subtraction...
Interactive Game
Interactive Game
Interactive Game
3:NA02C Addition and subtraction...
Interactive Game
3:NA20 The jump strategy
Interactive Game
3:NA21 Addition using Base 10...
Interactive Game
3:NA21C Addition using Base 10...
Interactive Game
3:NA22 Addition algorithm, no...
Interactive Game
3:NA22C Addition using Base 10...
Interactive Game
Interactive Game
3:NA30 More addition strategies
Interactive Game
3:NA31 Subtraction (blocks)
Interactive Game
3:NA31C The jump strategy
Interactive Game
3:NA32 Subtraction, no trading
Interactive Game
3:NA32C Relating addition and...
Interactive Game
3:NA33 Subtraction, no trading
Interactive Game
Interactive Game
3:NA34 Combining groups
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
Interactive Game
3:NA46 Shopping
Interactive Game
3:NA46C Giving change
Interactive Game
3:NA47 Representing money values
Interactive Game
3:NA48 Subtraction with trading
Interactive Game
3:NA48C Subtraction with trading
Interactive Game
3:NA49 Subtraction with trading
Interactive Game
3:NA49C Subtraction with trading
Interactive Game
3:NA50 Subtraction with trading
Interactive Game
Interactive Game
3:NA51C Addition and subtraction...
Interactive Game
3:NA52 Subtraction with trading
Interactive Game
3:NA59 Estimation with addition
Interactive Game
3:NA60 Estimation with...
Interactive Game
3:NA69 Inverse operations
Interactive Game
3:NA70 Addition and subtraction
Interactive Game
3:NA70C Addition and subtraction
Interactive Game | 874 | 3,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.737038 |
https://m.brighthubeducation.com/middle-school-math-lessons/128198-solve-problems-involving-scale-drawings-of-geometric-figures/ | 1,534,340,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210105.8/warc/CC-MAIN-20180815122304-20180815142304-00001.warc.gz | 742,807,772 | 28,831 | # Seventh Grade Math Lesson: Create a Scale Drawing of Your Dream House
By Nicole Hilsabeck
Invite your students to utilize their imaginations and create scale drawings to find real-life object sizes. Have your class design a dream home and then determine the square footage.
Objective: Students will make collages and use rulers to create scale drawings and find real-life dimensions of objects.
CCSS.Math.Content.7.G.A.1 Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale.
Note: students should be familiar with using rulers to find dimensions of two-dimensional objects and finding area of squares and rectangles.
Materials: Large drawing paper, rulers, pencils, home design magazines, scissors, gluesticks
## Step One: Introduction
Provide students with a variety of home design magazines (or ask students to bring in their own). Students will have 20 minutes to browse through the magazines and cut out ideas for their own “dream houses" that they would like to build someday.
After students have cut out their pictures, introduce the idea of drawing a floor plan for a house. Encourage students to make a pre-design collage, placing the ideas for their houses in the approximate layout of how they would like their homes to be designed.
After collages are complete, demonstrate how to draw a basic floor plan for a house, using a “bird’s eye" view and drawing the shapes of rooms with flat squares and rectangles (for this activity, you may want to discourage circles unless students are interested in an additional challenge).
## Step Two: Creation of Designs and Introduction to Scale
Give students time to draw their floor plans, asking them to label each room and to measure and note dimensions (in inches) of each room. Once your students have drawn their basic floor plans and measured the dimensions of the rooms, stop class for a brief discussion on scale.
Begin the discussion by asking students if they know the dimensions of their homes, and why they think home builders use blueprints. How do home builders know how to make homes the correct size? After students have shared their ideas, introduce a basic scale, drawing it on the board (example 1 inch = 3 feet). Using the sample floor plan drawn for the first portion of the lesson, demonstrate how to use the scale to calculate the actual dimensions of each room. You may want to demonstrate how to make a chart to show the scale in use:
## Step Three: Application of Skill
Give students time to apply their scales to find the dimensions of each room shown on their floor plans. Students will fill in their charts accordingly and write the square footage of each room under the label for each room on the blueprints.
Assessment: Collect student’s blueprints or charts and assess the accuracy of students’ calculations.
Extension: Offer students a chance to find the entire square footage of their dream homes, and research the prices of current homes for sale that match the general square footage of the homes they designed for this lesson. | 617 | 3,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-34 | latest | en | 0.933675 |
https://edurev.in/v/118122/UnderstandingPerimeter-of-a-Rectangle-Mensuration- | 1,726,176,783,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00882.warc.gz | 200,241,436 | 62,068 | Understanding: Perimeter of a Rectangle
# Understanding: Perimeter of a Rectangle Video Lecture | Mathematics (Maths) Class 6
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
## FAQs on Understanding: Perimeter of a Rectangle Video Lecture - Mathematics (Maths) Class 6
1. What is the formula for calculating the perimeter of a rectangle?
Ans. The formula for calculating the perimeter of a rectangle is P = 2(l + b), where P represents the perimeter, l represents the length, and b represents the breadth of the rectangle.
2. How do I find the length of a rectangle if the perimeter and breadth are given?
Ans. To find the length of a rectangle when the perimeter and breadth are given, you can use the formula: Length = Perimeter/2 - Breadth.
3. Can the perimeter of a rectangle be negative?
Ans. No, the perimeter of a rectangle cannot be negative. Perimeter is a measure of the distance around the shape, and distance cannot be negative.
4. If the length and breadth of a rectangle are the same, what is the shape called?
Ans. If the length and breadth of a rectangle are the same, the shape is called a square. A square is a special type of rectangle where all sides are equal.
5. How does the perimeter of a rectangle change if its length is doubled?
Ans. If the length of a rectangle is doubled, the perimeter will also be doubled. The perimeter of a rectangle is directly proportional to its length, so any change in the length will result in the same proportional change in the perimeter.
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
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# Trying to teach myself kotlin, any help would be great
Greenhorn
Posts: 14
• Number of slices to send:
Optional 'thank-you' note:
Any small challenges like making a small app in kotlin would be really helpful, I have been trying to build a small program that you can pick a few random numbers, don't repeat any numbers and prints the numbers in order ( 1..150)
Is this code effective and how could I improve the code?
import kotlin.random.Random
fun main(args: Array<String>) {
/** Random number picker with no repeating numbers.
* will display them in number order.
* Random numbers upto 150.
* Upto 150 numbers balls.
*
* randomList(max number int 1..) ( how many numbers/ balls)
**/
randomList( 27, 30)
}
fun randomList(maxNumber : Int, totalNumbers : Int){
if (totalNumbers > 150){
}else if (totalNumbers > maxNumber){
print("maxNumber has to be same or higher than totalNumbers")
}else{
var list = mutableListOf<Int>()
var totalNumbersMinus1 = totalNumbers-1
var randomList = (1..maxNumber).shuffled().take(totalNumbers)
for(I in 0..totalNumbersMinus1){
var number = randomList[I]
}
list.sort()
for(I in 0..totalNumbersMinus1){
print("\${list[I]} ")
}
}
}
Master Rancher
Posts: 3889
50
• 1
• Number of slices to send:
Optional 'thank-you' note:
Well, perhaps the first thing to do is compile and run your program - you will see that it does compile, which is good, and then you get an error message when you run it - which may or may not have been intentional, if you were testing what happens for different input values.
My IntelliJ shows "import kotlin.random.Random" greyed out - you don't need it. Maybe it was needed in some older version of Kotlin. But nowadays Random is directly available without that.
IntelliJ also shows messages indicating that each of your var declarations can actually be a val, since you never reassign a variable to be something else. This is good practice - use val rather than var whenever you can, to minimize unexpected changes. If you need something to be a var, fine, make it a var. But don't do it if you don't need it.
For the rest of the code...
You seem to be doing the same thing several different ways. The one I like best is this one:
Here you've accomplished most of what you want in one line - great! You don't need to copy another list with the same contents. And you don't need to worry about only getting totalNumbers elements from this list - it's already limited in length, to have only totalNumbers elements. So you just need sort it, and print each element. Elsewhere you use the sort() method, which is good, but won't work on randomList because it's immutable. You could fix thaqt with toMutableList():
But it's easier to replace sort() with sorted(). The latter creates a new immutable list, in sorted order, leaving the original immutable list alone:
But at this point, the randomList variable itself is no longer needed:
Jason Edwards
Greenhorn
Posts: 14
• Number of slices to send:
Optional 'thank-you' note:
Hi Mike Simmons,
Thanks for looking over my code and giving me some help.
Mike Simmons wrote:Well, perhaps the first thing to do is compile and run your program - you will see that it does compile, which is good, and then you get an error message when you run it - which may or may not have been intentional, if you were testing what happens for different input values.
I was paying around with inputs, testing my code and checking the errors.
Mike Simmons wrote:My IntelliJ shows "import kotlin.random.Random" greyed out - you don't need it. Maybe it was needed in some older version of Kotlin. But nowadays Random is directly available without that.
My workplace is quiet for the next 2 weeks. So I was using https://play.kotlinlang.org/ to build the app(spent about 3 hours on the code)
Mike Simmons wrote:IntelliJ also shows messages indicating that each of your var declarations can actually be a val, since you never reassign a variable to be something else. This is good practice - use val rather than var whenever you can, to minimize unexpected changes. If you need something to be a var, fine, make it a var. But don't do it if you don't need it.
Will do from now on.
Mike Simmons wrote:You seem to be doing the same thing several different ways. The one I like best is this one:
I really like this code but from what I can tell, its read only!, you can print the full list or an element, looked online for random list but could not find any post talking about .sorted() or .forEach({print("\$it ")})
so moved the numbers over to an array I could sort and edit the list.
Thank you for the help.
I really enjoy reading code like this, makes perfect sense.
Mike Simmons wrote:(1..maxNumber).shuffled().take(totalNumbers).sorted().forEach({print("\$it ")})
Jason Edwards
Greenhorn
Posts: 14
• Number of slices to send:
Optional 'thank-you' note:
Jason Edwards wrote:Hi Mike Simmons,
Thanks for looking over my code and giving me some help.
Well, perhaps the first thing to do is compile and run your program - you will see that it does compile, which is good, and then you get an error message when you run it - which may or may not have been intentional, if you were testing what happens for different input values.
I was paying around with inputs, testing my code and checking the errors.
My IntelliJ shows "import kotlin.random.Random" greyed out - you don't need it. Maybe it was needed in some older version of Kotlin. But nowadays Random is directly available without that.
My workplace is quiet for the next 2 weeks. So I was using https://play.kotlinlang.org/ to build the app(spent about 3 hours on the code)
IntelliJ also shows messages indicating that each of your var declarations can actually be a val, since you never reassign a variable to be something else. This is good practice - use val rather than var whenever you can, to minimize unexpected changes. If you need something to be a var, fine, make it a var. But don't do it if you don't need it.
Will do from now on.
You seem to be doing the same thing several different ways. The one I like best is this one:
I really like this code but from what I can tell, its read only!, you can print the full list or an element, looked online for random list but could not find any post talking about .sorted() or .forEach({print("\$it ")})
so moved the numbers over to an array I could sort and edit the list.
Thank you for the help.
I really enjoy reading code like this, makes perfect sense.
Don't get me started about those stupid light bulbs. | 1,687 | 7,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.722331 |
https://chemistry.stackexchange.com/questions/166993/why-is-the-standard-electrode-potential-positive-for-half-cells-that-are-easily | 1,723,105,024,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00147.warc.gz | 126,178,473 | 46,263 | # Why is the standard electrode potential positive for half cells that are easily reduced?
I was studying electrochemistry from my school textbook.
The cell potential in the book is defined as the difference between the electrode potentials of the cathode and the anode.$$E_\text{cell}=E_\text{right}-E_\text{left}$$
We know that electrode potential of an individual half-cell can't have an absolute value and thus we measure it taking the standard hydrogen electrode ($$\ce{Pt(s) | H2(g) | H+(aq)}$$, SHE) in relation that is assigned a zero potential at all temperatures.
As the electrons flow from the anode to the cathode (higher potential to lower potential), the cathode should possess a lower electrostatic potential than the anode.
That means that the electrode potential of a half cell when measured against the SHE should always be negative if the reaction is feasible, i.e., electrons flow from SHE to the cathode. Thus, the more negative the value of the electrode potential of the cathode, the more electrons will from the SHE, and the more the tendency of the substance of the cathode to get reduced.
But the values of standard electrode potentials of different half cells in my textbook and Wikipedia show that the electrode potentials of substances that get easily reduced is positive, e.g., $$E^\ominus = 2.87$$ V for $$\ce{F_2(g) +2e^-->2F^-}$$. Isn't this the complete opposite?
Shouldn't the values of electrode potentials of substances that are easily reduced be negative as we just discussed? What am I missing here?
• @Poutnik I'm unable to understand what you're trying to say. I have a simple question: electrons flow from higher potential to lower potential, in our case from anode (SHE) to the cathode, i.e., cathode has a lower potential. As the potential of SHE is taken to be zero, the potential of the cathode should be negative (lower) if the reaction is feasible. But for such half cells, we see that the electrode potential is positive. Why so? Commented Aug 9, 2022 at 13:21
• I have moved my comments to the answer. Commented Aug 10, 2022 at 8:22
A large number electrochemicals misconceptions will be solved if we take the electrode potential sign as the electrostatic sign of the galvanic cell with respect to the hydrogen electrode. Thus this sign will not be affected by how the half cell is written. This was the European understanding of the leading 20th century top-electrochemists. The Germans decided to set the hydrogen electrode at 0 V, the reference point. The Germans also decided to set the hydrogen electrode at 0 V, as the reference point. This is the take of many leading modern electrochemists including A.J. Bard.
Now let us take an electrode such as copper. Its half-cell electrode potential is +0.34 V vs. SHE. Now interpret the sign this way: If we make a galvanic cell of copper and SHE, the copper electrode will be *positively charged" = the cathode, and the hydrogen cell will behave as an anode.
In the same way, take the fluorine half-cell (hypothetical), and if we make a galvanic cell of fluorine and SHE, the fluorine electrode will be *positively charged" = the cathode, because its half-cell electrode is + 2.87 V.
In a galvanic cell, the current is from A(node) -> C(athode) i.e., A to C.
• One nagging detail that has bothered me: shouldn't the Faraday constant (and by extension the elementary charge of the electron) be negative? It would seem easier to visualize this distinction under that circumstance, but I can't recall seeing the Faraday constant written that way. Commented Aug 10, 2022 at 0:44
• @MikeSerfas, Good question. Please post a new question. I have not historically explored this but will search about it.
– ACR
Commented Aug 10, 2022 at 1:02
• @AChem Does that mean that in actuality, the cathode has a lower potential than SHE (thus negative), but just to make it easier to understand, we use the positive sign for feasible reactions? Commented Aug 10, 2022 at 7:44
• No, this will be the actual electrostatic sign.
– ACR
Commented Aug 10, 2022 at 13:35
• From the modern electronic current point of view, electrons flow from "lower" potential to "higher potential. Positive current flow (so called conventional current) flows from higher to lower potential.
– ACR
Commented Aug 11, 2022 at 18:46
Blame Ben Franklin or physicists for not knowing what + and - are and having terms like current flow! We chemists know better and think only of electrons altho we do lapse into positive ion flow in Li ion batteries and in acids. No one is perfect.
Oxidation = Loss of electrons; Reduction = Gain of electrons
Oxidation is at the ANODE; Electrons LEAVE the anode into the wire. Anions move to the anode.
Reduction is at the CATHODE; Electrons enter the cathode from the wire. Cations move to the cathode.
Those are the simple rules. Unfortunately the physicists still get a say! They defined volts backwards and we are stuck with it [but some references are unstuck so it still can be confusing, so be careful]. A positive voltage means a reaction proceeds as written and its deltaG is negative. Electrode potentials are written [usually] as reduction half reactions. To appease the physicists a positive voltage means the reaction proceeds as written; a negative voltage means it proceeds in reverse. These half reactions are for a reaction under standard conditions with a hydrogen electrode. To get any reaction simply reverse one of the half reactions with its sign and add the equations and the voltages, then correct for differences from standard conditions. If the voltage is positive the reaction goes as written; if negative reverse the equation and change the sign. [or else change the concentrations and the cell voltage. If the cell voltage is close to zero and the electrode reactions reversible the direction of reaction or the cell voltage is easily manipulated.]
The pertinent equation is: DeltaG = -EnF = -E[0]nF +RTlnQ Q is the reaction quotient with the form of the equilibrium constant Keq. At equilibrium deltaG = 0 so
E[0]nF = RTlnKeq = -deltaG[0]. As can be seen all these + and - signs can get confusing. [LOOK THIS UP IN A PCHEM TEXT or two] [The font this site uses substitutes a lower case o for the numeral zero.]
To sum it up a positive cell voltage corresponds to an equilibrium constant greater than 1. and a negative deltaG. The reaction proceeds as written. How far it goes depends on the voltage and the amount of material [the size of the battery]. At equilibrium the cell voltage is zero and the various concentrations satisfy the equilibrium constant
To react lithium and fluorine: galvanic cell: Li = Li+ + e- E= +3v; F- = 1/F2 +e- E= -3. We have two reduction reactions and need one oxidation and one reduction so must reverse one. Lets do Lithium e-+ Li+ = Li E= -3v. Put the two half reactions together Li+ + F- = Li + 1/2F2 E = -6V. Wow it is going nowhere. So we reverse the reaction. Li + 1/2F2 = Li+ + F-; E = 6V! an explosion!! Not so Good so we set up a cell with an anode, Li, a wire to a motor then to a cathode [who knows what, it can't react with F2], surrounded with F2 and a magic electrolyte to carry the Li+ and F- ions formed remotely and desperately trying to get to the cathode and anode respectively. To reverse this a potential greater than 6V is applied in reverse. I do not think that this particular cell has been made not even by accident.
Work out some simple cells such as a hydrogen ion concentration cell or electrolytic deposition of copper to get an idea what happens to cell voltages as concentrations change. If you get stuck on conventions or signs remember Li is the strongest reductant and F2 is [almost] the strongest oxidant use these two to determine if a different convention is being used for signs.
Easily reduced redox systems "suck" electrons from an electrode, charging it to the more positive equilibrium potential, at which the electron transfer rate is zero.
And vice versa for easily oxidized redox systems: They are pushing electrons to an electrode, charging it to the more negative equilibrium potential.
In galvanic cells, cathodes have higher potential than anodes, in contrary to electrolytic cells. By other words, when switching the cell mode, electrode names switch places, as the current switches direction.
E.g. Imagine the notoriously known Daniell cell:
$$\ce{Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)}$$
If we perform electrolysis on it, with the negative contact attached to $$\ce{Zn}$$, the cathode is the left, more negative electrode, as electrons come to $$\ce{Zn}$$ and reduce $$\ce{Zn^2+(aq)}$$.
If we use it as the galvanic cell, the cathode is the right, more positive electrode, as electrons come to $$\ce{Cu}$$ and reduce $$\ce{Cu^2+(aq)}$$.
SHE has the absolute alectrode potential $$\pu{+4.44 \pm 0.02 V}$$.
For Ox/Red redox pair and a formal reaction $$\ce{Ox + n/2 H2 -> n H+ + Red}$$, there is direct relation of the standard reaction Gibbs energy and the respective standard electrode potential $$\Delta_\mathrm{r} G^{\circ}=−nFE^{\circ}$$.
This is because of Gibbs free energy. The equation $$E^{\circ}=-\frac{\Delta_{\text{r}}G^{\circ}}{nF}$$ makes it clear that for spontaneous reactions $$(\Delta_{\text{r}}G^{\circ}<0)$$, the electrode potential is positive. Highly electronegative elements such as fluorine undergoes reduction very spontaneously and hence, the positive value. In contrast s- and p-block elements have negative electrode potentials as they tend to oxidize easily.
The reference of SHE allows one to determine if the given element can reduce $$\ce{H+}$$ to hydrogen gas. | 2,325 | 9,623 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.932725 |
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View Full Version : combining successive glRotate calls
malancha
03-11-2004, 07:58 PM
Hi, I'd like to know if it's possible to combine a sequence of successive glRotate calls into one glRotate call.
Let's assume I've stored the parameters used for each call and the order in which the calls are to be made.
Is there an algorithm or formula that achieves this?
mikael_aronsson
03-11-2004, 10:05 PM
Hi !
You will have to keep you own matrix and put the rotations in there, a matrix can contain any number of transformations (translate,rotate,scale) and you can then use glMultMatrix to get the combined transformation into OpenGL.
OpenGl itself does not have any suuport for this, but there are tons of tutorials and also lbraries available for this, try google.
Mikael
chemdog
03-11-2004, 10:41 PM
You want quaternions!
Usually only rotations on the same axis can combined together, and they needed to be successive, without other transformations between them.
Rotate x pi/2 Rotate x pi/6 Rotate y pi/12 Rotate x pi/2 Rotate z -pi/16 Rotate z pi/3 Rotate x pi/2
This can be combined into
Rotate x 2pi/3 Rotate y pi/12 Rotate x pi/2 Rotate z 13pi/48 Rotate x pi/2
If you would like to convert the rotations into quaternions, you can combine all the rotations into a single quaternion. Which the above can be reduced to
Quaternion (.65, .6, .7, 0.01)
then equivalently,
glRotate(167, .65, .6, .7)
The number above aren't perfect, but the procedure is sound.
malancha
03-11-2004, 11:24 PM
Perhaps I should make clear what exactly I'm trying to achieve. I have a plane which the user can rotate, translate, zoom, etc. Once the user is done, I need to be able to indicate the new orientation. This is important because the plane is such that it's easy to mix up the right and left sides, and I can't allow that. Also, I don't want to display the markers while the user plays around with the orientation.
Will it be enough to simply retrieve the Modelview matrix once the user is done, and then use glMultMatrix to determine where on the screen my markers need to be placed?
ZbuffeR
03-14-2004, 01:26 AM
Will it be enough to simply retrieve the Modelview matrix once the user is done, and then use glMultMatrix to determine where on the screen my markers need to be placed?
Yes, get the final matrix :
glGetDoublev(GL_MODELVIEW_MATRIX,double*);
And multmatrix it. | 613 | 2,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-30 | latest | en | 0.908584 |
fordatascientists.wordpress.com | 1,518,982,975,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.18/warc/CC-MAIN-20180218192636-20180218212636-00135.warc.gz | 677,600,356 | 16,167 | # Data Distributions
Welcome to data distribution, here we are going to talk about central tendency. Keywords used are Mean, Median, Mode, Normal Distribution, Skewness, Kurtosis and various types of Skewness & Kurtosis.
The three measures to identify central tendency(main behavior): the mean, median, and mode. These measures relate in different ways to different distributions either skewed or normal. Intelligence tests can also be measured using central tendency principles as well each has a distribution. Intelligence tests also are biased in material and need regular updates.
The mean is the average score or the sum of the scores divided by the number of different scores. The median is the middle number in a set, which divides a distribution in half. The mode is the most frequent score, which shows up the most often. A distribution is the curve of scores. A skewed distribution is a curve that has a long tail in one direction and has extreme scores that change the mean. The three measures mean, median and mode under a normal distribution are all the same. Also in a positively skewed distribution, the mean is the greatest number, as to the median and mode. An intelligence test measures the abilities of a person and distributes it as a score. Under the standard, most scores are within 15 points of the mean, being 85 and 115. In two normal distributions, there can be an overlap as some people score above and below the mean between the groups. To determine if a test is biased then the scores of two groups need to be measured to see if the test is biased or not.
## Mean, Median, Mode
How to use mean, median and mode to determine the shape of a distribution?
Two things once often we should start using to be DS, are
• Skewness
• Kurtosis
Definitions
Skewness/Kurtosis
Skewness is the degree of departure from symmetry of a distribution. A positively skewed distribution has a tail which is pulled in the positive direction. A negatively skewed distribution has a tail which is pulled in the negative direction.
Kurtosis is the degree of peakedness of a distribution. A normal distribution is a mesokurtic distribution. A pure leptokurtic distribution has a higher peak than the normal distribution and has heavier tails. A pure platykurtic distribution has a lower peak than a normal distribution and lighter tails.
## Normal Distribution Proportions
with 1 standard deviations distance, we cover 68.2% and with 2 standard deviations, we cover 68.2 + 27.2 = 95.4 (95%) of sample space.
Disclaimer: Images shown are not my own. Please be aware of copyrights before you re-publish contents. | 560 | 2,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-09 | latest | en | 0.939575 |
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List all the elements of the following sets : (i) A = {x : x is an odd natural number} (ii) B = {x : x is an integer, -1/2<x<9 2="" }="" (iii)="" c="{x" :="" x="" is="" an="" integer,="" x^2<=4" (iv)="" d="{x" a="" letter="" in="" the="" word="" "loyal"}="" (v)="" e="{x" i<="" b=""> </x<9>
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Latest from Doubtnut | 300 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-04 | latest | en | 0.774756 |
https://www.narviacademy.in/Topics/Competitive%20Exams/Quantitative%20Aptitude/Quantitative%20Aptitude%20Notes/time-speed-and-distance-quantitative-aptitude-exercise-7.php | 1,722,967,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00162.warc.gz | 719,473,627 | 5,607 | # Boat and Stream Aptitude Questions and Answers:
#### Overview:
Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Time Speed and Distance Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.
1. A boat goes downstream at the speed of $$25 \ km/hr$$. If the speed of stream is $$12 \ km/hr$$, then find the speed of boat in still water?
1. $$11 \ km/hr$$
2. $$12 \ km/hr$$
3. $$13 \ km/hr$$
4. $$14 \ km/hr$$
Answer: (c) $$13 \ km/hr$$
Solution: Given, speed of downstream $$(D_s) = 25 \ km/hr$$
speed of stream $$(S_s) = 12 \ km/hr$$
then speed of boat in still water, $$B_s = D_s - S_s$$ $$B_s = 25 - 12 = 13 \ km/hr$$
1. A swimmer goes downstream with the, speed of stream $$14 \ km/hr$$. If speed of the swimmer in still water is $$28 \ km/hr$$, then find the speed of downstream?
1. $$40 \ km/hr$$
2. $$42 \ km/hr$$
3. $$44 \ km/hr$$
4. $$45 \ km/hr$$
Answer: (b) $$42 \ km/hr$$
Solution: Given, speed of stream $$(S_s) = 14 \ km/hr$$
speed of swimmer in still water $$(B_s) = 28 \ km/hr$$
then speed of downstream, $$B_s = D_s - S_s$$ $$28 = D_s - 14 = 42 \ km/hr$$
1. A boat goes downstream with the speed of $$40 \ km/hr$$. If the speed of boat in still water is $$30 \ km/hr$$, then find the speed of stream?
1. $$10 \ km/hr$$
2. $$12 \ km/hr$$
3. $$13 \ km/hr$$
4. $$14 \ km/hr$$
Answer: (a) $$10 \ km/hr$$
Solution: Given, speed of downstream $$(D_s) = 40 \ km/hr$$
speed of boat in still water $$(B_s) = 30 \ km/hr$$ then, $$B_s = D_s - S_s$$ $$30 = 40 - S_s$$ $$S_s = 10 \ km/hr$$
1. A girl swims $$30 \ km$$ downstream in $$1 \ hour$$. If the speed of stream is $$5 \ km/hr$$, then find the speed of the girl in still water?
1. $$20 \ km/hr$$
2. $$22 \ km/hr$$
3. $$25 \ km/hr$$
4. $$27 \ km/hr$$
Answer: (c) $$25 \ km/hr$$
Solution: Given, speed of downstream $$(D_s) = \frac{30}{1} = 30 \ km/hr$$
speed of stream $$(S_s) = 5 \ km/hr$$
then speed of girl in still water, $$B_s = D_s - S_s$$ $$B_s = 30 - 5 = 25 \ km/hr$$
1. A boat goes upstream with the speed of $$18 \ km/hr$$. If speed of stream is $$5 \ km/hr$$, then find the speed of boat in still water?
1. $$23 \ km/hr$$
2. $$22 \ km/hr$$
3. $$21 \ km/hr$$
4. $$20 \ km/hr$$
Answer: (a) $$23 \ km/hr$$
Solution: Given, speed of upstream $$(U_s) = 18 \ km/hr$$
speed of stream $$(S_s) = 5 \ km/hr$$
then speed of boat in still water, $$B_s = U_s + S_s$$ $$B_s = 18 + 5 = 23 \ km/hr$$
1. A boat goes $$70 \ km$$ upstream, taking $$2 \ hours$$. If the speed of stream is $$20 \ km/hr$$, then find the speed of boat in still water?
1. $$52 \ km/hr$$
2. $$53 \ km/hr$$
3. $$54 \ km/hr$$
4. $$55 \ km/hr$$
Answer: (d) $$55 \ km/hr$$
Solution: Given, speed of upstream $$(U_s) = \frac{70}{2} = 35 \ km/hr$$
speed of stream $$(S_s) = 20 \ km/hr$$
then speed of boat in still water, $$B_s = U_s + S_s$$ $$B_s = 35 + 20 = 55 \ km/hr$$
1. If a boat goes $$12 \ km$$ downstream in $$36 \ minutes$$ and the speed of boat in still water is $$8 \ km/hr$$, then find the speed of stream?
1. $$10 \ km/hr$$
2. $$12 \ km/hr$$
3. $$13 \ km/hr$$
4. $$15 \ km/hr$$
Answer: (b) $$12 \ km/hr$$
Solution: Given, speed of downstream $$(D_s) = \frac{12}{36} \times 60 = 20 \ km/hr$$
speed of boat in still water $$(B_s) = 8 \ km/hr$$
then speed of stream, $$B_s = D_s - S_s$$ $$8 = 20 - S_s$$ $$S_s = 12 \ km/hr$$
1. If a man can swim upstream at the speed of $$7 \ km/hr$$ and speed of stream is $$3 \ km/hr$$, then find the speed of man in still water?
1. $$15 \ km/hr$$
2. $$13 \ km/hr$$
3. $$12 \ km/hr$$
4. $$10 \ km/hr$$
Answer: (d) $$10 \ km/hr$$
Solution: Given, speed of upstream $$(U_s) = 7 \ km/hr$$
speed of stream $$(S_s) = 3 \ km/hr$$
then speed of the man in still water, $$B_s = U_s + S_s$$ $$B_s = 7 + 3 = 10 \ km/hr$$
1. A women can swim $$18 \ km$$ downstream in $$10 \ minutes$$. If the speed of stream is $$80 \ km/hr$$, then find speed of the women in still water?
1. $$28 \ km/hr$$
2. $$25 \ km/hr$$
3. $$24 \ km/hr$$
4. $$22 \ km/hr$$
Answer: (a) $$28 \ km/hr$$
Solution: Given, speed of downstream $$(D_s) = \frac{18}{10} \times 60 = 108 \ km/hr$$
speed of stream $$(S_s) = 80 \ km/hr$$
then speed of the women in still water, $$B_s = D_s - S_s$$ $$B_s = 108 - 80 = 28 \ km/hr$$
1. A boat goes upstream with the speed of $$30 \ km/hr$$. If the speed of boat in still water is $$45 \ km/hr$$, then find the speed of stream?
1. $$12 \ km/hr$$
2. $$13 \ km/hr$$
3. $$14 \ km/hr$$
4. $$15 \ km/hr$$
Answer: (d) $$15 \ km/hr$$
Solution: Given, speed of upstream $$(U_s) = 30 \ km/hr$$
speed of boat in still water $$(B_s) = 45 \ km/hr$$
then speed of stream, $$B_s = U_s + S_s$$ $$45 = 30 + S_s$$ $$S_s = 15 \ km/hr$$ | 1,707 | 4,752 | {"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} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.832536 |
http://openstudy.com/updates/5572ade5e4b0bffbfefe70d3 | 1,476,992,754,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717783.68/warc/CC-MAIN-20161020183837-00474-ip-10-171-6-4.ec2.internal.warc.gz | 188,874,629 | 36,803 | UnkleRhaukus one year ago Traversing graphs
1. UnkleRhaukus
2. UnkleRhaukus
I know how to traverse: a tree (an undirected, acyclic graph, with one root), but how i am supposed to traverse: a cyclic graph with no root?
3. UnkleRhaukus
... maybe it helps to redraw the graph? |dw:1433579283033:dw|
4. UnkleRhaukus
I can't see how breadth, depth, or level make sense here
5. rational
|dw:1433580732682:dw|
6. rational
dequeue and grab all the adjacent vertices |dw:1433580852616:dw|
7. rational
push all the unseen vertices into Queue : |dw:1433580916917:dw|
8. rational
dequeue and grab all the adjacent vertices |dw:1433580986838:dw|
9. rational
push all the unseen vertices into Queue : |dw:1433581041015:dw|
10. rational
dequeue and grab all the adjacent vertices : |dw:1433581119983:dw|
11. rational
push all the unseen vertices into queue : |dw:1433581254223:dw|
12. rational
dequeue and grab all the adjacent vertices : |dw:1433581272392:dw|
13. rational
push all the unseen vertices into queue : |dw:1433581330168:dw|
14. rational
dequeue and grab all the adjacent vertices : |dw:1433581347294:dw|
15. rational
push all the unseen vertices into queue : |dw:1433581402680:dw|
16. rational
dequeue and grab all the adjacent vertices : the queue is empty, so we're done!
17. rational
|dw:1433581460670:dw|
18. UnkleRhaukus
19. rational
right! my mistake, but that wont change anything because both A and C are seen already
20. rational
since {E,D,C} are at same level, we will have 3! = 6 different paths using breadth first search
21. UnkleRhaukus
hmm, ok, should i start by making the adjacent list?
22. rational
you want to implement ?
23. UnkleRhaukus
24. rational
for representing graph is it ?
25. rational
this is a small graph so anything will do i guess
26. UnkleRhaukus
$\begin{array}{cccccc}&A&B&C&D&E\\ A&\infty&1&\infty&\infty&\infty\\B&\infty&\infty&1&1&1\\C&1&\infty&\infty& 1&\infty\\D&1&\infty&1&\infty&\infty\\E&\infty&\infty&\infty&\infty&\infty \end{array}$
27. UnkleRhaukus
hmm ok, i haven't yet been through the detail but that method seems like it works for breadth first traversal. what is the method for depth first traversal?
28. rational
for depth first, just replace queue with a stack
29. UnkleRhaukus
ok thanks !
30. UnkleRhaukus
i had no idea stack and queues could be used for anything like this, but it kinda makes perfect sense now
31. UnkleRhaukus
10. The Adjacency List A: {B} B: {C, D, E} C: {A, D} D: {A, C} E: {} Operation Queue Output __________________________________ enqueue(A) |A| dequeue | | A enqueue( adj(A) ) |B| A dequeue | | AB enqueue( adj(B) ) |E,D,C| AB dequeue 1 |E,D| ABC dequeue 1.1 |E| ABCD | | ABCDE. dequeue 1.2 |D| ABCE | | ABCED. dequeue 2 |C,E| ABD dequeue 2.1 |C| ABDE dequeue | | ABDEC. dequeue 2.2 |E| ABDC dequeue | | ABDCE. dequeue 3 |C,D| ABE dequeue 3.1 |C| ABED dequeue | | ABEDC. dequeue 3.2 |D| ABEC dequeue | | ABECD.
32. UnkleRhaukus
so the six possible breadth first traversals are: ABCDE (a) ABCED (b) ABDCE ABDEC (c) ABEDC ABECD so the answer to 10. is (d) all of the above [(a,b,c)]
33. UnkleRhaukus
but, i suppose i would have been able to guess from 1.2
34. rational
Nice!
35. UnkleRhaukus
i got these questions from a past paper of a subject i've almost completed [just final exam to go], seems like the course has changed quite a bit since 2012, ive learnt about graphs, digraphs, adjacency lists/matrices, trees, traversals, stacks, queues, etc. Now i can put all the ideas together!
36. UnkleRhaukus
37. UnkleRhaukus
Now, lets see what happens if add . . . |dw:1433594282292:dw|
38. UnkleRhaukus
Node  Adjacent Elements A: B B: C, D, E C: A, D D: A, C E: F
39. UnkleRhaukus
Is this right for breadth first? Operation Queue Output __________________________________ enqueue(A) |A| dequeue | | A enqueue( adj(A) ) |B| A dequeue | | AB enqueue( adj(B) ) |E,D,C| AB dequeue |E,D| ABC [1] dequeue |E| ABCD dequeue | | ABCDE enqueue( adj(E) ) |F| ABCDE dequeue | | ABCDEF. [other traversals are valid]
40. UnkleRhaukus
Is this right for breadth first? Operation Queue Output __________________________________ enqueue(A) |A| dequeue | | A enqueue( adj(A) ) |B| A dequeue | | AB enqueue( adj(B) ) |E,D,C| AB [1 of 6] dequeue |E,D| ABC dequeue |E| ABCD dequeue | | ABCDE enqueue( adj(E) ) |F| ABCDE dequeue | | ABCDEF.
41. UnkleRhaukus
breadth first ABCDEF, ABCEDF, ABDECF, ABDCEF, ABEDCF, ABECDF depth first ABCDEF, ABCEFD, ABDEFC, ABDCEF, ABEDCF, ABEFCD is this right? | 1,457 | 4,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-44 | longest | en | 0.665316 |
https://www.tutorialspoint.com/e-is-a-point-on-the-side-ad-produced-of-a-parallelogram-abcd-and-be-intersects-cd-at-f-show-that-abe-sim-cfb | 1,685,347,066,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644817.32/warc/CC-MAIN-20230529074001-20230529104001-00768.warc.gz | 1,166,097,670 | 11,662 | # $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $∆ABE \sim ∆CFB$.
Given:
$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$.
To do:
We have to show that $∆ABE \sim ∆CFB$.
Solution:
In the above figure, $ABCD$ is a parallelogram in which $E$ is a point on $AD$ produced and $BE$ intersects $CD$ at $F$.
In parallelogram $ABCD$,
$\angle A=\angle C$.......(i) (opposite angles)
In $\triangle ABE$ and $\triangle CFB$,
$\angle EAB=\angle BCF$ (Alternate angles)
$\angle ABE=\angle BFC$
Therefore, by AA criterion,
$\triangle ABE \sim \triangle CFB$
Hence proved.
Tutorialspoint
Simply Easy Learning | 253 | 750 | {"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} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.675582 |
http://mathhelpforum.com/algebra/162796-solving-equation.html | 1,481,168,747,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00242-ip-10-31-129-80.ec2.internal.warc.gz | 180,253,243 | 9,778 | 1. ## solving an equation
I am trying to solve (1/(2 sqrt(c))-1/3=0 can someone explain how to do this step by step? Thanks!
2. $\dfrac{1}{2\sqrt{c}}-\dfrac{1}{3}=0$
step 1: add $\frac{1}{3}$ to both sides.
step 2: multiply both sides by 3
step 3: multiply both sides by $\sqrt{c}$
step 4: square both sides to find c.
3. $\dfrac{1}{2\sqrt{c}}-\dfrac{1}{3}=0$
Step 1: Multiply both sides by 6 (the LCM of 2 and 3)
Step 2: Add 2 to both sides
Step 3: Take the reciprocal (flip the fraction)
Step 4: Multiply by 3
Step 5: Square both sides | 198 | 549 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2016-50 | longest | en | 0.884191 |
vipbettingtips.wordpress.com | 1,498,701,458,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00404.warc.gz | 858,764,657 | 30,356 | ### Analysis of line
So, what is it possible to see, giving a glance on the line of book-maker office and, mainly, what conclusions out of it tomake?
1. Difference of coefficients on a the same event for differentbook-makers.
Most lovers of rates choose one office and does notsquander time on comparison of coefficients. It isreasonable – as a rule, in the different offices of calculating sports betting odds verynear, because book-makers try to avoid a wide difference,not to give to the players to do the so-called “forks” (situation,when player, putting on the opposite events of the samematch for different book-makers, regardless of his endremains in a plus).
However, on distance even the small improvement ofcoefficients, at times, can change statistics with unprofitableon profitable, and taking into account that matches are,when an office is having the, special opinion and differencein coefficients can arrive at ten percents and more. It ispossible and quite to build the strategy exceptionally on thesearch of such rates.
Conclusion: at presence of accounts in different book-makercompanies, a player gets some advantage.
2. Difference between probabilities on an event oncoefficients and on your opinion.
It should be remembered that the task of line is being not ofexact probability on an event, and dividing of public opinioninto two approximately equal parts (on the sum of thepossible winning), in relation to one or another end. It isdone in order to avoid the accumulation of the put money onone side of event. In this case a book-maker can bearconsiderable losses, therefore, coefficients on the favouritesof match, majority of public puts on that, as a rule, are hardlyunderstated and rates on them on determination areunprofitable.
Thus, making a prognosis on a match, better at first to countcoefficients that you are distances on this match. For thispurpose, 100 is needed to divide into the probability of event,shown in percents, and if the coefficient of office appearedhigher, than got by you, id est sense to do a rate on this end.This strategy of rates is named value betting.
That is it better to present probability of winning of some”command And, it is needed to be set by a question: howmany matches from ten this command will be able to win ata “command”? Increasing the got number on 10 we will getprobability of event in percents.
Most difficult here – to expect the same probability of end.Undoubtedly, the workers of offices are very well able to do it,but they have to propose coefficients on hundreds ofmatches, while you are enough to choose from them only afew.
If, you discovered in coefficients on a match, some odditiesand does not understand them, then better do not do betting sports online odds football him.
3. Motion of line, i.e. change of coefficients on events in time.
Practically there are not events, that coefficients did notchange on 0.1-0.3, if motion of coefficient takes place on 0.5and more, maybe the result of this match is prejudgedbeforehand, i.e. simply it contractual match. Certainly,independently difficult enough to catch these motions, but inthe internet there are services that offer monitoring ofchange of lines of book-makers, for example, hot -odds.com.
If the starting line of book-maker represents his opinion onevents on that a line is given, then after the rates of players,a line represents opinion of players already, under influenceof that took place or another motion of line. Because as faras the accumulation of money on one of parties of event, abook-maker tries to bring over players to the rates onopposite. Thus, the main factor of motion of line i betting odds sport by players.
Far not always, but very often it is public opinion is faithful. Ifto put on the “proloaded” events on the changed coefficients,then a perceptible plus a player will not get. However, if hehave accounts in a few offices, then maybe, among themthere will be that in that coefficients remained primary. At thatrate there is sense to do a rate on this event, id est on that acoefficient fell down in other office.
So, it ensues from all said, that for a successful gameagainst a book-maker, it is desirable to have accounts in afew offices and analyse a line, but not to examine her only asthe final stage of forming of rate. In fact coefficient it is notonly some number by that a rate is multiplied in case ofwinning but also criterion for the choice of event betting news sports odds bookie horse . | 964 | 4,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-26 | longest | en | 0.948714 |
https://chemistry.stackexchange.com/questions/64180/how-do-lone-electrons-impact-the-charge-of-an-ion/65945 | 1,571,035,207,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00361.warc.gz | 504,049,381 | 31,810 | # How do lone electrons impact the charge of an ion?
This is a quite elementary question, but I still can't wrap my head around it.
I know how to calculate the formal charges of an ion or a molecule, but I don't understand what happens when oxygen (when sharing a single covalent bond with sulfur in a sulfate ion) has a lone electron that doesn't bond nor pair. I don't understand why this puts a -1 charge on the atom. What happens?
I've attached an image which explains what I'm trying to figure out:
See those two red dots? What do they mean? I understand that they symbolize the lack of a single electron needed to complete an octet. But why does it give the atom a negative charge?
The two red dots indicate the presence of two electrons, not the absence, if there were no electrons there it would be balanced. The colours of the electrons in the image indicate their origin: blue electrons originate with oxygen, black from sulfur and red from an outside source (eg. Potassium or H2). Ions have a charge because they have more or fewer electrons than the number of available protons. Sulfate has gained two extra electrons. The arrows in the second image indicate a 'dative covalent bond' where the electrons in the covalent bond all come from one atom rather than being shared, while the lines indicate a normal covalent bond (ie. an electron from each atom. The extra electrons added to the oxygen atoms to form a full orbital level in the normal covalent bond are responsible for the extra charge.
There's a nice visual depiction here: http://chem-net.blogspot.co.uk/2013/02/electron-dot-structure-of-sulfate-ion.html
The way this is normally taught (introduced) is as the concept of formal charge.
Think about the neutral oxygen atom, how many valence electrons does it have? Now look at the oxygen bonded to the Sulfur, how many electrons does it have if you consider the bonding ones to be shared?
-lone oxygen = 6
-bonded oxygen = 7
As it has one more electron than the neutral atom this gives it a negative charge. Each oxygen is -1 and if you do the same process on the sulfur you get +2 and the total (+2 + -4) is -2 which is the charge on the ion. | 494 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-43 | latest | en | 0.938425 |
https://www.jmp.com/support/help/13-2/Select_Rows_2.shtml | 1,718,894,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00831.warc.gz | 736,816,653 | 5,676 | Select All Rows selects (or highlights) all of the rows in a data table.
`dt << Select All Rows;`
If all rows are selected, you can deselect them all by using Invert Row Selection. This command reverses the selection state for each row, so that any selected rows are deselected, and any deselected rows are selected.
`dt << Invert Row Selection;`
Note: With the exception of Invert Row Selection, whose result depends on the current selection, any new selection message starts over with a new selection. If you already have certain rows selected and you then send a new message to select rows, all rows are first deselected.
`dt << Go To Row( 9 );`
To select specific rows in a data table based on their row number, use the Select Rows command. The argument to the command is a list of row numbers. For example, to select rows 1, 3, 5, and 7 of a data table, proceed as follows:
`dt << Select Rows( {1, 3, 5, 7} );`
`dt << Select Rows( Index( 7, 10 ) );`
`dt << Select Where( Any( Row() == Index( 7, 10 ) ) );`
To select rows according to data values, use Select Where, specifying a logical test inside the parentheses.
For example, using the Big Class.jmp sample data table, select the rows where the students’ age is greater than 13:
`dt = Open( "\$SAMPLE_DATA\Big Class.jmp" );`
`dt << Select Where( :age > 13 );`
`dt = Open( "\$SAMPLE_DATA\Big Class.jmp" );`
`col = Column( dt, 2 );`
`dt << Select Where( col[] < 14 );`
`dt = Open( "\$SAMPLE_DATA\Big Class.jmp" );`
`dt << Select Where( :age < 15 & :sex == "F" );`
To select a row without deselecting a previously selected row, combine << Select Where with << Select Where and the Current Selection("extend") argument. This is an alternative to using an OR statement.
`dt = Open( "\$SAMPLE_DATA\Big Class.jmp" );`
`dt << Select Where( :age == 14 );`
`dt << Select Where( :sex == "F", Current Selection( "extend" ) );`
`dt << Select Excluded;`
`dt << Select Hidden;`
`dt << Select Labeled;`
To select rows that are not excluded, hidden, or labeled, stack a select message and an invert selection message together in the same statement, or send the two messages sequentially:
`dt << Select Hidden << Invert Row Selection;`
`dt << Select Hidden;`
`dt << Invert Row Selection;`
To refer to a specific cell, assign a subscript to the cell’s row number. In the following example, the subscript [1] is used with the weight column. The formula then calculates the ratio between each height and the first value in the weight column.
`dt = Open( "\$SAMPLE_DATA\Big Class.jmp" );`
`New Column( "ratio", Formula( height / weight[1] ) );`
`dt << Select Randomly( number )`
`dt << Select Randomly( probability )`
`dt << Select Matching Cells;`
`// select matching cells in the current data table`
`dt << Select All Matching Cells;`
`// select matching cells in all open data tables.`
When you use column references to refer to column names in a Where statement, the column references need to be evaluated so they can be resolved to the proper data table. For example, in the following script, the parameters to the X(Xcol) and Y(Ycol) column references are linked to the data table in dt. However, the execution of the platform is associated with the Where subset data table. The script produces an error.
`dt = Open( "\$SAMPLE_DATA/Big Class.jmp" );`
`Ycol = Column( dt, "weight" ); // column reference to weight`
`Xcol = Column( dt, "height" ); // column reference to height`
`dt << Bivariate( Y( Ycol ), X( Xcol ), Fit Line(), Where( :sex == "F" ) );`
To evaluate the column names to the correct data table, use an Eval() expression or refer to the column names directly.
`dt = Open( "\$SAMPLE_DATA/Big Class.jmp" );`
`dt << Bivariate(`
` Y( Eval( Ycol ) ), // or Y( :weight )`
` X( Eval( Xcol ) ), // or X( :height )`
` Fit Line(),`
` Where( :sex == "F" )`
`);`
Help created on 9/19/2017 | 1,025 | 3,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.718369 |
https://www.philosophy-science-humanities-controversies.com/listview-details.php?id=285738&a=t&first_name=Hartry&author=Field&concept=Arbitrariness | 1,560,887,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998813.71/warc/CC-MAIN-20190618183446-20190618205446-00293.warc.gz | 860,606,073 | 5,391 | Hartry Field on Arbitrariness - Philosophy Dictionary of Arguments
# Philosophy Dictionary of Arguments
Arbitrariness: A. Arbitrariness is an everyday expression for a non-justified behavior or the refusal to give a reason for a behavior. For example, arbitrariness can arise in unfounded favor. B. In the narrower sense, arbitrariness is something subject to the will. Arbitrary action can be simulated by overriding regularities and thereby undermining expectability. See also conventions.
_____________
Annotation: The above characterizations of concepts are neither definitions nor exhausting presentations of problems related to them. Instead, they are intended to give a short introduction to the contributions below. – Lexicon of Arguments.
Author Item Summary Meta data
I 24
Identity/Identification/Field: in many areas, there is the problem of the continuous arbitrariness of identifications. - In mathematics, however, it is stronger than with physical objects.
I 181
Intensity relations between pairs or triples, etc. of points. - Advantage: that avoids attributing intensities to points and thus an arbitrary choice of a numerical scale for intensities.
- - -
III 32
Addition/Multiplication: not possible in Hilbert's geometry. - (Only with arbitrary zero and arbitrary 1) - Solution: intervals instead of points.
- - -
II 310
Non-Classical Degrees of Belief/Uncertainty/Field: E.g. that every "decision" about the power of the continuum is arbitrary is a good reason to not assume classical degrees of belief. - (Moderate non-classical logic: That some instances of the sentence cannot be asserted by the excluded third party).
- - -
III 31
Figure/Points/Field: no Platonist will identify real numbers with points on a physical line. - That would be too arbitrary ( "what line?"). - What should be zero - what is supposed to be 1?
III 32 f
Hilbert/Geometry/Axioms/Field: multiplication of intervals: not possible, because for that we would need an arbitrary "standard interval". - Solution: Comparing products of intervals. - generalization/Field: is then possible on products of spacetime intervals with scalar intervals. ((s) E.g. temperature difference, pressure difference). - Field: therefore, spacetime points must not be regarded as real numbers.
III 48
FieldVsTensor: arbitrary. - Solution/Field: simultaneity.
III 65
Def Equally Divided Region/Equally Split/Evenly Divided Evenly/Equidistance/Field: (all distances within the region equal: R: is a spacetime region all of whose points lie on a single line, and that for each point x of R the strict st-between (between in relation to spacetime) two points of R lies, there are points y and z of R, such that a) is exactly one point of R strictly st-between y and z, and that is x, and -b) xy P-Cong xz (Cong = congruent) - ((s) This avoids any arbitrary (length) units - E.g. "fewer" points in the corresponding interval or "the same number." - ((s) but not between temperature and space units (Which common measure?) - But definitely in mixed products - Then: "the mixed product... is smaller than the mixed product..." - Equidistance in each separate region: scalar/spatio-temporal.
III 79
Arbitrariness/Arbitrary/Scales Types/Scalar/Mass Density/Field: mass density is a very special scalar field which, due to its logarithmic structure, is "less arbitrary" than the scale for the gravitational potential. - ((s) >objectivity, >logarithm) - Logarithmic structures are less arbitrary. - Mass density: needs more fundamental concepts than other scalar fields. - Scalar field: E.g. height.
_____________
Explanation of symbols: Roman numerals indicate the source, arabic numerals indicate the page number. The corresponding books are indicated on the right hand side. ((s)…): Comment by the sender of the contribution.
The note [Author1]Vs[Author2] or [Author]Vs[term] is an addition from the Dictionary of Arguments. If a German edition is specified, the page numbers refer to this edition.
Field I
H. Field
Realism, Mathematics and Modality Oxford New York 1989
Field II
H. Field
Truth and the Absence of Fact Oxford New York 2001
Field III
H. Field
Science without numbers Princeton New Jersey 1980
Field IV
Hartry Field
"Realism and Relativism", The Journal of Philosophy, 76 (1982), pp. 553-67
In
Theories of Truth, Paul Horwich, Aldershot 1994
> Counter arguments against Field
Ed. Martin Schulz, access date 2019-06-18 | 1,034 | 4,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.88892 |
https://mathematica.stackexchange.com/questions/181825/how-can-i-animate-the-plot-of-solutions-to-this-system-with-different-initial-co/181828#181828 | 1,660,021,234,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00311.warc.gz | 379,479,280 | 65,789 | # How can I animate the plot of solutions to this system with different initial conditions?
I would like to plot the curves given by $\alpha(s) = (u(s), v(s))$, where $u$ and $v$ are like below.
Clear[u, v];
{u, v} = {u[t], v[t]} /.
NDSolve[{u'[t]^2 + v'[t]^2 == 1,
u'[t] v''[t] - u''[t] v'[t] == u[t] v'[t] - v[t] u'[t],
u'[0] == Sin[0.18], v'[0] == Cos[0.18]}, {u[t], v[t]}, {t, -7.5,
7.5}][[1]];
What I want to do is, for each different initial conditions $u'(0) = \cos(\theta)$, $v'(0) = \sin(\theta)$, $u'(t)v''(t) - u''(t)v'(t) = -c(u'(t)v(t) - u(t)v'(t))$ with $\theta$ varying in $[0, 2\pi]$ and $c$ varying in, say, $(0, 10]$, plot a new curve. I'm new to Mathematica so I still have a hard time doing these little things.
ClearAll[u, v, θ, c]
pndsv = ParametricNDSolveValue[{u'[t]^2 + v'[t]^2 == 1,
u'[t] v''[t] - u''[t] v'[t] == -c (-u[t] v'[t] + v[t] u'[t]),
u'[0] == Cos[θ], v'[0] == Sin[θ]}, {u, v}, {t, -7.5, 7.5}, {c, θ}]
An alternative visualization of the solution:
Manipulate[ParametricPlot[Evaluate@Through@pndsv[c, θ][t], {t, -7.5, 7.5},
PlotRange -> {{-3, 3}, {-3, 3}}], {c, 0, 10}, {θ, 0, 2 Pi}]
• Thanks! In the image you posted it appears I could type the parameters of $c$ and $\theta$, but when I run the code I only have slider controls. Why is that? Sep 13, 2018 at 23:08
• @MatheusAndrade, i used Autorun from the + menu to get the picture. You can also use the + button next to the slider to get the input field and other animation controls.
– kglr
Sep 13, 2018 at 23:22
Use ParametricNDSolveValue instead:
sol=ParametricNDSolveValue[
{
u'[t]^2 + v'[t]^2 == 1,
u'[t] v''[t] - u''[t] v'[t] == c (u[t] v'[t] - v[t] u'[t]),
u'[0] == Sin[θ],v'[0] == Cos[θ]
},
{u,v},
{t, -7.5, 7.5},
{c, θ}
];
Then, you can use Manipulate to explore the plot for different values of the parameters:
Manipulate[
Plot[{sol[c,θ][[1]][x],sol[c,θ][[2]][x]}, {x,-7.5,7.5}],
{{c, 1}, 0, 10},
{{θ, .18}, 0, 2Pi}
] | 795 | 1,930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-33 | latest | en | 0.702819 |
https://analyticsdefined.com/2017/08/ | 1,561,535,587,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000231.40/warc/CC-MAIN-20190626073946-20190626095946-00341.warc.gz | 341,255,043 | 92,239 | ## Using K-means to find the optimal nodal center for Radial Basis Function
Introduction In my previous article on “Introduction to the perceptron algorithm” we had seen how a single layer perceptron model can be used to classify an OR gate. But when the same model was used to classify a XOR gate it failed miserably. The problem was with linearity, i.e. if the classes are not linearly separable “Single layer perceptron model”…
## Understanding Softmax Regression with an example in R
Introduction to Softmax Regression We have commonly used many classification algorithms for binary classification. Now we will see a classification technique which is used to classify k classes. This technique is called softmax regression. Softmax regression is also called as multinomial logistic regression and it is a generalization of logistic regression. Softmax regression is used to model categorical dependent variables…
Regression
## Locally Weighted Regression (LWL)
Locally Weighted Regression (LWL) or LOWESS The basic assumption for a linear regression is that the data must be linearly distributed. But what if the data is not linearly distributed. Can we still apply the idea of regression? And the answer is ‘yes’… we can apply regression and it is called as locally weighted regression. We can apply LOESS or…
Conjoint Analysis
## Conjoint Analysis : From an HR analytics perspective
Introduction Conjoint analysis is a popular technique in the area of marketing. It gives valuable information about the voice of the consumer. We are often faced with situations in which we have to take decisions (purchase of goods, selection of a tourism package etc.) and we are required to evaluate multiple criteria while we arrive at the final decision. In almost…
Time Series Forecasting
## “Correlogram Analysis” Finding the order of Mean model in Time Series Analysis
Introduction to Corellogram Analysis To all those who know how to model a Time Series data using ARIMA ,you must have come across the term “Correlogram Analysis“,and to all those who don’t know what it is let me start with a basic definition. In the analysis of data, a correlogram is an image of correlation statistics. In time series analysis, a correlogram, also known as an autocorrelation…
## Plotting maps in R using ggmap
Introduction The objective is to explore ‘ggmap’ package in R and use this package to plot points on the map. Also, you can view other posts related to visualizations here. For this post, I’ll be using the map of India. Initially, I’ll try to explain some of the basic functions in ggmap and then I’ll explain by plotting different airports… | 563 | 2,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.886783 |
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Transposition as a CIPHER
This all about transcripting a text message. If the input string is: *s1 = 'My name is Sourav Mondal'*, then the output is: *s2...
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Given a string, containing several sentences, such as: 'I played piano. John played football. Anita went home. Are you safe?...
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Back to basics 17 - white space
Covering some basic topics I haven't seen elsewhere on Cody. Remove the trailing white spaces from the input variable
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Find the prime numbers in a vector
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A rectangle has a length of x centimeters and a width of w centimeters. Find the perimeter.
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UICBioE240 problem 1.15
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Given a vector x, return the indices to elements that will sum to exactly half of the sum of all elements. Example: Inpu...
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Spot the outlier
All points except for one lie on a line. Which one is the outlier? Example: You are given a list of x-y pairs in a column ...
meer dan 2 jaar ago | 1,469 | 5,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-10 | latest | en | 0.70615 |
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Find the base length of a parallelogram whose area is and height is cm.
• Acm
• Bcm
• Ccm
• Dcm
### Example
Find the base length of a parallelogram whose area is and height is cm.
### Solution
Use the formula for the base length of a parallelogram.
0
correct
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0
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https://www.themoneyillusion.com/the-best-economics-data-set-ever/ | 1,725,921,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00202.warc.gz | 1,010,958,606 | 35,965 | # The best economics data set ever (#1)
Update: People asked for a graph. Marcus Nunes has one for a very similar (but slightly different) data set.
I’m working on turning my blog into a book, and in order to do that I need to give readers an idea of how I ended up where I am today. One obvious need is to explain how I adopted a quantity theoretic approach to monetary analysis, rather than some alternative like the interest rate approach. For me it all goes back to the Great Inflation of the 1960s to the early 1980s.
As an aside, the quantity theory can be defined in several ways:
1. An X% rise in M will be associated with an X% rise in P
2. An X% rise in M will be associated with an X% rise in NGDP
3. An X% rise in M will cause both P and NGDP to be X% higher than otherwise, in the long run.
The third definition is probably the most accurate, and the first is the least accurate.
The following data set (from a Macroeconomics textbook by Robert Barro) is so rich in information, that we will spend many posts investigating all the implications. It shows average inflation, money growth and real GDP growth rates over 30 or 40-year periods around 1950-90, for 79 countries:
Right off the bat one notices the strong correlation between the growth rate of M (the monetary base) and P (the price level.) David Hume didn’t have this data set in 1752, but just using his brilliant mind he was able to figure out that if you double the money supply, the only long run effect is for prices to double. Money is just a measuring stick. For about 40 years Argentina and Brazil were doubling the money supply, on average, once every 14 months. And prices were doubling just as Hume predicted. All good, the Quantity Theory of Money (QTM) is triumphant.
Except it’s all downhill from here. I’ve just provided the best possible data set for convincing you of the QTM. Suppose I had only given you the bottom half of the data set? Now the correlation is much harder to see. Or suppose we’d looked at shorter time periods. Again, not so good. Or suppose we’d looked at countries at the zero bound? Now the QTM would have major problems.
The key to understanding the QTM is to hold two thoughts in your mind at the same time. In one sense the theory is logical, indeed blindingly obvious. It’s incredibly powerful, incredibly true. But in all sorts of situations it seems to fail. That’s what we need to figure out.
Before moving on, let’s remind ourselves why it’s the bedrock of monetary theory, and why all other theories fall short. In this data set we are doing the economic equivalent of when scientists expose objects to great heat, pressure, or speed, to get at the essential qualities. We’re looking at what happens with very fast money growth
No other model can explain the correlations we see. Yes, the growth in the money supply might have “root causes” elsewhere, such as budget deficits. But you can’t figure out that Brazil and Argentina would have 75% inflation for 40 years, whereas Iceland would have 19% and the US would have 4% by looking at budget deficits, you need the money supply growth rates to even get in the right ballpark. Note that some countries (the US in the 1970s) printed lots of money w/o big budget deficits.
Nor do interest rate models work. Ironically the only interest rate model that would even come close is NeoFisherism, as the nominal interest rates in these countries would also be highly correlated with inflation. But that model doesn’t tell you how you get those high nominal interest rates. Again, you need money supply data.
Nor will an exchange rate model work. Yes, the (depreciating) exchange rate for Brazil and Argentina was closely paralleling their inflation rate. They saw the local currency price of US dollars rise at around 70% per year over those 40 years. But that doesn’t explain how you cause the currency to depreciate so rapidly over 40 years. Again, you need money supply data. Both the Fisher effect and PPP are just appendages of the QTM.
Let’s finish today’s post with the first of several regressions that I’m going to give you–all provided by Patrick Horan of the Mercatus Center:
This is the Mona Lisa of macro regressions. The t-statistic on money growth is 45.2. Yup, I’d say there’s some truth to the QTM. The P-value? One over . . . umm . . . how many atoms are there in the universe? And the coefficient is pretty close to one, within two standard errors. When you raise the money supply at 75%/year for 40 years, you’ll get roughly 75% inflation.
Later we’ll see there’s a reason the coefficient is slightly greater than one. Can you guess? (It’s a very hard question.) But let’s finish up by noticing the coefficient on real GDP growth (delta Y). You’ve all been taught that economic growth is inflationary. The people at the Fed tell us that inflation will rise as we approach full employment. Maybe it will, but not because growth is inflationary. As you can see from the regression, economic growth is deflationary, indeed almost exactly as deflationary as money growth is inflationary. So are the Keynesians wrong?
Yes they are! And they are wrong in an interesting way. Let’s suppose their predictions turn out right, and inflation does rise as we approached full employment. Will I admit that I’m wrong? Of course not!! I’m an arrogant economic blogger. Instead I’ll claim that this bizarre outcome is proof of the Fed’s incompetence. They so botched monetary policy that they made inflation procyclical. Indeed they do this so often that some of my commenters think this is natural. Poor Mr. Ray Lopez found a dictionary somewhere that says inflation naturally falls during recessions and rises during booms.
And it’s all a myth. Don’t worry, we’ll explain the mystery of deflationary growth in the next post. And we’ll explain why the coefficient on money growth was a little bit bigger than one in the post after that. All our money/macro questions are answered in this data set, if we know where to look. Put on your David Hume thinking hat, you have lots more info to work with than he had. Indeed Milton Friedman became the second most famous economist of the 20th century mostly by figuring out how this data set allowed us to go “one derivative beyond Hume.”
PS. Here are the two “money quotes” (pun intended) from Friedman:
Double-digit inflation and double-digit interest rates, not the elegance of theoretical reasoning or the overwhelming persuasiveness of serried masses of statistics massaged through modern computers, explain the rediscovery of money.” (1975, p. 176.)
As I see it, we have advanced beyond Hume in two respects only; first, we now have a more secure grasp of the quantitative magnitudes involved; second, we have gone one derivative beyond Hume.” Friedman (1975, p. 177.)
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62 Responses to “The best economics data set ever (#1)”
1. ben
10. August 2015 at 18:11
“Maybe it will, but not because growth is inflationary.”
Our definition of “growth” is bogus. GDP is about money supply not wealth creation, so it’s hardly any wonder that people claim growth is inflationary because they *should* skip the incorrect term and say “money creation is inflationary”.
Vancouver, we are told, is booming. They create money via land speculation and then every now and then Ottawa has to shout “everyone devalue!” then off we go again. Meanwhile people decry Quebec for not “making money”.
Will the great age of un-reason ever end?
2. Jason Smith
10. August 2015 at 19:12
An ensemble of random markets gives you this same result:
http://informationtransfereconomics.blogspot.com/2015/06/the-quantity-theory-of-money-as.html
I show that the ensemble average (in angle brackets):
〈i〉= 〈m〉
where i is inflation and m is monetary base growth.
3. Peter
10. August 2015 at 19:57
I assume that lower demand for base money with higher inflation would make the coefficient larger than one.
4. Bob Murphy
10. August 2015 at 19:59
Scott, I’ve found the David Hume of our age!
5. E. Harding
10. August 2015 at 20:33
In the present article I am not taking a stand on whether price inflation will be high or low in 2008. However I will say this: If you are revising your inflation forecasts downward because of your expectation of sluggish economic growth, you might want to rethink that logic.
-Ah, but the person who didn’t rethink that logic would have been right, wouldn’t he have (even if for the wrong reasons)?
6. Philip George
10. August 2015 at 20:34
Just one question. You are trying to relate money to prices and NGDP.
Is money used to buy financial assets?
If not, with what are financial assets bought?
7. Aaron W
10. August 2015 at 20:56
It doesn’t have to be right away, but would you mind providing others with access to this database? Also, how was the data collected?
Not saying that any of your conclusions are wrong in the least, just interested in playing around with it. (A few graphs would also be much more helpful than trying to read through a giant table.)
8. Mark
10. August 2015 at 21:07
Which money aggregate does M represent here?
9. Ray Lopez
10. August 2015 at 21:50
Brilliant post! Our world-saving blogger found out that:
1) the two NGDP leaders, Brazil and Argentina, are at the top of his list of countries, while at the bottom is West Germany. Brazil has just been (as of yesterday) downgraded to fall to “junk bond” status, while Argentina is a textbook example of poor performance from initial promising conditions (the country is scarcely populated and rich in natural resources, like Norway is in a way). West Germany was torn asunder by war but outperforms. West Germany has sound money. Coincidence? Probably, though you could argue the society that values sound money is also sound in other areas, like economic productivity.
2) Some metaphysical nonsense is mentioned where Keynesians are right, but Sumner refuses to agree. Since IMO both monetarists and Keynesianists are both wrong (flip sides of the same coin) I won’t go there.
3) My name is dropped, always worth style points, along with Milt Friedman’s. Modesty prevents me from saying I’m a greater economist than M. Friedman, but since Friedman contributed nothing but rhetoric and misinformation (he himself says his greatest theoretical contribution was an obscure correction to the Cobbs production function), you could say that’s true.
I eagerly await Scott’s book…on the usual pirated torrent sites. It would actually be an honor for Sumner if some pirate thought his book is worthy enough to be copyright infringed. Somehow, I doubt I’ll see it however.
10. Benjamin Cole
10. August 2015 at 21:51
“[David Hume] was able to figure out that if you double the money supply, the only long-run effect is for prices to double.”–Scott Sumner.
Oaky, am I missing something?
What if transactions increase? Could not transactions double in the long run too?
Or is this Hume observation a “ceteris paribus” type of posit?
11. Lorenzo from Oz
10. August 2015 at 23:55
David Glasner might quibble on the last quote, as he holds that Friedman was wrong to follow Hume on the price specie flow mechanism (PSFM).
My view is the PSFM works in a network trade world (no global markets) but does not work in a mass trade world (with global prices). Hume looking backwards was correct; Adam Smith, with a better sense of the contemporary Atlantic economy, let his friend’s theory slide for good reason.
But that is perhaps a footnote to your major point, which I enjoyed.
12. Michael Byrnes
11. August 2015 at 02:29
Selgin has a good post on Friedman (with a link to a good appreciation of Friedman from David Ladler):
http://www.alt-m.org/2015/08/10/milton-friedman-monetary-freedom/
13. Juhani Huopainen
11. August 2015 at 02:48
Any chance of getting that data somewhere – or should I do OCR?
14. Honeyoak
11. August 2015 at 03:18
As someone who has experience with macro data, I can’t help but feel that some “adjustments” were made. Scott, I suggest you look at it the IMF datasets as they are quite comprehensive (https://www.imf.org/data#global)
15. ssumner
11. August 2015 at 03:42
Peter, That’s right, although the full explanation is a bit more involved—the change in the inflation rate is what matters. I’ll explain later
Bob, Good post. I suppose they’d say that inflation falls during the period of high unemployment, which is a lagged effect of recession.
Philip, The money used here is the base. Suppose you buy stocks with a personal check. Then base money is transferred from your bank to your stockbroker’s bank account.
Aaron, The data was collected by hand from an old textbook by Barro.
Mark, The base.
Ben, Hume said transactions rise in the short run.
Lorenzo, I’ll address fixed exchange rate regimes in a later post.
Juhani, What you see is all I have, it takes 10 minutes to enter it by hand.
HoneyOak, Can you be more specific? Every similar data set I’ve ever seen for the period shows essentially the same pattern. What do you find odd?
16. Ben
11. August 2015 at 03:53
Scott, don’t you think the RGDP / inflation stats are a bit endogenous? Also, can you post a scatter plot? Maybe it would be more fair to do natural log of your explanatory variables and see how the regression looks.
-Ben
17. Nick Rowe
11. August 2015 at 04:59
Suppose the truth is: y = bx + cz, but you estimate a simple regression y = Bx, so you have omitted variable bias.
Moving down your list of countries:
If the variance in x is very big, relative to the variance in z, this omitted variable bias won’t matter much. You get only a small bias in B, and the R-squared will be high.
If x and z are uncorrelated, you get no bias in B, but a lower R-squared.
If x and z are negatively correlated, you get a downward bias in B. In the limit, you get Milton Friedman’s thermostat result, where B = 0.
That’s my explanation for why the QTM performs worse and worse as you go down the list of countries.
18. benjamin cole
11. August 2015 at 05:07
Scott: Okay, that is what Hume said in his time and place in a relatively ststic economy…but today we have a globalized economy and capital gluts—it would seem an increase in demand or money supply would be met by an increase in trade and added productive capacity, and we see permanent increases in transactions…
This seems obvious to me…I must be missing something
19. ssumner
11. August 2015 at 05:11
Ben, As far as natural logs, these are already percentage changes.
Inflation is certainly endogenous, but I’ll later argue that long run growth is not related to money growth.
I’m not sure why people keep asking for a scatter plot, isn’t it obvious what it would look like? The points will all lie close to a 45 degree line.
Nick, That’s right, and I was going to point to Bretton Woods and inflation targeting as two examples of regimes where the coefficient on money would be biased for that reason.
20. ssumner
11. August 2015 at 05:12
Ben, Yes, Hume said that too, but only in the short run. On the long run money is neutral.
21. Njnnja
11. August 2015 at 05:35
Ben is correct. if you are going to do a regression on ratios you should take logs first. If you run off non log values then your large values are too large.
Also, instead of deltaX/X it should just be Xt / X0, the ratio of the ending value to the starting value (which is close because Xt/X0 = deltaX/X + 1), unless there is reason to believe that this relationship did not hold prior to X0.
If you can put the data in a copyable format and repost we can definitely tighten up the analysis.
22. ssumner
11. August 2015 at 05:55
Njnnja, Are you saying that the percentage changes should have been first differences of logs? If so, I agree. However it’s not going to make much difference, as it will effect the money growth and inflation in a similar fashion. And to be honest, I’m not sure whether Barro did that already, I am at home and don’t have the textbook with me.
If someone can get the textbook, they might be able to figure out where he got the data, and whether he used logs, but the easiest solution might just be to use data from the IMF or world bank, or something like that.
In any case, I’m nearly 100% certain that all the theoretical implications I draw from this data set would hold up with a more rigorous study. I’m not saying anything that’s even mildly controversial for people who have studied money demand. This is just an intro for outsiders.
23. Alex
11. August 2015 at 06:00
Scott,
Some related literature:
McCandless & Weber, Some Monetary Facts. https://www.minneapolisfed.org/research/quarterly-review/some-monetary-facts
Emiliano Basco, Laura D’Amato, Lorena Garegnani. Understanding the money – prices relationship under low and high inflation regimes: Argentina 1970-2005. http://www.bcra.gob.ar/Pdfs/Investigaciones/WP%202006%2013_i.pdf
And finally a plot showing Argentina´s fiscal deficit (with and without interest payments) and money printing.
https://pbs.twimg.com/media/CMIgUYdW8AAqF1j.jpg:large
24. John Hall
11. August 2015 at 06:21
With respect to your point about the coefficient on the real GDP term, I would think it would be relatively simple in terms of the AS/AD model. For instance, if you have a right-ward shift in long-term aggregate supply, then in equilbrium you would think that there is also a right-ward shift in short-term aggregate supply. In this case, the increase in Real GDP would be consistent with a decrease in inflation.
25. collin
11. August 2015 at 06:30
One interesting point on the chart is the top 7 nations were in South America which was the main area of US/Soviet blundering Foreign Policy until 1974 Oil Embargo made the Middle East our/Soviet blundering focal point. (Growing up in the 1970s/1980s I remember the nightly news had reports of South/Latin American coups once aweek and was not treated like a major event.) So how much of the correlation of money supply and inflation was a function of an era of consistent Civil War? How does the data change after 1990 or even 1980?
26. Njnnja
11. August 2015 at 06:52
Yes combining the 2 recommendations would get ln(Xt/X0) = ln(Xt) – ln(X0) = first difference of log values is the better metric.
I’m sure it won’t make much difference to the overall conclusion, but it should make a difference to the coefficients and especially the residuals. The residual error of 2.74 means that your 95% confidence interval corresponds to something like a window of 4% inflation (annual compounded) for a relatively low inflation country like the US – so anywhere from 2% to 6% compounded over 35 years. And second if something is worth doing it’s worth doing right so that other bloggers don’t have something to say what a dummy prof. Sumner is. 😛
27. ssumner
11. August 2015 at 06:58
Njnnja, Fair point. As you know, when dealing in ordinary percentages you can’t add inflation and real growth to get NGDP growth. If you look at the final column, which I had Patrick calculate, we assumed that Barro was suing ordinary percentages, so we adjusted the math accordingly. But perhaps Barro was already reporting first differences of logs, that would be more useful in a table structured to look vaguely like the Equation of Exchange.
28. ssumner
11. August 2015 at 07:03
Alex, Thanks for that info. What does the green line (MB) for Argentina represent? What units?
John, That’s right.
Collin, Inflation gets dramatically lower after 1990, but even today places like Argentina and Venezuela have very high inflation, and Brazil is a bit higher than western countries.
29. marcus nunes
11. August 2015 at 07:06
Scott, same data set used here to illustrate your “Quizz”;
https://thefaintofheart.wordpress.com/2013/03/26/scotts-quiz-illustrated/
30. honeyoak
11. August 2015 at 08:06
Scott, I used to work as an RA that involved researching macro data-sets used in publications. Without a fault (out of 16 ish) there was always some unaccounted filter or adjustment that was applied to the data by the academic researcher. I suspect that since there is little quality control in the standard publication process, the academics have a tendency to do what they like without sufficient documentation. Institutions such as the IMF, BEA, World Bank, and the OECD have strict data preparation/cleaning processes that ensure that at least one person (who is not an overworked phd student) has looked at the data. It could be that these adjustments do not impact your results however they may impact the specific examples that you choose to highlight the theory. Considering how easy it is these days (I would not feel comfortable excluding anything past 1990) I recommend that you collect a modern data set.
31. Ilya
11. August 2015 at 08:28
Scott, how do you know which direction the causation runs? A regression is not enough for that.
32. Andrew_FL
11. August 2015 at 09:08
That growth is deflationary is simply a mathematical consequence of the fact that we measure growth by taking NGDP and deflating it using a price index P. If P correlates with money growth, and money growth correlates with NGDP growth, “real” GDP growth *has* to be negatively correlated with inflation. It literally could not be otherwise. Saying saying RGDP%c negatively correlates with P%c is the same thing as saying (NGDP/P)%c negatively correlates with P. Uh, duh, you put P in the denominator right there.
However, if we defined growth as something other than increasing “real” GDP we might come up with reasons that aren’t tautological why growth should be deflationary. Those reasons will be correct, for the definition of growth they are associated with. But technically, they’re unnecessary. If we believe RGDP is a “real thing” then growth is deflationary by definition.
Personally I think RGDP is an illusory fiction, but I actually agree with the idea that “growth is deflationary”-in fact it’s something I’ve been saying for quite some time now. So I don’t want you to think I’m giving you a hard time, but you’re hardly discovering fire here.
33. ssumner
11. August 2015 at 09:13
Honeyoak, I assure you that you get almost exactly the same results with any data set from this period. I’ve seen others and the results are always like this.
I’m not interested in post-1990 data because of the widespread adoption of inflation targeting. I’ll explain that problem in a few days, in the 4th post of the series.
People need to focus on the big picture, none of these regression results are driven by quirks in the data, they are all 100% rock solid, and uncontroversial.
Ilya, For that you need theory. We have theory explaining M–>P, but not in the other direction. Plus you have some natural experiments, like when monetary easing in 1933 caused inflation.
34. Alex
11. August 2015 at 09:28
Scott,
The green line is the change in the monetary base (currency in circulation + bank reserves at the central bank) from t-1 to t (end of period values) divided by NGDP in t.
Alex.
35. James Alexander
11. August 2015 at 12:30
Barro has a page on his profile linking to all the datasets he uses.
http://rbarro.com/data-sets/
36. Beefcake the Mighty
11. August 2015 at 13:54
@Nick Rowe
The direction of the bias depends not only on the sign of the correlation between the regressors, but also on the sign of the coefficient. E.g. for c < 0 and x and z negatively correlated, the bias in B is positive.
Also, you can have low R-squared in a correctly specified model where the disturbance has much higher variance than the regressors.
37. Major.Freedom
11. August 2015 at 16:35
Actually this is a well written post.
Only quibble is the comment that money is a “measuring stick.”
Other than that, great from top to bottom.
38. Beefcake the Mighty
11. August 2015 at 16:50
@Nick Rowe
The direction of the bias depends not only on the sign of the correlation between the regressors, but also on the sign of the coefficient. E.g. for c < 0 and x and z negatively correlated, the bias in B is positive.
Also, you can have low R-squared in a correctly specified model where the disturbance has much higher variance than the regressors.
39. Ray Lopez
11. August 2015 at 17:14
Sumner’s “This is the Mona Lisa of macro regressions. The t-statistic on money growth is 45.2. Yup, I’d say there’s some truth to the QTM.” is misleading. The question is not whether there’s ‘some truth’ in the QTM but whether it predicts anything. It does not, simply cause velocity V on the LHS of the equation varies.
BTW, short term money neutrality (which Sumner disputes, though he concedes–IMO rather strangely–that long term money is neutral, which he should not concede but at least he’s a bit honest) also would give a “45 degree plot”. It’s in the definition of money neutrality.
40. Major.Freedom
11. August 2015 at 17:24
Ray:
“The question is not whether there’s ‘some truth’ in the QTM but whether it predicts anything. It does not, simply cause velocity V on the LHS of the equation varies.”
No, the question is whether there is some truth to the QTM. There is. There is no reason to expect V to fluctuate in the long run because of an increase in the quantity of money (subject to certain considerations of course).
If everyone’s money balances doubled, why would that lead to any permanent change in V? It shouldn’t, which is why despite all the myriad of factors that can affect V, despite the millions of possible choices people can make, we are still able to see some stable correlation between M and P in the long run, most of the time.
This is not to say that a rise in M causes a proportional rise in P, only that even with all the factors that might affect P, sound economic theory concludes that the cause of the rise in P over the long run, is the rise on M.
Surely you are not so misinformed so as to believe that the reason prices have risen since 1913, was caused by something other than the rise in M? Please wake up, you’re starting to disappoint.
41. Major.Freedom
11. August 2015 at 17:37
Ray,
Put it this way:
You agree that P = D/S.
Price equals demand (in dollars spent) divided by supply (quantity of goods).
This is a mathematical necessity.
There is no possible concievable way that prices can change OTHER than through a change in demand, or a change in supply, or both.
Agreed? Good.
Now we know that prices have risen quite substantially since 1913. Even the government’s fudged and constantly changing statistical methods designed to hide price inflation, has reported a significant rise in prices since 1913.
Agreed? Good.
So there are two and only two (actually three) reasons for why prices could have risen.
Either nominal demand has gone up, or supply has gone down, or a combination of the two.
Agreed? Good.
Now ask yourself if the cause for prices having risen since 1913 is only a fall in supply. That we can rule out without even thinking more than 2 seconds, since in order for prices to have risen that much by way of a fall in supply, would be if the country has undergone a collapse on par with the collapse of the Roman Empire. So you should agree that we can mathematically rule out falling supply as a cause.
This leaves rising demand in dollars as the ONLY cause. It is literally impossible for there to be any other cause, since there is only supply and demand, and we just ruled out supply.
The argument that the cause for the rise in prices since 1913 must be a rise in the quantity of money (and the concomitant volume of spending), IS the QTM.
Yes velocity can fluctuate. But there is no way in hell that the rise in prices since 1913 could have occurred only by a rise in velocity without any additional inflation in the supply of money. It is unthinkable. The rise MUST have been caused by an increase in M, period. That’s the QTM.
If you deny this, then you have no business discussing economics, no business anywhere but with your nose inside textbooks and on historical price trend reports, until it sinks in.
42. Ray Lopez
11. August 2015 at 17:51
@MF – I think you’re making a big mistake in dealing with aggregate figures, which is not your area of expertise. Your equation: P = D/S is not sourced, and I’ve never seen it in that form. At best you can say Demand and Supply are correlated by a price. And actually the Equation of Exchange does a good job in capturing this dynamic, but price and quantity are not divided into each other but multiplied by each other on the RHS.
Anyway, none of this math goes to my central thesis: we can agree that the inflation of money has produced changes in price, but so what? Money creation is neutral, and any inflation usually (unless you are foolish, like a senior citizen relying on a fixed income from government) is harmless: people will adjust. Short term, as Ben S. Bernanke says in his econometrics paper of 2003, it has between 3.2% to 13.2% effect out of 100% on the economy’s change, year to year. Most people would agree with Ben S. Bernanke that while this may be statistically significant, it is de facto trivial, don’t you agree?
43. Ray Lopez
11. August 2015 at 17:55
@myself – a correction – I see your attempt is D = P*S, which roughly correlated to the Equation of Exchange. Still, it does not address why this accounting identity has any relevance looking into the future. The issue is whether monetarism has any predictive effect. Looking backwards, monetarism via the Equation of Exchange simply says: “what is, is”. Hardly of any use. David Hume would agree. Sumner btw has conceded that long-term monetarism is useless, and him and the neo-Keynesians cling to the fantasy that 3.2% effect is a big deal. Trying to justify their existence.
44. Major.Freedom
11. August 2015 at 20:18
Ray,
“@MF – I think you’re making a big mistake in dealing with aggregate figures, which is not your area of expertise. Your equation: P = D/S is not sourced, and I’ve never seen it in that form. At best you can say Demand and Supply are correlated by a price. And actually the Equation of Exchange does a good job in capturing this dynamic, but price and quantity are not divided into each other but multiplied by each other on the RHS.”
No, that is incorrect.
If the demand for a supply of goods is \$100, and there are 100 units of that good, then the average price per unit must be \$1.
This is equivalent to P=D/S.
Price is not merely correlated with demand and supply. Price IS the ratio between money spent on a good or goods, to the supply of that good.
When you exchange \$5 in spending for 2 hamburgers, you are paying an average price of \$2.50 per hamburger.
The equation P=D/S works at both individual exchanges and all exchanges together. In the aggregate, which is actually an area of my expertise thank you very much, which is why I know to use it so sparingly, P=D/S can be understood as the price level equalling total dollars spent divided by all goods sold. The common denominator is an existant, a conception of goods being homogeneous.
Yes goods are not actually homogeneous, but it is absurd to say that we have no idea if prices in general are rising over time or falling over time. We know that prices in general are rising over time. As soon as you say that, you are presuming the equation P=D/S is meaningful, but not literal.
“Anyway, none of this math goes to my central thesis: we can agree that the inflation of money has produced changes in price, but so what?”
Your thesis is not the point here. The point here is an increase in the supply of money since 1913 is not only “a” factor that explains the rise in prices, but is the SOLE factor.
Do you deny that? Do you actually believe that the rise in prices since 1913 was caused by something other than an increase in the quantity of money?
If you do not deny this, then you are conceding that the QTM is correct.
“Money creation is neutral, and any inflation usually (unless you are foolish, like a senior citizen relying on a fixed income from government) is harmless”
No, I already explained to you multiple times how and why money is not neutral on any timescale.
You already conceded that money is not neutral when you said you believed the effect of central banking is in the range of 3.2% to 13.2%. That means money is de facto, effectively, non-neutral.
A neutral money is a contradiction in terms.
“Short term, as Ben S. Bernanke says in his econometrics paper of 2003, it has between 3.2% to 13.2% effect out of 100% on the economy’s change, year to year.”
Which means money is not neutral. Money having a non-zero effect, of anything higher than 0%, implies money neutrality. 3.2% is not zero. It is non zero.
“Most people would agree with Ben S. Bernanke that while this may be statistically significant, it is de facto trivial, don’t you agree?”
No, it is de facto non trivial. Higher than or equal to 3.2% is de facto non-neutral.
I already explained this, why on Earth are you asking me if I agree? Are you drunk?
“I see your attempt is D = P*S, which roughly correlated to the Equation of Exchange.”
No, I did not “attempt” to write D=P*S. If I attempted to write that, then I would have written that.
I meant to write P=D/S (which is mathematically equivalent), so as to show you that if a price or a set of prices changes, then it MUST be by way of either demand changing, or supply changing, or both.
There is no other way for prices to change.
Do you deny this? Answer the question.
“Still, it does not address why this accounting identity has any relevance looking into the future.”
You are still missing the point.
Economic theory is not meant to, nor is it capable to, predict the future.
The QTM, the law of marginal utility, the law of demand, these are economic principles that are always true, but that does not mean they have to allow us to predict our own future learning paths and our future actions.
Mathematical laws, laws of logic, economic laws, these are rationally grounded concepts that allow us to know what it even means to predict future economic phenomena. They are not obligated to predict my own future learning. If they could allow me to predict my own future learning, then I would have learned something before I actually learned it, a clear contradiction.
This contradiction is reality’s way of telling us that there are TWO classes of knowledge, not just one as you are implying by demanding that all knowledge be positivist/empirical.
“The issue is whether monetarism has any predictive effect.”
No, the issue the explanation for prices rising over the long run, given that production and supply is increasing in the long run.
The only explanation for that is an increase in the money supply.
But this cannot predict the future, because the QTM does not predict what the future supply of goods will be, and it does not predict what the future demand for money will be.
“Looking backwards, monetarism via the Equation of Exchange simply says: “what is, is”. Hardly of any use.”
Then mathematics and formal logic are of no use to you.
But that does not mean they are of no use to everyone.
Some of us find logic and deductive reasoning highly useful, not as predictive tools but as tools that help us understand what IS going on in the world.
“David Hume would agree.”
Ugh.
” Sumner btw has conceded that long-term monetarism is useless, and him and the neo-Keynesians cling to the fantasy that 3.2% effect is a big deal. Trying to justify their existence.”
Don’t care. You already conceded money is not neutral.
45. Major.Freedom
11. August 2015 at 20:20
Typo:
Meant to write:
“Money having a non-zero effect, of anything higher than 0%, implies money non neutrality.”
46. Dotsn
11. August 2015 at 20:48
US printing seems to activate spare capacity in far-flung places on the Pacific Rim. isn’t long-run growth probably improved when we accelerate 3rd world development and improve Indonesian IQ via nutrition, for example? these convergences r to be expected in the long run, maybe, but it seems like improving demand and therefore improving the returns to the easiest aspects of sound capitalism in poorer countries should make emergence from the frustratingly stable deadbeat political equilibria a bit easier, likelier
47. fran
12. August 2015 at 00:52
As a potential international reader of your future book, I’d like to let you know that I hope very much it is not going to be a U.S.-centric one on the explanation of the Great Recession. So, should you explain the Great Recession with the restrictive monetary policy by the Fed (which I would support), you should in my opinion keep in mind that that was a global phenomenon. So far, I have not found a satisfactory and compelling answer to the question why so many countries experienced a deep recession at the same time. Was it because they also had restrictive monetary policy or because of trade linkages? In my opinion, your book would gain a lot if it touches on this question. I am aware though, that this is hard to do and you probably know U.S. data best.
48. Ray Lopez
12. August 2015 at 07:48
@fran- indeed, good observation. Charles Kindleberger asked whether the smaller trade linkages of the late 1920s could have caused a Great Depression world wide, and concluded, as do I, that it was not the trade linkages per se that caused world depression, but the ‘contagion’ factor akin to shouting ‘fire’ in a crowded theatre. Sounds plausible and is also a reason that economics will never be a repeatable, exact science (much to the chagrin of our host).
49. Ray Lopez
12. August 2015 at 07:55
@MF – you think 3.2% out of 100% is important? Well then, if we ever discover \$1 M dollars, finder’s keepers, then you can have \$32k and I will get the remainder, \$968k. Fair’s fair.
As for your conceding that economics is only backwards looking, with no predictive power, I would agree, and while you may find economics interesting given this fact, I bet 96.8% of the population will ignore economists if that fact was accepted. And it’s one reason professional economists like Sumner pretend they can divine the economy short term; it gives them relevance to the 96.8% of the population that expects economists to predict the future.
50. ssumner
12. August 2015 at 08:03
Alex, Thanks. Note that this data point doesn’t, by itself, get us the inflation rate.
Thanks James, I don’t think that includes this data set, does it?
Dotsn, Always keep in mind that money is neutral in the long run.
fran, Very good point, I’ll try to remember that.
51. Philippe
12. August 2015 at 12:29
Scott,
“No other model can explain the correlations we see”
You could explain the correlation by choosing to read the MV=PQ equation from right to left, rather than from left to right, with MV being seen as the dependent variable and PQ being seen as the independent variable. This interpretation would also help to explain why the correlation you highlight sometimes breaks down.
52. Beefcake the Mighty
12. August 2015 at 12:43
Apparently, many people do not understand the difference between an equation and an identity.
53. W. Peden
12. August 2015 at 12:52
Beefcake the Mighty,
Which is weird, given that this is textbook stuff. Knowing the difference between the quantity theory and the equation of exchange really is the first thing you should learn to understand the QTM.
54. Philippe
12. August 2015 at 13:27
Scott defines the quantity theory as: “An X% rise in M will cause both P and NGDP to be X% higher than otherwise, in the long run.”
This amounts to assuming that the equation of exchange expresses a causality which goes from left to right, rather than from right to left.
He then says that “No other model can explain the correlations we see”. But you can actually explain the correlations by simply dropping the assumption that causation runs from left to right in the equation of exchange. Dropping this assumption can also help to explain why the correlations break down.
55. fran
12. August 2015 at 13:28
@Ray Lopez: Personally, I don’t find the pure contagion argument convincing. To me, this is at odds with efficient markets and rational expectations and it doesn’t really offer an in depth explanation. I can understand that in a crowded theatre a panic can be offset if someone screams fire. But it matters if the theatre is crowded or more than half empty, in which case everyone ‘knows’ the exits can be reached safely.
With regard to the Great Recession: I think that in many countries the stance of monetary policy could also have been restrictive at the same time prior to the crisis. The argument could go as follows: The steep rise of the oil prices at the time prior to the crises turned central banks around the world blind on the eye that should focus on business conditions. Instead, they only focused on inflation and by doing so, they ultimately made monetary policy conditions more restrictive than necessary. This is just a guess though, I have no data or other support for that claim. But it could be evaluated, f.e. by looking at how often monetary policy communications contain the word ‘oil’ prior to the crisis in different countries. (Maybe something for Scott and his book?)
56. Major.Freedom
12. August 2015 at 17:01
Ray:
“MF – you think 3.2% out of 100% is important? Well then, if we ever discover \$1 M dollars, finder’s keepers, then you can have \$32k and I will get the remainder, \$968k. Fair’s fair.”
That comment is ridiculously absurd. You are rhetorically asking me whether 3.2% is “important”, because YOU are the one saying that 3.2% to 13.2% (again, why do you keep saying 3.2% as if the lower bound is what it is on average? 3.2% to 13.2% means on average it is somewhere in the middle) is “de facto” unimportant.
You were trying to convince me to dismiss 3.2% to 13.2% as unimportant.
Now you’re saying on that basis I would have no problems with giving away the obverse of that?
What you should be saying if you’re going to put your money where your mouth is, is that because YOU don’t think 3.2% to 13.2% is important, then YOU should have no problems with giving me 3.2% to 13.2% of everything you earn.
Well? Fair’s fair right? According to you, you would be giving me “de facto” zero, right?
Oh what’s that? The percent 8%, which is very much in the range Bernanke found, is an “important” number of dollars to you after all? That it is not de facto zero?
As for whether 3.2% is “important”, all I am saying is that a range of 3.2% to 13.2%, IS BY YOUR OWN CRITERIA, and your own admission, merely sufficient evidence that money is not neutral.
You are just making an arbitrary assertion by denying a fact, totally ignoring the implication that if the range of 3.2% to 13.2% is not enough to make money non-neutral, then you would be having to explain what WOULD the range have to be to make money non-neutral, and importantly, which you had better address, is why THAT range, but not 3.2% to 13.2%.
Well?
You would be compelled to pull a range of numbers out of your derriere, because you have already left the realm of logic and science by unscientifically dismissing the range of 3.2% to 13.2%!
“As for your conceding that economics is only backwards looking, with no predictive power. I would agree…”
How in the world can I “concede” that to you when I was the one who tried to convince you of it after you said economics should be predictive?
I could only “concede” something to you if I previously held a particular opinion that you convinced me was wrong. Then I would have “conceded” something to you.
But in this case, now you’re actually agreeing with the argument that economics is not predictive?
Wtf!?
First you said the QTM should be dismissed because it doesn’t predict.
Now you’re saying I have “conceded” to you that economics is not predictive and that you agree!?!?
Huh!?!?
Ray your ideas are massively corrupted. If there was such a thing as a brain defragmenting or reformatting treatment, I would pay for your treatment.
“…and while you may find economics interesting given this fact, I bet 96.8% of the population will ignore economists if that fact was accepted. And it’s one reason professional economists like Sumner pretend they can divine the economy short term; it gives them relevance to the 96.8% of the population that expects economists to predict the future.”
YOU JUST SAID BEFORE that the QTM should be dismissed because it doesn’t predict. That to be useful, economics principles and ideas have to be able to predict.
Now you’re cavalierly guffawing at Sumner trying to do exactly what you just said economics should do?
Do you know how much of a chore it is to wade through the contradictions in your posts?
You say one thing, but then you say a totally opposite thing as if you never said what you said previously.
Do you think that would be fun?
57. Ray Lopez
12. August 2015 at 19:10
@Philippe – “you could explain the correlation by choosing to read the MV=PQ equation from right to left, rather than from left to right, with MV being seen as the dependent variable and PQ being seen as the independent variable” – right on brother! This is of course tantamount to saying that monetarism is dead, and the Fed follows the market. I concur.
@MF – due to tax reasons working offshore, I paid no US taxes last year, as my AGI was zero. Been that way for years. So you can have 3.2% to 13.2% of zero. And these numbers are from Ben S. Bernanke’s 2003 FAVOR paper, showing the Fed has very little influence over the economy. If Bernanke is saying this, you should be paying attention.
58. test » The Quantity Theory At The Extremes (#3)
13. August 2015 at 03:56
[…] the data set of 79 countries (in this post) there were 12 cases where inflation was higher than the money supply growth. In each case, real […]
59. W. Peden
13. August 2015 at 04:21
Philippe,
I think you need to re-examine the equation and also what Scott says before you go on discussing this topic.
13. August 2015 at 05:06
[…] 3. Money and inflation chart […]
61. ssumner
13. August 2015 at 05:42
Philippe, That’s not a model. For a model you need a causal mechanism. Furthermore, the advanced QTM can explain the various discrepancies better than any alternative.
And don’t forget that the QTM says that money supply increases will be inflationary if exogenous, even if done for many different reasons. No other theory gives you that.
62. Potpourri
15. August 2015 at 20:53
[…] Believe it or not, I actually like Scott Sumner’s recent series (1, 2, 3, and 4) using empirical data to discuss the quantity theory of […] | 11,127 | 46,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.948432 |
https://www.edaboard.com/threads/can-you-solve-this-interesting-problem.48859/#post-221487 | 1,642,500,832,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00038.warc.gz | 820,145,429 | 20,950 | # can you solve this interesting problem?
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#### wanily1983
##### Member level 5
space paper 377 ohms
hi,guys
when interviewed, i meet this interesting problem. the figure undermentioned is the resistor network, the question is calculating the Resistance between node A and node B.
#### flatulent
The standard solution is to use superposition. First put 1 Amp into node a and follow it through the network. Find the voltage difference between nodes A and B. Then do the same for node B with -1 Amp. Sum the two voltage differences. Then the resistance is the total voltage drop divided by 1 Amp.
#### mostafa0020
##### Member level 1
paralling an infinite number of resistors with a resistance R will result mathematically into an equivalent resistance value equals to 2/3 R, here is the conclusion :-
for a 3 resistors Xi,R,R parallel with a resistor R , the equivalent value will be :
=((Xi+2R)*R)/(Xi+3R)
For a series with X tends to zero as i tends to infinity , this results in final equivalent resistance value of 2*R/3.
Mostafa
sorry
the right side will result in 2R/3
the left the same
teh final will be 2R/6 // R = R/4.
My answer may needs other verifications
#### wanily1983
##### Member level 5
i am sorry for post the same topic twice.
by the way, Mostafa, "For a series with X tends to zero as i tends to infinity", what this mean? i am not clear about that. could you tell for detailed.
#### mostafa0020
##### Member level 1
ok wanily let's check this series i made to represent your case
from right meshes:-
R(i)=(((R(i-1)+2R)*R)/(R(i-1)+3R)
R(0) = R
Take R1 = 3/4 R
R2 = 11R/16
...
The values will decrese
getting R (∞) will be if we put X= R(i-1) tends to zero
resulting in R(∞) = 2/3 R
From left the same
both 2R/3 parallel results in 2R/6
Parallel with R
Resulting in R/4
It is fuzzy some, but , interesting, plz tell me who asked u this question and in which country , thx
sorry
i gave these valuse neglecting upper and lower meshes, taking into consideration the same reduction will result in 3 2R/3 resistors right // with R and 3 2R/3 left
means an equivalent parallel of R
the total equivalent resistance between A & B will be R/2.
### wanily1983
Points: 2
#### Mr.Cool
how is the answer not zero?
as each resistor gets paralleled with its neighbor, the equivalent resistance gets smaller. with an infinit # of resistors, that equivalent resistance is zero. on the left side of A/B and on the right side of A/B (and possibly above & below?)
Mr.Cool
#### flatulent
mostafa0020 got the correct answer. R/2
It is a lot easier to note that the array is infinite and that the currents injected as I described above go one quarter into each resistor connected to the node. They individually produce a (I/4)R voltage drop. Adding because of superposition gives a total of (I/2)R volts. Then the equivalent resistance is r=E/I or[( I/2)R]/I which sorts out to R/2
### wanily1983
Points: 2
#### Mr.Cool
i'll have to try that in PSpice
#### mostafa0020
##### Member level 1
Thankx flatulent, the same answer achieved..
#### wanily1983
##### Member level 5
i am a chinese student. the problem is from a Doctor of Berkeley and he is working in a chinese company now.
#### mostafa0020
##### Member level 1
Indeed Wanli I thank u and flatulent for this problem,it is an impressive one, i may test engineering students using this problem!!
#### wanily1983
##### Member level 5
hi,flatulent
i am still puzzled with your idea. Assume this condition that there ia only one resistorbetween node A and node B, using ur method,we can get the resistance between A and B is 2*R,which is not right.
#### flatulent
The method I used assumes that the current return is at infinity. In real world circuits any current source will have to have two connections at specific points.
In any circuit you can put a voltage or current source between any two points and calculate the current or voltage. Then divide to get the resistance.
The stated problem is the discrete equivalent of sheet resistance. The four point probe calculations are done with the method I suggested.
A piece of history: Before the days of computers, many clever methods were used to obtain solutions to problems. There was a special paper coated with a conducting chemical with a sheet resistance of 377 ohms per square. This was called "space paper" and used for transmission line impedance predictions. Two conductors of the shape and orientation of the cross section of the transmission line in question were pressed onto the space paper. The resistance between them was measured with the usual DC multimeter. The reading was the impedance of the transmission line.
The field plots of two dimensional electric fields could be done with a conductive liquid in a tray with a glass bottom. The metal pieces with the proper shape were put in the liquid and an AC voltage was put between them. A probe was put in the water to measure the voltage at each point. There was graph paper under the glass bottom. The readings were taken down by hand and later made into graphs on paper.
#### zorro
Now we can propose a variation on this theme: which is the resistance between two points located on a diagonal of the mesh?
Regards
Z
Status
Not open for further replies. | 1,297 | 5,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-05 | latest | en | 0.904278 |
https://www.airmilescalculator.com/distance/eze-to-irj/ | 1,680,331,969,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00424.warc.gz | 685,002,854 | 21,778 | # How far is La Rioja from Buenos Aires?
The distance between Buenos Aires (Buenos Aires Ministro Pistarini International Airport) and La Rioja (Capitán Vicente Almandos Almonacid Airport) is 612 miles / 985 kilometers / 532 nautical miles.
The driving distance from Buenos Aires (EZE) to La Rioja (IRJ) is 727 miles / 1170 kilometers, and travel time by car is about 12 hours 59 minutes.
612
Miles
985
Kilometers
532
Nautical miles
1 h 39 min
114 kg
## Distance from Buenos Aires to La Rioja
There are several ways to calculate the distance from Buenos Aires to La Rioja. Here are two standard methods:
Vincenty's formula (applied above)
• 612.147 miles
• 985.156 kilometers
• 531.941 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 611.994 miles
• 984.909 kilometers
• 531.808 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Buenos Aires to La Rioja?
The estimated flight time from Buenos Aires Ministro Pistarini International Airport to Capitán Vicente Almandos Almonacid Airport is 1 hour and 39 minutes.
## What is the time difference between Buenos Aires and La Rioja?
There is no time difference between Buenos Aires and La Rioja.
## Flight carbon footprint between Buenos Aires Ministro Pistarini International Airport (EZE) and Capitán Vicente Almandos Almonacid Airport (IRJ)
On average, flying from Buenos Aires to La Rioja generates about 114 kg of CO2 per passenger, and 114 kilograms equals 252 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Buenos Aires to La Rioja
See the map of the shortest flight path between Buenos Aires Ministro Pistarini International Airport (EZE) and Capitán Vicente Almandos Almonacid Airport (IRJ).
## Airport information
Origin Buenos Aires Ministro Pistarini International Airport
City: Buenos Aires
Country: Argentina
IATA Code: EZE
ICAO Code: SAEZ
Coordinates: 34°49′19″S, 58°32′8″W
Destination Capitán Vicente Almandos Almonacid Airport
City: La Rioja
Country: Argentina
IATA Code: IRJ
ICAO Code: SANL
Coordinates: 29°22′53″S, 66°47′44″W | 615 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.758086 |
https://www.ademcetinkaya.com/2023/12/where-will-clsk-stock-be-in-16-weeks.html | 1,713,699,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00351.warc.gz | 567,797,405 | 61,490 | Outlook: CleanSpark Inc. Common Stock is assigned short-term B1 & long-term B1 estimated rating.
AUC Score : What is AUC Score?
Short-Term Revised1 :
Dominant Strategy : Hold
Time series to forecast n: for Weeks2
Methodology : Modular Neural Network (CNN Layer)
Hypothesis Testing : Pearson Correlation
Surveillance : Major exchange and OTC
1The accuracy of the model is being monitored on a regular basis.(15-minute period)
2Time series is updated based on short-term trends.
## Summary
CleanSpark Inc. Common Stock prediction model is evaluated with Modular Neural Network (CNN Layer) and Pearson Correlation1,2,3,4 and it is concluded that the CLSK stock is predictable in the short/long term. CNN layers are a powerful tool for extracting features from images. They are able to learn to detect patterns in images that are not easily detected by humans. This makes them well-suited for a variety of MNN applications.5 According to price forecasts for 16 Weeks period, the dominant strategy among neural network is: Hold
## Key Points
1. Modular Neural Network (CNN Layer) for CLSK stock price prediction process.
2. Pearson Correlation
3. Market Risk
4. Trust metric by Neural Network
5. Can statistics predict the future?
## CLSK Stock Price Forecast
We consider CleanSpark Inc. Common Stock Decision Process with Modular Neural Network (CNN Layer) where A is the set of discrete actions of CLSK stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
Sample Set: Neural Network
Stock/Index: CLSK CleanSpark Inc. Common Stock
Time series to forecast: 16 Weeks
According to price forecasts, the dominant strategy among neural network is: Hold
F(Pearson Correlation)6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (CNN Layer)) X S(n):→ 16 Weeks $∑ i = 1 n a i$
n:Time series to forecast
p:Price signals of CLSK stock
j:Nash equilibria (Neural Network)
k:Dominated move of CLSK stock holders
a:Best response for CLSK target price
CNN layers are a powerful tool for extracting features from images. They are able to learn to detect patterns in images that are not easily detected by humans. This makes them well-suited for a variety of MNN applications.5 Pearson correlation, also known as Pearson's product-moment correlation, is a measure of the linear relationship between two variables. It is a statistical measure that assesses the strength and direction of a linear relationship between two variables. The sign of the correlation coefficient indicates the direction of the relationship, while the magnitude of the correlation coefficient indicates the strength of the relationship. A correlation coefficient of 0.9 indicates a strong positive correlation, while a correlation coefficient of 0.2 indicates a weak positive correlation.6,7
For further technical information as per how our model work we invite you to visit the article below:
How do PredictiveAI algorithms actually work?
### CLSK Stock Forecast (Buy or Sell) Strategic Interaction Table
Strategic Interaction Table Legend:
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
### Financial Data Adjustments for Modular Neural Network (CNN Layer) based CLSK Stock Prediction Model
1. If, at the date of initial application, it is impracticable (as defined in IAS 8) for an entity to assess a modified time value of money element in accordance with paragraphs B4.1.9B–B4.1.9D on the basis of the facts and circumstances that existed at the initial recognition of the financial asset, an entity shall assess the contractual cash flow characteristics of that financial asset on the basis of the facts and circumstances that existed at the initial recognition of the financial asset without taking into account the requirements related to the modification of the time value of money element in paragraphs B4.1.9B–B4.1.9D. (See also paragraph 42R of IFRS 7.)
2. Because the hedge accounting model is based on a general notion of offset between gains and losses on the hedging instrument and the hedged item, hedge effectiveness is determined not only by the economic relationship between those items (ie the changes in their underlyings) but also by the effect of credit risk on the value of both the hedging instrument and the hedged item. The effect of credit risk means that even if there is an economic relationship between the hedging instrument and the hedged item, the level of offset might become erratic. This can result from a change in the credit risk of either the hedging instrument or the hedged item that is of such a magnitude that the credit risk dominates the value changes that result from the economic relationship (ie the effect of the changes in the underlyings). A level of magnitude that gives rise to dominance is one that would result in the loss (or gain) from credit risk frustrating the effect of changes in the underlyings on the value of the hedging instrument or the hedged item, even if those changes were significant.
3. Paragraph 6.3.6 states that in consolidated financial statements the foreign currency risk of a highly probable forecast intragroup transaction may qualify as a hedged item in a cash flow hedge, provided that the transaction is denominated in a currency other than the functional currency of the entity entering into that transaction and that the foreign currency risk will affect consolidated profit or loss. For this purpose an entity can be a parent, subsidiary, associate, joint arrangement or branch. If the foreign currency risk of a forecast intragroup transaction does not affect consolidated profit or loss, the intragroup transaction cannot qualify as a hedged item. This is usually the case for royalty payments, interest payments or management charges between members of the same group, unless there is a related external transaction. However, when the foreign currency risk of a forecast intragroup transaction will affect consolidated profit or loss, the intragroup transaction can qualify as a hedged item. An example is forecast sales or purchases of inventories between members of the same group if there is an onward sale of the inventory to a party external to the group. Similarly, a forecast intragroup sale of plant and equipment from the group entity that manufactured it to a group entity that will use the plant and equipment in its operations may affect consolidated profit or loss. This could occur, for example, because the plant and equipment will be depreciated by the purchasing entity and the amount initially recognised for the plant and equipment may change if the forecast intragroup transaction is denominated in a currency other than the functional currency of the purchasing entity.
4. The assessment of whether an economic relationship exists includes an analysis of the possible behaviour of the hedging relationship during its term to ascertain whether it can be expected to meet the risk management objective. The mere existence of a statistical correlation between two variables does not, by itself, support a valid conclusion that an economic relationship exists.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
### CLSK CleanSpark Inc. Common Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*B1B1
Income StatementB2B2
Balance SheetB1Ba2
Leverage RatiosBa3Baa2
Cash FlowBaa2Caa2
Rates of Return and ProfitabilityCaa2C
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
## References
1. H. Khalil and J. Grizzle. Nonlinear systems, volume 3. Prentice hall Upper Saddle River, 2002.
2. E. Collins. Using Markov decision processes to optimize a nonlinear functional of the final distribution, with manufacturing applications. In Stochastic Modelling in Innovative Manufacturing, pages 30–45. Springer, 1997
3. Hartford J, Lewis G, Taddy M. 2016. Counterfactual prediction with deep instrumental variables networks. arXiv:1612.09596 [stat.AP]
4. R. Howard and J. Matheson. Risk sensitive Markov decision processes. Management Science, 18(7):356– 369, 1972
5. E. Collins. Using Markov decision processes to optimize a nonlinear functional of the final distribution, with manufacturing applications. In Stochastic Modelling in Innovative Manufacturing, pages 30–45. Springer, 1997
6. Zubizarreta JR. 2015. Stable weights that balance covariates for estimation with incomplete outcome data. J. Am. Stat. Assoc. 110:910–22
7. Imbens G, Wooldridge J. 2009. Recent developments in the econometrics of program evaluation. J. Econ. Lit. 47:5–86
Frequently Asked QuestionsQ: Is CLSK stock expected to rise?
A: CLSK stock prediction model is evaluated with Modular Neural Network (CNN Layer) and Pearson Correlation and it is concluded that dominant strategy for CLSK stock is Hold
Q: Is CLSK stock a buy or sell?
A: The dominant strategy among neural network is to Hold CLSK Stock.
Q: Is CleanSpark Inc. Common Stock stock a good investment?
A: The consensus rating for CleanSpark Inc. Common Stock is Hold and is assigned short-term B1 & long-term B1 estimated rating.
Q: What is the consensus rating of CLSK stock?
A: The consensus rating for CLSK is Hold.
Q: What is the forecast for CLSK stock?
A: CLSK target price forecast: Hold | 2,281 | 10,334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.870255 |
https://www.netlib.org/lapack/explore-html-3.6.1/d4/da2/zdrvac_8f_source.html | 1,719,235,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00125.warc.gz | 805,574,904 | 12,005 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
zdrvac.f
Go to the documentation of this file.
1 *> \brief \b ZDRVAC
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 * Definition:
9 * ===========
10 *
11 * SUBROUTINE ZDRVAC( DOTYPE, NM, MVAL, NNS, NSVAL, THRESH, NMAX,
12 * A, AFAC, B, X, WORK,
13 * RWORK, SWORK, NOUT )
14 *
15 * .. Scalar Arguments ..
16 * INTEGER NMAX, NM, NNS, NOUT
17 * DOUBLE PRECISION THRESH
18 * ..
19 * .. Array Arguments ..
20 * LOGICAL DOTYPE( * )
21 * INTEGER MVAL( * ), NSVAL( * )
22 * DOUBLE PRECISION RWORK( * )
23 * COMPLEX SWORK(*)
24 * COMPLEX*16 A( * ), AFAC( * ), B( * ),
25 * \$ WORK( * ), X( * )
26 * ..
27 *
28 *
29 *> \par Purpose:
30 * =============
31 *>
32 *> \verbatim
33 *>
34 *> ZDRVAC tests ZCPOSV.
35 *> \endverbatim
36 *
37 * Arguments:
38 * ==========
39 *
40 *> \param[in] DOTYPE
41 *> \verbatim
42 *> DOTYPE is LOGICAL array, dimension (NTYPES)
43 *> The matrix types to be used for testing. Matrices of type j
44 *> (for 1 <= j <= NTYPES) are used for testing if DOTYPE(j) =
45 *> .TRUE.; if DOTYPE(j) = .FALSE., then type j is not used.
46 *> \endverbatim
47 *>
48 *> \param[in] NM
49 *> \verbatim
50 *> NM is INTEGER
51 *> The number of values of N contained in the vector MVAL.
52 *> \endverbatim
53 *>
54 *> \param[in] MVAL
55 *> \verbatim
56 *> MVAL is INTEGER array, dimension (NM)
57 *> The values of the matrix dimension N.
58 *> \endverbatim
59 *>
60 *> \param[in] NNS
61 *> \verbatim
62 *> NNS is INTEGER
63 *> The number of values of NRHS contained in the vector NSVAL.
64 *> \endverbatim
65 *>
66 *> \param[in] NSVAL
67 *> \verbatim
68 *> NSVAL is INTEGER array, dimension (NNS)
69 *> The values of the number of right hand sides NRHS.
70 *> \endverbatim
71 *>
72 *> \param[in] THRESH
73 *> \verbatim
74 *> THRESH is DOUBLE PRECISION
75 *> The threshold value for the test ratios. A result is
76 *> included in the output file if RESULT >= THRESH. To have
77 *> every test ratio printed, use THRESH = 0.
78 *> \endverbatim
79 *>
80 *> \param[in] NMAX
81 *> \verbatim
82 *> NMAX is INTEGER
83 *> The maximum value permitted for N, used in dimensioning the
84 *> work arrays.
85 *> \endverbatim
86 *>
87 *> \param[out] A
88 *> \verbatim
89 *> A is COMPLEX*16 array, dimension (NMAX*NMAX)
90 *> \endverbatim
91 *>
92 *> \param[out] AFAC
93 *> \verbatim
94 *> AFAC is COMPLEX*16 array, dimension (NMAX*NMAX)
95 *> \endverbatim
96 *>
97 *> \param[out] B
98 *> \verbatim
99 *> B is COMPLEX*16 array, dimension (NMAX*NSMAX)
100 *> \endverbatim
101 *>
102 *> \param[out] X
103 *> \verbatim
104 *> X is COMPLEX*16 array, dimension (NMAX*NSMAX)
105 *> \endverbatim
106 *>
107 *> \param[out] WORK
108 *> \verbatim
109 *> WORK is COMPLEX*16 array, dimension
110 *> (NMAX*max(3,NSMAX))
111 *> \endverbatim
112 *>
113 *> \param[out] RWORK
114 *> \verbatim
115 *> RWORK is DOUBLE PRECISION array, dimension
116 *> (max(2*NMAX,2*NSMAX+NWORK))
117 *> \endverbatim
118 *>
119 *> \param[out] SWORK
120 *> \verbatim
121 *> SWORK is COMPLEX array, dimension
122 *> (NMAX*(NSMAX+NMAX))
123 *> \endverbatim
124 *>
125 *> \param[in] NOUT
126 *> \verbatim
127 *> NOUT is INTEGER
128 *> The unit number for output.
129 *> \endverbatim
130 *
131 * Authors:
132 * ========
133 *
134 *> \author Univ. of Tennessee
135 *> \author Univ. of California Berkeley
136 *> \author Univ. of Colorado Denver
137 *> \author NAG Ltd.
138 *
139 *> \date November 2011
140 *
141 *> \ingroup complex16_lin
142 *
143 * =====================================================================
144 SUBROUTINE zdrvac( DOTYPE, NM, MVAL, NNS, NSVAL, THRESH, NMAX,
145 \$ a, afac, b, x, work,
146 \$ rwork, swork, nout )
147 *
148 * -- LAPACK test routine (version 3.4.0) --
149 * -- LAPACK is a software package provided by Univ. of Tennessee, --
150 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
151 * November 2011
152 *
153 * .. Scalar Arguments ..
154 INTEGER NMAX, NM, NNS, NOUT
155 DOUBLE PRECISION THRESH
156 * ..
157 * .. Array Arguments ..
158 LOGICAL DOTYPE( * )
159 INTEGER MVAL( * ), NSVAL( * )
160 DOUBLE PRECISION RWORK( * )
161 COMPLEX SWORK(*)
162 COMPLEX*16 A( * ), AFAC( * ), B( * ),
163 \$ work( * ), x( * )
164 * ..
165 *
166 * =====================================================================
167 *
168 * .. Parameters ..
169 DOUBLE PRECISION ZERO
170 parameter ( zero = 0.0d+0 )
171 INTEGER NTYPES
172 parameter ( ntypes = 9 )
173 INTEGER NTESTS
174 parameter ( ntests = 1 )
175 * ..
176 * .. Local Scalars ..
177 LOGICAL ZEROT
178 CHARACTER DIST, TYPE, UPLO, XTYPE
179 CHARACTER*3 PATH
180 INTEGER I, IM, IMAT, INFO, IOFF, IRHS, IUPLO,
181 \$ izero, kl, ku, lda, mode, n,
182 \$ nerrs, nfail, nimat, nrhs, nrun
183 DOUBLE PRECISION ANORM, CNDNUM
184 * ..
185 * .. Local Arrays ..
186 CHARACTER UPLOS( 2 )
187 INTEGER ISEED( 4 ), ISEEDY( 4 )
188 DOUBLE PRECISION RESULT( ntests )
189 * ..
190 * .. Local Variables ..
191 INTEGER ITER, KASE
192 * ..
193 * .. External Subroutines ..
194 EXTERNAL alaerh, zlacpy, zlaipd,
195 \$ zlarhs, zlatb4, zlatms,
196 \$ zpot06, zcposv
197 * ..
198 * .. Intrinsic Functions ..
199 INTRINSIC dble, max, sqrt
200 * ..
201 * .. Scalars in Common ..
202 LOGICAL LERR, OK
203 CHARACTER*32 SRNAMT
204 INTEGER INFOT, NUNIT
205 * ..
206 * .. Common blocks ..
207 COMMON / infoc / infot, nunit, ok, lerr
208 COMMON / srnamc / srnamt
209 * ..
210 * .. Data statements ..
211 DATA iseedy / 1988, 1989, 1990, 1991 /
212 DATA uplos / 'U', 'L' /
213 * ..
214 * .. Executable Statements ..
215 *
216 * Initialize constants and the random number seed.
217 *
218 kase = 0
219 path( 1: 1 ) = 'Zomplex precision'
220 path( 2: 3 ) = 'PO'
221 nrun = 0
222 nfail = 0
223 nerrs = 0
224 DO 10 i = 1, 4
225 iseed( i ) = iseedy( i )
226 10 CONTINUE
227 *
228 infot = 0
229 *
230 * Do for each value of N in MVAL
231 *
232 DO 120 im = 1, nm
233 n = mval( im )
234 lda = max( n, 1 )
235 nimat = ntypes
236 IF( n.LE.0 )
237 \$ nimat = 1
238 *
239 DO 110 imat = 1, nimat
240 *
241 * Do the tests only if DOTYPE( IMAT ) is true.
242 *
243 IF( .NOT.dotype( imat ) )
244 \$ GO TO 110
245 *
246 * Skip types 3, 4, or 5 if the matrix size is too small.
247 *
248 zerot = imat.GE.3 .AND. imat.LE.5
249 IF( zerot .AND. n.LT.imat-2 )
250 \$ GO TO 110
251 *
252 * Do first for UPLO = 'U', then for UPLO = 'L'
253 *
254 DO 100 iuplo = 1, 2
255 uplo = uplos( iuplo )
256 *
257 * Set up parameters with ZLATB4 and generate a test matrix
258 * with ZLATMS.
259 *
260 CALL zlatb4( path, imat, n, n, TYPE, KL, KU, ANORM, MODE,
261 \$ cndnum, dist )
262 *
263 srnamt = 'ZLATMS'
264 CALL zlatms( n, n, dist, iseed, TYPE, RWORK, MODE,
265 \$ cndnum, anorm, kl, ku, uplo, a, lda, work,
266 \$ info )
267 *
268 * Check error code from ZLATMS.
269 *
270 IF( info.NE.0 ) THEN
271 CALL alaerh( path, 'ZLATMS', info, 0, uplo, n, n, -1,
272 \$ -1, -1, imat, nfail, nerrs, nout )
273 GO TO 100
274 END IF
275 *
276 * For types 3-5, zero one row and column of the matrix to
277 * test that INFO is returned correctly.
278 *
279 IF( zerot ) THEN
280 IF( imat.EQ.3 ) THEN
281 izero = 1
282 ELSE IF( imat.EQ.4 ) THEN
283 izero = n
284 ELSE
285 izero = n / 2 + 1
286 END IF
287 ioff = ( izero-1 )*lda
288 *
289 * Set row and column IZERO of A to 0.
290 *
291 IF( iuplo.EQ.1 ) THEN
292 DO 20 i = 1, izero - 1
293 a( ioff+i ) = zero
294 20 CONTINUE
295 ioff = ioff + izero
296 DO 30 i = izero, n
297 a( ioff ) = zero
298 ioff = ioff + lda
299 30 CONTINUE
300 ELSE
301 ioff = izero
302 DO 40 i = 1, izero - 1
303 a( ioff ) = zero
304 ioff = ioff + lda
305 40 CONTINUE
306 ioff = ioff - izero
307 DO 50 i = izero, n
308 a( ioff+i ) = zero
309 50 CONTINUE
310 END IF
311 ELSE
312 izero = 0
313 END IF
314 *
315 * Set the imaginary part of the diagonals.
316 *
317 CALL zlaipd( n, a, lda+1, 0 )
318 *
319 DO 60 irhs = 1, nns
320 nrhs = nsval( irhs )
321 xtype = 'N'
322 *
323 * Form an exact solution and set the right hand side.
324 *
325 srnamt = 'ZLARHS'
326 CALL zlarhs( path, xtype, uplo, ' ', n, n, kl, ku,
327 \$ nrhs, a, lda, x, lda, b, lda,
328 \$ iseed, info )
329 *
330 * Compute the L*L' or U'*U factorization of the
331 * matrix and solve the system.
332 *
333 srnamt = 'ZCPOSV '
334 kase = kase + 1
335 *
336 CALL zlacpy( 'All', n, n, a, lda, afac, lda)
337 *
338 CALL zcposv( uplo, n, nrhs, afac, lda, b, lda, x, lda,
339 \$ work, swork, rwork, iter, info )
340 *
341 IF (iter.LT.0) THEN
342 CALL zlacpy( 'All', n, n, a, lda, afac, lda )
343 ENDIF
344 *
345 * Check error code from ZCPOSV .
346 *
347 IF( info.NE.izero ) THEN
348 *
349 IF( nfail.EQ.0 .AND. nerrs.EQ.0 )
350 \$ CALL alahd( nout, path )
351 nerrs = nerrs + 1
352 *
353 IF( info.NE.izero .AND. izero.NE.0 ) THEN
354 WRITE( nout, fmt = 9988 )'ZCPOSV',info,izero,n,
355 \$ imat
356 ELSE
357 WRITE( nout, fmt = 9975 )'ZCPOSV',info,n,imat
358 END IF
359 END IF
360 *
361 * Skip the remaining test if the matrix is singular.
362 *
363 IF( info.NE.0 )
364 \$ GO TO 110
365 *
366 * Check the quality of the solution
367 *
368 CALL zlacpy( 'All', n, nrhs, b, lda, work, lda )
369 *
370 CALL zpot06( uplo, n, nrhs, a, lda, x, lda, work,
371 \$ lda, rwork, result( 1 ) )
372 *
373 * Check if the test passes the tesing.
374 * Print information about the tests that did not
375 * pass the testing.
376 *
377 * If iterative refinement has been used and claimed to
378 * be successful (ITER>0), we want
379 * NORM1(B - A*X)/(NORM1(A)*NORM1(X)*EPS*SRQT(N)) < 1
380 *
381 * If double precision has been used (ITER<0), we want
382 * NORM1(B - A*X)/(NORM1(A)*NORM1(X)*EPS) < THRES
383 * (Cf. the linear solver testing routines)
384 *
385 IF ((thresh.LE.0.0e+00)
386 \$ .OR.((iter.GE.0).AND.(n.GT.0)
387 \$ .AND.(result(1).GE.sqrt(dble(n))))
388 \$ .OR.((iter.LT.0).AND.(result(1).GE.thresh))) THEN
389 *
390 IF( nfail.EQ.0 .AND. nerrs.EQ.0 ) THEN
391 WRITE( nout, fmt = 8999 )'ZPO'
392 WRITE( nout, fmt = '( '' Matrix types:'' )' )
393 WRITE( nout, fmt = 8979 )
394 WRITE( nout, fmt = '( '' Test ratios:'' )' )
395 WRITE( nout, fmt = 8960 )1
396 WRITE( nout, fmt = '( '' Messages:'' )' )
397 END IF
398 *
399 WRITE( nout, fmt = 9998 )uplo, n, nrhs, imat, 1,
400 \$ result( 1 )
401 *
402 nfail = nfail + 1
403 *
404 END IF
405 *
406 nrun = nrun + 1
407 *
408 60 CONTINUE
409 100 CONTINUE
410 110 CONTINUE
411 120 CONTINUE
412 *
413 * Print a summary of the results.
414 *
415 IF( nfail.GT.0 ) THEN
416 WRITE( nout, fmt = 9996 )'ZCPOSV', nfail, nrun
417 ELSE
418 WRITE( nout, fmt = 9995 )'ZCPOSV', nrun
419 END IF
420 IF( nerrs.GT.0 ) THEN
421 WRITE( nout, fmt = 9994 )nerrs
422 END IF
423 *
424 9998 FORMAT( ' UPLO=''', a1, ''', N =', i5, ', NRHS=', i3, ', type ',
425 \$ i2, ', test(', i2, ') =', g12.5 )
426 9996 FORMAT( 1x, a6, ': ', i6, ' out of ', i6,
427 \$ ' tests failed to pass the threshold' )
428 9995 FORMAT( /1x, 'All tests for ', a6,
429 \$ ' routines passed the threshold ( ', i6, ' tests run)' )
430 9994 FORMAT( 6x, i6, ' error messages recorded' )
431 *
432 * SUBNAM, INFO, INFOE, N, IMAT
433 *
434 9988 FORMAT( ' *** ', a6, ' returned with INFO =', i5, ' instead of ',
435 \$ i5, / ' ==> N =', i5, ', type ',
436 \$ i2 )
437 *
438 * SUBNAM, INFO, N, IMAT
439 *
440 9975 FORMAT( ' *** Error code from ', a6, '=', i5, ' for M=', i5,
441 \$ ', type ', i2 )
442 8999 FORMAT( / 1x, a3, ': positive definite dense matrices' )
443 8979 FORMAT( 4x, '1. Diagonal', 24x, '7. Last n/2 columns zero', / 4x,
444 \$ '2. Upper triangular', 16x,
445 \$ '8. Random, CNDNUM = sqrt(0.1/EPS)', / 4x,
446 \$ '3. Lower triangular', 16x, '9. Random, CNDNUM = 0.1/EPS',
447 \$ / 4x, '4. Random, CNDNUM = 2', 13x,
448 \$ '10. Scaled near underflow', / 4x, '5. First column zero',
449 \$ 14x, '11. Scaled near overflow', / 4x,
450 \$ '6. Last column zero' )
451 8960 FORMAT( 3x, i2, ': norm_1( B - A * X ) / ',
452 \$ '( norm_1(A) * norm_1(X) * EPS * SQRT(N) ) > 1 if ITERREF',
453 \$ / 4x, 'or norm_1( B - A * X ) / ',
454 \$ '( norm_1(A) * norm_1(X) * EPS ) > THRES if ZPOTRF' )
455
456 RETURN
457 *
458 * End of ZDRVAC
459 *
460 END
subroutine alahd(IOUNIT, PATH)
ALAHD
Definition: alahd.f:95
subroutine alaerh(PATH, SUBNAM, INFO, INFOE, OPTS, M, N, KL, KU, N5, IMAT, NFAIL, NERRS, NOUT)
ALAERH
Definition: alaerh.f:149
subroutine zlacpy(UPLO, M, N, A, LDA, B, LDB)
ZLACPY copies all or part of one two-dimensional array to another.
Definition: zlacpy.f:105
subroutine zlarhs(PATH, XTYPE, UPLO, TRANS, M, N, KL, KU, NRHS, A, LDA, X, LDX, B, LDB, ISEED, INFO)
ZLARHS
Definition: zlarhs.f:211
subroutine zlatb4(PATH, IMAT, M, N, TYPE, KL, KU, ANORM, MODE, CNDNUM, DIST)
ZLATB4
Definition: zlatb4.f:123
subroutine zdrvac(DOTYPE, NM, MVAL, NNS, NSVAL, THRESH, NMAX, A, AFAC, B, X, WORK, RWORK, SWORK, NOUT)
ZDRVAC
Definition: zdrvac.f:147
subroutine zlaipd(N, A, INDA, VINDA)
ZLAIPD
Definition: zlaipd.f:85
subroutine zcposv(UPLO, N, NRHS, A, LDA, B, LDB, X, LDX, WORK, SWORK, RWORK, ITER, INFO)
ZCPOSV computes the solution to system of linear equations A * X = B for PO matrices ...
Definition: zcposv.f:211
subroutine zpot06(UPLO, N, NRHS, A, LDA, X, LDX, B, LDB, RWORK, RESID)
ZPOT06
Definition: zpot06.f:129
subroutine zlatms(M, N, DIST, ISEED, SYM, D, MODE, COND, DMAX, KL, KU, PACK, A, LDA, WORK, INFO)
ZLATMS
Definition: zlatms.f:334 | 5,264 | 13,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.295214 |
https://www.answers.com/Q/How_far_is_a_mile_in_meters | 1,563,937,082,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530250.98/warc/CC-MAIN-20190724020454-20190724042454-00090.warc.gz | 594,674,399 | 14,362 | # How far is a mile in meters?
1 mile = 1 609.344 meters
1 meter = 0.000621371192 mile | 32 | 88 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-30 | latest | en | 0.824357 |
http://www.enotes.com/homework-help/find-y-f-f-1-3x-form-y-ax-bx-c-where-b-c-real-429157 | 1,462,493,658,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861700245.92/warc/CC-MAIN-20160428164140-00053-ip-10-239-7-51.ec2.internal.warc.gz | 492,335,048 | 11,904 | # Find `y=f(f^(-1)(3x))` in the form of y=ax/(bx + c), where a,b and c are real constants. `f(x)=2log_e(x+1)` and ` f^(-1)(x)=e^(x/2) - 1` .
Posted on
You need to find first `f^(-1)(3x)` replacing `3x` for `x ` in equation of `f^(-1)(x)` , such that:
`f^(-1)(3x) = e^((3x)/2) - 1`
You need to find now `f(f^(-1)(3x))` replacing `f^(-1)(3x)` for x in equation of `f(x)` , such that:
`f(f^(-1)(3x)) = 2 ln (f^(-1)(3x) + 1)`
Replacing `e^((3x)/2) - 1` for `f^(-1)(3x)` yields:
`f(f^(-1)(3x)) = 2ln (e^((3x)/2) - 1 + 1)`
Reducing like terms yields:
`f(f^(-1)(3x)) = 2ln(e^((3x)/2))`
Using the following logarithmic identity `ln a^b = b*ln ` a, yields:
`f(f^(-1)(3x)) = 2*(3x)/2*ln e`
Since `ln e = 1` `=> f(f^(-1)(3x)) = 3x`
Since you need to put `f(f^(-1)(3x)) = y` in the form `y = (ax)/(bx+c)` , comparing the expression yields:
`3x = (ax)/(bx+c) => {(a = 3),(b = 0),(c = 1):}`
Hence, evaluating the function y, under the given conditions, yields:`y = (3x)/(0*x + 1)`
Posted on
Since we know `f(f^(-1)(x))=x` ( by definition of inverse)
Therefore
`y=f(f^(-1)(3x))=3x` (i)
`y=(ax)/(bx+c)` (ii)
from (i) and (ii)
a=3 , bx+c=1
a=3, c=1 and b=0
only possible solution.
y=`(3x)/(0x+1)` | 547 | 1,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.579167 |
https://origin.geeksforgeeks.org/c-scan-disk-scheduling-algorithm/ | 1,674,815,970,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494976.72/warc/CC-MAIN-20230127101040-20230127131040-00163.warc.gz | 467,568,504 | 38,169 | # C-SCAN Disk Scheduling Algorithm
• Difficulty Level : Medium
• Last Updated : 30 Nov, 2022
Prerequisite: Disk Scheduling Algorithms and SCAN Disk Scheduling Algorithm
Given an array of disk track numbers and initial head position, our task is to find the total number of seek operations done to access all the requested tracks if a C-SCAN disk scheduling algorithm is used.
### C-SCAN (Circular Elevator) Disk Scheduling Algorithm
The circular SCAN (C-SCAN) scheduling algorithm is a modified version of the SCAN disk scheduling algorithm that deals with the inefficiency of the SCAN algorithm by servicing the requests more uniformly. Like SCAN (Elevator Algorithm) C-SCAN moves the head from one end servicing all the requests to the other end. However, as soon as the head reaches the other end, it immediately returns to the beginning of the disk without servicing any requests on the return trip (see chart below) and starts servicing again once reaches the beginning. This is also known as the “Circular Elevator Algorithm” as it essentially treats the cylinders as a circular list that wraps around from the final cylinder to the first one.
#### Advantages of C-SCAN (Circular Elevator) Disk Scheduling Algorithm:
• Works well with moderate to heavy loads.
• It provides better response time and uniform waiting time.
#### Disadvantages of C-SCAN (Circular Elevator) Disk Scheduling Algorithm:
• May not be fair to service requests for tracks at the extreme end.
• It has more seek movements as compared to the SCAN Algorithm.
### Algorithm:
1. Let Request array represents an array storing indexes of tracks that have been requested in ascending order of their time of arrival. ‘head’ is the position of disk head.
2. The head services only in the right direction from 0 to the size of the disk.
3. While moving in the left direction do not service any of the tracks.
4. When we reach the beginning(left end) reverse the direction.
5. While moving in the right direction it services all tracks one by one.
6. While moving in the right direction calculate the absolute distance of the track from the head.
7. Increment the total seek count with this distance.
8. Currently serviced track position now becomes the new head position.
9. Go to step 6 until we reach the right end of the disk.
10. If we reach the right end of the disk reverse the direction and go to step 3 until all tracks in the request array have not been serviced.
Examples:
```Input:
Request sequence = {176, 79, 34, 60, 92, 11, 41, 114}
Initial head position = 50
Direction = right(We are moving from left to right)
Output:
Initial position of head: 50
Total number of seek operations = 389
Seek Sequence is
60
79
92
114
176
199
0
11
34
41```
The following chart shows the sequence in which requested tracks are serviced using SCAN.
Therefore, the total seek count is calculated as:
```= (60-50)+(79-60)+(92-79)
+(114-92)+(176-114)+(199-176)+(199-0)
+(11-0)+(34-11)+(41-34)
= 389```
### Implementation:
The implementation of C-SCAN algorithm is given below.
Note:
The distance variable is used to store the absolute distance between the head and current track position. disk_size is the size of the disk. Vectors left and right store all the request tracks on the left-hand side and the right-hand side of the initial head position respectively.
## C++
`// C++ program to demonstrate ` `// C-SCAN Disk Scheduling algorithm ` `#include ` `using` `namespace` `std; ` ` ` `// Code by Vikram Chaurasia ` ` ` `int` `size = 8; ` `int` `disk_size = 200; ` ` ` `void` `CSCAN(``int` `arr[], ``int` `head) ` `{ ` ` ``int` `seek_count = 0; ` ` ``int` `distance, cur_track; ` ` ``vector<``int``> left, right; ` ` ``vector<``int``> seek_sequence; ` ` ` ` ``// appending end values ` ` ``// which has to be visited ` ` ``// before reversing the direction ` ` ``left.push_back(0); ` ` ``right.push_back(disk_size - 1); ` ` ` ` ``// tracks on the left of the ` ` ``// head will be serviced when ` ` ``// once the head comes back ` ` ``// to the beginning (left end). ` ` ``for` `(``int` `i = 0; i < size; i++) { ` ` ``if` `(arr[i] < head) ` ` ``left.push_back(arr[i]); ` ` ``if` `(arr[i] > head) ` ` ``right.push_back(arr[i]); ` ` ``} ` ` ` ` ``// sorting left and right vectors ` ` ``std::sort(left.begin(), left.end()); ` ` ``std::sort(right.begin(), right.end()); ` ` ` ` ``// first service the requests ` ` ``// on the right side of the ` ` ``// head. ` ` ``for` `(``int` `i = 0; i < right.size(); i++) { ` ` ``cur_track = right[i]; ` ` ``// appending current track to seek sequence ` ` ``seek_sequence.push_back(cur_track); ` ` ` ` ``// calculate absolute distance ` ` ``distance = ``abs``(cur_track - head); ` ` ` ` ``// increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// accessed track is now new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``// once reached the right end ` ` ``// jump to the beginning. ` ` ``head = 0; ` ` ` ` ``// adding seek count for head returning from 199 to 0 ` ` ``seek_count += (disk_size - 1); ` ` ` ` ``// Now service the requests again ` ` ``// which are left. ` ` ``for` `(``int` `i = 0; i < left.size(); i++) { ` ` ``cur_track = left[i]; ` ` ` ` ``// appending current track to seek sequence ` ` ``seek_sequence.push_back(cur_track); ` ` ` ` ``// calculate absolute distance ` ` ``distance = ``abs``(cur_track - head); ` ` ` ` ``// increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// accessed track is now the new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``cout << ``"Total number of seek operations = "` ` ``<< seek_count << endl; ` ` ` ` ``cout << ``"Seek Sequence is"` `<< endl; ` ` ` ` ``for` `(``int` `i = 0; i < seek_sequence.size(); i++) { ` ` ``cout << seek_sequence[i] << endl; ` ` ``} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ``// request array ` ` ``int` `arr[size] = { 176, 79, 34, 60, 92, 11, 41, 114 }; ` ` ``int` `head = 50; ` ` ` ` ``cout << ``"Initial position of head: "` `<< head << endl; ` ` ``CSCAN(arr, head); ` ` ` ` ``return` `0; ` `}`
## Java
`// Java program to demonstrate ` `// C-SCAN Disk Scheduling algorithm ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``static` `int` `size = ``8``; ` ` ``static` `int` `disk_size = ``200``; ` ` ` ` ``public` `static` `void` `CSCAN(``int` `arr[], ``int` `head) ` ` ``{ ` ` ``int` `seek_count = ``0``; ` ` ``int` `distance, cur_track; ` ` ` ` ``Vector left = ``new` `Vector(); ` ` ``Vector right = ``new` `Vector(); ` ` ``Vector seek_sequence ` ` ``= ``new` `Vector(); ` ` ` ` ``// Appending end values which has ` ` ``// to be visited before reversing ` ` ``// the direction ` ` ``left.add(``0``); ` ` ``right.add(disk_size - ``1``); ` ` ` ` ``// Tracks on the left of the ` ` ``// head will be serviced when ` ` ``// once the head comes back ` ` ``// to the beginning (left end). ` ` ``for` `(``int` `i = ``0``; i < size; i++) { ` ` ``if` `(arr[i] < head) ` ` ``left.add(arr[i]); ` ` ``if` `(arr[i] > head) ` ` ``right.add(arr[i]); ` ` ``} ` ` ` ` ``// Sorting left and right vectors ` ` ``Collections.sort(left); ` ` ``Collections.sort(right); ` ` ` ` ``// First service the requests ` ` ``// on the right side of the ` ` ``// head. ` ` ``for` `(``int` `i = ``0``; i < right.size(); i++) { ` ` ``cur_track = right.get(i); ` ` ` ` ``// Appending current track to seek sequence ` ` ``seek_sequence.add(cur_track); ` ` ` ` ``// Calculate absolute distance ` ` ``distance = Math.abs(cur_track - head); ` ` ` ` ``// Increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// Accessed track is now new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``// Once reached the right end ` ` ``// jump to the beginning. ` ` ``head = ``0``; ` ` ` ` ``// adding seek count for head returning from 199 to ` ` ``// 0 ` ` ``seek_count += (disk_size - ``1``); ` ` ` ` ``// Now service the requests again ` ` ``// which are left. ` ` ``for` `(``int` `i = ``0``; i < left.size(); i++) { ` ` ``cur_track = left.get(i); ` ` ` ` ``// Appending current track to ` ` ``// seek sequence ` ` ``seek_sequence.add(cur_track); ` ` ` ` ``// Calculate absolute distance ` ` ``distance = Math.abs(cur_track - head); ` ` ` ` ``// Increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// Accessed track is now the new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``System.out.println(``"Total number of seek "` ` ``+ ``"operations = "` `+ seek_count); ` ` ` ` ``System.out.println(``"Seek Sequence is"``); ` ` ` ` ``for` `(``int` `i = ``0``; i < seek_sequence.size(); i++) { ` ` ``System.out.println(seek_sequence.get(i)); ` ` ``} ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String[] args) ``throws` `Exception ` ` ``{ ` ` ` ` ``// Request array ` ` ``int` `arr[] = { ``176``, ``79``, ``34``, ``60``, ``92``, ``11``, ``41``, ``114` `}; ` ` ``int` `head = ``50``; ` ` ` ` ``System.out.println(``"Initial position of head: "` ` ``+ head); ` ` ` ` ``CSCAN(arr, head); ` ` ``} ` `} ` ` ` `// This code is contributed by divyesh072019`
## Python3
`# Python3 program to demonstrate ` `# C-SCAN Disk Scheduling algorithm ` `size ``=` `8` `disk_size ``=` `200` ` ` ` ` `def` `CSCAN(arr, head): ` ` ` ` ``seek_count ``=` `0` ` ``distance ``=` `0` ` ``cur_track ``=` `0` ` ``left ``=` `[] ` ` ``right ``=` `[] ` ` ``seek_sequence ``=` `[] ` ` ` ` ``# Appending end values ` ` ``# which has to be visited ` ` ``# before reversing the direction ` ` ``left.append(``0``) ` ` ``right.append(disk_size ``-` `1``) ` ` ` ` ``# Tracks on the left of the ` ` ``# head will be serviced when ` ` ``# once the head comes back ` ` ``# to the beginning (left end). ` ` ``for` `i ``in` `range``(size): ` ` ``if` `(arr[i] < head): ` ` ``left.append(arr[i]) ` ` ``if` `(arr[i] > head): ` ` ``right.append(arr[i]) ` ` ` ` ``# Sorting left and right vectors ` ` ``left.sort() ` ` ``right.sort() ` ` ` ` ``# First service the requests ` ` ``# on the right side of the ` ` ``# head. ` ` ``for` `i ``in` `range``(``len``(right)): ` ` ``cur_track ``=` `right[i] ` ` ` ` ``# Appending current track ` ` ``# to seek sequence ` ` ``seek_sequence.append(cur_track) ` ` ` ` ``# Calculate absolute distance ` ` ``distance ``=` `abs``(cur_track ``-` `head) ` ` ` ` ``# Increase the total count ` ` ``seek_count ``+``=` `distance ` ` ` ` ``# Accessed track is now new head ` ` ``head ``=` `cur_track ` ` ` ` ``# Once reached the right end ` ` ``# jump to the beginning. ` ` ``head ``=` `0` ` ` ` ``# adding seek count for head returning from 199 to 0 ` ` ``seek_count ``+``=` `(disk_size ``-` `1``) ` ` ` ` ``# Now service the requests again ` ` ``# which are left. ` ` ``for` `i ``in` `range``(``len``(left)): ` ` ``cur_track ``=` `left[i] ` ` ` ` ``# Appending current track ` ` ``# to seek sequence ` ` ``seek_sequence.append(cur_track) ` ` ` ` ``# Calculate absolute distance ` ` ``distance ``=` `abs``(cur_track ``-` `head) ` ` ` ` ``# Increase the total count ` ` ``seek_count ``+``=` `distance ` ` ` ` ``# Accessed track is now the new head ` ` ``head ``=` `cur_track ` ` ` ` ``print``(``"Total number of seek operations ="``, ` ` ``seek_count) ` ` ``print``(``"Seek Sequence is"``) ` ` ``print``(``*``seek_sequence, sep``=``"\n"``) ` ` ` `# Driver code ` ` ` ` ` `# request array ` `arr ``=` `[``176``, ``79``, ``34``, ``60``, ` ` ``92``, ``11``, ``41``, ``114``] ` `head ``=` `50` ` ` `print``(``"Initial position of head:"``, head) ` ` ` `CSCAN(arr, head) ` ` ` `# This code is contributed by rag2127 `
## C#
`// C# program to demonstrate ` `// C-SCAN Disk Scheduling algorithm ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG { ` ` ` ` ``static` `int` `size = 8; ` ` ``static` `int` `disk_size = 200; ` ` ` ` ``static` `void` `CSCAN(``int``[] arr, ``int` `head) ` ` ``{ ` ` ``int` `seek_count = 0; ` ` ``int` `distance, cur_track; ` ` ` ` ``List<``int``> left = ``new` `List<``int``>(); ` ` ``List<``int``> right = ``new` `List<``int``>(); ` ` ``List<``int``> seek_sequence = ``new` `List<``int``>(); ` ` ` ` ``// Appending end values which has ` ` ``// to be visited before reversing ` ` ``// the direction ` ` ``left.Add(0); ` ` ``right.Add(disk_size - 1); ` ` ` ` ``// Tracks on the left of the ` ` ``// head will be serviced when ` ` ``// once the head comes back ` ` ``// to the beginning (left end). ` ` ``for` `(``int` `i = 0; i < size; i++) { ` ` ``if` `(arr[i] < head) ` ` ``left.Add(arr[i]); ` ` ``if` `(arr[i] > head) ` ` ``right.Add(arr[i]); ` ` ``} ` ` ` ` ``// Sorting left and right vectors ` ` ``left.Sort(); ` ` ``right.Sort(); ` ` ` ` ``// First service the requests ` ` ``// on the right side of the ` ` ``// head. ` ` ``for` `(``int` `i = 0; i < right.Count; i++) { ` ` ``cur_track = right[i]; ` ` ` ` ``// Appending current track to seek sequence ` ` ``seek_sequence.Add(cur_track); ` ` ` ` ``// Calculate absolute distance ` ` ``distance = Math.Abs(cur_track - head); ` ` ` ` ``// Increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// Accessed track is now new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``// Once reached the right end ` ` ``// jump to the beginning. ` ` ``head = 0; ` ` ` ` ``// adding seek count for head returning from 199 to ` ` ``// 0 ` ` ``seek_count += (disk_size - 1); ` ` ` ` ``// Now service the requests again ` ` ``// which are left. ` ` ``for` `(``int` `i = 0; i < left.Count; i++) { ` ` ``cur_track = left[i]; ` ` ` ` ``// Appending current track to ` ` ``// seek sequence ` ` ``seek_sequence.Add(cur_track); ` ` ` ` ``// Calculate absolute distance ` ` ``distance = Math.Abs(cur_track - head); ` ` ` ` ``// Increase the total count ` ` ``seek_count += distance; ` ` ` ` ``// Accessed track is now the new head ` ` ``head = cur_track; ` ` ``} ` ` ` ` ``Console.WriteLine(``"Total number of seek "` ` ``+ ``"operations = "` `+ seek_count); ` ` ` ` ``Console.WriteLine(``"Seek Sequence is"``); ` ` ` ` ``for` `(``int` `i = 0; i < seek_sequence.Count; i++) { ` ` ``Console.WriteLine(seek_sequence[i]); ` ` ``} ` ` ``} ` ` ` ` ``// Driver code ` ` ``static` `void` `Main() ` ` ``{ ` ` ` ` ``// Request array ` ` ``int``[] arr = { 176, 79, 34, 60, 92, 11, 41, 114 }; ` ` ``int` `head = 50; ` ` ` ` ``Console.WriteLine(``"Initial position of head: "` ` ``+ head); ` ` ` ` ``CSCAN(arr, head); ` ` ``} ` `} ` ` ` `// This code is contributed by divyeshrabadiya07`
## Javascript
``
Output
```Initial position of head: 50
Total number of seek operations = 389
Seek Sequence is
60
79
92
114
176
199
0
11
34
41```
My Personal Notes arrow_drop_up
Related Articles | 5,166 | 16,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.902555 |
https://lists.boost.org/boost-commit/2007/08/1720.php | 1,718,854,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861880.60/warc/CC-MAIN-20240620011821-20240620041821-00586.warc.gz | 313,163,771 | 3,975 | # Boost-Commit :
From: asutton_at_[hidden]
Date: 2007-08-27 12:01:47
Author: asutton
Date: 2007-08-27 12:01:46 EDT (Mon, 27 Aug 2007)
New Revision: 39006
URL: http://svn.boost.org/trac/boost/changeset/39006
Log:
Removed:
sandbox/SOC/2007/graphs/boost/graph/distance.hpp
Deleted: sandbox/SOC/2007/graphs/boost/graph/distance.hpp
==============================================================================
--- sandbox/SOC/2007/graphs/boost/graph/distance.hpp 2007-08-27 12:01:46 EDT (Mon, 27 Aug 2007)
+++ (empty file)
@@ -1,135 +0,0 @@
-// (C) Copyright Andrew Sutton 2007
-//
-// Use, modification and distribution are subject to the
-// Boost Software License, Version 1.0 (See accompanying file
-
-#ifndef BOOST_GRAPH_DISTANCE_HPP
-#define BOOST_GRAPH_DISTANCE_HPP
-
-#include <boost/graph/detail/combine_distances.hpp>
-
-namespace boost
-{
-
-
- template <typename Graph, typename DistanceMap>
- inline typename property_traits<DistanceMap>::value_type
- eccentricity(const Graph& g,
- DistanceMap dist)
- {
- typename property_traits<DistanceMap>::value_type ret(0);
- typename graph_traits<Graph>::vertex_iterator i, end;
- for(tie(i, end) = vertices(g); i != end; ++i) {
- ret = std::max(ret, dist[*i]);
- }
- return ret;
- }
-
- // The computation of eccentricities, radius and diameter are all
- // closely related. Basically, these computations can be run at
- // the same time - compute eccentricities of all vertices, and
- // the radius and diameter of the graph.
-
- template <typename Graph, typename DistanceMatrix, typename EccentricityMap>
- void
- eccentricities(const Graph& g, DistanceMatrix& dist, EccentricityMap ecc)
- {
- typename graph_traits<Graph>::vertex_iterator i, j, end;
- for(tie(i, end) = vertices(g); i != end; ++i) {
- // compute the max eccentricity "in-place"
- typename property_traits<EccentricityMap>::value_type& ei = ecc[*i];
- for(j = vertices(g).first; j != end; ++j) {
- ei = std::max(ei, dist[*i][*j]);
- }
- }
- }
-
- template <typename Graph, typename EccentricityMap>
- inline typename property_traits<EccentricityMap>::value_type
- radius(const Graph& g, EccentricityMap ecc)
- {
- typedef typename property_traits<EccentricityMap>::value_type eccentricity;
-
- eccentricity ret = ecc[*vertices(g).first];
- typename graph_traits<Graph>::vertex_iterator i, end;
- for(tie(i, end) = vertices(g); i != end; ++i) {
- ret = std::min(ret, ecc[*i]);
- }
- return ret;
- }
-
- template <typename Graph, typename EccentricityMap>
- inline typename property_traits<EccentricityMap>::value_type
- diameter(const Graph& g, EccentricityMap ecc)
- {
- typedef typename property_traits<EccentricityMap>::value_type eccentricity;
-
- eccentricity ret = ecc[*vertices(g).first];
- typename graph_traits<Graph>::vertex_iterator i, end;
- for(tie(i, end) = vertices(g); i != end; ++i) {
- ret = std::max(ret, ecc[*i]);
- }
- return ret;
- }
-
- // The following functions are pretty much gimmes once we've computed
- // some of the other properties (like eccentricities, radius, and
- // diameter).
-
- template <typename Graph, typename EccentricityMap, typename Inserter>
- inline void
- EccentricityMap ecc,
- Inserter ins,
- typename property_traits<EccentricityMap>::value_type level)
- {
- typename Graph::vertex_iterator i, end;
- for(tie(i, end) = vertices(g); i != end; ++i) {
- if(ecc[*i] == level) {
- *ins++ = *i;
- }
- }
- }
-
- template <typename Graph, typename EccentricityMap, typename Inserter>
- inline void
- center(const Graph& g,
- typename property_traits<EccentricityMap>::value_type r,
- EccentricityMap ecc,
- Inserter ins)
- {
- }
-
- template <typename Graph, typename EccentricityMap, typename Inserter>
- inline void
- center(const Graph& g,
- EccentricityMap ecc,
- Inserter ins)
- {
- }
-
-
- template <typename Graph, typename EccentricityMap, typename Inserter>
- inline void
- periphery(const Graph& g,
- typename property_traits<EccentricityMap>::value_type d,
- EccentricityMap ecc,
- Inserter ins)
- { | 1,110 | 3,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.391994 |
https://math.answers.com/math-and-arithmetic/How_many_meters_are_in_450cm | 1,721,861,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00265.warc.gz | 346,717,756 | 48,088 | 0
# How many meters are in 450cm?
Updated: 9/16/2023
Wiki User
9y ago
100cm = 1.00 meter
200cm = 2.00 meters
300cm = 3.00 meters
400cm = 4.00 meters
450cm = 4.50 meters
100 cm = 1 m
→ 450 cm = 450 ÷ 100 m = 4.5 m
Wiki User
9y ago
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Q: How many meters are in 450cm?
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Related questions
4.5m
### How many centimeters are in 4.50 meters?
4.50 meters = 450cm @100cm per meter.
### What is 450cm in meters?
450 centimeters is 4.5 meters.
### How long is 4.5 meters?
it is 4.5 metres long - 450cm
### How many meters 1 km and 450 cm?
1 km = 1000m 1 cm = 1/100m Therefore, 450cm=450*1/100m=4.5m Thus,1km and 450cm=1000m+4.5m=1004.5m
177.16 inches.
450cm
### What is 450 centimeters to 10 meters?
450cm is 4.5 metres. So there are 2 and 2/9 times in 10 metres
### What is 16 percent of 450cm?
16% of 450cm= 16% * 450= 0.16 * 450= 72cm
### How many cm does 4.5 m equal?
450cm (100cm per meter).
### What is 50 feet x 150 feet in square meters?
1ft=30cm 50ft=150cm=0.15m 150ft=450cm=0.45m 0.15X0.45= 0.0675 | 433 | 1,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.84983 |
https://www.jiskha.com/display.cgi?id=1325774330 | 1,503,136,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00408.warc.gz | 890,741,676 | 3,899 | # Physics
posted by .
A 200 g mass is attached to a spring of spring constant k. The spring is compressed 15 cm from its equilibrium value. When released the mass reaches a speed of 5 m/s. What is the spring constant (in N/m)?
• Physics -
w*(amplitude) = maximum velocity
Solve for w (in radians/second) and then use
w = sqrt(k/m) to get k, the spring constant
Amplitudes and velocities must be in meters and m/s. Mass must be in kg
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http://oeis.org/A162154 | 1,550,500,755,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486936.35/warc/CC-MAIN-20190218135032-20190218161032-00284.warc.gz | 200,863,246 | 3,733 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A162154 Odd terms are the number of consecutive prime numbers until a composite, even terms are the number of consecutive composite numbers until a prime. 2
2, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 5, 1, 1, 1, 5, 1, 3, 1, 1, 1, 3, 1, 5, 1, 5, 1, 1, 1, 5, 1, 3, 1, 1, 1, 5, 1, 3, 1, 5, 1, 7, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 13, 1, 3, 1, 5, 1, 1, 1, 9, 1, 1, 1, 5, 1, 5, 1, 3, 1, 5, 1, 5, 1, 1, 1, 9, 1, 1, 1, 3, 1, 1, 1, 11, 1, 11, 1, 3, 1, 1, 1, 3, 1, 5, 1, 1, 1, 9 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(1)=2 because there are two consecutive primes (2 and 3), a(2)=1 (4), a(6)=3 (8,9,10). LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 FORMULA a(2n)=A046933(n+1). - R. J. Mathar, Jun 27 2009 MATHEMATICA Join[{2}, Riffle[Last[#]-First[#]-1&/@Partition[Prime[Range[2, 60]], 2, 1], 1]] (* Harvey P. Dale, Jun 04 2012 *) CROSSREFS Sequence in context: A063669 A319734 A211005 * A134505 A076933 A071974 Adjacent sequences: A162151 A162152 A162153 * A162155 A162156 A162157 KEYWORD nonn AUTHOR Claudio Meller, Jun 26 2009 STATUS approved
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Last modified February 18 09:38 EST 2019. Contains 320249 sequences. (Running on oeis4.) | 691 | 1,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-09 | latest | en | 0.596888 |
https://tomcircle.wordpress.com/2013/03/31/logic-pascal-wager/ | 1,620,764,377,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00098.warc.gz | 573,470,938 | 27,706 | # Logic: Pascal Wager
Pascal Wager:
1. We can choose to believe God exists, or we can choose not to so believe.
2. If we reject God and act accordingly, we risk everlasting agony and torment if He does exist (Type I error in Statistics lingo) but enjoy fleeting earthly delights if He doesn’t exist.
3. If we accept God and act accordingly, we risk little if He doesn’t exist (Type II error) but enjoy endless heavenly bliss if He does exist.
4. It’s in our self-interest to accept God’s existence.
5. Therefore God exists!
Mathematical Proof:
Pascal assumed
Probability of God exists = p
Probability God doesn’t exist = 1-p
You lead 2 lives, either Worldly (世俗) or Piously （虔,诚) , you get rewards X, Y, infinity or Z, as shown in table below.
In Worldly Life, the Expectation in probability is
Ew = p.X + (1-p).Y
In pious life, the Expectation is
Ep = p.∞ + (1-p).Z
Regardless how big the rewards in X (negative, punishment by God) ,Y, Z (self-sacrifice, negative reward)
Ep is infinitely bigger than Ew.
(Ew is good only if p= 0)
Conclusion: no matter what, pious life is better. | 301 | 1,086 | {"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} | 3.4375 | 3 | CC-MAIN-2021-21 | latest | en | 0.845703 |
https://www.live2tech.com/how-to-solve-system-of-equations-word-problems-a-step-by-step-guide/ | 1,723,630,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00623.warc.gz | 669,849,827 | 43,288 | # How to Solve System of Equations Word Problems: A Step-by-Step Guide
Solving systems of equations word problems can seem like a daunting task, but it’s really just about breaking down the problem into smaller, manageable parts. First, you’ll need to convert the word problem into a set of mathematical equations. Then, use algebraic methods to solve these equations, whether by substitution, elimination, or matrix method. Once you have your solution, interpret it back into the context of the problem to ensure it makes sense.
## How to Solve System of Equations Word Problems
In this section, we’ll go through the steps needed to solve a system of equations word problem. By the end, you’ll know how to translate words into equations and use algebra to find the solution.
### Step 1: Read the Problem Carefully
The first step is to read the word problem thoroughly.
Taking the time to understand what the problem is asking will help you identify the variables and set up your equations correctly. Look for keywords that indicate mathematical operations, such as "total," "difference," or "product."
### Step 2: Identify the Variables
Step 2 involves identifying the unknowns in the problem and assigning variables to them.
For example, if the problem talks about the number of apples and oranges, you can let ( x ) represent the number of apples and ( y ) represent the number of oranges. This makes it easier to set up your equations.
### Step 3: Set Up the Equations
Step 3 is to write down the equations based on the relationships described in the problem.
Translate the words into mathematical expressions. If the problem states that the total number of fruits is 30 and there are twice as many apples as oranges, your equations would be ( x + y = 30 ) and ( x = 2y ).
### Step 4: Solve the Equations
Step 4 involves solving the equations using substitution or elimination methods.
If you have ( x = 2y ) and ( x + y = 30 ), you can substitute ( 2y ) for ( x ) in the second equation, giving you ( 2y + y = 30 ). Simplifying this gives ( 3y = 30 ), so ( y = 10 ).
### Step 5: Interpret the Solution
Step 5 is to interpret the solution back into the context of the problem.
You’ve found that ( y = 10 ). Since ( x = 2y ), it follows that ( x = 20 ). Therefore, there are 20 apples and 10 oranges. Make sure these numbers make sense in the context of the problem.
Once you complete these steps, you’ll have a solution to the word problem. It’s crucial to check your work to ensure everything adds up correctly.
## Tips for Solving System of Equations Word Problems
• Start by reading the problem multiple times to fully understand it.
• Highlight or underline key information and numbers.
• Always define your variables clearly.
• Write down all the equations before attempting to solve them.
• Double-check your solution to make sure it fits the context of the problem.
## Frequently Asked Questions
### What is a system of equations?
A system of equations is a set of two or more equations with the same variables. The solutions are the values that satisfy all equations in the system.
### What are the common methods to solve systems of equations?
The three common methods are substitution, which involves solving one equation for one variable and substituting that value in another equation; elimination, which involves adding or subtracting equations to eliminate a variable; and using matrices.
### How do I know which method to use?
The choice of method often depends on the problem. Substitution is useful when one equation is easily solvable for one variable, while elimination is useful when coefficients of a variable are easily aligned for adding or subtracting.
### Can you solve systems of equations graphically?
Yes, you can plot each equation on a graph. The intersection point(s) of the graphs represent the solution(s).
### What should I do if there is no solution?
If the equations represent parallel lines, there is no solution as the lines never intersect. This is known as an inconsistent system.
## Summary
1. Read the problem carefully.
2. Identify the variables.
3. Set up the equations.
4. Solve the equations.
5. Interpret the solution.
## Conclusion
Solving systems of equations word problems is like being a detective. You gather your clues (the words of the problem), translate them into equations, and use algebraic methods to uncover the solution. It’s a systematic approach that, with practice, becomes second nature.
Understanding how to translate a word problem into a system of equations and solve it is a valuable skill, not just for math class but for real-life situations. Think of it like cracking a code; once you have the key, you can unlock the solution to any problem. So, next time you come across a word problem, don’t panic. Break it down, set up your equations, and solve it step by step. With these tools in your math toolkit, you’re ready to tackle any challenge that comes your way. Happy solving! | 1,061 | 4,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-33 | latest | en | 0.919575 |
https://cs.stackexchange.com/questions/123907/why-does-a-dfa-have-multiple-final-states | 1,716,881,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00234.warc.gz | 161,441,349 | 45,423 | # Why does a DFA have multiple final states?
I am taking a course on programming language design and we got to the DFA part. It is known to be formally defined by a 5-tupple but they did not make it very clear to us why a DFA can have multiple final states. I know that the languages that it can accept are involved, but nothing else.
• The answer, strictly speaking, depends on the definition of a FA you're using. While they're equivalent, some authors say a FA must have at most one final state, while others allow them. As I said, it makes no real difference which you use. Apr 10, 2020 at 14:23
• Not sure why this was downvoted, it's a great conceptual question. Apr 10, 2020 at 14:39
• Apr 11, 2020 at 14:55
• Just to be sure, are you asking why it should be defined to have multiple final states? That's how we're interpreting the question. Apr 14, 2020 at 13:27
A DFA is a machine that reads in its input left to right, and, while reading, keeps track of its internal state. At the end, it has to decide whether to "accept" or "reject" the input based only on whatever internal state it has at the end. The final states are used to indicate which internal states should inform the machine to accept.
The reason we need to have multiple final states, then, is because we might want to accept the input in multiple different scenarios. Here is a simple example. Suppose we want to design a machine that accepts if the input is either ho, hoho, or hohoho (so we want to accept, in total, three possible input strings). Then the "state" of the machine can keep track of what letters we have seen so far: we have 7 states for [nothing], h, ho, hoh, hoho, hohoh, hohoho. If we get a letter that isn't going to be one of these strings (like if the input is haha or asdf), we need a different state to remember that the input was bad, and we can call that state [bad input]. So in total we have 8 states.
Now in this example, we want to accept three different strings: ho, hoho, and hohoho, so we need all three of those states to be final. It turns out that it would be impossible to accept these three strings if we have to have only one final state.
All states (8): [nothing], h, ho, hoh, hoho, hohoh, hohoho, [bad input]
Final states (3): ho, hoho, hohoho
In summary, multiple final states gives us the ability to accept multiple different possible patterns in the input. The above is one example of that, but there are many other examples where it is useful.
• Can't we design a big DFA that only one final state that has equal functionality with the multiple ones? Apr 10, 2020 at 17:38
• @kelalaka No that is not possible. In this example there is no way to have only one final state because the states for ho and hoho need to be different. Apr 10, 2020 at 19:05
• However, the longer answer is that designing such a DFA would be possible if you assume that all input strings are terminated with a special "end of input" symbol, like null-terminated strings. But in the usual definition of a DFA, this representation of strings is not used. Apr 10, 2020 at 19:06
• In this case it's very easy to prove that two different accepting states are needed: hoho and hohoho must both end in an accepting state. And they can't end in the same state, because processing ho from the state of hoho must end in an accepting state (because hohoho is in the language), but processing ho from the state of hohoho must end in a rejecting state (because hohohoho is not in the language. So we have two inputs that must end in different accepting states. Apr 11, 2020 at 15:00
Consider an arbitrary nondeterministic finite automaton.
If it has multiple final states, you can create an epsilon transition from all of the final states to one common final state, remove the "final state" mark from all of the previously final states so that you have only one final state. So, a nondeterministic finite automaton can very well work with just one final state.
However, if you explode a nondeterministic finite automaton into a deterministic finite automaton, you'll find that you no longer can have epsilon transitions. The states of the exploded deterministic finite automaton correspond to sets of states in the nondeterministic finite automaton.
You will find that you need to mark a state as "final state" if at least one of the NFA states in the set is the final state. There may be multiple such states.
So, in summary: an NFA can very well work with just one final state. A DFA can't, unless you want to restrict DFAs arbitrarily. By arbitrarily restricting DFAs, you are creating a situation where there are NFAs that cannot be converted to DFAs, and that there are regular expressions that cannot be accepted by any DFA.
Let me provide a characterization of what languages DFAs with a single accepting state can accept.
Proposition. A language $$L$$ over $$\Sigma$$ is accepted by a DFA with a single accepting state if there exist two regular prefix codes $$A,B$$ such that $$L = AB^*$$. Furthermore, this representation is unique, assuming $$L \neq \emptyset$$ and $$\epsilon \notin B$$.
(A prefix code is a set of words, none of which is a prefix of another one. It is regular if as a language it is regular.)
Proof. Let $$A$$ be the language of words which move the DFA from its initial state to its final state without transitioning through the final state (if the initial state is also final, $$A = \{\epsilon\}$$), and let $$B$$ be the language of non-empty words which move the DFA from its final state back to itself, again without transitioning through the final state.
The languages $$A,B$$ are clearly regular. To see that they are prefix-free, suppose to the contrary that $$x,y \in A$$, and $$x$$ is a proper prefix of $$y$$. Thus when reading $$y$$, the DFA passes through the accepting state upon reaching $$x$$. The case of $$B$$ is similar.
In the other direction, consider a minimal DFA for $$A$$. All words in $$A$$ are equivalent with respect to the Myhill–Nerode relation. Indeed, if $$x,y \in A$$ then $$x\epsilon,y\epsilon \in A$$ by $$xz,yz \notin A$$ for $$z \neq \epsilon$$ since $$A$$ is prefix-free. Thus the minimal DFA contains a unique accepting state. Similarly, the minimal DFA for $$B$$ contains a unique accepting state. Merging the accepting state of the former DFA with the initial DFA, we get the desired DFA.
Finally, let us show that the decomposition is unique. Given $$L = AB^*$$, we an extract $$A$$ as the set of words in $$L$$ which have no proper prefix in $$L$$. Taking any $$w \in L$$, we can extract $$B$$ as the set of words in $$w^{-1} L$$ which have no proper non-empty prefix in $$w^{-1} L$$. (This step fails if $$A = \emptyset$$.) $$\square$$
In particular, if $$L$$ contains two words $$x \neq y$$ such that $$x$$ is a prefix of $$y$$ then $$L$$ is infinite (since $$B$$ must contain a word other than $$\epsilon$$). This shows that $$\{ho,hoho,hohoho\}$$ requires more than one accepting state.
From Myhill–Nerode theory, it is known that the minimal number of accepting states equals the number of Myhill–Nerode equivalence classes of words in the language. In the example $$\{ho,hoho,hohoho\}$$, each word is in its own equivalence class, so exactly three accepting states are needed. (With some work, it should be possible to identify a combinatorial parameter which determines the minimal number of accepting states in any finite language.)
• There seems to be an error in the proposition: $ab^*a$ is accepted by a DFA with one final state, but is not of the given form. More generally any language with a "visible" end of string will be accepted by a DFA with one final state: e.g. $\# L \#$ for any regular language $L$ over $\Sigma \setminus \{\#\}$. Apr 11, 2020 at 15:41
• Right. That's what happens when a proof is not spelled out... Hopefully the current version is correct. I replaced "finite" with "regular". Apr 11, 2020 at 15:53
• +1. Just one other thing: I think the assumption $\epsilon \in B$ is not what you want, since that implies that $B = \{\epsilon\}$. I think you want $\epsilon \notin B$ instead (edited to fix). Apr 11, 2020 at 16:27
• I guess there is another edge case where $L$ is empty. Then $A$ must be empty but there are many possible choices for $B$. The case $L$ finite and nonempty corresponds to $B = \varnothing$, so that is OK. Apr 11, 2020 at 16:29
• Since $S^* = (S \cup \{\epsilon\})^*$, the exact status of $\epsilon \in B$ doesn't matter. Apr 11, 2020 at 16:39 | 2,134 | 8,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.960885 |
https://infomutt.com/c/co/complete_space.html | 1,675,278,148,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00632.warc.gz | 342,522,387 | 5,461 | Main Page | See live article | Alphabetical index
In mathematical analysis, a metric space M is said to be complete if every Cauchy sequence of points in M has a limit in M.
Intuitively, a space is complete if it "doesn't have any holes", if there aren't any "points missing". For instance, the rational numbers are not complete, because √2 is "missing". It is always possible to "fill all the holes", leading to the completion of a given space, as will be explained below.
## Examples
The space Q of rational numbers, with the standard metric given by the absolute value, is not complete. Consider for instance the sequence defined by x1 := 1 and xn+1 := xn/2 + 1/xn. This is a Cauchy sequence of rational numbers, but it does not converge towards any rational limit; in fact, it converges towards the irrational number √2, the square root of two.
The open interval (0,1), again with the absolute value metric, is not complete either. The sequence (1/2, 1/3, 1/4, 1/5, ...) is Cauchy, but does not have a limit in the space. However the closed interval [0,1] is complete; the sequence above has the limit 0 in this interval.
The space R of real numbers and the space C of complex numbers (with the metric given by the absolute value) are complete, and so is Euclidean space Rn. Other normed vector spaces may or may not be complete; those which are, are the Banach spaces.
The space Qp of p-adic numberss are complete for any prime number p. This space completes Q with the p-adic metric in the same way that R completes Q with the usual metric.
If S is an arbitrary set, then the set SN of all sequences in S becomes a complete metric space if we define the distance between the sequences (xn) and (yn) to be 1/N, where N is the smallest index for which xN is distinct from yN, or 0 if there is no such index. This space is homeomorphic to the product of a countable number of copies of the discrete space S.
## Some theorems
Every compact metric space is complete. In fact, a metric space is compact if and only if it is complete and totally bounded.
A subspace of a complete space is complete if and only if it is closed.
If X is a set and M is a complete metric space, then the set B(X,M) of all bounded functions f from X to M is a complete metric space. Here we define the distance in B(X,M) in terms of the distance in M as
If X is a topological space and M is a complete metric space, then the set Cb(X,M) consisting of all continuous bounded functions f from X to M is a closed subspace of B(X,M) and hence also complete.
The Baire category theorem says that every complete metric space is a Baire space. That is, the interior of a union of countably many nowhere dense subsets of the space is empty.
## Completion
For any metric space M, one can construct a complete metric space M' (which is also denoted as M with a bar over it), which contains M as a dense subspace. It has the following universal property: if N is any complete metric space and f is any uniformly continuous function from M to N, then there exists a unique uniformly continuous function f' from M' to N which extends f. The space M' is determined up to isometry by this property, and is called the completion of M.
The completion of M can be constructed as a set of equivalence classes of Cauchy sequences in M. For any two Cauchy sequences (xn)n and (yn)n in M, we may define their distance as
d(x,y) = limn d(xn,yn).
(This limit exists because the real numbers are complete.) This is only a pseudometric, not yet a metric, since two different Cauchy sequences may have the distance 0. But "having distance 0" is an equivalence relation on the set of all Cauchy sequences, and the set of equivalence classes is a metric space, the completion of M.
Cantor's contsruction of the real numbers is a special case of this; the real numbers are the completion of the rational numbers using the ordinary absolute value to measure distances. By using different notions of distance on the rationals, one obtains different incomplete metric spaces whose completions are the p-adic numberss.
If this completion procedure is applied to a normed vector space, one obtains a Banach space containing the original space as a dense subspace, and if it is applied to an inner product space, one obtains a Hilbert space containing the original space as a dense subspace.
## Topologically complete spaces
Note that completeness is a property of the metric and not of the topology, meaning that a complete metric space can be homeomorphic to a non-complete one. An example is given by the real numbers, which are complete but homeomorphic to the open interval (0,1), which is not complete. Another example is given by the irrational numbers, which are not complete as a subspace of the real numbers but are homeomorphic to NN (a special case of an example in Examples above).
In topology one considers topologically complete (or completely metrizable) spaces, spaces for which there exists at least one complete metric inducing the given topology. Completely metrizable spaces can be characterized as those spaces which can be written as an intersection of countably many open subsets of some complete metric space. Since the conclusion of the Baire category theorem is purely topological, it applies to these spaces as well.
## Generalisations
It is also possible to define the concept of completeness for uniform spaces using Cauchy netss instead of Cauchy sequences. If every Cauchy net has a limit in X, then X is called complete. One can also construct a completion for an arbitrary uniform space similar to the completion of metric spaces. The most general situation in which Cauchy nets apply is Cauchy spaces; these too have a notion of completeness and completion just like uniform spaces.
A topological space may be completely uniformisable without being completely metrisable; it is then still not topologically complete. | 1,328 | 5,919 | {"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} | 3.8125 | 4 | CC-MAIN-2023-06 | latest | en | 0.912599 |
http://www.bristol.ac.uk/maths/undergraduate/units1819/levelm7units/calculus-of-variations-34-mathm0015/ | 1,561,010,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999141.54/warc/CC-MAIN-20190620044948-20190620070948-00201.warc.gz | 208,532,050 | 7,259 | # Calculus of Variations 34
## Unit aims
To introduce students to calculus of variations and use it to solve basic problems arising in physics, mathematics and materials science.
## Unit description
Calculus of Variations is an important branch of optimization that deals with finding extrema of the functionals in certain functional spaces. It has deep relation with various fields in natural sciences, including differential geometry, ordinary and partial differential equations, materials science, mathematical biology, etc. It is one of the oldest and yet one of the most used tools for investigation of the problems involving free energy. The aim of this course is to present the basics of the calculus of variations, including 1D theory and its application to various problems arising in natural sciences.
## Learning objectives
After taking this unit, students will:
1. Understand the basics of the calculus of variations
2. Will be able to analyze and solve various variational problems arising in physics
## Syllabus
1. Basic concepts of the calculus of variations: Definitions: functionals, extremum, variations, function spaces. Necessary conditions for an extremum. Euler-Lagrange equations. Convexity and it's role in minimization. Minimization under constraints. Existence and nonexistence of minimizers. Basic examples: Brachistochrone problem, Isoperimetric problem, Geodesics on the surface.
2. Generalizations: Higher derivatives. Functions of several variables. Least action principle. Basic examples: vibrating rod, vibrating membrane.
3. Second variation and local minimality/stability: Second variation. Legendre Condition. Relation between local stability and local minimality.
4. Direct methods in the calculus of variations. Minimizing sequences. Ritz method and method of finite differences.
5. Hamilton-Jacobi theory. Geometric Optics. Eikonal. Hamilton-Jacobi equations (if time permits)
I M Gelfand and S V Fomin, Calculus of Variations, Prentice-Hall Bruce van
Brunt, The Calculus of Variations, Dover
Unit code: MATHM0015
Level of study: M/7
Credit points: 10
Teaching block (weeks): 2 (19-24)
Lecturer: Dr Yves Tourigny
## Pre-requisites
MATH20901 Multivariable Calculus and MATH20101 Ordinary Differential Equations 2
None
## Methods of teaching
Lectures & homeworks.
## Methods of Assessment
The pass mark for this unit is 50.
The final mark is calculated as follows:
• 100% from a 1 hour 30 minute exam in May/June
NOTE: Calculators are NOT allowed in the examination.
For information resit arrangements, please see the re-sit page on the intranet.
Please use these links for further information on relative weighting and marking criteria.
Further exam information can be found on the Maths Intranet. | 611 | 2,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-26 | latest | en | 0.862476 |
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/14%3A_Waves/14.10%3A_Sample_problems_and_solutions | 1,642,711,567,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00109.warc.gz | 454,364,759 | 25,474 | $$\require{cancel}$$
# 14.10: Sample problems and solutions
Exercise $$\PageIndex{1}$$
A clarinet can be modeled as an air column that is open at one end and closed at the other end, as in Figure $$\PageIndex{1}$$.
1. Draw the first three harmonics for a clarinet (draw the maximum displacement of the air molecules as a function of distance in the clarinet).
2. Find an expression for the wavelength of the $$n^{th}$$ harmonic for a clarinet of length $$L$$.
3. If a clarinet is $$60\text{cm}$$ long, what is the lowest frequency note it can produce?
a. The first three harmonics are are shown in Figure $$\PageIndex{2}$$.
b. The equation for a standing wave is: \begin{aligned} D(x,t)=2A\sin(kx)cos(\omega t)\end{aligned} We let the fixed end be at $$x=0$$. At the fixed end, the displacement is equal to zero. At the free end ($$x=L$$) the displacement is maximized. The first condition is always true. The second condition will be met when: \begin{aligned} \sin(kL)&=1\\ \therefore kL&=\pi/2,3\pi/2,...\\\end{aligned} This condition can be expressed as: \begin{aligned} kL&=\frac{(2n-1)\pi}{2}\\ \frac{2\pi L}{\lambda}&=\frac{(2n-1)\pi}{2}\\ \therefore \lambda&=\frac{4L}{2n-1}\end{aligned} where, in the second line, we used $$k=2\pi /\lambda$$. We can check that this formula works for the first three harmonics: \begin{aligned} n=1: \quad \lambda&=\frac{4L}{2(1)-1} \\ L&=\frac{1}{4}\lambda \\ n=2: \quad \lambda&=\frac{4L}{2(2)-1} \\ L&= \frac{3}{4}\lambda \\ n=3: \quad \lambda&=\frac{4L}{2(3)-1} \\ L&= \frac{5}{4}\lambda \end{aligned} Referring back to our diagram (Figure $$\PageIndex{2}$$), we can see that our formula holds true for the first three harmonics (i.e. for the first harmonic, the length of the clarinet is equal to $$1/4$$ of a wavelength, etc.)
c. We found that the wavelength for the $$n^{th}$$ wavelength is given by: \begin{aligned} \lambda=\frac{4L}{2n-1}\end{aligned} Writing $$\lambda$$ in terms of the velocity, $$v$$, and frequency, $$f$$, gives: \begin{aligned} \frac{v}{f}&=\frac{4L}{2n-1}\\ \therefore f&=\frac{v(2n-1)}{4L}\end{aligned} From this formula, we can see that, if we want to find the lowest frequency, we want $$n=1$$. The length of the clarinet is $$0.6\text{m}$$, and $$v$$ is the speed of sound in air which is $$343\text{m/s}$$ at room temperature. Using these values, the lowest frequency is: \begin{aligned} f&=\frac{(343\text{m/s})(2(1)-1)}{4(0.6\text{m})}\\ f&=143\text{Hz}\end{aligned} Discussion: This frequency is close to the $$D_3$$ note, which has a frequency of $$144\text{Hz}$$, so this answer makes sense. However, the value we found differs from the true value. Why might this be?
Exercise $$\PageIndex{2}$$
A pulse propagates down a rope of mass per unit length $$\mu_1$$ that is tied to a second rope with a mass per unit length $$\mu_2$$ (Figure $$\PageIndex{3}$$). The tensions in the ropes are equal in magnitude.
1. Write the displacements of the incident pulse, the reflected pulse, and the transmitted pulse in the form $$D(x,t)=D(a(t\pm x/v))$$, where $$a$$ is some constant that you need to determine, and the choice of $$+$$ or $$-$$ depends on the direction that the pulse is traveling in.
2. The reflection coefficient, $$R$$, is the ratio of the amplitude of the reflected pulse to the amplitude of the incident pulse. Using the boundary conditions, show that the reflection coefficient is given by: \begin{aligned} R=\frac{\sqrt{\mu_1}-\sqrt{\mu_2}}{\sqrt{\mu_1}+\sqrt{\mu_2}}\end{aligned}
Note: The boundary is the interface between the two ropes. By “using the boundary conditions”, we mean that you should think about what must be true at the boundary for this problem to make sense. Boundary conditions are often more obvious than you think!
a. We let the incident pulse move in the positive $$x$$ direction (Figure $$\PageIndex{4}$$), and set $$x=0$$ to be where the ropes connect.
The incident pulse (denoted by $$i$$) is a traveling wave, moving in one dimension in the positive $$x$$ direction. The incident pulse can thus be described by the function: \begin{aligned} D_I(x,t)=A_I\cos(k_1x-\omega t)\\\end{aligned} We will use the formulas $$k=2\pi/\lambda$$ and $$\omega=2\pi f$$ to rewrite this equation in the form $$D=(a(t\pm x/v))$$. The frequency, $$f$$, of the wave will be the same in both ropes. The velocity of the wave, and therefore its wavelength, depends on the mass density of the rope. Since the incident wave travels through the first rope ($$\mu_1$$), its velocity will be $$v_1$$ and its wavelength will be $$\lambda_1$$. The incident wave can thus be described by: \begin{aligned} D_I&=A_I\cos\left( \frac{2\pi}{\lambda_1}x-2\pi ft\right)\\ &=A_I\cos \left( 2\pi\left(\frac{1}{\lambda_1}x- ft\right)\right)\\ &=A_I\cos \left( 2\pi f\left(\frac{x}{v_1}- t\right)\right)\\ &=A_I\cos \left( -2\pi f\left(t-\frac{x}{v_1}\right)\right)\\ D_I&=A_I\cos \left(2\pi f\left(t-\frac{x}{v_1}\right)\right)\end{aligned} where we used $$v=f\lambda$$, and noted that $$\cos(-x)=\cos(x)$$.
The transmitted wave (denoted by the subscript $$T$$) will also travel in the positive $$x$$ direction, but its speed will be $$v_2$$, since it travels through the second rope: \begin{aligned} D_T&=A_T\cos \left( 2\pi f\left(t-\frac{x}{v_2}\right)\right)\end{aligned} The reflected wave (denoted by $$R$$) will travel in the $$-x$$ direction and at the same speed as the incident pulse. \begin{aligned} D_R&=A_R\cos \left( 2\pi f\left(t+\frac{x}{v_1}\right)\right)\end{aligned}
b. We will consider the boundary conditions at the interface between the two ropes. One boundary condition is that the rope must be continuous. As a result, the vertical displacement on the $$-x$$ side of the boundary must be the same as the vertical displacement on the $$+x$$ side of the boundary at every instant: \begin{aligned} D_{-x}&=D_{+x}\quad\text{at x=0}\end{aligned} The amplitude on the $$+x$$ side is equal to the amplitude of the transmitted pulse. For the $$-x$$ side of the boundary, we have to take into account that the incident and reflected pulses will superimpose (when the front of the incident pulse reaches the boundary, it will be reflected and interfere with the end of the incident pulse). This boundary condition can thus be expressed as: \begin{aligned} A_I+A_R&=A_T\end{aligned} The slope of the rope must also be continuous at the boundary. Since the incident and reflected pulses superimpose, and the principle of superposition states that the net displacement is the sum of the displacement of these two waves, we can write:
\begin{aligned} \frac{\partial}{\partial x}(D_I+D_R)\Bigr|_{x=0}&=\frac{\partial}{\partial x}D_T\Bigr|_{x=0}\\ \frac{\partial}{\partial x}D_I\Bigr|_{x=0}+\frac{\partial}{\partial x}D_R\Bigr|_{x=0}&=\frac{\partial}{\partial x}D_T\Bigr|_{x=0}\end{aligned}
Using our equations for the incident, transmitted, and reflected pulses found in part a), and taking the appropriate partial derivatives, this equation becomes:
\begin{aligned} (A_I/v_1) \sin \left(2\pi f\left( t-\frac{x}{v_1}\right)\right)\Bigr|_{x=0}+(-A_R/v_1) \sin \left( 2\pi f\left( t+\frac{x}{v_1}\right)\right)\Bigr|_{x=0}&=\A_T/v_2) \sin \left( 2\pi f\left( t-\frac{x}{v_2}\right)\right)\Bigr|_{x=0}\end{aligned} Evaluating at \(x=0 gives: \begin{aligned} (A_I/v_1) \sin (2\pi ft) +(-A_R/v_1) \sin (2\pi ft)&=(A_T/v_2) \sin (2\pi ft)\\ \frac{A_I}{v_1} -\frac{A_R}{v_1}&=\frac{A_T}{v_2} \end{aligned} Using our first condition, $$A_I+A_R=A_T$$, we get: \begin{aligned} \frac{A_I}{v_1} -\frac{A_R}{v_1}&=\frac{A_I}{v_2}+\frac{A_R}{v_2}\\\end{aligned} Now, we can rearrange to find the reflection coefficient, $$R=A_R/A_I$$: \begin{aligned} A_I\left( \frac{v_2-v_1}{v_1v_2}\right)&=A_R\left( \frac{v_2+v_1}{v_1v_2}\right)\\ R&=\frac{v_2-v_1}{v_2+v_1}\end{aligned} Since the velocities in the first and second rope are $$v_1=\sqrt{F_T/\mu_1}$$ and $$v_2=\sqrt{F_T/\mu_2}$$, respectively, the reflection coefficient can be written as: \begin{aligned} R&=\frac{\sqrt{\frac{F_T}{\mu_2}}-\sqrt{\frac{F_T}{\mu_1}}}{\sqrt{\frac{F_T}{\mu_2}}+\sqrt{\frac{F_T}{\mu_1}}}\\ &=\frac{\sqrt{F_T}}{\sqrt{F_T}}\cdot \frac{\frac{1}{\sqrt{\mu_2}}-\frac{1}{\sqrt{\mu_1}}}{\frac{1}{\sqrt{\mu_2}}+\frac{1}{\sqrt{\mu_1}}}\\ \therefore R&=\frac{\sqrt{\mu_1}-\sqrt{\mu_2}}{\sqrt{\mu_1}+\sqrt{\mu_2}}\end{aligned} as desired. | 2,708 | 8,300 | {"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} | 4.3125 | 4 | CC-MAIN-2022-05 | latest | en | 0.756732 |
https://fornoob.com/a-0-050-m-solution-of-the-weak-acid-ha-has-h3o-3-77-x-10-4-m-what-is-the-ka-for-the-acid/ | 1,669,910,254,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710829.5/warc/CC-MAIN-20221201153700-20221201183700-00445.warc.gz | 297,176,804 | 18,191 | # A 0.050 M solution of the weak acid HA has [H3O+] = 3.77 × 10-4 M. What is the Ka for the acid?
• HA <–> H^+1 + A^-1
Ka = [H+][A-]/[HA]
[H+] = 3.77×10^-4 M
therefore [A-] = 3.77×10^-4 M
and [HA] = 0.050 – 3.77×10^-4 = 0.0496 M
Ka = (3.77×10^-4)^2 / 0.0496 = 2.87 x 10^-6
Ka = 2.9×10^-6 | 161 | 294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.794099 |
https://elamarthistory.com/sculpture/often-asked-how-much-to-charge-for-a-custom-sculpture.html | 1,656,224,152,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037089.4/warc/CC-MAIN-20220626040948-20220626070948-00244.warc.gz | 273,159,505 | 16,039 | # Often asked: How Much To Charge For A Custom Sculpture?
There is just one major problem with commissioning – it costs a lot more than buying statues in stock. Commissioning a bronze statue can cost anywhere between \$10,000 to over \$100,000 depending on your specifications. The much more popular alternative is to buy ready-made.
## How much does a custom stone sculpture cost?
The stone sculptures range in price from \$5,000 to \$60,000. Pricing is based on the complexity of the design and the size of the piece. Granite, sandstone and marble demand special tools (see Tools and Stone) to carve and shape and require many hours of work to arrive at the finished piece.
## How do you price custom artwork?
Multiply the painting’s width by its length to arrive at the total size, in square inches. Then multiply that number by a set dollar amount that’s appropriate for your reputation. I currently use \$6 per square inch for oil paintings. Then calculate your cost of canvas and framing, and then double that number.
You might be interested: Question: What Is The Price Of Admission To Frederik Meijer Gardens & Sculpture Park?
## How should I price my art?
Pay yourself a reasonable hourly wage, add the cost of materials and make that your asking price. For example, if materials cost \$50, you take 20 hours to make the art, and you pay yourself \$20 an hour to make it, then you price the art at \$450 (\$20 X 20 hours + \$50 cost of materials).
## How much should I charge for art commissions?
5-7 hours * 10\$hr = \$50-\$70 for a full colored 1 character commission. For traditional, also figure in the cost of supplies it took for you to create the image, so you’d estimate the base time + cost of materials.
## How do you commission a statue?
How to Commission a Piece of Custom Artwork
1. Choosing the artist or gallery.
2. Holding several meetings with the artist (topics include reviewing themes and sketches, statue size, project budget, materials used, and projected deadline).
3. Reviewing a clay model of the sculpture and providing necessary feedback.
## How much is the Statue of Liberty worth?
With 31 tons of copper and 125 tons of steel, the scrap value of the Statue of Liberty comes in at \$227,610, far below two of the most expensive statues in the world. But that’s what happens when you use millions worth of gold and bronze.
## How do you price crafts?
Here’s the Craft Pricing Formula
1. Cost of Supplies + Labor + 10-15% Overhead = Total Costs.
2. Total Costs x 2 = Wholesale Price.
3. Wholesale Price x 2 = Retail Price.
## How do you price your work?
Business schools teach a standard formula for determining an hourly rate: Add up your labor and overhead costs, add the profit you want to earn, then divide the total by your hours worked. This is the minimum you must charge to pay your expenses, pay yourself a salary, and earn a profit.
You might be interested: Question: How To Build A Plane Sculpture Out Of Popsicle Sticks?
## How do you price handmade items?
In her Tips for Pricing your Handmade Goods blog on Craftsy, artesian entrepreneur Ashley Martineau suggests this formula:
1. Cost of supplies + \$10 per hour time spent = Price A.
2. Cost of supplies x 3 = Price B.
3. Price A + Price B divided by 2 (to get the average between these two prices) = Price C.
## How do you price art per square inch?
Obtain your square inch price by multiplying length times width and dividing your selling price by the total number of square inches. For example, a 12″ x 16″ painting that you sell for \$400 will cost \$2.08 per square inch. Subsequently, your 24 x 36″ painting will be \$2.08 x 864 square inches = \$1,797.
## What is a good hourly rate for an artist?
The average hourly wage for an Artist in the United States is \$27 as of September 27, 2021, but the range typically falls between \$23 and \$32. Hourly rate can vary widely depending on many important factors, including education, certifications, additional skills, the number of years you have spent in your profession.
## Is my art good enough to sell?
It’s important to realize that sales are not based on how good you think your drawings are. If you are getting validation in the form of likes, comments, and followers, you are good enough to be selling your work. But to get sales you actually have to make sales. Just making art is not enough.
## How do you price art for beginners?
So, if a piece took you 10 hours to make, you want to get \$15 per hour, and the materials cost you \$45, you could use \$195 as your starting point (10 times 15, plus 45). Cost of materials would include your canvas, paper, paint, ink, and so forth.
You might be interested: FAQ: How Is The Baroque Style In Sculpture And Painting Established By Bernini And Caravaggi?
## How do you determine the price of a product?
To calculate your product selling price, use the formula:
1. Selling price = cost price + profit margin.
2. Average selling price = total revenue earned by a product ÷ number of products sold. | 1,150 | 5,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-27 | longest | en | 0.934576 |
http://slideplayer.com/slide/4282824/ | 1,521,301,361,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645177.12/warc/CC-MAIN-20180317135816-20180317155816-00134.warc.gz | 283,848,645 | 23,180 | # 1 Multiple Frame Surveys Tracy Xu Kim Williamson Department of Statistical Science Southern Methodist University.
## Presentation on theme: "1 Multiple Frame Surveys Tracy Xu Kim Williamson Department of Statistical Science Southern Methodist University."— Presentation transcript:
1 Multiple Frame Surveys Tracy Xu Kim Williamson Department of Statistical Science Southern Methodist University
2 Multiple Frame Surveys Introduction Introduction – What is Multiple Frame Survey – What is Multiple Frame Survey Different estimators for population total Different estimators for population total Variance Estimators for those estimators Variance Estimators for those estimators Conclusion Conclusion References References
3 Introduction Hartley (1962) Hartley (1962) Multiple frame surveys refers to two or more frames that can cover a target population Multiple frame surveys refers to two or more frames that can cover a target population Very useful for sampling rare or hard-to-reach populations Very useful for sampling rare or hard-to-reach populations Dual frame design may result in considerable cost savings over a single frame design with comparable precision Dual frame design may result in considerable cost savings over a single frame design with comparable precision
4 Example 1 – Cost Reduction Agriculture [Hartley 1962, 1974] Agriculture [Hartley 1962, 1974] + List frame (incomplete, names, addresses) - Less costly + Area frame (complete, insensitive to changes) - Expensive to sample + Can achieve the same precision Linear Cost Function Linear Cost Function C = n A c A + n B c B
5 Example 2 – Rare Populations AIDS [Kalton and Anderson 1986] AIDS [Kalton and Anderson 1986] + Using a general population frame as well as std clinics, drug treatment centers, and hospitals Homeless [Iachan and Dennis 1993] Homeless [Iachan and Dennis 1993] + Frames: homeless shelters, soup kitchens, and street areas Alzheimer’s Alzheimer’s + Frames: general population and adult day-care centers
6 Issues to Consider Statisticians must address the following issues Statisticians must address the following issues + How should the information from the samples be combined to estimate samples be combined to estimate population quantities? population quantities? + How should variance estimates be calculated? calculated?
7 Notations Universe U = A U B = a U ab U b Universe U = A U B = a U ab U b N=# of elements in the population N=# of elements in the population N A = # of elements in Frame A N A = # of elements in Frame A N B = # of elements in Frame B N B = # of elements in Frame B N a = # of elements in Frame A, but not Frame B N a = # of elements in Frame A, but not Frame B N b = # of elements in Frame B, but not Frame A N b = # of elements in Frame B, but not Frame A N ab = # of elements in Frame A & Frame B N ab = # of elements in Frame A & Frame B S A = P{ i th element is in S} = π A i S A = P{ i th element is in S} = π A i Y = population total = Y a + Y b + Y ab Y = population total = Y a + Y b + Y ab
8 Estimators Hartley (H) Hartley (H) Fuller and Burmeister (FB) Fuller and Burmeister (FB) Single Frame estimators Single Frame estimators Pseudo-Maximum Likelihood (PML) Pseudo-Maximum Likelihood (PML)
9 Hartley & FB Estimator Minimizes the variance among the class of linear unbiased estimators of Y Minimizes the variance among the class of linear unbiased estimators of Y Have minimum variance for a single response Have minimum variance for a single response Use different set of weights for each response variable Use different set of weights for each response variable Disadvantages: Increased amount of calculations (uses covariances estimated by the data) and possible inconsistencies Disadvantages: Increased amount of calculations (uses covariances estimated by the data) and possible inconsistencies Estimators are not in general linear functions of y Estimators are not in general linear functions of y FB has the greatest asymptotic efficiency FB has the greatest asymptotic efficiency
10 Hartley & FB Estimator
11 Single Frame Estimators Bankier (1986), Kalton & Anderson (1986) and Skinner (1991) Bankier (1986), Kalton & Anderson (1986) and Skinner (1991) Treat all observations as if they had been sampled from a single frame with modified weights for observations in the intersections of frames Treat all observations as if they had been sampled from a single frame with modified weights for observations in the intersections of frames Do not use any auxiliary information about the population totals Do not use any auxiliary information about the population totals Linear in y Linear in y Other techniques may be applied: Regression Estimation and Ranking Ratio Estimation Other techniques may be applied: Regression Estimation and Ranking Ratio Estimation
12 Pseudo-Maximum Likelihood Estimator Skinner and Tao (1996) derived pseudo-ML(PML) estimator for dual frame survey that use the same set of weights for all items of y, similar to “single frame” estimators, and maintain efficiency. Skinner and Tao (1996) derived pseudo-ML(PML) estimator for dual frame survey that use the same set of weights for all items of y, similar to “single frame” estimators, and maintain efficiency. The idea of pseudo-MLE estimation is talked about in Roberts, Rao, Kumar (1987) and Skinner, Holt, and Smith (1989) in which a MLE estimator under simple random sampling is modified to achieve consistent estimation under complex designs. The idea of pseudo-MLE estimation is talked about in Roberts, Rao, Kumar (1987) and Skinner, Holt, and Smith (1989) in which a MLE estimator under simple random sampling is modified to achieve consistent estimation under complex designs.
13 The main advantages of PMLE are that it is design consistent and typically has a simple form. The main advantages of PMLE are that it is design consistent and typically has a simple form. The potential disadvantage is that it may not be asymptotically efficient, although it may be hoped that any loss of efficiency will tend to be small in practice. The potential disadvantage is that it may not be asymptotically efficient, although it may be hoped that any loss of efficiency will tend to be small in practice. Pseudo-Maximum Likelihood Estimator
14 Pseudo-MLE of Y is derived as Pseudo-MLE of Y is derived as and is the smallest root of the quadratic equation Pseudo-Maximum Likelihood Estimator
15 Extensive simulation was done to evaluate the performance of all the estimators in Sharon Lohr and J. N. K Rao(2005) paper Findings: Findings: In all the simulations, the PML method had either the smallest EMSE or an EMSE close to the minimum value. With its high efficiency and ease of computation, as well as the practical advantage of using the same set of weights for all response variables, the PML method appears to be a good choice for estimation in multiple frame survey. In all the simulations, the PML method had either the smallest EMSE or an EMSE close to the minimum value. With its high efficiency and ease of computation, as well as the practical advantage of using the same set of weights for all response variables, the PML method appears to be a good choice for estimation in multiple frame survey. Comparison of All Estimators
16 Findings Findings When Q>=3, the theoretically optimal Fuller- Burmeister and Hartley methods became unstable, because they require solving systems of equations using a large estimated covariance matrix. When Q>=3, the theoretically optimal Fuller- Burmeister and Hartley methods became unstable, because they require solving systems of equations using a large estimated covariance matrix. Comparison of All Estimators
17 Asymptotic Variance Under some conditions, the H, FB and PML estimators are all consistent estimators of the total. Under some conditions, the H, FB and PML estimators are all consistent estimators of the total. And And But neither H estimator or PML estimator is necessarily more efficient than the other.
18 Asymptotic Variance Sharon Lohr and J. N. K. Rao(2005) paper gives a general formula for the asymptotic variance for all above estimators, which can be used to construct optimal designs for multiple frame surveys. Sharon Lohr and J. N. K. Rao(2005) paper gives a general formula for the asymptotic variance for all above estimators, which can be used to construct optimal designs for multiple frame surveys.
19 Variance Estimators Two Methods: Skinner and Rao(1996) described a method for estimating the variance of using Taylor linearization. Skinner and Rao(1996) described a method for estimating the variance of using Taylor linearization. Lohr and Rao(2000) defined jackknife variance estimator for estimators from dual frame surveys and showed that jackknife variance estimator is asymptotically equivalent to the Taylor linearization variance estimator. Lohr and Rao(2000) defined jackknife variance estimator for estimators from dual frame surveys and showed that jackknife variance estimator is asymptotically equivalent to the Taylor linearization variance estimator.
20 Simulation results ( Lohr and Rao 2000) showed that in comparing the linearization estimator, full jackknife and modified jackknife estimators Simulation results ( Lohr and Rao 2000) showed that in comparing the linearization estimator, full jackknife and modified jackknife estimators 1. The jackknife estimator has exhibited smaller bias than the linearization estimator. 2. The relative bias of all three estimators of the variance tends to decrease as the sample size increase. 3. For the smaller sample sizes, the linearization and modified jackknife methods underestimate the EMSE. 4. Coverage probabilities, though similar for the three variance estimators, were slightly higher for the full jackknife. Variance Estimators
21 5. The jackknife methods are less stable than the linearization estimator of the variance as judged by the values of relative standard error. 6. For single frame estimator, the jackknife and linearization estimates of the variance coincide. 7. For the other estimators, both the linearization and modified jackknife estimates of the variance are biased downward. Variance Estimators
22 Conclusion Multiple Frame Surveys can be extremely beneficial when sampling rare populations and when a complete frame is very expensive to sample Multiple Frame Surveys can be extremely beneficial when sampling rare populations and when a complete frame is very expensive to sample Different estimators of the total are proposed. Choice of estimators will depend on survey design and complexity: FB is the most efficient, however due to additional calculations and complexity PML may be preferred Different estimators of the total are proposed. Choice of estimators will depend on survey design and complexity: FB is the most efficient, however due to additional calculations and complexity PML may be preferred
23 References H.O. Hartley (1974), “Multiple Frame Methodology and Selected Applications”, Sankhya, the Indian Journal of Statistics, Series C, 36, 99-118. H.O. Hartley (1974), “Multiple Frame Methodology and Selected Applications”, Sankhya, the Indian Journal of Statistics, Series C, 36, 99-118. C. J. Skinner and J. N. K. Rao(1996), “Estimation in Dual Frame Surveys with Complex Designs”, Journal of the American Statistical Association, 91, 349-356. C. J. Skinner and J. N. K. Rao(1996), “Estimation in Dual Frame Surveys with Complex Designs”, Journal of the American Statistical Association, 91, 349-356. Sharon L. Lohr and J.N.K. Rao(2000), “Inference from Dual Frame Surveys”, Journal of the American Statistical Association, 95, 2710280. Sharon L. Lohr and J.N.K. Rao(2000), “Inference from Dual Frame Surveys”, Journal of the American Statistical Association, 95, 2710280. Sharon L. Lohr and J. N. K. Rao(2006), “Estimation in Multiple-Frame Surveys”, Journal of the American Statistical Association (under revision). Sharon L. Lohr and J. N. K. Rao(2006), “Estimation in Multiple-Frame Surveys”, Journal of the American Statistical Association (under revision). J. Lessler and W. Kalsbeek (1992), Non-sampling Error in Surveys, John Wiley & Sons, Inc.
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Similar presentations | 2,670 | 12,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-13 | latest | en | 0.750566 |
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This is a python program to print all the numbers which are divisible by 3 and 5 from a given interger N. There are numerous ways we can write this program except that we need to check if the number is fully divisble by both 3 and 5.
Below is my code to write a python program to print all the numbers divisible by 3 and 5 −
lower = int(input("Enter lower range limit:"))
upper = int(input("Enter upper range limit:"))
for i in range(lower, upper+1):
if((i%3==0) & (i%5==0)):
print(i)
## Output
Enter lower range limit:0
Enter upper range limit:99
0
15
30
45
60
75
90
Above we try to print all the numbers between 0 and 99 which are divisble by 3 and 5. Same program can be used to print all the number between 0 and 1000 which are divisible by 3 and 5, we just need to alter our range and our output will be something like,
Enter lower range limit:0
Enter upper range limit:1000
0
15
30
45
60
75
90
105
120
135
150
165
180
195
....
....
915
930
945
960
975
990
In case we want to write a program which will print all the numbers in a range divisible by a given number not the fixed number like above, i just need to update by program like,
#Incase we want to print all number between a range divided by any given number
n = int(input("Enter the number to be divided by:"))
for i in range(lower, upper+1):
if(i%n==0):
print(i)
Below the steps to write above code −
• Take the lower and upper limit .i.e. the range from the user.
• Take the number to be divided by from the user. In case of our main problem, because we know that numbers(3 and 5), i write the 3 and 5 in the if statement only.
• Using a loop with &(and) operator statement(so that it print only those numbers which are divisble by both 3 & 5), prints all the factors which is divisible by the number.
• Exit.
karthikeya Boyini
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# LOG10 function
Returns the base-10 logarithm of a number.
## Syntax
LOG10(number)
Number is the positive real number for which you want the base-10 logarithm.
## Examples
Formula
Description (Result)
=LOG10(86)
Base-10 logarithm of 86 (1.934498451)
=LOG10(10)
Base-10 logarithm of 10 (1)
=LOG10(1E5)
Base-10 logarithm of 1E5 (5)
=LOG10(10^5)
Base-10 logarithm of 10^5 (5) | 135 | 405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.695235 |
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Message Subject Math: 6÷2(1+2) = ?
Poster Handle caper_26
Post Content
You do multiplication and division left to right:
6 divided by the quantity 2n
6 ÷ 2n , where n = 2+1
6 ÷ 2n = 3/n; n = 2+1
3/n = 3/(2+1) = 1
OR:
6 ÷ 2(2+1) = 6 ÷ (2(2)+2(1)) = 6 ÷(4+2) = 1
6 ÷ 2(2+1) = 1
6 ÷ (2+1)2 = 1
(6 ÷ 2)(2+1) = [(6÷2)(2) + (6÷2)1)] = [6 + 3] = 9
6(2+1) ÷ 2 = 9
WHY ???
Derive the equations showing all steps. Do not introduce unnecessary ( ) or *
6 ÷ 6 = 1
6 ÷ (4+2) = 1; Factor out the GCF (Greatest common factor of 2)
6 ÷ 2(2+1) = 1
Now for 9:
3 * 3 = 9
3 * (2+1) = 9
6/2 * (2+1) = 9
OR:
6 + 3 = 9
(6/2)2 + (6/2)1 = 9
(6/2)(2+1) = 9
Fractions as coefficients that use '/' , such as (6/2)n, must use parentheses. Otherwise, it is 6/2n which is 6/(2n). | 446 | 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} | 4 | 4 | CC-MAIN-2016-40 | longest | en | 0.773185 |
https://www.datacamp.com/community/tutorials/machine-learning-in-r?utm_campaign=News&utm_medium=Community&utm_source=DataCamp.com | 1,579,268,836,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589560.16/warc/CC-MAIN-20200117123339-20200117151339-00504.warc.gz | 844,115,736 | 49,831 | Tutorials
r programming
+1
# Machine Learning in R for beginners
This small tutorial is meant to introduce you to the basics of machine learning in R: it will show you how to use R to work with KNN.
## Introducing: Machine Learning in R
Machine learning is a branch in computer science that studies the design of algorithms that can learn. Typical machine learning tasks are concept learning, function learning or “predictive modeling”, clustering and finding predictive patterns. These tasks are learned through available data that were observed through experiences or instructions, for example. Machine learning hopes that including the experience into its tasks will eventually improve the learning. The ultimate goal is to improve the learning in such a way that it becomes automatic, so that humans like ourselves don’t need to interfere any more.
This small tutorial is meant to introduce you to the basics of machine learning in R: more specifically, it will show you how to use R to work with the well-known machine learning algorithm called “KNN” or k-nearest neighbors.
## Steps
If you’re interested in following a course, consider checking out our Introduction to Machine Learning with R or DataCamp’s Unsupervised Learning in R course!
## Using R For k-Nearest Neighbors (KNN)
The KNN or k-nearest neighbors algorithm is one of the simplest machine learning algorithms and is an example of instance-based learning, where new data are classified based on stored, labeled instances.
More specifically, the distance between the stored data and the new instance is calculated by means of some kind of a similarity measure. This similarity measure is typically expressed by a distance measure such as the Euclidean distance, cosine similarity or the Manhattan distance.
In other words, the similarity to the data that was already in the system is calculated for any new data point that you input into the system.
Then, you use this similarity value to perform predictive modeling. Predictive modeling is either classification, assigning a label or a class to the new instance, or regression, assigning a value to the new instance. Whether you classify or assign a value to the new instance depends of course on your how you compose your model with KNN.
The k-nearest neighbor algorithm adds to this basic algorithm that after the distance of the new point to all stored data points has been calculated, the distance values are sorted and the k-nearest neighbors are determined. The labels of these neighbors are gathered and a majority vote or weighted vote is used for classification or regression purposes.
In other words, the higher the score for a certain data point that was already stored, the more likely that the new instance will receive the same classification as that of the neighbor. In the case of regression, the value that will be assigned to the new data point is the mean of its k nearest neighbors.
## Step One. Get Your Data
Machine learning usually starts from observed data. You can take your own data set or browse through other sources to find one.
### Built-in Datasets of R
This tutorial uses the Iris data set, which is very well-known in the area of machine learning. This dataset is built into R, so you can take a look at this dataset by typing the following into your console:
eyJsYW5ndWFnZSI6InIiLCJzYW1wbGUiOiJpcmlzIn0=
### UC Irvine Machine Learning Repository
If you want to download the data set instead of using the one that is built into R, you can go to the UC Irvine Machine Learning Repository and look up the Iris data set.
Tip: don’t only check out the data folder of the Iris data set, but also take a look at the data description page!
Then, use the following command to load in the data:
eyJsYW5ndWFnZSI6InIiLCJzYW1wbGUiOiIjIFJlYWQgaW4gYGlyaXNgIGRhdGFcbmlyaXMgPC0gcmVhZC5jc3YodXJsKFwiaHR0cDovL2FyY2hpdmUuaWNzLnVjaS5lZHUvbWwvbWFjaGluZS1sZWFybmluZy1kYXRhYmFzZXMvaXJpcy9pcmlzLmRhdGFcIiksIFxuICAgICAgICAgICAgICAgICBoZWFkZXIgPSBGQUxTRSkgXG5cbiMgUHJpbnQgZmlyc3QgbGluZXNcbmhlYWQoaXJpcylcblxuIyBBZGQgY29sdW1uIG5hbWVzXG5uYW1lcyhpcmlzKSA8LSBjKFwiU2VwYWwuTGVuZ3RoXCIsIFwiU2VwYWwuV2lkdGhcIiwgXCJQZXRhbC5MZW5ndGhcIiwgXCJQZXRhbC5XaWR0aFwiLCBcIlNwZWNpZXNcIilcblxuIyBDaGVjayB0aGUgcmVzdWx0XG5pcmlzIn0=
The command reads the .csv or “Comma Separated Value” file from the website. The header argument has been put to FALSE, which means that the Iris data set from this source does not give you the attribute names of the data.
Instead of the attribute names, you might see strange column names such as “V1” or “V2” when you inspect the iris attribute with a function such as head(). Those are set at random.
To simplify working with the data set, it is a good idea to make the column names yourself: you can do this through the function names(), which gets or sets the names of an object. Concatenate the names of the attributes as you would like them to appear. In the code chunk above, you’ll have listed Sepal.Length, Sepal.Width, Petal.Length, Petal.Width and Species.
Once again, these names don’t come out of the blue: take a look at the description of the data set that is linked above; You’ll normally see all these names listed.
## Step Two. Know Your Data
Now that you have loaded the Iris data set into RStudio, you should try to get a thorough understanding of what your data is about. Just looking or reading about your data is certainly not enough to get started!
You need to get your hands dirty, explore and visualize your data set and even gather some more domain knowledge if you feel the data is way over your head.
Probably you’ll already have the domain knowledge that you need, but just as a reminder, all flowers contain a sepal and a petal. The sepal encloses the petals and is typically green and leaf-like, while the petals are typically colored leaves. For the iris flowers, this is just a little bit different, as you can see in the following picture:
### Initial Overview Of The Data Set
First, you can already try to get an idea of your data by making some graphs, such as histograms or boxplots. In this case, however, scatter plots can give you a great idea of what you’re dealing with: it can be interesting to see how much one variable is affected by another.
In other words, you want to see if there is any correlation between two variables.
You can make scatterplots with the ggvis package, for example.
Note that you first need to load the ggvis package:
# Load in ggvis
library(ggvis)
# Iris scatter plot
iris %>% ggvis(~Sepal.Length, ~Sepal.Width, fill = ~Species) %>% layer_points()
You see that there is a high correlation between the sepal length and the sepal width of the Setosa iris flowers, while the correlation is somewhat less high for the Virginica and Versicolor flowers: the data points are more spread out over the graph and don’t form a cluster like you can see in the case of the Setosa flowers.
The scatter plot that maps the petal length and the petal width tells a similar story:
iris %>% ggvis(~Petal.Length, ~Petal.Width, fill = ~Species) %>% layer_points()
You see that this graph indicates a positive correlation between the petal length and the petal width for all different species that are included into the Iris data set. Of course, you probably need to test this hypothesis a bit further if you want to be really sure of this:
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
You see that when you combined all three species, the correlation was a bit stronger than it is when you look at the different species separately: the overall correlation is 0.96, while for Versicolor this is 0.79. Setosa and Virginica, on the other hand, have correlations of petal length and width at 0.31 and 0.32 when you round up the numbers.
Tip: are you curious about ggvis, graphs or histograms in particular? Check out our histogram tutorial and/or ggvis course.
After a general visualized overview of the data, you can also view the data set by entering
eyJsYW5ndWFnZSI6InIiLCJzYW1wbGUiOiIjIFJldHVybiBhbGwgYGlyaXNgIGRhdGFcbmlyaXNcblxuIyBSZXR1cm4gZmlyc3QgNSBsaW5lcyBvZiBgaXJpc2BcbmhlYWQoaXJpcylcblxuIyBSZXR1cm4gc3RydWN0dXJlIG9mIGBpcmlzYFxuc3RyKGlyaXMpIn0=
However, as you will see from the result of this command, this really isn’t the best way to inspect your data set thoroughly: the data set takes up a lot of space in the console, which will impede you from forming a clear idea about your data. It is therefore a better idea to inspect the data set by executing head(iris) or str(iris).
Note that the last command will help you to clearly distinguish the data type num and the three levels of the Species attribute, which is a factor. This is very convenient, since many R machine learning classifiers require that the target feature is coded as a factor.
Remember that factor variables represent categorical variables in R. They can thus take on a limited number of different values.
A quick look at the Species attribute through tells you that the division of the species of flowers is 50-50-50. On the other hand, if you want to check the percentual division of the Species attribute, you can ask for a table of proportions:
eyJsYW5ndWFnZSI6InIiLCJzYW1wbGUiOiIjIERpdmlzaW9uIG9mIGBTcGVjaWVzYFxudGFibGUoaXJpcyRTcGVjaWVzKSBcblxuIyBQZXJjZW50dWFsIGRpdmlzaW9uIG9mIGBTcGVjaWVzYFxucm91bmQocHJvcC50YWJsZSh0YWJsZShpcmlzJFNwZWNpZXMpKSAqIDEwMCwgZGlnaXRzID0gMSkifQ==
Note that the round argument rounds the values of the first argument, prop.table(table(iris$Species))*100 to the specified number of digits, which is one digit after the decimal point. You can easily adjust this by changing the value of the digits argument. ### Profound Understanding Of Your Data Let’s not remain on this high-level overview of the data! R gives you the opportunity to go more in-depth with the summary() function. This will give you the minimum value, first quantile, median, mean, third quantile and maximum value of the data set Iris for numeric data types. For the class variable, the count of factors will be returned: 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 As you can see, the c() function is added to the original command: the columns petal width and sepal width are concatenated and a summary is then asked of just these two columns of the Iris data set. ## Step Three. Where To Go Now? After you have acquired a good understanding of your data, you have to decide on the use cases that would be relevant for your data set. In other words, you think about what your data set might teach you or what you think you can learn from your data. From there on, you can think about what kind of algorithms you would be able to apply to your data set in order to get the results that you think you can obtain. Tip: keep in mind that the more familiar you are with your data, the easier it will be to assess the use cases for your specific data set. The same also holds for finding the appropriate machine algorithm. For this tutorial, the Iris data set will be used for classification, which is an example of predictive modeling. The last attribute of the data set, Species, will be the target variable or the variable that you want to predict in this example. Note that you can also take one of the numerical classes as the target variable if you want to use KNN to do regression. ## Step Four. Prepare Your Workspace Many of the algorithms used in machine learning are not incorporated by default into R. You will most probably need to download the packages that you want to use when you want to get started with machine learning. Tip: got an idea of which learning algorithm you may use, but not of which package you want or need? You can find a pretty complete overview of all the packages that are used in R right here. To illustrate the KNN algorithm, this tutorial works with the package class: eyJsYW5ndWFnZSI6InIiLCJzYW1wbGUiOiJsaWJyYXJ5KC4uLi4uKSIsInNvbHV0aW9uIjoibGlicmFyeShjbGFzcykiLCJzY3QiOiJ0ZXN0X2Z1bmN0aW9uKFwibGlicmFyeVwiLCBhcmdzPVwicGFja2FnZVwiKVxuc3VjY2Vzc19tc2coXCJBd2Vzb21lIGpvYiFcIikifQ== If you don’t have this package yet, you can quickly and easily do so by typing the following line of code: install.packages("<package name>") Remember the nerd tip: if you’re not sure if you have this package, you can run the following command to find out! any(grepl("<name of your package>", installed.packages())) ## Step Five. Prepare Your Data After exploring your data and preparing your workspace, you can finally focus back on the task ahead: making a machine learning model. However, before you can do this, it’s important to also prepare your data. The following section will outline two ways in which you can do this: by normalizing your data (if necessary) and by splitting your data in training and testing sets. ### Normalization As a part of your data preparation, you might need to normalize your data so that its consistent. For this introductory tutorial, just remember that normalization makes it easier for the KNN algorithm to learn. There are two types of normalization: • example normalization is the adjustment of each example individually, while • feature normalization indicates that you adjust each feature in the same way across all examples. So when do you need to normalize your dataset? In short: when you suspect that the data is not consistent. You can easily see this when you go through the results of the summary() function. Look at the minimum and maximum values of all the (numerical) attributes. If you see that one attribute has a wide range of values, you will need to normalize your dataset, because this means that the distance will be dominated by this feature. For example, if your dataset has just two attributes, X and Y, and X has values that range from 1 to 1000, while Y has values that only go from 1 to 100, then Y’s influence on the distance function will usually be overpowered by X’s influence. When you normalize, you actually adjust the range of all features, so that distances between variables with larger ranges will not be over-emphasised. Tip: go back to the result of summary(iris) and try to figure out if normalization is necessary. The Iris data set doesn’t need to be normalized: the Sepal.Length attribute has values that go from 4.3 to 7.9 and Sepal.Width contains values from 2 to 4.4, while Petal.Length’s values range from 1 to 6.9 and Petal.Width goes from 0.1 to 2.5. All values of all attributes are contained within the range of 0.1 and 7.9, which you can consider acceptable. Nevertheless, it’s still a good idea to study normalization and its effect, especially if you’re new to machine learning. You can perform feature normalization, for example, by first making your own normalize() function. You can then use this argument in another command, where you put the results of the normalization in a data frame through as.data.frame() after the function lapply() returns a list of the same length as the data set that you give in. Each element of that list is the result of the application of the normalize argument to the data set that served as input: YourNormalizedDataSet <- as.data.frame(lapply(YourDataSet, normalize)) Test this in the DataCamp Light chunk below! 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 For the Iris dataset, you would have applied the normalize argument on the four numerical attributes of the Iris data set (Sepal.Length, Sepal.Width, Petal.Length, Petal.Width) and put the results in a data frame. Tip: to more thoroughly illustrate the effect of normalization on the data set, compare the following result to the summary of the Iris data set that was given in step two. ### Training And Test Sets In order to assess your model’s performance later, you will need to divide the data set into two parts: a training set and a test set. The first is used to train the system, while the second is used to evaluate the learned or trained system. In practice, the division of your data set into a test and a training sets is disjoint: the most common splitting choice is to take 2/3 of your original data set as the training set, while the 1/3 that remains will compose the test set. One last look on the data set teaches you that if you performed the division of both sets on the data set as is, you would get a training class with all species of “Setosa” and “Versicolor”, but none of “Virginica”. The model would therefore classify all unknown instances as either “Setosa” or “Versicolor”, as it would not be aware of the presence of a third species of flowers in the data. In short, you would get incorrect predictions for the test set. You thus need to make sure that all three classes of species are present in the training model. What’s more, the amount of instances of all three species needs to be more or less equal so that you do not favour one or the other class in your predictions. To make your training and test sets, you first set a seed. This is a number of R’s random number generator. The major advantage of setting a seed is that you can get the same sequence of random numbers whenever you supply the same seed in the random number generator. set.seed(1234) Then, you want to make sure that your Iris data set is shuffled and that you have an equal amount of each species in your training and test sets. You use the sample() function to take a sample with a size that is set as the number of rows of the Iris data set, or 150. You sample with replacement: you choose from a vector of 2 elements and assign either 1 or 2 to the 150 rows of the Iris data set. The assignment of the elements is subject to probability weights of 0.67 and 0.33. ind <- sample(2, nrow(iris), replace=TRUE, prob=c(0.67, 0.33)) Note that the replace argument is set to TRUE: this means that you assign a 1 or a 2 to a certain row and then reset the vector of 2 to its original state. This means that, for the next rows in your data set, you can either assign a 1 or a 2, each time again. The probability of choosing a 1 or a 2 should not be proportional to the weights amongst the remaining items, so you specify probability weights. Note also that, even though you don’t see it in the DataCamp Light chunk, the seed has still been set to 1234. Remember that you want your training set to be 2/3 of your original data set: that is why you assign “1” with a probability of 0.67 and the “2”s with a probability of 0.33 to the 150 sample rows. You can then use the sample that is stored in the variable ind to define your training and test sets: 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 Note that, in addition to the 2/3 and 1/3 proportions specified above, you don’t take into account all attributes to form the training and test sets. Specifically, you only take Sepal.Length, Sepal.Width, Petal.Length and Petal.Width. This is because you actually want to predict the fifth attribute, Species: it is your target variable. However, you do want to include it into the KNN algorithm, otherwise there will never be any prediction for it. You therefore need to store the class labels in factor vectors and divide them over the training and test sets: 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 ## Step Six. The Actual KNN Model ### Building Your Classifier After all these preparation steps, you have made sure that all your known (training) data is stored. No actual model or learning was performed up until this moment. Now, you want to find the k nearest neighbors of your training set. An easy way to do these two steps is by using the knn() function, which uses the Euclidian distance measure in order to find the k-nearest neighbours to your new, unknown instance. Here, the k parameter is one that you set yourself. As mentioned before, new instances are classified by looking at the majority vote or weighted vote. In case of classification, the data point with the highest score wins the battle and the unknown instance receives the label of that winning data point. If there is an equal amount of winners, the classification happens randomly. Note: the k parameter is often an odd number to avoid ties in the voting scores. To build your classifier, you need to take the knn() function and simply add some arguments to it, just like in this example: 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 You store into iris_pred the knn() function that takes as arguments the training set, the test set, the train labels and the amount of neighbours you want to find with this algorithm. The result of this function is a factor vector with the predicted classes for each row of the test data. Note that you don’t want to insert the test labels: these will be used to see if your model is good at predicting the actual classes of your instances! You see that when you inspect the the result, iris_pred, you’ll get back the factor vector with the predicted classes for each row of the test data. ## Step Seven. Evaluation of Your Model An essential next step in machine learning is the evaluation of your model’s performance. In other words, you want to analyze the degree of correctness of the model’s predictions. For a more abstract view, you can just compare the results of iris_pred to the test labels that you had defined earlier: 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 You see that the model makes reasonably accurate predictions, with the exception of one wrong classification in row 29, where “Versicolor” was predicted while the test label is “Virginica”. This is already some indication of your model’s performance, but you might want to go even deeper into your analysis. For this purpose, you can import the package gmodels: install.packages("package name") However, if you have already installed this package, you can simply enter library(gmodels) Then you can make a cross tabulation or a contingency table. This type of table is often used to understand the relationship between two variables. In this case, you want to understand how the classes of your test data, stored in iris.testLabels relate to your model that is stored in iris_pred: CrossTable(x = iris.testLabels, y = iris_pred, prop.chisq=FALSE) Note that the last argument prop.chisq indicates whether or not the chi-square contribution of each cell is included. The chi-square statistic is the sum of the contributions from each of the individual cells and is used to decide whether the difference between the observed and the expected values is significant. From this table, you can derive the number of correct and incorrect predictions: one instance from the testing set was labeled Versicolor by the model, while it was actually a flower of species Virginica. You can see this in the first row of the “Virginica” species in the iris.testLabels column. In all other cases, correct predictions were made. You can conclude that the model’s performance is good enough and that you don’t need to improve the model! ## Machine Learning in R with caret In the previous sections, you have gotten started with supervised learning in R via the KNN algorithm. As you might not have seen above, machine learning in R can get really complex, as there are various algorithms with various syntax, different parameters, etc. Maybe you’ll agree with me when I say that remembering the different package names for each algorithm can get quite difficult or that applying the syntax for each specific algorithm is just too much. That’s where the caret package can come in handy: it’s short for “Classification and Regression Training” and offers everything you need to know to solve supervised machine learning problems: it provides a uniform interface to a ton of machine learning algorithms. If you’re a bit familiar with Python machine learning, you might see similarities with scikit-learn! In the following, you’ll go through the steps as they have been outlined above, but this time, you’ll make use of caret to classify your data. Note that you have already done a lot of work if you’ve followed the steps as they were outlined above: you already have a hold on your data, you have explored it, prepared your workspace, etc. Now it’s time to preprocess your data with caret! As you have done before, you can study the effect of the normalization, but you’ll see this later on in the tutorial. You already know what’s next! Let’s split up the data in a training and test set. In this case, though, you handle things a little bit differently: you split up the data based on the labels that you find in iris$Species. Also, the ratio is in this case set at 75-25 for the training and test sets.
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
You’re all set to go and train models now! But, as you might remember, caret is an extremely large project that includes a lot of algorithms. If you’re in doubt on what algorithms are included in the project, you can get a list of all of them. Pull up the list by running names(getModelInfo()), just like the code chunk below demonstrates. Next, pick an algorithm and train a model with the train() function:
eyJsYW5ndWFnZSI6InIiLCJwcmVfZXhlcmNpc2VfY29kZSI6ImxpYnJhcnkoY2FyZXQpXG5zZXQuc2VlZCgxMjM0KVxuaW5kZXggPC0gY3JlYXRlRGF0YVBhcnRpdGlvbihpcmlzJFNwZWNpZXMsIHA9MC43NSwgbGlzdD1GQUxTRSlcbmlyaXMudHJhaW5pbmcgPC0gaXJpc1tpbmRleCxdXG5pcmlzLnRlc3QgPC0gaXJpc1staW5kZXgsXSIsInNhbXBsZSI6IiMgT3ZlcnZpZXcgb2YgYWxnb3Mgc3VwcG9ydGVkIGJ5IGNhcmV0XG5uYW1lcyhnZXRNb2RlbEluZm8oKSlcblxuIyBUcmFpbiBhIG1vZGVsXG5tb2RlbF9rbm4gPC0gdHJhaW4oaXJpcy50cmFpbmluZ1ssIDE6NF0sIGlyaXMudHJhaW5pbmdbLCA1XSwgbWV0aG9kPSdrbm4nKSJ9
Note that making other models is extremely simple when you have gotten this far; You just have to change the method argument, just like in this example:
model_cart <- train(iris.training[, 1:4], iris.training[, 5], method='rpart2')
Now that you have trained your model, it’s time to predict the labels of the test set that you have just made and evaluate how the model has done on your data:
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
Additionally, you can try to perform the same test as before, to examine the effect of preprocessing, such as scaling and centering, on your model. Run the following code chunk:
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
## Move On To Big Data
Congratulations! You’ve made it through this tutorial!
This tutorial was primarily concerned with performing basic machine learning algorithm KNN with the help of R. The Iris data set that was used was small and overviewable; Not only did you see how you can perform all of the steps by yourself, but you’ve also seen how you can easily make use of a uniform interface, such as the one that caret offers, to spark your machine learning.
But you can do so much more!
If you have experimented enough with the basics presented in this tutorial and other machine learning algorithms, you might want to find it interesting to go further into R and data analysis. | 13,808 | 36,948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-05 | longest | en | 0.946968 |
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# Solving and Graphing Systems of Equations and Inequalities
I need to know how to graph the following inequalities:
3x + y = or <5
x-y= or <2
y= or >5
Solve the system by graphing:
x-y=2
3x-3y=6
3x-y=1
3x-y=2
Solve the system of linear inequalities by graphing:
x-2y= or <4
x->1
#### Solution Summary
Systems of equations and inequalities are solved and graphed. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.
\$2.19 | 160 | 543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | longest | en | 0.935863 |
https://www.quantopian.com/posts/how-relevant-is-style-risk-if-common-returns-are-relatively-small | 1,575,874,905,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00459.warc.gz | 843,638,939 | 23,512 | How relevant is Style risk if common returns are relatively small?
(I think this was asked in one of the risk megathreads a while ago by Grant, but it wasn't answered and I'm wondering about it now:)
Let's say I have a mean-reversion style algo that has 4.5% annualized specific return and 0.5% annualized common return, with a specific Sharpe ratio of 2.
How problematic is it if it has a 'Short Term Reversal' style risk exposure of let's say 50% (> required 40%)? The whole point of the algo is to use mean reversion, so some style risk is expected, and if 95% of the returns are 'pure' alpha that's not too bad right? (I can see why 50% risk on the total returns would be more problematic.)
Just wondering if any progress was made on this philosophical question in the mean time :).
17 responses
You are focusing on the annualized returns where I think the role of risk pertains more to volatility. As such, I would be curious what the vol of your specific and common returns are. Since your exposure is high I would imagine it's quite a lot. So if that is a static (over time) high exposure I would still argue that the strategy takes on unnecessary risk. Or put another way, 50% (or whatever) of your risk budget is allocated to something that is already well known and an investor will not pay you a lot of money for.
As such, I would try and make your mean-reversion signal orthogonal to the existing one and just capture how your factor differs. You can try this (experimental) code snippet that Max wrote:
class OrthogonalAlpha(CustomFactor):
window_length=504
def compute(self, today, assets, out, alpha, symbol, *risk_exposures):
regressor = LinearRegression(fit_intercept=True)
risk_exposures = np.array(risk_exposures)
residuals = np.zeros((1, risk_exposures.shape[-1]))
for i in range(risk_exposures.shape[-1]):
X = risk_exposures[:,:,i].T
X = np.nan_to_num(X)
sectors = X[:, :11][:, ~(np.all(X[:,:11]==0., axis=0))]
if sectors.shape[-1] == 0:
residuals[:,i] = 0
continue
styles = X[:,11:]
y = alpha[:,i].T
y = np.nan_to_num(y)
# Do the sector regression
regressor.fit(sectors, y)
sector_resid = regressor.intercept_ + \
(regressor.predict(sectors) - y)
# Do the style regression on the residuals of the sector regression
regressor.fit(styles, sector_resid)
residuals[:,i] = regressor.intercept_ + \
(regressor.predict(styles) - y)[-1]
out[:] = residuals[-1]
which you would call like:
orthogonal_alpha = OrthogonalAlpha( inputs=[my_custom_mean_reversion_factor], mask=universe )
As I said, this is still experimental so it might not work but I would be curious if you try it what the outcome is.
Disclaimer
The material on this website is provided for informational purposes only and does not constitute an offer to sell, a solicitation to buy, or a recommendation or endorsement for any security or strategy, nor does it constitute an offer to provide investment advisory services by Quantopian. In addition, the material offers no opinion with respect to the suitability of any security or specific investment. No information contained herein should be regarded as a suggestion to engage in or refrain from any investment-related course of action as none of Quantopian nor any of its affiliates is undertaking to provide investment advice, act as an adviser to any plan or entity subject to the Employee Retirement Income Security Act of 1974, as amended, individual retirement account or individual retirement annuity, or give advice in a fiduciary capacity with respect to the materials presented herein. If you are an individual retirement or other investor, contact your financial advisor or other fiduciary unrelated to Quantopian about whether any given investment idea, strategy, product or service described herein may be appropriate for your circumstances. All investments involve risk, including loss of principal. Quantopian makes no guarantees as to the accuracy or completeness of the views expressed in the website. The views are subject to change, and may have become unreliable for various reasons, including changes in market conditions or economic circumstances.
Would be very interesting indeed, bummer I can't get it to work (also had to make my CustomFactor window_safe=True):
TypeError: compute() takes at least 6 arguments (5 given)
I've attached a backtest notebook so you can look at volatility, I couldn't really see the part where it shows the difference between common and specific vola.
Note that I managed to bring the style risk down to ~35% by aiming for a [-0.10,0.10] constraint in the optimizer.
7
Notebook previews are currently unavailable.
Maybe a stupid question but how is this different from setting the style exposure to close to 0 using the Risk API?
Joakim: I'm not quite sure to be honest, so it warrants exploring / thinking about this further. One scenario were the difference might become more apparent is if you imagine to have multiple factors, you would orthogonalize each one individually, combine, and then do risk optimization on the aggregate vs just doing risk optimization at the end. Probably requires some experimentation as to what is preferable.
Thanks Thomas, and for the code snippet. Will play around with it.
Ivory, Joakim: I've included a NB here that has a helper function that I use to create these orthogonal factors more easily. It is still super experimental, but it won't give you that compute error.
29
Notebook previews are currently unavailable.
Disclaimer
The material on this website is provided for informational purposes only and does not constitute an offer to sell, a solicitation to buy, or a recommendation or endorsement for any security or strategy, nor does it constitute an offer to provide investment advisory services by Quantopian. In addition, the material offers no opinion with respect to the suitability of any security or specific investment. No information contained herein should be regarded as a suggestion to engage in or refrain from any investment-related course of action as none of Quantopian nor any of its affiliates is undertaking to provide investment advice, act as an adviser to any plan or entity subject to the Employee Retirement Income Security Act of 1974, as amended, individual retirement account or individual retirement annuity, or give advice in a fiduciary capacity with respect to the materials presented herein. If you are an individual retirement or other investor, contact your financial advisor or other fiduciary unrelated to Quantopian about whether any given investment idea, strategy, product or service described herein may be appropriate for your circumstances. All investments involve risk, including loss of principal. Quantopian makes no guarantees as to the accuracy or completeness of the views expressed in the website. The views are subject to change, and may have become unreliable for various reasons, including changes in market conditions or economic circumstances.
Thanks!
An initial test using Alphalens suggests alpha goes from 0.074 to -0.012 by applying orthoganalization to a simple mean-reversion factor. I kind of expected it to leave some alpha with the same sign, since the backtest showed just a small part of the returns can be explained by the common mean reversion factor.
Did anyone else try this?
We did notice that it often has quite a detrimental effect, not quite sure why yet. But it makes sense if you apply it to a factor that's already in the risk-model. A good test would be a factor that should be largely independent like maybe a Quality factor.
how is this different from setting the style exposure to close to 0 using the Risk API
Interesting point by Joakim above. I have to say, the whole apparatus of the risk model, its implementation in risk_loading_pipeline and:
risk_model_exposure = opt.experimental.RiskModelExposure(
)
constraints.append(risk_model_exposure)
has been a matter of faith for me, mostly. By some magic, it does seem to work, but I don't understand it well enough, given its importance in shifting money from Q to my wallet.
It seems that the Optimize API should be sufficient, no? I'm wondering if one could achieve the same end result by calling the Optimize API with TargetWeights and calculate_optimal_portfolio, and returning the resulting portfolio weight vector? In other words, can the Optimize API be re-purposed to better align individual alpha factors with the desired risk profile, prior to their combination? Alternatively, one could write a custom implementation using CVXPY.
For example, on a given alpha factor, if we run the Optimize API on it, and constrain only short_term_reversal we should be left with a residual that contains everything except the 14-day RSI (to within a specified tolerance), right?
Another observation is that all of this relies on the past being predictive of the future. For example, I've found that depending on the stock universe, the effectiveness of SimpleBeta and the Optimize API vary. Using a trailing 260-day computation of beta isn't sufficient; for example, I need to bias the beta exposure, to achieve beta ~ 0:
MIN_BETA_EXPOSURE = -0.2
MAX_BETA_EXPOSURE = -0.2
My hypothesis is that the trailing beta doesn't persist into the future. I suspect that a similar problem is at play in the application of the risk model; it only works ideally under the assumption that the future equals the past.
One thing to try is to include the z-scored style risk factors in the alpha combination, but instead of adding them with the alpha factors, subtract them. I tried this once with the RSI style risk factor, and it seemed to work well.
The above code doesn't do the right thing, but this snippet should work:
def orthogonalize_alpha(pos_pct, risk_loadings, return_coeffs=False):
from sklearn import linear_model
import pandas as pd
pos_dt = pos_dt.reindex(overlap_sids)
clf = linear_model.LinearRegression()
return resid, clf.coef_
if isinstance(pos_pct, pd.DataFrame):
pos_resid = {}
for dt in pos_pct.index:
continue
pos_resid[dt], coeffs.loc[dt] = _run_regression(pos_pct.loc[dt].dropna(),
pos_resid = pd.concat(pos_resid)
coeffs = pd.DataFrame(coeffs).T
elif isinstance(pos_pct, pd.Series):
else:
raise ValueError('Pass either DataFrame or Series.')
if return_coeffs:
return pos_resid, coeffs
else:
return pos_resid
I haven't turned this into a factor, but if you do, please post it here.
I've been trying to implement this code snippet into the L/S Equity template strategy, but I get the below error message:
There was a runtime error.
AttributeError:'Index' object has no attribute 'levels'
Line: 91 inorthogonalize_alpha
Would anyone be able to help?
6
Backtest from to with initial capital
Total Returns
--
Alpha
--
Beta
--
Sharpe
--
Sortino
--
Max Drawdown
--
Benchmark Returns
--
Volatility
--
Returns 1 Month 3 Month 6 Month 12 Month
Alpha 1 Month 3 Month 6 Month 12 Month
Beta 1 Month 3 Month 6 Month 12 Month
Sharpe 1 Month 3 Month 6 Month 12 Month
Sortino 1 Month 3 Month 6 Month 12 Month
Volatility 1 Month 3 Month 6 Month 12 Month
Max Drawdown 1 Month 3 Month 6 Month 12 Month
import quantopian.algorithm as algo
import quantopian.optimize as opt
from quantopian.pipeline import Pipeline
from quantopian.pipeline.factors import SimpleMovingAverage
from quantopian.pipeline.data.psychsignal import stocktwits
from quantopian.pipeline.data import Fundamentals
# Constraint Parameters
MAX_GROSS_LEVERAGE = 1.0
TOTAL_POSITIONS = 600
MAX_SHORT_POSITION_SIZE = 2.0 / TOTAL_POSITIONS
MAX_LONG_POSITION_SIZE = 2.0 / TOTAL_POSITIONS
def initialize(context):
algo.attach_pipeline(make_pipeline(), 'long_short_equity_template')
algo.schedule_function(func=rebalance,
date_rule=algo.date_rules.week_start(),
time_rule=algo.time_rules.market_open(hours=0, minutes=30),
half_days=True)
algo.schedule_function(func=record_vars,
date_rule=algo.date_rules.every_day(),
time_rule=algo.time_rules.market_close(),
half_days=True)
def make_pipeline():
value = Fundamentals.ebit.latest / Fundamentals.enterprise_value.latest
quality = Fundamentals.roe.latest
sentiment_score = SimpleMovingAverage(
inputs=[stocktwits.bull_minus_bear],
window_length=3,
)
value_winsorized = value.winsorize(min_percentile=0.05, max_percentile=0.95)
quality_winsorized = quality.winsorize(min_percentile=0.05, max_percentile=0.95)
sentiment_score_winsorized = sentiment_score.winsorize(min_percentile=0.05, max_percentile=0.95)
combined_factor = (
value_winsorized.zscore() +
quality_winsorized.zscore() +
sentiment_score_winsorized.zscore()
)
long_short_screen = (longs | shorts)
pipe = Pipeline(
columns={
'longs': longs,
'shorts': shorts,
'combined_factor': combined_factor
},
screen=long_short_screen
)
return pipe
context.pipeline_data = algo.pipeline_output('long_short_equity_template')
def record_vars(context, data):
algo.record(num_positions=len(context.portfolio.positions))
# Below function from Thomas Wiecki to orthogonalize the alpha factor
from sklearn import linear_model
import pandas as pd
pos_resid = {}
for dt in pos_pct.index:
continue
pos_dt = pos_pct.loc[dt].dropna()
if len(overlap_sids) == 0:
continue
pos_dt = pos_dt.reindex(overlap_sids)
clf = linear_model.LinearRegression()
coeffs.loc[dt] = clf.coef_
pos_resid = pd.concat(pos_resid)
coeffs = pd.DataFrame(coeffs).T
return pos_resid, coeffs
def rebalance(context, data):
pipeline_data = context.pipeline_data
# Changed from MaximizeAlpha to normalized weigths using TargetWeights instead:
alpha_weight = pipeline_data['combined_factor']
alpha_weight_norm = alpha_weight / alpha_weight.abs().sum()
objective = opt.TargetWeights(orth_alpha)
constraints = []
constraints.append(opt.MaxGrossExposure(MAX_GROSS_LEVERAGE))
constraints.append(opt.DollarNeutral())
neutralize_risk_factors = opt.experimental.RiskModelExposure(
version=0
)
constraints.append(neutralize_risk_factors)
constraints.append(
opt.PositionConcentration.with_equal_bounds(
min=-MAX_SHORT_POSITION_SIZE,
max=MAX_LONG_POSITION_SIZE
))
algo.order_optimal_portfolio(
objective=objective,
constraints=constraints
)
There was a runtime error.
Hi Joakim,
Try using risk_loadings.index.get_level_values(0) instead. However there is other errors since, unles I am mistken, the function seems to have been wrote for an 'pos_pct' matrix and not a vector.
Thanks Mathieu, I appreciate your reply. I'm afraid this one might be a too tough nut for me to crack unfortunately. I'll put it on hold for now. I can't even pronounce 'orthogonalize' but it seems like very useful code for anyone looking for 'pure alpha.'
I think this fixes it. I also updated the function above.
21
Backtest from to with initial capital
Total Returns
--
Alpha
--
Beta
--
Sharpe
--
Sortino
--
Max Drawdown
--
Benchmark Returns
--
Volatility
--
Returns 1 Month 3 Month 6 Month 12 Month
Alpha 1 Month 3 Month 6 Month 12 Month
Beta 1 Month 3 Month 6 Month 12 Month
Sharpe 1 Month 3 Month 6 Month 12 Month
Sortino 1 Month 3 Month 6 Month 12 Month
Volatility 1 Month 3 Month 6 Month 12 Month
Max Drawdown 1 Month 3 Month 6 Month 12 Month
import quantopian.algorithm as algo
import quantopian.optimize as opt
from quantopian.pipeline import Pipeline
from quantopian.pipeline.factors import SimpleMovingAverage
from quantopian.pipeline.data.psychsignal import stocktwits
from quantopian.pipeline.data import Fundamentals
# Constraint Parameters
MAX_GROSS_LEVERAGE = 1.0
TOTAL_POSITIONS = 600
MAX_SHORT_POSITION_SIZE = 2.0 / TOTAL_POSITIONS
MAX_LONG_POSITION_SIZE = 2.0 / TOTAL_POSITIONS
def initialize(context):
algo.attach_pipeline(make_pipeline(), 'long_short_equity_template')
algo.schedule_function(func=rebalance,
date_rule=algo.date_rules.week_start(),
time_rule=algo.time_rules.market_open(hours=0, minutes=30),
half_days=True)
algo.schedule_function(func=record_vars,
date_rule=algo.date_rules.every_day(),
time_rule=algo.time_rules.market_close(),
half_days=True)
def make_pipeline():
value = Fundamentals.ebit.latest / Fundamentals.enterprise_value.latest
quality = Fundamentals.roe.latest
sentiment_score = SimpleMovingAverage(
inputs=[stocktwits.bull_minus_bear],
window_length=3,
)
value_winsorized = value.winsorize(min_percentile=0.05, max_percentile=0.95)
quality_winsorized = quality.winsorize(min_percentile=0.05, max_percentile=0.95)
sentiment_score_winsorized = sentiment_score.winsorize(min_percentile=0.05, max_percentile=0.95)
combined_factor = (
value_winsorized.zscore() +
quality_winsorized.zscore() +
sentiment_score_winsorized.zscore()
)
long_short_screen = (longs | shorts)
pipe = Pipeline(
columns={
'longs': longs,
'shorts': shorts,
'combined_factor': combined_factor
},
screen=long_short_screen
)
return pipe
context.pipeline_data = algo.pipeline_output('long_short_equity_template')
def record_vars(context, data):
algo.record(num_positions=len(context.portfolio.positions))
# Below function from Thomas Wiecki to orthogonalize the alpha factor
from sklearn import linear_model
import pandas as pd
pos_dt = pos_dt.reindex(overlap_sids)
clf = linear_model.LinearRegression()
return resid, clf.coef_
if isinstance(pos_pct, pd.DataFrame):
pos_resid = {}
for dt in pos_pct.index:
continue
pos_resid[dt], coeffs.loc[dt] = _run_regression(pos_pct.loc[dt].dropna(),
pos_resid = pd.concat(pos_resid)
coeffs = pd.DataFrame(coeffs).T
elif isinstance(pos_pct, pd.Series):
else:
raise ValueError('Pass either DataFrame or Series.')
if return_coeffs:
return pos_resid, coeffs
else:
return pos_resid
def rebalance(context, data):
pipeline_data = context.pipeline_data
# Changed from MaximizeAlpha to normalized weigths using TargetWeights instead:
alpha_weight = pipeline_data['combined_factor']
alpha_weight_norm = alpha_weight / alpha_weight.abs().sum()
objective = opt.TargetWeights(orth_alpha)
constraints = []
constraints.append(opt.MaxGrossExposure(MAX_GROSS_LEVERAGE))
constraints.append(opt.DollarNeutral())
neutralize_risk_factors = opt.experimental.RiskModelExposure(
version=0
)
constraints.append(neutralize_risk_factors)
constraints.append(
opt.PositionConcentration.with_equal_bounds(
min=-MAX_SHORT_POSITION_SIZE,
max=MAX_LONG_POSITION_SIZE
))
algo.order_optimal_portfolio(
objective=objective,
constraints=constraints
)
There was a runtime error.
That's awesome, thanks heaps! Appears to be working.
Here's a version of Thomas' code above that doesn't error with NaN values (and forces them stay NaN). Potentially useful if you want NaNs to persist in your signals for whatever reason:
def orthogonalize_alpha(pos_pct, risk_loadings, return_coeffs=False):
# Keep NaNs at NaN
isnan = pos_dt.isnull()
pos_dt_nonans = pos_dt[~isnan].copy()
# Get the overlapping SIDs for each case
# Subset the data sets for each case
pos_dt = pos_dt.loc[overlap_sids]
pos_dt_nonans = pos_dt_nonans.loc[overlap_sids_nonans]
# Fit the regression to non-NaN data
try:
clf = LinearRegression()
except: # All data points are NaN
return pos_dt, None
# Perform the regression
# Keep NaNs at NaN
resid[isnan] = np.nan
return resid, clf.coef_
if return_coeffs:
return pos_resid, coeffs
else:
return pos_resid
I've been exploring how to individually orthogonalize each of my factors and then later on combined the two for a final orthogonalize process. However, while the below code works, I am not sure if it's the best approach. In this case, I'm just adding the two values together. These steps can be performed in either: def before_trading_start(context, data): or in your def rebalance(context, data): . Any suggestions?
alpha_weight_A = pipeline_data['alpha_a']
alpha_weight_norm_A = alpha_weight_A / alpha_weight_A.abs().sum()
alpha_weight_C = pipeline_data['alpha_c']
alpha_weight_norm_C = alpha_weight_C / alpha_weight_C.abs().sum() | 4,555 | 19,883 | {"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} | 2.859375 | 3 | CC-MAIN-2019-51 | longest | en | 0.935747 |
https://aimsciences.org/article/doi/10.3934/ipi.2014.8.507 | 1,566,581,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00205.warc.gz | 349,363,359 | 21,893 | # American Institute of Mathematical Sciences
May 2014, 8(2): 507-535. doi: 10.3934/ipi.2014.8.507
## A stable method solving the total variation dictionary model with $L^\infty$ constraints
1 Institute of Microelectronics, Chinese Academy of Sciences, Beijing, China 2 MAP5, Université Paris Descartes, Paris, 75006, France 3 School of Computer and Information Technology, Beijing Jiaotong University, Beijing, China 4 Department of Mathematics, Hong Kong Baptist University, Kowloon Tong, Hong Kong, China
Received November 2011 Revised May 2013 Published May 2014
Image restoration plays an important role in image processing, and numerous approaches have been proposed to tackle this problem. This paper presents a modified model for image restoration, that is based on a combination of Total Variation and Dictionary approaches. Since the well-known TV regularization is non-differentiable, the proposed method utilizes its dual formulation instead of its approximation in order to exactly preserve its properties. The data-fidelity term combines the one commonly used in image restoration and a wavelet thresholding based term. Then, the resulting optimization problem is solved via a first-order primal-dual algorithm. Numerical experiments demonstrate the good performance of the proposed model. In a last variant, we replace the classical TV by the nonlocal TV regularization, which results in a much higher quality of restoration.
Citation: Liyan Ma, Lionel Moisan, Jian Yu, Tieyong Zeng. A stable method solving the total variation dictionary model with $L^\infty$ constraints. Inverse Problems & Imaging, 2014, 8 (2) : 507-535. doi: 10.3934/ipi.2014.8.507
##### References:
[1] M. Afonso, J. Bioucas-Dias and M. Figueiredo, Fast image recovery using variable splitting and constrained optimization,, IEEE Trans. Image Process., 19 (2010), 2345. doi: 10.1109/TIP.2010.2047910. Google Scholar [2] L. Ambrosio, N. Fusco and D. Pallara, Functions of Bounded Variation and Free Discontinuity Problem,, Oxford, (2000). Google Scholar [3] G. Aubert and P. Kornprobst, Mathematical Problems in Image Processing: Partial Differential Equations and the Calculus of Variations,, Applied Mathematical Sciences, (2006). Google Scholar [4] A. Beck and M. Teboulle, A fast iterative shrinkage-thresholding algorithm for linear inverse problems,, SIAM J. Imag. Sci., 2 (2009), 183. doi: 10.1137/080716542. Google Scholar [5] M. Bertalmio, G. Sapiro, V. Caselles and C. Ballester, Image inpainting,, in Proc. SIGGRAPH, (2000), 417. doi: 10.1145/344779.344972. Google Scholar [6] S. Boyd and L. Vandenberghe, Convex Optimization,, Cambridge University Press, (2004). doi: 10.1017/CBO9780511804441. Google Scholar [7] J. P. Boyle and R. L. Dykstra, A method for finding projections onto the intersection of convex sets in Hilbert spaces,, Lecture Notes in Statistics, 37 (1986), 28. doi: 10.1007/978-1-4613-9940-7_3. Google Scholar [8] X. Bresson, A Short Note for Nonlocal TV Minimization,, Technical Report, (2009). Google Scholar [9] X. Bresson and T. F. Chan, Fast dual minimization of the vectorial total variation norm and applications to color image processing,, Inverse Problems and Imaging, 2 (2008), 455. doi: 10.3934/ipi.2008.2.455. Google Scholar [10] A. Buades, B. Coll and J. M. Morel, A review of image denoising algorithms, with a new one,, Multiscale Model. Simul., 4 (2005), 490. doi: 10.1137/040616024. Google Scholar [11] A. Bugeau, M. Bertalmio, V. Caselles and G. Sapiro, A Comprehensive Framework for Image Inpainting,, IEEE Trans. Image Process., 19 (2010), 2634. doi: 10.1109/TIP.2010.2049240. Google Scholar [12] E. Candes and F. Guo, A new multiscale transform, minimum total variation synthesis: Application to edge-preserving image reconstruction,, Signal Processing, 82 (2002), 1519. Google Scholar [13] A. Chambolle, An algorithm for total variation minimization and applications,, J. Math. Imag. Vis., 20 (2004), 89. doi: 10.1023/B:JMIV.0000011321.19549.88. Google Scholar [14] A. Chambolle, R. DeVore, N.-Y. Lee and B. Lucier, Nonlinear wavelet image processing: Variational problems, compression, and noise removal through wavelet shrinkage,, IEEE Trans. Image Process., 7 (1998), 319. doi: 10.1109/83.661182. Google Scholar [15] A. Chambolle and T. Pock, A First-Order Primal-Dual Algorithm for Convex Problems with Applications to Imaging,, J. Math. Imaging Vis., 40 (2011), 120. doi: 10.1007/s10851-010-0251-1. Google Scholar [16] T. F. Chan, G. H. Golub and P. Mulet, A nonlinear primal-dual method for total variation-based image restoration,, SIAM J. Sci. Comp., 20 (1999), 1964. doi: 10.1137/S1064827596299767. Google Scholar [17] T. F. Chan, A. M. Yip and F. E. Park, Simultaneous total variation image inpainting and blind deconvolution,, Int. J. of Imaging Systems and Technology, 15 (2005), 92. doi: 10.1002/ima.20041. Google Scholar [18] T. Chan and J. Shen, Image Processing and Analysis: Variational, PDE, Wavelet, and Stochastic Methods,, Society for Industrial and Applied Mathematics (SIAM), (2005). doi: 10.1137/1.9780898717877. Google Scholar [19] C. Chaux, J. C. Pesquet and N. Pustelnik, Nested iterative algorithms for convex constrained image recovery problems,, SIAM J. Imag. Sci., 2 (2009), 730. doi: 10.1137/080727749. Google Scholar [20] P. L. Combettes and V. R. Wajs, Signal recovery by proximal forward backward splitting,, Multiscale Model. Simul., 4 (2005), 1168. doi: 10.1137/050626090. Google Scholar [21] I. Daubechies, Ten Lectures on Wavelets,, SIAM Publ., (1992). doi: 10.1137/1.9781611970104. Google Scholar [22] I. Daubechies, M. Defriese and C. D. Mol, An iterative thresholding algorithm for linear inverse problems with a sparsity constraint,, Commun.Pure Appl. Math., 57 (2004), 1413. doi: 10.1002/cpa.20042. Google Scholar [23] D. L. Donoho and I. M. Johnstone, Ideal spatial adaptation by wavelet shrinkage,, Biometrika, 81 (1994), 425. doi: 10.1093/biomet/81.3.425. Google Scholar [24] D. L. Donoho and I. M. Johnstone, Adapting to unknown smoothness via wavelet shrinkage,, J. Amer. Statist. Assoc., 90 (1995), 1200. doi: 10.1080/01621459.1995.10476626. Google Scholar [25] I. Ekeland and R. Temam, Convex Analysis and Variational Problems,, Studies Math. Appl., (1976). Google Scholar [26] D. Gabay, Applications of the method of multipliers to variational inequalities,, in Augmented Lagrangian Methods: Applications to the Solution of Boundary-Valued Problems, (1983), 299. Google Scholar [27] G. Gilboa and S. Osher, Nonlocal operators with applications to image processing,, Multiscale Model. Simul., 7 (2008), 1005. doi: 10.1137/070698592. Google Scholar [28] T. Goldstein and S. Osher, The split Bregman method for l1 regularized problems,, SIAM J. Imag. Sci., 2 (2009), 323. doi: 10.1137/080725891. Google Scholar [29] S. Lintner and F. Malgouyres, Solving a variational image restoration model which involves contraints,, Inverse. Probl., 20 (2004), 815. doi: 10.1088/0266-5611/20/3/010. Google Scholar [30] J. Liu, X-C. Tai, H. Huang and Z. Huan, A weighted dictionary learning model for denoising images corrupted by mixed noise,, IEEE Trans. on Image Process, 22 (2013), 1108. doi: 10.1109/TIP.2012.2227766. Google Scholar [31] F. Malgouyres, A framework for image deblurring using wavelet packet bases,, Appl. and Comp. Harmonic Analysis, 12 (2002), 309. doi: 10.1006/acha.2002.0379. Google Scholar [32] F. Malgouyres, Mathematical analysis of a model which combines total variation and wavelets for image restoration,, Journal of Information Processes, 2 (2002), 1. Google Scholar [33] F. Malgouyres, Minimizing the total variation under a general convex constraint for image restoration,, IEEE Trans. on Image Process, 11 (2002), 1450. doi: 10.1109/TIP.2002.806241. Google Scholar [34] S. Masnou and J.-M. Morel, Level lines based disocclusion,, Int. Conf. on Image Processing, 3 (1998), 259. doi: 10.1109/ICIP.1998.999016. Google Scholar [35] C. A. Micchelli, L. Shen and Y. Xu, Proximity algorithms for image models: Denoising,, Inverse Probl., 27 (2011), 45009. doi: 10.1088/0266-5611/27/4/045009. Google Scholar [36] J.-J. Moreau, Fonctions convexes duales et points proximaux dans un espace hilbertien,, C.R. Acad. Sci. Paris Ser. A Math, 255 (1962), 2897. Google Scholar [37] J.-J. Moreau, Proximité et dualité dans un espace hilbertien,, Bull. Soc. Math. France, 93 (1965), 273. Google Scholar [38] Y. Nesterov, A method of solving a convex programming problem with convergence rate $O(1/k^2)$, (Russian),, Dokl. Akad. Nauk SSSR, 269 (1983), 543. Google Scholar [39] M. Ng, W. Fan and X. Yuan, Inexact alternating direction methods for image recovery,, SIAM J. Sci. Comp., 33 (2011), 1643. doi: 10.1137/100807697. Google Scholar [40] G. Peyre, S. Bougleux and L. Cohen, Non-local regularization of inverse problems,, Inverse Problems and Imaging, 5 (2011), 511. doi: 10.3934/ipi.2011.5.511. Google Scholar [41] L. Rudin S. Osher and E. Fatemi, Nonlinear total variation based noise removal algorithms,, Physica D, 60 (1992), 259. Google Scholar [42] S. Setzer, Operator splittings, Bregman methods and frame shrinkage in image processing,, Int. J. Comput. Vis., 92 (2011), 265. doi: 10.1007/s11263-010-0357-3. Google Scholar [43] G. Steidl, J.Weickert, T. Brox, P. Mrázek and M. Welk, On the equivalence of soft wavelet shrinkage, total variation diffusion, total variation regularization, and sides,, SIAM J. Numer. Anal., 42 (2004), 686. doi: 10.1137/S0036142903422429. Google Scholar [44] X.-C. Tai and C. Wu, Augmented Lagrangian method, dual methods and split Bregman iteration for ROF model,, SSVM 2009, 42 (2009), 502. Google Scholar [45] A. Tikhonov and V. Arsenin, Solution of Ill-Posed Problems,, Winston and Sons, (1977). Google Scholar [46] C. Wu, J. Zhang and X.-C. Tai, Augmented Lagrangian method for total variation restoration with non-quadratic fidelity,, Inverse Problems and Imaging, 5 (2011), 237. doi: 10.3934/ipi.2011.5.237. Google Scholar [47] C. Zalinescu, Convex Analysis in General Vector Spaces,, Singapore: World Scientific, (2002). doi: 10.1142/9789812777096. Google Scholar [48] T. Zeng, Incorporating known features into a total variation dictionary model for source separation,, Int. Conf. on Image Processing, (2008), 577. doi: 10.1109/ICIP.2008.4711820. Google Scholar [49] T. Zeng and F. Malgouyres, Using Gabor dictionaries in a TV-$L^\infty$ model for denoising,, Int. Conf. on Acoust. Speech and Signal Proc., (2006), 865. Google Scholar [50] T. Zeng and M. K. Ng, On the total variation dictionary model,, IEEE trans. on Image Process., 19 (2010), 821. doi: 10.1109/TIP.2009.2034701. Google Scholar [51] X. Zhang, M. Burger, X. Bresson and S. Osher, Bregmanized Nonlocal Regularization for Deconvolution and Sparse Reconstruction,, SIAM J. Imag. Sci., 3 (2010), 253. doi: 10.1137/090746379. Google Scholar [52] M. Zhu and T. Chan, An Efficient Primal-Dual Hybrid Gradient Algorithm for Total Variation Image Restoration,, UCLA CAM Report 08-34 (2008)., (2008), 08. Google Scholar
show all references
##### References:
[1] M. Afonso, J. Bioucas-Dias and M. Figueiredo, Fast image recovery using variable splitting and constrained optimization,, IEEE Trans. Image Process., 19 (2010), 2345. doi: 10.1109/TIP.2010.2047910. Google Scholar [2] L. Ambrosio, N. Fusco and D. Pallara, Functions of Bounded Variation and Free Discontinuity Problem,, Oxford, (2000). Google Scholar [3] G. Aubert and P. Kornprobst, Mathematical Problems in Image Processing: Partial Differential Equations and the Calculus of Variations,, Applied Mathematical Sciences, (2006). Google Scholar [4] A. Beck and M. Teboulle, A fast iterative shrinkage-thresholding algorithm for linear inverse problems,, SIAM J. Imag. Sci., 2 (2009), 183. doi: 10.1137/080716542. Google Scholar [5] M. Bertalmio, G. Sapiro, V. Caselles and C. Ballester, Image inpainting,, in Proc. SIGGRAPH, (2000), 417. doi: 10.1145/344779.344972. Google Scholar [6] S. Boyd and L. Vandenberghe, Convex Optimization,, Cambridge University Press, (2004). doi: 10.1017/CBO9780511804441. Google Scholar [7] J. P. Boyle and R. L. Dykstra, A method for finding projections onto the intersection of convex sets in Hilbert spaces,, Lecture Notes in Statistics, 37 (1986), 28. doi: 10.1007/978-1-4613-9940-7_3. Google Scholar [8] X. Bresson, A Short Note for Nonlocal TV Minimization,, Technical Report, (2009). Google Scholar [9] X. Bresson and T. F. Chan, Fast dual minimization of the vectorial total variation norm and applications to color image processing,, Inverse Problems and Imaging, 2 (2008), 455. doi: 10.3934/ipi.2008.2.455. Google Scholar [10] A. Buades, B. Coll and J. M. Morel, A review of image denoising algorithms, with a new one,, Multiscale Model. Simul., 4 (2005), 490. doi: 10.1137/040616024. Google Scholar [11] A. Bugeau, M. Bertalmio, V. Caselles and G. Sapiro, A Comprehensive Framework for Image Inpainting,, IEEE Trans. Image Process., 19 (2010), 2634. doi: 10.1109/TIP.2010.2049240. Google Scholar [12] E. Candes and F. Guo, A new multiscale transform, minimum total variation synthesis: Application to edge-preserving image reconstruction,, Signal Processing, 82 (2002), 1519. Google Scholar [13] A. Chambolle, An algorithm for total variation minimization and applications,, J. Math. Imag. Vis., 20 (2004), 89. doi: 10.1023/B:JMIV.0000011321.19549.88. Google Scholar [14] A. Chambolle, R. DeVore, N.-Y. Lee and B. Lucier, Nonlinear wavelet image processing: Variational problems, compression, and noise removal through wavelet shrinkage,, IEEE Trans. Image Process., 7 (1998), 319. doi: 10.1109/83.661182. Google Scholar [15] A. Chambolle and T. Pock, A First-Order Primal-Dual Algorithm for Convex Problems with Applications to Imaging,, J. Math. Imaging Vis., 40 (2011), 120. doi: 10.1007/s10851-010-0251-1. Google Scholar [16] T. F. Chan, G. H. Golub and P. Mulet, A nonlinear primal-dual method for total variation-based image restoration,, SIAM J. Sci. Comp., 20 (1999), 1964. doi: 10.1137/S1064827596299767. Google Scholar [17] T. F. Chan, A. M. Yip and F. E. Park, Simultaneous total variation image inpainting and blind deconvolution,, Int. J. of Imaging Systems and Technology, 15 (2005), 92. doi: 10.1002/ima.20041. Google Scholar [18] T. Chan and J. Shen, Image Processing and Analysis: Variational, PDE, Wavelet, and Stochastic Methods,, Society for Industrial and Applied Mathematics (SIAM), (2005). doi: 10.1137/1.9780898717877. Google Scholar [19] C. Chaux, J. C. Pesquet and N. Pustelnik, Nested iterative algorithms for convex constrained image recovery problems,, SIAM J. Imag. Sci., 2 (2009), 730. doi: 10.1137/080727749. Google Scholar [20] P. L. Combettes and V. R. Wajs, Signal recovery by proximal forward backward splitting,, Multiscale Model. Simul., 4 (2005), 1168. doi: 10.1137/050626090. Google Scholar [21] I. Daubechies, Ten Lectures on Wavelets,, SIAM Publ., (1992). doi: 10.1137/1.9781611970104. Google Scholar [22] I. Daubechies, M. Defriese and C. D. Mol, An iterative thresholding algorithm for linear inverse problems with a sparsity constraint,, Commun.Pure Appl. Math., 57 (2004), 1413. doi: 10.1002/cpa.20042. Google Scholar [23] D. L. Donoho and I. M. Johnstone, Ideal spatial adaptation by wavelet shrinkage,, Biometrika, 81 (1994), 425. doi: 10.1093/biomet/81.3.425. Google Scholar [24] D. L. Donoho and I. M. Johnstone, Adapting to unknown smoothness via wavelet shrinkage,, J. Amer. Statist. Assoc., 90 (1995), 1200. doi: 10.1080/01621459.1995.10476626. Google Scholar [25] I. Ekeland and R. Temam, Convex Analysis and Variational Problems,, Studies Math. Appl., (1976). Google Scholar [26] D. Gabay, Applications of the method of multipliers to variational inequalities,, in Augmented Lagrangian Methods: Applications to the Solution of Boundary-Valued Problems, (1983), 299. Google Scholar [27] G. Gilboa and S. Osher, Nonlocal operators with applications to image processing,, Multiscale Model. Simul., 7 (2008), 1005. doi: 10.1137/070698592. Google Scholar [28] T. Goldstein and S. Osher, The split Bregman method for l1 regularized problems,, SIAM J. Imag. Sci., 2 (2009), 323. doi: 10.1137/080725891. Google Scholar [29] S. Lintner and F. Malgouyres, Solving a variational image restoration model which involves contraints,, Inverse. Probl., 20 (2004), 815. doi: 10.1088/0266-5611/20/3/010. Google Scholar [30] J. Liu, X-C. Tai, H. Huang and Z. Huan, A weighted dictionary learning model for denoising images corrupted by mixed noise,, IEEE Trans. on Image Process, 22 (2013), 1108. doi: 10.1109/TIP.2012.2227766. Google Scholar [31] F. Malgouyres, A framework for image deblurring using wavelet packet bases,, Appl. and Comp. Harmonic Analysis, 12 (2002), 309. doi: 10.1006/acha.2002.0379. Google Scholar [32] F. Malgouyres, Mathematical analysis of a model which combines total variation and wavelets for image restoration,, Journal of Information Processes, 2 (2002), 1. Google Scholar [33] F. Malgouyres, Minimizing the total variation under a general convex constraint for image restoration,, IEEE Trans. on Image Process, 11 (2002), 1450. doi: 10.1109/TIP.2002.806241. Google Scholar [34] S. Masnou and J.-M. Morel, Level lines based disocclusion,, Int. Conf. on Image Processing, 3 (1998), 259. doi: 10.1109/ICIP.1998.999016. Google Scholar [35] C. A. Micchelli, L. Shen and Y. Xu, Proximity algorithms for image models: Denoising,, Inverse Probl., 27 (2011), 45009. doi: 10.1088/0266-5611/27/4/045009. Google Scholar [36] J.-J. Moreau, Fonctions convexes duales et points proximaux dans un espace hilbertien,, C.R. Acad. Sci. Paris Ser. A Math, 255 (1962), 2897. Google Scholar [37] J.-J. Moreau, Proximité et dualité dans un espace hilbertien,, Bull. Soc. Math. France, 93 (1965), 273. Google Scholar [38] Y. Nesterov, A method of solving a convex programming problem with convergence rate $O(1/k^2)$, (Russian),, Dokl. Akad. Nauk SSSR, 269 (1983), 543. Google Scholar [39] M. Ng, W. Fan and X. Yuan, Inexact alternating direction methods for image recovery,, SIAM J. Sci. Comp., 33 (2011), 1643. doi: 10.1137/100807697. Google Scholar [40] G. Peyre, S. Bougleux and L. Cohen, Non-local regularization of inverse problems,, Inverse Problems and Imaging, 5 (2011), 511. doi: 10.3934/ipi.2011.5.511. Google Scholar [41] L. Rudin S. Osher and E. Fatemi, Nonlinear total variation based noise removal algorithms,, Physica D, 60 (1992), 259. Google Scholar [42] S. Setzer, Operator splittings, Bregman methods and frame shrinkage in image processing,, Int. J. Comput. Vis., 92 (2011), 265. doi: 10.1007/s11263-010-0357-3. Google Scholar [43] G. Steidl, J.Weickert, T. Brox, P. Mrázek and M. Welk, On the equivalence of soft wavelet shrinkage, total variation diffusion, total variation regularization, and sides,, SIAM J. Numer. Anal., 42 (2004), 686. doi: 10.1137/S0036142903422429. Google Scholar [44] X.-C. Tai and C. Wu, Augmented Lagrangian method, dual methods and split Bregman iteration for ROF model,, SSVM 2009, 42 (2009), 502. Google Scholar [45] A. Tikhonov and V. Arsenin, Solution of Ill-Posed Problems,, Winston and Sons, (1977). Google Scholar [46] C. Wu, J. Zhang and X.-C. Tai, Augmented Lagrangian method for total variation restoration with non-quadratic fidelity,, Inverse Problems and Imaging, 5 (2011), 237. doi: 10.3934/ipi.2011.5.237. Google Scholar [47] C. Zalinescu, Convex Analysis in General Vector Spaces,, Singapore: World Scientific, (2002). doi: 10.1142/9789812777096. Google Scholar [48] T. Zeng, Incorporating known features into a total variation dictionary model for source separation,, Int. Conf. on Image Processing, (2008), 577. doi: 10.1109/ICIP.2008.4711820. Google Scholar [49] T. Zeng and F. Malgouyres, Using Gabor dictionaries in a TV-$L^\infty$ model for denoising,, Int. Conf. on Acoust. Speech and Signal Proc., (2006), 865. Google Scholar [50] T. Zeng and M. K. Ng, On the total variation dictionary model,, IEEE trans. on Image Process., 19 (2010), 821. doi: 10.1109/TIP.2009.2034701. Google Scholar [51] X. Zhang, M. Burger, X. Bresson and S. Osher, Bregmanized Nonlocal Regularization for Deconvolution and Sparse Reconstruction,, SIAM J. Imag. Sci., 3 (2010), 253. doi: 10.1137/090746379. Google Scholar [52] M. Zhu and T. Chan, An Efficient Primal-Dual Hybrid Gradient Algorithm for Total Variation Image Restoration,, UCLA CAM Report 08-34 (2008)., (2008), 08. Google Scholar
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Practical strategy
From: Walter Trice Address: wgt@world.std.com Date: 31 July 1995 Subject: Re: 2 away - 2 away, another aspect Forum: rec.games.backgammon Google: DCLBqn.Crv@world.std.com
```Harald Wittmann asks:
> Why should loner care about gammons? I think it should play like in
> 1-point matches (playing for the highest entire winning probability).
> Is this correct? Any comments?
>
> I thought a little bit longer and did some calculations. Assuming that
> loner's rating is 2000 and it wins 58.5% of the 1-point matches against
> a player with rating 1700, I got (using the rating formula) loner has to
> win 61.9% of the 2-point matches to held its rating.
> I was surprised! Never possible, I thought.
In practice most players do not double on time in a 2 point match
and you can win more 2-pointers than 1-pointers against a
substantially weaker player. There are 3 ways to gain equity from
not doubling:
(1) Opponent loses his market;
(2) he takes a drop;
(3) he drops a take.
I would not forgo these opportunities for equity theft just to
pick up small gains that only occur 2% of the time, which is
usually the situation if you make the first 'optimal' double.
My practical rule is that I prefer not to double until I get
to a position in which an error is possible, and against a
weak opponent such positions are very common!
It is also worth considering that take-points are very different
when one player can win an even position 58.5% of the time. Gammon
rates will also be different -- gammons might be something like
30% of the stronger player's wins but only 15% for the weaker
player. This means that at 2-away/1-away the stronger player
would still have about a 40% chance of winning the match but for
the weaker player it would only be 20%. Thus some racing positions
that are drops for money would still be takes for the underdog.
But positional complexity is also a factor, and there are
conceivably positions so difficult to play that the weaker
player would have to drop at this score though they would be
easy takes between equal opponents.
In short, it becomes marvelously complicated when your weaker
opponent doesn't know when to double. The investment needed
to discover this fact is very small because the weaker player
should ABSOLUTELY double as soon as he thinks he MIGHT have
a market losing sequence.
Walter Trice
```
Match Play at 2-away/2-away
Basic strategy (Darse Billings, Feb 1995)
Counterexample? (Jim Williams+, Mar 1998)
Do you need an advantage to cube? (Keene Marin+, Feb 2006)
Double immediately? (Chuck Bower, Oct 1998)
Ever too good to double? (Kit Woolsey, July 1995)
Minimum game winning chances to double (Walter Trice, Mar 1999)
Practical strategy (Walter Trice, July 1995)
Practical strategy (Albert Steg+, Feb 1995)
Proof for doubling immediately (Robert Koca+, May 1994)
Proof of doubling with market losers (Walter Trice+, July 2001)
Sample game (Ron Karr, Dec 1996) | 777 | 2,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.974767 |
https://kidsjoybox.com/75-divided-by-5-using-long-division/ | 1,721,292,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00659.warc.gz | 289,452,947 | 21,347 | What is 75 divided by 5 using Long Division?
Want to calculate 75 divided by 5 using long division method?
The process of long division is always useful to us when we do not have a calculator or need to solve sums in different cases. Hence we are going to learn the method of long division with a simple example of 75 divided by 5 through the process of long division.
What is 75 divided by 5? The Quotient is 15 and the remainder is 0
Basic concepts of long division
1) The number 75 is the dividend in this case as it is the number which we are about to divide in this case.
2) Similarly, since 5 is the number that is used to divide the greater number 75, it is called the divisor.
3) A divisor always sits on the left of the dividend during the process of long division.
4) A quotient is a value that you will find after dividing the dividend and the remainder is the value that is left behind at the end of all the division steps.
Also read: What is 30/12 as a Mixed Fraction?
Calculations to find 75 divided by 5 using long division
Step 1
First, you need to check whether 5 goes in 7 and how many times with the help of the multiplicative table of 5. Hence we check that 5×1=5, 5×2=10, and so on. Since five can go only one time in the first digit that is 7 we write 1 in the quotient.
Step 2
Now you need to subtract 5 from 7 and see that the result comes as 2.
Step 3
Now you need to bring down 5 so that we now have 25 as the new dividend to be divided in the next step.
Step 4
In the next step now you need to check how many times 5 will go in 25. With the help of the multiplicative table, it is seen that 5×5=25.
Hence,
Hence we see that the final result is 15 and the remainder is 0.
Other methods to calculate using long division
1) You may also use a calculator where you need to type 75÷5 and the results will be shown as 15.
2) The answer can also be expressed in the form of the mixed fraction where 15 is the whole number and 0 is the numerator and 5 is the denominator such that the arrangement will look like 15 0/5.
3) As we know that the whole fraction now becomes 0 since the numerator is 0, we are only left with the whole number 15.
Also read: How to convert 13/16 to decimal? | 571 | 2,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-30 | latest | en | 0.951223 |
https://sports.answers.com/team-sports/How_many_timeouts_does_each_team_have_per_overtime_period | 1,726,438,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00394.warc.gz | 486,540,371 | 48,309 | 0
# How many timeouts does each team have per overtime period?
Updated: 12/16/2022
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Related questions
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One for each team
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You get 3 timeouts per half, and 2 timeouts for each overtime.
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### How many time outs is one team allowed in one half in basketball?
For the NBA, each team has six timeouts for each game, four of which are mandatory and are used at specific times in each of the four quarters. Each team also has two 20-second timeouts which can be used (one in each half, and they cannot be carried over or saved). In overtime, each team receives three full timeouts. For College, in non-televised games each team gets 4-75 second and 2-30 second timeouts per game. In televised games each team gets 1-60 second and 4-30 second timeouts, with an unused 60 second and a maximum of 3 unused 30 second timeouts carrying over to the second half.
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In an NBA game, each team is given six timeouts. They get four 60-second timeouts and two 20-second timeouts.
### How many minutes does a collage game consists?
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1
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### Whow many time out are in the nba?
Each team is entitled to six (6) charged timeouts during regulation play. Each team is limited to no more than three (3) timeouts in the fourth quarter and no more than two (2) timeouts in the last two minutes of regulation play. (This is in addition to one 20-second timeout per half.) In overtime periods each team shall be allowed three (3) 60-second timeouts regardless of the number of timeouts called or remaining during regulation play or previous overtimes. There must be two 100-second timeouts in the first and third periods and three 100-second timeouts in the second and fourth periods. If neither team has taken a timeout prior to 5:59 of the first or third period, it shall be mandatory for the Official Scorer to take it at the first dead ball and charge it to the home team. If no subsequent timeouts are taken prior to 2:59, it shall be mandatory for the Official Scorer to take it and charge it to the team not previously charged. If neither team has taken a timeout prior to 8:59 of the second or fourth period, a mandatory timeout will be called by the Official Scorer and charged to neither team. If there are no subsequent timeouts taken prior to 5:59, it shall be mandatory for the Official Scorer to take it at the first dead ball and charge it to the home team. If no subsequent timeouts are taken prior to 2:59, it shall be mandatory for the Official Scorer to take it and charge it to the team not previously charged. The Official Scorer shall notify a team when it has been charged with a mandatory timeout. Any additional timeouts in a period beyond those which are mandatory shall be 60 seconds. No regular or mandatory timeout shall be granted to the defensive team during an official's suspension-of-play for (1) a delay-of-game warning, (2) retrieving an errant ball, (3) an inadvertent whistle, or (4) any other unusual circumstance.
### How many time outs are allowed in a NBA game?
in the first to quarters each team is allowed to have 3 timeouts. This is the same for the third and fourth quarter's. A team can lose their timeouts if the coach unsuccessfully challenges a play. | 922 | 4,069 | {"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} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.955903 |
https://www.physicsforums.com/threads/groups-and-subgroups.297527/ | 1,553,442,467,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00044.warc.gz | 905,264,541 | 17,365 | # Groups and Subgroups (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### duki
1. The problem statement, all variables and given/known data
Recall that Vm is the set of all invertibles in Z/m
a) List the elements in V15
b) Find all the subgroups of V15
c) Is V15 cyclic? why?
2. Relevant equations
3. The attempt at a solution
From my notes:
a) V15 = {1, 2, 4, 7, 8, 11, 13, 14}
b) {2,4,8,1}{4,1}{7,4,13,1}{11,1}{14,13}{1}
c) No, no generator for whole group.
To be honest, I don't understand how any of these answers were found. I'm lost
#### duki
If I'm listing the invertibles, are they the relatively prime numbers to 15?
#### sutupidmath
there should be a theorem that states that: in Z_m a is said to be an invertibile if and only if gcd(a,m)=1.
proof:
let gcd(a,m)=1. we want to show that a is invertible. First let's see what does it mean for a to be invertible?
a is said to be invertible if there exist a b such that [a]=1.
Now, since gcd(a,m)=1=> that ax+my=1, where x,y are integers.
=>ax-1=(-y)m=> ax=1mod(m) =>[a][x]=[1], so a is invertible.
Now, if a is invertible, it means that there exists some b such that [a]=[1]=> ab=1mod(m)=> ab-1=km=> ab+(-k)m=1=> gcd(a,m)=1, so a and m are relatively prime.
Note:[a] means a mod m
Then a group G, of ord m, is said to be cyclic if there is an element g of order m. In other words, G is cyclic if [g]=G, that is if there is an element such that [g]={g^m:m integer} is the group G itself.
But, in Z_m, there is also a theorem i believe, that says sth like this: a is a generator of V_m if and only if gcd(a,m)=1
Last edited:
#### duki
Oohhhh ok thanks!
Can you provide any help on B) or C)?
#### sutupidmath
Well, if G is a group and g is any element of that group, then the group generated by g, i.e [g] is also a subgroup of G. so this is one way of finding some subgroups of G. For part C) i already gave u a hint in my previous post. What is the order of V_15?
Is there any a in V_15 such that (a,ord{V_15})=1 ?
#### duki
I'm sorry, I don't understand. Do you think you could use {2,4,8,1} as an example?
#### duki
But, in Z_m, there is also a theorem i believe, that says sth like this: a is a generator of V_m if and only if gcd(a,m)=1
How does this differ from an invertible?
#### sutupidmath
How does this differ from an invertible?
It doesn't. Every invertible element is a generator. Why? if a is an invertible element of V_m,what it means is gcd(a,m)=1. So, this means that m is not an integral multiple of a. hence, a^m=1 or ma=1(when dealing with additive groups). i.e. order of a will be m, which would mean that a is a generator as well.
#### sutupidmath
I'm sorry, I don't understand. Do you think you could use {2,4,8,1} as an example?
Well, assuming that this is a group, then it would be cyclic as well. WHy, the order of this group is simply the nr. of its elements. so ord =5. But, since 5 is a prime nr. it means that every nr. smaller than 5, is relatively prime to it, and also every non integral multiple of it.
#### duki
Why is the order 5 instead of 4?
#### sutupidmath
Why is the order 5 instead of 4?
Well, because obviously,i have forgotten to count!
In that case it would mean that it is not cyclic, since none of the elements is relatively prime with 4. 1 is not a generator, since it is the identity(iif the operator is multiplication) and thus 1^m =1 for any m.
#### duki
lol.
I still don't understand how this subgroup was found (part b). Maybe that part of my brain just doesn't work. Can you try one more time to explain it?
#### duki
how do you generate a subgroup from a group? specifically {2,4,8,1} from {0,1,2,4,7,8,10,13,14}
#### sutupidmath
[2]={2^k:k in Z}={2^1=2,2^2=4,2^3=8,2^4=16=1}={2,4,8,1}
notice that 16=1mod(15)
now take 4
[4]={4^k:k in Z}={4^1=4,4^2=16=1}={4,1}
As soon as we reach the multiplicative identity, 1, the elements start to repeat.
Now you do the same with others.
#### duki
great! you have no clue how much you've helped... i've been so upset the past couple of hours. I have a test in the morning and was going to freak if i didn't figure these out.
Could you maybe explain how to find a generator one more time?
#### duki
I made a 89 on my abstract algebra test (should have been a 90, but w/e >.>) !!
Thank you!!
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• Solo and co-op problem solving | 1,405 | 4,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-13 | latest | en | 0.906144 |
https://support.sas.com/documentation/cdl/en/statug/66103/HTML/default/statug_glmpower_details08.htm | 1,627,587,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00511.warc.gz | 564,532,199 | 7,210 | # The GLMPOWER Procedure
#### Contrasts in Fixed-Effect Univariate Models
The univariate linear model has the form
where is the vector of responses, is the design matrix, is the vector of model parameters corresponding to the columns of , and is an vector of errors with
In PROC GLMPOWER, the model parameters are not specified directly, but rather indirectly as , which represents either conjectured response means or typical response values for each design profile. The values are manifested as the dependent variable in the MODEL statement. The vector is obtained from according to the least squares equation,
Note that, in general, there is not a 1-to-1 mapping between and . Many different scenarios for might lead to the same . If you specify with the intention of representing cell means, keep in mind that PROC GLMPOWER allows scenarios that are not valid cell means according to the model specified in the MODEL statement. For example, if exhibits an interaction effect but the corresponding interaction term is left out of the model, then the cell means () derived from differ from . In particular, the cell means thus derived are the projection of onto the model space.
It is convenient in power analysis to parameterize the design matrix in three parts, , defined as follows:
1. The essence design matrix is the collection of unique rows of . Its rows are sometimes referred to as design profiles. Here, is defined simply as the number of unique rows of .
2. The weight vector reveals the relative proportions of design profiles. Row i of is to be included in the design times for every times row j is included. The weights are assumed to be standardized (that is, sum up to 1).
3. The total sample size is N. This is the number of rows in . If you gather copies of the row of , for , then you end up with .
It is useful to express the crossproduct matrix in terms of these three parts,
since this factors out the portion (N) depending on sample size and the portion () depending only on the design structure.
A general linear hypothesis for the univariate model has the form
where is an contrast matrix (assumed to be full rank) and is the null value (usually just a vector of zeros). Note that effect tests are just contrasts that use special forms of . Thus, this scheme covers both effect tests and custom contrasts.
The test statistic is
where
where . Note that if has full rank.
Under , . Under , F is distributed as with noncentrality
Muller and Peterson (1984) give the exact power of the test as
Sample size is computed by inverting the power equation.
See Muller and Benignus (1992) and O’Brien and Shieh (1992) for additional discussion. | 564 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.923 |
https://ambitiousbaba.com/ibps-clerk-pre-reasoning-ability-quiz-4/ | 1,669,756,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00679.warc.gz | 128,638,966 | 51,143 | # IBPS Clerk Pre Reasoning Ability Quiz – 4
## IBPS Clerk Pre Reasoning Ability Quiz
Aspirants have a strong possibility of scoring well in the Reasoning Ability section if they practice quality questions on a regular basis. This section takes the least amount of time if the practice is done every day in a dedicated manner. In this article, we have come up with the IBPS Clerk Reasoning Ability Quiz to help you prepare better. Candidates will be provided with a detailed solution for each question in this IBPS Clerk Reasoning Ability Quiz. This IBPS Clerk Reasoning Ability Quiz includes a variety of questions ranging in difficulty from easy to tough. This IBPS Clerk Reasoning Ability Quiz is totally FREE. This IBPS Clerk Reasoning Ability Quiz has important Reasoning Ability Questions and Answers that will help you improve your exam score. Aspirants must practice this IBPS Clerk Reasoning Ability Quiz in order to be able to answer questions quickly and efficiently in upcoming exams.
Directions (1-5): Study the following information carefully to answer these questions carefully.
There are eight persons namely A, B, P, R, T, X, V and Z sits around the circular table and all of them faces inside of the table. Also, each of them likes a different number i.e. 5, 7, 18, 8, 10, 16, 21 and 15, but not necessarily in the same order.
The person, who like 21 number sits second to the right of A. V sits opposite to T. Immediate neighbor of A sits opposite to B, who likes a number which is a perfect cube. X sits third to the left of P, who likes that number which is multiple of 5. A like 15 number and sits second to the right of that person, who like 18 number. Immediate neighbor of P likes 8 number. T likes 16 number and sits next to P. V sits immediate right of Z, who sits opposite to R. The number like by P is 2 times the number like by V.
1. How many persons sits between A and the one, who like 16 number, when counted anti-clock direction from A?
(a) Four
(b) Three
(c) Two
(d) Five
(e) None of these
Ans. c
2. Who sits third to the right of T?
(a) A
(b) X
(c) R
(d) Z
(e) None of these
Ans. d
3. Which of the following number like by the person, who sits opposite to the one, who likes 8 number?
(a) 8
(b) 7
(c) 15
(d) 21
(e) None of these
Ans. b
4. Four of the following are alike in a certain way, then which of the following does not follow the same pattern?
(a) X
(b) B
(c) P
(d) T
(e) Z
Ans. a
5. Who sits second to the left of X?
(a) R
(b) Z
(c) T
(d) A
(e) None of these
Ans. e
Directions (6-10): Study the following series carefully and answer the questions.
6 1 2 3 5 7 8 9 7 1 3 5 7 4 6 9 5 8 6 1 4 2 6 3 9 7 5 9 4 2 1 8
6. Which of the following digit is at 8th to the right of the one which is 7th from the left end?
(a) 9
(b) 7
(c) 1
(d) 6
(e) None of these
Ans. d
7. Which of the following digit is exactly between the one, which is 16th from left end and which is 9th from right end?
(a) 9
(b) 1
(c) 5
(d) 6
(e) None of these
Ans. b
8. How many even digits are immediately preceded and succeeded by an odd number?
(a) Three
(b) More than three
(c) One
(d) Two
(e) None of these
Ans. d
9. How many odd digits are there which are immediately succeeded by a perfect square?
(a) Three
(b) Four
(c) Five
(d) Six
(e) None of these
Ans. d
10. How many 7’s are in the above statement there which are immediately preceded by 5?
(a) Three
(b) More than three
(c) One
(d) Two
(e) None of these
Ans. d
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3 | 1,209 | 4,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.943134 |
https://s4be.cochrane.org/blog/2018/11/29/what-is-heterogeneity/ | 1,726,003,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00237.warc.gz | 467,680,704 | 13,797 | A network for students interested in evidence-based health care
# Heterogeneity: what is it and why does it matter?
Posted on 29th November 2018 by
Tutorials and Fundamentals
Heterogeneity is not something to be afraid of, it just means that there is variability in your data. So, if one brings together different studies for analysing them or doing a meta-analysis, it is clear that there will be differences found. The opposite of heterogeneity is homogeneity meaning that all studies show the same effect.
It is important to note that there are different types of heterogeneity:
• Clinical: Differences in participants, interventions or outcomes
• Methodological: Differences in study design, risk of bias
• Statistical: Variation in intervention effects or results
We are interested in these differences because they can indicate that our intervention may not be working in the same way every time it’s used. By investigating these differences, you can reach a much greater understanding of what factors influence the intervention, and what result you can expect next time the intervention is implemented.
Although clinical and methodological heterogeneity are important, this blog will be focusing on statistical heterogeneity.
## How to identify and measure heterogeneity
### Eyeball test
In your forest plot, have a look at overlapping confidence intervals, rather than on which side your effect estimates are. Whether the results are on either side of the line of no effect may not affect your assessment of whether heterogeneity is present, but it may influence your assessment of whether the heterogeneity matters.
With this in mind, take a look at the graph below and decide which plot is more homogeneous.
Of course, the more homogeneous one is the plot number 1 . The confidence intervals are all overlapping and in addition to that, all studies favour the control intervention.
For the people who love to measure things instead of just eyeballing them, don’t worry, there are still some statistical methods to help you seize the concept of heterogeneity.
### Chi-squared (χ²) test
This test assumes the null hypothesis that all the studies are homogeneous, or that each study is measuring an identical effect, and gives us a p-value to test this hypothesis. If the p-value of the test is low we can reject the hypothesis and heterogeneity is present.
Because the test is often not sensitive enough and the wrong exclusion of heterogeneity happens quickly, a lot of scientists use a p-value of < 0.1 instead of < 0.05 as the cut-off.
### I²
This test was developed by Professor Julian Higgins and has a theory to measure the extent of heterogeneity rather than stating if it is present or not.
Thresholds for the interpretation of I² can be misleading, since the importance of inconsistency depends on several factors. A rough guide to interpretation is as follows:
• 0% to 40%: might not be important
• 30% to 60%: moderate heterogeneity
• 50% to 90%: substantial heterogeneity
• 75% to 100%: considerable heterogeneity
To understand the theory above have a look at the following example.
We can see that the p-value of the chi-squared test is 0.11, confirming the null hypothesis and thus suggesting homogeneity. However, by looking at the interventions we can already see some heterogeneity in the results. Furthermore, the I² Value is 51% suggesting moderate to substantial heterogeneity.
This is a good example of how the χ² test can be misleading when there are only a few studies in the meta-analysis.
## How to deal with heterogeneity?
Once you have detected variability in your results you need to deal with it. Here are some steps on how you can treat this issue:
• Check your data for mistakes – Go back and see if you maybe typed in something wrong
• Don’t do a meta-analysis if heterogeneity is too high – Not every systematic review needs a meta-analysis
• Explore heterogeneity – This can be done by subgroup analysis or meta-regression
• Perform a random effects meta-analysis – Bear in mind that this approach is for heterogeneity that cannot be explained because it’s due to chance
• Changing the effect measures – Let’s say you use the Risk Difference and have high heterogeneity, then try out Risk Ratio or Odds Ratio
## References
(1) Fletcher, J. What is heterogeneity and is it important? BMJ 2007; 334 :94
(2) Deeks JJ, Higgins JPT, Altman DG (editors). Chapter 9: Analysing data and undertaking meta-analyses. In: Higgins JPT, Green S (editors). Cochrane Handbook for Systematic Reviews of Interventions Version 5.1.0 [updated March 2011]. The Cochrane Collaboration, 2011. Available from www.cochrane-handbook.org.
## No Comments on Heterogeneity: what is it and why does it matter?
• Lucija Anušić
Dear Mr. Sieber,
I am involved in a coordinate-based meta-analysis of neuroimaging studies and would like to know what are the ways of assessing heterogeneity if the only information I have from primary studies are peak coordinates and sample size?
Thank you very much,
Lucija
10th November 2021 at 2:54 pm | 1,095 | 5,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.932514 |
https://www.my-mooc.com/fr/mooc/essential-statistics-for-data-analysis-using-excel/ | 1,537,796,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160454.88/warc/CC-MAIN-20180924125835-20180924150235-00316.warc.gz | 795,882,653 | 22,127 | list 6 séquences
assignment Niveau : Introductif
label Informatique & Programmation
chat_bubble_outline Langue : Anglais
- /5
Avis de la communauté
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## Les infos clés
credit_card Formation gratuite
verified_user Certification payante
timer 18 heures de cours
## En résumé
Gain a solid understanding of statistics and basic probability, using Excel, and build on your data analysis and data science foundation.
What you'll learn
• Descriptive statistics
• Basic probability
• Random variables
• Sampling and confidence intervals
• Hypothesis testing
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dns
## Le programme
Before taking this course, you should be familiar with organizing and summarizing data using Excel analytic tools, such as tables, pivot tables, and pivot charts. You should also be comfortable (or willing to try) creating complex formulas and visualizations. Want to start with the basics? Check out DAT205x: Introduction to Data Analysis using Excel. As you learn these concepts and get more experience with this powerful tool that can be extremely helpful in your journey as a data analyst or data scientist, you may want to also take the third course in our series, DAT206x Analyzing and Visualizing Data with Excel. This course includes excerpts from Microsoft Excel 2016: Data Analysis and Business Modeling from Microsoft Press and authored by course instructor Wayne Winston.
Course Syllabus
Skip Syllabus DescriptionModule 1: Descriptive Statistics
You will learn how to describe data using charts and basic statistical measures. Full use will be made of the new histograms, Pareto charts, Boxplots, and Treemap and Sunburst charts in Excel 2016.
Module 2: Basic Probability
You will learn basic probability including the law of complements, independent events, conditional probability and Bayes Theorem.
Module 3: Random Variables
You will learn how to find the mean and variance of random variables and then learn about the binomial, Poisson, and Normal random variables. We close with a discussion of the beautiful and important Central Limit Theorem.
Module 4: Sampling and Confidence Intervals
You will learn the mechanics of sampling, point estimation, and interval estimation of population parameters.
Module 5: Hypothesis Testing
You will learn null and alternative hypotheses, Type I and Type II error, One sample tests for means and proportions, Tests for difference between means of two populations, and the Chi Square Test for Independence.
record_voice_over
## Les intervenants
Liberty J. Munson
Microsoft
Matthew Minton
Senior Content Publishing Manager
Microsoft
Wayne Winston
Professor Emeritus of Decision Sciences at the Kelly School of Business
Indiana University
store
## Le concepteur
Microsoft Corporation est une entreprise d'informatique et de micro-informatique multinationale américaine, fondée par Bill Gates et Paul Allen. Son activité principale consiste à développer et vendre des systèmes d’exploitation et des logiciels.
assistant
## La plateforme
EdX est une plateforme d'apprentissage en ligne (dite FLOT ou MOOC). Elle héberge et met gratuitement à disposition des cours en ligne de niveau universitaire à travers le monde entier. Elle mène également des recherches sur l'apprentissage en ligne et la façon dont les utilisateurs utilisent celle-ci. Elle est à but non lucratif et la plateforme utilise un logiciel open source.
EdX a été fondée par le Massachusetts Institute of Technology et par l'université Harvard en mai 2012. En 2014, environ 50 écoles, associations et organisations internationales offrent ou projettent d'offrir des cours sur EdX. En juillet 2014, elle avait plus de 2,5 millions d'utilisateurs suivant plus de 200 cours en ligne.
Les deux universités américaines qui financent la plateforme ont investi 60 millions USD dans son développement. La plateforme France Université Numérique utilise la technologie openedX, supportée par Google.
Quelle note donnez-vous à cette ressource ?
Contenu
0/5
Plateforme
0/5
Animation
0/5 | 896 | 4,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.754559 |
https://www.askiitians.com/forums/Algebra/if-the-letters-of-the-word-master-are-permuted-in_78321.htm | 1,680,323,726,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00635.warc.gz | 697,429,837 | 32,260 | # If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the rank of the word '' REMAST ''
SHAIK AASIF AHAMED
askIITians Faculty 74 Points
8 years ago
Hello student,
The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words begin with A is 5! = 120
The number of words begin with E is 5! = 120
The number of words begin with M is 5! =120
The number of words begin with RA is 4! = 24
The number of words begin with REA is 3! = 6
The next word is REMAST .
∴Rank of the word REMAST - 3 (120) + 24 + 6 + 1 = 360 + 31 = 391
Thanks and Regards
Shaik Aasif | 211 | 675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.869452 |
https://docs.google.com/spreadsheets/d/1iwCh0oInliPkSWC5GuK9jSnD0OJYvmdX7G8plEY0r_g/edit?usp=sharing | 1,726,499,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00708.warc.gz | 178,023,988 | 65,712 | ABCDEFGH
1
2
Conversion Value Calculator for Music Schools
3
Instructions: Fill out the cells highlighted in yellow. Try and be as accurate as possible, and think about both the high and low ends of the spectrum when estimating averages.
4
5
Average price of a lesson (in dollars)\$65.00(Typically between \$50-90)Lifetime Value (LTV) of a full-time student:\$4,680
6
Average # of lessons in a month4(Typically between 2-6 lessons/month)Trial Lesson Value:\$3,276
7
Average tenure of students (in years)1.5(Typically between 0.5-3 years)Lead Form Value:\$936
8
If a student books a trial lesson, what percent of them enroll full-time?70%(In percentages, typically between 60-70%)Phone Call Value:\$468
9
If a student or parent fills out the info form, what percent of them enroll full-time?20%(In percentages, typically between 15-30%. Consider that a significant portion of phone calls are likely existing customers, and not prospects.)
10
If a student or parent calls the studio, what percent of them enroll full-time?10%(In percentages, typically between 5-15%. Consider that a significant portion of phone calls are likely existing customers, and not prospects.)
11 | 292 | 1,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.912061 |
saborcatrachorestaurant.com | 1,722,690,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00507.warc.gz | 410,490,132 | 15,197 | # The Math Behind Online Slots
The biggest myth when it comes to online slot machines is dafa casino that you’ll never be able to win money by playing them. Although this may be true in certain cases, there is actually an algorithm behind the game and it’s absolutely possible to earn money from the long run. You must stop believing this and try playing these games – there is no need to worry even if you don’t succeed on the first go. The chances of winning are against the players, but you’ll still make money.
The mathematical rules behind slot games are quite simple. A bet that is red has a 50/50 chance to win, and theoretically, it will. It will double your wager should it happen. A single number bet will pay 36x your stake. This is a lower volatility wager. While lower volatility bets will yield lower winnings while higher volatility bets give you much higher winnings. These games are mathematically the same and it’s difficult to predict which one will yield the best results.
You can increase the odds of winning by understanding how to maximize the Return to Player Ratio. This will enable you to calculate your chances of winning. You should never use mathematical strategies to predict the outcome, because you’ll never be able to ever win every time! You can select games with a high RTP in order to lower the risk of losing money and earn more. If you don’t know how to play online slot machines it is best to conduct some research.
No matter what strategy you choose to use when playing online slots There is always a chance and strategy. Therefore, you must stick to your game plan and not bet more than you’re able to lose. The odds of winning are not identical to playing in casinos. The best genesis method to play online slot machines is to only invest small amounts of money. As you gain experience, your stake can be increased and you will win more often.
When choosing the most suitable online slot machines, it’s crucial to take into consideration the Return to Player (RTP). It’s the percentage of each machine’s payback over time. A 96 percent RTP is the best option if you’re willing to lose \$100 on a slot machine. The more high the RTP is, the greater the chances of winning the game. If you’re willing only a few dollars at one time, you can try roulette or blackjack.
The returns to player value (RTP), value of a slot machine indicates how much you could earn over the course of time. It’s usually expressed as a percentage. For instance, a slot that has 96% RTP will give you \$96 for every \$100 you wager. These numbers are speculative and basing on the exact math that is used in casinos. This is the reason online slot machines are so popular and frequently played.
Online slots offer a better return to player than other games at casinos. The better the RTP, If you don’t win, you can make losses and still win money. You can play online slot machines by selecting games with the highest RTP. Usually you’ll come across a slot with a high reward to player ratio. Understanding how these games function can help you win a lot.
The RTP is an excellent tool to increase your odds of winning and increase your chances of winning. The RTP can be utilized to your advantage, in contrast to other games in casinos. You could increase your chances of win by choosing games that have a high rate of return. A high RTP could increase your chances of winning, which is something you may not think of. You can increase your chances of hitting the jackpot by learning how to determine a slot return to player ratio.
Slot machines are extremely popular and easy to play. However there are some common mistakes that players make. Although there are many strategies that you can use to succeed, luck is the main factor. There are certain slots that have higher RTP than others, but it’s important to keep in mind that you’ll be relying on luck in online slots to earn money. The higher the RTP is, the greater your chances of winning. The game is all about luck. The more you can do to improve your odds of winning the more you can win. | 846 | 4,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.942988 |
https://documen.tv/question/a-car-starts-from-a-full-top-and-in-20-seconds-is-traveling-10-m-per-second-what-is-the-accelera-16807863-81/ | 1,726,573,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00688.warc.gz | 195,914,284 | 16,198 | A car starts from a full top and in 20 seconds is traveling 10 m per second. what is the acceleration
Question
A car starts from a full top and in 20 seconds is traveling 10 m per second. what is the acceleration
in progress 0
3 years 2021-08-31T05:21:46+00:00 1 Answers 8 views 0
1. Answer: acceleration is 0.5 m/s^2
Explanation: From the question, we are given the following parameters
Time t = 20s
Initial velocity U = 0
Final velocity V = 10 m/s
Using first equation of motion,
V = U + at
Where a = acceleration
Substitute all the parameters into the formula to get acceleration a
10 = 0 + 20a
10 = 20a
Make a the subject of formula
a = 10/20 = 0.5 m/s^2
The acceleration is 0.5 m/s^2 | 216 | 707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.833802 |
https://kullabs.com/other/other/ledger-2/balancing-and-closing-of-ledger-accounts | 1,632,391,855,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00405.warc.gz | 384,932,209 | 44,823 | Balancing and Closing of Ledger Accounts
Subject: Other
Overview
The ledger accounts are balanced and closed after all transactions occurred during have been posted therein. Generally, the accounts are balanced and closed at the end of an accounting period. This note has information about balancing and closing of ledger accounts.
Balancing and Closing of Ledger Accounts
The ledger accounts are balanced and closed after all transactions occurred during have been posted therein. Generally, the accounts are balanced and closed at the end of an accounting period. However, they may also be balanced and closed as and when required by the business. The balance of an account is the difference between the total of all debit items and credit items appearing in the account. The following steps are taken while balancing/ closing a ledger account:
• Prepare all required accounts in the ledger book.
• Post all the journal entries in the concerned accounts.
• Total the two sides of the accounts separately by noting the difference.
• Put the difference on the shorter side of the account to make the two sides equal.
• Against the difference, write the words 'To Balance c/d' on the debit side or 'By Balance b/d' on credit side, whatever is the case in the particulars column of the concerned account. The balance used to close the account is called closing balance.
• Write the totals in two amount columns opposite each other and draw two parallel lines below the lines.
• Bring down the amount of balance on the opposite side of the account for the next accounting period. Against the amount, write the words 'To balance b/d' on the debit side or 'By Balance b/d' on the credit side, whatever is the case in the particulars column of the concerned account. The balance used to open the account is called opening balance. The balance with the words 'To balance b/d' is called debit balance and the balance with the words 'By Balance b/d' is called credit balance.
Significance of ledger account balances
The balances of different ledger accounts carry significant meanings. The significant meanings of different types of accounts are as follows:
1. Personal accounts
Personal accounts relate to the persons of different kinds. A person can be either the receiver or the giver of the benefits. Therefore, the different balances of personal accounts signify the following meanings:
• Debit balance
If the personal account has the debit balance, it signifies that the person received more benefits from the business than what s/he has given benefits to it. Therefore, the person is debtor of the business.
• Credit balance
If the personal account shows the credit balance, it implies that the person has given more benefits to the business than what s/he has received from it. Therefore, the person is the creditor of the business.
2. Real accounts
Real accounts relate to assets of different kinds. An asset either comes into the business through purchases or goes out of it through the sale or write off. Since assets can go out from the business unless there is any, it never shows credit balance. Therefore, real accounts have only debit balances.
• Debit balance
If the real account shows the debit balance, it signifies that the business has the particular assets of a certain value on the day of balancing the account.
3. Nominal accounts
Nominal accounts relate to expenses and incomes. Therefore, different balances of nominal accounts show the following significance:
• Debit balance
If the nominal account shows the debit balance, it implies that the business has made expenses of a particular amount on a certain head during a period.
• Credit balance
If the nominal account shows the credit balance, it signifies that the business has earned incomes of a particular amount on a certain head
Things to remember
• The ledger accounts are balanced and closed after all transactions occurred during have been posted therein.
• Generally, the accounts are balanced and closed at the end of an accounting period.
• The balance of an account is the difference between the total of all debit items and credit items appearing in the account.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society. | 854 | 4,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | longest | en | 0.948709 |
https://wiki.math.uoi.gr/index.php/Introduction_to_Computational_Mathematics_(MAE742A) | 1,701,947,589,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00635.warc.gz | 694,579,992 | 8,580 | Introduction to Computational Mathematics (MAE742A)
General
School School of Science Department of Mathematics Undergraduate MAE742A 7 Introduction to Computational Mathematics Lectures (Weekly Teaching Hours: 3, Credits: 6) Special Background - Greek Yes (in English) See eCourse, the Learning Management System maintained by the University of Ioannina.
Learning Outcomes
Learning outcomes Science is based on two major pillars, both theoretical and experimental. However, over the last few decades scientific computing has emerged and recognized as the third pillar of science. Now, in most scientific disciplines, theoretical and experimental studies are linked to computer analysis. In order for the graduate student to be able to stand with claims in the modern scientific and work environment, knowledge in computational techniques is considered a necessary qualification. The course aims to introduce the student into the field of computational mathematics, emphasizing the implementation of numerical methods using computers. The student will be able to familiarize himself with Matlab and Python programming languages, the most widespread for performing scientific calculations. Working autonomously and in groups, the student will be required to implement computational methods related to the fields of numerical analysis and numerical linear algebra. Specifically, the objectives of this laboratory course are: Familiarity with Matlab and Python programming languages to implement numerical methods and graphical design of the numerical solutions Implementation of polynomial interpolation and function approximation Apply numerical integration Solving linear and nonlinear equations Solving systems of linear equations Study of direct and iterative methods. The course aims to enable the student to: Search, analyze and synthesize data and information, using the available technologies Work autonomously Work in a team Promote free, creative and inductive thinking
Syllabus
• Vector and matrix definition and calculations
• Basic commands and functions
• Graphic representation of the numerical results
• Polynomial interpolation: Lagrange Method, Newton's Method
• Numerical integration: Simple and generalized types of numerical integration, rectangular rule, trapezoid rule, Simpson rule, Gauss integration
• Numerical solution of non-linear equations: iterative methods, bisection method, fixed point method, Newton's method
• Numerical solution of linear systems - Direct methods: Gauss elimination, LU decomposition.
Teaching and Learning Methods - Evaluation
Delivery
In the laboratory
Use of Information and Communications Technology Use of scientific computing software packages
Teaching Methods
Lectures 39
Study of bibliography 39
Laboratory exercises 39
Home exercises (project) 33
Course total 150
Student Performance Evaluation
• Weekly assignments
• Final project
• Written examination at the end of the semester
Attached Bibliography
See the official Eudoxus site or the local repository of Eudoxus lists per academic year, which is maintained by the Department of Mathematics. Books and other resources, not provided by Eudoxus:
• Introduction to Numerical Analysis, G.D. Akrivis, V.A. Dougalis, 2010 (in Greek).
• Numerical Linear Algebra, V. Dougalis, D. Noutsos, A. (in Greek).
• A Primer on Scientific Programming with Python, H. P. Langtangen, Springer-Verlag Berlin Heidelberg, 5 Edition, 2016.
• Programming for Computations- MATLAB/Octave, S. Linge, H. P. Langtangen, Springer International Publishing, 2016 (in Greek). | 695 | 3,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.884151 |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-9-review-exercises-page-441/7 | 1,723,552,129,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00315.warc.gz | 609,401,312 | 12,337 | ## Elementary Geometry for College Students (7th Edition)
l$\approx$9.43 cm
l$^2$=8$^2$+5$^2$ l$^2$=64+25=89 l=$\sqrt {89}$ l$\approx$9.43 | 63 | 139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.588309 |
https://www.coursehero.com/file/p6c6e/b-compete-for-resources-a-n-1-a-n-k-1-a-n-k-3-a-n-b-n-b-n-1-b-n-k-2-b-n-k-4-a-n/ | 1,542,663,145,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746112.65/warc/CC-MAIN-20181119212731-20181119234731-00094.warc.gz | 827,013,668 | 62,318 | difference eqns
# B compete for resources a n 1 a n k 1 a n k 3 a n b n
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b compete for resources a n + 1 = a n + k 1 a n - k 3 a n b n b n + 1 = b n + k 2 b n - k 4 a n b n Predator-prey – species b eats species a a n + 1 = a n + k 1 a n - k 3 a n b n b n + 1 = b n - k 2 b n + k 4 a n b n War! a n + 1 = a n - k 3 b n b n + 1 = b n - k 4 a n Phil Hasnip Mathematical Modelling
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Introduction Discrete systems Population analysis Population analysis There are lots of questions we might want to ask about how these models behave, e.g.: What is the long-time behaviour? How sensitive are the solutions to the initial conditions? Can we have sustainable hunting/farming? Phil Hasnip Mathematical Modelling
Introduction Discrete systems Population analysis Long-time behaviour Phil Hasnip Mathematical Modelling
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Introduction Discrete systems Population analysis Malthus Recall the simplest model we looked at, the Malthus model: a n + 1 = a n + ka n = ( 1 + k ) a n = ra n How does its behaviour depend on r ? r = 0 a n + 1 = 0 r = 1 a n + 1 = a n r < 0 oscillatory | r | < 1 decay | r | > 1 growth Phil Hasnip Mathematical Modelling
Introduction Discrete systems Population analysis Malthus What is the equilibrium value? At equilibrium: a n + 1 = a n r = 1 or a n = 0 Phil Hasnip Mathematical Modelling
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Introduction Discrete systems Population analysis Savings account Back to our savings account. Same as Malthus! Include regular withdrawls: a n + 1 = ra n + b Equilibrium: a n + 1 = a n a n = b 1 - r Phil Hasnip Mathematical Modelling
Introduction Discrete systems Population analysis Savings account Equilibrium: a n = b 1 - r r = 1 no equilibrium Otherwise an equilibrium a exists Are the equilibria all the same?
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https://leanprover-community.github.io/archive/stream/116395-maths/topic/nat.2Epow.20and.20ring.html | 1,621,327,752,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00262.warc.gz | 378,792,900 | 3,845 | ## Stream: maths
### Topic: nat.pow and ring
#### Kevin Buzzard (Jun 09 2018 at 01:19):
import tactic.ring
example (d : ℕ) : d ^ 2 + (2 * d + 1) = (d + 1) ^ 2 :=
begin
unfold has_pow.pow monoid.pow nat.pow,
ring
end
#### Kevin Buzzard (Jun 09 2018 at 01:19):
Could I have done that in one line with ring? [using some options or something]
#### Andrew Ashworth (Jun 09 2018 at 01:45):
[deleted - incorrect information]
#### Kevin Buzzard (Jun 09 2018 at 01:52):
So I could make an even cooler ring tactic by writing a tactic which tries to do those unfolds and then applies ring?
#### Kevin Buzzard (Jun 09 2018 at 01:52):
Is life that easy?
#### Andrew Ashworth (Jun 09 2018 at 01:58):
[deleted - incorrect information]
#### Andrew Ashworth (Jun 09 2018 at 01:59):
[deleted - incorrect information]
#### Mario Carneiro (Jun 09 2018 at 05:24):
ring should handle powers... it automatically handles ring like operations that make sense as polynomial expressions, although it can't handle x^n for nonconstant n
#### Mario Carneiro (Jun 09 2018 at 05:26):
in particular it has optimizations for sparse polynomials like x^100 + x, which requires interpreting ^
#### Andrew Ashworth (Jun 09 2018 at 06:25):
that's pretty sweet! I didn't expect that you'd put that much effort into the tactic. thanks for writing it!
#### Kevin Buzzard (Jun 09 2018 at 10:41):
Yes thanks very much indeed for writing it. It is an essential part of the "mathematician's interface" to Lean. Writing it was I'm sure nontrivial but at the end of the day, as I know I've said before, if a mathematician can't prove things like the example above with one or two lines then they will never take to Lean.
#### Kevin Buzzard (Jun 09 2018 at 10:42):
Just to be clear -- in the example above ring falls without the initial unfolding
#### Patrick Massot (Jun 09 2018 at 10:43):
The ring tactic is already very useful but it has bugs
#### Johan Commelin (Jun 09 2018 at 14:51):
How hard would it be to state a theorem about the ring tactic, and prove that the implementation is compliant? Then we are sure we won't have bugs. But I guess that the meta stuff makes this complicated.
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