Member: Nauseated

$\text {Calculate the compression load in terms of the cross-sectional area.}\\ \displaystyle {3^2*\pi = 28.27 {in}^2} \Rightarrow \dfrac{150 {~kilopounds}}{28.27 {in}^2} = 5.305 {~ksi}\\\ \text {Calculate the longitudinal compression }\\ \dfrac {5.305}{8000}*12 in = 0.0079575{in}\\$

$\text {Calculate the expansion in the radial direction.}\\ \dfrac{5.305}{8000}(0.35)*6{in} = 0.00139256 {in}$

$\text {Final length } (12 - 0.00796) = 11.99204in\\ \text {Final diameter } ( 6.0 + 6.00139) = 6.00139in$

Source:

Formulae Esoterica

Processus algorythmic

Twelfth edition

Morgan le Fay

NauseatedMar 13, 2016

+1038

+10

Mathbitch, you obviously know something, but your primary purpose seems to be to complain about members and guests not following your interpretation of the diaphanous rules for posting on the forum. You are a perpetual complainer extraordinaire, whose need to bÍtch is insatiable.

You have an OCD personality that is contrasted in hypocrisy when you make irrelevant posts consistent with the very ideas you complain about. You are an aberrant personality who pleads for posters to make relevant math related posts, only to respond to them with caustic retorts, irrelevant drivel, and, very often, errant math.

Mathbitch you are an intellectual pygmy, who covers your runt-hood with complaints and bitching, while waiting to reach the status of imbecile.

This is my MathbÍtch presented in all its glory and splendor.

NauseatedMar 9, 2016

#16

+1038

Errata:

Corrected to indicate exponent in term of general expansion

$- \binom{6}{5}g^{(y)} + 1$

Corrected odd term subtraction

$\left(\dfrac{6^{11}}{11!}\right) - 6 \left(\dfrac{5^{11}}{11!}\right) + 15 \left(\dfrac{4^{11}}{11!}\right) - 20 \left(\dfrac{3^{11}}{11!}\right) + 15 \left(\dfrac{2^{11}}{11!}\right) -6\left(\dfrac{1^{11}}{11!}\right)$

------------

Wolfram alpha link for

Sum[Binomial[6, i] (-1)^i (6 - i)^11, {i, 0, 10}]

https://www.wolframalpha.com/input/?i=sum+%28binom%286,i%29*%28-1%29^i*%286-i%29^11%29+from+i%3D0+to+10

NauseatedMar 2, 2016

#15

+1038

Your logic is excellent, Melody.

This helps to explain the formula referenced here.

http://web2.0calc.com/questions/this-is-a-continuation-of-an-earlier-counting-question-that-nauseated-and-geno-answered

$\displaystyle \sum \limits_{i=0}^{N-1} *(-1)^i*\binom{N}{i}* (N-i)^k\\ \text {Current problem}\\ \displaystyle \sum \limits_{i=0}^{10} *(-1)^i*\binom{N}{i}* (N-i)^{11} = 129230640$

This formula is a modification of the formula to generate Stirling numbers of the second kind.

$\displaystyle \text{Sterling number -2}^{nd} {~type:~} \dfrac{1}{N!}\sum \limits_{i=0}^{N-1} *(-1)^i*\binom{N}{i}* (N-i)^{k}$

The modification is to multiply by N!.

Wolfram input

StirlingS2[11, 6] = 179487

Multiplying by 6! Returns value above.

StirlingS2[11, 6]*6! = 129320640

Stirling numbers of the second kind generate solutions to bąll and box distributions where the bąlls are unique and the boxes are identical.

This die rolling problem is equivalent to a bąll (die roll) and box (die face) problem where both the bąlls and boxes are unique, requiring an increase in the counts of solutions by a factor of 6!

General form:

$\dfrac{11!}{(B_1!)*(B_2!)*(B_3!)*(B_4!)*( B_5!)* (B_6!)} = \text{unique distributions}\\ \text{where } {B_1+B_2+B_3+ B_4+ B_5+B_6 =11} \text{ with } {B_i} \ne 0$

Expanded this by using a generating function for polynomials

[g^y - 1]^6 (The minus 1 correlates to having no empty box).

The general polynomial expansion is

$g^{(6y)} - \binom{6}{1}g^{(5y)} + \binom{6}{2}g^{(4y)} - \binom{6}{3}g^{(3y)} + \binom{6}{4}g^{(2y)} - \binom{6}{5}{g(y)} + 1$

$\text{Now in terms of }~ y^{11} \text{ for the 11 rolls of the die.}$

$\left(\dfrac{6^{11}}{11!}\right) + 6 \left(\dfrac{5^{11}}{11!}\right) + 15 \left(\dfrac{4^{11}}{11!}\right) + 20 \left(\dfrac{3^{11}}{11!}\right) + 15 \left(\dfrac{2^{11}}{11!}\right) +6\left(\dfrac{1^{11}}{11!}\right)$

$\text{Factor out the } \left(\dfrac{1}{11!}\right)\\ \left(\dfrac{1}{11!}\right) [(6^{11} - 6(5^{11}) + 15(4^{11}) - 20(3^{11}) + 15(2^{11}) - 6(1^{11})]\\ = \left(\dfrac{1}{11!}\right) \left(129230640\right) \\$

Note the similarity to the modification of the sterling number formula, where (6!) is factored out.

$\dfrac{129230640}{11^6} = 0.0.3562064186099 \\ \text { (Probability of rolling at least one of each number (1-6) for 11 rolls of a fair die)}$

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A Markov matrix may also be used to solve this type of problem.

NauseatedMar 2, 2016

+1038

+10

Lick your nose and that will attract their attention. IF you can lick your eyebrows then they will be lined up for you.

NauseatedFeb 23, 2016

#16

+1038

LMFAO This is goddamnn funny

NauseatedFeb 23, 2016

+1038

+10

Look at the graph and note that strain C has the least virulence in March. Use this point in time and calculate the weighted average to determine the efficacy of the vaccine.

To calculate the weighted average, take the (March) prevalence of each strain and multiply it by the efficacy of the vaccine in the table. Then add these products.

$(0.35 * 0.23) + (0.13 * 0.25) + (0.76 * 0.13) + (0.68 * 0.39) =0.447 ~|~(47.7\%)$

---

Now if only we could develope a vaccine for CDD.

NauseatedFeb 15, 2016

+1038

+15

How do you think that the moderators should behave differently?

What you like us to do differently.

The other moderators are fine the way they are. However, for you, Melody, there is something I’d like you to do differently. I would like you to change the way you present my Valentine's greeting this year. Last year’s Valentine was kind of “ho hum,” not much spirit. It was just plain, bland. It was so bland it didn’t even exist.

This year, I would like a post with heart-felt romance when you ask me to be your Valentine. It doesn’t need to be too over-the-top, just give it more spirit and sincerity.

Here’s an example:

To my dearest Nauseated, I beseech thee from the bottom of my heart to be my Valentine this year. Heart-felt pleas, with pretty please and sugars on top. :)

You can use this or write your own. You have more than 3 weeks to compose a glorious Valentine greeting. That’s long enough to make it a spare time project.

Thank you for the invite. This should be helpful and I know it is interesting.

NauseatedJan 18, 2016

+1038

+11

Melody, you left my name off the list of great answers. I’m sure it was just an oversight and not troll-discrimination or anything like that. Now, I know you will say I didn’t answer the question, but I did reply to the one that did, pointing out the CDD in its very virulent – spooky dumbness at a distance – form. I do this as a public service and that has the same or even greater value as answering the question.

BTW I did answer it. It was a major chore but I managed.

NauseatedJan 15, 2016

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