All Answers 358120 Answers

i want that neck.....

Thanks Bertie,

This is another that I would like to spend more time on as well. :)

Hi anon,

That Guy has got a really long neck LOL

You can upload Gif's like this one directly onto the forum :)

Why don't you try to do that?   :D

Good MG

I want you to learn it too :)

If the variance is 16 then the standard deviation is 4

49 is one standard deviation above the mean.

50% of scores are above the mean and

68%/2 = 34% are between the mean and 1 standard dev above the mean.

SO

(50-34)% = 16% of scores are above 49    

you made a mistake anon.  Would you like to fix it with another post?

By factoring out all the squared numbers.  :)

I am not sure what you want either :/

Maybe this :/

$$\\\frac{6}{7}-9 \\\\
=\frac{6}{7}-1-8 \\\\=-\frac{1}{7}-8\\\\=-8-\frac{1}{7}\\\\
=-8\frac{1}{7}$$

It is hard to understand what you are trying to find. Can you state the problem in another way?

Each place to the right of the decimal represents a division by ten of the previous digit, so .0 is 0/10, .07 is 7/100, etc.

.075 is 75/100

@@ End of Day Wrap   Sun 5/4/15    Sydney, Australia    Time 11:55pm   ♪ ♫

Hi everyone,

It was beautifully quiet today - I wish we had more quiet days. 

Still there are always some great questions and answers.  Today our wonderful answerers were Alan, Geno3141, CPhill, MathGod1, Princessnauwisa and Bertie.  Thanks all  

Interest posts:

Mon 6/4/15

FTJ means "For the Juniors"

Welcome to the Web2 forum Princess

1,219,326,236,500,524,895,589,688,165,822,763,742,208,159,487,018,872,856,201,160,948,321,827,935,253,950,758,389,877,040,172,268

Hi MG

If the asker means   f*45, it should always be written as  45f  

The number must be written first if you are going to have an invisable multiply sign. :)

Here's a verification of the original 19 computer 23 peripheral problem.

The 19 computers are each going to have a peripheral assigned to them, the remaining 4 peripherals can be assigned as (1) all 4 to a single computer, (2) 2 to one computer and 2 to another, (3) 3 to one computer and 1 to another, (4) 2 to one computer and the other 2 to a computer each, (5) each one of the 4 to a single computer.

The number of connection setups for each of these is as follows.

(1) 23C5*19! = 33649*19!

(2) 23C3*20C3*19! / 2! = 1009470*19!

(3) 23C4*19C2*19! = 1514205*19!

(4) 23C3*20C2*18C2*19! / 2! = 25741485*19!

(5) 23C2*21C2*19C2*17C2*19! / 4! = 51482970*19!

Add those together and you get 79781779*19! = 9,705,062,517,250,244,272,128,000 which is the original answer as stated.

I've confirmed that this is the number given by the formula, (but I still don't know how it does it). Maybe later ?

Now Alan, I want to look at yours :)

This first one is very obvious, thank you          

$$\prod_{n=2}^K(2n-1)$$

or as:

Now this I don't  get.  

$$\frac{2^{K+1}\Gamma(K+\frac{1}{2})}{2\sqrt{\pi}}
=\frac{2^k(K-\frac{1}{2})!}{\sqrt{\pi}}\;\;?$$                  Where did $$\sqrt{\pi}$$      appear from?

where Γ() is the Gamma function

$$\Gamma (t)=\int_0^{\infty}x^{t-1}e^{-x}dx$$

It doesn't really matter, I  think this is all be above me.            

7*45=315

Answer

We don't work.

We share our knowledge and apply that knowledge for the future.

Pi = to many numbers that go on forever!

You'll find all number combinations there!!!

Thanks Bertie,   

You know I was really proud of myself getting that far on my own. 

I chewed on it for ages!

I have repeated your work here because I was having problems with it but by the time I had written it all out in LaTex I finally understood,  but I decided just to leave the LaTex here.   

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

So the number of ways that 2k people can be paired off is:

$$3*5*7*9*.....*(2k-1)\\\\
=3*5*7*9*.....*(2k-3)*(2k-1)\\\\
=\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2*4*6*8*...(2k-2)}\\\\
=\frac{1*2*3*4*.....*(2k-3)(2k-2)(2k-1)}{2^{k-1}[1*2*3*4*...(k-1)]}\\\\
=\frac{(2k-1)!}{2^{k-1}(k-1)!}$$

THIS IS GREAT!     

I JUST HAD A THOUGHT.

Can this be extended to seperating into other groups Like 3N separated into groups of 3? 

A project for the future perhaps. 

$$\\\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{k}.$$

The object of the exercise is to show that this is equal to zero if N>k and equal to N! if N=k.

Begin with

$$(x-y)^{N}=\left(\begin{array}{c} N \\ 0 \end{array} \right)x^{N}-\left(\begin{array}{c} N \\ 1 \end{array} \right)x^{N-1}y+\left(\begin{array}{c} N \\ 2 \end{array} \right)x^{N-2}y^{2}- \dots (-1)^{N}\left(\begin{array}{c} N \\ N \end{array} \right)y^{N}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)x^{N-\imath}y^{\imath}$$

Differentiating this wrt x,

$$N(x-y)^{N-1}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)x^{N-\imath-1}y^{\imath}..............(1)$$

and substituting x =  y = 1,

$$0=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)$$                 This proves the result for k = 1.

Multiplying (1) by x and differentiating again,

$$N(x-y)^{N-1}+N(N-1)x(x-y)^{N-2}=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{2}x^{N-\imath-1}y^{\imath}..............(2)$$

and substituting x = y = 1,

$$0=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{2}$$ which is the result for k=2.

Proceeding in this way, (multiply by x, differentiate wrt x and substitute x = y = 1), gets the results for k = 3 ,4,...,  until we arrive at k = N. For that, we find

$$\text{ (a whole load of terms, each containing the factor (x-y))} +N!x=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{N}x^{N-\imath-1}y^{\imath}$$

And now, putting x = y = 1,

$$N!=\displaystyle \sum^{N-1}_{\imath=0}(-1)^{\imath}\left(\begin{array}{c} N \\ \imath\end{array} \right)(N-\imath)^{N}$$ which is the N=k result.

Been away for a few days, but to continue ... .

Melody, N is the number of computers, k the number of peripherals, not the other way round.

To repeat what I said earlier,

if N>k (more computers than peripherals), there is no possible connection setup. A peripheral can be connected to just a single computer, so there are not enough peripherals to go round. In this case the formula returns a value of zero. A proof of this is in the next post.

If N=k then it's simply one peripheral to one computer. There are N! ways of doing this. The formula produces this value.

If N<k, (more peripherals than computers), two or more peripherals will be connected to one or more computers. There will be many ways of doing this and the formula (apparently) produces the number of possibles connection setups. I still don't have a proof of this, but I do have a verification for the 19,23 case. I shall post that later.