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3*5*7*...*(2k-1)

= 2*3*4*5*6*7*...*(2k-2)*(2k-1) / (2*4*6*...*(2k-2))

= (2k-1)! / (2^(k-1)*1*2*3*...*(k-1))

= (2k-1)! / (2^(k-1)*(k-1)!)

I used the graphing facilities in MATLAB.  Unfortunately, this is a commercial program, not free open-source.

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You can't - there is no equal sign.  :)

$$(1+3/4)*29=203/4=50.75$$

That  is a great answer Geno.  I had forgotten that that is how it is done.  

For a binomial distribution with an estimated success rate of 0.477 from a sample of 90:

μ  =  np  =  90(0.477)  = 42.93

σ  =  √(npq)  =  √[(90)(0.477)(0.523)]  =  √(22.)  =  4.74

Using the normal approximation to the binomial with a continuity correction:

lower z-score  =  (28.5 - 42.93)/4.74  =  -3.044

upper z-score  =  (29.5 - 42.93)/4.74  =  -2.833

Probability of existing between those two z-scores:  0.0011

A P-score that low allows one to reject the claim that this data matches Benford's law.

You have to look at the gretaest common factors of the denomonators. In other words, what number do 4 and 6 have in common? By quickly going through the numbers in your head, you can see that the GCF is 24. After this step, you can convert each one into a fraction

The probability of getting one tail with one coin is 1/2.

The probability of getting two tails with two coins is 1/2 x 1/2 = 1/4.

The probability of getting three tails with three coins is 1/2 x 1/2 x 1/2 = 1/8.

pi equle 3.14

You can register and become a member......nobody "works" here.......at least.....they haven't sent me any paychecks, yet.....LOL!!!!!

Use atan( )

 0=4(e^(2x)) - e^(-x)    rearrange

e^(-x)  = 4e^2x   (1)

1 /e^(x)  = 4e^(2x)  multilpy both sides by e^x   and divide both sides by  4

1/4 = e^(3x)   take the Ln of both sides

Ln(1/4) = Ln e^(3x)   and, by a log property, we can write

Ln(1/4)  = 3x Lne    and Lne = 1  ....so we have

Ln(1/4)  = 3x     divide both sides by 3

Ln(1/4)/3  = x =  about -0.462

Here's the graph of both sides of (1) above showing the approximate solution.......

The smallest possible number is 0000 and the largest is 9999; this means that there are 10,000 possibilities.

Picking 5 numbers from 41 can be done in C(41,5) = 749,398 ways

And choosing another number from 31 can be done in 31 ways.....so....

749,398 x 31  = 23,231,338  ways

So  1 / 23,231,338  = 0.0000000430453037  = 0.00000430453037% chance of winning on one ticket.....not too good, is it ???

Nice, MG1  !!!

 4 = 0.5x*(5-y)^2 and 1 = 0.5x*(3-y)^2

Rearanging both equations, we have

x = 8/(5- y)^2      and  x = 2/(3-y)^2

This implies that

 8/(5- y)^2  =  2/(3-y)^2      divide both sides by 2

4/(5 - y)^2  = 1/(3-y)^2       cross-multiply  and simplify

4(y^2 - 6y + 9) = y^2 - 10y + 25

4y^2 - 24y + 36  = y^2 - 10y + 25

3y^2 - 14y + 11  = 0    factor

(3y -11) (y - 1) = 0     set each factor to 0

y = 11/3    and y  = 1

And when y = 11/3, x =  9/2        

And when y = 1, x = 1/2

Here's the graph of both    x = 8/(5- y)^2      and  x = 2/(3-y)^2   and the points of intersection

2.5 / 10 =0.25

In addition to Melody, Colin is the only one who isn't accusing anyone else.

Almost as if he's nervous in guilt.

Lol

f(x) = e^(2x) - e^(-x)

f ' (x) = 2e^(2x) + e^(-x)

9+10=21 as 3x3=9 2x5=10 so 3x2=6 3x5=15 15+6=21!

are you that dumb that u don't know it?you won't evenknow what is 9+10

9+10=21

IT IS 55% TRAIN OF THOUGHT!

thanks a lot