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I understand but the computer says its not 5/16

The only ways you can get an even sum is if 

exactly 4 or 2 or 0 dice are even.

4 even     1/(2^4) = 1/16

2even      [1/(2^4) ]* 4!/(2!2!)   = 3/16

0even      1/(2^4) = 1/16

1+3+1=5

So I get a total probability of   5/16

There are 10C5 / 2 ways to split the kids up = 126

If Jenny and Kenny are togther and Lenny is not then there are   7C3 ways to choose the other three in L and K's group.

7C3=35

So I get     35/126

so 4 even + 2 even + 0 even?

50% chance of getting an even on one die

4 even     1/(2^4)

2even      [1/(2^4) ]* 4!/(2!2!)

0even      1/(2^4)

Now add them together.  That's what I think.

(s,t) = (-7/3,2)

f(0) = -72.

The area of the triangle is 20*sqrt(26).

Thanks very much for clarifying Qube73  :)

How did you get that?

This is easy if you use the discriminant.

There are 44 ways.

(a) Each boy received 132 dollars.

(b) The average amoutn of money was 108 dollars.

The answer is sqrt(13).

The answer is -8.

The answer is 7 dollars and 50 cents.

The probability is 3/40.

The answer is 172 miles.

Thanks for the correction. It's all right.

  !