Rom

\(c_{reg}(n) = 40n \\ c_{special}(n) = 150 + 25n\)

a)   \(c_{reg}(n) = c_{special}(n)\\ 40n=150+25n\\ 15n=150\\ n=10\)

b) you can do this

c) use the special for 10 people or greater.  See part (a)

the sum and product symbols are not defined to use fractional limits.  The index is only ever an integer.

what you can do is multiply the limit by whatever integer is in the denominator of the limit, in this case 2

and then divide the index by that scale factor in the body of the product term

to expand on this a bit permutations of the same digits are indistinguishable.

5,5 is just 5,5, same with 5,5,5, or 5,5,5,5.  They all have only a single permutation.

Whereas n different digits would have n! permutations.

When you go through and take the product of the possibilities like I did it assumes all permutations

are distinguishable.  That's why you then have to divide the result by n!, in this case 2! as there are 2 5's.

If there were 3 5's I'd divide by 3! etc.

CoolStuffYT is correct but here's a more rigorous analysis

mean = (19x9 + 20x6 + 21x7 + 22x3 + 23x2+ 24x1 + 28x2 + 36x1 + 68x1 + 18x4)/(9+6+7+3+2+1+2+1+1+4)

mean = 403/18 ~22.39

4 choices for the 1st digit

4 choices for the 2nd digit

3 choices for the 3rd digit

2 choices for the 4th digit

However because there are 2 5's the number will be halved because of division by 2!

4x4x3x2/2 = 4x4x3 = 48 different 5 digit numbers

\(\dfrac{\dfrac{4}{2x-1}+\dfrac 1 4}{x+1}=\)

\(\dfrac{\dfrac{4\cdot 4 + 2x-1}{4(2x-1)}}{x+1} =\)

\(\dfrac{15+2x}{4(2x-1)(x+1)}\)

\(-1=\sqrt{3x-2}-\sqrt{4x+1}\\ \sqrt{4x+1}-1 = \sqrt{3x-2}\\ (4x+1)-2\sqrt{4x+1}+1 = 3x-2\\ 4x+2 - 2\sqrt{4x+1} = 3x-2\\ x+4=2\sqrt{4x+1}\\ x^2+8x+16 = 4(4x+1) = 16x+4\\ x^2 -8x + 12 = 0\\ (x-6)(x-2) = 0\\ x=6,~2\)

Let's check these solutions

\(\sqrt{18-2} - \sqrt{24+5} = 4 - 5 = -1\\ \sqrt{6-2}-\sqrt{8+1} = 2-3 = -1\)

Both solutions are valid.

36 total marbles

a) P[green] = 8/36 = 2/9

b) P[red] = 10/36 = 5/18

c) P[!yellow] = 1 - P[yellow] = 1 - 12/36 = 1-1/3 = 2/3

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