Rom

5) Just follow the same ideas as in 4

4)

\(\text{the primes that can be rolled are }2,3,5,7,11\\ P[\text{roll a prime}] = \dfrac{1+2+4+6+2}{36} = \dfrac{5}{12}\\ E_A = 15 \cdot \dfrac{5}{12} = \dfrac{25}{4}\\ \text{The multiples of 6 that can be rolled are }6,12\\ P[\text{roll a multiple of 6}]=\dfrac{5+1}{36}=\dfrac 1 6\\ E_B = 42 \cdot \dfrac 1 6 = 7\\ E_B = 7 > \dfrac{25}{4} = E_A\)

2) It should be obvious that the number of people in the group must be a factor of 28

3) \(E=2P[\text{odd}]-6P[\text{even}]\\ P[\text{odd}]=P[1]+P[3]+P[5]=\dfrac 3 4\\ P[\text{even}] = P[6] = \dfrac 1 4\\ E= 2\cdot \dfrac 3 4 - 6 \cdot \dfrac 1 4 = 0\)

\(1)~ \text{The possible values are }\\ 6 = (2,4)\\ 8=(4,4)\\ 9=(2,7)\\ 11=(4,7)\)

\(P[6]=\dfrac{\dbinom{1}{1}\dbinom{2}{1}}{\dbinom{4}{2}}=\dfrac 1 3\\ P[8] = \dfrac{\dbinom{2}{2}}{\dbinom{4}{2}}= \dfrac 1 6\\ P[9] = \dfrac{\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 6\\ P[11]=\dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 3\)

\(\text{We want a distribution of }\left(\dfrac 1 3,\dfrac 1 3,\dfrac 1 3\right) \text{ on cone sizes}\\ \text{And choice D is the only one that accomplishes this}\)

Looks like 10

\(\left( \begin{array}{c} x^4-x \\ x^4-3 x^3+3 x^2-x \\ x^4-x^3 \\ x^4 \\ x^4-2 x^3+x^2 \\ x^4+x^3+x^2 \\ x^4-4 x^3+6 x^2-4 x+1 \\ x^4-x^3-x+1 \\ x^4+x^3+x^2+x+1 \\ x^4+2 x^3+3 x^2+2 x+1 \\ \end{array} \right)\)

I'm going to assume that the pairs aren't labeled, and that order within the pairs and triplet don't matter.

\(\text{Let's form the triplet first}\\ \text{There are }\dbinom{3}{1}=3 \text{ ways to choose the male }\\ \text{and }\dbinom{4}{2}=6 \text{ ways to choose the females for a total of 18 ways}\\ \text{next we form a pair. There are }\dbinom{2}{1}=2 \text{ ways to choose the male}\\ \text{and }\dbinom{2}{1}=2 \text{ ways to choose the female for a total of 4 ways}\\ \text{but the two pairs are unlabeled so we have to divide the total arrangments by }2!=2\\ \text{Thus there are }\dfrac{18\cdot 4}{2}=36 \text{ ways to arrange the people as required.}\)

\(\text{We're talking a 6 digit number that is made up of the digits 0-6, 8-9}\\ \text{For the first digit we have 8 choices}\\ \text{For the remaining 5 digits we have 9 choices each}\\ 8 \cdot 9^5 = 472392\)

flip it about the x axis, then flip it about the y axis

\(\text{Linear growth is a straight line with a single slope}\\ \text{It has the form }L(t) = m t + L_0 \text{ here we have}\\ L_0=2000\\ L(4)=10125 = 4m + 2000\\ m = \dfrac{8125}{4}\\ L(t) = \dfrac{8125}{4}t+2000,~t \text{ in months}\)

I'll let you fill out the table

To answer their question take a look at the difference in earnings between adjacent months.