Rom

$slope = \dfrac{y_2-y_1}{x_2-x_1} \\ -3 = \dfrac{y-2}{9-5} = \dfrac{y-2}{4}\\ y-2=-12 \\ y = -10$

two right triangles.. you can figure it out

$\text{we normalize the data by finding }\\ z = \dfrac{X - \mu}{\sigma}\\ z_1 = \dfrac{120-180}{30} = -2\\ z_2 = \dfrac{240-180}{30} = 2\\ \text{using the 68, 95, 99.7 rule we have that}\\ P[-2 \leq z \leq 2] = 0.9545 \\ \text{we want the probability outside this area so we have}\\ P([z<-2) \wedge (2 < 2)] = 1-0.9545 = 0.0455$

$u = -7 - 2v\\ 3u = 4v - 25 \\ 3(-7-2v)=4v-25\\ -21-6v=4v-25\\ 4=10v \\ v=\dfrac 2 5 \\ u = -7 - 2\left(\dfrac 2 5\right) = -\dfrac{39}{5}$

the probabilities must sum up to 1

$1 = P[1]+P[2]+P[3]+P[4]+P[5] = \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6} + k\\ \\ 1 = \dfrac{20+15+12+10}{60}+k\\ \\ 1 = \dfrac{57}{60}+k\\ \\ k=\dfrac{3}{60} = \dfrac{1}{20}$

It's pretty straightforward.

P(X) must sum to 1.  You're given the first 4 values.  Determine the 5th.  The rest are 0.

$E[X] = \sum~x p(x) = \sum \limits_{x=1}^4~x\cdot \dfrac{1}{x+2}+5k$

Use the value of $k=P(5)$ and evaluate that sum.

what a bizarre question...

Well let's get the equation of the body in standard form so we can see what we're dealing with.

$x^2+y^2 - 10x - 12y + 45 = 0\\ \text{complete the two squares on the left hand side}\\ (x-5)^2 - 25 + (y-6)^2 - 36 + 45 = 0 \\ \\ (x-5)^2 + (y-6)^2 = 16 \\ \\ \text{so the body is a circle centered at }(5,6) \text{ with a radius of 4 cm}\\ \\ \text{The overall height is 14cm, 8cm of this is the body, so the diameter of the head is 6cm}\\ \text{And the head must be centered at }(5, 11), \text{ i.e. one radius above the top of the body}\\ \\ \text{so the equation of the head centered at }(5,11) \text{ with radius 3 is}\\ (x-5)^2 + (y-11)^2 = 9$

$P[\text{not get a hit in next 3 at bats}] = \\ \\ \left(P[\text{not get a hit in an at bat}]\right)^3 =\\ \\ \large \left(1-\frac 1 4\right)^3 =\left(\frac 3 4\right)^3 = \frac{27}{64}$

This does assume that consecutive at bats are independent.

12

$\left( \begin{array}{c} \{15\} \\ \{5,10\} \\ \{1,4,10\} \\ \{1,6,8\} \\ \{2,3,10\} \\ \{2,5,8\} \\ \{3,4,8\} \\ \{4,5,6\} \\ \{1,2,4,8\} \\ \{1,3,5,6\} \\ \{2,3,4,6\} \\ \{1,2,3,4,5\} \\ \end{array} \right)$

This is all correct, well done.