Rom

you left out the lines A'C' and AC

the line A'D' is supposed to be solid

the line AD is supposed to be solid

the lines A'E' and  D'E' are supposed to be only dashed

AE and DE are supposed to be dashed only

since this is a 90 degree triangle we can use trig functions

$\sin(35^\circ)=\dfrac{x}{15}\\ x = 15 \sin(35^\circ)~cm$

I'll let you plug and chug those numbers

c) 6 feet is 72 inches.  So total up the bins 71-72 and less.

10+7+6+3 = 26

d) 6' 2" is 74 inches.  So only count the last bin which has 2 in it.

a) $\left( \begin{array}{cc} 4 & \{2,3,6,6,7,7,8,9,9\} \\ 5 & \{0,1,1,1,1,1,2,2,4,4,4,4,4,5,5,5,5,6,6,6,7,7,7,7,8\} \\ 6 & \{0,1,1,1,2,4,4,5,8,9\} \\ \end{array} \right)$

b) just count up the leafs of 5.  There are 25 of them

c) just look at the largest leaf of the largest stem.  You correctly got 69.

$\mathbb{Z} \subset \mathbb{Q} \\ \\ \text{simply says that the integers are a subset of the rationals}$

$\text{The strike zone looks to be }18 \times 24 = 432~sq~in \text{ in area}\\ \text{each of the triangles is }\dfrac{6^2}{2} = 18~sq~in~\text{in area} \\ \text{Thus the probability of hitting any of those triangles is }\\ p=\dfrac{18}{432} = \dfrac{1}{24}$

$\text{The throws are independent so we just multiply the individual probabilities together to}\\ \text{get the probability of all 4 events occurring}\\ p = \left(\dfrac{1}{24}\right)^4 = \dfrac{1}{331776}$

a) $p = \dfrac{1}{13} \dfrac{1}{2} = \dfrac{1}{26}$

b) $p = \dfrac{1}{2}\dfrac{1}{4} = \dfrac{1}{8}$

c) $p = \dfrac{1}{13} \dfrac{3}{13} = \dfrac{3}{169}$

d) $p = \dfrac{1}{52}\dfrac{1}{13} = \dfrac{1}{676}$

a) $p = \dfrac{8}{32}\dfrac{6}{31} = \dfrac{3}{62}$

b) $p = \dfrac{6}{32}\dfrac{6}{31}= \dfrac{36}{992} = \dfrac{9}{248}$

c) $p=\dfrac{6}{32}\dfrac{5}{31} = \dfrac{15}{496}$

so we need the integers, -4 < n < 2, i.e.

-3, -2, -1, 0, 1

$P[\text{random student is female}] = \dfrac{22}{22+18} = \dfrac{22}{40}=\dfrac{11}{20}$

$P[\text{random student is Mathematics major}] = \dfrac{18}{40} = \dfrac{9}{20}$