Rom

\(\text{Since you are replacing the marbles after each draw the total number of green marbles}\\ \text{you select has a binomial distribution with parameters }n=5,~p=\dfrac 3 5\\ \)

\(P[2] = \dbinom{5}{2} \left(\dfrac 3 5\right)^2 \left(\dfrac 2 5\right)^3 = \dfrac{144}{625}\)

\(V = \pi r^2 h\\ r = \dfrac{6}{2}~in = \dfrac 1 4 ~ft\\ V = \pi \left(\dfrac 1 4\right)^2 \cdot 5 = \dfrac{5\pi}{16}\\ $60\times \dfrac{5\pi}{16} = $\dfrac{75\pi}{4} \approx $58.90\)

you fixed it :D

there's probably a number or two in there that are both perfect squares and perfect cubes

\(P[\text{# is neither perfect cube or square}] = 1 - P[\text{it is}] = \\ 1 \\ -P[\text{# is a perfect cube}]\\ -P[\text{# is a perfect square}] \\ +P[\text{# is both a perfect cube and perfect square}]\)

\(P[\text{perfect cube}]=\dfrac{5}{150}=\dfrac{1}{30}\\ P[\text{perfect square}] = \dfrac{12}{150} = \dfrac{2}{25}\\ P[\text{both}] = \dfrac{2}{150}\)

\(P[\text{# is neither cube or square}] = 1 - \dfrac{1}{30}-\dfrac{2}{25}+\dfrac{2}{150}= \dfrac{9}{10}\)

\(A = \pi r^2 \approx 3.14 r^2 = 314 ~in^2\\ r^2 = 100~in^2\\ r= 10~in\)

\(25^x - 57 = 43\\ 25^x = 100\\ 5^{2x} = 100\\ 2x\log(5) = \log(100) = 2\\ x = \dfrac{1}{\log(5)}\)

\(U=(A\cap W) \cup (A \cap W)^\prime\\ (A \cap W) \cap (A \cap W)^\prime = \emptyset \Rightarrow \\ |U| = |(A\cap W)|+|(A \cap W)^\prime|\\ |A\cap W| =3\\ |(A\cap W)^\prime| = |(A^\prime \cup W^\prime)| = 25\\ |U| = 3 + 25 = 28 \)

as far as I can see the other details aren't needed.

آیا شما یکی دارید؟

\(P(\theta) = \{\cos(\theta),~\sin(\theta)\}\\ P\left(\dfrac{3\pi}{4}\right) = \left\{\cos\left(\dfrac{3\pi}{4}\right),~\sin\left(\dfrac{3\pi}{4}\right)\right\} = \left\{-\dfrac{\sqrt{2}}{2},~\dfrac{\sqrt{2}}{2}\right\}\)

\(\sum \limits_{k=1}^{40}~2k-1 = \\ \left(2\sum \limits_{k=1}^{40}~k\right) - 40 = \\ 40\cdot 41-40 = 40^2 = 1600\)