Rom

$E[R] = 5((0.4)\cdot 0 + (0.25)\cdot 4 + (0.35)\cdot 10)~in = 22.5~in$

$x^2 + m x + n = 0,~(m,n)\in \mathbb{Z} \\ \text{the only possible value for x is }x = -3\\ x^2 + m x + n = (x+3)^2 = x^2 + 6x + 9\\ m = 6$

$n \text{ identical objects can be placed into }r \text{ distinguishable boxes in }\\ N=\dbinom{n+r-1}{r-1} \text{ ways}\\ \text{In this problem }n=4,~r=3\\ N=\dbinom{6}{2}=15 \text{ ways}$

$P[\text{product is even}]=1-P[\text{product is odd}] = 1 - P[\text{all three dice roll odd}]$

$P[\text{1 die rolls odd}]=\dfrac 1 2 \\ P[\text{product is odd}]=P[\text{all 3 dice roll odd}]=\left(\dfrac 1 2\right)^3 = \dfrac 1 8\\ P[\text{product is even}]=1 - \dfrac 1 8 = \dfrac 7 8$

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Are the teams the same size?

$P[\text{3rd 6 on roll 10}]=P[\text{2 6's in 9 rolls and a 6 on the 10th roll}=\\ \dbinom{9}{2}\left(\dfrac 1 6\right)^2 \left(\dfrac 5 6\right)^7 \cdot \dfrac 1 6 = \dfrac{78125}{1679616}$

$C = 2 \pi r\\ A = \pi r^2 = \dfrac{C^2}{4\pi}\\ C= \sqrt{4\pi A} = \sqrt{1600\pi}=40\sqrt{\pi}~cm$