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Leya [2.2K]
3 years ago
12

How do I find the length of this?

Mathematics
2 answers:
olga55 [171]3 years ago
4 0
Volume is length times width times height so you would take 10656/(24*18.5)=24 inches
Nookie1986 [14]3 years ago
3 0
Use the formula V = Bh and plug in the given numbers. So the formula would be 10,656 = l*24*18 1/2 and you would get length = 24
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Let Y be a random variable with a density function given by
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From the given density function we find the distribution function,

F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y

(a)

F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)

\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1

\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}

(b)

F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y

\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2

\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}

(c)

F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})

\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3

\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}

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3 years ago
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