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cluponka [151]
3 years ago
10

Solve the equation.

Mathematics
1 answer:
Sphinxa [80]3 years ago
6 0

Answer:

x=-5

Step-by-step explanation:

4-2(x+7)=3(x+5)

Distribute:

4-2*(x)-2*(7)=3*(x)3*(5)

Simplify

4-2x-14=3x+15

Simplify each side by combining like terms

-2x-10=3x+15

Add 10 to both sides

-2x-10+10=3x+15+10\\-2x=3x+25

Subtract 3x from both sides

-2x-3x=3x-3x+25\\-5x=25

Divide both sides by -5

\frac{-5x}{-5} =\frac{25}{-5} \\x=-5

Done

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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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ANSWER
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Ratio of 8th graders to 7th graders = 3 : 2

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_______________________________

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