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3 years ago

PLS HELP ME ASAP FOR 12 AND 14!! (MUST SHOW WORK!!) + LOTS OF POINTS!!

Mathematics

2 answers:

nignag [31]3 years ago

5 0

R for ?
14) 96x1=24r
96=24r
r= 4

Advocard [28]3 years ago

3 0

Both of these problems are proportions. To solve proportions you cross multiply.
#12
4/(-1/2) = 6/x
4x = (-1/2)*6
4x = -3
Divide both sides by 4
x = -3/4
#14
96/24 = x/1
24x = 96
Divide both sides by 24
x = 4

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Express the solution of, x^3 + 2x^2 < x + 2, using interval notation.

Rudiy27

x³ + 2x² - x - 2 < 0

x³ - x + 2x² - 2 < 0

x(x² - 1) - 2(x² - 1) < 0

(x - 2)(x² - 1) < 0

(x - 2)(x - 1)(x + 1) < 0

by chacking we get the solution

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3 years ago

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algol13

Answer:

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Step-by-step explanation:

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Read 2 more answers

Let Y be a random variable with a density function given by

Neporo4naja [7]

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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zloy xaker [14]

Answer:

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Step-by-step explanation:

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