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4 years ago

Help! I don’t understand 7 & 8!

Mathematics

2 answers:

Advocard [28]4 years ago

8 0

Hi!

For #7 V stands for volume, H stands for height, L stands for length, and W stands for width. So the formula V = lwh means volume is equal to length times width times height.

For #8 Ax+By=c is a standard formula with which you plug in the appropriate number in for A and B and C to solve for x or y.

-ASIAX Frequent Answerer

german4 years ago

7 0

Unless there are numbers involved, it’s just asking to rearrange the equation. 7. W = V/(lh) and 8. X = (C -By)/A

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Help me with my homework idk how to do it

QveST [7]

Answer:

You have to make an equation

Make everything equal to 180 degrees because a triangle makes 180 degrees

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Step-by-step explanation:

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3 years ago

It cost 17 and $.70 send 170 in text messages how many text messages did you send if you spent $19.10?

ZanzabumX [31]

Answer:

184

Step-by-step explanation:

17.70 / 170 = .104 per text

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6 0

3 years ago

Celeste wants to have her hair cut and permed and also go to lunch. She knows she will need $68. The perm cost twice as much as

lutik1710 [3]

I'm pretty sure it's $42

4 0

3 years ago

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

4 years ago

Solve the system of equations

Brums [2.3K]

5x+4y = 73
-4x + 4y = -8

5x + 4y -(-4x+4y) = 73 -(-8)
5x + 4y + 4x - 4y = 81
5x + 4x + 4y - 4y = 81
9x = 81
x = 81÷9
x = 9

Substitute x = 9 into any of the equations to find y.

5(9) + 4y = 73
45 + 4y = 73
4y = 73 - 45
4y = 28
y = 28÷4
y=7

Answers :
x = 9
y = 7

8 0

3 years ago

Read 2 more answers