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3 years ago

What is the measure of angle C? 25° 30° 60° 75°

2 answers:

5 0

Hello, My Dear Friend!

The Correct Answer To This Question Is 100%:

"Option" D". "<span>75°"

Therefore, That Is The Correct Answer!^^^

</span>I Hope my answer has come to your Help. Thank you for posting your question here in $Brainly.$ We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

$(Ps. Mark As Brainliest IF Helped!)$

$-UniteTheSeven, UniteTheLeague, Justice League!!! :D$

<span> $-TheOneAboveAll :)$ </span>

VLD [36.1K]3 years ago

5 0

Answer:

75°

Step-by-step explanation:

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What is the vertex of the graph of y = –x2

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The parabola goes down(makes a sad face) because it is negative. The answer is 0,0

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3 years ago

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Which equation represents a line which is parallel to the line 2x+y=-8?A. y=− 1/2 x−4B. y=−2x+5C. y=2x+3D. y= 1/2x+2

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Two lines that are parallel have the same slope. In its slope-intersect form, we can write the equation of a line with slope m and y-intercept b as:

$y=mx+b$

Step 1

Write the given equation in slope-intercept form and identify its slope m.

$\begin{gathered} 2x+y=-8 \\ \\ 2x+y-2x=-2x-8 \\ \\ y=-2x-8 \end{gathered}$

Thus:

$m=-2$

Step 2

Find the equation with the same slope m = -2. We need to identify which of them has -2 multiplying the variable x.

Answer

From the given options, the only one with the same slope m = -2, therefore parallel to the given line, is:

$y=-2x+5\text{ (option B)}$

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1 year ago

Let X and Y have the joint density f(x, y) = e −y , for 0 ≤ x ≤ y. (a) Find Cov(X, Y ) and the correlation of X and Y . (b) Find

adoni [48]

a. I assume the following definitions for covariance and correlation:

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

$\mathrm{Corr}[X,Y]=\dfrac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}}$

Recall that

$E[g(X,Y)]=\displaystyle\iint_{\Bbb R^2}g(x,y)f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

where $f_{X,Y}$ is the joint density, which allows us to easily compute the necessary expectations (a.k.a. first moments):

$E[XY]=\displaystyle\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$

$E[X]=\displaystyle\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$

$E[Y]=\displaystyle\int_0^\infty\int_0^yye^{-y}\,\mathrm dx=2$

Also, recall that the variance of a random variable $X$ is defined by

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We use the previous fact to find the second moments:

$E[X^2]=\displaystyle\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$

$E[Y^2]=\displaystyle\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$

Then the variances are

$\mathrm{Var}[X]=2-1^2=1$

$\mathrm{Var}[Y]=6-2^2=2$

Putting everything together, we find the covariance to be

$\mathrm{Cov}[X,Y]=3-1\cdot2\implies\boxed{\mathrm{Cov}[X,Y]=1}$

and the correlation to be

$\mathrm{Corr}[X,Y]=\dfrac1{\sqrt{1\cdot2}}\implies\boxed{\mathrm{Corr}[X,Y]=\dfrac1{\sqrt2}}$

b. To find the conditional expectations, first find the conditional densities. Recall that

$f_{X,Y}=f_{X\mid Y}(x\mid y)f_Y(y)=f_{Y\mid X}(y\mid x)f_X(x)$

where $f_{X\mid Y}$ is the conditional density of $X$ given $Y$ , and $f_X$ is the marginal density of $X$ .

The law of total probability gives us a way to obtain the marginal densities:

$f_X(x)=\displaystyle\int_x^\infty e^{-y}\,\mathrm dy=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^ye^{-y}\,\mathrm dx=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$

Then it follows that the conditional densities are

$f_{X\mid Y}(x\mid y)=\begin{cases}\frac1y&\text{for }0\le x$

$f_{Y\mid X}(y\mid x)=\begin{cases}e^{x-y}&\text{for }0\le x$

Then the conditional expectations are

$E[X\mid Y=y]=\displaystyle\int_0^y\frac xy\,\mathrm dy\implies\boxed{E[X\mid Y=y]=\frac y2}$

$E[Y\mid X=x]=\displaystyle\int_x^\infty ye^{x-y}\,\mathrm dy\implies\boxed{E[Y\mid X=x]=x+1}$

c. I don't know which theorems are mentioned here, but it's probably safe to assume they are the laws of total expectation (LTE) and variance (LTV), which say

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

$\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]$

We've found that $E[X\mid Y]=\frac Y2$ and $E[Y\mid X]=X+1$ , so that by the LTE,

$E[X]=E[E[X\mid Y]]=E\left[\dfrac Y2\right]\implies E[Y]=2E[X]$

$E[Y]=E[E[Y\mid X]]=E[X+1]\implies E[Y]=E[X]+1$

$\implies2E[X]=E[X]+1\implies\boxed{E[X]=1}$

Next, we have

$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\dfrac{Y^2}3-\left(\dfrac Y2\right)^2\implies\mathrm{Var}[X\mid Y]=\dfrac{Y^2}{12}$

where the second moment is computed via

$E[X^2\mid Y=y]=\displaystyle\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$

In turn, this gives

$E\left[\dfrac{Y^2}{12}\right]=\displaystyle\int_0^\infty\int_0^y\frac{y^2e^{-y}}{12}\,\mathrm dx\,\mathrm dy\implies E[\mathrm{Var}[X\mid Y]]=\frac12$

$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\dfrac Y2\right]=\dfrac{\mathrm{Var}[Y]}4\implies\mathrm{Var}[E[X\mid Y]]=\dfrac12$

$\implies\mathrm{Var}[X]=\dfrac12+\dfrac12\implies\boxed{\mathrm{Var}[X]=1}$

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3 years ago

A city park commission received a donation of playground equipment from a parents' organization. The area of the playground need

soldi70 [24.7K]

A = (P/4)(P/2 - P/4)
= (P/4)^2

Therefore it is a square
Playground
256 yd^2
P least perimeter

256 = (P/4)^2
P/4 = √256
= 16
P = 4*16
= 64

The least fencing is 64 yards.

Hope I helped!

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4 years ago

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A 50% increase followed by a 33% decrease leaves you with a constant 17% increase. It is less than the original.

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