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3 years ago

Hi!!!

Mathematics

1 answer:

jekas [21]3 years ago

3 0

Answer:

im srry but ion kno how to do this-bryan spieker

Step-by-step explanation:

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For 15 points Select the correct answer. What is |-14|?

Tasya [4]

Answer: 14

Step-by-step explanation: The absolute value is always positive.

7 0

3 years ago

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Calculate 85/5 - 12*3 + 4*3 =

Leokris [45]

Answer:-7

Step-by-step explanation:

85/5 = 17

12*3=36

4*3=12

17-36+12

-19+12

-7

4 0

3 years ago

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What is the m∠ABC? 1)m∠ABC = 60° 2)m∠ABC = 67° 3)m∠ABC = 120° 4)m∠ABC = 127°

docker41 [41]

we are given

m∠BCD =67

and m∠BDC=60

we know that

m∠ABC is exterior angle

m∠BCD and m∠BDC are interior angles

exterior angle is sum of interior angles

so, we can write it as

m∠ABC=m∠BCD+m∠BDC

now, we can plug values

and we get

m∠ABC=60+67

m∠ABC=127

so, option-4.........Answer

7 0

4 years ago

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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

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

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

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

5 0

3 years ago

Simplify the expression x^2+3x-28/x^2-7x+12

EleoNora [17]

X^2 + 3x - 28 factors to (X+7)(X-4)
X^2 - 7X + 12 factors to (X-4)(X-3)
Now we see that the equation is [(X+7)(X-4)]/[(X-4)(X-3)]
By dividing the top and bottom by (X-4), we see that the equation equals (X+7)/(X-3).

6 0

3 years ago

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