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solong [7]
2 years ago
12

What is the volume of this right triangular prism? please help!!!! 40 points!! ill mark brainliest ​

Mathematics
1 answer:
OleMash [197]2 years ago
3 0

Answer:

180

Step-by-step explanation:

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What fraction is equivalent to 2/5
Semenov [28]

Answer:

4/10

Step-by-step explanation:

8 0
4 years 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}

5 0
3 years ago
Determine if the Mean Value Theorem for Integrals applies to the function f of x equals the square root of x minus 1 on the inte
german

The theorem does not apply.

1. Set the function f(x)=sqrt(x-1) equal to 0 to find what values of x make f(x) defined. So:

x - 1 is < or = to 0

x > or = to 0

2.) find the values where x is defined:

x - 1 > or = to 0

x > or = to 0

Domain = all values of x that define f(x)

[1,infinity)

{x|x> or = to 1}

DNE because 1 is not in the domain because (1-1)^1/2 = 0.

4 0
3 years ago
Find the whole if the part is 69 and the percent is 12.
wlad13 [49]

Answer:

\frac{ (69 \times 12) }{100} \\  \\  =  \frac{828}{100}  \\  \\  = 8.28

5 0
3 years ago
Read 2 more answers
Types of submarines and the depths are below sea level they can safely reach are down below are shown in the table below. (The p
Damm [24]

Answer:

D

Step-by-step explanation:

The greatest, farthest depths listed are the Guided Missile and Scientific submarines' depths, 850 and 925 feet. So, The answer is D, Guided Missile and Scientific.

Good luck ^^ Hope this helps :)

6 0
3 years ago
Read 2 more answers
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