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erastova [34]
3 years ago
6

Let X and Y be joint continuous random variables with joint density function fpx, yq " # e ´y y 0 ă x ă y, 0 ă y ă 8 0 otherwise

Compute ErX2 | Y " ys. Hint: Start with ErX2 | Y " ys " ż y 0 x 2 fX|Y px |
Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

For X, the parameter

is λX = 1/2 and for Y it is λY = 1. Using the formula sheet, Mean is

E[Z] = E[X] + 2E[Y ] = 4. Variance is V ar(Z) = V ar(X) + 4V ar(Y ) = 8.

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What is the distance from point Yto wx in the figure below?
vovikov84 [41]

Answer:

C

Step-by-step explanation:

The distance YZ is sqrt(34^2-30^2)=16

6 0
3 years ago
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What is the value of x? I don’t understand this please help
Monica [59]

Answer:

54

Step-by-step explanation:

A tangent line makes a right angle with the radius to the point of tangency. Hence this triangle is a right triangle, and x° is the complement of 36°:

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3 years ago
Which expressions are equivalent to Negative 9 (two-thirds x + 1)? Check all that apply.
Sav [38]

<u><em>Answer:</em></u>

<u><em /></u>-9(\frac{2}{3}x) + 9(1)\\  \\-6x + 9

<u><em>Explanation:</em></u>

<u>Before we begin, remember the following rules:</u>

<u>1- Distribution property:</u>

<u />a(b+c) = ab+ac

<u>2- Simplification of fractions:</u>

<u />\frac{xy}{yz}=\frac{x}{z}

<u>3- Signs in multiplication:</u>

+ve * +ve = +ve

-ve * -ve = +ve

+ve * -ve = -ve

<u>Now, for the given problem, we have:</u>

<u></u>9 (\frac{2}{3}x + 1)

<u>Starting with the distributive property:</u>

<u />-9 (\frac{2}{3}x +1) =  -9(\frac{2}{3}x) - (-9)(1) = -9(\frac{2}{3}x) + 9(1)

..................>This corresponds to option 1

<u>Now, we simplify the output from the above step:</u>

<u />-9(\frac{2}{3}x) + 9 = \frac{-9*2}{3}x + 9 = -6x+9

................> This corresponds to option 5

Hope this helps :)

4 0
3 years ago
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Is the solution of b x 5/6 = 25 greater than or less than 25
azamat

Answer:

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Step-by-step explanation:

Simplest way to find out is divide.

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8 0
3 years ago
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Kipish [7]

Answer:

SSS Triangle Congruence.

8 0
3 years ago
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