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Gnom [1K]
3 years ago
15

g Let X and Y be two independent standard normal random variables, i.e., each follows the distribution N (0, 1). Now, define two

new random variables as Find cov(Z, W)
Mathematics
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer: hello your question is incomplete below is the complete question

Let X and Y be two independent standard normal random variables, i.e., each follows the distribution N (0, 1). Now, define two new random variables as Z = 11 - X + X²Y, W = 3-Y. find cov ( Z, W )

answer : Cov ( Z, W ) = -1

Step-by-step explanation:

Z = 11 - X + X²Y

w = 3 - Y

Cov ( Z, W ) = Cov ( 11 - x + x^2 Y , 3 - Y )

  = Cov ( 11, 3 ) - Cov ( 11, Y ) - Cov(x,3) + cov (x,y) + cov(x^2y, 3) -cov (x^2y, y )

attached below is the remaining part of the detailed solution

hence

Cov ( Z, W ) = -1

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Describe how (2 cubed) (2 superscript negative 4) can be simplified.
irina [24]

Answer:

1/2

Step-by-step explanation:

Given:

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3 years ago
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8 0
3 years ago
Read 2 more answers
Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.
il63 [147K]

Answer:

The minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

Step-by-step explanation:

Given function is f(x) = 0.15(x + 1)^2 - 3

We need to find the maximum value or the minimum value for the function.

Now, differentiate f(x) = 0.15(x + 1)^2 - 3  w.r.t x.

f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\

f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)

Now, we will equate f'(x)=0 to find critical point.

0.3(x+1)=0\\x=-1

Plug this critical point in to the function f(x) = 0.15(x + 1)^2 - 3  we get,

f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3

Also, f''(x)=0.3 which is positive, We have minimum value.

So, the minimum value for  f(x) = 0.15(x + 1)^2 - 3 is -3.

7 0
3 years ago
a swimming pool is 30 feet long 20 feet wide and 6 feet deep .There are 7.5 gallons of water in 1 cubic foot.How many gallons wi
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4 0
3 years ago
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