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3 years ago

4. Please help thanks

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

7 0

I got A I hope this helps

katrin [286]3 years ago

4 0

I got 48(D)

i found this by using pythagorean therom to find the hypotenuse (20) then added them all up (20+16+12) which = 48

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Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

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Recall that, by definition

$E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy$

Also, we are given that $C = 50+3Y+9Y^2$ . Recall the following properties of the expected value. If X,Y are random variables, then

$E(a+bX+cY) = a+bE(X)+cE(Y)$

Then, using this property we have that $E[C] = 50+3E[Y]+ 9E[Y^2]$ .

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

$E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3$

$E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}$

Then

$E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152$

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