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kirill115 [55]
3 years ago
9

(1 pt) let x be an exponential random variable with parameter \lambda = 4, and let y be the random variable defined by y = 2 e^x

. compute the probability density function of y:
Mathematics
1 answer:
kupik [55]3 years ago
4 0
Y=2e^X\iff X=\ln\dfrac Y2

Note that x\in[0,\infty), so that y\in[2,\infty).

F_Y(y)=\mathbb P(Y\le y)=\mathbb P(2e^X\le y)=\mathbb P\left(X\le\ln\dfrac y2\right)=F_X\left(\ln\dfrac y2\right)

F_X(x)=1-e^{-4x}
\implies F_Y(y)=1-e^{-4\ln\frac y2}=1-\dfrac{16}{y^4}

\implies f_Y(y)=\dfrac{\mathrm dF_Y(y)}{\mathrm dy}=\dfrac{64}{y^5}
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  B.) y=csc x

Step-by-step explanation:

The cosecant function has a minimum magnitude of 1, so its range excludes any values in the range -1 < y < 1.

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3 years ago
A pole that is 3.4 m tall casts a shadow that is 1.54 m long. At the same time, a nearby building casts a shadow that is 35.75 m
inna [77]

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The building would be 78.92 meters tall.

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3 years ago
A movie collector has 105 movie posters and 90 band posters if she sells 21 movie posters then how many band posters should she
kotykmax [81]

Hi there! :)

<u>Answer:</u>

In order to maintain the same ratio, she should sell 18 band posters.


<u>Step-by-step explanation:</u>

<u>Initial ratio:</u>

Ratio is: 105 movie posters for 90 band posters = <u>105 : 90</u>


She sells 21 movie posters, which means that we are left with 105 movies minus the 21 she sold.

105 - 21 = 84 → She now has <u>84</u> movie posters.


In order to answer your question, you'll need to use the cross product method:

105 movie posters → 90 band posters

84 movie posters  → X

<u>(90 × 84)</u> ÷ 105 = X

<u>7,560 ÷ 105</u> = X

72 = X ⇒ Number of band posters she should have left to maintain the same ratio.

She has 90 band posters :

90 <u>- X</u> = 72

Add "X" to each side of the equation → 72 + X = 72 + X

90 = <u>72</u> + X

Subtract 72 from each side of the equation → 90 - 72 = 18

<u>18 = X</u>


There you go! I really hope this helped, if there's anything just let me know! :)

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