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3 years ago

12

The perimeter of the figure below is 54.9 in. Find the length of the missing side.

1 answer:

Hoochie [10]3 years ago

8 0

7.3in have a nice dayyyyyy

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A stone is dropped of a 1296-ft-cliff. The height of the stone above the ground is given by the equation h= - 16t^2+1296, where

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When stone hits the ground, it's height will be zero, and since we're finding the time that's required for the stone to hit the ground, we can set h = 0 and solve for t.

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2 years ago

On its municipal website, the city of Tulsa states that the rate it charges per 5 CCF of residential water is $21.62. How do the

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Answer:

hello your question is incomplete attached below is the complete question

answer :

H0 : <em> u </em>= 21.62 ( Null hypothesis )

H1 : <em> u</em> = 21.62 ( alternative hypothesis )

Step-by-step explanation:

Formulating hypothesis that can be used to determine whether the population mean rate at 5 CCF of residential water discharged by U.S public utilities differs from $21.62

H0 : <em> u </em>= 21.62 ( Null hypothesis )

H1 : <em> u</em> = 21.62 ( alternative hypothesis )

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3 years ago

Which is not a property of countercurrent multiplication?

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It is opposed by the vasa recta.

Step-by-step explanation:

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2 years ago

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To rent a canoe costs $15 for the first hour and $12 for each additional hour or fraction of an hour. Which point is NOT include

pogonyaev

Answer:

A

Step-by-step explanation:

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3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

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3 years ago