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3 years ago

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Helpppppppp pleaseeeee fasttt

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Marta_Voda [28]3 years ago

5 0

Answer:

where is the question

Step-by-step explanation:

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Is 9.55x10 to the 3rd power greater than less than or equal to 5,900

erastovalidia [21]

Answer:

See below.

Step-by-step explanation:

9.55 * 10^3 = 9550.

So it is greater than 5,900.

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3 years ago

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Zepler [3.9K]

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3 years ago

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If Alejandro walks 3 miles in 43 minutes, then how far will Alejandro walk in 93 minutes?

LiRa [457]

Answer:

he will walk about 8 miles

Step-by-step explanation:

4 0

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

6 0

3 years ago

(a) Find an angle between 0 and 2π that is coterminal with −19π10.

Katarina [22]

Answer:

$\dfrac{\pi}{10}$ .

Step-by-step explanation:

All coterminal angles of an angle $\theta$ are defined as

$\theta +2n\pi$ or $\theta + n360^{\circ}$

where, n is an integer.

The given angle is

$\theta=-\dfrac{19\pi}{10}$

So, all coterminal angles of an angle $\theta$ are

$-\dfrac{19\pi}{10}+2n\pi$

For n=1,

$\Rightarrow -\dfrac{19\pi}{10}+2(1)\pi$

$\Rightarrow -\dfrac{19\pi}{10}+2\pi$

$\Rightarrow \dfrac{-19\pi+20\pi}{10}$

$\Rightarrow \dfrac{1\pi}{10}$

Since, $\dfrac{\pi}{10}$ between 0 and 2π, therefore, the required coterminal angle is $\dfrac{\pi}{10}$ .

6 0

2 years ago