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3 years ago

Plz I need it now I am stuck

Mathematics

1 answer:

vredina [299]3 years ago

5 0

Answer:

116

Step-by-step explanation:

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IrinaVladis [17]

How sad, that you didn't include the list of choices.

The volume of the cone is (1/3) x (π) x (radius²) .

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4 years ago

Find the area of this circle. Use 3.14 for pie?

pshichka [43]

Answer:

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Step-by-step explanation:

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3 years ago

Solve 5.7y - 3.9y I is not smart lolz

denpristay [2]

5.7 of anything minus 3.9 of the same thing leaves 1.8 of them.

8 0

4 years ago

Solve for xxx. Enter the solutions from least to greatest. x^2 +7=43x 2 +7=43x, squared, plus, 7, equals, 43 \text{lesser }x =le

vaieri [72.5K]

Answer:

Step-by-step explanation:

x²+7=43x

x²-43x+7=0

$x=\frac{43 \pm \sqrt{(43)^2-4(1)(7)} }{2(1)} \\=\frac{43 \pm \sqrt{1849-28} }{2} \\=\frac{43 \pm \sqrt{1821} }{2} \\\approx \frac{43 \pm 42.673}{2}\\\approx 0.1635,42.8365$

4 0

4 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

4 years ago