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3 years ago

PLEASE HELP ME I WILL MARK BRAINLIEST (ITS DUE IN A HOUR)

Mathematics

1 answer:

lesantik [10]3 years ago

5 0

its b/13=5 and then to solve you multiply both sides by 13 to get b=65

so 65 bean bags

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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

Write the sum of the numbers as the product of their gcf and another sum 58+14

AnnyKZ [126]

I do not know but i am going to still glith the server if you need help go to google

4 0

3 years ago

The play director spent 190190190 hours preparing for a play. That time included attending 353535 rehearsals that took varying a

dybincka [34]

Answer:

The equation $35x+93\frac{3}{4} =190$ gives average time spent on 35 rehearsals.

Step-by-step explanation:

We are supposed to find that what question does the equation $35x+93\frac{3}{4} =190$ finds answer of.

We can see that 35x represents time spent on 35 rehearsals and $93\frac{3}{4}$ is time spent on other responsibilities related to play. The sum of these times equals to total time spent on preparing the play.

Now let us solve our equation step by step.

$35x+\frac{375}{4} =190$

After subtracting $93\frac{3}{4}$ hours from 190 hours we will get time spent on 35 rehearsals.

$35x =190-\frac{375}{4}$

$35x =\frac{760-375}{4}$

$35x =\frac{385}{4}$

Time spent on 35 rehearsals is 96.25 hours and we are told that each rehearsal took different amount of time. Dividing 96.25 by 35 we will get average time spent on each rehearsal.

$x =\frac{96.25}{35}=2.75$