I need help on this question.

Nathan plants equal sized squares of sod in his Front yard each square has an area of 6 ft.² Nathan plans a total of 1000 square

Monica [59]

6x1000=6000ft. your answer is 6000ft.

8 0

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

6 0

4 years ago

A part selected for testing is equally likely to have been produced on any one of six cutting tools.

JulijaS [17]

Answer:

a)S= (1,2,3,4,5,6)

b) 1/6 or 16.67%

c) 1/3 or 33.33%

d) 5/6 or 83.33%

Step-by-step explanation:

a) Sample space is the list of all the possible sample in this its either tool 1,2,3,4,5 or 6

b) probability of selecting from tool 1 is equally likely to picking from any cutting tool therefore it

1/6 or 16.7%

c) probability of selecting from any tool is 16.7% so probability of selecting from 3 or 5 = 16.67% + 16.67% = 33.3%

d) if the part is not from tool 4 then its from tool 1, or tool 2 or tool 3 or tool 5 or tool 6 therefore its

1/6+1/6+1/6+1/6+1/6 = 5/6 or 8.33%

8 0

3 years ago

ssue bought two pairs of jeans and a belt that costs 6.95.the tax on the items was 5.85 sue paid the cashier$70.00how much money

zubka84 [21]

I think the answer is 57.10

8 0

3 years ago

20 POINTS. Given cos θ=4/5, find cot(θ).

nydimaria [60]

$\displaystyle\\\cos\theta=\frac{4}{5}\\\\ \sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\left(\frac{4}{5}\right)^2} =\sqrt{1-\frac{16}{25}} =\sqrt{\frac{9}{25}} =\frac{3}{5}~~\text{(If } \theta \in \text{quadrant 1)}}\\\\\\ \cot(\theta)=\frac{\cos\theta}{\sin\theta}=\frac{~~\dfrac{4}{5}~~}{\dfrac{3}{5}}=\frac{4}{5}\times\frac{5}{3}=\frac{4}{3}\\\\\\\boxed{\cot(\theta)=\frac{4}{3}}$

.

3 0

3 years ago