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3 years ago

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Frank starts off with $ 120 to spend on his shirts and shoes. Frank buys 3 shirts for $15.50 each. Shoes cost $32.50 each per pa

ir. how many pairs of shoes did frank buy?

1 answer:

evablogger [386]3 years ago

4 0

Shirts: 3 x 15.50 = 46.50

120 - 46.50 = 73.50 left to spend on shoes

73.50/32.50(cost per pair) = 2.26....so he was able to purchase 2 pairs of shoes.

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Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

3 years ago

Solve for x 3x-1=9x+2

ANEK [815]

3x - 1 = 9x + 2

3x (-3x) - 1 (-2) = 9x (-3x) + 2 (-2)

-3 = 6x

-3/6 = 6x/6

-1/2 = x

hope thsi helps

4 0

3 years ago

Read 2 more answers

What is the following quotient? 5/√11-√3

kati45 [8]

ANSWER

$\frac{5( \sqrt{11} + \sqrt{3} )} {8}$

EXPLANATION

The given rational function is

$\frac{5}{ \sqrt{11} - \sqrt{3}}$

We need to rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of

$\sqrt{11} - \sqrt{3}$

which is

$\sqrt{11} + \sqrt{3}$

When we rationalize we obtain:

$\frac{5( \sqrt{11} + \sqrt{3} )}{(\sqrt{11} - \sqrt{3} )( \sqrt{11} + \sqrt{3} )}$

The denominator is now a difference of two squares:

$(a - b)(a + b) = {a}^{2} - {b}^{2}$

We apply this property to get

$\frac{5( \sqrt{11} + \sqrt{3} )}{( \sqrt{11}) ^{2} - ( \sqrt{3}) ^{2} )}$

$\frac{5( \sqrt{11} + \sqrt{3} )}{11 - 3}$

This simplifies to

$\frac{5( \sqrt{11} + \sqrt{3} )} {8}$

Or

$\frac{5 } {8}\sqrt{11} + \frac{5 } {8}\sqrt{3}$

8 0

3 years ago

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Simplify the following expression 4w x 3w x 2y x 2y

Anestetic [448]

Answer:

48w^2y^2

I used a simplify calculator I hope this is correct.

4 0

3 years ago

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13, 5, 4, 9, 7, 14, 4 The deviations are _____.

fenix001 [56]

Answer:

B."5, -3, -4, 1, -1, 6, -4"

Step-by-step explanation:

We are given that

13,5,4,9,7,14,4

We have to find the deviation.

Mean= $\frac{sum\;of\;data}{total\;number\;of\;data}$

Using the formula

$Mean,\mu=\frac{13+5+4+9+7+14+4}{7}$

$Mean,\mu=\frac{56}{7}=8$

Deviation= $x_i-\mu$

$x_i-\mu$

13 5

5 -3

4 - 4

9 1

7 -1

14 6

4 - 4

Hence, option B is correct.

7 0

3 years ago