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shtirl [24]
3 years ago
12

Eli is moving to a new city. He packs his belongings in cube-shaped boxes that each have a volume of one cubic foot. He places t

he boxes in a storage unit that is shaped like a rectangular prism. The floor of the storage unit can be completely covered by a layer of 150 boxes. If 6 layers of boxes can completely fill the storage unit, what is its volume?
Mathematics
1 answer:
klemol [59]3 years ago
4 0

Answer:

jnnm mnn m

Step-by-step explanation:

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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
What is the "k" value of the following function? Is it 3, -2, -3, or 2?
snow_tiger [21]

Answer:

-3

Step-by-step explanation:

we khow that the vertex form of a function is written this way :

f(x)=a(x-h)²+k

  • the coordinates of the vertex are (h,k) so from the graph we can deduce that k = -3

5 0
3 years ago
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