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goblinko [34]
11 months ago
14

Jennifer and Jane are best friends. They placed a map of their town on a coordinate gridand found the point at which each of the

ir house lies. If Jennifer's house lies at (2, 12)and Jane's house is at (16, 2) and they wanted to meet in the middle, what are thecoordinates of the place they should meet?
Mathematics
1 answer:
koban [17]11 months ago
8 0

Given:

Jennifer and Jane are best friends. They placed a map of their town on a coordinate grid and found the point at which each of their house lies. If Jennifer's house lies at (2, 12) and Jane's house is at (16, 2) and they wanted to meet in the middle.

Required:

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When writing an standard number in scientific notation, the first factor must have a value ≥ 1 and <
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Thats true a number has to be greater than 1 and less than 10.
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3 years ago
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.

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3 years ago
How would you write the variable expression “5 less than n”?
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Answer: n-5

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Read 2 more answers
) There are 8 children in Sophie's preschool class. During free time yesterday, 1 of them
Klio2033 [76]

Using it's concept, it is found that there is a 0.125 = 12.5% experimental probability that a randomly selected preschooler would choose to read books today.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

For an experimental probability, these numbers of outcomes are taken from previous trials.

In this problem, in the previous trial, one out of eight students read a book, hence:

p = 1/8 = 0.125 = 12.5%.

There is a 0.125 = 12.5% experimental probability that a randomly selected preschooler would choose to read books today.

More can be learned about probabilities at brainly.com/question/14398287

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2 years ago
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