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Juli2301 [7.4K]
2 years ago
6

Joe works for a charity. Each year the charity gives money to different projects. The charity has a total of eight thousand nine

hundred and sixty-eight pounds to give to the projects this year.
Mathematics
1 answer:
baherus [9]2 years ago
7 0

Answer:

Each =\pounds 1121

Step-by-step explanation:

See comment for complete question.

Given

Total =  \pounds 8968

Projects = 8

Required

Determine the amount allotted to each

From the question, we understand that each project gets each allocation.

So, the amount of each is:

Each =\frac{Total}{Projects}

Each =\frac{\pounds 8968}{8}

Each =\pounds 1121

<em>Each project gets £1121</em>

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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.

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3 years ago
Three dumpsters are delivering sand to a golf course
timofeeve [1]

Answer:

10,000 tons of sand is being delivered.

Step-by-step explanation:

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2 years ago
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If x = 3 and y = -1, then 3x2 + 2xy - 5 = ? <br> A: 18 <br> B: 24 <br> C: 12 <br> D: 16
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x = 3 , \ y = -1 \\ \\ 3x^2 + 2xy - 5 = 3\cdot 3^2 + 2\cdot 3\cdot (-1)-5=\\ \\=3\cdot 9 -6-5=27-11=16 \\ \\ Answer : \ D: \ \ 16


8 0
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Pls help due in 40 minutes Pls I will give brainliest But pls only right answers
defon

Answer:

6. 15.2

Step-by-step explanation:

60.8 divided by 4 so the constant is 15.2 because 15.2 x 4 is 60.80

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Answer:

194.8 miles

Step-by-step explanation:

510 - 315.2 = 194.8

Hope this helps!

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