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Neko [114]
3 years ago
11

A lottery game requires that 3 different numbers are picked from 1 to 9 if someone picks all 3 winning numbers the person wins 9

million complete parts a and b
Mathematics
1 answer:
adell [148]3 years ago
6 0
I need the questions!!
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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
If there are 52 ounces of cereal in 2 boxes, how man ounces are there in the box?
Firdavs [7]
The answer is 26. 26 ounces per box o cereal.
3 0
3 years ago
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Let r = ⟨7, 1⟩, s = ⟨–6, 2⟩, and t = ⟨–9, –3⟩. What is 10r + s – t? ⟨55, 9⟩ ⟨55, 15⟩ ⟨73, 6⟩ ⟨73, 15⟩
sergiy2304 [10]

Answer: d. 73,15

Step-by-step explanation:

5 0
3 years ago
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Describe the end behavior of the given function f(x)=2/3x-2
Ipatiy [6.2K]

Answer:

The end behavior of f(x)=2/3x-2 is: as x->+ infinity, f(x)->+ infinity

as x->- infinity, f(x)->- infinity

Step-by-step explanation:

When you are asked about the end behavior of a function, look to see where the function is traveling on the graph. For instance, this graph is linear, so you should look to see if the slope is positive or negative. This linear function is positive, so as x is reaching positive infinity the f(x) would also be reaching positive infinity. As x is reaching negative infinity, f(x) would also be reaching negative infinity. The end behavior of a function describes the trend of the graph on the left and right side of the x- axis. (As x approaches negative infinity and as x approaches positive infinity).  

4 0
3 years ago
Consider that x=5 and y=7. Is the sum of x + y rational, irrational, or neither irrational nor rational!
nasty-shy [4]
The sum of y+x, or 5+7, is rational.  This is because each integer(non fraction/decimal)is rational, it is also rational.
5 0
4 years ago
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