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Zinaida [17]
2 years ago
10

Help me please and thx

Mathematics
1 answer:
Olenka [21]2 years ago
3 0
Y = kx, so y/x is constant

you want y such that
y/12 = -2/-8
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Solve the following <br> x^2-2x-24=0<br> X^2-11+18=0
Marrrta [24]

Answer:

x=-11

Step-by-step explanation:

6 0
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.

3 0
3 years ago
The airplane in the scale drawing has a wingspan of a feet and a tailspan of b feet. if the tailspan in the drawing is 2 inches,
Elena L [17]

30 feet is correct. Since a = 3b, the wingspan is 6 inches in the drawing, so it is 6x5 = 30 feet on the plane.


7 0
3 years ago
Read 2 more answers
What is a proportion
Anit [1.1K]
<span>A number considered in comparative relation to a whole.</span>
5 0
3 years ago
Read 2 more answers
Which pair of expressions below are equivalent? a. −4−8 and 4+8 b. −4−8 and −4+8 c. −4−8 and −4−(−8) d. −4−8 and −4+(−8)
omeli [17]

Answer:

d. −4−8 and −4+(−8)

Step-by-step explanation:

+ - is equivalent to -

6 0
3 years ago
Read 2 more answers
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