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3 years ago

F(x) = x ^ 2 + 5x + 6

Mathematics

1 answer:

yaroslaw [1]3 years ago

3 0

A)306

b)-294

this is the answer to the two questions on the provided picture

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In which segment is the function constant

Debora [2.8K]

need more details pls

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4 years ago

HELP ASAP AGAIN PLSS!!!!! MAKRING BRAINLIST!! MATH!

jonny [76]

Answer:

200 I think I'm not entirely sure

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2 years ago

Read 2 more answers

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

How do I do this I really don't understand anything here

Lera25 [3.4K]

The red arrows show that the lines are parallel, but using the stuff given you can make an equation.
its hard to explain without drawing but starting from the 60 if you can make a Z you will get alternate angles and that mean that the given equation MUST be equal to 60. So,

8x-4=60

8x=60+4

8x=64

x=64/8

x=8 (is your answer)

hope you understand

8 0

4 years ago

X + 3= 8x -11

lapo4ka [179]

The answer is x=2

explanation:

2+3=5

8 times 2 is 16, minus 11=5

PLEASE MARK AS BRAINLIEST :)

6 0

3 years ago