All categories

3 years ago

Simplify the following: (25 ×−716 ) + ( 25 × 28)

Mathematics

2 answers:

kvv77 [185]3 years ago

7 0

Answer:

- 17900 + 700

answer to the equation would be -17,200

Makovka662 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

(25 ×−716 ) + ( 25 × 28)

25(28 - 716)

25(- 688)

-17,200

You might be interested in

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

Calculate 50% of 200

OlgaM077 [116]

Answer:

100

Step-by-step explanation:

hopes this helps

5 0

3 years ago

Read 2 more answers

Determine the slope of the line that has the following coordinates: (-3,6) (4,-2)

AveGali [126]

-8/7 is the correct answer.

6 0

3 years ago

What is 2+2*56*89/5+69

Natalija [7]

Answer:

2064.6 is the answer

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How much 75 octane gas and 90 octane gas should be blended to make 15 gallons of 88 octane gas

Lisa [10]

75x + 90(15-x)=88(15)

75x+1350-90x = 1320

-15x + 1350 = 1320

-15x= -30

x=2

the amount of 75 octane is 2 gallons
the amount of 90 octane is 13 gallons

Hope this helps.

3 0

3 years ago