Answer in factored form 13+14m+m^2

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

s344n2d4d5 [400]

The various answers to the question are:

To answer 90% of calls instantly, the organization needs four extension lines.
The average number of extension lines that will be busy is Four
For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$$

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$$

Generally, the equation for is mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

$P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$$

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

$P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$$

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

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Measures of Center and Variability

pickupchik [31]

1. The mean of all the numbers is 41.
5. Median : 52

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Which of the following is the solution set of the given equation?<br><br> (x - 3) - 2(x + 6) = -5

Vlad1618 [11]

Answer: x=−10

Step-by-step explanation:

x−3−2(x+6)=−5

x+−3+(−2)(x)+(−2)(6)=−5(Distribute)

x+−3+−2x+−12=−5

(x+−2x)+(−3+−12)=−5(Combine Like Terms)

−x+−15=−5

−x−15=−5

−x−15+15=−5+15

−x=10

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3 years ago

the cost of printing a store flyer is 440 for 1000 copies and 880 for 2000 copies. write a linear equation for modeling this dat

Anna35 [415]

Since they tell us that this is linear, having a constant rate of change, we can express this as a line:

y=mx+b, where m=slope (change in y divided by change in x) and b=y-intercept (value of y when x=0)

First find the slope, or m, which mathematically is:

m=(y2-y1)/(x2-x1), in this case:

m=(880-440)/(2000-1000)

m=440/1000

m=0.44, so far our line is:

y=0.44x+b, now we can use either data point to solve for b, I'll use (1000,440)

440=0.44(1000)+b

440=440+b

0=b, so our line is just:

y=0.44x

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Find the solution of y = 2x + 5 for x = 1.

Mumz [18]

Answer:

1.)

y = 2x + 5 for x = 1

y = 2(1) + 5

y = 2 + 5

y = 7

2.) The function rule that relates the total cost of the party to the number of children n.

TCP = $60 + $4(n)

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