Please help!<br> [(a-ab)/a^2] divided by [(a-1)/a^3]

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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2 years ago

Please help :) what is the area of the triangle

Flura [38]

Answer:

30 units ^2

Step-by-step explanation:

The area of a triangle is

A = 1/2 bh

The base is on the left b = 10

The height is perpendicular to the base = 6

A = 1/2 * 10 * 6

= 30 units ^2

4 0

4 years ago

Which represents a unit rate

Mice21 [21]

Answer:

Step-by-step explanation:

When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. If you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.

3 0

3 years ago

How do you find the measure of an angle?

Ymorist [56]

Answer:

use the corner ofa paper to see if its obtuse it should be long and wide if its acute it should be a corne

Step-by-step explanation:

3 0

3 years ago

A scatterplot is produced to compare the number of hours that students study to their test scores. There are 25 data points, eac

yKpoI14uk [10]

The question is incomplete, the complete question is;

A scatterplot is produced to compare the number of hours that students study to their test scores. There are 25 data points, each representing a different student. The scatterplot shows a grouping of points rising from left to right. Which statement could be true?

a. There is no relationship between the number of hours that students study and their test scores because the scatterplot does not have a cluster.

b. As the number of hours of studying increases, test scores decrease because the scatterplot has a cluster that increases from left to right.

c. As the number of hours of studying increases, test scores increase because the scatterplot has a cluster that increases from left to right.

d. The number of hours that students study is equal to the test scores because the scatterplot does not show a cluster.

Answer:

c. As the number of hours of studying increases, test scores increase because the scatterplot has a cluster that increases from left to right.

Step-by-step explanation:

Though we can't see the scatter plot but it makes sense from option c that as the number of hours of studying increases, test scores increase.

That option even stated that the cluster of points increases from left to right. If the scatterplot has a cluster that increases from left to right, then it follows that the test scores actually increases positively with the number of study hours.

5 0

3 years ago

Please help! [(a-ab)/a^2] divided by [(a-1)/a^3]