Write the solution as a coordinate pair. SHOW YOUR WORK!!!!!

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

Mumz [18]

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.

Read more about extension lines

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8 0

1 year ago

Please solve and explain steps for this problem.

Helen [10]

This means Square root (^1/2). this is kind of hard to read. anyways. 32^(1/2) separates into 4^(1/2) and 8^(1/2), which further separates into 4^(1/2) and 2^(1/2), root 4 becomes 2, their are two root 4's, so you get 2 x 2, and you are left with 2^(1/2) now why go to all this trouble. because now you can multiply the 4 you created (2x2) times, the 7... giving you 28*2^(1/2) now subtract it from the the other one with root 2. -5*2^(1/2), giving you 23*2^(1/2)-...idk what that last bit is. if its a odd number then this is the end of the problem, if you can get it to root 2. then do that and simplify.

7 0

2 years ago

A number line going from 0 to 4. Numbers 0, 1, 2, 3, and 4 are marked on the line. Marks divide the spaces between each number i

olga nikolaevna [1]

Answer:

8 pieces of string

Step-by-step explanation:

Given that :

Total yards of string = 4

Length in yd of each string = 1/2 yards

Number of pieces of string Ms. Cole has :

Total. Yards / length in yard of each string

4 / (1/2)

4 * (2 /1)

= 8/1

= 8 pieces of string.

6 0

2 years ago

Help please<br><br> 3 + 3 + 2 x 2 x 1 = ?

kramer

3 + 3 + 2 x 2 x 1 = ?

3 + 3 = 6

2 x 2 = 4

6 + 4 = 10

10 x 1 = 10

Your answer is <u>10</u>

hope this helps!

3 0

2 years ago

What’s the gcf of 18k and 15k cubed?

Contact [7]

Answer:

The GCF for the variable part is

k

Step-by-step explanation:

Since

18

k

,

15

k

3

contain both numbers and variables, there are two steps to find the GCF (HCF). Find GCF for the numeric part then find GCF for the variable part.

Steps to find the GCF for

18

k

,

15

k

3

:

1. Find the GCF for the numerical part

18

,

15

2. Find the GCF for the variable part

k

1

,

k

3

3. Multiply the values together

Find the common factors for the numerical part:

18

,

15

The factors for

18

are

1

,

2

,

3

,

6

,

9

,

18

.

Tap for more steps...

1

,

2

,

3

,

6

,

9

,

18

The factors for

15

are

1

,

3

,

5

,

15

.

Tap for more steps...

1

,

3

,

5

,

15

List all the factors for

18

,

15

to find the common factors.

18

:

1

,

2

,

3

,

6

,

9

,

18

15

:

1

,

3

,

5

,

15

The common factors for

18

,

15

are

1

,

3

.

1

,

3

The GCF for the numerical part is

3

.

GCF

Numerical

=

3

Next, find the common factors for the variable part:

k

,

k

3

The factor for

k

1

is

k

itself.

k

The factors for

k

3

are

k

⋅

k

⋅

k

.

k

⋅

k

⋅

k

List all the factors for

k

1

,

k

3

to find the common factors.

k

1

=

k

3

=

k

⋅

k

⋅

k

The common factor for the variables

k

1

,

k

3

is

k

.

k

The GCF for the variable part is

k

.

GCF

Variable

=

k

Multiply the GCF of the numerical part

3

and the GCF of the variable part

k

.

3

k

3 0

2 years ago