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denpristay [2]
3 years ago
14

$25,000 $45,000 $55,000 $25,000 $35,000 $1,000,000 $35,000 $45,000 $55,000 $25,000 Which of these is the MOST reasonable represe

ntation of the annual salary of a typical employee of the company?
Mathematics
1 answer:
MaRussiya [10]3 years ago
4 0
45,000 would be the normal pay for a typical employee
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Complete the table for the value of each digit in 123.
artcher [175]

Answer:

1,2,3,4,5,6,7,8,9,10

2,4,6,8,10,12,14,16,18,20

3,6,9,12,15,18,21,24,27,30

Step-by-step explanation:

6 0
3 years ago
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.

Read more about signal

brainly.com/question/14699772

#SPJ1

3 0
1 year ago
A cylinder has a height of 13 feet and a radius of 11 feet. What is its volume? Use a ~ 3.14
ArbitrLikvidat [17]

Answer:

900

Step-by-step explanation:

V=πr2h

3.14 x 11= 34.54

34.54 x 2= 69.08

69.08x 13= 898.04

898.04 rounded to nearest hundreth is 900

8 0
3 years ago
A rectangular cooler made of MycoFoam is 13 inches wide, 7 inches tall, and 7 inches long. what is its surface area?
yarga [219]
To find area u have to multiply the sides so it should be 7x7 or 13x7 it shouldnt be 13x7x7
6 0
3 years ago
Determine the intervals on which the function is increasing, decreasing, and constant
max2010maxim [7]
I believe it's increasing on all fronts, because if you start from the right, you see that the y values always increase, hence they are increasing. They do it for when x<0 and when x>0. So, it should be increasing on all real numbers
3 0
3 years ago
Read 2 more answers
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