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cluponka [151]
3 years ago
7

Please help me figure out these two problems!! Thanks

Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0

Answer:

A) eight legs per one spider (56/7)

B) one pretzel is 2.25 (6.75/3)

Step-by-step explanation:

56 divided by 7 is 8

6.75 divided by 3 is 2.25

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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
2 years ago
Can some on help me please
kati45 [8]
3 feet = 36 inches 4 yards = 13 feet 12 feet = 4 yards
5 0
3 years ago
Read 2 more answers
What is the value of z?<br> 117°
mario62 [17]
Hi I would love to help but the picture is black
6 0
2 years ago
Eight less than the product of 7 and a number is 3. Use the variable w for the unknown number
Tpy6a [65]
W = unknown number
7w - 8 = 3    (product of 7 and a number, w, subtract 8) is 3
7w = 11
w = 11/7


8 0
3 years ago
Read 2 more answers
Solve for x<br> 2(−4x+3)+3x+3=−16
Flauer [41]

Answer:

x=5

Step-by-step explanation:

distribute first

2(-4x+3)

-8x+6

-8x+6+3x+3=-16

combine like terms

then get the variable alone

-5x+9=-16

-9. -9

<u>-5x</u><u>=</u><u>-25</u>

-5. -5

x=5

6 0
3 years ago
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