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Aleksandr [31]
2 years ago
11

Y=3/2x+6 parallel to (-2 -4)

Mathematics
1 answer:
Aleonysh [2.5K]2 years ago
4 0

Answer:

no it's not parallel to that.

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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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1 year ago
At the end of each year, how much money would you earn in interest if you invested $200 and earned 5.5% simple interest?
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\bf \qquad \textit{Simple Interest Earned}\\\\&#10;I = Prt\qquad &#10;\begin{cases}&#10;I=\textit{interest earned}\\&#10;P=\textit{original amount deposited}\to& \$200\\&#10;r=rate\to 5\%\to \frac{5}{100}\to &0.055\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;I=200\cdot 0.055\cdot 1
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Which of the following is the midsegment for A ACE?
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Answer:

A. DB

Step-by-step explanation:

The line BD is the mid segment because it connects the midpoints of line segment AC and CE

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. Simplify the sum. (2u3 + 6u2 + 2) + (7u3 – 7u + 4)
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Answer:

9u^3 + 6u^2 - 7u + 6

Step-by-step explanation:

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What are the factors of58
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<span>1,2, 29,and 58 are the factors of 58.</span>
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