Rewrite using standard notation : 8.07 x 10^3

1 answer:

4 0

8070, because when it is a positive exponent you have to move the exponent to the right. When it is a negative exponent, you would move it to the left. Hope this helps :)

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Write 82,700,000 in scientific notation.

aleksandr82 [10.1K]

Answer:

8.27 x 10^7

Step-by-step explanation:

Move the decimal over 7 times

8 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 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.

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

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.