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3 years ago

What is 3123532488697 x 5676576876896786

Mathematics

2 answers:

son4ous [18]3 years ago

7 0

3123532488697 x 5676576876896786= 17,730,972,299,573,261,776,980,672,842

horrorfan [7]3 years ago

4 0

The answer to ur question is 1.77309723E28.
Hope this helps and can i get brainliest answer.

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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 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.

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1 year ago

The diagram below, circumscribed about quadrilateral DEFG What is the value of x

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3 years ago

Suppose the claim size of an auto collision insurance, X, is uniformly distributed on the interval $1,000 to $10,000. What is th

timurjin [86]

Answer:

35.35%

Step-by-step explanation:

If there were no deductibles, the expected claim payment would be:

$E(X) = \frac{10,000 +1,000}{2} \\E(X) =\$5,500$

If the collision insurance claim is under $2,000, then the insurer would not pay anything, but if X > $2,000, then the insurer would pay X - $2,000. The new expected value is:

$E_2(X)=\frac{2,000-1,000}{10,000-1,000} *0+\frac{10,000-2,000}{10,000-1,000}*\frac{(2,000-2,000)+(10,000-2,000)}{2} \\E_2(X)=\frac{8}{9}*\frac{0+8,000}{2}\\ E_2(X)=\$3,555.56$