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2 years ago

The planets close to the sun are

Mathematics

2 answers:

solmaris [256]2 years ago

8 0

Answer:

small and gaseous.

kiruha [24]2 years ago

5 0

It would be A small and rocky and the planets names are Mercury, Venus, Earth, and Mars

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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.

Read more about signal

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3 0

1 year ago

Question Is In The Picture

Alexxandr [17]

Answer:

Step-by-step explanation:

Pythagorean theorem states that a^2 + b^2 = c^2

this theorem only works on triangles with an angle of 90 degrees or a right triangle.

18^2 + 24^2 = 900

the square root of 900 is 30

I did the square root of 900 because the square root is the 'opposite' of squaring which is what you need to do in order to find 'c' is the Pythagorean theorem. 'a' is 18, and 'b' is 24

Answer is 30

5 0

2 years ago

Helppp please use previous question

4vir4ik [10]

I will answer if u post the ?s in a coment

8 0

3 years ago

Lexie began making posters for the school awards banquet. She can make 6 posters per minute. Zach joined her 10 minutes after sh

Snowcat [4.5K]

10 * 6 = 60 by the time Zach joins so 125-60 = 65
Now 6 (lexie) +7 (zach) = 13 (together) and 65/13 = 5.
6*5 = 30 and 7*5 = 35. 35+30 = 65
Your answer is 5.

7 0

3 years ago

Daniel buys a block of clay for an art project. The block is shaped like a cube with edges lengths of 10 inches. Daniel decides

Andre45 [30]

Answer:

=14.92...

14 the maximum number of spheres that Daniel can make from the chunk of clay.

Step-by-step explanation:

10^3=1000 in^3

1000/2=500 vol of one congruent chunk.

4/3 * pi * r^3 = vol of a sphere

500/(4/3 * pi * 2^3)

=14.92...

5 0

2 years ago