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3 years ago

HELPPP IM PANICKINGGGG

Mathematics

1 answer:

Ray Of Light [21]3 years ago

7 0

Hello!

We are given that the measure of one angle is equal to 115 degrees (for convenience, we’ll refer to this angle as ∠W). Looking at the image above, we can see that ∠W and ∠Y are supplementary angles. Supplementary angles are two angles whose measures add to a sum of 180 degrees, also known as a straight line. This relationship can be expressed using the following formula:

A1 + A2 = 180

Now insert any known values provided by the image above:

(115) + (∠Y) = 180

Subtract 115 from both sides of the equation:

∠Y = 65

We have now proven that ∠Y has a measure of 65 degrees. Now, looking at the image again, we can see that ∠W and ∠X are vertical angles. Vertical angles are two angles that lie on opposite sides of two intersecting lines. A pair of vertical angles is always equal in measure. This relationship can be expressed using the following formula:

A1 = A2

Now insert any known values provided by the image above:

(∠X) = (115)

We have now proven that ∠X has a measure of 115 degrees.

I hope this helps!

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$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$

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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 !}}$$\approx 0.3425$$

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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

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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

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