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2 years ago

Write an inequality that represents the graph.

Mathematics

1 answer:

Anika [276]2 years ago

5 0

Answer:

$y - 2x - 1 = 0$

Step-by-step explanation:

the slope is :

$\tan( \alpha ) = 4 \div 2 = 2$

at a pt on graph , when

$x = 1 \\ y = 3 \\$

therefore, coordinates are (1,3)

The equation is :

$(y - 3) = slope \times (x - 1) \\ (y - 3) = 2 \times (x - 1) \\ y - 3 = 2x - 2 \\ y - 2x - 1 = 0$

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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

2 years ago

Answer the questions below.

DedPeter [7]

Answer:

63 m

Step-by-step explanation:

Perimeter is 276, 2L + 2W = 276, so L + W = 138

L = 75

138 - 75 = W

63 m

3 0

2 years ago

Write 31/11 as a decimal. If necessary, use a bar to indicate with digit or group of digits repeats

Andrews [41]

Answer:

2.81

Step-by-step explanation:

.81 has a bar over it.

5 0

3 years ago

The half-life of a certain substance is 20 years. How much of a 100 gram sample will be left after 20 years?

cluponka [151]

$\bf \textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &100\\ t=\textit{elapsed time}\dotfill &20\\ h=\textit{half-life}\dotfill &20 \end{cases} \\\\\\ A=100\left( \frac{1}{2} \right)^{\frac{20}{20}}\implies A=100\left( \frac{1}{2} \right)^1\implies A=50$

5 0

3 years ago

Read 2 more answers

Plz help with my math!!!

morpeh [17]

Set up a proportion with the similar sides. Then you should find what KB is. Sorry I couldnt answer directly. Im actually doing math right now

7 0

4 years ago