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3 years ago

PLEASE HELP! giving BRAINLIEST AND GIVE CORRECT ANSWER AND EXPLAIN!

Mathematics

1 answer:

Ghella [55]3 years ago

3 0

so the answer is 20 square centimeter.

Hope this will help u...

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The sum of two integers is (-32). Their quotient is 7. What are the integers?

Aneli [31]

<h3>Answer: -28 and -4</h3>

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Explanation:

Let x and y be the two integers such that |x| > |y|. This is another way of saying that x is further away from 0 compared to y.

Their quotient being 7 means x/y = 7 which rearranges to x = 7y

The sum of the integers is -32 which allows us to say....

x+y = -32

7y+y = -32

8y = -32

y = -32/8

y = -4

Use this y value to find x

x = 7y

x = 7*(-4)

x = -28

The two integers are -28 and -4.

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.

Read more about signal

brainly.com/question/14699772

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3 0

2 years ago

The sum of a number (n) and 14 is 72. Which equation shows this relationship

Luden [163]

The sum is of n and 14 can be written as:

n+14=72

4 0

4 years ago

Did I graph this linear equation correctly?

devlian [24]

Answer:

Step-by-step explanation:

The question asks you to graph the inequality

The steps of drawing the inequality are the same steps of drawing the equation except the area of the solution

→ The solution of the equation is just all the points lie on the line

→ The solution of the inequality is one of these cases:

All the points on the line and on the area down the line if the inequality y ≤ ......

All the points on the line and on the area over the line if the inequality y ≥ .......

All the points on the area down the line if the inequality y < ......

All the points on the area over the line if the inequality y > .......

∵ The inequality is 4x - 3y ≤ 6

∴ The equation of the line is 4x - 3y = 6

Your drawing is right, BUT you must shade the area over the line because the solution of the inequality is all the points on the line and over the line

The coefficient of y is -3 which means we will divide the sides of the inequality by a negative number, so sign the inequality must e reversed

I attached the figure of the inequality to show you the explanation above.

The red shaded is the area of the solution of the inequality.