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2 years ago

8. Mary and Todd went on a bike trip. The first day they biked 17 miles, and the second

Mathematics

2 answers:

Georgia [21]2 years ago

6 0

Answer:

They traveled 72 miles in all.

Step-by-step explanation:

sweet [91]2 years ago

3 0

Answer:

The answer is 72.

Step-by-step explanation:

217+229+314+215=975

Hope you have A Great day!

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What is the geometric mean of 3 and 15

Helen [10]

Trick question. There is not one. Due to the fact that 3 is a factor of 15 and if you factor it you will get five which is a third of 15.

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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

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1 year ago

What is the measure if each interior angle of a 15-gon

olya-2409 [2.1K]

Answer:

The sum of the interior angles of the 15-gon $=\:2340^0$

Each interior angle of the regular polygon $=\:156^0$

Step-by-step explanation:

As we know that

In any convex polygon, if we may start at one vertex and draw the diagonals to all the other vertices, we would form triangles,

The number of triangles thus formed would always 2 LESS than the number of sides.

The sum of measure of the angles of any triangle is 180°.

Thus,

The sum of the interior angles of the 15-gon will be:

$13\:\times\:180\:=\:2340^0$

Also

15-gon is regular, it means this total $2340^0$ is shared in equal proportion among the 15 interior angles.

And

Each interior angle of the regular polygon will be: $\frac{2340^{0} }{15}=156^{0}$

Therefore, we conclude that:

The sum of the interior angles of the 15-gon $=\:2340^0$

Each interior angle of the regular polygon $=\:156^0$

Keywords: regular polygon, 15-gon, triangle

Learn more about polygon from brainly.com/question/11932357

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3 years ago

Mr. Martin drove 290 miles last week This week he drove 125 miles more than last week. How many miles did Mr. Martin drive this

DanielleElmas [232]

Answer:

Mr. Martin drove 415 miles this week.

Step-by-step explanation:

The question says Mr. Martin drove 125 miles more than <u>last week</u>. How many miles did Mr. Martin drive this week?

How many miles did he drive last week?

Remember: 290 miles is the miles he drove last week.

The problem says he drove 125 miles AND 290 miles this week. So you take the SUM of them to see how many miles he drove together.

290 + 125 = 415

6 0

3 years ago

What is the table of 75

jarptica [38.1K]

Step-by-step explanation:

75×1=75

75×2=150

75×3=225

75×4=300

75×5=375

75×6=450

75×7=525

75×8=600

75×9=675

75×10=750

4 0

3 years ago