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3 years ago

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To kick off the summer, Colton is planning a "School's Out!" party for all his friends. He and his mom pick up a giant sub and a

vat of root beer from Subs & Suds. When they get to the park, they cut the sub into 18 equal sections. Each section is 1/3 of a foot long.

1 answer:

marshall27 [118]3 years ago

4 0

Answer:

(a)6 feet

(b)24 sections

Step-by-step explanation:

(a)We are required to find the length of the Giant Sub.

If each section is 1/3 of a foot long and there are 18 equal parts.

Length of the Giant Sub =

Length per Section X Number of Sections

= 1/3 X 18 = 6 feet

(b)If each section is 1/4 of a feet long

Length of the Giant Sub =

Length per Section X Number of Sections

6= 1/4 X Number of Sections

Number of Sections =6 X 4 =24 sections

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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)

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.

b)

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

c)

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.

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2 years ago

What is 91/125 as a percent

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Answer: 72.8%

Step-by-step explanation:

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140 fluid ounces increased by 45%

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Hello!

"Increase" means to add. We must find out what 45% of 140 fluid ounces is, and to do that we must first convert the percentage to a decimal and then multiply it by the amount of fluid ounces.

I know, that's probably a lot to take in! :D

45% ÷ 100 = 0.45 as a decimal

0.45 × 140 = 63

Add this to the amount of fluid ounces:

140 + 63 = 203

Final Answer:

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George drives at 45mi/h. write two expressions for the number of miles George travels in h hours

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45h=miles in h hours

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use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set

Greeley [361]

Answer:

Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

Step-by-step explanation:

x y

3 470

4 416

5 403

Analyzing Option A:

Considering the equation

$y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:$

$y=64.02$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:$

$y=673.18$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:$

$y=1348.06$

Analyzing Option B:

$y=32.86\:\left(x\right)^2-379.14\left(x\right)$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)$

$y=-841.68$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)$

$\:y=-990.8$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)$

$\:y=-1074.2$

Analyzing Option C:

Considering the equation

$y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:$

$y=527.46$

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:$

$y=378.34$

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:$

$y=294.94$

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

$y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86$

From (3, 470), putting x = 3

$y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:$

$y=-13426.82$

From (4, 470), putting x = 4

$y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86$

$y=-23389.94$

From (5, 403), putting x = 5

$y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:$

$y=-36091.34$

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

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