I forgot how to do these problems can you help me

A height of 12 and a volume of 1.356.48 what is the radius of the cylinder?

alexgriva [62]

Answer:

11.99 ft

Step-by-step explanation:

The formula for the volume of a cyl. of radius r and height h is V = πr²h.

Solving this immediately for h yields:

V

h = ----------

πr²

Inserting the known quantities results in:

V 1356.48 ft³

h = ---------- = ---------------------- Note that r = (1/2)d, so r = (1/2)(12 ft) = 6 ft

πr² 3.14159 (6 ft)²

= 11.994 ft

The height of the cyl. is 11.99 ft (to the nearest hundredth foot)

8 0

3 years ago

Mid-West Publishing Company publishes college textbooks. The company operates an 800 telephone number whereby potential adopters

Mumz [18]

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.

Read more about extension lines

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8 0

2 years ago

Sarah, a self employed blogger, has adjusted gross earnings of $45,162 this year. Find the amount of social security and Medicar

maksim [4K]

Answer:

Social Security = 5600.09, Medicare = 1309.70

Step-by-step explanation:

8 0

3 years ago

Which expression is equivalent???<br><br> Please help!!

My name is Ann [436]

Answer:

answer is d.-3/2x +5

Step-by-step explanation:

this was tough but I did it. First you multiply all the terms with 1/4. Then just simplify the terms by breaking them down and then apply normal addition and subtraction. Simple. Thank you

8 0

3 years ago

The length of the terrarium is 14.5 inches and the width is 7 inches. what is the height of the terrarium?

mote1985 [20]

The height is: 7.5 inches. Since the terrarium is 14.5 inches in total, and the width is 7 inches, the remaining inches makes up the height and that is 7.5.

3 0

3 years ago