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Andrei [34K]
3 years ago
5

The company is building a scale model of the theater's main show tank for an investor's presentation. Each dimension will be mad

e 1/6 of the original dimension to accommodate the mock-up in the presentation room. What is the volume of the smaller mock-up tank?
Mathematics
2 answers:
soldi70 [24.7K]3 years ago
6 0

Answer:

1047.2 ft³

Step-by-step explanation:

The following information is missing:

<em>The main show tank has a radius of 60 feet and forms a quarter sphere where the bottom of the pool is spherical and the top of the pool is flat. (Imagine cutting a sphere in half vertically and then cutting it in half horizontally)</em>

The only dimension needed for the main show tank is its radius. Then the model radius will be 1/6 * 60 = 10 feet

The volume of a sphere is:

V = 4/3 * π * r³

For a quarter sphere it is<em>:</em>

V = 4/3 * π * r³ * 1/4

V = 1/3 * π * r³

For the smaller mock-up tank:

V = 1/3 * π * 10³

V = 1047.2 ft³

Ede4ka [16]3 years ago
5 0

Answer:

This means the mock-up tank has a volume 216 times smaller than the initial tank.

Step-by-step explanation:

The ratio between volume is given by

\frac{V_2}{V_1}=\frac{(1/6)L^{3} }{L^{3}}  =\frac{1}{216}

This means the mock-up tank has a volume 216 times smaller than the initial tank.

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The various answers to the question are:

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<h3>How many extension lines should be used if the company wants to handle 90% of the calls immediately?</h3>

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A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

$$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$

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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$

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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$

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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.

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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 !}} $\approx 0.0624$

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