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kondor19780726 [428]
2 years ago
9

William has a 26 -liter glass tank. First, he wants to put some marbles in it, all of the same volume. Then, he wants to fill th

e tank with water until it's completely full. If he uses 85 marbles, he will have to add 20.9 liters of water.
What is the volume of each marble?

How much water is necessary if William uses 200 marbles?
Mathematics
1 answer:
madreJ [45]2 years ago
4 0

Subtract the volume of the water from the volume of the tank to get the total volume of the marbles.

26 liters - 20.9 liters = 5.1 liters

He has 20.9 liters of water.

He has 85 marbles that add up to 5.1 liters of volume.

To find the volume of one marble, we divide 5.1 liters by 85.

(5.1 liters)/85 = 0.06 liters

The volume of each marble is 0.06 liters.

If he uses 200 marbles, the the marbles will occupy 200 * 0.06 liters = 12 liters.

The volume of the tank is 26 liters. 12 liters of volume is occupied by the marbles. The rest of the volume is water.

26 liters - 12 liters = 14 liters

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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 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 !}}$

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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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For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

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