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omeli [17]
2 years ago
7

Two of your friends go bowling. One friend rents a pair of bowling shoes for $3 and bowls 3 games. The other friend brings his o

wn bowling shoes, bowls 4 games and buys a soda for $0.50. Both friends spend the same amount of money. Show how you can write and solve an equation to find the cost of one game.
Mathematics
1 answer:
Nesterboy [21]2 years ago
4 0

Answer: The cost of each game is $2.50

Step-by-step explanation:

Let's find the equations for each friend:

Friend 1:

Pays $3 for the shoes, and plays 3 games, then if X is the cost of each game, Friend 1 pays a total of:

$3 + 3*X

Friend 2:

This friend buys a soda for $0.50, and he plays 4 games, remember that the cost of each game was X. then this friend pays:

$0.50 + 4*X

And we know that both friends pay the exact same amount, then we can write:

$3 + 3*X = $0.50 + 4*X

And solve this for X.

We need to isolate X, then we can move all the terms with X to the right, and all the terms without X to the left:

$3 - $0.50 = 4*X - 3*X

$2.50 = (4 - 3)*X = X

This means the cost of each game is $2.50

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The various answers to the question are:

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a)

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

Use the calculation for busy k servers.

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

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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$

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.

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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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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.

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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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