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Tresset [83]
3 years ago
14

Glen is making accessories for the soccer team. He uses 771.78 inches of fabric on headbands for 36 players and 2 coaches. He al

so uses 339.48 inches of fabric on wristbands for just the players. How much fabric was used on a headband and wristband for each player?
Mathematics
1 answer:
Nataly [62]3 years ago
6 0

Answer:29.74 inches

Step-by-step explanation:

Given

it takes 771.78 inches fabric for 36 player and 2 coaches headband

i.e. 1 headband takes around=\frac{771.78}{36+2}=20.31\ inches

it takes 339.48 inches fabric for player wristband

one wristband takes =\frac{339.48}{36}=9.43\ inches

For each Player =20.31+9.43=29.74\ inches are required

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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.

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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

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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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