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Solnce55 [7]
3 years ago
7

In the service department of the Glenn-Mark Auto Agency, mechanics requiring parts for auto repair or service present their requ

est forms at the parts department counter. The parts clerk fills a request while the mechanic waits. On average everyday around 210 mechanics arrive in a random fashion over 10-hour working period. It takes the clerk average 2 minutes to fill a request. Assuming the service time and inter-arrival time is random and exponentially distributed. a) What is the average time mechanics spend waiting and having requests filled under the current system
Mathematics
1 answer:
Anton [14]3 years ago
6 0

Answer:

the average time spend is 0.111

Step-by-step explanation:

The computation of the average time spend is as follows;

Arrival rate is

= 210 ÷ 10

= 21 per hour

Now the service rate is

= 60 ÷ service time

= 60 ÷ 2 minutes

= 30 per minute

And, finally the average time spend is

= 1 ÷ (service rate - arrival rate)

= 1 ÷ (30 - 21)

= 0.111

Hence, the average time spend is 0.111

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This question is incomplete because it was not written properly

Complete Question

A teacher gave his class two quizzes. 80% of the class passed the first quiz, but only 60% of the class passed both quizzes. What percent of those who passed the first one passed the second quiz? (2 points)

a) 20%

b) 40%

c) 60%

d) 75%

Answer:

d) 75%

Step-by-step explanation:

We would be solving this question using conditional probability.

Let us represent the percentage of those who passed the first quiz as A = 80%

and

Those who passed the first quiz as B = unknown

Those who passed the first and second quiz as A and B = 60%

The formula for conditional probability is given as

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