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jeka57 [31]
4 years ago
14

An example of a frictionally unemployed individual is a. Sylvia, who quit her job to spend more time with her children b. Rod, t

he lifeguard, who cannot find a job because the temperature is too low in October c. Ileana, a college student who quits her job to return to school d. Steve, an individual does not have skills to keep his job as an aerospace engineer e. Samantha, who quits her job to look for better one
Business
1 answer:
mote1985 [20]4 years ago
8 0

Answer:

e. Samantha, who quits her job to look for a better one

Explanation:

This is a topic in economics and business that seeks to test your understanding of business cycles. This particular question is on frictional unemployment.

The main feature of frictional unemployment is that there is someone or a group of people who are actively looking for work. They remain unemployed until they find work.

The other answers in the question point to people who are not looking for work and thus do not make up part of the frictionally unemployed population group.

E.g Sylvia quit her job to spend more time with the kids, the college student who quit work to return to school etc... all these are not actively looking for job.

I hope this helps you understand the question better and you can solve similar questions

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Answer:

12.55 days

Explanation:

<em><u>Provided information </u></em>

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Standard deviation of inter-arrival time= 1 day

The coefficient of variations

<u>Inter-arrival coefficient of variation </u>

C_{vi}=\frac {\sigma}{T} where \sigma is standard deviation of inter-arrival time, T is inter-arrival time and C_v is coefficient of variation of inter-arrival time

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<u>Production time coefficient of variation </u>

C_{vp}=\frac {2.7}{1.8}=1.5

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U=\frac {T}{n*T_i} where T is the time of production, n is number of employees, U is utilization, T_i is inter-arrival time

U=\frac {1.8}{2*1}=0.9

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<u>Expected average waiting time </u>

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Where T_e is expected average waiting time and the other symbols as already defined

Substituting 1.5 for C_{vp}, 1 for C_{vi}, 0.9 for U, 2 for n, 1 for T_iand 1.8 for T

T_e=(\frac {1.8}{2*1})*0.5(1^{2}+1.5^{2})*(\frac{0.9^{\sqrt{2(2+1)}-1}}{1-0.9})

T_e=0.9*1.625*8.583709=12.55367 days  and rounding off to 2 decimal places we obtain 12.55 days

Therefore, expected duration between order received and beginning of production is approximately 12.55 days

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