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MAVERICK [17]
2 years ago
11

Eva, the owner of eva's second time around wedding dresses, currently has five dresses to be altered, shown in the order in whic

h they arrived: job processing time (hrs) due (hrs from now) v 3 5 w 1 1 x 4 9 y 2 3 z 5 7 if eva uses the first come, first served priority rule to schedule these jobs, what will be the average completion time? 5 hours 7 hours 7.2 hours 3 hours 8 hours
Mathematics
2 answers:
ira [324]2 years ago
6 0

Answer:

8 hours

Step-by-step explanation:

Let the times be:

dress 1 = 3

dress 2 = 4

dress 3 =8

dress 4 = 10

dress 5 = 15

Therefore, the average processing time to work on the dress is:

\frac{3+4+8+10+15}{5}\\ = 8 hours

Vikentia [17]2 years ago
3 0

Answer:

  8 hours

Step-by-step explanation:

Job v starts at time 0 and completes at time 3.

Job w starts at time 1 and completes at time 4.

Job x starts at time 4 and completes at time 8.

Job y starts at time 8 and completes at time 10.

Job z starts at time 10 and completes at time 15.

The average completion time is (3 +4 +8 +10 +15)/5 = 8 . . . hours.

_____

<em>Comment on the schedule</em>

Jobs v and x are the only ones completed by the time due. The latest is 8 hours late. If they are scheduled in order of due time, jobs v, w, and y can be completed by the due time; the others will be 4 and 6 hours late.

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1) A family consisting of three persons—A, B, and C—goes to a medical clinic that always has a doctor at each of stations 1, 2,
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Step-by-step explanation:

Let us record the station number 1, 2 or 3 for each family member A, B or C.

I am attaching a table containing total outcomes. Outcomes are presented along rows while the assigned station to each member is written along columns. For ease of understanding, 1 3 2 in the table should be interpreted as family member A being assigned to station 1, member B to station 3 and member C to station number 2, respectively.

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We know that, probability can be defined as,

PROBABITILY = \frac{NUMBER\;OF\;DESIRED\;OUTCOMES}{TOTAL\;NUMBER\;OF\;OUTCOMES}

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Cases for all members being assigned to same station are as follows:

[1 1 1], [2 2 2], [3 3 3] (outcome number 1, 14 and 27 in the table).

Therefore,

PROBABILITY\;(Case\;a) = \frac{3}{27}\\\\PROBABILITY\;(Case\;a) = 0.111

b) At Most Two Members Assigned to the Same Station.

It means that maximum of 2 members can have the same station. Cases for this situation are as follows:

[1 1 2], [1 1 3], [1 2 1], [1 2 2], [1 3 1], [1 3 3], [2 1 1], [2 1 2], [2 2 1], [2 2 3], [2 3 2],

[2 3 3], [3 1 1], [3 1 3], [3 2 2], [3 2 3], [3 3 1], [3 3 2]

(outcome number 2, 3, 4, 5, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 25 and 26 in the table).

Therefore,

PROBABILITY\;(Case\;b) = \frac{18}{27}\\\\PROBABILITY\;(Case\;b) = 0.666

c) All Members Assigned to a Different Station.

For this scenario, we have the following results:

[1 2 3], [1 3 2], [2 1 3], [2 3 1], [3 1 2], [3 2 1] (outcome number 6, 8, 12, 16, 20 and 22 in the table).

Therefore,

PROBABILITY\;(Case\;c) = \frac{6}{27}\\\\PROBABILITY\;(Case\;c) = 0.222

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