Answer:
case 2 with two workers is the optimal decision.
Step-by-step explanation:
Case 1—One worker:A= 3/hour Poisson, ¡x =5/hour exponential The average number of machines in the system isL = - 3. = 4 = lJr machines' ix-A 5 - 3 2 2Downtime cost is $25 X 1.5 = $37.50 per hour; repair cost is $4.00 per hour; and total cost per hour for 1worker is $37.50 + $4.00
= $41.50.Downtime (1.5 X $25) = $37.50 Labor (1 worker X $4) = 4.00
$41.50
Case 2—Two workers: K = 3, pl= 7L= r= = 0.75 machine1 p. -A 7 - 3Downtime (0.75 X $25) = S J 8.75Labor (2 workers X S4.00) = 8.00S26.75Case III—Three workers:A= 3, p= 8L= ——r = 5- ^= § = 0.60 machinepi -A 8 - 3 5Downtime (0.60 X $25) = $15.00 Labor (3 workers X $4) = 12.00 $27.00
Comparing the costs for one, two, three workers, we see that case 2 with two workers is the optimal decision.
As division is the inverse of multiplication, the rules for division are the same as the rules for multiplication. So when multiplying and dividing positive and negative numbers remember this: If the signs are the same the answer is positive, if the signs are different the answer is negative.
prob( it lands on open end) = 8/20 = 2/5
prob ( lands on closed end) = 4/20 = 1/5
The second choice is correct.
We must find the common denominator.
The list of multiples of 4: 0, 4, 8, 12, 16, 20, ...
The list of multiples of 3: 0, 3, 6, 9, 12, 15, 18, ...
The common denominator is 12.
Therefore:
<h3>Answer: 17/12 = 1 5/12</h3>
