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.
Answer:
Step-by-step explanation:
"<u><em>The size of the tank</em></u> a fish collector needs <u><em>depends</em></u> on <em>how many fish he has</em>."
Dependent variable is the size of the tank.
<em>Independent variable</em> is how many fish a fish collector has.
Answer:
The answer is 63.31
Step-by-step explanation:
Answer:
Step-by-step explanation:
For a
2 + x = 15
x = 15 -2
x = 13
For B
x - 12 = 14
x = 12 + 14
x = 26
For C
3x = 12
x = 12/3
x = 4
Hope it helps:)
