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:
3.825
Step-by-step explanation:
First, you have to move the decimal two over on 45%. If you move the decimal over two spaces to the right, it would be 0.45. In math, of means to multiply, so then you multiply 0.45 and 8.5 to get your answer of 3.825.
Answer:
I think it is C
Step-by-step explanation:
This is going to be 4 remainders one.
Answer:
X = -5
Step-by-step explanation:
Subtract one from both sides
- 3 = 3x / 5
multiply by 5 to remove fraction
-15 = 3x
divide by 3
x = -5
