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:
1. 81
2. 108 1/2
3. 189 1/2
Step-by-step explanation:
Simplified: -6m-6
Factored: 6(m+1)
We know that: <span>3(x-1)^2-162=0
or (x-1)</span>²= 162:3
and (x-1)²= 54
we <span>take the square root of both sides
* x-1=</span>√54= 3√6 or x=1+3√6
* x-1= -<span>√54= -3√6 or x=1-3√6
This equation has 2 solutions</span>
Answer:
264 cu
Step-by-step explanation:
This is because you have to add it from all sides, the first prism is 144 cu and then the second one is, 120, and when you add both of those together you get 264
<u><em>Thus the answer is....264</em></u>
