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:
A. 8 x 100
D. 8 x 10 x 10 x 10
Step-by-step explanation:
When you do the original equation it is 912
these 2 are close but don't go over
3 4/5 rounds to 4
5 1/6 rounds to 5
4 x 5 = 20
I hope this helped! :)
12x-3y=3
y=4x-1
12x-3(4x-1)=3
12x-12x-3=3
24x=6
x=4
y=4(4)-1
y=15
x=4; y=15
Here are the steps. I hope I do them easy enough for you to understand it
-3/4+1/8
=-3/4+1/8
=-6/8+1/8
=6+1
-------
8
And the answer is -5/8 (The decimal is -0.625)
I hoped I helped :)
