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.
144 divided by 260 is 60%
We immediately note that 3 and 9 are multiples. Therefore, they are factorable. 9 goes into 3 three times, and 3 goes into three 1 time. Therefore, we can take out three from the whole expression to get 3(c - 3).
Answer:
x=15°
Step-by-step explanation:
We can see angle 6x-5 has a corresponding angle above it, which is congruent to it. That angle is supplementary with angle 7x-10. We can tell because it forms a straight angle with it. Supplementary angles add to 180°.
(7x-10)+(6x-5)=180
7x-10+6x-5=180
13x-15=180
13x=195
x=15
Ok so
18 - 1 = 17
17 - 2 = 15
15/3 = 5
she can carry 5 books
