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.
Given:
The figure.
To find:
The segment bisector of MN and value of MN.
Solution:
From the given figure it is clear that ray RP,i.e.,
is the segment bisector of MN because it divides segment MN in two equal parts.
Now,


Since,
is the segment bisector of MN, therefore,



Therefore, the length of MN is
.
A) Deposits: +34.98; +51.02; +51.22
Withdrawals: -5.23; -5.22; -7.89
B) -7.89, -5.23, -5.22, +34.98, +51.02, +51.22
The smallest amount will be the largest withdrawal, as it takes away from the account; the largest amount will be the largest deposit, as it adds to the account.
C) She would have $216.88.
98 + 34.98 = 132.98 + 51.02 = 184.00 + 51.22 = 235.22 - 7.89 = 227.33 - 5.23 = 222.10 - 5.22 = 216.88