Answer:
x>6
Step-by-step explanation:
2x > 16-4
x > 12/2
x > 6
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:
18 in
Step-by-step explanation:
Answer:
Total money left with Jamie = £13
Step-by-step explanation:
Jamie received pocket money = £26
He spent on sweets = 10% of total pocket money
Expense on magazines = 25% of the total money
Expense on games = 15% of the pocket money
Total expenditure on sweets, magazines and games = 
=
Or 50% of the pocket money he received
Therefore, money left with Jamie = 50% of total money he received as pocket money
= 
= £13
Step-by-step explanation:
2⁵ˣ = 8ˣ⁺²
2⁵ˣ = (2³)ˣ⁺²
2⁵ˣ = 2³ˣ⁺⁶
5x = 3x + 6
2x = 6
x = 3