Answer:
60 weeks
Step-by-step explanation:
Factor both numbers 12 and 15:
Find LCM(12,15). The common factor is only 3, multiply ths common factor by all factors that remain. Then
This means that the fewest number of weeks that will pass before both machines are cleaned again in the same week is 60 weeks.
Answer:
a. 0.1536
b. 0.9728
Step-by-step explanation:
The probability that a component fails, P(Y) = 0.2
The number of components in the system = 4
The number of components required for the subsystem to operate = 2
a. By binomial theorem, we have;
The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;
P(Y = 2) = × 0.2² × 0.8² = 0.1536
The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536
b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working
Therefore, we have;
P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)
P(Y = 0) = × 0.2⁰ × 0.8⁴ = 0.4096
P(Y = 1) = × 0.2¹ × 0.8³ = 0.4096
P(Y = 2) = × 0.2² × 0.8² = 0.1536
∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728
The probability that the subsystem operates longer than 1,000 hours = 0.9728