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Complete Question
The life of a semiconductor laser at a constant power is normally distributed with a mean of 7,000 hours and a standard deviation of 600 hours. If three lasers are used in a product and they are assumed to fail independently, the probability that all three are still operating after 7,000 hours is closest to? Assuming percentile = 95%
Answer:
0.125
Step-by-step explanation:
Assuming for 95%
z score for 95th percentile = 1.645
We find the Probability using z table.
P(z = 1.645) = P( x ≤ 7000)
= P(x<Z) = 0.95
After 7000 hours = P > 7000
= 1 - P(x < 7000)
= 1 - 0.95
= 0.05
If three lasers are used in a product and they are assumed to fail independently, the probability that all three are still operating after 7,000 hours is calculated as:
(P > 7000)³
(0.05)³ = 0.125
A) Because the 80 is by itself that would be the start up fee.
B) We are told x is the number of months. Since the X is being multiplied by 30, we know that would be the total monthly cost. This is being added to the 80, which does not have an exponent, so we know this is a single cost, which would be the start up fee.
C) Copying the same format as the given equation above, change the numbers:
f(x) = 20 + 35x
D) I used the same format as the first equation, which meant replacing the start up cost from the original one ( 80) with the start up of the new one (20). Then I changed the monthly cost from the original one (30) with the monthly cost of the new one (35).
E) Replace x in each equation with 8 and calculate the cost of each:
80 + 30(8) = 80 + 240 = $320
20 + 35(8) = 20 + 280 = $300
The second club (club B) is the cheaper option for her.
I think it would be A because if you multiply the square by a 30 thinking there all the same space between it will give u 120 square yards
The coefficient in (2/3q-3/4) is 2/3 and in (-1/6q-r) is -1/6
