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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
N=-13/4 because you subtract 3/4 from -5/2
Fraction is 7/27 or 0.259
The correct answer is A.
You have direct variation if x and y are modeled by the equation
In this case, m is the constant of proportionality. So, if the constant has to be 2, the equation becomes
A side note: Actually, option C has a constant of proportionality of two as well, except the roles of x and y are interchanged. I chose option A because usually you want the y = mx form, but the names of the variables are obviously meaningless.
Divide 555/.26 & you’ll get your original price
