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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
Answer:
or y = 0.25x + 0.5
Step-by-step explanation:
You need to try change the subject from x to y.
x = 4y - 2
We first move -2 to the left, making the equation
x + 2 = 4y i.e.
4y = x + 2
We then divide both sides by 4.
or y = 0.25x + 0.5
We now have successfully made y the subject, and effectively solving the equation.
To answer this, we substitute 12 ft to the f(h) in the given equation,
f(h) = -8t² + 8t + 12 = 12
Subtracting 12 from both sides of the equation will give us an answer of,
-8t² + 8t = 0
Then, we transpose 8 to the other side and divide the equation by -8 and t, we get an answer of t = 1 second.
