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

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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%

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

ok then what happened.

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: