Not sure how simplified you need it...
8ab-3b+8
This question not incomplete
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
If you take the square root of a number squared number then they cancel each other out and the number stays the same i.e. √[(4)^2] would equal 4.
In this problem the square root and numbered squared cancel out to leave the problem as -2a.
The solution of this problem is -2a
Multiply all of the numbers together: 5 x 2 x 8 = 10 x 8 = 80 ways
x = number of hours
want to find probability (P) x >= 13
x is N(14,1) transform to N(0,1) using z = (x - mean) / standard deviation so can look up probability using standard normal probability table.
P(x >= 13) = P( z > (13 - 14)/1) = P(z > -1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413
To convert that to percentage, multiply 100, to get 84.13%
