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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
You should plug the x and y values into the original equations to get your b value. You should get a b value of 7. Your new equation should be y=2x+7
Answer:
1
Step-by-step explanation:
Solve. Remember to follow PEMDAS.
PEMDAS is the order of operation, and =
Parenthesis
Exponent (& Roots)
Multiplication
Division
Addition
Subtraction
First, divide 54 with -6:
54/(-6) = -9
Next, combine the terms.
10 + (-9) = 10 - 9 = 1
1 is your answer.
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5
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15 would be in simplest form
