All are relations except for relation 3
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
Answer:
the graph is as illustrated above
The absolute value of x is shifting 3 units up
Answer:
R ≈ 8.31
Step-by-step explanation:
Filling in the given values ...
... pV = nRT
... 2·4.155 = 1·R·1
... 8.31 = R
Answer:
8p +56 +16q
Step-by-step explanation:
8 (p+7+2q)
If we distribute the 8 into everything in the parenthesis, we get:
8p +56 +16q
