1. C (2,2)
2. H (-9,-5)
3. B (-2,6)
4. E (8,-4)
5. G (-4,-2)
6. ( 6,8)
7. A (-8,4)
8. F (4,-9)
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: umm wheres y at? after the Equal sign?
Step-by-step explanation:
Answer:
log 5
Step-by-step explanation:
We can rewrite this by using a property of logs
log a - log b = log (a/b)
log(10) - log(2)
log(10/2)
log 5
Answer:
A
Step-by-step explanation:
