Answer: 49/60
Step-by-step explanation:
(1/3+2/5)-1/4-(5/6-7/6)
Primero (5/6-7/6)
5/6-7/6=-2/6
1/3+2/5-1/4-(-2/6)
1/3+2/5-1/4+2/6
2/6=1/3
1/3+2/5-1/4+1/3
2/3+2/5-1/4
2/3=10/15
2/5=6/15
10/15+6/15-1/4
16/15-1/4
16/15=64/60
1/4=15/60
64/60-15/60=49/60
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
10 is the answer
why are you struggling
so simple
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It is a very high probability. If you role two dice no matter what you will get 12 or less.
Answer:
7
Step-by-step explanation:
Substitute x = -4 :
f(-4) = 3 - (-4)
= 3 + 4
= 7
