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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
The mean increases when the outlier is removed
Answer:
BD = 3 units
Step-by-step explanation:
Since, AD is an angle bisector of ∠BAC,
m∠BAD = m∠CAD = 20°
CD = 3 units
In ΔACD and ΔABD,
m∠BAD = m∠CAD = 20° (Given)
AD ≅ AD [Reflexive property]
Therefore, by H-A property of congruence both the triangles will be congruent.
And by CPCTC,
CD ≅ BD = 3 units
Answer:
62
Step-by-step explanation:
(-8)^2-2
64-2 = 62
