F(5) = -1 would be the correct answer.
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
1) tan 41°= 30cm divide by y cm
y= 30 divide by tan-1 41°
1: answer: 4^6 ( reduce the fraction with 4^3)
2: answer: 1 ( any nonzero expression raised to the power of 0 equals 1 )
3: answer 3^8
steps: (3^2)^4 = 3^2×4 ( to raise a power to another Power multiply the exponents)
3^2×4 = 3^8 ( multiply the 2 and the 4 exponent to get a 8 as a exponent )
