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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
Answer:
p = ½ (x₁ + x₂)
q = a (x₁x₂ − ¼ (x₁ + x₂)²)
Step-by-step explanation:
y = a (x − x₁) (x − x₂)
Expand:
y = a (x² − x₁x − x₂x + x₁x₂)
y = a (x² − (x₁ + x₂)x + x₁x₂)
Distribute a to the first two terms:
y = a (x² − (x₁ + x₂)x) + ax₁x₂
Complete the square:
y = a (x² − (x₁ + x₂)x + ¼(x₁ + x₂)²) + ax₁x₂ − ¼ a(x₁ + x₂)²
y = a (x − ½ (x₁ + x₂))² + a (x₁x₂ − ¼ (x₁ + x₂)²)
Therefore:
p = ½ (x₁ + x₂)
q = a (x₁x₂ − ¼ (x₁ + x₂)²)
Answer:
0.45454545454, also known as 0.45 repeated
Step-by-step explanation:
Answer: Down
Step-by-step explanation:
We know that the graph of y
=
x
4 has both arms pointing up.
Now, since the leading coefficient of f
(
x
) is negative, both arms are pointing down. Thus the right side is pointing down.
