Answer:
t=27/4
Step-by-step explanation:
Answer:
0.9177
Step-by-step explanation:
let us first represent the two failure modes with respect to time as follows
R₁(t) for external conditions
R₂(t) for wear out condition ( Wiebull )
Now,

where t = time in years = 1,
n = failure rate constant = 0.07
Also,

where t = time in years = 1
where Q = characteristic life in years = 10
and B = the shape parameter = 1.8
Substituting values into equation 1

Substituting values into equation 2

let the <em>system reliability </em>for a design life of one year be Rs(t)
hence,
Rs(t) = R1(t) * R2(t)
t = 1
![Rs(1) = [e^{-0.07} ] * [e^{-0.0158} ] = 0.917713](https://tex.z-dn.net/?f=Rs%281%29%20%3D%20%5Be%5E%7B-0.07%7D%20%5D%20%2A%20%5Be%5E%7B-0.0158%7D%20%5D%20%3D%200.917713)
Rs(1) = 0.9177 (approx to four decimal places)
D
If you multiply 3 by 10 and add a zero onto the ten in the question that is 100 in half an hour, all you need to do then is multiply that answer by two to get your answer.
Answer:
D.
Explanation:
PEMDAS method. First do the parentheses. Multiply first, then subtract 12 from that number. Once finished with the parentheses, multiply by 8.
Answer is answer 17 plus 78