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,
![R1(t) = e^{-nt} .....1](https://tex.z-dn.net/?f=R1%28t%29%20%3D%20e%5E%7B-nt%7D%20.....1)
where t = time in years = 1,
n = failure rate constant = 0.07
Also,
![R2(t)=e^{-(\frac{t}{Q} )^{B} }......2](https://tex.z-dn.net/?f=R2%28t%29%3De%5E%7B-%28%5Cfrac%7Bt%7D%7BQ%7D%20%29%5E%7BB%7D%20%7D......2)
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
![R1(t) = e^{-(0.07)(1)} \\\\R1(t) = e^{-0.07}](https://tex.z-dn.net/?f=R1%28t%29%20%3D%20e%5E%7B-%280.07%29%281%29%7D%20%5C%5C%5C%5CR1%28t%29%20%3D%20e%5E%7B-0.07%7D)
Substituting values into equation 2
![R2(t)=e^{-(\frac{1}{10} )^{1.8} }\\\\R2(t)=e^{-(0.1)}^{1.8} }\\\\R2(t)=e^{-0.0158}](https://tex.z-dn.net/?f=R2%28t%29%3De%5E%7B-%28%5Cfrac%7B1%7D%7B10%7D%20%29%5E%7B1.8%7D%20%7D%5C%5C%5C%5CR2%28t%29%3De%5E%7B-%280.1%29%7D%5E%7B1.8%7D%20%7D%5C%5C%5C%5CR2%28t%29%3De%5E%7B-0.0158%7D)
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)
We don't know which side your triangle is sitting on, so the 'height'
could be any one of three values.
Here are the two simple tools that tell you everything you want to
know about a 30°-60°-90° triangle. These are worth memorizing:
-- The side opposite the 30° angle is 1/2 of the hypotenuse.
-- The side opposite the 60° angle is 1/2 of the hypotenuse times √3 .
I memorized them exactly 60 years ago. They've been very useful,
and as far as I know, they haven't changed.
Answer:
See attached graph
Step-by-step explanation: