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)
Answer:
one thousand, seven hundred twenty-eight
Do 2*3 then do the answer divided by 18 to get the answrr
(2) 2x
Bc you’re adding the imaginary one infront of the x.
(1) X squared + 25
Distribution
(3) 15x-30
Distribution
(1) x to the 6 power
You add the powers when they are being multiplied
(2) 2x squared + 7x + 12
Foil method
(2) 9x + 20
Adding Like terms
(4) 12x + 25
Adding and multiplying like terms
(2) 19x + 7
Multiplying then adding like terms.
I hope this helps ;)
Mark me brainliest please! ;D
Writing an equation would be most helpful, but depending on the situation drawing a diagram or reading a table could work better.
