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:
A
Step-by-step explanation:
given
= 
using the method of cross- multiplication then
8P = 42 ( divide both sides by 8 )
P =
= 5.25
They are not the same thing.
Let
x be the price for a pound of ham
y the price for a pound of cheese
<span>1.3x +0.8y = 11.28 ....(1)
1.5x +1.2y = 14.76 ....(2)
Lets multiply (1) by -1.2
Lets multiply (2) by 0.8
-1.56x -0.96y = -13.536 .....(3)
</span> 1.2x + 0.96y = 11.808 .....(4)
If we add (3) and (4)
-0.36x = -1.728 ............> x = 4.8
We substitute in (1)
1.3(4.8) +0.8y = 11.28
y = 6.3
Price per pound of ham = x = 4.8
Price per pound of cheese = y = 6.3