Answer:
a) The probability of falling in the warranty period is 11.6%.
b) The warranty period need to be 19771 km to ensure that no more than 5 per cent of tires fail in the warranty period.
Step-by-step explanation:
a) To fall within the warranty period, the tire has to fail before the 20,000 km.
To calculate the probability, we first calculate the z-value:

Then, the probability of falling in the warranty period is:

b) To calculate this we have to go on from a P(z<z₁)=0.05. This happens for z=-1.645.
This corresponds to a value X of:
The warranty period need to be 19771 km to ensure that no more than 5 per cent of tires fail in the warranty period.
Answer:
338.4Step-by-step explanation:
If you do 80 percent of 188, its 150.4. Then, you add 150.4 to 188. thats 338.4
solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)