(1) a manufacturer of tires believes that their mean time to failure (mttf) is 20 613 km. they also believe that time to failure (ttf) is normally distributed with a standard deviation of 512 km (a) if the manufacturer wants to honor a 20 000 km warranty, what is the probability that any tire will fail within the warranty period? (b) what does the warranty period need to be to ensure that no more than 5 per cent of tires fail in the warranty period

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: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere

change of stock price/initial stock price on friday * total stock value = change of stock value

0.89/73.73 *x = 42.72

x=3539.04 = total stock value at initial stock price on friday .

total stock value at initial stock price on friday/ initial stock price on friday = number of shares

3539.04/73.73 = 48 shares

they own 48 shares of this stock