C. assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.

Step-by-step explanation:

The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

This is the question of normal distribution:

First w calculate the value of Z corresponding to X = 60,000 miles

We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles

Now, for Z, we know that:

Z = (x-μ)/σ

Z = (60,000 - 50,000)/5,000

Z = 2

Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.

P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)

P(X > 60,000) = 1 - 0.9772

P(X > 60,000) = 0.0228

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

idk

step-by-step explanation: