A tire manufacturer believes that the life of its tires follow a normal distribution with a mean of 46,000 miles and a standard deviation of 4,000 miles. What mileage can he guarantee each tire to last so that 99% of the tires last longer than the guaranteed lifetime?

Guarantee of Tyre = 36,680 miles

Step-by-step explanation:

Given:

Mean = 46,000 miles

Standard deviation = 4,000 miles

Computation:

Assume;

x = guaranteed lifetime

P(X > x) = 99%

So,

P(X > x) = 0.99

So,

P(Z > z) = 0.99

So,

1 - P(Z > z )  = P(Z < z)

P(Z < z) = 0.01

So, in z table value of z = -2.33

x = μ + [z x σ]

x = 46,000 + [-2.33 x 4000]

x = 36,680

Guarantee of Tyre = 36,680 miles

step-by-step explanation:

the shape,the size and the distance

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5