7. The lifetime of a certain brand of tires is approximately normally distributed, with a mean of 40,000 miles and a standard deviation of 2,500 miles under normal driving conditions.

Tire wear is greatly affected by road, weather and driver conditions along with proper

maintenance of the tires. A driver who aggressively accelerates or makes quick stops, for

example, will wear out a tire much more quickly. The brand carries a warranty of 33,000 miles

under normal driving conditions, i. e., the company will replace a tire if it wears out before this

mileage limit is reached.

0.0025551

Step-by-step explanation:

Given that:

Mean (m) = 40000

Standard deviation (s) = 2500

P(x < 33,000) :

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (33000 - 40000) / 2500 = - 2.8

p(Z < -2.8) = 0.0025551 ( Z probability calculator)

Probability of Tyre wagering out before mileage limit is reached = 0.0025551

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine