The lives of a premium sports car's brakes are normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. What is the probability that such brakes last between 54,000 and 66,000 miles? Group of answer choices

86.64% probability that such brakes last between 54,000 and 66,000 miles

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

What is the probability that such brakes last between 54,000 and 66,000 miles?

This is the pvalue of Z when X = 66000 subtracted by the pvalue of Z when X = 54000.

X = 66000

has a pvalue of 0.9332

X = 54000

has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that such brakes last between 54,000 and 66,000 miles

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )

hope this is what you were asking for!

step-by-step explanation: