The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 2900 miles. What is the probability a particular tire of this brand will last longer than 57,100 miles

84.13% probability a particular tire of this brand will last longer than 57,100 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 a particular tire of this brand will last longer than 57,100 miles

This is 1 subtracted by the pvalue of Z when X = 57100. So

has a pvalue of 0.1587

1 - 0.1587 = 0.8413

84.13% probability a particular tire of this brand will last longer than 57,100 miles

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5