The average lifespan of a set of tires is 38,000 miles, with a standard deviation of 1500 miles. What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Step-by-step explanation:

Problems of normally distributed samples are solved using 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 problem, we have that:

What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

This is the pvalue of X when X = 44000 subtracted by the pvalue of Z when X = 32000. So

X = 44000

has a pvalue of 1.

X = 32000

has a pvalue of 0.

1 - 0 = 100%

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

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hope this !

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)