In January of 2011, the U. S. saw an increase in gas prices. Imagine the average price per gallon was $3.12 with a standard deviation of $0.27, according to a source such as AAA (Automobile Association of America) that tracks gas prices. You are on your long mid-semester break, so you and some friends decide to go on a 3000-mile road trip. You record the price of gas each of the 10 times you fill up your tank, and you compute an average price per gallon of $3.16. What percent of other sample means, based on 10 gas stations, would be greater than the one you observed

31.92%

Step-by-step explanation:

We are given;

Population mean; μ = $3.12

Sample mean; x¯ = $3.16

Sample size; n = 10

Standard deviation; σ = $0.27

Z-score formula is; z = (x¯ - μ)/(σ/√n)

z = (3.16 - 3.12)/(0.27/√10)

z = 0.04/(0.08538)

z ≈ 0.47

Now, the percent of other sample means, based on 10 gas stations, that would be greater than the one observed is;

P(x¯ > 3.12) = 1 - P(z < 0.47)

From z-table attached P(z < 0.47) = 0.68082

Thus;

P(z > 0.47) = 1 - 0.68082

P(z > 0.47) ≈ 0.3192

This expressed in percentage is 31.92%

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

the answer is ! ?