A chemical company claims that its new additives increases gasmileage. As a pilot study, the company send 36 cars fueled withthe new additive on a road trip from Las Vegas to LA. Without the additive, those same cars are known to average 25mpg with a standard deviation of 2.4 mpg. Suppose it turns out that the 36 cars averaged 25.8 mpg with theadditive. Is this additive effective

Check Explanation

Step-by-step explanation:

Normally the mean mileage for the 36 cars without the additive is 25 miles per gallon of gas with a standard deviation of 2.4 mpg

With the additive, the mileage improves to 25.8 mpg.

The sample mean without additive is 25 mpg

The standard deviation of the distribution of sample means for mileage without additive = σₓ = (σ/√n)

where σ = normal standard deviation = 2.4 mpg

n = Sample size = 36

σₓ = (σ/√n) = (2.4/√36)

σₓ = 0.4 mpg

So, a normal distribution of sample means for mileage without gas should have 95% of the possible values between 2 standard deviations of the mean. That is, within 24.2 mpg and 25.8 mpg.

Therefore, a mileage of 25.8 mpg with the additive added cannot point at a significant effectiveness of the additive as this data has shown that 25.8 mpg is within the possible values for sample means for mileage without the use of additives.

Although, it does point at an improved performance with the additive because a sample mean of 25.8 mpg without the additive is on the outskirts of possible sample mean values.

Hope this Helps!!!

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%

the answer is a.

step-by-step explanation:

in three games of golf, alex shot two under par four times.