Answer:
-4 • (x - 5) • (2x + 3)
Step-by-step explanation:
Answer:
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!!!
Hope this help to answer you question.
a. Your answer is correct.
b. Divide the given value of sigma by √n where n = sample size = 7. Then perform the same calculation you did for part a. The probabiilty is 0.4427.
