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adambbogard1589
31.03.2020 •
Mathematics
A certain car model has a mean gas mileage of 31 miles per gallon (mpg) with astandard deviation 3 mpg. A pizza delivery company buys 43 of these cars. What is theprobability that the average mileage of the fleet is greater than 30.7 mpg?
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Ответ:
Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
Step-by-step explanation:
We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.
A pizza delivery company buys 43 of these cars.
Let
= sample average mileage of the fleet
The z-score probability distribution of sample average is given by;
Z =
~ N(0,1)
where,
= mean gas mileage = 31 miles per gallon (mpg)
n = sample of cars = 43
So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P(
> 30.7 mpg)
P(
> 30.7 mpg) = P(
>
) = P(Z > -0.66) = P(Z < 0.66)
= 0.7454
Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454.
Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
Ответ:
16 km due west
Step-by-step explanation:
The bearing of the school p from school q is 16 km due west.
To find the bearing of school q from school p, we have to find the direction that the school q is with respect to school p.
Since p is directly west of q, then it implies that q must be directly east of p.
We now know the direction.
Since the distance from q to p is exactly the same as the distance from p to q, then, the distance from p to q is 16 km.
Hence, the bearing of q from p is 16 km due west.