Answer:
Mean 160
Standard deviation 2.63
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation 
In this problem, we have that:
Find the mean and standard deviation of the sampling distribution of sample means with sample size n = 58.
Mean 160
Standard deviation 
Answer:
Standard form: 2c^4 + 6c^2 - c
Degree: 4th
Leading coefficient: 2
Classification: Trinomial
Step-by-step explanation:
In standard form, you have to put the term with the highest exponent first.
With finding the degree, you just look at the highest exponent and that's the degree. 4 is the highest exponent, so it is to the 4th degree.
The leading coefficient is the number attached to the highest exponent. 4 is the highest and the number attached to it is 2.
Classification goes by how many terms there are. Since there are 3, it's trinomial. Tri as in three.
Answer: the probability that a truck drives between 166 and 177 miles in a day is 0.0187
Step-by-step explanation:
Since mileage of trucks per day is distributed normally, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = mileage of truck
µ = mean mileage
σ = standard deviation
From the information given,
µ = 100 miles per day
σ = 37 miles miles per day
The probability that a truck drives between 166 and 177 miles in a day is expressed as
P(166 ≤ x ≤ 177)
For x = 166
z = (166 - 100)/37 = 1.78
Looking at the normal distribution table, the probability corresponding to the z score is 0.9625
For x = 177
z = (177 - 100)/37 = 2.08
Looking at the normal distribution table, the probability corresponding to the z score is 0.9812
Therefore,
P(166 ≤ x ≤ 177) = 0.9812 - 0.9625 = 0.0187
since the smaller tank has a volume of 720π, then the larger tank has a volume of 3 times that much or 3*720π, or 2160π, and its height is the same as the smaller one's, 20.
