Answer:
The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 16 - 1 = 15
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.7531
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 23.29 - 0.34 = 22.95 mpg
The upper end of the interval is the sample mean added to M. So it is 23.29 + 0.34 = 23.63 mpg
The 90% confidence interval for the mean combined fuel economy for Ford Explorers is between 22.95 and 23.63 mpg.
<span>Finding the square root.</span>
we know that
In the right triangle ABC
∠
∠
------> by complementary angles
so
Step 
Find the measure of angle B
we know that
in this problem
so
Step 
Find the measure of angle A
therefore
the answer is the option
35.5° and 54.5°
Just factor. y = (x-1)(x-1)
Answer: The probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.
Explanation:
Step 1: Estimate the standard error. Standard error can be calcualted by dividing the standard deviation by the square root of the sample size:
So, Standard Error is 0.08 million or $80,000.
Step 2: Next, estimate the mean is how many standard errors below the population mean $1 million.
-6.250 means that $1 million is siz standard errors away from the mean. Since, the value is too far from the bell-shaped normal distribution curve that nearly 100% of the values are greater than it.
Therefore, we can say that because 100% values are greater than it, probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.
