Answer:
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
Step-by-step explanation:
We have the standard deviation for the differences, 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 = 8 - 1 = 7
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.8946
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 9.582 - 10.67 = -1.088
The upper end of the interval is the sample mean added to M. So it is 9.582 + 10.67 = 20.252
The 90% confidence interval to estimate the average difference in scores between the two courses is (-1.088, 20.252).
Let the score on her seventh test be "x".
SO,

∴ He must get 104% on his seventh test to get an average score of 90%
Answer:
Find out what x should be the apply by the power of 2 then you should be good
Answer:
I = $ 1,937.50
Equation:
I = Prt
Calculation:
First, converting R percent to r a decimal
r = R/100 = 3.875%/100 = 0.03875 per year,
then, solving our equation
I = 10000 × 0.03875 × 5 = 1937.5
I = $ 1,937.50
The simple interest accumulated
on a principal of $ 10,000.00
at a rate of 3.875% per year
for 5 years is $ 1,937.50.
Step-by-step explanation:
please mark brainliest:)
Answer: Hey, your answer should be 372, if i’m wrong, im so sorry.
Step-by-step explanation: