The answer is 180.add 40 to its self or multiply by 4 then add 20 for the .5 portion, 40x4=160 160+20=180.
hoped this helped =)
Answer:
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
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 = 14 - 1 = 13
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of
. So we have T = 3.0123
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 664.14 - 224.85 = $439.29
The upper end of the interval is the sample mean added to M. So it is 664.14 + 224.85 = $888.99.
The 99% confidence interval for the true mean checking account balance for local customers is ($439.29, $888.99).
I would think it is b sorry if im wrong
Answer: x=5
Step-by-step explanation:
If you break the fraction into a whole number it would be 5 and 10 minus 5 is 5. so x=5