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).
Answer: the opposite answer is 35F
Step-by-step explanation: the C goes to a F
The next step would be:
-0.4 -0.4
explanation:
You would need to move the variables to one side so minus the 0.4 would move the numerical number to the other side which leave the variable to the other side.
Hope this makes sense :)
Answer:
SOLVE FOR X, Y
x=12
y=−6
Step-by-step explanation:
Answer:

Step-by-step explanation:
Alice and Bill are planning to have three children
We have to find the probability that all three of their children will be girl.
Sample space =S={BBB,BBG,BGB,GBB,GGG,GBG,GGB,BGG}
Total =8
Number of cases in which all three children are girls={GGG}
Probability,P(E)=
Using the formula
Then , the probability that all three of their children will be girls=