Mathematics
30.08.2021 11:18
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A savings account starts with $231 25. After 8 years of continuous compounding at

A savings account starts with $231 25. After 8 years of continuous compounding at an interest rate the account has $1850.
What is the interest rate percentage?
Round the answer to the nearest hundredth.
Enter your answer in the box
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smallsbjs
smallsbjs
4,8(58 marks)

87.5%

Step-by-step explanation:

Took the k12 math test

wonderwonder2748
wonderwonder2748
4,9(35 marks)

The 98% confidence interval for the mean of the population is (59, 68.2).

Step-by-step explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation.

Sample mean:

\overline{x} = \frac{55+64+58+61+69+64+59+69+72+65}{10} = 63.6

Sample standard deviation:

s = \sqrt{\frac{(55-63.6)^2+(64-63.6)^2+(58-63.6)^2+(61-63.6)^2+(69-63.6)^2+(64-63.6)^2+(59-63.6)^2+(69-63.6)^2+(72-63.6)^2+(65-63.6)^2}{10}} = 5.142

Confidence interval:

We have the standard deviation for the sample, and thus, we use the t-distribution.

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 = 10 - 1 = 9

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.98}{2} = 0.99. So we have T = 2.821

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.821\frac{5.142}{\sqrt{10}} = 4.6

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 63.6 - 4.6 = 59

The upper end of the interval is the sample mean added to M. So it is 63.6 + 4.6 = 68.2.

The 98% confidence interval for the mean of the population is (59, 68.2).

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