Mathematics
04.04.2022 08:29
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PLEASE HELP THIS IS TIMED! The table shows the dates,times, and high and low tide

PLEASE HELP THIS IS TIMED! The table shows the dates,times, and high and low tide heights.

On what date and time was the lowest tide?

A) Thursday, Feb 7, at 2:29 p.m

B) Friday, Feb 7, at 9:21 p.m

C) Friday, Feb 8, at 2:24 a.m

D) Friday, Feb 8 at 3:16 p.m
PLEASE HELP THIS IS TIMED!

The table shows the dates,times, and high and low tide heights.
On wha
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Lilshawn1
Lilshawn1
4,8(62 marks)

the answer is C hope this helps you

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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