On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to work with. The mean and standard deviation of these 10 temperatures were 98.44oF and 0.30oF, respectively. Construct a 95% confidence interval for the mean of all body temperatures.

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution 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 = 10 - 1 = 9

95% 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 . So we have T = 2.2622

The margin of error is:

M = T*s = 2.2622*0.3 = 0.68

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.44 - 0.68 = 97.76 ºF

The upper end of the interval is the sample mean added to M. So it is 98.44 + 0.68 = 99.12 ºF

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.