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Answer:
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
Answer:
(x) =
Step-by-step explanation:
let y = g(x) and rearrange making x the subject, that is
y = 4x - 11 ( add 11 to both sides )
y + 11 = 4x ( divide both sides by 4 )
= x
Change y back into terms of x , with x = (x) , thus
(x) =