11. A certain brand of margarine was analyzed to determine the level of polyunsaturated fatty acid (in percent). A sample of 6 packages had an average of 16.98 and sample standard deviation of 0.31. Assuming normality, a 99% confidence interval for the true mean of fatty acid level is:

(16.47, 17.49)

Step-by-step explanation:

Formula for confidence interval for the true mean if population stanmdard deviation is unknown:

, where = sample mean

n= sample size

s= sample standard deviation

= Two tailed critical value.

We assume that the level of polyunsaturated fatty acid is normally distributed.

Given,

n= 6

degree of freedom = n-1 =5

= 16.98

s= 0.31

significance level =1-0.99=0.01

Two tailed t- value for degree of freedom of 5 and significance level of 0.01 =    [by student's t-table]

Now , the 99% confidence interval for the true mean of fatty acid level is:

Hence, a 99% confidence interval for the true mean of fatty acid level is: (16.47, 17.49)

idk bring the function back   im not smart

step-by-step explanation:

answer: city needs 2.4 inches of rainfall more to match the record of 40 inches.

step-by-step explanation:

since we have given that

total annual rainfall a city record = 40 inches

this year, total rainfall recorded = 37.6 inches

number of inches of rainfall   is needed to match the record of 40 inches is given by

hence, city needs 2.4 inches of rainfall more to match the record of 40 inches.