The percent body fat in a random sample of 36 men aged 20 to 29 has a sample mean of 14.42. find a 99% confidence interval for the mean percent body fat in all men aged 20 to 29. assume that percent body fat follows a normal distribution, with a standard deviation of 6.95.
a) (0.8, 28.04)b) (12.15, 16.69)c) (12.51, 16.33)d) (12.07, 16.77)

Step-by-step explanation:

The confidence interval for population mean is given by :-

Given : Sample size :

Sample mean :

Standard deviation :

Significance level :

Critical value :

Now, the 99% confidence interval for the mean percent body fat in all men aged 20 to 29 will be :-

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

hello, sorry being so rude after all that fighting i found it kind of funny, so as me saying sorry here is an answer!

since the table showed her that   2km = 2000m.

to determine, how many km in 2000m, she would divide by 1000.

= 2000 / 1000 = 2 km.

similarly, for 8000m = 8000 / 1000 = 8 km

cheers to savings

~

your pal papaguy