Abear biologist is planning a survey of female kodiak grizzly bears to estimate the percent that are pregnant. examining a bear for pregnancy is risky business (duh) and the researcher doesn't want to survey too many or too few bears (just right). how many bears must be surveyed to be 99% confident that the measured pregnancy rate (proportion pregnant) is within no more than 5 percentage points of the true value? based on an initial survey of 16 bears she expects a pregnancy rate of 32%.

 Hence the value of n should be 580.

Step-by-step explanation:

Consider the provided information.

It is given that e = 0.05

p = 32% = 0.32

The confidence interval is 99%

for 99% CI critical Z = 2.580

We can calculate the sample size  by using the formula.

Substitute the respective values

Hence the value of n should be 580.

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

what are the things "represented below"?

step-by-step explanation:

nobody can if you dont give the choices