At a certain clinic, 2 percent of all pap smears show signs of abnormality. (a) What is the expected number of pap smears that must be inspected before the first abnormal one is found? Expected number of pap smears

The expected amount of pap smears that must be inspected before the first abnormal one is found is 50

Step-by-step explanation:

If 2 percent of all pap smears show signs of abnormality, then the probability that a pap smear is abnormal is 0.02.

Let X be the amount of pap smears needed before the first abnormal case is found. X has geometric ditribution with parameter p = 0.02. The mean of X, in other words, the expected amount of cases that must be inspected before the first abnormal one appears, is 1/p = 1/0.02 = 50.

you can not drow on this

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet