Six men and four women are waiting to be interviewed for jobs. If they are all selected in random order, find the probability that no man will be interviewed until at least two women have been interviewed.

The correct answer is 0.1714 .

Step-by-step explanation:

There are 6 men and 4 women to be interviewed for jobs.

Total number of arrangements in which they can be called for the interview process is 10!.

Number of ways any two women are interviewed before any man is 4 × 3 × 8!. = 12 × 8!.

Number of ways any three women are interviewed before any man is 4 × 3 × 2 × 7!. = 24 × 7!.

Number of ways all the women are interviewed before any man is 4! × 6!.

Required number of ways in which at least two women being interviewed before any man is given by 12 × 8! + 24 × 7! + 4! × 6! = 864 × 6!.

Required probability is 864 × 6! ÷ 10! = 0.1714

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

a: inscribed angles

step-by-step explanation:

i reviewed this question with a peer and teacher, seems to be correct, me later