Seven women and nine men are on the faculty in the mathematics department at a school.

a) How many ways are there to select a committee of five members of the department if at least one woman must be on the committee?

3,507ways

Step-by-step explanation:

Combination has to do with selection.

Since there are 7 women and nine men on the committee, in order to select a committee of five members of the department if at least one woman must be on the committee this means the number of women that must be part of the committee must be minimum of a woman.

Generally, to select r object from a total of n object is represented as nCr = n!/(n-r)!r!

Owing to that condition, the selection will done this way;

7C1 × 9C4 = 7×21 = 147ways (1 woman on the committee)

7C2 × 9C3 = 21 × 84 = 1764ways (2women on the committee

7C3 × 9C2 = 35× 36 = 1260ways (3women on the committee)

7C4 × 9C1 = 35 × 9 = 315ways (4women on the committee)

7C5 × 9C0 = 21 × 1 = 21ways (5women on the committee)

The total number of ways the committee can be formed is gotten by adding all individual selections i.e

147+1764+1260+315+21

= 3,507ways

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

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