A group of 30 people, 12 females and 18 males, are electing 5 representatives. a. How many ways can the 5 representatives be chosen from the group?

b. How many ways can the 5 representatives be 2 males and 3 females?

c. What is the probability that the representatives end up being 2 males and 3 females if they are randomly selected? Enter your answer as a decimal rounded to the nearest thousandth. (3 decimal places).

a) Combination of 5 out of 30:

b) Combination of 2 males:

Combination of 3 females:

The number of ways with 2 males and 3 females:

c) Required probability is:

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