The varsity soccer team has 20 players. Three of the players are trained to be goalies while the remaining 17 can play any position. Only 11 of the players can be on the field at once. (a) We want to find the number of possible groups of 11 players the coach can choose. Is this a
permutation or a combination?
(b) If the coach wanted to choose her 11 starters at random by drawing names from a hat, how many possible groups of 11 starters could she choose?
(c) The coach wants to make sure there is exactly one goalie on the field. How many ways can the
coach choose a lineup of 11 players if exactly 1 player must be a goalie?

Combination.

167960

58344

Step-by-step explanation:

A.)

Choosing 11 players from 20 ; here the order in which the players are chosen does not matter, hence it is a combination problem.

2.)

Using nCr

n = 20 ; selection, r = 11

nCr = n! / (n-r)!r!

20C11 = 20! / 9!11!

20C11 = 167960

3.)

1 goalie 10 players

3C1 * 17C10

3 * 19448

= 58,344 ways

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p