At a game show, there are 8 people (including you and your friend) in the front row. The host randomly chooses 3 people from the front row to be contestants. The order in which they are chosen does not matter. There are ^8C^3 = 56 total ways to choose the 3 contestants. What is the probability that you and your friend are both chosen?​

6 / 56

Step-by-step explanation:

Total Number of people = 8

Number of people to be chosen = 3

Hence, total possible outcomes = 8C3 (since order does not matter)

Recall:

nCr = n! ÷ (n-r)!r!

8C3 = 8! ÷ 5!3!

8C3 = (8*7*6) / (3*2*1) = 56

Required outcome :

I and my friend plus one other person

I and my friend = 2C2

1 person from the remaining 6 = 6C1

Required outcome :

2C2 * 6C1 = 1 * 6 = 6

P = required outcome / Total possible outcomes

= 6 / 56

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

according to egenuity the answer is

step-by-step explanation:

d. 344.22