NinjaAnswer

(3/7x + 4/7) - (1/3x -1/2) Combine Like Terms

3/7x + 4/7 - 1/3x +1/2

2/21x + 15/14

Just a lot of smart ninjas out there. Not naming any tho.........​​

The probability that it rains and Sheila attends is (0.4)(0.2) = 0.08 . The probability that it doesn't rain and Sheila attends is (o.6)(0.8) .So the overall probability that Sheila attends is 0.08 + 0.48 = 0.56 = 56% .

Theo = t

Callie = c

t = 2c

c = 14

t = 2 * 14

t = 28

Ok!

There are 4 ways to choose which 3 days it rains. It would be 

((3/4)^3)(1/4) = 27/256 because each day has a 3/4 chance of rain and 1/4 chance of sunshine.

For each of the 4 ways this can happen, that would be:

4 * (27/256) 

27/64

You can position all of the wombats in 7! ways BUT you have over counted the 3 hairy wombats. So the answer is:

7!/3! = 840 

@CPhill, you are wrong. Everything was correct up to this(From your post:)

Note.....the only remaining team is the one where both are on it and we can pick any 3 of the other 10 players  = C (10,3)   =  120

BUT...If you look at the question, it says that Bob and Yogi refuse to play together.

The Screamers are coached by Coach Yellsalot. The Screamers have 12 players, but two of them, Bob and Yogi, refuse to play together. How many starting lineups (of 5 players) can Coach Yellsalot make, if the starting lineup can't contain both Bob and Yogi? (The order of the 5 players in the lineup does not matter; that is, two lineups are the same if they consist of the same 5 players.)

There are 6^2=36 possibilities of rolls.

There are only three of them that sum less than or equal to 3. (1,1), (1,2), (2,1). That means 3/36 = 1/12 don't work. The ones that ARE greater than 3 are 1 - 1/12 = 11/12

There are 16 cards in a deck of 52 which is an ace or a diamond or both. So that means the probability of NOT choosing an ace or a diamond or both on both turns is:

((52-16)/52)^2 

(36/52)^2

(9/13)^2

81/169

This means that the probability of choosing an ace or a diamond or both on both turns is:

1 - 81/169

88/169