One pile of cards contains the number 2 through 6 in red hearts. A second pile of cards contains the numbers 4 through 8 in black spades. Each pile of cards has been randomly shuffled. If one card from each pile is chosen at the same time, what is the probability that the sum will be less than 12?

19/25.

Step-by-step explanation:

The total possible number of combinations of 2 cards from the 2 piles = 5*5 = 25.

Possible ways to get sum of 12 or more =

4,8   5,7   5,8   6,6   6,7  6,8  =  6 ways.

So the number of ways to get sum of < 12 = 25 - 6 = 19.

So the required probability = 19/25.

I think the probability  is 19 out of 25    = 19/25, but I am not sure so please check by step by step solution.

Step-by-step explanation:   5  29  12  38

Hearts 2, 3, 4, 5, 6

Spades 4, 5, 6, 7, 8

Find the probability that the sum of any two randomly cards is less than 12.

I count 5²  = 25  possibilites

I found 5 + 5 + 4 + 3 + 2 combinaitons of cards whose sum is less than 12

             sum of 19 combinations less than 12

               19 /25 is the probability of the sum being less than 12

     hearts   + spade < 12

       the 2 + any shade      5

       the 3 + any spade      5

       the 4 + the four lowest spades   4

       the 5 + the three lowest spades   3

       the 6 + the four lowest spades   2

c.35

step-by-step explanation:

8+13=21

21+8=29

29+6=35

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.