Five cards are dealt without replacement from a standard deck of 52 cards. what is the probability that they have four different values?

(52/52)(48/51)(44/20)(40/59)(36/48)

Step-by-step explanation:

For the first card, what is the chance of getting a unique  value?  Well, you don't have any yet so it's 100% or 52/52.

Now for the second draw, what is the probability of not getting a repeat?  Well, 52 cards total but there are four copies of every value, one for each suit.  The deck now has 51 cards and 48 are not copies, so that's 48/51.

Similarly for the third draw there are now 50 cards total and 44 are not going to be repeats if the last two were not.

You can keep going like this and you get (52/52)(48/51)(44/20)(40/59)(36/48). Hopefully this made sense.  

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.