Aset of 36 cards is numbered with the positive integers from 1 to 36. if the cards are shuffled and one is chosen at random, what is the probability that the number on the card is a multiple of both 4 and 6?

1/12.

Step-by-step explanation:

The numbers divisible by 6 are 6, 12, 18, 24, 30 and 36.

Of these  12, 24 and 36 are also divisible by 4.

So the required probability is 3/36

= 1/12.

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.