Aset of 48 cards is numbered with positive integers from 1 to 48. 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 6 and 8?

The probability of randomly drawing a card that is a multiple of both 6 and 8 is:

P = 1/24 = 0.42

Assuming that all the cards have the same probability of being randomly drawn, we can conclude that the probability of randomly drawing a number that is multiple of both 6 and 8 is equal to the quotient between the number of cards that are multiples of both 6 and 8, and the total number of cards on the deck.

The multiples of 6 are:

6*1 = 6

6*2 = 12

6*3 = 18

6*4 = 24

6*5 = 30

6*6 = 36

6*7 = 42

6*8 = 48

The multiples of 8 are:

8*1 = 8

8*2 = 16

8*3 = 24

8*4 = 32

8*5 = 40

8*6 = 42

The common multiples are: 24 and 42.

Then out of 48 cards, there are only two that are multiples of both 6 and 8, then the probability of randomly choosing one of these two is:

P = 2/48 = 1/24 = 0.042

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multiples of 6 = 6, 12, 18, 24 , 30, 36 ,42 , 48

multiples of 8 = 8, 16, 24,32,40,48

 24 & 48 are the common multiples

so there is a 2 out of 48 chance which is 2/48 which reduces to 1/24 probability

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61