Suppose that two cards are randomly selected from a standard 52-card deck. (a) what is the probability that the first card is a club and the second card is a club if the sampling is done without replacement? (b) what is the probability that the first card is a club and the second card is a club if the sampling is done with replacement?

(a) 1/ 17  (b) 1/16.

Step-by-step explanation:

(a) Probability ( first car d is a club) = 13/52 = 1/4

Probability( second card is a club) = 12/51 .

So the probability ( first and second are clubs) = 1/4 * 12/51

= 12/ 204 = 1/17 (answer).

(b)  Probability ( first car d is a club) = 13/52 = 1/4

Probability ( second card is a club) = 13 / 52 = 1/4

So the probability ( first and second are clubs) = 1/4 * 1/4

= 1/16. (answer).

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1