There are 12 pieces of candy in a bowl. 6 are mints, 4 are Butterscotch, and 2 are Cinnamon. What is the probability of Mr. Binder pulling
Butterscotch from the bowl, putting it back, and
then choosing mint? P(Butterscotch, then mint)

1/6

Step-by-step explanation:

total = 12

number of butterscotch = 4

number of mint = 6

Since he replaces the mint, these events are independent events.

p(butterscotch, then mint) = p(butterscotch) * p(mint)

p(event) = (number of desired outcomes)/(total number of outcomes)

p(butterscotch) = 4/12 = 1/3

p(mint) = 6/12 = 1/2

p(butterscotch, then mint) = p(butterscotch) * p(mint)

p(butterscotch, then mint) = 1/3 * 1/2

p(butterscotch, then mint) = 1/6

1/6

true

explanation:

{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