Randy wants to either ride share to work or drive his own car to work, but it is impossible for Randy to ride share and drive his own car in one trip. If the probability that Randy ride shares is 0.22, and the probability that Randy drives his own car is 0.42, what is the probability that Randy ride shares or drives his own car to work?

The probability that Randy ride shares or drives his own car to work is 0.64

Step-by-step explanation:

Given;

let the probability that Randy ride shares be A

let the probability that Randy drives his own car be B

P(A) = 0.22

P(B) = 0.42

The probability that Randy ride shares or drives his own car to work;

P = P(A) + P(B)

P = 0.22 + 0.42

P = 0.64

Therefore, the probability that Randy ride shares or drives his own car to work is 0.64

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.

{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