There is a bag filled with 5 blue and 4 red marbles. A marble is taken at random from the bag, the colour is noted and then it is replaced.
Another marble is taken at random.
What is the probability of getting 2 that are different in colour?

 40/81

Work Shown:

B = selecting 1 blue

R = selecting 1 red

P(B) = 5/9, since there are 5 blue out of 5+4 = 9 total

P(R) = 4/9, since there are 4 red out of 9 total

P(2 blue) = P(B)*P(B) = (5/9)*(5/9) = 25/81

P(2 red) = P(R)*P(R) = (4/9)*(4/9) = 16/81

The last two equations are valid because we are sampling with replacement. Each selection is independent.

P(2 same color) = P(2 blue OR 2 red)

P(2 same color) = P(2 blue) + P(2 red)

P(2 same color) = 25/81 + 16/81

P(2 same color) = (25+16)/81

P(2 same color) = 41/81

P(2 different color) = 1 - P(2 same color)

P(2 different color) = 1 - 41/81

P(2 different color) = 81/81 - 41/81

P(2 different color) = (81-41)/81

P(2 different color) = 40/81

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

i can't read the graph

step-by-step explanation: