PLEASE HELP A bag contains red and blue marbles, such that the probability of drawing a blue marble is 3/8 An experiment consists of drawing a marble, replacing it, and drawing another marble. The two draws are independent. A random variable assigns the number of blue marbles to each outcome. What is the probability of drawing one blue marble?

15/64

Step-by-step explanation:

3 blue

5 red marbles.

since the marble is put back in, the denominator will not change,

so to draw ONE marble, the chances would be

chances of picking a blue marble 3/8

chances of picking a red marble 5/8

3/8 * 5/8 - 15/64

please mark me .

The probability of drawing one blue marble is 3/8 because you have three blue marbles out of a total of eight marbles. The probability of drawing a second blue marble is also 3/8 but since you have to draw both marbles consecutively you have to multiply the probabilities of drawing each marble. So,

3/8•3/8 = 9/64.

Now you need to change it to a percent.

We do this by

9/64 x 100 = 14% approx

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.