Abag contains 6 red balls, 4 green balls, and 3 blue balls. if we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

Step-by-step explanation:

Given : A bag contains 6 red balls, 4 green balls, and 3 blue balls.

Total balls = 6+4+3=13

Probability of drawing first ball as green :

If the first ball will be green, then the total balls left in bag = 13-1=12 and number of red balls remains the same.

Now, The conditional probability of drawing a red ball given that first ball was green :-

Now,  the probability that the first ball will be green and the second will be red will be :-

Hence, the required probability =

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

50x5=$250

step-by-step explanation: