A bag contains 15 blue marbles, 5 red marbles, and 11 green marbles. How many blue marbles must be removed from the 31 marbles in the bag so that the probability of randomly drawing a blue marble is ?

A bag contains 15 blue marbles, 5 red marbles, and 11 green marbles. How many blue marbles must be removed

from the 31 marbles in the bag so that the probability of randomly drawing a blue marble is 1/3?

7 blue marbles

Step-by-step explanation:

15 blue

5 red

11 green

Total = 31

Number of Blue marbles that must be removed to obtain a probability of :

P(blue) = 1/3

Probability = required outcome / Total possible outcomes

Let number of blue marble to be removed = x

Number of Blue becomes : 15 - x

Total becomes : 31 - x

P(blue) = 1/3

1/3 = (15 - x) / (31 - x)

1/3 * (31 - x) = 15 - x

(31 - x) / 3 = 15 - x

31 - x = 45 - 3x

-x + 3x = 45 - 31

2x = 14

x = 14 / 2

x = 7

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!

answer: just subtact the units and then add the non negative : happy to

step-by-step explanation: