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

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The solution is given in the image below

Here,

tip=18%of $32

tip=(18/100)*32

=0.18*32

=$5.76

Total payment=32+5.76=$37.76

The answer is in the image 

30% of 420 are first year students

first year students = (30/100 )*420

=126 students

Student who are not in first year = 420-126

=294 students