Because 1/3 are blue and 1/2 are red and the rest are green we can calculate that 1/6 of the marbles are green.
1/3 = 2/6
1/2 = 3/6
2/6 + 3/6 = 5/6 which leaves 1/6 green.
Since there are 7 green marbles we know there must be 7 x 6 total.
The answer is 42 marbles.
10 Green marbles, 13 Blue marbles.
Let X = the number of green marbles. X+3 = the number of blue marbles. X + (X+3) = 23 2X + 3 = 23 2X = 20 X = 10 or the number of green marbles.
a+b=23 b=a+3 a+b a+(a+3)=23 a+a=23-3 2a=20 a=10 green marbles=10 blue marbles =23-10 =13
There are at least 11 green marbles in the bag.
hypergeom. f(1;13,3,1) * f(1;12,5,1)
10 Green marbles, 13 Blue marbles.
He will have 13 blue marbles and 10 green marbles.
He has 10 green marbles.
Let X = the number of green marbles. X+3 = the number of blue marbles. X + (X+3) = 23 2X + 3 = 23 2X = 20 X = 10 or the number of green marbles.
a+b=23 b=a+3 a+b a+(a+3)=23 a+a=23-3 2a=20 a=10 green marbles=10 blue marbles =23-10 =13
The ratios are: red : green = 2 : 3 = (2×3) : (3×3) = 6 : 9 green : blue = 9 : 4 → ratio of red : green : blue = 6 : 9 : 4 There are 6 + 9 + 4 = 19 parts 76 marbles ÷ 19 parts = 4 marbles per part → red: 6 parts = 6 × 4 marbles per part = 24 red marbles → green: 9 parts = 9 × 4 marbles per part = 36 green marbles → blue: 4 parts = 4 × 4 marbles per part = 16 blue marbles To check: red + green + blue = 24 marbles + 36 marbles + 16 marbles = 76 marbles in the bag.
There are at least 11 green marbles in the bag.
it is 6/9 simplifyyou get 2/3.
8:6
The fraction is 3/15.
The theoretical probability of randomly drawing a green marble can be calculated by dividing the number of green marbles by the total number of marbles in the bag. In this case, there are 12 green marbles out of a total of 5 red marbles + 8 blue marbles + 12 green marbles, which is 25 marbles in total. Therefore, the theoretical probability of drawing a green marble is 12/25 or 48%.
0No blue marbles in the bag.