Q: Jerry has a total of 23 marbles The marbles are either blue or green He has three more blue marbles than green marbles How many blue marbles does he have?

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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

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.

It is not explicitly stated in the question, but it is assumed that you draw one marble from each bag. In this case, you have unrelated sequential probability, similar to tossing three coins. The answer is 0.53 or 0.125.

12 white marbles from (7+3+12) = 22 marblesChance of a white marble on first pick = 12/22 = 6/11.Chance of a white marble on second and third picks are the same, as the marble is replaced.So, the chance of a white marble three times is 6/11 * 6/11 * 6/11 = 216/1331 = approximately 16.23%

answer is 7

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He will have 13 blue marbles and 10 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

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.

He has 10 green marbles.

It is 1248/57120 = 13/595 = 0.0218 approx.

Number of possibilities for one category / Total of all possibilities. For example, if I had a bag of marbles where there are three white marbles and two black marbles. The probability of pulling out a white marble is how many white marbles are in the bag which is: three. But the total of things you can draw out of the bag can either be one of the three white marbles or one of the two black marbles. 3 white marbles+ 2 Black marbles= five marbles. Possibility is 3/5 for drawing a white marble.

There are 286 possible sets of three, (any 3 from 13). There are 28 pairs of reds and 5 possible greens so probability is 5 x 28 ie 140 out of 286 which is just under half (0.4895).

Three sixteenths, or 19%.

at least 39 blue marbles

3/16

11/16

It is 15/50 = 0.3