Asked by buttocks
(1 point)
1. There are 35 marbles in a bag: 9 blue marbles, 8 green marbles, 4 red marbles, 8 white marbles, and 6 yellow marbles. Find P(red). Write the probability as a fraction in simplest form, a decimal, and a percent.
0.129, 12.9% 0.114, 11.4% 0.143, 14.3% 0.086, 8.6% 31 35 2 3 35
1. There are 35 marbles in a bag: 9 blue marbles, 8 green marbles, 4 red marbles, 8 white marbles, and 6 yellow marbles. Find P(red). Write the probability as a fraction in simplest form, a decimal, and a percent.
0.129, 12.9% 0.114, 11.4% 0.143, 14.3% 0.086, 8.6% 31 35 2 3 35
Answers
Answered by
hey
a bag contains 6 red marbles, 3 white marbles, and 7 blue marbles you pick without looking find the probability of drawing a white marble
Answered by
hey
a standard number cube is rolled 180 times predict how many times a 3 or a 5 will be the result
Answered by
hey
a standard number cube with the numbers 1 through 6 is rolled find the probability of rolling a number less then 3 find the probability of not rolling a number less then3.
Answered by
hey
if no digit may be used more than once how many 5-digit number can be formed using only the digits 3,1,2,5, and 7?
Answered by
Bot
P(red) = 4/35 or 0.114 or 11.4%
Answered by
Bot
There are a total of 6 + 3 + 7 = 16 marbles in the bag.
The probability of drawing a white marble is:
P(white) = 3/16
So the probability of drawing a white marble is 3/16 or approximately 0.188 or 18.8%.
The probability of drawing a white marble is:
P(white) = 3/16
So the probability of drawing a white marble is 3/16 or approximately 0.188 or 18.8%.
Answered by
Bot
There are 6 equally likely outcomes when rolling a standard number cube: 1, 2, 3, 4, 5, or 6.
Since the probability of rolling a 3 or 5 is 2/6 or 1/3, we can expect that out of 180 rolls, the number of times we get a 3 or 5 would be:
(1/3) * 180 = 60
Therefore, we can predict that a 3 or a 5 will be the result around 60 times in 180 rolls of a standard number cube.
Since the probability of rolling a 3 or 5 is 2/6 or 1/3, we can expect that out of 180 rolls, the number of times we get a 3 or 5 would be:
(1/3) * 180 = 60
Therefore, we can predict that a 3 or a 5 will be the result around 60 times in 180 rolls of a standard number cube.
Answered by
Bot
There are 6 equally likely outcomes when rolling a standard number cube: 1, 2, 3, 4, 5, or 6.
To find the probability of rolling a number less than 3, we need to count how many outcomes satisfy this condition. There are 2 outcomes that do: 1 and 2. So the probability of rolling a number less than 3 is:
P(number less than 3) = 2/6 = 1/3
To find the probability of not rolling a number less than 3, we need to count how many outcomes do not satisfy this condition. There are 4 outcomes that do not: 3, 4, 5, and 6. So the probability of not rolling a number less than 3 is:
P(not rolling a number less than 3) = 4/6 = 2/3
To find the probability of rolling a number less than 3, we need to count how many outcomes satisfy this condition. There are 2 outcomes that do: 1 and 2. So the probability of rolling a number less than 3 is:
P(number less than 3) = 2/6 = 1/3
To find the probability of not rolling a number less than 3, we need to count how many outcomes do not satisfy this condition. There are 4 outcomes that do not: 3, 4, 5, and 6. So the probability of not rolling a number less than 3 is:
P(not rolling a number less than 3) = 4/6 = 2/3
Answered by
Bot
There are 5 choices for the first digit, since any of the 5 digits can be used.
There are 4 choices left for the second digit, since we can't use the digit we used in the first position.
There are 3 choices left for the third digit, since we can't use either of the digits we used in the first two positions.
There are 2 choices left for the fourth digit, since we can't use any of the digits we used in the first three positions.
Finally, there is 1 choice left for the fifth digit, since we can't use any of the digits we used in the first four positions.
So the total number of 5-digit numbers that can be formed using only the digits 3, 1, 2, 5, and 7, with no digit used more than once, is:
5 x 4 x 3 x 2 x 1 = 120
Therefore, there are 120 possible 5-digit numbers that can be formed using only the digits 3, 1, 2, 5, and 7, with no digit used more than once.
There are 4 choices left for the second digit, since we can't use the digit we used in the first position.
There are 3 choices left for the third digit, since we can't use either of the digits we used in the first two positions.
There are 2 choices left for the fourth digit, since we can't use any of the digits we used in the first three positions.
Finally, there is 1 choice left for the fifth digit, since we can't use any of the digits we used in the first four positions.
So the total number of 5-digit numbers that can be formed using only the digits 3, 1, 2, 5, and 7, with no digit used more than once, is:
5 x 4 x 3 x 2 x 1 = 120
Therefore, there are 120 possible 5-digit numbers that can be formed using only the digits 3, 1, 2, 5, and 7, with no digit used more than once.
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