15/100=0.15
Part1: Finding probability of getting sum as a perfect square. Maximum sum of both the dice is (6+6) equal to 12. Up to 12, the perfect squares are: 1, 4 and 9. Getting a sum of 1 from two dice is not possible. So, we are left with 4 and 9. To get 4, the combination can be: (2,2) or (1,3) or (3,1). This means, to get the sum as 4, the probability is [3/36]. To get 9, the combination can be: (3,6) or (6,3) or (5,4) or (4,5). This means, to get the sum as 9, the probability is [4/36]. Therefore,the total probability of getting the sum as a perfect square is: [(3/36)+(4/36)]=[7/36]. Part2: Finding the probability of getting sum as an even number. The possible even numbers can be 2, 4, 6, 8, 10 and 12. But, as 4 is already considered in part1, it should be ignored in this case. The probability of getting sum as 2 is: [1/36] The probability of getting sum as 6 is: [5/36] The probability of getting sum as 8 is: [5/36] The probability of getting sum as 10 is: [3/36] By adding all the above, the probability of getting sum as an even number (ignoring 4) is: [(1/36)+(5/36)+(5/36)+(3/36)]=[14/36]. From part 1 and part 2, we get the total probability as [(7/36)+(14/36)]=[7/12]=0.583333.
The probability that 2 people have the same number is 2 out of 10
The factors of 10 are the numbers that divide 10 evenly: 1, 2, 5 and 10. To answer your question, you have to figure out what the probability of rolling one of these numbers is on a number cube.
Empirical means by observation, so empirical probability, or experimental probability, is the probability that is observed in a set of trials. For example, if you flip a coin ten times and get seven heads, your empirical probability is 7 in 10. This is different than the theoretical probability, which for a fair coin is 5 in 10, but that result will only be approximated by the empirical results, and then only with a larger number of trials.
depends on the numbers on the spinner. if 1 thru 10, 7/10
"Still" implies that the original number is a square number. In that case, the answer is as follows: There is no number such that it is a perfect sqiuare and that the number increased (or decreased) by 10 is also a perfect square. And if you do not limit it to perfect square then every non-negative number is a square with the number that is 10 more also being a square.
no, 10 is not a perfect square. in order for a number to be a perfect square, you have to see if the numbers that are multiplied to get it are the same. for example: 2x2=4; 4 is a perfect square. 12x12=144; 144 is also a perfect square 5x2=10 or 10x1=10. 10 isn't a perfect square because 5 and 2, and 10 and 1, are different numbers.
There are 90 two-digit numbers from 10 to 99. Of those, 6 are perfect squares (16, 25, 36, 49, 64, and 81) and 2 are perfect cubes (27 and 64). Each perfect square or root has a probability of 1 in 90 in being drawn.
perfect square numbers are 16 and 25
No. The smallest 2 digit number is 10 the smallest 3 digit number is 100 - so the question is whether (100-10) is a perfect square? No - 90 is not a perfect square. 8x8 is 64 and 9x9 is 91, not 90.
36 is the only perfect square amongst those listed.
The square root of any non-perfect square is an irrational number. Since sqrt(1000) = 10 sqrt(10), it is irrational.
15/100=0.15
It is not a perfect square.
It is not a perfect square.
Part1: Finding probability of getting sum as a perfect square. Maximum sum of both the dice is (6+6) equal to 12. Up to 12, the perfect squares are: 1, 4 and 9. Getting a sum of 1 from two dice is not possible. So, we are left with 4 and 9. To get 4, the combination can be: (2,2) or (1,3) or (3,1). This means, to get the sum as 4, the probability is [3/36]. To get 9, the combination can be: (3,6) or (6,3) or (5,4) or (4,5). This means, to get the sum as 9, the probability is [4/36]. Therefore,the total probability of getting the sum as a perfect square is: [(3/36)+(4/36)]=[7/36]. Part2: Finding the probability of getting sum as an even number. The possible even numbers can be 2, 4, 6, 8, 10 and 12. But, as 4 is already considered in part1, it should be ignored in this case. The probability of getting sum as 2 is: [1/36] The probability of getting sum as 6 is: [5/36] The probability of getting sum as 8 is: [5/36] The probability of getting sum as 10 is: [3/36] By adding all the above, the probability of getting sum as an even number (ignoring 4) is: [(1/36)+(5/36)+(5/36)+(3/36)]=[14/36]. From part 1 and part 2, we get the total probability as [(7/36)+(14/36)]=[7/12]=0.583333.