2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,
58,60,62,64,66,68,70,72,74,76,78,80,82.84,86,88,90,92,94,96,98,and 100
Start with 2 and keep adding 2. JHC!
1.00 or 100%. Look at the possible outcomes: 1, 2, 3, 4, 5, 6. In this range, what are the even numbers: 2, 4, 6. What are the odd: 1, 3, 5. So the set of (even or odd) numbers is the union of odd numbers and even numbers. This is the same as the set of possible outcomes. Therefore it will happen, so 100 percent chance (or 1.00)
If you chose the number out of an equal numberof odd and even numbers then it would be 50% chance. Like if you chose an odd number out of the numbers from 1 to 100
50 half of 100 is 50
Assuming then that there are 100 numbers, 1-100, the probability of the number 23 being randomly picked out of 100 is: 1/100 or 0.01.
There are 6 numbers on the die; 3 even. Therefore P(Even) = 3/6 or 1/2.
even numbers 1-100
In Java:System.out.println("Even numbers")for (int i = 2; i
50
101
There are 100 even numbers between 1 and 200 (inclusive).
101
-99
Every second number is Even. So there are 50 even numbers from 1 to 100 if you include 100. If you do not include 100 in the count there are obviously 49. The wording of the question is not clear on whether 100 is to be included or not.
The only even prime number (not only from 1 to 100, but in general) is 2. All other even numbers are multiples of 2, and therefore, by definition, not prime numbers.
To find how many numbers between 1 and 100 inclusive are divisible by 2, we can identify the even numbers in that range. The even numbers from 1 to 100 are 2, 4, 6, ..., 100. This forms an arithmetic sequence where the first term is 2, the last term is 100, and the common difference is 2. There are 50 even numbers (2, 4, 6, ..., 100), so there are 50 numbers between 1 and 100 that are divisible by 2.
n*(n+1)=sum 100*(100+1)=10100
Every number has 1 as a multiple. 1*100 = 100 1*7 = 7 As 1 is an odd number there are no numbers with only even multiples.