It is 1 - Prob(No even numbers in 10 throws) = 1 - [Prob(Not an even number)]10 = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024 or 0.9990 So you are almost certain to get an even number.
numbers which are divisible by both an even number and odd number greater than 1: 6, 10, 12, 14, 18, ...
Greg has the same number of 1-blocks, 10-blocks and 100 blocks. He models an even number that is less than 400. What is the number
-- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number plus an extra ' 1 '. -- Add them. You get (an even number) + (an even number) + ( 2 ). That's an even number. ======================== Another mind-bending , brain-busting way to look at it : -- The first odd number is some even number plus an extra ' 1 '. -- The second odd number is some even number minus ' 1 '. -- Add them. You get (an even number) + (an even number) + ( [1 - 1] or zero ). That's an even number.
Any number evenly divisible by 2 is an even number
It is 1 - Prob(No even numbers in 10 throws) = 1 - [Prob(Not an even number)]10 = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024 or 0.9990 So you are almost certain to get an even number.
2 is aa number and 1/4 is the 1 to 10 even
10 is an even number.
If Greg has the same number of 1-blocks 10-blocks and 100-blocks to model an even number that is less than 400 what is the number?
10 because it's an even number !
1. Is 10 a composite number?2. Is 3 a composite number?3. Is 17 a composite number?4. Is 24 a composite number?5. Is 31 a composite number?6. Is 38 a composite number?
numbers which are divisible by both an even number and odd number greater than 1: 6, 10, 12, 14, 18, ...
The number is called a "googol" (1 x 10^100). The even larger number (googolplex) is 10^googol which could never actually be written out.
Greg has the same number of 1-blocks, 10-blocks and 100 blocks. He models an even number that is less than 400. What is the number
6, 10, 14 etc
It is even because you can divide 10 by 2
No. The sum of 1 odd number and 1 even number must always be odd.