1 and a quarter.
no,there will be no 5 but there will be 4.number 3 is attack of the zines
number one legend number 2 anniversary number 3 underworld number 4 tr 2 number 5 tomb raider 3 in my opinion
It's a 5 book series & number 4 comes out 10/12/09
This is a combination problem because you have to assess a subgroup without a particular order. I find it easiest to first deduce what the total number of arrangements are and then subtract from there. Firstly, There are a total of 9 balls getting assigned to a group of 4. So, to get our total, we use 9C4, which equals 9!/(4!*5!)=126. This means there are a total of 126 arrangements regardless of order. However, this number also includes the combination of 4 like-colored balls we're told to avoid. To calculate the number of arrangements for 4 like-colored balls we set up a separate subgroup for each color. For red balls the possible arrangement are 5C4, which is 5!/(4!*1!)=5. For blue balls the possible arrangements are 4C4, which is 4!/(4!*0!)=1 (remember that 0!=1). So the total number of like-colored arrangements is 5+1=6. Now we just subtract the exceptions from the total to get the answer...126-6=120
49 over 5 as a mixed number is 9 4/5.
It is -1 and 4/5 as a mixed number
It is 5 and 4/5 as a mixed number
4 over 4
the answer is 1 4 over 5
4
21/5 written as a mixed number is 4 and 1/5
4 and 3/5
2 4/5
7 4/5
It is 2 and 4/5
It is: 5 and 1/4