Jill is playing a game in which a number cube is rolled two times. The sum of the rolls is recorded. What is the sample space fo
r the game? (PLEASE HELP)!!!!! a. 1,2,3,4,5,6,
b. 2,3,4,5,6,7,8,9,10,11,12
c. 1,2,3,4,5,6,7,8,9,10,11,12
d. 1,2,3,4,5,6,7,8,9,10,12,15,16,18,20,24,25,30,36
When you roll a cube once, the smallest number it can show is 1, and the biggest one is 6.
So when you roll it twice, the smallest possible sum of both results is 2, and the biggest one is 12. . . . . . Just like list 'b' .
An <u>example of a problem</u> where I <em>would not</em> group the addends differently is:
3+2+4.
An <u>example of a problem</u> where I <em>would</em> group the addends differently is:
2+5+8.
Explanation:
In the <u>first problem</u>, I would not group the addends differently before adding. This is because I cannot make 5 or 10 out of any of the numbers. We group addends together to make "easier" numbers for us to add, such as 5 and 10. If we cannot do that, there is no reason to regroup the addends.
In the <u>second problem</u>, I would regroup like this: 2+8+5
This is because 2+8=10, which makes the problem "easier" for us to add. Since we can get a number like this that shortens the process, we can regroup the addends.
a) if she cuts them into halves, she will have 20 slices with 5 left over, who will get an extra slice while who doesn't? it can't work
b) if she cuts them into thirds, she will have 30 slices with 15 left over, that is enough to give each player two slices with none left over, it works!!
c) if she cuts them into quarters, she would have 40 slices with 25 left over. this means some players would get three slices, while others would only get two. it doesn't work.
