It is the same. Even though you put a 0 in the end can't change the number
Answer: The loser's card shows 6.
Explanation: Let's start by naming the first student A and the second student B.
Since the product of A and B are either 12, 15, or 18, let's list every single possibility, the first number being A's number and the second number being B's number.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
4 3
5 3
6 2
6 3
9 2
12 1
15 1
18 1
Now, the information says that A doesn't know what B has, so we can immediately cross off all of the combinations that have the integer appearing once and once ONLY off, because if it happened once only, A would know of it straight away. Now, our sample space becomes much smaller.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
6 2
6 3
Using this same logic, we know that we can cross off all of the digits that occur only once in B's column.
2 6
3 6
Now, A definitely knows what number B has because there is only one number left in B. Hence, we can conclude that the loser, B, has the integer 6.
Answer:
I believe it's the 3rd option:
(-1, 8), (0, 10), (1,12), (2, 14)
Step-by-step explanation:
Answer:
a number line that goes from 0 to 16. the whiskers range from 1 to 12, and the box ranges from 3 to 9. a line divides the box at 7.
