Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
Answer:
1. (1,1)
2.(2,-6)
3.Not sure.
4.No solution
5.(-3,4)
6. (6,-4)
7.(7,2)
8.(-7,-12)
9.(9,-1)
10.(2,1)
Step-by-step explanation:
Hope that helps bud!:)
The answer is B because on the graph you start at $3.00 then you add $5.00 per hour so it goes up 5/1 per hour on the graph.
Answer:
y=21.14
Step-by-step explanation:
x=6 because after 6 weeks
y= 3.27(6)+ 1.52
y=19.62 + 1.52
y=21.14
This is not possible. Why not? Because the smallest the variance can get is 0.
Recall that 's' represents the standard deviation, so s^2 is the variance. It basically measures how spread out the values are. The higher the variance, the more spread out the data. You can think of it as "average distance from the mean". If the variance is 0, then all of the values are at the same point. So you could have a list like {2,2,2,2,2} which has variance 0. We cannot get any smaller variance than that. If your teacher insists all the values in the list are different, then the variance will be greater than 0.
