Answer:
There are 35 students in the class
Step-by-step explanation:
We would divide 21 by 60%
21/0.6 = 35 students.
Let's compare apples to apples and oranges to oranges: convert each given proper fraction into its decimal counterpart:
3 63/80 => 3 .788 approx.
3 1/5 => 3.2
3 11/20 => 3.55
It's now an easy matter to arrange these numbers from least to greatest:
3.2, 3.55, 3.79, or
3 1/5, 3 11/20, 3 63/80
So let's start out by labeling Ethan as X
Since 85 is 1/3 of what Ethan drove, that means Ethan drove 3 times of 85.
85= (3x) -21
85 -21 = 3x
60=3x
20 = x
True , is the answer ! :)
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.
