13.23 is the answer for that equation
How sad, that you didn't include the list of choices.
The volume of the cone is (1/3) x (π) x (radius²) .
The volume of the sphere is (4/3) x (π) x (radius³) .
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:
ex:Speed (S) is the ratio of the distance (D) covered to the time (t) taken.
That is, S = D/t
Suppose Andrew ran a distance D1 in 1 hour (3600 seconds) at a Speed, say S1, we have
S1 = D1/t
We can then say he ran a distance
D1 = t × S1
= 3600S1
Similarly, let's say Karleigh ran a distance
D2 = t × S2
= 3600S2
Let us compare these two, you will notice that the bigger number between S1 and S2 is going to determine the bigger number between D1 and D2.
Let's choose random numbers for S1 and S2 for clarity, say S1 = 5, S2 = 10
D1 = 3600 × 5
= 18000
D2 = 3600 × 10
= 36000
This makes D2 bigger than D1. this is an example i found on the internet.
Step-by-
hope this helps, good luck
