Answer:
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.
Answer:
2 fruit for every 3 monsters
Step-by-step explanation:
Answer:
Approx 61.554 feet
Step-by-step explanation:
Since we are dealing with the opposite and adjacent sides, we will use tangent as our ratio.
Since the angle of elevation is 72 degrees, the left side of our equation is 
.
The right side of the equation is opposite over adjacent.
Set these equal to each other.
Simplify.
x is approximately 61.554.
Answer:
No
Step-by-step explanation:
To determine if (- 5, - 5) is a solution
Substitute x = - 5 into the inequality and compare answer to y
- 2(- 5) + 4 = 10 + 4 = 14 > - 5 ← the y- coordinate
Thus (- 5, - 5) is not a solution to the inequality
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.
