Answer:
X ≤ 50
Step-by-step explanation:
Since the additional fee for a piece of luggage is only charged if it weighs more than 50 lbs, than Michael's suitcase weight must be less than, or equal to, 50 lbs. Using 'X' as the weight of Michael's suitcase in pounds:
X ≤ 50
'X' is less than or equal to 50 lbs
1) So the ratio of Machine A to B is x + 16 : x (x being the paper printed in machine b).
2) This is in 4 minutes so every minute it will be 152.
2) So x + 16 + x = 152, or 2x + 16 = 152.
3) So to find x, we can subtract 16 from both sides. 2x = 136
4) Now we just divide both sides by 2 to get x = 68.
5) So machine A will be 68 + 16 = 84.
= 84
⭐ Answered by Foxzy0⭐
⭐ Brainliest would be appreciated, I'm trying to reach genius! ⭐
⭐ If you have questions, leave a comment, I'm happy to help! ⭐
So first start out by writing an expression for the cost of the child and the adult separately.
Child:
6 + 1r ($6 + $1 per ride)
Adult:
10 + 1.5r ($10 + $1.50 per ride)
to find how much more the adult will spend, just do Adult Expression - Child Expression which will be:
10 + 1.5r - (6 + 1r) just simplify this
For Part B, use your equation from part A where r = 7
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.
