Answer:
186 Miles per hour.
Step-by-step explanation:
558/3
Answer:
- 273 mL of 5%
- 117 mL of 15%
Step-by-step explanation:
Let q represent the quantity of 15% dressing used. Then the amount of 5% dressing is (390 -q). The amount of vinegar in the mix is ...
0.15q + 0.05(390 -q) = 0.08(390)
0.10q = 31.2 -19.5 = 11.7 . . . . . . subtract 0.05(390) and simplify
q = 117 . . . . . . . . . . . . . . . . . . multiply by 10
390-q = 273
The chef should use 273 mL of the first brand (5% vinegar) and 117 mL of the second brand (15% vinegar).
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<em>Additional comment</em>
You may have noticed that the value of q is (0.08 -0.05)/(0.10 -0.05)×390. The fraction of the mix that is the highest contributor is the ratio of the difference between the mix value and least contributor, divided by the difference between the contributors: (8-5)/(15-5) = 3/10, the fraction that is 15% vinegar. This is the generic solution to mixture problems.
First one is 7.73 (23.19/3)
Second one is 14.04 ((19.71/7.3) x 5.2)
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:
4
Step-by-step explanation:
because The square root of 16 is 4, which is a rational number.