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Leni [432]
3 years ago
9

How to convert improper fraction to mixed numbers

Mathematics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

Step-by-step explanation:This is an improper fraction: 243/10 to convert it to a mixed number just do 243 divided by 10 which equals 24, so put that number as a whole number in front of the fraction, then do 24 times 10 which equals 240, then since 240 is 3 less than 243 the answer is 24 and 3/10.

If you need, I can try to explain it better.

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Tuesday’s Hugh temperature was 4 degrees less than Monday’s high temperature
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What is the domain of the following ordered pairs?
Vlad [161]

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3 years ago
We put 200 balls into 100 boxes such that every box got at least 1 ball and at most 100 balls. Prove that there are some boxes t
Vinvika [58]

Explanation:

Lets first show that al least 50 of the boxes contail at most 2 balls.

If there were 50 + k boxes with 3 balls or more, then we should have 100 - (50 + k) = 50 - k balls with 1 ball or 2. However in those 50 + k boxes with 3 or more balls we have alredy at least 3*(50+k) = 150 + 3k balls in them, and the amount of balls remaining is, as a result, at most 200 - (150 + 3k) = 50 - 3k, which cant be fit in 50 - k balls if we put at least 1 on each.

Therefore, there are at least 50 boxes with 1 or 2 balls. Whithin those boxes, we can obtain any number of balls selecting the appropiate boxes. Lets assume that we want M balls, and we have A boxes with 2 balls and B boxes with 1 ball, we have this possibilities (M equal or less than 2A + B, the total number of balls):

  • If M > 2A, then we pick all boxes with 2 balls (A in total) and M - 2A boxes with 1 ball. We have 2*A + (M-2A) = M. We are able to pick M - 2A boxes because B ≥ M - 2A.
  • If M ≤ 2A, and it is even, then we pick M/2 boxes with 2 balls.
  • If M ≤ 2A and it is odd, then we pick (M-1)/2 boxes with 2 balls and 1 box with 1 ball (if all boxes contain 2 balls or more, then we could pick 50 boxes with 2 balls because at least 50 boxes contain 1 or 2 balls; so we can assume that at least 1 box contain one single ball).

Lets call C the sum of the balls in the boxes with 1 or 2 balls. C should be at least 50. The argument made previously shows that we can pick boxes of 1 or 2 balls that cover any number of balls below to C. This means that we can obtain any number below 50; furthermore, if C is equal or greater than 100, then the problem is alredy solved. Lets suppose that C is lower than 100.  This means that the other boxes contain more than 100 balls in total.

Since we cant put more than 100 balls in one single box, then there should be a combination of boxes with 3 or more balls that contain between 50 and 100 balls. If that is not the case, then lets call L the biggest number of balls below 50 that we can obtain with boxes with 3 or more balls. Since the sum of all balls is bigger than 100, then there should be a box outside those we use to obtain L with 3 or more balls. Since L was the biggest number we could obtain below 50, and we are supposing that we cant obtain any number between 50 and 100, then that box should have more than 50 balls. Which means that that box alone could be used to obtain a number between 50 and 100. This is a contradiction.

The paragraph above shows that we can make a combination of boxes with 3 or more balls which combined number of balls is a number N between 50 and 100. Since we can make any number between 0 and 50 with boxes, for example 100 - N, with boxes of 1 or 2 balls, then we should be able to make exactly 100 balls using the boxes we have available.

I hope that works for you!

4 0
3 years ago
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