course Mth 151 zޜջӦywzݷassignment #001
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14:00:24 `q001. Note that there are 4 questions in this assignment. `q001. Let A stand for the collection of all whole numbers which have at least one even digit (e.g., 237, 864, 6, 3972 are in the collection, while 397, 135, 1, 9937 are not). Let A ' stand for the collection of all whole numbers which are not in the collection A. Let B stand for the collection { 3, 8, 35, 89, 104, 357, 4321 }. What numbers do B and A have in common? What numbers do B and A' have in common?
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RESPONSE --> A: 237, 864, 6, 3972 A': 397, 135, 1, 9937 B: 3, 8, 35, 89, 104, 357, 4321 A and B have these numbers in common: 6 and 8 They are both just one number that is even. Other then that, there are no similarities that I can see, unless one could say that 864 and 35 are similar because they both have all of the same type of number (even and odd). There are more similiarities between B and A' however. They both have a single number that is odd (A':1, B:3). And they have a three digit number that is all odd (A': 397, B:357). If one wanted to look deep enough, they would see that A': 135 and B: 35 are similiar simply because they both have 35 in them. I suppose I could go deeper into it the more I think about similarities, like multiples of, or prime numbers. But I stuck along the lines of finding number that were similar to each other in terms of the amount of even or odd numbers in the whole numbers. I am sure that I answered the question deep enough. confidence assessment: 2
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14:02:41 Of the numbers in B, 8, 89, 104, 4321 each have at least one even digit and so are common to both sets. 3 is odd, both of the digits in the number 35 are odd, as are all three digits in the number 357. Both of these numbers are therefore in A ' .
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RESPONSE --> I don't believe that I looked far enough into the problem, or examined every possibility of a similarity. In my opinion, I limited my mind's capabilities, and will continue to take more time looking into future problems of this nature, as I completely missed that those numbers in B have at least one even digit and were common to both sets. Or maybe I did come to that conclusion, but was more scattered about it. self critique assessment: 2
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14:05:46 `q002. I have in a room 8 people with dark hair brown, 2 people with bright red hair, and 9 people with light brown or blonde hair. Nobody has more than one hair color. Is it possible that there are exactly 17 people in the room?
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RESPONSE --> No, there would be 19. 8 people with dark hair brown, 2 with bright red, and 9 with light brown or blonde hair. Well if nobody has more then one hair color, one assumes that the 9 people in the room ARE blonde since there is already brown in the room. So after adding all these people together, with the different hair colors, you get 19 people, adding brown, red, and blonde together. I am confident of this answer, perhaps I'm not looking deep enough into it, as it seems I'm missing something. confidence assessment: 2
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14:06:04 If we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17.
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RESPONSE --> OK self critique assessment: 3
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14:08:13 `q003. I have in a room 6 people with dark hair and 10 people with blue eyes. There are only 14 people in the room. But 10 + 6 = 16, which is more than 14. How can this be?
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RESPONSE --> Two of people that have dark hair, could also have blue eyes, so that would give us 14, because if you subtract 2 from 10, and use the 2 to describe the amount of people with black hair and blue eyes, that would be 8 + 6, which gives us 14 people. I am confident of my answer, as I utilized similar techniques in the previous question. confidence assessment: 3
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14:08:23 The key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people.
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RESPONSE --> OK self critique assessment: 3
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14:12:10 `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red?
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RESPONSE --> Simple. If there are 20 cubical red blocks, and only 30 of the blocks are red, that leaves us with only one solution. It has to be that 10 of the cylindrical blocks are red, there is no other answer, part of the question is answered already simply by saying 20 cubical blocks are red, all you have to do is add 10 to achieve the amount of 30. I am sure of this answer, at first I was looking at it from terms of cubical and cylindrical, which confused me at first, until I simply looked at the problem in terms of numbers instead of colors and shapes. confidence assessment: 3
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14:12:17 Of the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks.
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RESPONSE --> OK self critique assessment: 3
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|晁V assignment #002 002. Representing Sets Liberal Arts Mathematics I 09-02-2008