course mth 151 rG׃oИ謸lն{xassignment #001
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18:23:54 `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 --> If A are all whole numbers in a collection which have an even number in them, (e.g. 8,104) then the answer the question one is. set of (8,89,104,4321) is the common of A and B. Taken that A' would be numbers not listed in A. Then looking at whole numbers, and the collection A. Then subtracting the numbers not listed in A would leave A'. The solution the the second question would be. the set of (3,35,357). confidence assessment: 2
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18:27:15 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 --> The first sentence of the problem did worry me. Then reading on into the problem i found a simpler problem. I found to same solution, but wasn't sure of the correct termiolgy. self critique assessment: 1
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18:55:55 `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 --> I started by listing the three set using whole numbers, and then listing and common numbers, and looking at the prime of thoose set. comeing to the conclusion of no. confidence assessment: 1
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18:57:49 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 --> I used sets to come to the conclusion of no also. but my reasoning came form more of a logical quess, then the use of a mathmatical idea. self critique assessment: 1
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19:02:48 `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 --> I have no real reason for the result of the question. confidence assessment: 0
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19:03:49 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 --> Okay self critique assessment: 0
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19:11:23 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. I had accidently clicked on 10. To explain the solution i used basic dedution. given that one set of red block =30 and 20 are cubical, the that would leave 10 cylindrical blocks. self critique assessment: 1
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