course mth151 Mr. Smith, i don't necessarly agree with the answer to q003 from assignment 1. I thought my answer was correct. ģ➱RǸSassignment #001
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15:47:55 `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 --> B and A have the numbers {8,89,104,4321} in common because {8, 89, 104, 4321} from B are the only whole numbers in the B set that have at least one even digit, therefore these same numbers can be found on the A set because they have at least one even digit and they are a whole number. B and A' have the numbers { 3, 35, 357} in common because these are all odd whole numbers. A' contains only whole number's that are not in A , therefore number's found in A cannot have any even digits. The number's in B that are odd cannot be in A, so therefore can be common with A'. confidence assessment: 1
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15:48:34 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 --> ok self critique assessment: 2
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15:59:17 `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 --> If 8 people have dark brown hair, 2 people have bright red hair and 9 people have light brown or blonde hair and nobody has more than one hair color, the only possible way that there can be exactly 17 people in the room is if the person counting people counted some people more then once. The numbers given 8, 2, 9, add up to equal 19. The statement indicates 19 people, so there cannot be 17 people in the room. confidence assessment: 1
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15:59:48 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: 2
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16:06:51 `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 --> It is possible to have just 14 people in the room. 10 of those 14 people have blue eyes. 6 of those 14 people have brown hair. That means, 6 of those 14 people have dark hair and blue eyes while 4 of the other 8 people also have blue eyes but have a different hair color. Therefore 10 people have blue eyes. confidence assessment: 3
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16:07:58 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 --> self critique assessment: 2
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16:14:34 `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 --> answer to q.004 There are 30 red blocks as stated. Of the 30 red blocks there are 20 red blocks that are cubical. That leaves only 10 red blocks that are left that could be cylindrical. There are no other shapes in the box so the remaining 10 red blocks that are not cubical have to be cylindrical. confidence assessment: 3
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16:14:59 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: 2
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