course Mth 151
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20:40:04 Query 2.3.15 (Y ^ Z')U X, univ={a,..g}, X={a,c,e,g}, Y = {a,b,c}, Z = {b, ..., f} What is the set (Y ^ Z')U X?
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RESPONSE --> (Y^Z' )U X? ({a} ^ {a,c,e,g}) U X {a} U {a,c,e,g} = {a,c,e,g}
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20:40:14 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'. So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. **
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RESPONSE --> ok
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20:42:49 Give the intersection of the two sets Y and Z'
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RESPONSE --> Y ^ Z' ? {a,b,c} ^ {a,g} = {a}
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20:43:02 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.**
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RESPONSE --> ok
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20:45:01 Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')
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RESPONSE --> The set of all elements that are in A but not in B, or in B but not in A.
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20:45:29 ** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'. STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A. INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. **
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RESPONSE --> ok
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20:47:04 2.3.51 always or not always true: n(A U B) = n(A)+n(B)
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RESPONSE --> not always true.
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20:48:24 ** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_. Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection. } MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true. The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B. The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5. The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6. The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. **
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RESPONSE --> Ok, I new that from an example we done on the CD. I just didn't explain how I got my answer.
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20:55:08 Query 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'.
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RESPONSE --> (X^Y)'? ({1,3,5} ^ {1,2,3}) ({1,3})' ={2,4,5} X' U Y' {2,4} U {4,5} = {2,4,5}
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20:55:13 ** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}. (X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5} The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. **
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RESPONSE --> ok
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21:00:08 2.3.72 A = {3,6,9,12}, B = {6,8}.
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RESPONSE --> 2.3 # 72 A x B= {3,6}, {3,8}, {6,6}, {6,8}, {9,6}, {9,8}, {12,6}, {12,8} B x A= {6,3}, {6,6}, {6,9}, {6,12}, {8,3}, {8,6}, {8,9}, {8,12}
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21:00:48 ** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)} (B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)} How is n(A x B) related to n(A) and n(B)? n(S) stands for the number of elements in the set S, i.e., its cardinality. n(A x B) = n(A) * n(B) **
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RESPONSE --> ok
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21:02:06 2.3.84 Shade A U B
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RESPONSE --> 2.3 # 84 All of A and B are shaded.
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21:03:20 ** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. **
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RESPONSE --> ok
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21:38:36 Query 2.3.100 Shade (A' ^ B) ^ C
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RESPONSE --> The only section that is shaded is the top,right-hand section of C, where only B and C join.
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21:39:11 ** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A **
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RESPONSE --> ok
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21:44:39 Describe the shading of the set (A ^ B)' U C.
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RESPONSE --> Every region of c would be shaded, except the region wher a,b,and c intersect.
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21:48:30 ** All of C would be shaded because we have a union with C, which will include all of C. Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded. Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. **
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RESPONSE --> ok, I had that shaded on my paper, but i didn't list the union, instead I listed the intersection. But I have it correct on my work. On my paper, the ony region that is not shaded, is the (A^B)-C.
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21:51:35 2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region.
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RESPONSE --> A - (B U C)
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21:52:10 ** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. **
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RESPONSE --> ok
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course Mth 151 I worked the query ass. #3 and submitted it. Then I worked the query ass. #4 and when I went to find the file it wasn't there. Then I remembered that if I completed the ass in the same night it stays on the same file. So my second submitt form does have both assignments on it. I'll remember next time not to submit but once if I'm going to complete them both in on the same day.
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20:40:04 Query 2.3.15 (Y ^ Z')U X, univ={a,..g}, X={a,c,e,g}, Y = {a,b,c}, Z = {b, ..., f} What is the set (Y ^ Z')U X?
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RESPONSE --> (Y^Z' )U X? ({a} ^ {a,c,e,g}) U X {a} U {a,c,e,g} = {a,c,e,g}
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20:40:14 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'. So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. **
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RESPONSE --> ok
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20:42:49 Give the intersection of the two sets Y and Z'
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RESPONSE --> Y ^ Z' ? {a,b,c} ^ {a,g} = {a}
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20:43:02 **Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.**
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RESPONSE --> ok
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20:45:01 Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')
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RESPONSE --> The set of all elements that are in A but not in B, or in B but not in A.
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20:45:29 ** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'. STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A. INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. **
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RESPONSE --> ok
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20:47:04 2.3.51 always or not always true: n(A U B) = n(A)+n(B)
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RESPONSE --> not always true.
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20:48:24 ** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_. Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection. } MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true. The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B. The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5. The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6. The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. **
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RESPONSE --> Ok, I new that from an example we done on the CD. I just didn't explain how I got my answer.
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20:55:08 Query 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'.
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RESPONSE --> (X^Y)'? ({1,3,5} ^ {1,2,3}) ({1,3})' ={2,4,5} X' U Y' {2,4} U {4,5} = {2,4,5}
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20:55:13 ** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}. (X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5} The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. **
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RESPONSE --> ok
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21:00:08 2.3.72 A = {3,6,9,12}, B = {6,8}.
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RESPONSE --> 2.3 # 72 A x B= {3,6}, {3,8}, {6,6}, {6,8}, {9,6}, {9,8}, {12,6}, {12,8} B x A= {6,3}, {6,6}, {6,9}, {6,12}, {8,3}, {8,6}, {8,9}, {8,12}
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21:00:48 ** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)} (B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)} How is n(A x B) related to n(A) and n(B)? n(S) stands for the number of elements in the set S, i.e., its cardinality. n(A x B) = n(A) * n(B) **
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RESPONSE --> ok
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21:02:06 2.3.84 Shade A U B
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RESPONSE --> 2.3 # 84 All of A and B are shaded.
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21:03:20 ** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. **
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RESPONSE --> ok
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21:38:36 Query 2.3.100 Shade (A' ^ B) ^ C
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RESPONSE --> The only section that is shaded is the top,right-hand section of C, where only B and C join.
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21:39:11 ** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A **
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RESPONSE --> ok
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21:44:39 Describe the shading of the set (A ^ B)' U C.
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RESPONSE --> Every region of c would be shaded, except the region wher a,b,and c intersect.
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21:48:30 ** All of C would be shaded because we have a union with C, which will include all of C. Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded. Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. **
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RESPONSE --> ok, I had that shaded on my paper, but i didn't list the union, instead I listed the intersection. But I have it correct on my work. On my paper, the ony region that is not shaded, is the (A^B)-C.
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21:51:35 2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region.
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RESPONSE --> A - (B U C)
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21:52:10 ** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. **
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RESPONSE --> ok
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侻ĒVݕ assignment #004 űpZ Liberal Arts Mathematics I 01-30-2006
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22:01:18 2.4.12 n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10
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RESPONSE --> 2.4 # 12 region A - B =15 region B - A =18 the region outside of A U B is 7
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22:02:07 ** In terms of the picture (2 circles, linked, representing the two sets) there are 28 in B and 10 in A ^ B so there are 18 in the region of B outside of A--this is the region B-A. There are 25 outside of A, and 18 of these are accounted for in this region of B. Everything else outside of A must therefore also be outside of B, so there are 25-18=7 elements in the region outside of both A and B. A ' U B ' consists of everything that is either outside of A or outside of B, or both. The only region that's not part of A ' U B ' is therefore the intersection A ^ B, since everything in this region is inside both sets. A' U B' is therefore everything but the region A ^ B which is common to both A and B. This includes the 18 elements in B that aren't in A and the 7 outside both A and B. This leaves 40 - 18 - 7 = 15 in the region of A that doesn't include any of B. This region is the region A - B you are looking for. **
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RESPONSE --> I got my answers the same way.
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22:02:55 query 2.4.18 wrote and produced 2, wrote 5, produced 7 &&&& How many did he write but not produce?
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RESPONSE --> 3
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22:03:22 ** You need to count the two he wrote and produced among those he wrote, and also among those he produced. He only wrote 5, two of which he also produced. So he wrote only 3 without producing them. In terms of the circles you might have a set A with 5 elements (representing what he wrote), B with 7 elements (representing what he produced) and A ^ B with 2 elements. This leaves 3 elements in the single region A - B and 5 elements in the single region B - A. The 3 elements in B - A would be the answer to the question. **
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RESPONSE --> ok. This is the same way i got my answer.
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22:03:53 2.4.24 9 fat red r, 18 thn brown r, 2 fat red h, 6 thin red r, 26 fat r, 5 thin red h, 37 fat, 7 thin brown ......!!!!!!!!...................................
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RESPONSE --> ok
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22:10:13 ** Here's my solution. Tell me if there is anything you disagree with (I'm not infallible) or don't understand. incidental: 18 thin brown roosters, 7 thin brown hens, 6 thin red hens and the 6 thin roosters which aren't fat (out of the 50-26=24 thin roosters 18 are brown so 6 are red) adds up to 37 thin chickens How many chickens are fat? 37 as given How many chickens are red? 22: 9 fat red roosters, 6 thin red roosters, 5 thin red hens, 2 fat red hens. How many chickens are male? 50: 9 fat red roosters are counted among the 26 fat roosters so the remaining 17 fat roosters are brown; then there are 18 thin brown roosters and 6 thin red roosters; the number of roosters therefore adds up to 9 + 18 + 6 + 17 = 50 How many chickens are fat not male? 26 of the 37 fat chickens are male, leaving 11 female How many chickens are brown not fat? 25: 18 thin brown roosters, 7 thin brown hens adds up to 25 thin brown chickens How many chickens are red and fat? 11: 9 fat red roosters and 2 fat red hens.**
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RESPONSE --> After much thinking, I got the same numbers you did, except I never figured out how to put this in a venn diagram, unless you don't use all the information.
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|dREܝVacͳy挻 assignment #004 űpZ Liberal Arts Mathematics I 01-30-2006
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22:11:27 2.4.12 n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10
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RESPONSE -->
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22:11:37 ** In terms of the picture (2 circles, linked, representing the two sets) there are 28 in B and 10 in A ^ B so there are 18 in the region of B outside of A--this is the region B-A. There are 25 outside of A, and 18 of these are accounted for in this region of B. Everything else outside of A must therefore also be outside of B, so there are 25-18=7 elements in the region outside of both A and B. A ' U B ' consists of everything that is either outside of A or outside of B, or both. The only region that's not part of A ' U B ' is therefore the intersection A ^ B, since everything in this region is inside both sets. A' U B' is therefore everything but the region A ^ B which is common to both A and B. This includes the 18 elements in B that aren't in A and the 7 outside both A and B. This leaves 40 - 18 - 7 = 15 in the region of A that doesn't include any of B. This region is the region A - B you are looking for. **
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RESPONSE -->
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22:11:39 query 2.4.18 wrote and produced 2, wrote 5, produced 7 &&&& How many did he write but not produce?
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RESPONSE -->
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22:11:43 ** You need to count the two he wrote and produced among those he wrote, and also among those he produced. He only wrote 5, two of which he also produced. So he wrote only 3 without producing them. In terms of the circles you might have a set A with 5 elements (representing what he wrote), B with 7 elements (representing what he produced) and A ^ B with 2 elements. This leaves 3 elements in the single region A - B and 5 elements in the single region B - A. The 3 elements in B - A would be the answer to the question. **
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RESPONSE -->
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22:11:46 2.4.24 9 fat red r, 18 thn brown r, 2 fat red h, 6 thin red r, 26 fat r, 5 thin red h, 37 fat, 7 thin brown ......!!!!!!!!...................................
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RESPONSE -->
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22:11:49 ** Here's my solution. Tell me if there is anything you disagree with (I'm not infallible) or don't understand. incidental: 18 thin brown roosters, 7 thin brown hens, 6 thin red hens and the 6 thin roosters which aren't fat (out of the 50-26=24 thin roosters 18 are brown so 6 are red) adds up to 37 thin chickens How many chickens are fat? 37 as given How many chickens are red? 22: 9 fat red roosters, 6 thin red roosters, 5 thin red hens, 2 fat red hens. How many chickens are male? 50: 9 fat red roosters are counted among the 26 fat roosters so the remaining 17 fat roosters are brown; then there are 18 thin brown roosters and 6 thin red roosters; the number of roosters therefore adds up to 9 + 18 + 6 + 17 = 50 How many chickens are fat not male? 26 of the 37 fat chickens are male, leaving 11 female How many chickens are brown not fat? 25: 18 thin brown roosters, 7 thin brown hens adds up to 25 thin brown chickens How many chickens are red and fat? 11: 9 fat red roosters and 2 fat red hens.**
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RESPONSE -->
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course Mth 151
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22:30:50 `q001. Note that there are 5 questions in this assignment. Again we have a total of 35 people in a room. Of these, 20 have dark hair and 15 have bright eyes. There are 8 people with dark hair and bright eyes. Let A stand for the collection of people who have dark hair and B for the collection who have bright eyes. The Intersection of these two collections is denoted A ^ B, and stands for the collection of all people who have both dark hair and bright eyes. The Union of these two collections is denoted A U B, and stands for the collection of all people who have at least one of these characteristics. In terms of the diagram you made for the preceding problem, describe the collection A ^ B and the collection A U B. Give the number of people in each of these collections (these numbers are designated by the notation n ( A ^ B) and n(A U B) ). Refer to the diagrams you have made.
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RESPONSE --> n(A^B) N(20^15)= n8 n(A u B ) n(20) + n(15)= n35
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22:32:59 The collection A ^ B consists of all the people with both dark hair and bright eyes, which corresponds to the overlap between the two circles (region I). There are 8 people in this overlap, so we say n(A ^ B) = 8. The collection A U B consists of all the people who have least one of the characteristics. This would include the 12 people with dark hair but not bright eyes, located in the first circle but outside the overlap (region II); plus the 7 people with bright eyes but not dark hair, located in the second circle but outside the overlap (region III); plus the 8 people with both characteristics, located in the overlap (region I). Thus we include the 12 + 8 + 7 = 27 people who might be located anywhere within the two circles.
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RESPONSE --> Ok, this is the answers that I got on my Venn diagram i drew for this problem, but to answer the question i just added 20 and 15 which isn't write. i had the 12,7 and 8 on my diagram!
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22:35:10 `q002. Continuing the preceding example, we let A' stand for the people who are not in the collection A, and we let B' stand for the people who are not in the collection B. What are the characteristics of the people in A', and what characterizes people in B' ? What are n(A ') and n(B '), the numbers of people in A' and B' ?
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RESPONSE --> n(A' )= 7 n(B' )=12
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22:35:41 Of the 35 people, those in A' are those outside of A. Since A consists of all the dark-haired people, A' consists of all the people lacking dark hair. This includes the 8 people outside of both circles (people having neither dark hair nor bright eyes, region IV) and the 7 people in the second circle but outside the overlap (people having bright eyes but not dark hair, region III). n(A ' ) is therefore 8 + 7 = 15. Since B consists of all the bright-eyed people, B' consists of all the people lacking bright eyes. This would include the 8 people outside both circles (region IV), all of whom lack both dark hair and bright eyes, and the 12 people in the first circle but outside the overlap (region II), who have dark hair but not bright eyes. n ( B ' ) is therefore 12 + 8 = 20.
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RESPONSE --> ok I didn't add everything.
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22:41:49 `q003. ( A U B ) ' stands for the everyone outside A U B, and ( A ^ B ) ' stands for everyone outside A ^ B. What characterizes the people in each of these collections, and how many people are there in each?
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RESPONSE --> (A ^ B) ' = 12 dark haired people and 7 people with bright eyes. (A u B) ' = 8 people with neither dark hair or bright eyes.
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22:42:05 A U B consists of everyone having at least one of the characteristics (dark hair, bright eyes), and is represented by the numbers in the two circles (regions I, II, III). ( A U B ) ' consists of the people who do not have at least one of the characteristics, and is represented by the number outside both circles (region IV). This number is 8, representing the 8 people who have neither dark hair nor bright eyes. A ^ B stands for all the people with both of the two characteristics (represented by the overlap, region I), so ( A ^ B ) ' stands for all the people who do not have both of the two characteristics (represented by everything outside region I, or regions II, III and IV). [ Note that (A ^ B)' is not the same as the collection of people who have neither characteristic. Anyone who does not have both characteristics will be in ( A ^ B ) ' . ] ( A ^ B )' must include those who have neither characteristic, and also those who have only one of the characteristics. The 8 people outside both circles, the 12 people in the first circle but outside the overlap, and the 7 people in the second circle but outside the overlap all lack at least one characteristic to, so these 8 + 12 + 7 = 27 people make up( A ^ B ) '.
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RESPONSE --> ok
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22:48:52 `q004. How many people are in A ' U B ', and how could those people be characterized? Answer the same for A ' ^ B '.
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RESPONSE --> A' U B' = 12 + 7 + 8= 27 ; 12-dark hair, 7-bright eyes, 8 with neither dark hair nir brown eyes.
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22:49:56 A ' U B ' consists of all the people who are in at least one of the sets A ' or B '. A ' consists of all the people who do not have dark hair, represented by every region of the diagram which does not include any of A. This will include the 7 people in B who are outside the overlapping region, and the 8 people who are outside of both A and B (regions III and IV. Since A consists of regions I and II, A' consists of regions III and IV). B ' consists of all the people who do not have bright eyes, represented by every region of the diagram which does not include any of B (regions II and IV). This will include the 12 people in A but outside the overlap, and the 8 people outside of both A and B. Thus A ' U B ' consists of everyone in at least one of A ' or B ', including the 7 people in B but outside the overlap (region III), the 12 people in A let outside the overlap (region II), and the 8 people outside of both A and B (region IV). These will be the people who lack at least one of the characteristics dark hair and/or bright eyes. Thus n(A' U B') = 7 + 12 + 8 = 27. Note that these are the same 27 people who are in ( A ^ B ) '. So at least in this case, ( A ^ B ) ' = A ' U B '. A ' ^ B ' consists of all the people in both A ' and B '. As before A ' includes the 7 people in B but not A (region III) as well as the 8 people outside both A and B (region IV), and B ' includes the 12 people in A but not B (region II) as well as the 8 people outside both A and B (region IV). The people in both A ' and B ' will be the 8 people outside both A and B, those who have neither dark hair nor bright eyes. We note that this is the same as the set ( A U B ) ', so at least for the present case we see that ( A ' ^ B ' = ( A U B ) '.
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RESPONSE --> ok
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22:51:34 `q005. Succinctly describe the relationships between ( A U B ) ', A ' U B ', (A ^ B) ' and A ' ^ B '.
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RESPONSE --> These all contain the elements in either A or B or both, ot the elements outside of A and B.?
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22:52:20 ( A U B ) ' = A ' ^ B ' and ( A ^ B ) ' = A ' U B '. The collection outside of the union A U B is the intersection A ' ^ B ', and the collection outside the intersection A ^ B is the union A ' U B '. The ' operation changes union to intersection and intersection to union.
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RESPONSE --> Ok, i wasn't for sure what this question was asking for, but now i understand.
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PB؍ Student Name: assignment #004
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23:01:32 Note that there are 4 questions in this assignment. `q001. From the collection of letters a, b, c, d, e, how many smaller collections having at least one element may be formed?
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RESPONSE --> 30
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23:01:48 We will list the original collection by placing its elements between braces: { a, b, c, d, e }. The collection {a, b, c, d} is a smaller collection obtained by eliminating e from the original collection. Similarly we can eliminate d or c or b or a to get the 4-element collections {a, b, c, e}, {a, b, d, e}, { a, c, d, e} and {b, c, d, e}. Alternatively we could simply include either a or b or c or d or e in a 1-element collection, obtaining {a}, {b}, {c}, {d} and {e}. It should be clear that these are the only ways to form collections of 1 or 4 elements. To form a collection of 2 elements we could include a and one other element, obtaining { a, b}, { a, c }, { a, d } and { a, e }. Or we could include b and one other element (excluding a, since we already have the collection { a, b } which is identical to the collection { b, a } since it has exactly the same elements). We obtain { b, c }, { b, d } and { b, e }. {}Or we could include c and one other element (other than a or b, since these have already been listed) to obtain { c, d } and { c, e }. Finally we could include d and the only other element left, e, to get { d, e}. This gives us a complete listing of the 10 sets we can form with 2 of the original elements. This leaves us the 3-element sets, which can be formed by excluding the 2-element sets. Working in reverse order, we can exclude { d, e } to get { a, b, c }, or { c, e } to get { a, b, d }, etc.. The remaining sets we get in this fashion are { a, b, e}, { a, c, d }, { a, c, e}, { a, d, e}, { b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}. We thus have 10 three-element sets. The total number of smaller sets containing at least one element is therefore 5 + 5 + 10 + 10 = 30.
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RESPONSE --> ok
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23:04:43 `q002. A one-to-one correspondence between two sets is a rule that associates each element of the each with exactly one element of the other. A natural one-to-one correspondence between the sets { a, b, c } and { 1, 2, 3 } would be to associate a with 1, b with 2, c with 3. This correspondence might be represented as [ a <--> 1, b <--> 2, c <--> 3 ]. This isn't the only possible one-to-one correspondence between these sets. Another might be [ a <--> 2, b <--> 1, c <--> 3 ]. In each case, every element of each set is associated with exactly one element of the other. Another correspondence between the sets might be [ a <--> 3, b<-->2, c<-->3 ]. This correspondence is not one-to-one. In what way does it fail to be a one-to-one correspondence (remember that a one-to-one correspondence is one in which every element of each set is associated with exactly one element of the other).
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RESPONSE --> It used the element 3 twice(a<-->3 and c<-->3).
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23:04:58 [ a <--> 3, b<-->2, c<-->3 ] fails to be a one-to-one correspondence for two reasons. In the first place, 3 is associated with a and with c, and every element of each set is to be associated with exactly one element of the other. 3 is associated with two elements of the other set. It also fails because the element 1 of the second set is not associated with anything in the first set.
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RESPONSE --> ok
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23:16:38 `q003. There are four possible one-to-one correspondences between the sets {a, b, c} and {1, 2, 3} which were not described in the preceding exercise. One of them would be [ a <--> 3, b <--> 2, c <--> 1 ]. What are the other three possible one-to-one correspondences?
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RESPONSE --> [a <--> 2, b <--> 3, c <--> 1] [a <--> 3, b <--> 1, c <--> 2] [a <--> 2, b <--> 1, c <--> 3] [a <--> 1, b <--> 3, c <--> 2]
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23:17:36 If we designate the correspondence [ a <--> 1, b <--> 2, c <--> 3 ] as the '123' correspondence, [a <--> 2, b <--> 1, c <--> 3 ] as the '213' correspondence and [a <--> 3, b <--> 2, c <--> 1 ] as the '321' correspondence, in each case listing the numbers associated with a, b, c in that order, we see that the remaining three correspondences could be designated 132, 231 and 312. These correspondences could of course be written out as [ a <--> 1, b <--> 3, c <--> 2 ], [ a <--> 2, b <--> 3, c <--> 1 ] and [ a <--> 3, b <--> 1, c <--> 2 ]. Note that 123, 132, 213, 231, 312, 321 represent the six ways of rearranging the digits 1, 2, 3 into a 3-digit number, listed in increasing order.
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RESPONSE --> I listed an extra to make sure I wasn't listing one from the previous exercise.
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23:20:12 `q004. Explain why it is not possible to put the sets { a, b, c} and {1, 2, 3, 4} into a one-to-one correspondence.
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RESPONSE --> A one-to-one corresponence has to have the same number of elements as numbers. Set {a,b,c} has 3 elements, and set {1,2,3,4} has 4 elements.
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23:20:18 One set has 3 elements and the other has 4 elements. A 1-to-1 correspondence has to match each element of each set with exactly one element of the other. It would not be possible to find four different elements of the first set to match with the four elements of the second.
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RESPONSE --> ok
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