course Mth 151 HƗ{x҇JPַassignment #003
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12:41:49 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= (a,c) U (a,c,e,g) (Y ^ Z') U X = (a,c,e,g) confidence assessment: 2
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12:42:07 **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 --> I understand and got it correct. self critique assessment: 2
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12:44:01 Give the intersection of the two sets Y and Z'
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RESPONSE --> Y = (a,b,c) Z' = (a, c, g) Y ^ Z' = (a,c,g) confidence assessment: 2
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xF assignment #003 003. `Query 3 College Algebra 09-16-2007
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12:48:25 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 = (a,b,c) Z' = (a,g) X= (a,c,e,g) (Y ^ Z') U X = (a,c,e,g) confidence assessment: 2
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12:48:44 **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 --> I understand and got it correct. self critique assessment: 2
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12:51:11 Give the intersection of the two sets Y and Z'
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RESPONSE --> Y = (a,b,c) Z' = (a,g) Y ^ Z' = a confidence assessment: 2
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12:51:29 **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 --> I understand and got it correct. self critique assessment: 2
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13:11:16 Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')
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RESPONSE --> The elements common in the set of A and the complement of B or the elements in set B and set A. confidence assessment: 2
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13:12:12 ** 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 --> self critique assessment:
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13:12:53 2.3.51 always or not always true: n(A U B) = n(A)+n(B)
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RESPONSE --> Not always true. confidence assessment: 2
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13:13:33 ** 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 --> I understand,. self critique assessment: 2
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13:20:09 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)' = X = (1,3,5) Y = (1,2,3) X' = (2,4) Y' = (4,5) (X ^ Y )' = (4) X' U Y' = X' = (2,4) Y' = (4,5) X' U Y ' = (2,4,5) confidence assessment: 2
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13:23:06 ** 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 --> I missed the first part but I see where I went wrong. I used the complement of each element instead of doing the problem in the parentheses first and then using the complement.I got the second part correct. self critique assessment: 2
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13:26:08 2.3.72 A = {3,6,9,12}, B = {6,8}.
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RESPONSE --> 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)} confidence assessment: 2
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13:26:37 ** (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 --> self critique assessment:
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13:28:36 2.3.84 Shade A U B
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RESPONSE --> This would be all the elements belonging to either set. The shading would include region II, region III, and region IV. confidence assessment: 2
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13:28:58 ** 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 --> I understand and got this correct,. self critique assessment: 2
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13:39:56 Query 2.3.100 Shade (A' ^ B) ^ C
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RESPONSE --> A' = everything outside of A including region I, region VI, VII, VIII. B = everything inside B including region III, region IV, region VII, region VIII. C= everything inside of C including region IV, region V, region VI, region VII. (A'^ B) ^ C = Region 7. confidence assessment: 2
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13:40:50 ** 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 --> The way I labeled my diagram (using the fig. 16 in ex. 12 of 2.3) was this area. self critique assessment: 2
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13:49:37 Describe the shading of the set (A ^ B)' U C.
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RESPONSE --> A = everything inside A including overlapping of B and C. B = everything inside B including overlapping of A and C. A = regions 2,3,4,5 B = regions 3,4,7,8 (A ^ B)' = {2,5,7,8} C = everything inside C including overlapping of A, and B. C = regions 4,5,6,7 (A ^ B)' U C = (2,4,5,6,7,8) confidence assessment: 2
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13:51:12 ** 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 --> I understand and I got it correct using the regions in Fig. 16, pg. 73. self critique assessment: 2
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13:56:22 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 --> (B' ^ C') ^ A. confidence assessment: 2
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13:57:00 ** 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 --> I understand and got it correct. self critique assessment: 2
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Ѻ֯ϬYTLٌ assignment #005 005. `Query 5 College Algebra 09-16-2007
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17:50:40 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> n({9, 12, 15, 18, 21, 24, 27, 30, 33, 36}) Cardinal number = 10 confidence assessment: 2
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17:50:56 ** There are 10 numbers in the set: 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 **
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RESPONSE --> I understand and got it correct. self critique assessment: 2
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17:53:25 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> n({2, 4, 6, 8, 10.....} No (aleph null) confidence assessment: 2
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17:53:54 ** {x | x is an even integer } indicates the set of ALL possible values of the variable x which are even integers. Anything that satisfies the description is in the set. This is therefore the set of even integers, which is infinite. Since this set can be put into 1-1 correspondence with the counting numbers its cardinality is aleph-null. **
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RESPONSE --> I understand and got it correct. self critique assessment: 2
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18:02:05 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> stallone bogart diCaprio dawson rocky blaine stallone bogart diCaprio rocky blaine dawson stallone bogart diCaprio blaine dawson rocky stallone bogart diCaprio dawson blaine rocky stallone bogart diCaprio blaine rocky dawson 5 different corresp. confidence assessment: 2
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18:02:31 ** Listing them in order, according to the order of listing in the set. We have: [ {S,D},{B,R},{Dic.,BL}] , [{S,bl},{B,D},{Dic.,R}], [{S,R},{B,Bl},{dic.,D}] [ {S,D},{B,DL},{Dic.,R}], [{S,bl},{B,R},{Dic.,D}], [{S,R},{B,D},{dic.,B1}] for a total of six. Reasoning it out, there are three choices for the character paired with Stallone, which leaves two for the character to pair with Bogart, leaving only one choice for the character to pair with diCaprio. **
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RESPONSE --> I see where I missed one. self critique assessment: 2
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18:04:17 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> -17 -22 -27 -32 -37....... 1 2 3 4 5....... confidence assessment: 2
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18:05:24 **You have to describe the 1-1 correspondence, including the rule for the nth number. A complete description might be 1 <-> -17, 2 <-> -22, 3 <-> -27, ..., n <-> -12 + 5 * n. You have to give a rule for the description. n <-> -12 - 5 * n is the rule. Note that we jump by -5 each time, hence the -5n. To get -17 when n=1, we need to start with -12. THE REASONING PROCESS TO GET THE FORMULA: The numbers in the first set decrease by 5 each time so you need -5n. The n=1 number must be -17. -5 * 1 = -5. You need to subtract 12 from -5 to get -17. So the formula is -5 n - 12. **
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RESPONSE --> I understand. I didn't write out that we add -5 to each counting number. self critique assessment: 2
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18:06:14 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> I am not certain how to do this. confidence assessment: 0
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18:07:40 ** This is a pretty tough question. One way of describing the correspondence (you will probably need to do the construction to understand): Sketch a straight line from the top of the blue line at the right to the top of the blue line at the left, extending this line until it meets the dotted line. Call this meeting point P. Then for any point on the shorter blue line we can draw a straight line from P to that point and extend it to a point of the longer blue line, and in our 1-1 correspondence we match the point on the shorter line with the point on the longer. From any point on the longer blue line we can draw a straight line to P; the point on the longer line will be associated with the point we meet on the shorter. We match these two points. If the two points on the long line are different, the straight lines will be different so the points on the shorter line will be different. Thus each point on the longer line is matched with just one point of the shorter line. We can in fact do this for any point of either line. So any point of either line can be matched with any point of the other, and if the points are different on one line they are different on the other. We therefore have defined a one-to-one correspondence. **
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RESPONSE --> I am still not quite clear on this. self critique assessment: 2
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㦢JȾߥ~ assignment #004 004. `Query 4 College Algebra 09-16-2007
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20:35:30 2.4.12 n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10
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RESPONSE --> A = 15 B = 18 A U B = 10 15 + 10 - 18 = 7 U = 7 confidence assessment: 1
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20:36:00 ** 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 understand and I got it correct. self critique assessment: 2
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20:40:42 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 --> A = 3 (wrote) B = 5 ( produced) A U B = 2 (wrote and produced) = 10 projects total 10 - 5 - 2 = 3 projects that he wrote but did not produce. confidence assessment: 2
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20:41:12 ** 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 --> I understand. self critique assessment: 2
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21:08:03 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 --> A = 9, B = 18, C = 5, D = 2, E = 9, F = 26, G = 6, U = 7 fat = 26 + 9 + 2 = 37 red = 5 + 2 +9 + 6 = 22 male = 18 + 26 + 9 + 6 = 49 fat, not male = 9 + 2 = 11 brown, not fat = 18 + 7 = 25 red & fat = 9 + 2 = 11 confidence assessment: 1
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21:10:25 ** 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 --> I only came up with 49 males but I see where I went wrong. I forgot to subtract the 9 from the 26. self critique assessment: 2
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