#$&* course 151 9/10 9 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'. So (Y ^ Z') U X = {a, c, e, g}, the set of all elements which lie in at least one of the sets (Y ^ Z') U X. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The Z’ almost got me on that one He intersectionis ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qGive the intersection of the two sets Y and Z' YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Z' ={a,g} because they are the only elements of the U set that is not contained in the Z sets Y={a,b,c} Y^Z’={a} because it is the only one contained in both confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a**Z' = {a,g}, the set of all elements of the universal set not in Z. Y ^ Z' = {a}, since a is the only element common to both Y and Z'.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery 2.3.32 (formerly 2.3.30). This was not assigned, but you answered a series of similar questions and should be able to give a reasonable answer to this one: Describe in words (A ^ B' ) U (B ^ A') YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All elements in A and not in B or all the elements in B and not in A confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** a description, not using a lot of set-theoretic terms, of (A ^ B' ) U (B ^ A') would be, all the elements that are in A and not in B, or that are not in A and are in B Or you might want to say something like 'elements which are in A but not B OR which are in B but not A'. STUDENT SOLUTION WITH INSTRUCTOR COMMENT:everything that is in set A and not in set B or everything that is in set B and is not in set A. INSTRUCTOR COMMENT: I'd avoid the use of 'everything' unless the word is necessary to the description. Otherwise it's likely to be misleading. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sure if I had the and , or in the right places ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q2.3.53 (formerly 2.3.51) Is it always or not always true that n(A U B) = n(A)+n(B)? This was not among the assigned questions but having completed the assignment you should be able to answer this. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This can be either true or false all its saying is that the number of elements of A + the number of elements of B will = to the total number of elements but it doesn’t account for intersections of the two sets confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** This conclusion is contradicted by many examples, including the one of the dark-haired and bright-eyed people in the q_a_. Basically n(A U B) isn't equal to n(A) + n(B) if there are some elements which are in both sets--i.e., in the intersection. } MORE DETAIL: The statement can be either true or false, depending on the sets A and B; it is not always true. The statement n(A U B) = n(A)+n(B) means that the number of elements in A U B is equal to the sum of the number of elements in A and the number of elements in B. The statement would be true for A = { c, f } and B = { a, g, h} because A U B would be { a, c, f, g, h} so n(A U B) = 5, and n(A) + n(B) = 2 + 3 = 5. The statement would not be true for A = { c, f, g } and B = { a, g, h} because A U B would be the same as before so n(AUB) = 5, while n(A) + n(B) = 3 + 3 = 6. The precise condition for which the statement is true is that A and B have nothing in common. In that case n(A U B) = n(A) + n(B). A more precise mathematical way to state this is to say that n(A U B) = n(A) + n(B) if and only if the intersection A ^ B of the two sets is empty. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Find the intersection of X^Y = {1,3} Now to find(X^Y)’={2,4,5} X’= {2,4} Y’= {4,5} X’ U Y’ = {2,4,5} confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** X ^ Y = {1,3} so (X ^ Y) ' = {1,3}' = {2, 4, 5}. (X ' U Y ' ) = {2, 4} U {4, 5} = {2, 4, 5} The two resulting sets are equal so a reasonable conjecture would be that (X ^ Y)' = X' U Y'. ** STUDENT QUESTION: Where did the 4 come from? INSTRUCTOR RESPONSE: I believe this problem, as stated in the text, indicates that the universal set is {1, 2, 3, 4, 5}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q2.3.72 A = {3,6,9,12}, B = {6,8}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A X B= {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6),(12,8)} All elements of A are in an order pair with all elements of B B X A = {(8,3),(8,6),(8,9),(8,12),(6,3),(6,6),(6,9),(6,12)} All elements of B are in an order pair with all elements of A confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** (A X B) = {(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6), (12,8)} (B X A) = (6,3),(6,6),(6,9),(6,12),(8,3),(8,6),(8,9),(8,12)} How is n(A x B) related to n(A) and n(B)? n(S) stands for the number of elements in the set S, i.e., its cardinality. n(A x B) = n(A) * n(B) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q2.3.84 Shade A U B All contained in A and all contained in B is shaded all else is not YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** everything in A and everything in B would be shaded. The rest of the universal set (the region outside A and B but still in the rectangle) wouldn't be. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery 2.3.100 Shade (A' ^ B) ^ C YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All not contained in A intersected with B and also intersected with C will be shaded confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** you would have to shade every region that lies outside of A and also inside B and also inside C. This would be the single region in the overlap of B and C but not including any part of A. Another way to put it: the region common to B and C, but not including any of A ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That is what I was trying to say ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery 2.3.108. Describe the shading of the set (A ^ B)' U C. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Shade what is not included in the intersection of (A^ B) and all that is contained in C Even the part that intersects the (A^B) intersection confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** All of C would be shaded because we have a union with C, which will include all of C. Every region outside A ^ B would also be shaded. A ^ B is the 'overlap' region where A and B meet, and only this 'overlap' would not be part of (A ^ B) '. The 'large' parts of A and B, as well as everything outside of A and B, would therefore be shaded. Combining this with the shading of C the only the part of the diagram not shaded would be that part of the 'overlap' of A and B which is not part of C. ** STUDENT QUESTION I think I understand because the ‘ was outside the ( ) then only the answer to A^B would be prime. And so my answer is wrong to the extent that the larger regions of A &B would also be shaded, but had it been (AUB)’ no part of either A or B would have been Shaded? INSTRUCTOR RESPONSE Exactly. Very good question, which you answered very well. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to draw this one out on a piece of paper ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q2.3.114 Largest area of A shaded (sets A,B,C). Write a description using A, B, C, subset, union, intersection symbols, ', - for the shaded region. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (B’ ^ C’)^ A confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** Student Answer and Instructor Response: (B'^C')^A Instructor Response: Good. Another alternative would be A - (B U C ), and others are mentioned below. COMMON ERROR: A ^ (B' U C') INSTRUCTOR COMMENT: This is close but A ^ (B' U C') would contain all of B ^ C, including a part that's not shaded. A ^ (B U C)' would be one correct answer. ** If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 004. `Query 4 ********************************************* Question: `q2.4.13 (formerly 2.4.12) This was not assigned but you answered similar questions and should be able to answer this one: n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10. What is n(A - B)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n(A U B) = n(A) +n(B) – n(A ^ B) there are 28 units of B and 10 are shared with A there are 25 units outside of A and 18 are taken by B that only leaves 7 that is out of both A and B n(A-B)= 40-18-7=15 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** 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. Thus n(A - B) = 40 - 18 - 7 = 15.** Supplementary comments: For example, with (A' U B'), you ask the following questions in order: What regions are in A? What regions are therefore in A'? What regions are in B? What regions are therefore in B'? So, what regions are in A' U B'? If you can break a question down to a series of simpler questions, you can figure out just about anything. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I found this one very tricky I had to really think about it ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qquery 2.4.18 wrote and produced 2, wrote 5, produced 7 &&&& How many did he write but not produce? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: He wrote 5 but only produce 2 so he wrote 3 that he didn’t produce confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q2.4.25 (formerly 2.4.24) 9 fat red r, 18 thin brown r, 2 fat red h, 6 thin red r, 26 fat r, 5 thin red h, 37 fat, 7 thin brown hens. ......!!!!!!!!................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a) 37 fat b) 9+2+6+5 = 22 red c) 18+6+26= 50 males d) 37-26=11 fat female e) 18+7=25 thin brown f) 9+2=11 red and fat confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.** I got the same thing as you did so two thumbs up "