course Mth 151 The second one, assignment 8 made me me feel stupid. I don't know if it's because it's late or I just don't know how to process things like that. I'm working on it. ۉҎY٠assignment #001 001. `Query 1 College Algebra 01-19-2008
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12:41:11 `questionNumber 10000 Query 2.1.12 counting #'s 4 to 14 List the elements of the set.
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RESPONSE --> 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 confidence assessment: 2
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12:41:19 `questionNumber 10000 **A list of the elements would just be 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. **
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RESPONSE --> self critique assessment: 3
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12:42:37 `questionNumber 10000 query 2.1.24 listing for set of presidents between LBJ and Clinton
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RESPONSE --> Nixon, Gerald, Carter, Reagan, Bush confidence assessment: 3
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12:44:22 `questionNumber 10000 ** A listing would be {}{}{Lyndon Johnson, Richard Nixon, Gerald Ford, Jimmy Carter, Ronald Regan, George HW Bush, William Clinton}.{}{}Set-builder notation is {x|x is a president who served between Lyndon Johnson and William Clinton} x is a variable and the condition 'x is a president who served between Lyndon Johnson and William Clinton' tells you what possible things the variable can be. COMMON ERROR: It's incorrect to say {x | x is the set of presidents who served between Johnson and Clinton}. x is a president, not a set of presidents. Should be {x|x is a president who served between Lyndon Johnson and William Clinton} **
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RESPONSE --> self critique assessment: 3
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12:45:16 `questionNumber 10000 2.1.40 finite or infinite: set of rat #'s 0 to 1
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RESPONSE --> finite confidence assessment: 3
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12:47:43 `questionNumber 10000 ** Rational numbers have form p/q, where p and q are integers. Numbers like 5/8, 57/31, -3/5, -57843/7843, etc. The subset {1/2, 1/3, 1/4, 1/5, ... } is just by itself an infinite set of rational numbers between 0 and 1. Then you have things like 348/937, and 39827389871 / 4982743789, and a whole infinite bunch of others. There are thus infinitely many rational numbers in any interval of the real line. COMMON MISCONCEPTION: finite, because it doesn't go on forever Rational numbers have form p/q, where p and q are integers. Numbers like 5/8, 57/31, -3/5, -57843/7843, etc. Not all of these lie between 0 and 1, of course. **
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RESPONSE --> self critique assessment: 3
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12:52:10 `questionNumber 10000 2.1.48 n(A), A={x|x is a U.S. senator} What is n(A) and why?
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RESPONSE --> n(A) is 1 because there is only one senator in a set. confidence assessment: 1
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12:53:01 `questionNumber 10000 ** n(A) stands for the number of elements in the set--in this case for the number of senators. There are 100, 2 from each State. So n(A) = 100. **
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RESPONSE --> Yeah it's Saturday I'm an idiot. self critique assessment: 3
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12:54:34 `questionNumber 10000 query 2.1.54 {x|x is neagtive number}
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RESPONSE --> Infinite confidence assessment: 0
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12:55:21 01-19-2008 12:55:21 ** This notation means all possible values of x such that x is a negative number. The question is whether the set is well-defined or not. It is in fact well-defined because there is a definite way to decide whether a given object is an element of the set, because there is a definite way to determine whether an object is a negative number or not. ALTERNATIVE ANSWER: The set is well-defined because you have a criterion by which you can definitely decide whether something is or is not in the set. **
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NOTES ------->
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12:55:25 `questionNumber 10000 ** This notation means all possible values of x such that x is a negative number. The question is whether the set is well-defined or not. It is in fact well-defined because there is a definite way to decide whether a given object is an element of the set, because there is a definite way to determine whether an object is a negative number or not. ALTERNATIVE ANSWER: The set is well-defined because you have a criterion by which you can definitely decide whether something is or is not in the set. **
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RESPONSE --> self critique assessment: 3
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13:00:44 `questionNumber 10000 2.1.74 (formerly 2.1.72) This was not assigned, but you should be able to answer based on your work on similar problems: It is or is it not true that 2 is not not subset of {7,6,5,4}?
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RESPONSE --> True it is not a subset confidence assessment: 2
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13:00:59 `questionNumber 10000 ** The statement is that 2 is not a subset. The statement is true because 2 isn't even a set, it's just a number. {2} is a set and could be a subset of something. 2 is just a number; it isn't a set so it can't be a subset of anything. The usual answer is that 2 isn't a subset because 2 isn't in the set. However that's not the correct reason. The correct reason is that 2 isn't a set and a subset must be a set. COMMON MISCONCEPTION: the statement says that 2 is not a subset, not that it is not an element of the set. So the reason it's not a subset is that 2 isn't a set at all, so it can't be a subset of anything. **
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RESPONSE --> self critique assessment: 3
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13:04:43 `questionNumber 10000 2.1.86 (formerly 2.1.84). This was not assigned but you did answer several questions related to the sets C={4,10,12}, B={2,4,8,10}, and should be able to answer this. Is it true or false that every element of C is also an element of B? Be sure to include your reasoning.
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RESPONSE --> False, 12 is not included. confidence assessment: 2
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13:04:58 `questionNumber 10000 ** Specifically it's false because the element 12 is in C but not in B. **
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RESPONSE --> self critique assessment: 3
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ȴSo˃`Ƌ assignment #002 002. `Query 2 College Algebra 01-19-2008
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13:10:06 `questionNumber 20000 2.2.24 U={a,b,...,g}, A={a,e}, B={a,b,e,f,g}, C={b,f,g}, D={d,e} Is C ps U? Is the statement true or false and why?
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RESPONSE --> Yes, they contain all the same elements. confidence assessment: 1
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13:10:20 `questionNumber 20000 ** It's true because all elements of C are in the universal set, and because there are elements of U that aren't in C. You have to have both conditions, since a proper subset cannot be identical to other set. **
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RESPONSE --> self critique assessment: 3
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13:13:59 `questionNumber 20000 Query 2.2.30 phi s D
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RESPONSE --> confidence assessment: 0
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13:16:40 `questionNumber 20000 ** Note that You should be responding to problem 2.2.30 from the homework you worked out on paper. The shorthand notation is for my reference and won't always make sense to you. For clarification, though, the symbol for the empty set is the Greek letter phi. One set is a subset of another if every element of that set is in the other. To show that a set isn't a subset of another you have to show something in that set that isn't in the other. There's nothing in the empty set so you can never do this--you can never show that it has something the other set doesn't. So you can never say that the empty set isn't a subset of another set. Thus the empty set is a subset of any given set, and in particular it's a subset of D. ALTERNATIVE ANSWER: As the text tells you, the empty set is a subset of every set. ANOTHER ALTERNATIVE Every element of the empty set is in D because there is no element in the empty set available to lie outside of D. ONE MORE ALTERNATIVE: The empty set is a subset of every set. Any element in an empty set is in any set, since there's nothing in the empty set to contradict that statement. **
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RESPONSE --> self critique assessment: 3
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13:24:11 `questionNumber 20000 2.2.33 D not s B Is the statement true or false and why?
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RESPONSE --> True, because it is not in B's set. confidence assessment: 2
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13:38:55 `questionNumber 20000 ** D is a subset of B if every element of D is an element of B-i.e., if D doesn't contain anything that B doesn't also contain. The statement says that D is not a subset of B. This will be so if D contains at least one element that B doesn't. **
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RESPONSE --> self critique assessment: 3
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13:42:19 `questionNumber 20000 2.2.36 there are exactly 31 subsets of B
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RESPONSE --> False confidence assessment: 1
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13:42:48 `questionNumber 20000 ** If a set has n elements then is has 2^n subsets, all but one of which are proper subsets. B has 5 elements so it has 2^5 = 32 subsets. So the statement is false. There are exactly 31 proper subsets of B, but there are 32 subsets of B. **
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RESPONSE --> self critique assessment: 3
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13:45:56 `questionNumber 20000 Query 2.2.40 there are exactly 127 proper subsets of U Is the statement true or false and why?
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RESPONSE --> True, because there are 7 items in the set. confidence assessment: 3
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13:46:17 `questionNumber 20000 ** The set is not a proper subset of itself, and the set itself is contained in the 2^n = 2^7 = 128 subsets of this 7-element set. This leaves 128-1 = 127 proper subsets. **
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RESPONSE --> self critique assessment: 3
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13:47:57 `questionNumber 20000 Query 2.2.48 U={1,2,...,10}, complement of {2,5,7,9,10} What is the complement of the given set?
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RESPONSE --> confidence assessment: 0
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13:48:08 `questionNumber 20000 ** the complement is {1,3,4,6,8}, the set of all elements in U that aren't in the given set. **
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RESPONSE --> self critique assessment: 3
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13:51:14 `questionNumber 20000 query 2.2.63 in how many ways can 3 of the five people A, B, C, D, E gather in a suite?
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RESPONSE --> In 10 ways. confidence assessment: 3
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13:51:23 `questionNumber 20000 ** The answer here would consist of a list of all 3-element subsets: {a,b,c}, {a,b,d}, {a,b,e}, {a,c,d} etc. There are ten such subsets. Using a,b,c,d,e to stand for the names, we can list them in alphabetical order: {a,b,c), {a,b,d}, {a,b,e}, {a,c,d}, {a,c,e}, {a,d,e|, {b,c,d}, {b,c,e}, {b,d,e}, {c, d, e}**
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RESPONSE --> self critique assessment: 3
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17:28:07 `questionNumber 30000 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 --> a is the only one in the set. confidence assessment: 1
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17:28:54 `questionNumber 30000 **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 --> self critique assessment: 3
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17:30:19 `questionNumber 30000 Give the intersection of the two sets Y and Z'
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RESPONSE --> b and c. confidence assessment: 3
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17:31:26 `questionNumber 30000 **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 --> self critique assessment: 3
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17:33:10 `questionNumber 30000 Query 2.3.30 describe in words (A ^ B' ) U (B ^ A')
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RESPONSE --> Not in A or B. confidence assessment: 1
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17:33:49 `questionNumber 30000 ** 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: 3
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17:37:43 `questionNumber 30000 ** 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 --> . self critique assessment: 3
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17:40:14 `questionNumber 30000 Query 2.3.60 X = {1,3,5}, Y = {1,2,3}. Find (X ^ Y)' and X' U Y'.
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RESPONSE --> 5 and 2. confidence assessment: 1
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17:41:19 `questionNumber 30000 ** 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 --> self critique assessment: 3
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17:53:19 `questionNumber 30000 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: 3
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17:53:31 `questionNumber 30000 ** (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: 3
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17:55:36 `questionNumber 30000 2.3.84 Shade A U B
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RESPONSE --> everything but u would be shaded. confidence assessment: 2
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17:56:19 `questionNumber 30000 ** 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 --> self critique assessment: 3
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17:58:20 `questionNumber 30000 Query 2.3.100 Shade (A' ^ B) ^ C
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RESPONSE --> All of B is shaded except the overlap with A. Everything is shaded in C except its overlap with A confidence assessment: 2
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17:58:40 `questionNumber 30000 ** 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 --> self critique assessment: 3
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17:59:36 `questionNumber 30000 Describe the shading of the set (A ^ B)' U C.
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RESPONSE --> Nothing would be shaded in the regions. confidence assessment: 2
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18:00:27 `questionNumber 30000 ** 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 --> self critique assessment: 3
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18:02:03 `questionNumber 30000 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' confidence assessment: 2
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18:02:40 `questionNumber 30000 ** 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 --> self critique assessment: 3
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XϓqFZ assignment #004 004. `Query 4 College Algebra 01-30-2008
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18:06:35 `questionNumber 40000 2.4.12 n(A') = 25, n(B) = 28, n(A' U B') = 40, n(A ^ B) = 10
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RESPONSE --> A = 10 B = 10, 28 A' U B' = 25, 40 confidence assessment: 2
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18:07:02 `questionNumber 40000 ** 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 --> self critique assessment: 3
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18:08:56 `questionNumber 40000 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 --> 5 confidence assessment: 3
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18:09:18 `questionNumber 40000 ** 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 --> self critique assessment: 3
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}x~ƿ߿| assignment #005 005. `Query 5 College Algebra 02-06-2008
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11:01:08 `questionNumber 50000 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> There are aleph-null of them. confidence assessment: 2
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11:01:40 `questionNumber 50000 ** There are 10 numbers in the set: 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 **
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RESPONSE --> self critique assessment: 3
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11:02:21 `questionNumber 50000 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> There are aleph-null of them. confidence assessment: 1
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11:02:38 `questionNumber 50000 ** {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 --> self critique assessment: 3
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11:03:33 `questionNumber 50000 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> 3 confidence assessment: 2
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11:05:27 `questionNumber 50000 ** 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 --> self critique assessment: 2
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11:10:45 `questionNumber 50000 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> (-17,1) (-22, 2) (-27, 3) (-5n) confidence assessment: 2
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11:11:00 `questionNumber 50000 **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 --> self critique assessment: 3
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11:12:30 `questionNumber 50000 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> confidence assessment: 0
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11:13:29 `questionNumber 50000 ** 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 --> self critique assessment: 3
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רT֍qΥ[zz assignment #006 006. `Query 6 College Algebra 02-06-2008
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11:15:53 `questionNumber 60000 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> Inductive confidence assessment: 2
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11:16:01 `questionNumber 60000 ** The argument is inductive, because it attempts to argue from a pattern. **
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RESPONSE --> self critique assessment: 3
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11:17:33 `questionNumber 60000 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> Deductive confidence assessment: 2
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11:17:45 `questionNumber 60000 ** this argument is deductive--the conclusions follow inescapably from the premises. 'all men' is general; 'Socrates' is specific. This goes general to specific and is therefore deductive. COMMON ERROR: because it is based on a fact, or concrete evidence. Fact isn't the key; the key is logical inevitability. The argument could be 'all men are idiots, Socrates is an man, therefore Socrates is an idiot'. The argument is every bit as logical as before. The only test for correctness of an argument is that the conclusions follow from the premises. It's irrelevant to the logic whether the premises are in fact true. **
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RESPONSE --> self critique assessment: 3
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11:18:31 `questionNumber 60000 Query 1.1.20 1 / 3, 3 / 5, 5/7, ... Probable next element.
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RESPONSE --> 7/9 confidence assessment: 3
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11:18:48 `questionNumber 60000 **The numbers 1, 3, 5, 7 are odd numbers. We note that the numerators consist of the odd numbers, each in its turn. The denominator for any given fraction is the next odd number after the numerator. Since the last member listed is 5/7, with numerator 5, the next member will have numerator 7; its denominator will be the next odd number 9, and the fraction will be 7/9. There are other ways of seeing the pattern. We could see that we use every odd number in its turn, and that the numerator of one member is the denominator of the preceding member. Alternatively we might simply note that the numerator and denominator of the next member are always 2 greater than the numerator and denominator of the present member. **
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RESPONSE --> self critique assessment: 3
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11:20:01 `questionNumber 60000 Query 1.1.23 1, 8, 27, 64, ... Probable next element.
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RESPONSE --> 216 confidence assessment: 3
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11:20:29 `questionNumber 60000 ** This is the sequence of cubes. 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125. The next element is 6^3 = 216. Successive differences also work: 1 8 27 64 125 .. 216 7 19 37 61 .. 91 12 18 24 .. 30 6 6 .. 6 **
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RESPONSE --> self critique assessment: 3
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11:22:11 `questionNumber 60000 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> 11111 X 11111 = 1,234,554,321 confidence assessment: 3
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11:22:19 `questionNumber 60000 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> self critique assessment: 3
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11:23:46 `questionNumber 60000 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> I think it would always add another one. confidence assessment: 2
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11:24:05 `questionNumber 60000 ** You could think forward to the next few products: What happens after you get 12345678987654321? Is there any reason to expect that the sequence could continue in the same manner? The middle three digits in this example are 8, 9 and 8. The logical next step would have 9, 10, 9, but now you would have 9109 in the middle and the symmetry of the number would be destroyed. There is every reason to expect that the pattern would also be destroyed. **
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RESPONSE --> self critique assessment: 3
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11:26:44 `questionNumber 60000 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> confidence assessment: 0
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11:27:36 `questionNumber 60000 ** Pair up the first and last, second and second to last, etc.. You'll thus pair up 1 and 2000, 2 and 1999, 3 and 1998, etc.. Each pair of numbers totals 2001. Since there are 2000 numbers there are 1000 pairs. So the sum is 2001 * 1000 = 2,001,000 **
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RESPONSE --> self critique assessment: 3
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11:30:55 `questionNumber 60000 Query 1.1.57 142857 * 1, 2, 3, 4, 5, 6. What happens with 7? Give your solution to the problem as stated in the text.
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RESPONSE --> 1,42857 ; 285714 ; 428,571 ; 571,428 ; 714,285 ; 857,142 ; 999,999 They all contain the same numbers until you get to seven. confidence assessment: 2
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11:31:15 `questionNumber 60000 ** Multiplying we get 142857*1=142857 142857*2= 285714 142857*3= 428571 142857*4=571428 142857*5= 714285 142857*6=857142. Each of these results contains the same set of digits {1, 2, 4, 5, 7, 8} as the number 1428785. The digits just occur in different order in each product. We might expect that this pattern continues if we multiply by 7, but 142875*7=999999, which breaks the pattern. **
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RESPONSE --> self critique assessment: 3
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11:31:47 `questionNumber 60000 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> It doesn't always work confidence assessment: 3
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11:31:57 `questionNumber 60000 ** Inductive reasoning would have led us to expect that the pattern continues for multiplication by 7. Inductive reasoning is often correct it is not reliable. Apparent patterns can be broken. **
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RESPONSE --> self critique assessment: 3
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kqMksߩ assignment #007 007. `Query 7 College Algebra 02-10-2008
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18:58:35 `questionNumber 70000 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> 5, 376 confidence assessment: 2
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18:59:56 `questionNumber 70000 ** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. **
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RESPONSE --> I started off with the wrong number, I put a 51 instead of a 57 self critique assessment: 3
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C倉ࢽv}S| assignment #007 007. `Query 7 College Algebra 02-10-2008
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23:04:25 `questionNumber 70000 Query 1.2.6 seq 2, 51, 220, 575, 1230, 2317 ... by successive differences
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RESPONSE --> 3,992 confidence assessment: 2
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23:04:32 `questionNumber 70000 ** If the sequence is 2, 57, 220, 575, 1230, 2317, ... then we have: 2, 57, 220, 575, 1230, 2317, # 3992 55, 163, 355, 655, 1087, # 1675 108, 192, 300, 432, # 588 84, 108, 132, # 156 24, 24, The final results, after the # signs, are obtained by adding the number in the row just below, in the following order: Line (4) becomes 132+24=156 Line (3) becomes 432+156=588 Line (2) becomes 1087+588=1675 Line (1) becomes 2317+1675=3992 The next term is 3992. **
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RESPONSE --> self critique assessment: 3
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23:08:02 `questionNumber 70000 1.2.18 1^2 + 1 = 2^2 - 2; 2^2 + 2 = 3^2 - 3; 3^2 + 3 = etc.
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RESPONSE --> 4^2 + 4 = 5^2 - 5 It is true because they would both equal 20 confidence assessment: 3
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23:08:09 `questionNumber 70000 ** The next equation in the sequence would be 4^2 + 4 = 5^2 - 5 The verification is as follows: 4^2 + 4 = 5^2 - 5 simplifies to give you 16 + 4 = 25 - 5 or 20 = 20 **
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RESPONSE --> self critique assessment: 3
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23:09:43 `questionNumber 70000 1.2.30 state in words (1 + 2 + ... + n ) ^ 2 = 1^3 + 2^3 + ... + n^3
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RESPONSE --> confidence assessment: 0
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23:09:58 `questionNumber 70000 ** the equation says that the square of the sum of the first n counting numbers is equal to the sum of their cubes **
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RESPONSE --> self critique assessment: 3
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23:13:13 `questionNumber 70000 1.2.36 1 st triangular # div by 3, remainder; then 2d etc. Pattern.
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RESPONSE --> .3333... 1 2 3.333... 5 I don't see a pattern really. confidence assessment: 1
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23:14:01 `questionNumber 70000 ** The triangular numbers are 1, 3, 6, 10, 15, 21, . . . . We divide these by 3 and get the sequence of remainders. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0. It turns out that the sequence continues as a string of 1,0,0 's. At this point that is an inductive pattern, but remmeber that the sequence of triangular numbers continues by adding successively larger and larger numbers to the members of the sequence. Since the number added always increases by 1, and since every third number added is a multiple of 3, is isn't too difficult to see how the sequence of remainders comes about and to see why it continues as it does. COMMON ERROR: .3333333,1,2,3.3333333,etc. INSTRUCTOR CORRECTION: You need the remainders, not the decimal equivalents. When you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. Therefore the remainders are 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. COMMON ERROR: 1/3, 1, 2, 3 1/3 CORRECTION: These are the quotients. You need the remainders. If you get 1/3 that means the remainder is 1; same if you get 3 1/3. If you just getting whole number (like 1 or 2 in your calculations) the remainder is 0. In other words, when you divide 1 by 3 you get 0 with remainder 1 (3 goes into 1 zero times with 1 left over). 3 divided by 3 gives you 1 with remainder 0. 6 divded by 3 is 2 with remainder 0. 10 divided by 3 is 3 with remainder 1. The remainders form a sequence 1,0,0,1,0,0 and the sequence continues as a string of 1,0,0 's. **
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RESPONSE --> self critique assessment: 3
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23:24:36 `questionNumber 70000 1.2.48 use formula to find the 12 th octagonal number. Explain in detail how you used the formula to find this number.
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RESPONSE --> confidence assessment: 0
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23:25:45 `questionNumber 70000 ** The pattern to formulas for triangular, square, pentagonal, hexagonal, heptagonal and octagonal numbers is as follows: Triangular numbers: n / 2 * [ n + 1 ] note that this is the same as Gauss' formula Square numbers: n / 2 * [ 2n + 0 ] or just n^2 Pentagonal #'s: n / 2 * [ 3n - 1 ] Hexagonal #'s: n / 2 * [ 4n - 2 ] Heptagonal #'s: n / 2 * [ 5n - 3 ] Octagonal #'s: n / 2 * [ 6n - 4 ] The coefficient of n in the bracketed term starts with 1 and increases by 1 each time, and the +1 in the first bracketed term decreases by 1 each time. You will need to know these formulas for the test. The last formula is for octagonal numbers. To get n = 12 octangonal number use n/2 * [ 6n - 4 ] to get 12 / 2 * [ 6 * 12 - 4 ] = 6 * 68 = 408. **
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RESPONSE --> self critique assessment: 0
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d˙߮dcJԕ}|X assignment #008 008. `Query 8 College Algebra 02-10-2008
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23:34:48 `questionNumber 80000 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
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RESPONSE --> confidence assessment: 0
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23:35:38 `questionNumber 80000 ** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. **
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RESPONSE --> self critique assessment: 3
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23:37:09 `questionNumber 80000 Query 1.3.10 divide clock into segments each with same total
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RESPONSE --> I'm not quite sure what this is asking. You could divide into 12 equal segments. confidence assessment: 1
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23:37:37 `questionNumber 80000 ** The total of all numbers on the clock is 78. So the numbers in the three sections have to each add up to 1/3 * 78 = 26. This works if we can divide the clock into sections including 11, 12, 1, 2; 3, 4, 9, 10; 5, 6, 7, 8. The numbers in each section add up to 26. To divide the clock into such sections the lines would be horizontal, the first from just beneath 11 to just beneath 2 and the second from just above 5 to just above 8. Horizontal lines are the trick. You might have to draw this to see how it works. **
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RESPONSE --> self critique assessment: 3
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23:42:29 `questionNumber 80000 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
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RESPONSE --> 32 X 5 = 160 160 \ 2 + 80 confidence assessment: 1
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23:42:59 `questionNumber 80000 ** You have to work this one backwards. If they were left with 35 on Friday they had 70 at the beginning (after bringing in the 32) on Friday, so they had 70 - 32 = 38 at the end on Thursday. So after bringing in the 32 they had 2 * 38 = 76 at the beginning of Thursday, which means they had 76 - 32 = 44 before the 32 were added. So they had 44 Wednesday night ... etc. **
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RESPONSE --> self critique assessment: 3
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23:46:01 `questionNumber 80000 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
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RESPONSE --> I'm sorry I'm just not getting the way these questions are asked, I understand them when they are explained though. It should be able to get out of the well, with 1 ft to spare. confidence assessment: 0
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23:46:43 `questionNumber 80000 ** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. **
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RESPONSE --> I didn't know that we were suppose to be measuring days. self critique assessment: 2
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23:49:32 `questionNumber 80000 Query 1.3.48 How many ways to pay 15 cents?
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RESPONSE --> 3 nickles 15 pennies 2 dimes and nickel a dime and 5 pennies a nickle and 10 pennies 2 nickels and 5 pennies 6 confidence assessment: 2
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23:49:43 `questionNumber 80000 ** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies **
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RESPONSE --> self critique assessment: 3
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23:53:06 `questionNumber 80000 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
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RESPONSE --> I don't know. confidence assessment: 0
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23:53:21 `questionNumber 80000 ** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. **
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RESPONSE --> self critique assessment: 3
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