course MTH 151 ›Þ©š¦éþŠn”’~–ÓñÒ™NÍÓassignment #005
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12:32:25 Query 2.5.12 n({9, 12, 15, ..., 36})
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RESPONSE --> The missing numbers in the set are {18,21,24,27,30,33}. confidence assessment: 2
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12:32:51 ** There are 10 numbers in the set: 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 **
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RESPONSE --> correct self critique assessment: 3
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12:34:19 Query 2.5.18 n({x | x is an even integer }
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RESPONSE --> There could be many solutions to this, but one could be {2,4,,6,8,10,...} confidence assessment: 2
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12:34:37 ** {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 --> correct self critique assessment: 3
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12:44:07 Query 2.5.18 how many diff corresp between {stallone, bogart, diCaprio} and {dawson, rocky, blaine}?
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RESPONSE --> [{stallone, dawson},{bogart, rocky}, {dicaprio, blaine}] [{stallone, rocky},{bogart, blaine}, {dicaprio, dawson}] [{stallone, blaine}, {bogart, dawson}, {dicaprio, rocky}] [{stallone, dawson}, {bogart, blaine}, {dicaprio, rocky}] [{stallone, rocky}, {bogart, dawson}, {dicaprio,blaine}] [{stallone, blaine}, {bogart, rocky}, {dicaprio, dawson}] I found 6 different corresp. confidence assessment: 2
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12:44:16 ** 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 --> correct self critique assessment: 3
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12:48:54 2.5.36 1-1 corresp between counting #'s and {-17, -22, -27, ...}
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RESPONSE --> since each number is a negative number and then increases by 5 we would have to add -5 each time. so the formula should be -5n-12= -17. confidence assessment: 1
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12:49:34 **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 --> correct self critique assessment: 3
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12:53:06 2.5.42 show two vert lines, diff lengths have same # of points
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RESPONSE --> I am not sure how to answer this because wouldn't it depend on the size of the lines? confidence assessment: 0
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12:54:34 ** 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 tried to draw this out but I am still uncertain how to do this. I will keep studying so I can figure this out. self critique assessment: 2
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’ûÅùòÏvÆøÆ‹®ÒÝä¯óືÀ¬DËšÏLyÕ¶ assignment #006 006. `Query 6 College Algebra 02-03-2009
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12:57:50 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> It would be inductive because it is drawing a conclusion based on repetition. confidence assessment: 2
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12:57:57 ** The argument is inductive, because it attempts to argue from a pattern. **
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RESPONSE --> correct self critique assessment: 3
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12:59:10 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> Deductive because it says ""all"". confidence assessment: 2
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12:59:17 ** 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 --> correct self critique assessment: 3
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13:01:39 Query 1.1.20 1 / 3, 3 / 5, 5/7, ... Probable next element.
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RESPONSE --> the next element would be 7/9 because the denominator becomes the numerator and the next odd number will become the new denominator. confidence assessment: 2
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13:01:47 **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 --> correct self critique assessment: 3
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13:08:02 Query 1.1.23 1, 8, 27, 64, ... Probable next element.
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RESPONSE --> the next would be 125 because 5x5x5=125. each number is being multiplied 3 times starting with 1x1x1=1, 2x2x2=8, 3x3x3=27 etc. confidence assessment: 1
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13:09:42 ** 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 --> i got this correct but should have said 6x6x6=216 and i only went up to 5. self critique assessment: 3
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13:13:14 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> i would use inductive reasoning and continue the pattern. 11111*11111=123454321. confidence assessment: 1
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13:13:22 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> correct self critique assessment: 3
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13:14:11 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> yes because that is the pattern. confidence assessment: 1
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13:15:18 ** 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 --> i understand how it could it end because the numbers start at one and goes up to nine then back down to one. self critique assessment: 2
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13:19:27 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> the sum should be 2,001,000. there are 1000 pairs of sums of 2001 which would be 1000*2001=2,001,000. confidence assessment: 2
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13:19:37 ** 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 --> correct self critique assessment: 3
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13:28:36 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 --> 142857*1= 142857 142857*2= 285714 142857*3= 428571 142857*4= 571428 142857*5= 714285 142857*6= 857142 142857*7= 999999 the numbers in 1-6 are just switched around but 7 has all nines. confidence assessment: 1
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13:29:11 ** 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 --> correct self critique assessment: 3
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13:29:49 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> this shows that just because the pattern has been repeated several times does not mean it will continue that way. confidence assessment: 2
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13:30:03 ** 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 --> correct self critique assessment: 3
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