course MTH 152 o䀘У}assignment #001
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13:50:38 query 11.1.6 {Andy, Bill, Kathy, David, Evelyn}. In how many ways can a secretary, president and treasuer be selected if the secretary must be female and the others male?
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RESPONSE --> AC AE BC BE DC DE CA CB CD EA EB ED = 12 ways
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13:51:15 ** Using letters for the names, there are 12 possibilities: kab, kba, kdb, kbd, kda, kad, edb, ebd, eba, eab, eda, ead. There are two women, so two possibilities for the first person selected. The other two will be selected from among the three men, so there are 3 possibilities for the second person chosen, leaving 2 possibilities for the third. The number of possiblities is therefore 2 * 3 * 2 = 12. **
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13:52:58 query 11.1.12,18 In how many ways can the total of two dice equal 5?
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RESPONSE --> 2
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13:53:57 ** Listing possibilities on first then second die you can get 1,4, or 2,3 or 3,2 or 4,1. There are Four ways. **
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RESPONSE --> I'm not sure how I got 3, but I understand why there are 4 whys
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13:55:18 In how many ways can the total of two dice equal 11?
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RESPONSE --> 2 6,5 and 5,6
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13:55:41 ** STUDENT SOLUTION AND INSTRUCTOR RESPONSE: There is only 1 way the two dice can equal 11 and that is if one lands on 5 and the other on 6 INSTRUCTOR RESPONSE: There's a first die and a second. You could imagine that they are painted different colors to distinguish them. You can get 5 on the first and 6 on the second, or vice versa. So there are two ways. **
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14:16:20 ** If you look at the figure you see that it forms a pentagon in the middle (if you are standing at the very center you would be within this pentagon). Each side of the pentagon is the side of a unique triangle; the five triangles formed in this way are the 'spikes' of the star. Each side of the pentagon is also part of a longer segment running from one point of the start to another. This longer segment is part of a larger triangle whose vertices are the two points of the star and the vertex of the pentagon which lies opposite this side of the pentagon. There are no other triangles, so we have 5 + 5 = 10 triangles. *&*&, BDE and CDE. Each of these is a possible triangle, but not all of these necessarily form triangles, and even if they all do not all the triangles will be part of the star. You count the number which do form triangles and for which the triangles are in fact part of the star. **
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RESPONSE --> I understand that there a five from the point of the star and five more within that form from the center of the tringle out to the points. This equals ten.
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14:17:36 query 11.1.40 4 x 4 grid of squares, how many squares in the figure?
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RESPONSE --> 25 16 single and 9 formed with 4 of the singles
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14:18:52 ** I think there would be 16 small 1 x 1 squares, then 9 larger 2 x 2 squares (each would be made up of four of the small squares), 4 even larger 3 x 3 squares (each made up of nin small squares) and one 4 x 4 square (comprising the whole grid), for a total of 30 squares. Do you agree? **
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RESPONSE --> Ok, I missed the ones made up of 9, but I see them now.
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14:19:49 query 11.1.50 In how many ways can 30 be written as sum of two primes?
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RESPONSE --> 4 ways 29+1 23+7 19+11 17+13
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14:20:41 **STUDENT SOLTION AND INSTRUCTOR COMMENT: There are 4 ways 30 can be written as the sum of two prime numbers: 29 + 1 19 + 11 23 + 7 17 + 13 INSTRUCTOR COMMENT: Good, but 1 isn't a prime number. It only has one divisor. **
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14:21:28 query 11.1.60 four adjacent switches; how many settings if no two adj can be off and no two adj can be on
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RESPONSE --> 2 0-1-0-1 1-0-0-1
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14:21:28 query 11.1.60 four adjacent switches; how many settings if no two adj can be off and no two adj can be on
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14:22:37 ** There are a total of 16 settings but only two have the given property of alternating off and on. If the first switch is off then the second is on so the third is off so the fourth is on. If the first is off then then the second is on and the third is off so the fourth is on. So the two possibilies are off-on-off-on and on-off-on-off. If we use 0's and 1's to represent these possibilities they are written 0101 and 1010. **
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RESPONSE --> I understand that the 2 are 0-1-0-1 and 1-0-1-0
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14:23:31 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No, it was pretty staight forward from the problems I worked for homework.
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sPŭeΏωޱ assignment #002 002. `query 2 Liberal Arts Mathematics II 01-26-2008
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20:49:21 query 11.2.12 find 10! / [ 4! (10-4)! ] without calculator
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RESPONSE --> The books question for number 12 is 6!?(6-40)! the answer to this is 720/2 or 360 The answer to this one is 3628800/720 or 5040
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20:53:23 ** 10! / [ 4! * (10-4) ! ] can be simplified to get 10! / ( 4! * 6! ). This gives you 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] . The numerator and denominator could be multiplied out but it's easier and more instructive to divide out like terms. Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1). Every factor of the denominator divides into the numerator without remainder: Divide 4 into 8, divide 3 into 9 and 2 into 10 and you get 5 * 3 * 2 * 7 = 210. NOTE ON WHAT NOT TO DO: You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divided 3628800 by 16480. But that would process would lose accuracy and be ridiculously long for something like 100 ! / ( 30! * 70!). Much better to simply divide out like factors until the denominator goes away. **
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RESPONSE --> I used the bottom method, but I made note not to do so and to use the method shown here by breaking down the factorals and dividing them
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20:55:03 query 11.2.25 3 switches in a row; fund count prin to find # of possible settings
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RESPONSE --> 23!/6! 2! 2!= 4.151586701E12
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20:57:31 ** There are two possible settings for the first switch, two for the second, two for the third. The setting of one switch in independent of the setting of any other switch so the fundamental counting principle holds. There are therefore 2 * 2 * 2 = 8 possible setting for the three switches. COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways. INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. **
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RESPONSE --> I think that I entered a different problem. I entered problem number 25 which is n!/(n-r)! where n= 23 and r= 4
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20:59:32 query 11.2.27 If no two adjacent switches are off why does the fundamental counting principle not apply?
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RESPONSE --> it would be impossible to set the problem up to give the correct answer 3*2*1 would not be correct.
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20:59:44 ** The reason is that the Fund. Counting Principle requires that the events be independent. Here we have the state of one switch influencing the state of its neighbors (neither neighbor can be the same as that switch). The Fund. Counting Principle requires that the events be independent. **
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21:00:31 query 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}?
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RESPONSE --> 2*3*3= 18
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21:00:36 ** Using the box method: 1st can be any of the three so the first number of possibilities is 3 2nd number can also be any of the three so the second number of possibilities is 3 The last digit must be odd, so there are only 2 choices for it. We therefore have 3*3*2=18 possible combinations.
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21:02:01 query 11.2.50 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups
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RESPONSE --> This question was not on my homework, but I think what you are looking for is 10*4*6*3= 720
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21:02:05 ** A setup consists of a guitar, a case, an amp and a processor. There are 10 choices for the guitar, 4 for the case, 6 for the amp and 3 for the processor. So there are 10 * 4 * 6 * 3 = 720 possible setups. **
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21:03:16 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I think a couple of and questions may have been off from what the book had, but I was able to figure them out.
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