course Mth 152 I had some confusion with the problem numbers you actually wanted us to perform on this query for Assignments 1 and 2 of Chapter 11 Sections 11.1 and 11.2. Could you please clarify why the numbers are not matching up with the correct problem. ??b??~??·?·?}{????????·assignment #001001. `query 1
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18:01:13 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 --> 2 women * 1 position = 2 3 men * 2 positions = 6 2 *6 = 12 possibilities ACB ACD BCD BCA DCA AEB AED BEA DEA BCD DCB BED DEB
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18:01:34 ** 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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RESPONSE --> My answer is correct.
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18:04:37 query 11.1.12,18 In how many ways can the total of two dice equal 5?
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RESPONSE --> Q012 4 possible ways (1,4) (2,3) (3,2) (4,1) Q018 2 possible ways (5,6) (6,5)
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18:05:03 ** 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 --> My answer is correct.
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18:05:37 In how many ways can the total of two dice equal 11?
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RESPONSE --> 2 possible ways (5,6) (6,5)
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18:05:54 ** 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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RESPONSE --> My answer is correct.
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18:08:19 query 11.1.36 5-pointed star, number of complete triangles How many complete triangles are there in the star and how did you arrive at this number?
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RESPONSE --> 5+5=10 I counted the 5 outside small triangle and then each of the small triangle also formed a larger inside triangle.
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18:09:03 ** 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 --> My answer is correct.
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18:09:48 query 11.1.40 4 x 4 grid of squares, how many squares in the figure?
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RESPONSE --> There are 25 squares. 16+4+4+1=25
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18:09:48 query 11.1.40 4 x 4 grid of squares, how many squares in the figure?
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RESPONSE -->
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18:11:49 ** 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 --> Yes , I agree, When I counted I left out the 4 3x3 squares.
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18:12:41 query 11.1.50 In how many ways can 30 be written as sum of two primes?
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RESPONSE --> 3 possible ways: 23+7=30 19+11=30 17+13=30
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18:13:20 **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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RESPONSE --> 1 is not a prime number, there should only be 3 ways.
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18:14:02 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 different settings. 0-1-0-1 1-0-1-0
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18:14:19 ** 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 --> My answer is correct.
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18:16:12 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I could not come up with the correct answer to question 65 according to the book answer of 883. This was the question that required you to look at the match sticks that formed a rectangular grid. Could you please explain the correct process for obtaining this answer.
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18:17:30 ** STUDENT COMMENT: No suprises and it's early so i'm reaching for insight as a child reaches for a warm bottle of milk I would like the answers to all the problems I worked in Assignment 11.1. I was surprised that you only ask for a few. I could not answer 11.1. 63 - What is a Cartesain plane? I could not find it in the text. INSTRUCTOR RESPONSE: I ask for selected answers so you can submit work quickly and efficiently. I don't provide answers to all questions, since the text provides answers to most of the odd-numbered questions. Between those answers and and comments provided here, most people get enough feedback to be confident in the rest of their work. Also I don't want people to get in the habit of 'working backward' from the answer to the solution. If you want to send in your work on other problems, including a full descripton of your reasoning, I'm always glad to look at them. You would have to make those problems self-contained (tell me enough about the problem so I know what the problem is), since I don't always respond from the place where I have my copy of the text. The Cartesian Plane is a plane defined by an x axis and a y axis, on which you can specify points by their coordinates. **
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RESPONSE --> ok
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?·???F?????a?}??z·? assignment #002 002. `query 2 Liberal Arts Mathematics II 01-25-2009
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18:26:30 query 11.2.12 find 10! / [ 4! (10-4)! ] without calculator
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RESPONSE --> This question is not problem 12, it is problem 14. Since I am unsure which you wanted because 14 was not an assigned problem I am submitting both. q012 6! / (6-4)! = 6! / 2! = 6*5*4*3*2! / 2! = 360 q014 10! / 4! (10-4)! = 10! / 4! (6)! = 10! / 24! = (10! / 24!) / 2 = 5! / 12! = 5! / 12*11*10*9*8*7*6*5! = 1/3,991,680 I am unsure about the answer for question 14.
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18:29:40 ** 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 --> My answer on question14 was wrong, I do now understand the process I should have performed.
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18:34:33 query 11.2.25 3 switches in a row; fund count prin to find # of possible settings
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RESPONSE --> Question 25 is not the question that is listed. I am unsure what question the listed question is, it does in some ways resemble question number 33 but I am not sure. Question 25 is : n! / (n-r)!, whwere n=23 and r=10 23! / (23-10)! = 23! / 13! = 23*22*21*20*19*18*17*16*15*14*13 / 13! = 4.1515867008*10^12
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18:35:20 ** 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 --> Please let me know if the actual question 25 is correct.
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18:38:38 query 11.2.27 If no two adjacent switches are off why does the fundamental counting principle not apply?
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RESPONSE --> This is question number 33: The answer to this question is: The fundamental counting principle does not apply to the statement: Assume that no two adjacent switches, on a panel containing three on/off switches in a row, can both be off. It does not apply because it contains a restriction that ""no two adjacent switches can both be off"", therefore it does not satisfy the uniformity criterion.
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18:38:56 ** 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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RESPONSE --> My answer is correct.
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18:40:04 query 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}?
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RESPONSE --> This is question number 42: 2*3 = 6
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18:41:41 ** 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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RESPONSE --> My answer was incorrect, I do see the correct way to get the answer to this problem.
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18:43:54 query 11.2.50 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups
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RESPONSE --> This is question 56: 10*4*6*3 = 720 possible setups
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18:44:03 ** 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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RESPONSE --> My answer is correct.
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18:45:35 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I am unclear on why the numbers listed on this program do match those in the textbook. On several questions I was unsure of which question to answer. Could you please clarify?
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