#$&* course 152 9/10 5 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution:: ** We can list the possibilities, or we can analyze the numbers. First we list: Using letters for the names, there are 12 possibilities (note that the secretary must be either k or e, the others chosen from a, b, d. The secretary, president and treasurer are listed in said order): eab, ead, eba, ebd, eda, edb, kab, kad, kba, kbd, kda, kdb. Next we analyze the numbers: There are two women, so two possibilities for the first person selected. The other two will be selected from among the three men, so there are 3 possibilities for the second person chosen, leaving 2 possibilities for the third. The number of possibilities is therefore 2 * 3 * 2 = 12. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.1.12 and 18. In how many ways can the total of two dice equal 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 4 1,4 and 2,3 amd 4,1 and 3,2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** Listing possibilities on first then second die you can get 1,4, or 2,3 or 3,2 or 4,1. There are Four ways. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* question: In how many ways can the total of two dice equal 11? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 2 5,6 and 6,5 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** STUDENT SOLUTION AND INSTRUCTOR RESPONSE: There is only 1 way the two dice can equal 11 and that is if one lands on 5 and the other on 6 INSTRUCTOR RESPONSE: There's a first die and a second. You could imagine that they are painted different colors to distinguish them. You can get 5 on the first and 6 on the second, or vice versa. So there are two ways. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: In order to answer this question you first need to draw a star 10 You get 5 small and 5 larger triangles confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:: ** 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. STUDENT COMMENT I am sorry but I cannot see but 8 triangles the 5 spikes , but only 3 larger triangles that incorporate two spikes and the vertex just as you discussed. What am I missing????? INSTRUCTOR RESPONSE: I know your work well, and you are seldom wrong, so I went back to take another look, just to be sure. The figure fools the eye, and it fooled mine enough that I was just about convinced. I clearly saw just three larger triangles and couldn't make myself see five. So I started out my response believing that I had a long-standing error in the solution to this problem. And it took me awhile to convince myself that I was really right in the first place. Here's the reasoning that led me back to my original solution. My eye still doesn't really want to believe it, but I can draw the picture, and if I look at one vertex of the pentagon at a time, I can see it. I'm more convinced than ever that we can't believe our eyes. Here goes. You might want to draw the picture in order to follow the labels: If the points of the 'star' are A, B, C, D and E, in order as we go around the 'star', then each of these points is connected to exactly two of the others, and no point is connected to either of its 'nearest neighbors'. So A is connected to C and D B is connected to D and E C is connected to A and E D is connected to A and B E is connected to B and C. This gives us 10 line segments, namely AC, AD, BD, BE, CE, CA, DA, DB, EB and ED, each potentially the side of a triangle. However these 10 segments are redundant, as follows: AC AD BD BE CA (same segment as AC) CE DA (same segment as AD) DB (same segment as BD) EB (same segment as BE) EC (same segment as CE). Thus we have only five segments connecting the points of the star. Each of these segments forms the longest side of an iscosceles triangle, two of whose vertices are the two points of the 'star', and the third of which is a vertex of the pentagon. The pentagon has five vertices. If you look at each vertex in turn, you will see that it is the vertex of a triangle. So in addition to the five 'points' on the star, there are five triangles, one for each vertex. This raises the total number of triangles to 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I knew this one because I had worked this type of problem before ------------------------------------------------ Self-critique rating #$&* question 11.1.40 4 x 4 grid of squares, how many squares in the figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 4 x 4 grid which has 16 small 1x1 squares and 9 larger 2x2 squares and 4 3x3 squares and 1 4x4 square So to find the answer just add 16+9+4+1=30 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** I count 16 small 1 x 1 squares, then 9 larger 2 x 2 squares (each would be made up of four of the small squares), 4 even larger 3 x 3 squares (each made up of nin small squares) and one 4 x 4 square (comprising the whole grid), for a total of 30 squares. Do you agree? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.1.50 In how many ways can 30 be written as sum of two primes? Your Solution: 11+19=30 13+17=30 23+7=30 30 can be written 3 ways using primes confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: **STUDENT SOLTION AND INSTRUCTOR COMMENT: There are 4 ways 30 can be written as the sum of two prime numbers: • 30 = 29 + 1 (instructor note: this is not a sum of two primes) • 30 = 19 + 11 • 30 = 23 + 7 • 30 = 17 + 13 INSTRUCTOR COMMENT: Good, but 1 isn't a prime number. It only has one divisor. The rest of your answers are correct. All sums give you 30, and 7, 11, 13, 17, 19 and 23 are all prime numbers.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.1.60 four adjacent switches; how many settings if no two adj can be off and no two adj can be on YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 2 because there are only 4 switches and no two adj to each other can be on at the same time and no two adj can be off at the same time so you get on off on off, or off on off on confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** There are a total of 16 settings but only two have the given property of alternating off and on. If the first switch is off then the second is on so the third is off so the fourth is on. If the first is off then the second is on and the third is off so the fourth is on. So the two possibilities are off-on-off-on and on-off-on-off. If we use 0 to represent ‘off’ and 1 to represent ‘on’ these possibilities they are written 0101 and 1010. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question Add comments on any surprises or insights you experienced as a result of this assignment. ** STUDENT COMMENT: No surprises and it's early so i'm reaching for insight as a child reaches for a warm bottle of milk Your comments or questions: This section really made me think outside of the box, but I have worked problems like these before and feel confident in my ability to be successful in this course Some previous student comments and questions: 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: The question you ask about the Cartesian Plane is a good one and I’ll be glad to answer below, but first let me address your request for answers to all questions. • 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 the comments provided here, most students get enough feedback to be confident in the rest of their work. Another reason I don’t provide answers to all questions is that want students to learn to work ‘forward’ through the problems, which won’t be the case if all answers are provided. If they were most students would fall into the habit of 'working backward' from the answer to the solution. • If you have a question on a specific text problem or on anything else, you should submit it using the Submit Question form from the General Information page http://vhcc2.vhcc.edu/dsmith/geninfo/. A direct link to the question form is http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm . To answer your question about the Cartesian Plane: The Cartesian Plane can be thought of as a plane defined by an x axis and a y axis, on which you can specify points by their coordinates. For example the point (5, 9) can be found by starting at the origin (0, 0) and moving 5 units along the positive x axis, then moving 9 units in the direction parallel to the positive y axis. The above is probably sufficient for the work you are doing at this point in your course. A more specific definition: The x axis and the y axis are mutually perpendicular (i.e., at right angles with one another). The x axis is traditionally oriented in the horizontal direction, the y axis in the vertical direction, and is right-handed. The idea of right-handedness is defined in 3 dimensions, but if the x axis is directed ‘toward the right’ and the y axis is ‘up’, then the system will be right-handed. If the x axis was directed ‘toward the right’ and the y axis was ‘down’ the system would be left’-handed’. In a right-handed system, if the y axis is rotated 90 degrees in what we perceive as the ‘right-hand’ direction for rotation (the direction in which you turn a standard right-handed screw or bolt to tighten it), it will then coincide with the x-axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 002. `query 2 ------------------------------------------------ Self-critique rating #$&* question 11.2.12 find 10! / [ 4! (10-4)! ] without calculator YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 10! / [4! (10-4)!] First thing I will have to factor the numbers n!=n(n-1)(n-2)…2*1 Ensure you factor properly 10! = 10*9*8*7*6*5*4*3*2*1=3628800 / [4! (10-4)!] = 4*3*2*1(6!) = 4*3*2*1(6*5*4*3*2*1) =24*720=17280 3628800/17280= 210 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** Starting with 10! / [ 4! * (10-4) ! ], we replace (10 – 4) by 6 to get 10! / ( 4! * 6! ). Writing out the factorials our expression becomes 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / [ ( 4 * 3 * 2 * 1) * ( 6 * 5 * 4 * 3 * 2 * 1) ] . The numerator and denominator could be multiplied separatedly then divided out but it's easier and more instructive to divide out like terms. Dividing ( 6 * 5 * 4 * 3 * 2 * 1) in the numerator by the same expression ( 6 * 5 * 4 * 3 * 2 * 1) in the denominator leaves us 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1). Every factor of the denominator divides into a number in the numerator without remainder: • Divide 4 in the denominator into 8 in the numerator, • divide 3 in the denominator into 9 in the numerator and • divide 2 in the denominator into 10 in the numerator and you end up with • 5 * 3 * 2 * 7 = 210. NOTE ON WHAT NOT TO DO: You could figure out that 10! = 3628800, and that 4! * 6! = 24 * 720 = 16480, then finally divide 3628800 by 16480. But that process would lose accuracy and be ridiculously long for an expression like 100 ! / ( 30! * 70!). For this calculation the numbers involved would be hundreds of digits long. Much better to simply divide out like factors until the denominator goes away, as it always will with expressions of this form. (form n ! / (r ! * (n – r)! ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did mine in a long division problem and got the correct answer but I do see where factoring like expressions will be a necessity on larger numbers ------------------------------------------------ Self-critique rating #$&* question 11.2.31 (10th edition #25) 3 switches in a row; fund count prin to find # of possible settings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since there are no restrictions they can be either on or off which says for the first there are 2 choices also for the second and the third. 2*2*2=8 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** There are two possible settings for the first switch, two for the second, two for the third. The setting of one switch is independent of the setting of any other switch so the fundamental counting principle holds. There are therefore 2 * 2 * 2 = 8 possible setting for the three switches. COMMON ERROR: There are six possible settings and I used fundamental counting principle : first choice 3 ways, second choice 2 ways and third choice 1 way or 3 times 2 times 1 equals 6 ways. INSTRUCTOR CRITIQUE: You're choosing states of the switches and there are only two states on each. No single choice has 3 possibilities. Also the setting of each switch is independent of the settings of the others, so the number of possibilities on one choice is independent of the number of possibilities on the other. Each of the 3 choices therefore has 2 possibilities. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.2.31 (10th edition #27) If no two adjacent switches are off why does the fundamental counting principle not apply? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The fundamental Counting Principle states that each event has to be independent of the other and this is not the case no two can be off adj to the other which makes one dependent on another. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** The reason the principle doesn’t apply is that the Fund. Counting Principle requires that the events be independent. Here the state of one switch influences the state of its neighbors (neither neighbor can be the same as that switch). So the choices are not independent. The Fund. Counting Principle requires that the events be independent. Since this is not the case the principle does not apply. ** STUDENT QUESTION I thought for sure I had this right, but apparently I am confused. A switch is considered independent when it is concerned with the placement of the other switches. It is dependent if it is not concerned. These terms seem backward to me. INSTRUCTOR RESPONSE The given condition is that no two adjacent switches can be off. So for example if the first switch is off, the second switch cannot also be off. Or if the middle switch is off, neither of the other two can be off. Thus in some circumstances the state of one switch has an influence on the state of the other switches, meaning that its state is dependent on the state of the other switch, and the Fundamental Counting Principle does not apply. I believe that in this case, you did simply reverse the meanings of 'dependent' and 'independent'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.2.36 How many odd 3-digit #'s from the set {3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: First I need to find out how many numbers can go in as the first number 3 and then the second can be 3 but the third number can only be 2 because I am trying to find out how many 3 digit odd numbers I can make from the set {3,4,5} 3*3*2=18 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: ** Using the box method, where in this case we draw three boxes and put one number in each: The 1st box can be filled with any of the three so the first number of possibilities is 3 The 2nd box can also be filled with any of the three so the second number of possibilities is 3 The last digit must be odd, so there are only 2 choices for the third box. By the Fundamental Counting Principle we therefore have 3*3*2=18 possible combinations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* question 11.2.56 (10th edition 50) 10 guitars, 4 cases, 6 amps, 3 processors; # possible setups YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: First I find out how many choices I have for each element in the setup which contain a guitar, a case, an amp, and a processor 10*4*6*3 720 possibilities confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:: A setup consists of a guitar, a case, an amp and a processor. There are 10 choices for the guitar, 4 choices for the case, 6 choices for the amp and 3 choices for the processor. So by the Fundamental Counting Principle there are 10 * 4 * 6 * 3 = 720 possible setups. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): You may add comments on any surprises or insights you experienced as a result of this assignment: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I found this section pretty easy to do but I must admit this is not the first time I’ve seen it. "