MA Exam 2 - Spring 2007

Transcription

1 MA Exam 2 - Spring 2007 Call me. Directions: Answer the following questions. Show all work. An answer with no work receives NO credit. Saying I did it on my calculator does NOT constitute work. Guessing and checking is not an acceptable problem-solving method. (2 points each) Answer questions 1-5 with either TRUE or FALSE. 1. FALSE The equation (3 4) 5 = (4 3) 5 illustrates the associative property of multiplication. 2. TRUE Consider the group formed by all rotations and all re ections of a square that leave the square in the same relative position. This is an example of a noncommutative group. 3. FALSE? = f?g 4. FALSE If two sets are equivalent, then they are equal. 5. FALSE Writing the set of all counting numbers less than or equal to 5 as f1; 2; 3; 4; 5g is an example of writing a set using set-builder notation. (2 points each) For questions 6-10, choose the letter that corresponds to the best answer. 6. C Which of the following sets is in nite? (a) the number of people in this room (b) the number of stars in the universe (c) the set of even counting numbers (d) the set of odd counting numbers less than 1,000,000,000,000,000,000,000,000 (e) the number of times the Cubs have won the World Series 7. D Let A = findiana, Iowa, Kentucky, Missouri, Wisconsing : (These are the states that have a land border with Illinois.) Which of the following statements is NOT true? (a) New York 62 A (b)? A (c) Illinois 2 A 0 (d) fiowag 6 A (e) findiana, Iowa, Kentucky, Missouri, Wisconsing A 8. E Which of the following is NOT a requirement for a group? (a) the operation is associative (b) the operation is closed (c) there is an identity (d) each element has an inverse (e) the operation is commutative 9. B The letters A, B, C, D, E, F, G are written in this order, over and over again. (So, the sequence starts out ABCDEFGABCDEFGABC...) What is the 4861st letter in the sequence? (a) A (b) C (c) D (d) F (e) G 10. E At a fast food restaurant, you can order a hamburger with or without any of the following condiments: ketchup, mustard, mayonnaise, pickles, onions, tomato. How many possible ways are there to order your hamburger? (a) 6 (b) 12 (c) 32 (d) 63 (e) 64

2 11. (4 points each) Let U = fa; b; c; d; e; f; g; h; i; j; kg ; A = fa; c; d; f; g; ig ; B = fb; c; d; f; gg ; C = fa; b; f; i; jg. Find: (a) A 0 = fb; e; h; j; kg (b) B \ C 0 (c) ja [ Bj (d) (C \ B) [ (A \ B 0 ) C 0 = fc; d; e; g; h; kg B \ C 0 = fc; d; gg A [ B = fa; b; c; d; f; g; ig ja [ Bj = 7 C \ B = fb; fg B 0 = fa; e; h; i; j; kg A \ B 0 = fa; ig (C \ B) [ (A \ B 0 ) = fa; b; f; ig 12. (12 points) Using Venn diagrams, determine if A[(B [ C) 0 = (A [ B 0 )\(A [ C 0 ) : Be sure to clearly shade in your nal region for each set. A B A [ (B [ C) 0 - Everything shaded at least once is nal region C A B (A [ B 0 ) \ (A [ C 0 ) - Double-hatched region is nal region C 13. (16 points) A survey of 400 homes was done, and the following information was collected. 133 had blacktop driveways.

3 142 had under one acre of land. 135 had city water. 50 had a blacktop driveway and under one acre of land. 64 had a blacktop driveway and city water. 61 had under one acre of land and city water. 29 had a blacktop driveway, under one acre of land, and city water. Blacktop 1 acre City Water (a) How many homes had under one acre of land and city water, but not a blacktop drive-way? 32 (b) How many homes had only city water? 39 (c) How many homes had a blacktop driveway or city water? 204 (d) How many homes had a blacktop driveway or under one acre of land, but not city water? 129 (e) How many homes had exactly two of these items? 88 (f) How many homes had none of these items? (10 points) Assuming the set fv, W, X, Y, Zg is a commutative group under the given operation, ll in the missing entries in the operation table. V W X Y Z V Z V W X Y W V W X Y Z X W X Y Z V Y X Y Z V W Z Y Z V W X 15. (10 points) Suppose is an operation de ned on the positive integers, meaning the only numbers we plug in for a and b are positive integers. De ne a b = 4a 7b:

4 (a) Compute 4 2 and 2 4: 4 2 = 4 (4) 7 (2) = = 4 (2) 7 (4) = 20 (b) Is closed over the set of positive integers? Why or why not? No, because we took two positive integers, 2 and 4, and got a negative answer out. (c) Is commutative? Why or why not? No, because 4 2 6= 2 4 (d) Compute 1 (2 3) and (1 2) 3: 1 (2 3) = 1 (4 (2) 7 (3)) = 1 ( 13) = 4 (1) 7 ( 13) = 95 (e) Is associative? Why or why not? No, because 1 (2 3) 6= (1 2) 3: (1 2) 3 = (4 (1) 7 (2)) 3 = 10 3 = 4 ( 10) + 7 (3) = (16 points) Answer the following about UPC s and ISBN s. (a) Is the UPC correct? Why or why not? 3 (0) + 1 (4) + 3 (2) + 1 (1) + 3 (9) + 1 (6) + 3 (3) + 1 (4) + 3 (9) + 1 (0) + 3 (3) + 1 (2) = 95 Since 95 is not divisible by 10, this UPC is not correct. (b) Find the check digit for the UPC ?. We need 3 (0) + 1 (7) + 3 (1) + 1 (9) + 3 (4) + 1 (6) + 3 (4) + 1 (0) + 3 (0) + 1 (2) + 3 (4) + 1 (?) to be a multiple of 10. Since 3 (0) + 1 (7) + 3 (1) + 1 (9) + 3 (4) + 1 (6) + 3 (4) + 1 (0) + 3 (0) + 1 (2) + 3 (4) + 1 (?) = 63+? We need? = 7: So, the check digit is 7. (c) Is the ISBN X correct? Why or why not? 10 (0) + 9 (4) + 8 (1) + 7 (0) + 6 (6) + 5 (3) + 4 (1) + 3 (9) + 2 (9) + 1 (10) = 154 Since 154 is divisible by 11, this ISBN is correct.

5 (d) Find the check digit for the ISBN ?. We need 10 (0) + 9 (6) + 8 (1) + 7 (8) + 6 (5) + 5 (1) + 4 (4) + 3 (7) + 2 (2) + 1 (?) to be a multiple of 11. We have 10 (0) + 9 (6) + 8 (1) + 7 (8) + 6 (5) + 5 (1) + 4 (4) + 3 (7) + 2 (2) = 194 When we divide 194 by 11, we get 17 with a remainder of 7. So, we see that if we add 4 to 194, we will get a multiple of 11. So, the check digit is 4.