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# In other words the general point of b is described by

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In other words, the general point of B is described by the five param- eters a , b , x 2 , x 4 , x 6 , and so B is five dimensional. Therefore the correct answer is (f) . square Solution of problem 1.6: Cleraly D can be distorted into O by rounding- off the corners on the left side of D . Thus, the correct answer must be (b) . We can also easily rule out the other answers directly. Indeed, if we delete a point from the bottom part of one of the legs of A , we will get two connected pieces. But removing any point from either D or O we get a single piece. Therefore A can not be distorted into either D or O . Similarly, removing a point from M results in two pieces, and so M can not be distorted into either D or O . Finally, if we delete the top apex of A we will get one piece, whereas when we remove any point from M we will get two pieces. So A can not be distorted into M . The correct answer is (b) . square 14

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Solution of problem 1.7: If we cut the two hole pretzel along the indi- cated circle we will obtain two connected pieces each with one hole and one boundary. Thus the correct choice is (d) . square Solution of problem 1.8: (1) A connected planar graph has Euler char- acteristic 2. So if the graph has 54 vertices, 177 edges, and F faces, then it follows that 54 - 177 + F = 2, or that F = 125. Therefore (1) is True. (2) A planar graph with 7 vertices, 4 edges, and 7 faces has Euler characteristic 7 - 4 + 7 = 10. By the Euler characteristic theorem such a graph must have 10 - 1 = 9 connected components. Therefore (2) is False. (3) If a planar graph has 2 connected pieces then it has Euler charac- teristic 3. If the graph also has 12 edges, 7 vertices, and F faces, then 3 = 7 - 12 + F . In other words, we must have that F = 8. Therefore (3) is True. The correct answer is (b) . square Solution of problem 1.9: Since the pond can sustain 300 fish and the ini- tial population is 1200 fish, it follows that the initial population density is P 1 = 1200 300 = 4 . Similarly, the population density after two years is P 2 = 600 300 = 2 . The Verhulst model gives the relation P 2 P 1 = 1 + c (1 - P 1 ) , 15
or equivalently 2 4 = 1 - 3 c. This gives c = 1 6 Applying againthe Verhulst model we get P 3 P 2 = 1 + c (1 - P 2 ) , or equivalently P 3 2 = 1 + 1 6 (1 - 2) . Solving for P 3 we get P 3 = 5 / 3. If x is the fish population in three years, then x obeys x/ 300 = 5 / 3, and so x = 500. The correct answer is (d) . square Solution of problem 1.10: If we roll three numbers that add up to 6, none of those numbers can be bigger than 4. Furthermore if one of the numbers is 4 then the other two must be 1 and 1. Similarly, if one of the numbers is 3, then the other two must be 1 and 2. Finally, if none of the numbers is equal to 3 or 4, then all three numbers must be equal to 2. So the outcomes of the roll that show three numbers adding up to 6 are 411, 141, 114, 312, 321, 132, 231, 123, 213, 222. Since the total number of outcomes of the roll is 6 · 6 · 6 = 216 the probability of getting three numbers that add to 6 is 10 / 216 = 5 / 108.

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