Thinking Mathematically by Robert Blitzer Thinking

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Thinking Mathematically by Robert Blitzer Thinking Powered By Docstoc
					  Thinking
Mathematically
       MATH1101
 Comprehensive Mathematics
    Homework Solutions
        Chapter 1
Problems
             Chapter 1, Section 1
     In Exercises 1-30, identify a pattern in each list of numbers.
     Then use this pattern to find the next number. (More than
     one pattern might exist, so it is possible that there is more
     than one correct answer.)

     1. 8, 12, 16, 20, 24, ____       2. 19, 24, 29, 34, 39, _____
     3. 37, 32, 27, 22, 17, _____     4. 33, 29, 25, 21, 17, _____

1.    Answer: 28 (keep adding 4 to the previous.)
2.    Answer: 44 (keep adding 5 to the previous.)
3.    Answer: 12 (keep subtracting 5 from the previous.)
4.    Answer: 13 (keep subtracting 4 from the previous.)
Chapter 1, Section 1
                        square,
                        triangle,
                        circle...


                          rotate 90º
                          counterclockwise

           d d d      alphabetical: a, b, c, d, ... and
           d d        add an additional letter each
                      time.

                   Alternate triangles and squares.
                   Add little line going down from
                   bottom each time.
           Chapter 1, Section 1




35.
      5      12          100    2          n
      20     48          400    8          4n
      28     56          408    16         4n+8
      14     28          204    8          2n+4
      10     24          200    4          2n
      four examples :                algebraic solution:
      By inductive reasoning:        By deductive reasoning:
         twice the number            twice the number
           Chapter 1, Section 1




36.
      5      12 100      2               n
      15     36 300      6               3n
      21     42 306      12              3n+6
      7      14 102      4               n+2
      2       2 2        2               n+2 – n = 2
         four examples : by    algebraic solution: by
      inductive reasoning: 2   deductive reasoning: 2
           Chapter 1, Section 1




37.
      5      12          100    2             n
      10     17          105    7             n+5
      20     34          210    14            2n+10
      16     30          206    10            n+3
      8      15          103    5             n+3 – n = 3
      3       3             3   3
         four examples : by          algebraic solution: by
      inductive reasoning: 3         deductive reasoning: 3
           Chapter 1, Section 1




38.
      5      12          100    2              n
      8      15          103    5              n+3
      16     30          206    10             2n+6
      20     34          210    14             2n+10
      10     17          105    7              n+5
       5      5             5   5              5
         four examples : by          algebraic solution: by
      inductive reasoning: 5         deductive reasoning: 5
           Chapter 1, Section 1
                                          51. The problem states that in
                                          general, the sum of the
                                          numbers from 1 to n is equal to
                                          n( n  1)
                                                    . So if n is 100 and the
                                              2
                                          formula holds, the sum up to
                                                            100(101)
                                          100 would be               .
                                                                2
                                          This is deductive reasoning.



52. The HMO studied only 200 patients . The
conclusion drawn generalizes for the public in general
on the basis of these 200 patients. Therefore this is
inductive reasoning.
Chapter 1, Section 1
         53. We are generalizing
         about all college students
         on the basis of a sample
         of 1200 of them. This is
         inductive reasoning.
        Chapter 1, Section 1


54. We are reasoning that because the policy
exists and because I did turn my report in a
day late, that my grade will be reduced by
one letter grade. This is deductive reasoning.
           Chapter 1, Section 2




The key here is to NOT use a calculator. By rounding
off, the six items cost about $3, $6, $20, $2, $12 and $0.
Adding these numbers together in your highly
mathematical brain, you get $43. So these items cost
around $43. If you whip out your calculator, they come
to exactly $43.45. Not bad for a quick guess.
            Chapter 1, Section 2




24. $4 + $8 + $29 + $4 + $13 + $1 = $59. Calculator: $59.22
25. Estimate 40 hours per week, $20 and 50 weeks per year. That
    gives us around $40,000 per year. Calculator: $40,560.
26. 40 ∙ $30 ∙ 50 = $60,000 per year. Calculator: $62,088
27. Estimate $600 month, 3 years and 10 months a year. You get
    about $18,000, but clearly more than that. Calculator: $21,780.
              Chapter 1, Section 2



31. About $60,000 per year, about 40 ∙ 50 = 2000 hours per year. That
    gives us about $30 per hour. Calculator: $29.57.
32. About $40,000 per year, about 40 ∙ 50 = 2000 hours per year. That
    gives us about $20 per hour. Calculator: 18.73 per hour.
33. If there about 400 days in a year and about 25 hours in a day, there
    are about 10,000 hours in a year. About 80 years would be about
    800,000 hours. Calculator: 701,676 (not including leap years.)
34. Similarly, about half the years, about half the hours, about 400,000
    hours. Calculator: 353,028.
        Chapter 1, Section 2




43. About half of approximately 200 million Americans.
    So, about 100,000,000. (100 million).
44. About ¼ of approximately 200 million Americans. So,
    about 50,000,000. (50 million).
         Chapter 1, Section 2




47. a) From 2002-2004, two years, TV went from $222 to $248 or up by
$26. That averages to $13 a year.
    b) Assuming a steady rise, the cost in 2010, six years after 2004,
would be about $248 + 6 ∙ $13 = $248 + $78 = about $316.
48. a) By similar reasoning, $172 - $135 = $37. About $18.50 a year.
    b) In 2010, $172 + 6 ∙ $18.50 = $172 + $111 = about $283.
              Chapter 1, Section 2




49. a) 1960. a bit above 4000 cigarettes, maybe 4100 to 4200.
    b) Looks like 1940 to 1950 was the greatest increase.
    c) Around 1930, consumption was around 1500 cigarettes.
50. a) Minimum is clearly in 1910. Seems to be around 100 cigarettes.
    b) Greatest drop-off from 1980 to 1990
    c) Around 700 to 800 cigarettes in 1920.
              Chapter 1, Section 2




51. a) In 1998: about $13,800. In 2005, (7 years) about $20,000.
$6,200/7 is about $900.
     b) C ≈ $14,000 + 900x, where x is years after 1998.
     c) C ≈ $14,000 + 900 ∙ 14 ≈ $26,600

				
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