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   M 4/3/06 
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   Classes resume. 
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   T 4/4/06 
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   HW due:
  §11-6 #25, 26; §11-7 #1, 3, 5, 9. 
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   W 4/5/06 
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   HW due:
  First, finish up yesterday’s HW if you have not already done so. Make sure that
  your answers follow the 3-step “formula/plug-ins/result” standard that we
  discussed. If you did some of this during class yesterday, so much the
  better. Then write §11-8 #2-24 even, 25, 26. Show work, but you may find your
  calculator’s “!” symbol (found under MATH PRB 4) to be helpful for checking
  your answers. Problems 17-23 odd are done below to help you see what is
  required. 
   
    
   
   
   
   
   
   
   
   
   
   
   
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   Th 4/6/06 
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   HW due:
  Read §11-10; write §11-10 #4-52 eoe, plus 53. 
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   F 4/7/06 
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   “Practice Test” on Chapter 11. 
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   weekend 
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   Please visit the following
  links in order to learn from your mistakes and prepare for Monday’s test. Smart student hint: Use the other
  period’s test to give yourself extra practice. Find a quiet room and work the
  problems under time pressure. The format is similar, but you will notice that
  the problems are switched around and altered a bit. 
   
  E period: 
  -- Blank copy 
  -- Answer key with statistics 
   
  F period: 
  -- Blank copy 
  -- Answer key with statistics 
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   M 4/10/06 
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   “Real Test” on Chapter 11. This test will be similar to Friday’s test and will
  begin after 5-10 minutes of Q&A at the start of the period. The higher of
  your two scores will be recorded. If you missed school on Friday, even for an
  excused absence, then today’s test will be the one that counts. Please learn
  as much as you can from Friday’s test! 
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   T 4/11/06 
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   No class (Diversity Day). 
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   W 4/12/06 
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   HW due:
  Read §§12-1 and 12-2; write §12-1 #1-13 odd, §12-2 #1. Check your answers for
  accuracy. Show work in the manner illustrated below. 
   
  2. P(total ³ 10) = P(total
  = 10, 11, or 12) = P( (5, 5) or (4,
  6) or (6,4) or (5, 6) or (6, 5) or (6, 6) ) = 6/36 = 1/6 
   
  8. P(3 < total < 7) = P(total = 4, 5, or 6) = P( (2, 2) or (1, 3) or (3, 1) or (1,
  4) or (4, 1) or (2, 3) or (3, 2) or (3, 3) or (2, 4) or (4, 2) or (1, 5) or
  (5, 1) ) = 12/36 = 1/3 
   
  10. P(total = 13) = P(impossible outcome) = 0 
   
  2(h) in second set: P(penny T or
  nickel T) = 1 – P(complementary
  event) = 1 – P(neither is tails) 
  = 1 – P(penny H and nickel H) = 1 –
  (½ · ½) = 3/4 
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   Th 4/13/06 
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   HW due:
  Read §12-3; write §12-3 #1-13 odd. You may show less writing this time than
  you did yesterday. Hint: You can
  use the symbols Ù, Ú, and ~ to stand for
  “and,” “or,” and “not,” respectively. For example, in #4, you could show your
  work as follows: 
   
  4a. # of ways(dog Ú cat) = #(dog) + #(cat) = 37 + 15 = 52 
  4b. #of ways(dog Ù cat) = #(dog) · #(cat) = 37 · 52 = 555 
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   F 4/14/06 
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   HW due:
  Read §12-4 plus pp. 651-652 of §12-5; write §12-4 #9, 11, 15, 21, §12-5 #17,
  20, 22, 26, 27. 
   
  Bonus (2 points): Compute the probability
  of a full house on the deal in 5-card draw poker. Be prepared to explain your
  work and your answer. Write your answer in terms of permutations and
  combinations. For example, the earlier problem we analyzed could be written
  as P(two pair) = 13C2 · 4C2 · 4C2 · 44C1/(52C5). 
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   M 4/17/06 
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   HW due:
  Read §12-6; write §12-6 #1, 3, and the following probability problem. 
   
  Probability problem: A fair coin is
  flipped 10 times. What is the probability that a run of at least 4 heads in
  row or 4 tails in a row occurs? Write up your work and your answer. We will
  compare answers in class. 
   
  Analytic solution to Friday’s bonus:
  (13P2)(4C2)(4C3)/(52C5) = 3744/2,598,960 » .00144, approximately 3 times in every 2000 hands. 
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   T 4/18/06 
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   HW due: 
   
  1. Use a truth table to prove the tautology that ~(P Þ ~Q) Û (P Ù Q). 
   
  2. Read §13-1; write §13-1 #1-10 all. Extremely rough sketches are
  acceptable, but be sure to label your axes with English description, variable letter, and units. Also show at
  least one labeled tick mark on each axis. 
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   W 4/19/06 
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   HW due: 
   
  1. Read §13-2 twice and make reading notes that include questions. If you
  have no questions, I will administer a short quiz to verify that you
  understand everything. This assignment is required from everyone, since it
  was mentioned in class that we would be proceeding into §13-2. 
   
  2. Use a truth table to prove that ~(A
  Ú (B Ú ~C)) Û ~A Ù ~B Ù C. Since
  this was not posted before 2:30, I will grant extra credit for those who do
  this. Everyone should attempt it, however. 
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   Th 4/20/06 
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   HW due:
  Read §13-3; prepare #13-#36 for quick oral presentation; show written work
  for §13-3 #4-36 eoe. You will not
  be able to use notes or calculator during the oral presentation of #13-#36. 
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   F 4/21/06 
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   Quiz (possibly written, possibly oral) on §13-3. For example, if you were asked to compute sec 5p/3, you would say, “Hmmm, that’s asking for the
  reciprocal of the cosine of 5p/3. Since the angle’s in the fourth quadrant, just a little past
  270º, the cosine would be .5 and the reciprocal would have to be 2.” 
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   M 4/24/06 
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   Phi Beta Kappa Day (no
  school). 
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   T 4/25/06 
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   Quiz Redux.
  Anyone who did not earn a score of 10 last Friday will have another shot today
  or tomorrow. 
   
  HW due: Read blue box on p. 731,
  changing the word “An” to “The” in each case; read §13-5; write §13-3 #47,
  64; write §13-4 #16-18 all, #21-31 odd. 
   
  Important: For the problems in
  §13-4, show estimations before beginning, and show at least as much detail in
  your work as is shown below. (A circle sketch is also recommended.) 
   
  5. sec 12.3º = 1/(cos 12.3º) 
  Estimate: Since 12.3º is a small
  angle, cos 12.3º is close to 1. A good guess is .95, since we know that the
  cosine decreases very slowly until we get past about 30º. Take reciprocal to
  estimate 1/.95 » 1.05 as the answer. 
   
  Approx. by calc.: 1/(cos 12.3º) » 1/.977 » 1.023. 
   
  28. q = cot–1
  1.452 = the angle whose tangent is 1/1.452 
  Estimate: Since 1.452 » 1.5, we are talking about an angle whose tangent is
  approx. 2/3. Since tangent means “slope,” we have an angle somewhat less than
  45º. (Remember, 45º gives a slope of 1.) I would guesstimate about 35º,
  which is about .6 radians. 
   
  Approx. by calc.: tan–1
  (1/1.452) » .603 in radian mode,
  or approx. 34.555º. Then hit 2nd ANGLE 4 ENTER to convert to 34º33'.
  [DMS stands for Degrees, Minutes, Seconds. This is a fast way to convert an
  answer in “plain degrees” to degrees, minutes, and seconds.] 
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   W 4/26/06 
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   HW due:
  Patch up any missing problems in your old homework. Then work problems on pp.
  703-704 #T1abcd, T3ab, T4 all, T6a; p. 795 #C1, p. 798 #T1. 
   
  Note: On the test tomorrow, you
  will be required to solve the problems like those on p. 795 and p. 798 without
  a calculator. Practice your speed and accuracy. 
   
  In class: Review for test. 
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   Th 4/27/06 
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   Test on Chapter 12 (first 6 sections only), Truth
  Tables, and Circular Function Basics.
  Here are the answers to yesterday’s review problems to help you study: 
   
  pp. 703-704 
  T1a. 8P8 = 40320 
      b. We could have GBGBGBGB or BGBGBGBG. Either way,
  there are 4 · 4 · 3 · 3 · 2 · 2 · 1 · 1 = 576 ways to fill the slots. Total =
  1152. 
      c. 8P5
  = 6720 
      d. You must choose 3 boys and then 2 girls to be part
  of the line, then permute those children. This is a 3-step process, where
  choosing 3 boys can be done in 4C3 = 4 ways, choosing 2 girls can be done in 4C2 = 6 ways, and permuting
  the chosen children can be done in 5P5 = 120 ways. Total = 4 · 6 · 120 = 2880. 
    
  T3a. P(kid 1 muddy Ù kid 2 muddy Ù kid 3 muddy Ù . . . Ù kid 8 muddy) = P(kid
  1 muddy) · P(kid 2 muddy) · P(kid 3 muddy) · . . . · P(kid 8 muddy) = (0.3)8 » 0.00006561. 
      b. P(kid 1
  not muddy Ù kid 2 not muddy Ù . . . Ù kid 8 not muddy) = (0.7)8 » 0.0576. The reasoning is exactly the same as
  in part (a), except that we use 0.7 instead of 0.3, since P(kid does not get a muddy shirt) = 1
  – P(kid gets muddy shirt) = 1 – 0.3
  = 0.7. 
   
  T4a. 0.8 
      b. 0.7 
      c. 0.56 
      d. P(~mudface
  Ú ~mudshirt) = P(~(mudface Ù mudshirt)) by DeMorgan’s Law. The second
  probability is 1 – P(mudface Ù mudshirt) = 1 – (0.2)(0.3) = 1 – 0.06 = 0.94. 
      e. 0.06 
   
  T6ai. 0.98 
        ii. 0.97 
        iii. (0.98)(0.97) = 0.9506 
        iv. (0.02)(0.03) = 0.0006 
   
  pp. 795, 798 
   
  C1a. –(Ö3)/2 
       b. –1 
       c. –1 
       d. –2 
       e. DNE 
       f. 3p/4 
       g. 5p/6 
       h. p/3 
       i. DNE 
       j. 2.5 
   
  T1a. (Ö3)/2 
       b. –(Ö3)/3 
       c. p/3 (same problem as C1h) 
       d. 5p/6 (same problem as C1g) 
       e. –(Ö74)/7, since this is like a geometry problem with
  hypotenuse of Ö74 
       f. Solve proportion p/180 = 3/(no. of deg.) to get no. of degrees = 540/p. 
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   F 4/28/06 
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   HW due:
  Write §14-1 #3-23 eoo, 28; §14-2 #1-5 odd. Reading notes are optional, since
  the material in the text is duplicative of information already covered in
  class. However, if you are stuck, you should certainly read over the text
  and/or examples in order to help get “unstuck.” 
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