course Mth 271 002. `Query 2
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Given Solution: (10, 75) (20, 60) (60, 30). `a Continue to the next question ** Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31? ********************************************* Your solution: Given Solution: `a Continue to the next question ** Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)? ********************************************* Your solution: (10, 75) (20, 60) (60, 30).
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Given Solution: `a A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ********************************************* Question: `qWhat is the first equation you got when you substituted into the form of a quadratic? ********************************************* Your solution: 75= A(10)^2 + B(10) + C
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given Solution: `a STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self- critique Rating:ok ********************************************* Question: `qWhat is the second equation you got when you substituted into the form of a quadratic? ********************************************* Your solution: 60= A(20)^2 + B ( 20) + C Given Solution: `a STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self -critique Rating:ok ********************************************* Question: `qWhat is the third equation you got when you substituted into the form of a quadratic? ********************************************* Your solution: 30= A(60)^2 + B(60) +C Given Solution: `a STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qWhat multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c? ********************************************* Your solution: I subtracted the first equation from the second equation. -15= 300 A +10B Given Solution: `a STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c. By doing this, I obtained my first new equation 3200a + 40b = -30. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self -critique Rating:ok ********************************************* Question: `qTo get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation? ********************************************* Your solution: I subtracted the second equation from the third one. -30= 3200A +40B Solution: `a STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `qWhich variable did you eliminate from these two equations, and what was its value? ********************************************* Your solution: I eliminated b. To do this I multiplied the first equation by 4: -60=1200A + 40B Minus -30 = 3200A+40B Equals -30 = -2000A A= .015 Given Solution: `a STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qWhat equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable? ********************************************* Your solution: -60 = 1200(.015) +40B B= -1.95 Given Solution: `a STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self critique Rating: ok ********************************************* Question: `qWhat is the value of c obtained from substituting into one of the original equations? ********************************************* Your solution: C=93
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given Solution: `a STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qWhat is the resulting quadratic model? ********************************************* Your solution: Y=.015x^2 + -1.95x + 93
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Given Solution: `a STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qWhat did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times? ********************************************* Your solution: Substitute 0, 10, and 20 in for x For 0: pred: 93 For 10: pred: 75 For 20: pred: 60 given Solution: `a STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat was your average deviation? ********************************************* Your solution: Given Solution: `a STUDENT SOLUTION CONTINUED: My average deviation was .6 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ********************************************* Question: `qIs there a pattern to your deviations? ********************************************* Your solution: No obvious pattern to the deviations
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Given Solution: `a STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `qHave you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process? ********************************************* Your solution: yes Given Solution: `a STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me. ********************************************* Your solution: mostly Given Solution: `a STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!! INSTRUCTOR COMMENT: OK, I'm convinced. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self critique Rating: ok ********************************************* Question: `qQuery Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs. ********************************************* Your solution: I got this data from version 2 on the randomized problems. clock t 5.2 10.4 15.6 20.8 26 31.2 depth 80.1 70.6 62.2 55 48.9 44.2 Clock times are in seconds, depths in cm. (5.2, 80.1) (10.4, 70.6) (15.6, 62.2) (20.8, 55) (26, 48.9) (31.2, 44.2) Given Solution: `a STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems. (5.3, 63.7) (10.6. 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? ********************************************* Your solution: (5.2, 80.1) (20.8, 55) (31.2, 44.2) given Solution: `a STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qGive the first of your three equations. ********************************************* Your solution: (5.2, 80.1) gives you 80.1= (5.2)^2A+(5.2)B+C 80.1= 27.04A + 5.2B +C Given Solution: `a STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ok Question: `qGive the second of your three equations. ********************************************* Your solution: (20.8, 55) gives you 55= (2o.8)^2A + (20.8)B +C 55= 432.6A + 20.8B + C
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Given Solution: `a STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok self-critique Rating: ok ********************************************* Question: `qGive the third of your three equations. ********************************************* Your solution: ( 31.2, 44.2) 44.2 = (31.2)^2A + (31.2)B + C 44.2 = 973.44A + 31.2B +C Given Solution: `a STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. ** Self-critique (if necessary): Self-critique Rating: ok Question: `qGive the first of the equations you got when you eliminated c. ********************************************* Your solution: -25.1= 405.6A + 15.6B Given Solution: `a STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. ** Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `qGive the second of the equations you got when you eliminated c. ********************************************* Your solution: -10.8 = 540.8A + 10.4B Given Solution: `a ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ********************************************* Question: `qExplain how you solved for one of the variables. ********************************************* Your solution: You get rid of variable B. Multiply first equation by 10.4 Multiply second equation by 15.6 -261.04 = 4218.24A + 162.24B - (-168.48 = 8436.48A +162.24B) Then you are left with the equation: -92.56= -4218.24A Solve for A A=.0219 Given Solution: `a STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. ** Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qWhat values did you get for a and b? ********************************************* Your solution: A= .0219 B= -2.178 Given Solution: `a STUDENT SOLUTION CONTINUED: a = .0165, b = -2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok Question: `qWhat did you then get for c? ********************************************* Your solution: C= 90.827584 Given Solution: `a STUDENT SOLUTION CONTINUED: c = 73.4 ** Self-critique (if necessary): Self-critique Rating:ok Question: `qWhat is your function model? ********************************************* Your solution: Y= .0219x^2 + -2.178x + 90.827584
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Given Solution: `a STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWhat is your depth prediction for the given clock time (give clock time also)? ********************************************* Your solution: The given clock time was 45 seconds and my depth prediction was 37.1575
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Given Solution: `a STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok Question: `qWhat clock time corresponds to the given depth (give depth also)? ********************************************* Your solution: If the given depth is 60 then y=60 therefore, 60= .0219x^2+-2.178x+90.827584 X = 17.085993 and x = 82.366062 Given Solution: `a The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation: 68 = .01t^2 - 1.6t + 126 using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. ** Self-critique (if necessary): Self critique Rating: ok Question: `qCompletion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs. ********************************************* Your solution: (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). Given Solution: `a STUDENT SOLUTION: Grade vs. percent of assignments reviewed (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? ********************************************* Your solution: (20, 2.118034) (50, 2.767767) (100, 3.5)
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Given Solution: `a STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qGive the first of your three equations. ********************************************* Your solution: 2.118034=400A + 20B + C
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Given Solution: `a STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `qGive the second of your three equations. ********************************************* Your solution: 2.767767 = 2500A+50B+C
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Given Solution: `a STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ok ********************************************* Question: `qGive the third of your three equations. ********************************************* Your solution: 3.5 =10000A+100B + C
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Given Solution: `a STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qGive the first of the equations you got when you eliminated c. ********************************************* Your solution: I subtracted the first equation from the second equation. .649733= 2100A+30B
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Given Solution: `a STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qGive the second of the equations you got when you eliminated c. ********************************************* Your solution: I subtracted the second equation from the third one. .73224= 7500A +50B given Solution: `a STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qExplain how you solved for one of the variables. ********************************************* Your solution: multiplied the first equation by 5 and multiplied the second equation by 3 And subtracted Given Solution: `a STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the variable b. In order to do this, I multiplied the first new equation by 80 and the second new equation by -50. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qWhat values did you get for a and b? ********************************************* Your solution: A= -.0000876638 B = .01727
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given Solution: `a STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 ** Self-critique (if necessary)self critique Rating:ok ********************************************* Question: `qWhat did you then get for c? ********************************************* Your solution: C=1.773
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Given Solution: `a STUDENT SOLUTION CONTINUED: c = 1.773. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): se sef-critique Rating: ********************************************* Question: `qWhat is your function model? ********************************************* Your solution: Y= (.00008)x^2 + (.01727)x +1.773
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Given Solution: `a STUDENT Answer: ->->->->->->->->->->->-> y = (0) x^2 + (.01727)x + 1.773 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)? Your solutionGiven Solution: that depends on the wanted average If you wanted 3.6 grade average then: 3.6= -.00008x^2+ .01727x + 1.773 `a The precise solution depends on the model desired average. For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have 3.3 = -.00028 x^2 + .06 x + .5. This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0. We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility. To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range. In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: question: `qWhat grade average corresponds to the given percent of review (give grade average also)? ********************************************* Your solution: if you wanted the value for when x is 80: y= -.00008(80)^2+ .01727(80) + 1.773 Given Solution: `a Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qHow well does your model fit the data (support your answer)? ********************************************* Your solution: overall accurate but not exact
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given Solution: `a You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qillumination vs. distance Give your data in the form of illumination vs. distance ordered pairs. ********************************************* Your solution: : (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)
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Given Solution: `a STUDENT SOLUTION: (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? ********************************************* Your solution: (2, 264.4411) (4, 61.01488) (8, 16.27232)
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Given Solution: `a STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qGive the first of your three equations. ********************************************* Your solution: 4A + 2B + C = 264.4411
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Given Solution: `a STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qGive the second of your three equations. ********************************************* Your solution: 16A + 4B + C = 61.01488
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Given Solution: `a STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self critique Rating:ok ********************************************* Question: `qGive the third of your three equations. ********************************************* Your solution: 64A + 8B + C = 16.27232
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Given Solution: `a STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qGive the first of the equations you got when you eliminated c. ********************************************* Your solution: 48A + 4B = -44.74256
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Given Solution: `a STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `qGive the second of the equations you got when you eliminated c. ********************************************* Your solution: 60A + 6B = -248.16878
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given Solution: `a STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qExplain how you solved for one of the variables. ********************************************* Your solution: eliminated variable b: multiplied first equation by 4 and multiplied the second equation by -6
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Given Solution: `a STUDENT SOLUTION CONTINUED: I solved for a by eliminating the variable b. I multiplied the first new equation by 4 and the second new equation by -6 ** Self-critique (if necessary) Self-critique Rating:ok ********************************************* Question: `qWhat values did you get for a and b? ********************************************* Your solution: A=15.088 B=-192.24
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given Solution: `a STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating:ok ********************************************* Question: `qWhat did you then get for c? ********************************************* Your solution: C= 588.5691
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Given Solution: `a STUDENT SOLUTION CONTINUED: c = 588.5691** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qWhat is your function model? ********************************************* Your solution: Y= (15.088) x^2 - (192.24)x + 588.5691
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Given Solution: `a STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): self-critique Rating: ********************************************* Question: `qWhat is your illumination prediction for the given distance (give distance also)? ********************************************* Your solution: 319.61 w/m^2
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Given Solution: `a STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun. My illumination prediction was 319.61 w/m^2, obtained by evaluating my function model for x = 1.6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `qWhat distances correspond to the given illumination range (give illumination range also)? ********************************************* Your solution: Depends, place the value in for y and solve for r
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Given Solution: `a The precise solution depends on the model and the range of averages. For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations 25=9.4 r^2 - 139 r + 500 and 100 =9.4 r^2 - 139 r + 500 Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data. The solutions which correspond to the data are r = 3.9 when y = 100 and r = 5.4 when y = 25. So when the distance x has range 50% - 69% if the illumination range is 25 to 100. Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qppCal1 Section 0.2 EXTRA QUESTION. What is the midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?{}{}What is the midpoint between the points (3, 8) and (7, 12)? ********************************************* Your solution: Midpoint = ( x1+x2 , y1 + y2 ) 2 2 (3+7)/2 = 5 (8+12)/2 = 10 Midpoint= (5,10)
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Given Solution: `a You are given two points. The points each have two coordinates. You have to average the x coordinates to get the x coordinate of the midpoint, then average the y coordinates to get the y coordinate of the midpoint. For example if the points are (3, 8) and (7, 12), the average of the x coordinates is (3 + 7) / 2 = 5 and the average of the y coordinates is (8 + 12) / 2 = 10 so the coordinates of the midpoint are (5, 10). ** Self-critique (if necessary): ?/?? how is it 9.5 ???
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Given Solution: `a abs(a) >= b translates to a >= b OR a <= -b. In this case abs(3x+1) > 4 gives you 3x + 1 >= 4 OR 3x + 1 <= -4, which on solution for x gives x >= 1 OR x < = -5/3. ** ********************************************* Your solution:
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Given Solution: `a the given inequality is equivalent to the two inequalities 3x+1 >= 4 and 3x+1 =< -4. The solution to the first is x >= 1. The solution to the second is x <= -5/3. Thus the solution is x >= 1 OR x <= -5/3. COMMON ERROR: -5/3 > x > 1 INSTRUCTOR COMMENT: It isn't possible for -5/3 to be greater than a quantity and to have that same quantity > 1. Had the inequality read |3x+1|<4 you could have translated it to -4 < 3x+1 <4, but you can't reverse these inequalities without getting the contradiction pointed out here. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ok ********************************************* Question: `q0.2.24 (was 0.2.16 solve abs(2x+1)<5. What inequality or inequalities did you get from the given inequality, and are these 'and' or 'or' inequalities? Give your solution. ********************************************* Your solution: 2x +1 < 5 and 2x+1> -5 X<2 and x >-3
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given Solution: `a abs(a) < b means a < b AND -b < -a so from the given inequality abs(2x+1) < 5 you get -5 < 2x+1 AND 2x+1 < 5. These can be combined into the form -5 < 2x+1 < 5 and solved to get your subsequent result. Subtracting 1 from all expressions gives us -6 < 2x < 4, then dividing through by 2 we get -3 < x < 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q0.2.5 (was 0.2.23 describe [-2,2 ] Your solution Midpoint = (a+b)/2 (-2+2 )/ 2 = 0 Midpoint is x=0
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Given Solution: `a The interval [-2, 2] is centered at the midpoint between x=-2 and x=2. You can calculate this midpoint as (-2 + 2) / 2 = 0. It is also clear from a graph of the interval that it is centered at x = 0 The center is at 0. The distance to each endpoint is 2. The interval is | x - center | < distance to endpoints. So the interval here is | x - 0 | < 2, or just | x | < 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q0.2.10 (was 0.2.28 describe [-7,-1] ********************************************* Your solution: (-7+-1)/2 = -4 Midpoint at x= -4
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Given Solution: `a the interval is centered at -4 (midpt between -7 and -1). The distance from the center of the interval to -7 is 3, and the distance from the center of the interval to -1 is 3. This translates to the inequality | x - (-4) | < 3, which simplifies to give us | x + 4 | < 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q0.2.12 (was 0.2.30) describe (-infinity, 20) U (24, infinity) ********************************************* Your solution: Abs{x-22} > 2 x-22 >2 x -22< -2 x>24 x< 20
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Given Solution: `a 22 is at the center of the interval. The endpoints are 2 units from the midpoint, and are not included. Everything that lies more than 2 units from 22 is in one of the intervals, and everything in either of the intervals lies at least 2 units from 22. So the inequality that describes this union of two intervals is | x - 22 | > 2. ** Self-critique (if necessary): Self-critique Rating: ok Question: `q0.2.42 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1 ********************************************* Your solution: W -57.5 <7.5 and w-57.5>7.5 W< 65 and w> 50 given Solution: `a The inequality is translated as -1<=(w-57.5)/7.5<=1. Multiplying through by 7.5 we get -7.5<=w-57.5<=7.5 Now add 57.5 to all expressions to get -7.5 + 57.5 <= x <= 7.5 + 57.5 or 50 < x < 65, which tells you that the dogs weigh between 50 and 65 pounds. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:ok ********************************************* Question: `q0.2.40 (was 0.2.38 stocks vary from 33 1/8 by no more than 2. What absolute value inequality or inequalities correspond(s) to this prediction? ********************************************* Your solution: distance is d<=2 Distance between price and 33 1/8 is abs{p-33 1/8} Abs{p- 33 1/8} <=2
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Given Solution: `a this statement says that the 'distance' between a stock price and 33 1/8 must not be more than 2, so this distance is <= 2 The distance between a price p and 33 1/8 is | p - 33 1/8 |. The desired inequality is therefore | p = 33 1/8 | < = 2. ** Self-critique (if necessary):