course MTH 163 002. `query2
.............................................
Given Solution: ** Continue to the next question ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Estimating from the graph, the ordered pairs might be: (7, 77) (19, 61) (31, 46) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Continue to the next question ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I chose (10, 75) , (30, 49) , and (70, 26). Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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 nswers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps, you should not expect that the numbers given here will be the same as the numbers you obtained when you solved the problem.) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the first equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100a + 10b + c = 75 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the second equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 900a + 30b + c = 49 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the third equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4900a + 70b + c = 26 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 900a + 30b + c = 49 - 100a + 10b + c = 75 ---------------------- 800a + 20b = -26 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4900a + 70b + c = 26 - 900a + 30b + c = 49 ---------------------- 4000a + 40b = -23 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I eliminated b, the simplest of the two to eliminate, and solved for a. My solution for parameter a was 4000a + 40b = -23 800a + 20b = -26 (* 2) 4000a + 40b = -23 -1600a + 40b = -52 ------------------ 2400a = 29 a = 0.012083 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Substituting 0.012083 for a in one of the previous equations to find b: 800(0.012083) + 20b = -26 9.6664 + 20b = -26 20b = -35.6664 b = -1.7833 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the value of c obtained from substituting into one of the original equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100(0.012083) + 10(-1.7833) + c = 75 1.2083 - 17.833 + c = 75 -16.6247 + c = 75 c = 91.625 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the resulting quadratic model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: temperature = 0.012083 t^2 - 1.7833 t + 91.625 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Time Temp Prediction Deviation 0 90 91.625 -1.625 10 75 75.0003 -0.0003 20 60 60.7922 -0.7922 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat was your average deviation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average of the deviations is 0.39525 degrees. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: My average deviation was .6 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok ********************************************* Question: `qIs there a pattern to your deviations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The deviations seem to increase from -1.625 until reaching 2.3325 and then decreased again down to -0.0007. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have studied the process and understand it pretty well. My only real trouble seems to come in the form of common math mistakes (e.g. losing track of pos/neg notations). I seemed to keep ending up with negative numbers in the sqrt() function of the quadratic formula also, which I'm not sure if that was the way the equations worked out or if I made a mistake somewhere along the line. I believe I understand the concepts and the process quite well, however. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have memorized everything involving the modeling process for quadratic equations. I made enough minor mathematical mistakes and came up with incorrect solutions often enough that I had to repeat each exercise a few times, so trust me... it's imprinted in my gray matter. My future involves a lot of precalculus, calculus, and other higher mathematics, so I won't soon forget it. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used simulated data for the experiment. Here are my ordered pairs: (3.6, 75.2) (7.2, 70.4) (10.8, 65.2) (14.4, 61.2) (18, 57.7) (21.6, 55.4) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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: 3 ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (7.2, 70.4) (14.4, 61.2) (18, 57.7) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 51.84a + 7.2b + c = 70.4 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 207.36a + 14.4b + c = 61.2 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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: 3 ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 324.00a + 18.0b + c = 57.7 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 324.00a + 18.0b + c = 57.7 -(207.36a + 14.4b + c = 61.2) ----------------------------- 96.64a + 3.6b = -3.5 Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 207.36a + 14.4b + c = 61.2 - (51.84a + 7.2b + c = 70.4) ----------------------------- 155.52a + 7.2b = -9.2 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using these two equations, we eleminate b to solve for a, multiplying the second equation by 2 to make b equal in each equation.: 155.52a + 7.2b = -9.2 -(96.64a + 3.6b = -3.5) (* 2) ----------------------- 155.52a + 7.2b = -9.2 - (193.28a + 7.2b = -7.0) ------------------------- -37.76a = -2.2 a = 0.0583 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = 0.0583, as found above. We substitute a into a previous equation to find b: 96.64(0.0583) + 3.6b = -3.5 5.634112 + 3.6b = -3.5 3.6b = -9.134112 b = -2.54 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Now with values for parameters a and b, we can solve for c using one of the original 3 quadratic equations: 51.84(0.0583) + 7.2(-2.54) + c = 70.4 3.022272 - 18.288 + c = 70.4 -15.265728 + c = 70.4 c = 85.7 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: c = 73.4 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 0.0583t^2 - 2.54t + 85.7 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your depth prediction for the given clock time (give clock time also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For a clock time of 46 seconds: y = 0.0583(46)^2 - 2.54(46) + 85.7 y = 123.3628 - 116.84 + 85.7 y = 92.2228cm This solution did not appear to fit with the data. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat clock time corresponds to the given depth (give depth also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine clock time when water depth is 14 cm, we use our model, solving for t when y = 14cm, using the quadratic formula. We must first put the equation into standard form: 0.0583x^2 - 2.54x + 85.7 = 14 And set the solution equal to zero by subtracting 14 from each side: 0.0583x^2 - 2.54x + 71.7 = 0 We may now use the quadratic formula to solve for time: t = [-b +- sqrt(b^2-4ac)] / (2a) t = [-(-2.54) +- sqrt( (-2.54)^2 - 4(0.0583)(71.7))] / ( 2(0.0583) ) t = [2.54 +- sqrt(-2.54^2-4*0.0583*71.7)] / (2*0.0583) t = [ -2.54 +- sqrt( 6.4516 - 16.72044 ) ] / 0.1166 t = [ -2.54 +- sqrt( -10.26884 ) ] / 0.1166 At this point, we have a negative number inside of our sqrt() function. I believe this means there's no real world solution to the problem since it will involve imaginary numbers. I will attempt to solve the equation anyway, using the property of square roots: sqrt( -10.26884 ) = sqrt(10.26884) * sqrt(-1) or 3.20450308160251 i: t = ( -2.54 +- 3.20450308160251i ) / 0.1166 t = -2.54 + 3.20450308160251i / 0.1166 or -2.54 - 3.20450308160251i / 0.1166 t = 0.66450308160251i / 0.1166 or -5.74450308160251i / 0.1166 t = 5.69899726931827i or -49.2667502710336i Assuming that neither of the numbers are imaginary, a time of 5.7 sec with a depth of 14cm would not fit with our given data set. Since our other solution, -49.3 is a negative, it would also not fit with our data, since time cannot be negative. Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** INSTRUCTOR COMMENT: The exercise should have specified a depth. 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): I got a negative value inside my sqrt() function, so I think I may have gone wrong somewhere along the line. I tried to muddle through it with imaginary numbers, but my results did not fit with the data. Self-critique Rating: 3 ********************************************* Question: `qCompletion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (20, 2.118034) (50, 2.767767) (90, 3.371708) Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 400a + 20b + c = 2.118034 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2500a + 50b + c = 2.767767 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8100a + 90b + c = 3.371708 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8100a + 90b + c = 3.371708 -(2500a + 50b + c = 2.767767) ------------------------------ 5600a + 40b = 0.603941 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2500a + 50b + c = 2.767767 - (400a + 20b + c = 2.118034) ------------------------------ 2100a + 30b = 0.649733 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the two equations, we multiply each to set the b parameter equal to one another, and then subtract, as thus: 5600a + 40b = 0.603941 (* 30) 2100a + 30b = 0.649733 (* 40) ---------------------- 168000a + 120b = 18.11823 - (84000a + 120b = 25.98932) ---------------------------- 84000a = -7.87109 a = -0.00009370345239 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = -0.0000937 b = 0.0282 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 400(-0.0000937) + 20(0.0282) + c = 2.118034 0.52652 + c = 2.118034 c = 1.591514 Rounded to 1.59. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: c = 1.773. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = -0.0000937x^2 + 0.0282x + 1.59 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** y = -.0000876638 x^2 + (.01727)x + 1.773 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the precent of classes reviewed for GPAs of 3.0 and 4.0, we must use the quadratic formula to solve for x (where y = 3.0 and y = 4.0). To do this, we must first set the equation equal to zero for each x value: 3.0 = -0.0000937x^2 + 0.0282x + 1.59 or -0.0000937x^2 + 0.0282x + 1.59 = 3.0 Subtracting 3.0 from each side, we achieve: -0.0000937x^2 + 0.0282x -1.41 = 0 4.0 = -0.0000937x^2 + 0.0282x + 1.59 or -0.0000937x^2 + 0.0282x + 1.59 = 4.0 Subtracting 4.0 from each side, we achieve: -0.0000937x^2 + 0.0282x -2.41 = 0 Let's solve the equation for y = 3.0: x = [-b +- sqrt(b^2-4ac)] / (2a) x = [-(0.0282) +- sqrt(0.0282^2 - 4 * -0.0000937 * -1.41 )] / (2 * -0.0000937) x = [ -0.0282 +- sqrt( 0.000266772 ) ] / -0.0001874 x = ( -0.0282 +- 0.01633315646162) / -0.0001874 x = ( -0.0282 + 0.01633315646162 ) / -0.0001874 or ( -0.0282 - 0.01633315646162 ) / -0.0001874 x = -0.01186684353838 / -0.0001874 or -0.04453315646162 / -0.0001874 x = 63.3236047939168 or 237.636907479296 Given the available data points, the solution 63.3% makes the most sense and appears to fit in with our data with a GPA of 3.0, since at 60% we have a GPA of around 2.9 and at 70% the GPA averages around 3.1. 63% at 3.0GPA would fit in between these points. Now, let's solve the equation for y = 4.0: x = [-b +- sqrt(b^2-4ac)] / (2a) x = [-(0.0282) +- sqrt(0.0282^2 - 4 * -0.0000937 * -2.41 )] / (2 * -0.0000937) x = [ -0.0282 +- sqrt( 0.00079524 - 0.000903268 ) ] / -0.0001874 x = [ -0.0282 +- sqrt(-0.000108028) ] / -0.0001874 We appear to have a negative number inside of our sqrt() function, which will result in an imaginary number. I assume this will give us no real world solution for the problem. I will attempt to solve the equation and evaluate the results anyway: x = ( -0.0282 + 0.01039365190873i ) / -0.0001874 or ( -0.0282 - 0.01039365190873i ) / -0.0001874 x = -0.01780634809128i / -0.0001874 or -0.03859365190873i / -0.0001874 x = 95.0178660153681i or 205.942646257898i Given that students who studied 100% of assignments achieved a 3.5 GPA, students who studied 95% of assignments would not fit with a 4.0 GPA. 205% seems to be too high of a number to fit with the data. Unless students studied each assignments more than 2 times, which is possible, they might achieve a GPA of 4.0, but this ""real world"" scenario would not fit with our model. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): I was concerned that since I got a negative in my sqrt() function when attempting to solve for a 4.0 average I had made some mistake. I'm glad to see that wasn't the case. Self-critique Rating: 3 ********************************************* Question: `qWhat grade average corresponds to the given percent of review (give grade average also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For a student who reviews 80% of the material, we could predict their GPA by substiting a value of 80 for x into our model to find y (GPA): y = -0.0000937(80)^2 + 0.0282(80) + 1.59 y = -0.59968 + 2.256 + 1.59 y = 3.24632 This solution fits fairly well with the given solution for (80%, 3.236068 GPA). Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qHow well does your model fit the data (support your answer)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The model fits the data quite well. The only exception was determining a percentage of studied material for a student with a 4.0 GPA. Since the data is dealing with an average GPA of students who studied 100% of the material, unless all of those students scored a 4.0, this seems to be acceptable. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qillumination vs. distance Give your data in the form of illumination vs. distance ordered pairs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2, 264.4411) (6, 25.91537) (9, 11.28082) Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4a + 2b + c = 264.4411 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 36a + 6b + c = 25.91537 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 81a + 9b + c = 11.28082 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 45a + 3b = -14.63455 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 32a + 4b = -238.52573 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I eliminated b to solve for a. To do so, I had to multiply the first equation by 4 and the second by 3: Eliminate b: 45a + 3b = -14.63455 (* 4) 32a + 4b = -238.52573 (* 3) 180a + 12b = -58.5382 - (96a + 12b = -715.57719) -------------------------- 84a = 657.03899 a = 7.82189273809523 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = 7.82189273809523 rounded to 7.822 b = -122.206574404762 rounded to -122.2 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c = 477.566677857142 rounded to 477.6 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: c = 588.5691** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 7.822x^2 - 122.2x + 477.6 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is your illumination prediction for the given distance (give distance also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using our model, we simply 1.6AU for x: y = 7.82189273809523(1.6)^2 - 122.206574404762(1.6) + 477.566677857142 y = 20.0240454095238 - 195.53051904762 + 477.566677857142 y = 302.060204219047 This solution appears to fit in with our data set. Since illumination at 2.0 AU is 264.4411 W/m^2, at 1.6AU, an illumination factor of 302.060204219047 W/m^2 seems fairly in line with the other data. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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): ok Self-critique Rating: 3 ********************************************* Question: `qWhat distances correspond to the given illumination range (give illumination range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine which range of distances in AU, using our model, y = 7.822x^2 - 122.2x + 477.6, we would first substitute 25 for y (the illumination in W/m^2) and then set the equation equal to zero for use with the quadractic formula: 25 = 7.822x^2 - 122.2x + 477.6 or 7.822x^2 - 122.2x + 477.6 = 25 7.822x^2 - 122.2x + 452.6 = 0 And do the same for 100 W/m^2 100= 7.822x^2 - 122.2x + 477.6 or 7.822x^2 - 122.2x + 477.6 = 100 7.822x^2 - 122.2x + 377.6 = 0 We then solve each equation for x using the quadratic formula x = [-b +- sqrt(b^2-4ac)] / (2a): x = [ -(122.2) +- sqrt( (-122.2)^2 - 4(7.822)(452.6 ) ] / (2 (7.822) ) x = [ -122.2 +- sqrt( 14932.84 - 14,160.9488 ) ] / 15.644 x = [ -122.2 +- sqrt( 771.8912 ) ] / 15.644 x = ( -122.2 + 27.7829300110698 ) / 15.644 or ( -122.2 - 27.7829300110698 ) / 15.644 x = -94.4170699889302 / 15.644 or -149.98293001107 / 15.644 x = -6.03535348944837 or -9.58724942540719 We have arrived at two negative values for our possible solutions for x. Since we are looking for distances in AU, this solution does not fit with our data. A response of 6.03535348944837 AU would possible fit with the data for 25 W/m^2 Solving for 100: x = [ -(122.2) +- sqrt( (-122.2)^2 - 4(7.822)(377.6) ) ] / (2 (7.822) ) x = [ -122.2 +- sqrt( 3,118.4912 ) ] / 15.644 x = (-122.2 + 55.8434526153246) / 15.644 or (-122.2 - 55.8434526153246) / 15.644 x = -4.2416611726333 or -11.3809417422222 Again, we arrive at two negative numbers, which indicates no real world solution. Even if these numbers were positive, they would not fit the model for an illumination rating of 100 W/m^2. The number would have to be somewhere between 3 and 4 AU. To estimate the answer to the original question, the range of distances with illumination ratings between 25 to 100 W/m^2, the solution would be roughly in the range of 3 to 6 AU. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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 3.9 - 5.4 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. **"