#$&* course Mth 173 1/23 11:22 p.m. 001. `query1
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Given Solution: ** Continue to the next question ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ??? the question says according to my 'graph', does this mean i was just to graph points, and get a rough estimate, or do it as follows my equation was Temperature = (1/60)t^2-2t+(280/3) at clock time 7, temperature is about 80.15 at clock time 19, temperature is about 61.35 at clock time 31, temperature is about 47.35
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Given Solution: ** Continue to the next question ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: time vs temp (10,75),(20,60),(50,35) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* Question: `qWhat is the first equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 75=a10^2+b10+c OR 75= 100a+10b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is the second equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 60=a20^2+b20+c OR 60=400a+20b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* Question: `qWhat is the third equation you got when you substituted into the form of a quadratic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 35=a50^2+b50+c OR 35=2500a+50b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the multiples would only have to be of -1 (which i dont see this as a 'multiple', just as the negative equivalent) since c always has a coefficient of 1 I used the negative equivalent of 65=400a+20b+c to get -65=-400a-20b-c against 35=2500a+50b+c I obtained -25=2100a+30b I confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: negative equivalent of 75=100a+10b+c which is -75=-100a-10b-c against 35=2500a+50b+c and obtained -40=2400a+40b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I personally always choose to elimante variable b. The reason is high school our teacher always removed b, so it just stuck in my head. i found a = 1/60 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I substituted it in -40=2400a+40b i got b= -2 confidence rating #$&*: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? I noticed in my equation, that for more precise answerws, i could leave as fractions, because a few answers when multiplied by a, were answer/60, which could be reduced to answer(divide by 20)/3, and that my c was also 280/3, and that i could multiply b by 3 to get banswer/3, and just do the adding and subtracting of numerators, and then divide by 3. Or that i could multiply b by 60, and c by 20, if a* answer couldnt be reduced from something/60 ???question in short, is it alright to choose fractions over decimals (i find them to be easier to work with in a lot of cases) ------------------------------------------------ Self-critique Rating: OK
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Given Solution: ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is the resulting quadratic model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Temperature=(1/60)t^2-2t+(280/3) (temperature in Celsius) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Time Prediction Deviance 0 280/3 OR ~98.33 ~3.33 positive 10 75 0 (point used in model) 20 60 0 (point used in model) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat was your average deviation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ~2.16 (add all the deviations up, and divide by number of deviations (including 0 deviation)) confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: My average deviation was .6 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIs there a pattern to your deviations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think I would need more points to be sure, it looks like it is trying to form one though, I do notice two points are both ~3.33 off So for whats given, no pattern can be discerned. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qHave you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have glanced, but not perused. If I was asked a question on it, better chance I would not be able to tell the answer right away. My focus was understanding the formula, and using it to a table. When looking at those, it distracted my attention. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should go back and read them after understand what my focus was, to have a better concept. ------------------------------------------------ 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: from your charts, no I havent. This is what I will remember forever (assuming proper health, no mental disorders, etc) quadratic formula is used for 3 pieces of data. quadratic formula will not always work for a graph. the model you make is a prediction, expect deviation. from observed data, pick 3 points, it is wise to pick points spread out across the data. set up 3 equations based on the basic formula. (y=ax^2+bx+c) you will then do the exponent for the three equations pick one, and then find the negative equivalent of the other two you did not pick. take the first negative equivalent, and add it to the one you picked, then do the same with the second. this gives you two equations without c I then pick b, and find a way to make the equations have a elimination of b, if nothing else, multiply equation 1's b with equation 2, and equation 2's b with equation 1, and you will either be there, or just have to find the negative equivalent of one of the equations. put the equations together to solve for a. plug a back in to one of the equations you elimated c in to solve for b. then plug both a and b into an original equation you made to solve for c. now plug this in your standard equation. this is my prediction model.
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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: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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???? Is there a reason you ask depth vs clock time instead time vs depth, to me this goes against (x,y) it seems you are asking (y,x) are you asking to just make sure we are paying attention
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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): The student response you showed has it as x then y, but you said depth vs time, in my mind depth vs time say y vs x, which means to display y first, and then show x against it. Do i just look at this wrong, does depth vs time mean (x,y)? ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (time,depth) (3.6,75.2), (10.8,65.2), (18,57.7) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5.2=12.96a+3.6b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 65.2=116.64a+10.8b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 57.7=324a+18b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -17.5=311.04a+14.4b confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = - 14. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -7.5=207.36a+7.2b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for the b's, I noticed one equation was half the other. I also know that to eliminate one has to be positive, and the other negative, so i multiplied the one that was half the other by -2, before putting the two equations together. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a= ~.0241127 b= ~-1.736112 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c= ~81.13753 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: c = 73.4 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: formula Depth= 0.0241127t^2-1.736112t+81.13753 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat is your depth prediction for the given clock time (give clock time also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for 46 seconds: 0.0241127(46)^2-1.736112(46)+81.13753 51.0224732-79.861152+81.13753 Depth = ~52.19885cm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): noticed our depths were far apart ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat clock time corresponds to the given depth (give depth also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for 14cm 14 = 0.0241127t^2-1.736112t+81.13753 0 =0.0241127t^2-1.736112t+67.13753 using quadratic formula I ran across the square root of a negative number. My model did not cover that depth confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???you said vs clock time, there is no clock time, I assume you meant % of assignments reviewed Average =(-1.197493/8400)t^2+0.0334802679t+1.47002219 ####I noted that when you say your data, i read it as you want (y,x) but you showed (x,y) as given solution so i will do this ?????Is it alright for me to use a graphing calculator to store values for a, b, and c, to solve equations easier Predicted results (0,~1.47) (10,~1.79) (20,~2.08) (30,~2.35) (40,~2.58) (50,~2.79) (60,~2.97) (70,~3.12) (80,~3.24) (90,~3.33) (100,~3.39) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION: Grade vs. percent of assignments reviewed (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I assumed you wanted our model prediction, since you already gave us the values of the table ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Pairs picked (10, 1.790569), (40, 2.581139), (80,3.236068) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.790569=100a+10b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.581139=1600a+40b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3.236068=6400a+80b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.445499=6300a+70b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0.654929=4800a+40b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I always eliminate b first. I didnt see any small way to make them the same, so i said 70b and 40b, both can be divided by 10 for 7 and 4, and multiplied the equation with 40b by -7 and the 70b by 4. I then added the two equations together to elimate b. I multiplied by a -7 to get one b negative and the other positive so they would cancel out. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a=(-1.197493/8400) b=.0334802679 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c=1.47002219 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: c = 1.773. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average =(-1.197493/8400)t^2+0.0334802679t+1.47002219 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** y = -.0000876638 x^2 + (.01727)x + 1.773 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for 3.0 0= (-1.197493/8400)t^2+0.0334802679t-1.52997781 (-.0334802679+-sqrt(.0334802679^2-4(-1.197493/8400)(-1.52997781)))/2(-1.197493/8400) (-.0334802679+-sqrt(.0011209283-4(-1.197493/8400)(-1.52997781)))/-.000285117381 (-.0334802679+-sqrt(.0011209283- -.000570234762(-1.52997781)))/-.000285117381 (-.0334802679+-sqrt(.0011209283-.0008724465322)/-.000285117381 (-.0334802679+-sqrt.(0002484817678))/-.000285117381 (-.0334802679+- .0157633045)/-.000285117381 -.0177169634/-.000285117381 = 62.13919102 -.0492435724/-.000285117381= 172.7133303 172 is past our 100% mark so it must be ~62% for 3.0 for 4.0 the square root will become negative, meaning not possible in reality. and if you look at the range, even at 100% the actual table doesnt go past 3.5, so I would say impossible. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat grade average corresponds to the given percent of review (give grade average also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for 80%, I used this data point, so 3.236068 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qHow well does your model fit the data (support your answer)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It fits it decently well, from a quick glance, no deviation is even goes up to .2, except for at 0%. But for reality, a gpa even .01 is a somewhat big deal difference, so that would depend on the person if your ""estimate"" is close. Mathematically the data fits well though. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* 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 #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION: (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat three points on your graph did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Points: (1,935.1395),(3,105.1209),(9,11.28082) confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the first of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 935.1395=1a+1b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the second of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 105.1209=9a+3b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the third of your three equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 11.28082=81a+9b+c confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the first of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 923.85868=-80a-8b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qGive the second of the equations you got when you eliminated c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 830.0186=-8a-2b confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qExplain how you solved for one of the variables. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I always eliminate b, and noticed 8 is a multiple of 2, so just multiplied 2 by -4 to get +8, since the other equation's b was -8 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `qWhat values did you get for a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a=49.92116083 b=-614.6939434 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat did you then get for c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c=1499.912283 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: c = 588.5691** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Illumination= 49.92116083t^2-614.6939434t+1499.912283 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is your illumination prediction for the given distance (give distance also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: at 1.6 Earth distances from the sun. 49.92116083(1.6)^2-614.6939434(1.6)+1499.912283 49.92116083(2.56)-983.5103094+1499.912283 127.7981717-983.5103094+1499.912283 -855.7121377+1499.912283 644.2001453 at 1.6 Earth distances the predicted Illumination of coment will be 644.2001453W/m^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `qWhat distances correspond to the given illumination range (give illumination range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At what range of distances from the sun would the illumination be comfortable for reading, if reading comfort occurs in the range from 25 to 100 Watts per square meter? 25 = 49.92116083t^2-614.6939434t+1499.912283 0 =49.92116083t^2-614.6939434t+1474.912283 using the quadratic formula (614.693943+-sqrt(377848.6441-294517.3331))/99.84232166 (614.693943+-sqrt(83331.311))/99.84232166 (614.693943+-288.6716318)/99.84232166 903.3655748/99.84232166= 9.047922362 326.0223112/99.84232166 = 3.265371896 ???im not sure how im to figure which data point to throw away, 9 just seems way to far off, so i chose the other value 3.265371896
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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. **" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat distances correspond to the given illumination range (give illumination range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At what range of distances from the sun would the illumination be comfortable for reading, if reading comfort occurs in the range from 25 to 100 Watts per square meter? 25 = 49.92116083t^2-614.6939434t+1499.912283 0 =49.92116083t^2-614.6939434t+1474.912283 using the quadratic formula (614.693943+-sqrt(377848.6441-294517.3331))/99.84232166 (614.693943+-sqrt(83331.311))/99.84232166 (614.693943+-288.6716318)/99.84232166 903.3655748/99.84232166= 9.047922362 326.0223112/99.84232166 = 3.265371896 ???im not sure how im to figure which data point to throw away, 9 just seems way to far off, so i chose the other value 3.265371896
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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. **" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `qWhat distances correspond to the given illumination range (give illumination range also)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At what range of distances from the sun would the illumination be comfortable for reading, if reading comfort occurs in the range from 25 to 100 Watts per square meter? 25 = 49.92116083t^2-614.6939434t+1499.912283 0 =49.92116083t^2-614.6939434t+1474.912283 using the quadratic formula (614.693943+-sqrt(377848.6441-294517.3331))/99.84232166 (614.693943+-sqrt(83331.311))/99.84232166 (614.693943+-288.6716318)/99.84232166 903.3655748/99.84232166= 9.047922362 326.0223112/99.84232166 = 3.265371896 ???im not sure how im to figure which data point to throw away, 9 just seems way to far off, so i chose the other value 3.265371896
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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. **" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!