course Mth 271 Yzvg}wQjjyassignment #002
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17:53:11 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> (95, 0), (60, 20), (41, 40) confidence assessment: 3
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17:53:19 Continue to the next question **
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RESPONSE --> self critique assessment: 3
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17:55:26 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> According to my graph, the temperature at clock times 7, 19, and 31 would be approximately 80, 62, and 48. confidence assessment: 2
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17:55:31 Continue to the next question **
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RESPONSE --> self critique assessment: 3
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17:55:39 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> confidence assessment:
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17:56:38 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> I used the points (60, 20), (41, 40) and (30, 60) confidence assessment: 3
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17:57:16 A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). **
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RESPONSE --> self critique assessment: 3
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17:58:56 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My first equation was 400a + 20b + c = 60 confidence assessment: 3
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17:59:04 STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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RESPONSE --> self critique assessment: 3
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17:59:55 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My second equation for point (41, 40) was 1600a + 40b + c = 41 confidence assessment: 3
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18:00:02 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> self critique assessment: 3
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18:00:48 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My third equation for point (30, 60) was 3600a + 60b + c = 30 confidence assessment: 3
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18:00:55 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> self critique assessment: 3
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18:02:35 What 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?
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RESPONSE --> I added the equations from the first two points to get the equation 1200a + 20b = -19 confidence assessment: 3
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18:02:47 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. **
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RESPONSE --> self critique assessment: 3
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18:03:47 To 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?
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RESPONSE --> For my second equation, I added the the second and third equations to get 2000a + 20b = -11 confidence assessment: 3
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18:03:56 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**
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RESPONSE --> self critique assessment: 3
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18:05:07 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> For these 2 equations I eliminated b and solved for a and found that a = .01 confidence assessment: 3
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18:05:26 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. **
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RESPONSE --> self critique assessment: 3
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18:07:35 What 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?
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RESPONSE --> When I substituted I got the equation 1200(.01) + 20 b = -19. I solved for b and got b = -1.55 confidence assessment: 3
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18:07:49 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. **
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RESPONSE --> self critique assessment: 3
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18:09:55 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> I substituted into the equation for the first point 400(.01) + 20(-1.55) + c = 60 and found that c = 87 confidence assessment: 3
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18:10:12 STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> self critique assessment: 3
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18:11:25 What is the resulting quadratic model?
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RESPONSE --> The resulting quadratic model is y = .01t^2 - 1.55t + 87 confidence assessment: 3
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18:11:35 STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> self critique assessment: 3
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18:16:20 What 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?
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RESPONSE --> For the first, second, and third times, my quadratic model gave me 87, 72.5, and 60. The deviations were 8, 2.5, and 0. confidence assessment: 3
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18:16:39 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. **
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RESPONSE --> self critique assessment: 3
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18:21:56 What was your average deviation?
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RESPONSE --> My average deviation was 1.25 confidence assessment: 3
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18:22:01 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> self critique assessment: 3
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18:23:00 Is there a pattern to your deviations?
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RESPONSE --> The only pattern I see, is that earlier in the experiment, the deviation is positive, but as time goes on, the deviation becomes negative. confidence assessment: 3
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18:23:11 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. **
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RESPONSE --> self critique assessment: 3
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18:24:09 Have 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?
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RESPONSE --> Yes, I have studied and completely understand the process. confidence assessment: 3
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18:24:19 STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> self critique assessment: 3
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18:24:49 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> No, I have not completely memorized the steps of the modeling process yet. confidence assessment: 3
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18:25:59 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. **
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RESPONSE --> I think printing out the steps to refresh my memory often would be a pretty good idea. self critique assessment: 2
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18:34:33 Query Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.
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RESPONSE --> I didnt have a two liter bottle to use or any type of cylinder that i didnt mind poking a hole in. I used the data found on one of the class notes videos. I got the ordered pairs of (90, 0), (80, 11), (70, 22), (60, 34), (50, 48), (40, 63), (30, 81), (20, 101), (10, 128). confidence assessment: 3
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18:40:44 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). **
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RESPONSE --> Earlier today, I tried accessing the randomized problem link to get the simulated data, but the link wouldnt work. I just tried the link again and it worked. I selected version 5 and got the data (82.1, 2.8) (76.8, 5.6) (72.7, 8.4) (69.8, 11.2) (66.8, 14) (63.7, 16.8) self critique assessment: 2
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18:41:21 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the first, second, and fourth points. confidence assessment: 3
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18:41:27 STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)**
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RESPONSE --> self critique assessment: 3
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18:45:00 Give the first of your three equations.
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RESPONSE --> My first equation for point (82.1, 2.8) was 7.84a + 2.8b + c = 82.1 confidence assessment: 3
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18:45:12 STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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RESPONSE --> self critique assessment: 3
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18:46:45 Give the second of your three equations.
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RESPONSE --> My equation for point (76.8, 5.6) was 31.36a + 5.6b + c = 76.8 confidence assessment: 3
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18:46:53 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> self critique assessment: 3
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18:48:27 Give the third of your three equations.
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RESPONSE --> My third equation for point (69.8, 11.2) was 125.44a + 11.2b + c = 69.8 confidence assessment: 3
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18:48:45 STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> self critique assessment: 3
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18:50:44 Give the first of the equations you got when you eliminated c.
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RESPONSE --> The first equation I got when I eliminated c was 23.52a + 2.8b = -5.3 confidence assessment: 3
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18:50:50 STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> self critique assessment: 3
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18:53:10 Give the second of the equations you got when you eliminated c.
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RESPONSE --> The second equation I got after eliminating c was 94.08a + 5.6b = -7 confidence assessment: 3
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18:53:18 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> self critique assessment: 3
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18:57:54 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied the first equation by 2 because that would give me equal values for b in the two equations. I then subtracted the first equation from the second, solved for a, and found that a rounds to .0765 confidence assessment: 3
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18:58:03 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. **
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RESPONSE --> self critique assessment: 3
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19:00:36 What values did you get for a and b?
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RESPONSE --> a rounds to .07 and b rounds to -2.5 confidence assessment: 3
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19:00:52 STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> self critique assessment: 3
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19:04:36 What did you then get for c?
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RESPONSE --> c is approximately equal to 88.6 confidence assessment: 3
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19:04:43 STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> self critique assessment: 3
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19:06:15 What is your function model?
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RESPONSE --> My function model is y = .07t^2 - 2.5t + 88.6 confidence assessment: 3
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19:06:24 STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> self critique assessment: 3
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19:06:51 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> ? confidence assessment: 0
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19:07:22 STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> I dont understand what the question was asking for. self critique assessment: 2
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19:27:08 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> I used the depth of 70 and used the quadratic formula to find the time of 10.6 seconds. confidence assessment: 1
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19:27:22 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. **
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RESPONSE --> self critique assessment: 3
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11:55:58 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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RESPONSE --> im not sure what data the question is talking about confidence assessment: 0
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11:57:29 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). **
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RESPONSE --> i understand now. the question said clock time vs. grade but the actual data is percent of assignments reveiwed vs. grade. self critique assessment: 2
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12:01:51 What three points on your graph did you use as a basis for your model?
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RESPONSE --> i used points (20, 2.118034) (60, 2.936492) (90, 3.371708) confidence assessment: 3
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12:01:58 STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> self critique assessment: 3
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12:04:08 Give the first of your three equations.
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RESPONSE --> 2.118034 = 400a + 20b + c confidence assessment: 3
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12:04:52 STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> self critique assessment: 3
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12:06:23 Give the second of your three equations.
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RESPONSE --> 3600a + 60b + c = 2.936492 confidence assessment: 3
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12:06:29 STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> self critique assessment: 3
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12:07:30 Give the third of your three equations.
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RESPONSE --> 8100a + 90b + c = 3.371708 confidence assessment: 3
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12:07:39 STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> self critique assessment: 3
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12:09:38 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I subtracted the first equation from the second and got 3200a + 40b = .818458 confidence assessment: 3
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12:09:44 STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> self critique assessment: 3
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12:11:45 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I subtracted the second equation from the third and got 4500a + 30b = .435216 confidence assessment: 3
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12:11:53 STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> self critique assessment: 3
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12:15:50 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied the first equation by 3 and the second equation by 4 in order to get the b values equal. I then subtracted the first equation from the second confidence assessment: 3
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12:16:02 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. **
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RESPONSE --> self critique assessment: 3
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12:19:47 What values did you get for a and b?
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RESPONSE --> I rounded using significant numbers and got a = .000085 b = .013661 confidence assessment: 2
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12:19:57 STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> self critique assessment: 3
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12:22:11 What did you then get for c?
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RESPONSE --> c = 1.810814 confidence assessment: 2
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12:22:17 STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> self critique assessment: 3
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12:23:32 What is your function model?
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RESPONSE --> y = .000085x^2 + .013661b + 1.810814 confidence assessment: 3
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12:23:49 STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE --> self critique assessment: 3
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12:29:56 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> i dont understand confidence assessment: 0
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12:31:24 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. **
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RESPONSE --> i understand. i chose a number that gave me the square root of a negative number self critique assessment: 2
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12:33:30 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> I chose the percent review of 45 and got a grade average of 2.597684 confidence assessment: 2
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12:33:41 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. **
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RESPONSE --> self critique assessment: 3
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12:50:45 How well does your model fit the data (support your answer)?
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RESPONSE --> My model did not fit the data very well because I had an average deviation of .1942271 confidence assessment: 2
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12:50:54 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. **
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RESPONSE --> self critique assessment: 3
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13:48:02 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (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 assessment: 3
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13:48:15 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)**
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RESPONSE --> self critique assessment: 3
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13:49:24 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used points (2, 264.4411) (3, 105.1209) (7, 19.92772) confidence assessment: 3
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13:49:29 STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> self critique assessment: 3
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13:50:04 Give the first of your three equations.
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RESPONSE --> 4a + 2b + c = 264.4411 confidence assessment: 3
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13:50:10 STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> self critique assessment:
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13:50:44 Give the second of your three equations.
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RESPONSE --> 9a + 3b + c = 105.1209 confidence assessment: 3
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13:50:50 STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> self critique assessment: 3
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13:51:22 Give the third of your three equations.
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RESPONSE --> 49a + 7b + c = 19.92772 confidence assessment: 3
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13:51:29 STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> self critique assessment: 3
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13:52:01 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 5a + b = -159.3202 confidence assessment: 3
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13:52:07 STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> self critique assessment: 3
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13:52:35 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 40a + 4b = -85.19318 confidence assessment: 3
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13:52:41 STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> self critique assessment: 3
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13:53:17 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 **
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RESPONSE --> I multiplied the first equation by 4 and eliminated the b values. self critique assessment: 3
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13:54:03 What values did you get for a and b?
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RESPONSE --> a = 27.604381 b = -297.342105 confidence assessment: 3
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13:54:13 STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> self critique assessment: 3
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13:54:39 What did you then get for c?
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RESPONSE --> c = 748.707786 confidence assessment: 3
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13:54:43 STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> self critique assessment: 3
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13:56:05 What is your function model?
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RESPONSE --> y = 27.604381x^2 - 297.342105x + 748.707786 confidence assessment: 3
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13:56:19 STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> 3 self critique assessment: 3
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13:57:12 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> For the distance of 1.6, I got the illumination of 564.4626814 confidence assessment: 3
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13:57:37 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. **
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RESPONSE --> self critique assessment: 3
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13:59:09 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> For the illumination range of 25-100 w/m^2 I got the distances ranging from 7.055946165 through 3.039200446 confidence assessment: 3
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13:59:50 The precise solution depends on the model and the range of averages. For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations 25=9.4 r^2 - 139 r + 500 and 100 =9.4 r^2 - 139 r + 500 Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data. The solutions which correspond to the data are r = 3.9 when y = 100 and r = 5.4 when y = 25. So when the distance x has range 50% - 69% if the illumination range is 25 to 100. Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **
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RESPONSE --> self critique assessment: 3
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14:06:24 ppCal1 Section 0.2 EXTRA QUESTION. What is the midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?{}{}What is the midpoint between the points (3, 8) and (7, 12)?
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RESPONSE --> The midpoint is the average of two numbers with the two numbers as endpoints. My interval is [7, 21]. The midpoint is 14. I added the two numbers together and divided by 2 to find the midpoint. The midpoint between the points (3, 8) and (7, 12) is the point (5, 10). confidence assessment: 3
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14:06:47 You are given two points. The points each have two coordinates. You have to average the x coordinates to get the x coordinate of the midpoint, then average the y coordinates to get the y coordinate of the midpoint. For example if the points are (3, 8) and (7, 12), the average of the x coordinates is (3 + 7) / 2 = 5 and the average of the y coordinates is (7 + 12) / 2 = 9.5 so the coordinates of the midpoint are (5, 9.5). **
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RESPONSE --> 3 self critique assessment:
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14:09:36 0.2.24 (was 0.2.14 solve abs(3x+1) >=4
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RESPONSE --> The inequality must be changed into 2 inequalities 3x + 1 >= 4 or 3x + 1 <= -4 I then solve algebraically and find x >= 1 or x <= -5/3 confidence assessment: 2
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14:09:47 abs(a) >= b translates to a >= b OR a <= -b. In this case abs(3x+1) > 4 gives you 3x + 1 >= 4 OR 3x + 1 <= -4, which on solution for x gives x >= 1 OR x < = -5/3. **
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RESPONSE --> self critique assessment: 3
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14:10:20 the given inequality is equivalent to the two inequalities 3x+1 >= 4 and 3x+1 =< -4. The solution to the first is x >= 1. The solution to the second is x <= -5/3. Thus the solution is x >= 1 OR x <= -5/3. COMMON ERROR: -5/3 > x > 1 INSTRUCTOR COMMENT: It isn't possible for -5/3 to be greater than a quantity and to have that same quantity > 1. Had the inequality read |3x+1|<4 you could have translated it to -4 < 3x+1 <4, but you can't reverse these inequalities without getting the contradiction pointed out here. **
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RESPONSE --> self critique assessment: 3
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14:13:13 0.2.24 (was 0.2.16 solve abs(2x+1)<5. What inequality or inequalities did you get from the given inequality, and are these 'and' or 'or' inequalities? Give your solution.
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RESPONSE --> I get -5 < 2x + 1 < 5 -6 < 2x < 4 -3 < x < 2 These are ""and"" inequalities confidence assessment: 3
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14:13:25 abs(a) < b means a < b AND -b < -a so from the given inequality abs(2x+1) < 5 you get -5 < 2x+1 AND 2x+1 < 5. These can be combined into the form -5 < 2x+1 < 5 and solved to get your subsequent result. Subtracting 1 from all expressions gives us -6 < 2x < 4, then dividing through by 2 we get -3 < x < 2. **
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RESPONSE --> self critique assessment: 3
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14:14:49 0.2.7 (was 0.2.23 describe [-2,2 ]
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RESPONSE --> abs(x)<=2 confidence assessment: 2
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14:15:09 The interval [-2, 2] is centered at the midpoint between x=-2 and x=2. You can calculate this midpoint as (-2 + 2) / 2 = 0. It is also clear from a graph of the interval that it is centered at x = 0 The center is at 0. The distance to each endpoint is 2. The interval is | x - center | < distance to endpoints. So the interval here is | x - 0 | < 2, or just | x | < 2. **
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RESPONSE --> self critique assessment: 3
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14:19:38 0.2.22 (was 0.2.28 describe [-7,-1]
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RESPONSE --> abs(x + 4) < 3 confidence assessment: 2
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14:20:14 the interval is centered at -4 (midpt between -7 and -1). The distance from the center of the interval to -7 is 3, and the distance from the center of the interval to -1 is 3. This translates to the inequality | x - (-4) | < 3, which simplifies to give us | x + 4 | < 3. **
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RESPONSE --> self critique assessment: 3
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14:22:05 0.2.14 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE --> abs(x -22) > 2 confidence assessment: 2
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14:22:16 22 is at the center of the interval. The endpoints are 2 units from the midpoint, and are not included. Everything that lies more than 2 units from 22 is in one of the intervals, and everything in either of the intervals lies at least 2 units from 22. So the inequality that describes this union of two intervals is | x - 22 | > 2. **
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RESPONSE --> self critique assessment: 3
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14:23:34 0.2.44 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> [50, 65] confidence assessment: 3
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14:23:46 The inequality is translated as -1<=(w-57.5)/7.5<=1. Multiplying through by 7.5 we get -7.5<=w-57.5<=7.5 Now add 57.5 to all expressions to get -7.5 + 57.5 <= x <= 7.5 + 57.5 or 50 < x < 65, which tells you that the dogs weigh between 50 and 65 pounds. **
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RESPONSE --> self critique assessment: 3
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14:25:17 0.2.42 (was 0.2.38 stocks vary from 33 1/8 by no more than 2. What absolute value inequality or inequalities correspond(s) to this prediction?
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RESPONSE --> I have use the letter p to represent the price of stock abs( p - 33 1/8 ) <= 2 confidence assessment: 3
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14:25:33 this statement says that the 'distance' between a stock price and 33 1/8 must not be more than 2, so this distance is <= 2 The distance between a price p and 33 1/8 is | p - 33 1/8 |. The desired inequality is therefore | p = 33 1/8 | < = 2. **
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RESPONSE --> self critique assessment: 3
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