course Mth 271 ӨhЃWJꁡassignment #002
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13:19:29 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 --> (0,91) (20,60) (40,41) confidence assessment: 3
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13:26:05 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7,79) (19,61) (31,48) time in minutes, temp in degrees celsius confidence assessment: 3
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13:26:12 Continue to the next question **
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RESPONSE --> ok self critique assessment: 3
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13:26:43 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (20,60) (30,49) (40,41) confidence assessment: 3
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13:27:10 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 --> yes i could have chosen better points self critique assessment: 3
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13:27:39 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 60=400a +20b+ c confidence assessment: 3
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13:27:49 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 --> ok self critique assessment: 3
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13:28:16 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 49=900a+30b+c confidence assessment: 3
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13:28:22 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok self critique assessment: 3
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13:28:38 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 41=1600a+40b+c confidence assessment: 3
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13:28:43 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok self critique assessment: 3
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13:31:04 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 --> equation 2 - equation 3 gives 500a+10b= -11 eq 3-eq2 gives 700a+10b= -8 -(500a+10b=-11) -500a - 10b=11 + 700a +10b= -8 this gives 200a=3, a= .015 confidence assessment: 3
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13:31:12 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 --> ok self critique assessment: 3
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13:31:27 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 --> see above confidence assessment: 3
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13:32:23 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 --> ok self critique assessment: 3
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13:32:34 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> a=.015 confidence assessment: 3
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13:32:45 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 --> ok self critique assessment: 3
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13:33:22 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 --> 500(.015)+ 10b= -11 b= -1.85 confidence assessment: 3
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13:33:29 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 --> ok self critique assessment: 3
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13:33:37 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=91 confidence assessment: 3
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13:33:43 STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok self critique assessment: 3
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13:34:15 What is the resulting quadratic model?
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RESPONSE --> y= .015t^2 -1.85t +91 confidence assessment: 3
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13:34:26 STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> ok self critique assessment: 3
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13:35:27 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 --> 91, 74, and 60 degrees respectively deviations 4,1, and 0 confidence assessment: 3
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13:35:45 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 --> ok self critique assessment: 3
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13:36:04 What was your average deviation?
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RESPONSE --> avg deviation -1.125 confidence assessment: 3
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13:36:11 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok self critique assessment:
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13:36:31 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> no as sure about how to do avg deviation self critique assessment: 1
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13:37:15 Is there a pattern to your deviations?
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RESPONSE --> well i have 4,1,0,0,0, -1, -1, -9 so i would say yes confidence assessment: 3
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13:37:24 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 --> yup self critique assessment: 3
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13:37:33 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 confidence assessment: 3
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13:37:40 STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok self critique assessment: 3
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13:38:08 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> well i will confidence assessment: 3
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13:38:19 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 --> ok self critique assessment: 3
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13:40:12 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 --> time vs depth 5.3 63.7 10.6 54.8 15.9 46 21.2 37.7 26.5 32 31.8 26.6 confidence assessment: 3
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13:40:21 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 --> ok self critique assessment: 3
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13:42:14 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (5.3, 63.7) (15.9, 46) (31.8, 26.6) confidence assessment: 3
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13:42:28 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 --> ok self critique assessment: 3
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13:44:18 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 --> 63.7= 28.09a+5.3b+c self critique assessment: 3
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13:45:03 Give the second of your three equations.
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RESPONSE --> 46=252.81a+46b+c confidence assessment: 3
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13:45:39 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> i mixed up the points its 15.9 b, not 46 b self critique assessment: 3
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13:46:26 Give the third of your three equations.
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RESPONSE --> 26.6= 1011.24a+ 31.8b+c confidence assessment: 3
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13:46:33 STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok self critique assessment: 3
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14:04:27 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 224.72a+ 10.6b= -17.7 confidence assessment: 3
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14:04:36 STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok self critique assessment: 3
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14:05:02 Give the second of the equations you got when you eliminated c.
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RESPONSE --> eq3-2 759.14a+ 15.9b= -19.4 confidence assessment: 3
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14:05:06 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok self critique assessment: 3
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14:24:15 Explain how you solved for one of the variables.
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RESPONSE --> 15.9 (224.72a+ 10.6b) 10.6 (759.14a+ 15.9b) 3573.048a+ 168.54b= -17.7 - 8046.884a+ 168.54b= -19.4 which gives -4473.832a= 1.7 a= -3.8 x 10^-4 confidence assessment: 3
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14:24:34 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 --> ok self critique assessment: 3
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14:30:34 What values did you get for a and b?
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RESPONSE --> a= -3.8 x 10^-4 b=-1.66 confidence assessment: 3
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14:30:41 STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok self critique assessment: 3
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14:30:55 What did you then get for c?
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RESPONSE --> c=72.51 confidence assessment: 3
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14:31:03 STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok self critique assessment: 3
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14:32:05 What is your function model?
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RESPONSE --> y= -3.8 x10^-4t^2 -1.66t + 72.51 confidence assessment: 3
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14:32:16 STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok self critique assessment: 3
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14:34:40 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> no clock time given, so i picked 10 seconds (10, 56) s, cm confidence assessment: 3
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14:34:48 STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok self critique assessment: 3
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14:46:17 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> again,no given depth that i can find d= 30 t= -4394.74, which we throw out b/c can't have neg time answer is t= 26.32 s confidence assessment: 3
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14:46:26 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 --> ok self critique assessment: 3
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14:48:05 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 --> the data i have is % of assignments reviewed vs grade avg confidence assessment: 3
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14:48:28 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 --> same data as this student used self critique assessment: 3
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14:49:10 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (20, 2.118034) (50, 2.767767) (80, 3.236068) confidence assessment: 3
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14:49:14 STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok self critique assessment: 3
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14:49:48 Give the first of your three equations.
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RESPONSE --> 2.118034= 400a+20b+c confidence assessment: 3
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14:49:54 STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> ok self critique assessment: 3
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14:50:14 Give the second of your three equations.
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RESPONSE --> 2.767767=2500a+50b+c confidence assessment: 3
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14:50:19 STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> ok self critique assessment: 3
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14:50:39 Give the third of your three equations.
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RESPONSE --> 3.23606=6400a+80b+c confidence assessment: 3
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14:50:44 STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> ok self critique assessment: 3
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14:51:09 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq2-1 2100a+30b= .649733 confidence assessment: 3
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14:51:14 STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> ok self critique assessment: 3
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14:51:43 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 3900a+30b=.468301 eq3-2 confidence assessment: 3
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14:51:48 STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok self critique assessment: 3
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14:53:13 Explain how you solved for one of the variables.
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RESPONSE --> multiplied 1st eq by neg one, then add the two equations u get 1800a= -.181432 a=1.007955x10^-4 confidence assessment: 3
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14:53:18 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 --> ok self critique assessment: 3
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14:53:43 What values did you get for a and b?
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RESPONSE --> a= -1.007955x10^-4 b= .014602 confidence assessment: 3
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14:53:50 STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> ok self critique assessment: 3
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14:54:13 What did you then get for c?
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RESPONSE --> c= 1.7856758 confidence assessment: 3
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14:54:20 STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> ok self critique assessment: 3
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14:55:06 What is your function model?
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RESPONSE --> y= 1.007955x0^-4 t^2 +.014602t+ 1.7856758 confidence assessment: 3
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14:55:12 STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE --> ok self critique assessment: 3
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14:59:04 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> using the quad formula i end up w/ a non real answer due to a negative under the sq root. but by guesstimating i get (59.5, 3.0) and (80, 3.5) confidence assessment: 3
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14:59:49 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 --> yes i understand but my math does not correlate self critique assessment: 3
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14:59:59 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> can't find the answer confidence assessment: 0
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15:00:13 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 --> ok self critique assessment: 3
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15:00:48 How well does your model fit the data (support your answer)?
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RESPONSE --> my model does not fit the data well- there must be errors in the math used to develop the model as the quadratic formula will not work here confidence assessment: 0
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15:01:01 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 --> ok self critique assessment: 3
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15:04:35 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> distance illum 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 AU w/m^2 confidence assessment: 3
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15:04:41 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 --> yup self critique assessment: 3
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15:06:13 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (2, 264.4411) (6, 25.91537) (10, 9.484465) confidence assessment: 3
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15:06:19 STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> ok self critique assessment: 3
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15:07:03 Give the first of your three equations.
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RESPONSE --> 264.4411+4a+2b+c confidence assessment: 3
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15:07:08 STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> ok self critique assessment: 3
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15:07:31 Give the second of your three equations.
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RESPONSE --> 25.91537=36a+6b+c confidence assessment: 3
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15:07:35 STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> ok self critique assessment: 3
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15:07:52 Give the third of your three equations.
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RESPONSE --> 9.484465=100a+10b+c confidence assessment: 3
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15:07:56 STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> ok self critique assessment: 3
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15:08:29 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq2-1 32a+4b= -238.52573 confidence assessment: 3
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15:08:33 STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> ok self critique assessment: 3
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15:09:03 Give the second of the equations you got when you eliminated c.
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RESPONSE --> eq3-2 64a+4b=-16.430905 confidence assessment: 3
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15:09:08 STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> ok self critique assessment: 3
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15:09:33 Explain how you solved for one of the variables.
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RESPONSE --> subtract and you get 32a=222.094825 confidence assessment: 3
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15:09:38 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 --> ok self critique assessment: 3
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15:10:05 What values did you get for a and b?
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RESPONSE --> a= 6.9405 b= -115.0869 confidence assessment: 3
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15:10:11 STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> ok self critique assessment: 3
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15:10:23 What did you then get for c?
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RESPONSE --> c=466.8499 confidence assessment: 3
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15:10:29 STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> ok self critique assessment: 3
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15:11:05 What is your function model?
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RESPONSE --> y=6.9405t^2 -115.0869t= 466.8499 confidence assessment: 3
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15:11:11 STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> ok self critique assessment: 3
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15:11:53 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> (1.6, 300.48) AU, w/m^2 confidence assessment: 3
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15:12:05 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 --> ok self critique assessment: 3
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15:21:23 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> (6.04, 25) (4.305, 100) Au, w/m^2 confidence assessment: 3
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15:21:57 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 --> yup self critique assessment: 3
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15:23:28 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 --> add the points divide by 2 (-4.6, -1.3) M=-2.95 dont know confidence assessment: 2
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15:23:51 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 --> okay that makes sense self critique assessment: 3
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15:24:49 0.2.24 (was 0.2.14 solve abs(3x+1) >=4
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RESPONSE --> 3x+1 >_ 4 x>_ 1 i dont know about the rest, as to how to set up the other equation, i cant remember confidence assessment: 0
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15:26:01 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 --> i can do the math, but i just didnt know how to set up the second equation. self critique assessment: 3
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15:26:21 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 --> yes i understand that it is or self critique assessment: 3
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15:30:18 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 --> -5/2
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15:31:11 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 --> i copied the second equation down wrong self critique assessment: 3
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15:31:36 0.2.7 (was 0.2.23 describe [-2,2 ]
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RESPONSE --> i dont understand these problems at all! confidence assessment: 0
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15:33:20 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 --> this is explained much better here than in the book self critique assessment: 2
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15:35:49 0.2.22 (was 0.2.28 describe [-7,-1]
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RESPONSE --> -7+-1= -8/2=-4 midpt distance to one endpoint is 3, and other endpoint is 3 abs value of x-4<3 confidence assessment: 1
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15:36:06 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 --> ok im beginning to get it self critique assessment: 3
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15:37:24 0.2.14 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE --> no idea confidence assessment: 0
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15:37:59 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 --> the U threw me off self critique assessment: 3
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15:39:30 0.2.44 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> i really have no idea confidence assessment: 0
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15:41:13 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 --> again i understand the math, just not how to set up the equation at the beginning self critique assessment: 3
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15:43:03 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 --> $31.8<=x<=$35.8 confidence assessment: 3
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15:44:11 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 --> i would never have thought of this answer self critique assessment: 0
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