course Math 271 ??~??????k?d?assignment #002
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17:29:52 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 --> hopefully you are talking about the potaoto temps and times. the first one was (118, 0) the third one was (91.96194, 40) the fifth one was (73.29324, 80)
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17:35:33 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 at clock time 7, the temp would be approximately 111, at clock time 19 the temp would be approx 103 and last at 31 the approx temp would be 95.
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17:37:25 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> The points I used as a basis for my quadriatic model was (20, 104) , (60, 82) and (100, 66).
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17:38:18 ** 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 --> good, I did that by having a spread of 40 between each time.
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17:40:52 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My first equation for the points (20, 104) is 104 = a(20)^2 + 20b + c or simplified 104 = 400a + 20b + c
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17:41:00 ** 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
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17:42:07 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 the points (60, 82) is 82 = a(60)^2 + 60b + c or simplified 82 = 3600a + 60b + c
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17:42:21 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok
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17:43:31 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 the points (100, 66) is 66 = a(100)^2 + 100b + c or simplified 66 = 10,000a + 100b + c
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17:43:42 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok
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17:47:25 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 multiplied the first equation by negative 1 and added to the second equation to eliminate c. 82 = 3600a + 60b + c -104 = -400a - 20b -c which gives you the new equation of -22 = 3200a + 40b
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17:47:41 ** 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
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17:50:20 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 --> i subtracted the second equation from the first to eliminate c and get my second equation as follows: 66 = 10000a + 100b + c -82 = -3600a -60b - c which creates my new equation of -16 = 6400a + 40b
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17:52:45 ** 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
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17:56:21 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b next by multiplying the second entire equation by negative one and then adding the second equation to the first equation -22 = 3200a + 40b 16 = -6400a - 40b so now I have -6 = -3200a or a = .001875
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17:56:29 ** 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
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17:58:41 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 .001875 for a in the equation of -16 = 6400a + 40b -16 = 6400(.01875) + 40b -16 = 12 + 40b -28 = 40b b = -.7
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17:58:47 ** 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
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18:05:09 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> when I sbstituted a (.001875) and b(-.7) into the equation this is what I had 82 =3600(.001875) + 60(-.7) + c 82 = 6.75 - 42 + c 82 = -35.25 + c c = 82+ 35.25 c = 117.25
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18:05:18 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok
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18:06:34 What is the resulting quadratic model?
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RESPONSE --> My resulting quadratic equation is y = .001875(t)^2 -.7t + 117.25
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18:06:39 ** 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
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18:13:15 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 time of zero my model gave me 117.25 which is a deviation of -.75 (table had 118) I know I am going to have larger deviations than if I used the exact numbers but since I was learning this I rounded the table times. If I had used the exact table times the deviation would be a lot lower the second point of 20 gave me 104 a deviation from the table of .0004 because the table had 103.9006 the third point of 40 gave me 92.25 which is a deviation of .28806
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18:14:44 ** 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
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18:16:47 What was your average deviation?
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RESPONSE --> .2033 was my average deviation
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18:16:53 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok
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18:17:38 Is there a pattern to your deviations?
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RESPONSE --> Seems like closer the points I used the smaller the deviation.
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18:17:59 ** 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 --> oh, ok
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18:18:54 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 this very much and I do understand the process.
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18:19:00 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok
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18:23:01 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> Believe me, I have spent extra time on this to make sure I understand it and this is why it has taken alot more time than I had planned. I am now planning a great deal more time for this class. I was doing 2 hours a night, and Saturday about 3, but now I am doing 3 hours a night and approximately 8 on Saturday. I know I am behind but hopefully this next week I'll get caught up. I am learning this and understanding it, but it is taking me a long time. I want to do this. I am going to do this.
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18:23:07 ** 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
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18:31:11 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 --> ok, I hope you are talking about my flow model that I did with a pop bottle. my pairs were depth, time (depth is in cm) 15.5, 10 12.5, 20 8.5, 30 6, 40 4, 50 2.5, 60 1, 70 .5, 80
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18:31:21 ** 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
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18:34:07 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the points (30, 8.5), (50, 4) and (70, 1) as the basis for my model. I rounded the 8.5 to 9 to keep me from going insane.
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18:34:23 ** 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
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18:38:30 Give the first of your three equations.
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RESPONSE --> the first equation is 9 = a(30)^2 +30b + c or simplified 9 = 900a + 30b +c
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18:39:25 ** 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 --> oops my point was (30, 9)
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18:41:01 Give the second of your three equations.
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RESPONSE --> my second equation for the points of (50, 4) is 4 = a(50)^2 + 50b +c or 4 = 2500a + 50b +c
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18:41:08 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok
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18:42:46 Give the third of your three equations.
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RESPONSE --> the point of (70, 1) gives me the equation 1 = a(70)^2 + 70b + c
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18:42:50 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok
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19:00:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> when i eliminated c from the first two equations, I subtracted the first equation from the second equation 4 = 2500a +50b +c -9 -900a -30b -c which leaves -5 = 1600a + 20b
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19:00:44 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok
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19:02:50 Give the second of the equations you got when you eliminated c.
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RESPONSE --> when i eliminated c from the second two equations, I subtracted the second equation from the third equation 1 = 4900a +70b +c -4 -2500a -50b -c which leaves -3 = 2400a + 20b
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19:02:54 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok
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19:08:37 Explain how you solved for one of the variables.
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RESPONSE --> since the multiple of b was the same I just subtracted the equation of -5 = 1600a +20b from the equation of -3 = 2400a +20b. this gives us 2 = 800a or a = .0025
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19:08:49 ** 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
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19:09:09 What values did you get for a and b?
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RESPONSE --> a is .0025 and b is -.45
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19:09:12 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE -->
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19:09:22 What did you then get for c?
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RESPONSE --> c = 20.25
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19:09:26 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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19:10:38 What is your function model?
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RESPONSE --> my function model is y = .0025(t)^2 + (-.45)t + 20.25
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19:10:40 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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19:14:38 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> clock time 40 was 6 from my model it is y = .0025(40)^2 + (-.45)40 + 20.25 y= .0025(1600) + (-18) +20.25 y = 4 -18 +20.25 y = 6.25 pretty close
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19:14:47 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE -->
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19:15:28 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> the actual time was 6 that corresponded to the time of 40 and my model predicted 6.25
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19:16:22 ** 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 --> ok
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19:19:30 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 --> i really get confused on listing these i am listing these in order percent reviewed, grade ( 10, 1.36495) (40, 2.388479) (70, 3.083704)
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22:29:42 ** 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 must have had a different version, maybe the wrong one but all the ones I have are (0, .9258546) (10, 1.36495) (20, 1.750993) (30, 2090227) (40, 2.388479) (50, 2.650655) (60, 2.881118) (70, 3.083704) (80, 3.261786) (90, 3.418326) 100, 3.555931)
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22:31:19 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the following points for my model (10, 1.36495) (40, 2.388479) (70, 3.083704)
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22:31:25 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok
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22:32:06 Give the first of your three equations.
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RESPONSE --> My first equation is 1.36495 = 100a + 10b + c
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22:32:09 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE -->
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22:33:19 Give the second of your three equations.
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RESPONSE --> my second equation is for points (40, 2.388479) is 2.388479 = 1600a + 40b + c
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22:33:22 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE -->
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22:34:19 Give the third of your three equations.
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RESPONSE --> my final equation is for (70, 3.083704) which is 3.083704 = 4900a + 70b + c
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22:34:21 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE -->
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22:35:31 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 1.023529 = 1500a +30b
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22:35:37 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE -->
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22:36:31 Give the second of the equations you got when you eliminated c.
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RESPONSE --> my second equation after i eliminated c is .695225 = 330a + 30b
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22:36:41 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok
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22:43:12 Explain how you solved for one of the variables.
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RESPONSE --> since the two equations both had 30b it made it easy to subtract the one equation from the other to solve for a .695225 = 3300a + 30b -1.023529 = -1500 -30b which gives me: -.328304 = 1800a divide both sides by 1800 and get a = .000182391
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22:43:25 ** 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
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22:55:55 What values did you get for a and b?
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RESPONSE --> a = -.000182391 oops on that last page I forgot to put the negative sign. I inserted a into the equation .695225 = 330a + 30b .695225 = 330(-.000182391) + 30b .695225 = -.6018903 + 30b add .6018903 to both sides 30b = 1.2033806 then divide both sides by 30 b = .04012602
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22:56:00 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE -->
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22:56:15 What did you then get for c?
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RESPONSE --> c = .9819289
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22:56:17 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE -->
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22:57:52 What is your function model?
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RESPONSE --> my function model is y = -.000182391(t)^2 + .04012602b + .9819289
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22:57:56 ** STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE -->
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23:06:51 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> for the three i did or for all? the three points that for the point (10, 1.36495) my model predicted 1.40142 for the point (40, 2.388479) my model predicted 2.3023441 for the point (70, 30.83704) my model predicted 2.8970344 all pretty close
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00:07:24 ** 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 --> ok for 3.3 it would be 3.3 = -.000182391(x)^2 + .04012602x + .9819289 so to solve subtract 3.3 from each side to get -.000182391 x^2 + .04012602x + 2.31080711 = 0 i used the quadratic equation of x =[ -b + or - Sq(B^2 -4AC) ] / 2A This one I think I messed up the numbers at b or c or something , I just can't get it to work, I didn't round off the numbers
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00:08:07 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> i had 3.3 and 157% pretty sure it was wrong
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00:32:40 ** 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 --> i went back and reworked my problem again, and i found that a is -.000182391 b is .04323717667 and c is .950817333 so for this 75 percent it would be y = -.000182391(5625) + .04323717667(75) + .950817333 y = -1.015949375 + 3.24278825 + .950817333 y = 3.177646208
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00:33:03 How well does your model fit the data (support your answer)?
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RESPONSE --> it comes fairly close to the model
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00:33:20 ** 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
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00:36:02 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> the data for illumination is: (1, 1280) (2, 320) (3, 142.222) (4, 80) (5, 51.2) (6, 35.5556) (7, 26.12245) (8, 20) (9, 15.80247) (10, 12.8)
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00:36:06 ** 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 -->
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00:41:54 What three points on your graph did you use as a basis for your model?
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RESPONSE --> the three points i used are (2, 320) (4, 80) and (8, 20)
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00:42:06 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> ok
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00:42:27 Give the first of your three equations.
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RESPONSE --> 320 = 4a + 2b + c
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00:42:34 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE -->
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00:42:52 Give the second of your three equations.
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RESPONSE --> 80 = 16a + 4b + c
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00:42:54 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE -->
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00:43:24 Give the third of your three equations.
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RESPONSE --> 20 = 64a + 8b + c
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00:43:27 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE -->
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00:45:54 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -240 = 12a +2b
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00:45:56 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE -->
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00:46:16 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -60 = 48a +4b
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00:46:21 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE -->
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00:51:38 Explain how you solved for one of the variables.
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RESPONSE --> took the first equation and multiplyied by -2 then subtracted from the second -2 (-240) = -2(12a) + -2(2b) which equals 480 = -24a - 4b -60 = 48a + 4b 480 = -24a - 4b so 420 =24a divide both sides by 24 a = 17.5
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00:51:44 ** 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 -->
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00:52:05 What values did you get for a and b?
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RESPONSE --> a = 17.5 b = -225
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00:52:07 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE -->
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00:53:22 What did you then get for c?
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RESPONSE --> c = 700
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00:53:25 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE -->
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00:54:28 What is your function model?
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RESPONSE --> y = 17.5t^2 + (-225)t + 700
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00:54:30 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE -->
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00:57:49 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> my prediction for the distance of 6 is negative 6770 i just don't think this one works because of the extreme numbers, or I'm just blind to it
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01:01:54 ** 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 --> my distance was 6 and i just realized i missed the math so this time I only got negative 20 17.5 (6)^2 -225(6) + 700 = y 17.5 (36) -225(6) + 700 = y 630 - 1350 + 700 = y y = -20
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01:02:09 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE -->
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01:04:18 ** 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 --> I understand the concept, I am still working on it could it be the difference in the points I picked?
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01:09:47 ppCal1 Section 0.2 #6 midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?
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RESPONSE --> to find the midpoints you add them then divide by 2 the midpoint problems in my book are .02 # 35 #35 [7, 21] 7 + 21 =28/2 = 14 the midpoint of this one is 14
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01:13:41 ** 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 --> ok, the ones I had only have one set of points but I understand, but in your example, why didn't you divide (8 + 12) by 2 to find the average of the y coordinates. wouldn't you match 3 and 7 and then 8 and 12 so the coordinates of the midpoint would be (5, 10)?
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01:20:30 Section 0.2 #14 solve abs(3x+1) >=4
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RESPONSE --> x >=1 0r x>= -2
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01:20:52 ** 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 --> ok
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01:21:13 ** 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 --> ok
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01:32:36 Section 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 --> the inequality is an ""and"" the answer is: -3 < x < 2
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01:32:53 ** 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 --> ok
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01:40:23 Section 0.2 #23 describe [-2,2 ]
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RESPONSE --> This problem isn't in my book, but the point (-2, 2) would be negative two pints on the x axis and up positive 2 points on the y axis y+ + + + point (-2,2) o 2+ + ++++++++++++++++++++++++++++++ -2
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01:41:19 ** 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 --> I guess it is getting late, I read that completely wrong.
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01:42:05 Section 0.2 #28 describe [-7,-1]
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RESPONSE --> the midpoint would be -7 + -1 / 2 = -4
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01:42:41 ** 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
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17:46:27 Section 0.2 #30 describe (-infinity, 20) U (24, infinity)
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RESPONSE --> Abs(x -2) > 2
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17:46:35 ** 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 --> ok
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17:48:31 Query AppCal1 Section 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> 50<= x <= 65
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17:50:13 ** 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 --> ok
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17:51:39 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 --> abs (p - 33 1/8) <=2
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17:53:59 ** 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 don't understand p = 33 1/8 looks like to me the abs value of p minus 33 1/8 would be equal to or less than 2
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"
course Math 271 ??~??????k?d?assignment #002
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17:29:52 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 --> hopefully you are talking about the potaoto temps and times. the first one was (118, 0) the third one was (91.96194, 40) the fifth one was (73.29324, 80)
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17:35:33 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 at clock time 7, the temp would be approximately 111, at clock time 19 the temp would be approx 103 and last at 31 the approx temp would be 95.
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17:37:25 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> The points I used as a basis for my quadriatic model was (20, 104) , (60, 82) and (100, 66).
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17:38:18 ** 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 --> good, I did that by having a spread of 40 between each time.
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17:40:52 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My first equation for the points (20, 104) is 104 = a(20)^2 + 20b + c or simplified 104 = 400a + 20b + c
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17:41:00 ** 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
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17:42:07 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 the points (60, 82) is 82 = a(60)^2 + 60b + c or simplified 82 = 3600a + 60b + c
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17:42:21 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok
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17:43:31 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 the points (100, 66) is 66 = a(100)^2 + 100b + c or simplified 66 = 10,000a + 100b + c
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17:43:42 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok
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17:47:25 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 multiplied the first equation by negative 1 and added to the second equation to eliminate c. 82 = 3600a + 60b + c -104 = -400a - 20b -c which gives you the new equation of -22 = 3200a + 40b
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17:47:41 ** 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
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17:50:20 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 --> i subtracted the second equation from the first to eliminate c and get my second equation as follows: 66 = 10000a + 100b + c -82 = -3600a -60b - c which creates my new equation of -16 = 6400a + 40b
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17:52:45 ** 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
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17:56:21 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b next by multiplying the second entire equation by negative one and then adding the second equation to the first equation -22 = 3200a + 40b 16 = -6400a - 40b so now I have -6 = -3200a or a = .001875
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17:56:29 ** 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
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17:58:41 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 .001875 for a in the equation of -16 = 6400a + 40b -16 = 6400(.01875) + 40b -16 = 12 + 40b -28 = 40b b = -.7
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17:58:47 ** 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
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18:05:09 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> when I sbstituted a (.001875) and b(-.7) into the equation this is what I had 82 =3600(.001875) + 60(-.7) + c 82 = 6.75 - 42 + c 82 = -35.25 + c c = 82+ 35.25 c = 117.25
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18:05:18 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok
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18:06:34 What is the resulting quadratic model?
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RESPONSE --> My resulting quadratic equation is y = .001875(t)^2 -.7t + 117.25
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18:06:39 ** 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
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18:13:15 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 time of zero my model gave me 117.25 which is a deviation of -.75 (table had 118) I know I am going to have larger deviations than if I used the exact numbers but since I was learning this I rounded the table times. If I had used the exact table times the deviation would be a lot lower the second point of 20 gave me 104 a deviation from the table of .0004 because the table had 103.9006 the third point of 40 gave me 92.25 which is a deviation of .28806
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18:14:44 ** 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
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18:16:47 What was your average deviation?
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RESPONSE --> .2033 was my average deviation
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18:16:53 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok
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18:17:38 Is there a pattern to your deviations?
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RESPONSE --> Seems like closer the points I used the smaller the deviation.
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18:17:59 ** 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 --> oh, ok
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18:18:54 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 this very much and I do understand the process.
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18:19:00 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok
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18:23:01 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> Believe me, I have spent extra time on this to make sure I understand it and this is why it has taken alot more time than I had planned. I am now planning a great deal more time for this class. I was doing 2 hours a night, and Saturday about 3, but now I am doing 3 hours a night and approximately 8 on Saturday. I know I am behind but hopefully this next week I'll get caught up. I am learning this and understanding it, but it is taking me a long time. I want to do this. I am going to do this.
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18:23:07 ** 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
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18:31:11 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 --> ok, I hope you are talking about my flow model that I did with a pop bottle. my pairs were depth, time (depth is in cm) 15.5, 10 12.5, 20 8.5, 30 6, 40 4, 50 2.5, 60 1, 70 .5, 80
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18:31:21 ** 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
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18:34:07 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the points (30, 8.5), (50, 4) and (70, 1) as the basis for my model. I rounded the 8.5 to 9 to keep me from going insane.
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18:34:23 ** 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
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18:38:30 Give the first of your three equations.
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RESPONSE --> the first equation is 9 = a(30)^2 +30b + c or simplified 9 = 900a + 30b +c
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18:39:25 ** 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 --> oops my point was (30, 9)
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18:41:01 Give the second of your three equations.
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RESPONSE --> my second equation for the points of (50, 4) is 4 = a(50)^2 + 50b +c or 4 = 2500a + 50b +c
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18:41:08 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok
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18:42:46 Give the third of your three equations.
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RESPONSE --> the point of (70, 1) gives me the equation 1 = a(70)^2 + 70b + c
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18:42:50 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok
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19:00:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> when i eliminated c from the first two equations, I subtracted the first equation from the second equation 4 = 2500a +50b +c -9 -900a -30b -c which leaves -5 = 1600a + 20b
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19:00:44 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok
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19:02:50 Give the second of the equations you got when you eliminated c.
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RESPONSE --> when i eliminated c from the second two equations, I subtracted the second equation from the third equation 1 = 4900a +70b +c -4 -2500a -50b -c which leaves -3 = 2400a + 20b
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19:02:54 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok
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19:08:37 Explain how you solved for one of the variables.
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RESPONSE --> since the multiple of b was the same I just subtracted the equation of -5 = 1600a +20b from the equation of -3 = 2400a +20b. this gives us 2 = 800a or a = .0025
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19:08:49 ** 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
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19:09:09 What values did you get for a and b?
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RESPONSE --> a is .0025 and b is -.45
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19:09:12 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE -->
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19:09:22 What did you then get for c?
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RESPONSE --> c = 20.25
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19:09:26 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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19:10:38 What is your function model?
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RESPONSE --> my function model is y = .0025(t)^2 + (-.45)t + 20.25
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19:10:40 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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19:14:38 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> clock time 40 was 6 from my model it is y = .0025(40)^2 + (-.45)40 + 20.25 y= .0025(1600) + (-18) +20.25 y = 4 -18 +20.25 y = 6.25 pretty close
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19:14:47 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE -->
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19:15:28 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> the actual time was 6 that corresponded to the time of 40 and my model predicted 6.25
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19:16:22 ** 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 --> ok
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19:19:30 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 --> i really get confused on listing these i am listing these in order percent reviewed, grade ( 10, 1.36495) (40, 2.388479) (70, 3.083704)
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22:29:42 ** 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 must have had a different version, maybe the wrong one but all the ones I have are (0, .9258546) (10, 1.36495) (20, 1.750993) (30, 2090227) (40, 2.388479) (50, 2.650655) (60, 2.881118) (70, 3.083704) (80, 3.261786) (90, 3.418326) 100, 3.555931)
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22:31:19 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the following points for my model (10, 1.36495) (40, 2.388479) (70, 3.083704)
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22:31:25 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok
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22:32:06 Give the first of your three equations.
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RESPONSE --> My first equation is 1.36495 = 100a + 10b + c
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22:32:09 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE -->
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22:33:19 Give the second of your three equations.
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RESPONSE --> my second equation is for points (40, 2.388479) is 2.388479 = 1600a + 40b + c
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22:33:22 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE -->
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22:34:19 Give the third of your three equations.
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RESPONSE --> my final equation is for (70, 3.083704) which is 3.083704 = 4900a + 70b + c
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22:34:21 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE -->
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22:35:31 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 1.023529 = 1500a +30b
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22:35:37 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE -->
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22:36:31 Give the second of the equations you got when you eliminated c.
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RESPONSE --> my second equation after i eliminated c is .695225 = 330a + 30b
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22:36:41 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok
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22:43:12 Explain how you solved for one of the variables.
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RESPONSE --> since the two equations both had 30b it made it easy to subtract the one equation from the other to solve for a .695225 = 3300a + 30b -1.023529 = -1500 -30b which gives me: -.328304 = 1800a divide both sides by 1800 and get a = .000182391
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22:43:25 ** 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
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22:55:55 What values did you get for a and b?
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RESPONSE --> a = -.000182391 oops on that last page I forgot to put the negative sign. I inserted a into the equation .695225 = 330a + 30b .695225 = 330(-.000182391) + 30b .695225 = -.6018903 + 30b add .6018903 to both sides 30b = 1.2033806 then divide both sides by 30 b = .04012602
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22:56:00 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE -->
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22:56:15 What did you then get for c?
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RESPONSE --> c = .9819289
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22:56:17 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE -->
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22:57:52 What is your function model?
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RESPONSE --> my function model is y = -.000182391(t)^2 + .04012602b + .9819289
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22:57:56 ** STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE -->
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23:06:51 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> for the three i did or for all? the three points that for the point (10, 1.36495) my model predicted 1.40142 for the point (40, 2.388479) my model predicted 2.3023441 for the point (70, 30.83704) my model predicted 2.8970344 all pretty close
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00:07:24 ** 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 --> ok for 3.3 it would be 3.3 = -.000182391(x)^2 + .04012602x + .9819289 so to solve subtract 3.3 from each side to get -.000182391 x^2 + .04012602x + 2.31080711 = 0 i used the quadratic equation of x =[ -b + or - Sq(B^2 -4AC) ] / 2A This one I think I messed up the numbers at b or c or something , I just can't get it to work, I didn't round off the numbers
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00:08:07 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> i had 3.3 and 157% pretty sure it was wrong
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00:32:40 ** 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 --> i went back and reworked my problem again, and i found that a is -.000182391 b is .04323717667 and c is .950817333 so for this 75 percent it would be y = -.000182391(5625) + .04323717667(75) + .950817333 y = -1.015949375 + 3.24278825 + .950817333 y = 3.177646208
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00:33:03 How well does your model fit the data (support your answer)?
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RESPONSE --> it comes fairly close to the model
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00:33:20 ** 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
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00:36:02 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> the data for illumination is: (1, 1280) (2, 320) (3, 142.222) (4, 80) (5, 51.2) (6, 35.5556) (7, 26.12245) (8, 20) (9, 15.80247) (10, 12.8)
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00:36:06 ** 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 -->
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00:41:54 What three points on your graph did you use as a basis for your model?
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RESPONSE --> the three points i used are (2, 320) (4, 80) and (8, 20)
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00:42:06 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> ok
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00:42:27 Give the first of your three equations.
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RESPONSE --> 320 = 4a + 2b + c
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00:42:34 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE -->
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00:42:52 Give the second of your three equations.
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RESPONSE --> 80 = 16a + 4b + c
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00:42:54 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE -->
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00:43:24 Give the third of your three equations.
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RESPONSE --> 20 = 64a + 8b + c
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00:43:27 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE -->
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00:45:54 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -240 = 12a +2b
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00:45:56 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE -->
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00:46:16 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -60 = 48a +4b
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00:46:21 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE -->
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00:51:38 Explain how you solved for one of the variables.
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RESPONSE --> took the first equation and multiplyied by -2 then subtracted from the second -2 (-240) = -2(12a) + -2(2b) which equals 480 = -24a - 4b -60 = 48a + 4b 480 = -24a - 4b so 420 =24a divide both sides by 24 a = 17.5
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00:51:44 ** 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 -->
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00:52:05 What values did you get for a and b?
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RESPONSE --> a = 17.5 b = -225
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00:52:07 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE -->
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00:53:22 What did you then get for c?
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RESPONSE --> c = 700
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00:53:25 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE -->
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00:54:28 What is your function model?
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RESPONSE --> y = 17.5t^2 + (-225)t + 700
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00:54:30 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE -->
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00:57:49 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> my prediction for the distance of 6 is negative 6770 i just don't think this one works because of the extreme numbers, or I'm just blind to it
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01:01:54 ** 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 --> my distance was 6 and i just realized i missed the math so this time I only got negative 20 17.5 (6)^2 -225(6) + 700 = y 17.5 (36) -225(6) + 700 = y 630 - 1350 + 700 = y y = -20
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01:02:09 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE -->
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01:04:18 ** 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 --> I understand the concept, I am still working on it could it be the difference in the points I picked?
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01:09:47 ppCal1 Section 0.2 #6 midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?
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RESPONSE --> to find the midpoints you add them then divide by 2 the midpoint problems in my book are .02 # 35 #35 [7, 21] 7 + 21 =28/2 = 14 the midpoint of this one is 14
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01:13:41 ** 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 --> ok, the ones I had only have one set of points but I understand, but in your example, why didn't you divide (8 + 12) by 2 to find the average of the y coordinates. wouldn't you match 3 and 7 and then 8 and 12 so the coordinates of the midpoint would be (5, 10)?
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01:20:30 Section 0.2 #14 solve abs(3x+1) >=4
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RESPONSE --> x >=1 0r x>= -2
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01:20:52 ** 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 --> ok
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01:21:13 ** 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 --> ok
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01:32:36 Section 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 --> the inequality is an ""and"" the answer is: -3 < x < 2
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01:32:53 ** 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 --> ok
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01:40:23 Section 0.2 #23 describe [-2,2 ]
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RESPONSE --> This problem isn't in my book, but the point (-2, 2) would be negative two pints on the x axis and up positive 2 points on the y axis y+ + + + point (-2,2) o 2+ + ++++++++++++++++++++++++++++++ -2
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01:41:19 ** 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 --> I guess it is getting late, I read that completely wrong.
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01:42:05 Section 0.2 #28 describe [-7,-1]
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RESPONSE --> the midpoint would be -7 + -1 / 2 = -4
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01:42:41 ** 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
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17:46:27 Section 0.2 #30 describe (-infinity, 20) U (24, infinity)
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RESPONSE --> Abs(x -2) > 2
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17:46:35 ** 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 --> ok
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17:48:31 Query AppCal1 Section 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> 50<= x <= 65
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17:50:13 ** 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 --> ok
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17:51:39 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 --> abs (p - 33 1/8) <=2
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17:53:59 ** 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 don't understand p = 33 1/8 looks like to me the abs value of p minus 33 1/8 would be equal to or less than 2
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