course 271 6/18 ښB슖Eassignment #002
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06:24:47 The three points on the graph are (3, 5000), (7, 5300) and (12, 5500). The rise between the first point and the second is from 5000 to 5300, or 300, and the run is from 3 to 7, or 4, so the slope is 300 / 4 = 75. Note that the 300 represents $300 and the 4 represents 4 months, so the slope represents $300 / (4 months) = $75 / month, which is the average rate of change during the first time interval. The rise between the second point and the third is from 5300 to 5500, or 200, and the run from 7 to 12 is 5, so the slope is 200 / 5 = 40. This slope represents the $40/month average rate of change during the second time interval. Click on 'Next Picture' to see the graph.
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RESPONSE --> The slope between March and july will be 300/4 or 75. The slope between July and december is 200 divided by 5, or 40. self critique assessment: 3
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06:26:42 `q003. To what extent do you think your graph, consisting of 3 points with straight line segments between them, accurately depicts the detailed behavior of the stocks over the 9-month period?
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RESPONSE --> They are the same. With a straight line. Not very, There would most likely be more variation in the market. confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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06:27:39 `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period?
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RESPONSE --> In a general sense yes, or at least we know the outcome of each period. confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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{jޥ[¬ assignment #002 002. `Query 2 Applied Calculus I 06-18-2009 iK㮇譽|yI assignment #002 002. `Query 2 Applied Calculus I 06-18-2009
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19:17:59 Continue to the next question **
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RESPONSE --> 95,60,41 self critique assessment: 3
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19:18:39 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> 80, 60, 50 confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:19:32 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> 30, 49 40, 41 50, 35 confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:21:07 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 --> 49=900a+30b+c self critique assessment: 3
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19:21:51 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> 41=a1600+40b+c self critique assessment: 3
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19:22:32 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> 35=a2500+50b+c self critique assessment: 3
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19:23:28 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 --> In this particular problem, I found c by using the y intercept instead of eliminating self critique assessment: 3
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19:24:01 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:26:45 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 --> b=-1633/840 confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:27:02 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> 95 confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:29:05 STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> y=(23/1680)x^2+(1633/840)x+95 self critique assessment: 3
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19:31:38 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 --> 95 77.93 62.42 0 2.43 2.42 self critique assessment: 3
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19:31:58 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> .05 self critique assessment: 3
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19:32:30 Is there a pattern to your deviations?
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RESPONSE --> yes, they decrease and then increase confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:33:26 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 --> For the most part yes. It was difficult with the number which we were given to put together the equations, but not impossible confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:34:00 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I doubt it, seriously...but ill try confidence assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:38:07 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 --> 0, 100 1,90 2,80 3,70 4,60 5,50 6,40 7,30 8,20 The randomized data wasnt available at the time I was working the problems, so I made up my own self critique assessment: 3
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19:39:43 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 --> 0,100 3,70 6,40 self critique assessment: 3
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19:41:09 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 --> 0=0+0+100 self critique assessment:
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19:43:11 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> 3=a4900+b70+c self critique assessment:
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19:44:18 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Again, I used the y intercept to eliminate this step confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:45:24 Explain how you solved for one of the variables.
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RESPONSE --> I used the process of addition to solve simultaneous equations confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:46:11 What values did you get for a and b?
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RESPONSE --> a=0 b=-10 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:46:22 What did you then get for c?
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RESPONSE --> 100 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:46:58 What is your function model?
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RESPONSE --> y=0x^2+-10x+100 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:48:29 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> -4500@46 seconds confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:48:45 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> 14@ about 8.5 seconds confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:50:40 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 --> 0,1 10,1.79 20,2.11 30,2.36 40,2.58 50,2.76 etc... confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:52:05 What three points on your graph did you use as a basis for your model?
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RESPONSE --> 10,1.79 60,2.93 90,3.37 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:53:30 Give the first of your three equations.
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RESPONSE --> 1.79=100x+10x+c confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:55:03 Give the second of your three equations.
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RESPONSE --> 2.93=3600a+60b+c confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:55:46 Give the third of your three equations.
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RESPONSE --> 3.37=8100a+90b+c confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:56:04 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Again, I used the y intercept to find c confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:56:55 Explain how you solved for one of the variables.
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RESPONSE --> I used the process of addition, which was ridiculous considering the numbers involved confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:57:23 What values did you get for a and b?
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RESPONSE --> a=.00035 b=.0755 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:57:37 What did you then get for c?
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RESPONSE --> 1 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:58:55 What is your function model?
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RESPONSE --> y=.00035x^2+.0755x+1 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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19:59:31 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> 3.0=23.5 4.0=35.5 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:00:50 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 7.83 for 80 percent confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:01:03 How well does your model fit the data (support your answer)?
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RESPONSE --> Not well at all confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:01:59 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> 1,935 2,264 3,105 4,61 5,43 6,25 etc... confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:02:40 What three points on your graph did you use as a basis for your model?
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RESPONSE --> 1, 935 5, 43 10, 9.4 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:03:41 Give the first of your three equations.
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RESPONSE --> 935=a+b+c 43=25a+5b+c 9.4=100a+10b+c confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:04:53 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -925.6=99a+9b -33.6=75a+5b confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:05:58 Explain how you solved for one of the variables.
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RESPONSE --> I used the addition method confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:06:34 What values did you get for a and b?
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RESPONSE --> a=24 b=-367 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:06:46 What did you then get for c?
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RESPONSE --> 1278 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:07:38 What is your function model?
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RESPONSE --> y=24x^2+-367x+1278 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:08:04 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> 752.24@1.6 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:08:34 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> around 4.6 to 5.2 for 25 to 100 confidence assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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20:12:26 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 --> The point halfway between. 5, 10 self critique assessment:
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20:16:25 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 --> absolute value of x is greater than or equal to 1 or less than or equal to 5/3 self critique assessment:
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20:18:50 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 --> x is greater than or equal to -3 and less than or equal to 2 self critique assessment:
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20:19:35 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 --> absolute value of x is less than 2 self critique assessment:
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20:22:13 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 --> absolute value of x+4 is <3 self critique assessment:
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20:24:14 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 --> absolute value of x-22 is > 2 self critique assessment:
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20:27:55 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 --> x is less than 75 and greater than 50, the given equation is (w-68.5/2.7)<=1 self critique assessment:
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20:29:19 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 --> abs(p-33.15)<2 self critique assessment:
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