course Mth 158 ?????{???v?n?i????assignment #020 020. `query 20 College Algebra 07-26-2008
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06:10:13 2.6.8 (was 2.5.6). graph like basic stretched cubic centered around (20,20) How well does the graph appear to indicate a linear relation? Describe any significant deviation of the data from its best-fit linear approximation.
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RESPONSE --> It does indicate a linear relationship. confidence assessment: 1
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06:10:42 ** The graph is curved and in fact changes its concavity. The data points will lie first above the best-fit straight line, then as the straight line passes through the data set the data points will lie below this line. **
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RESPONSE --> Ok self critique assessment: 3
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06:20:01 2.5.12. x = 5, 10, ..., 25; y = 2, 4, 7, 11, 18 **** What two points did you select on the line you graphed, and what is the equation of the line through those points? **** What is the equation of the best-fit line and how well does the line fit the data? Describe any systematic deviation of the line from the best-fit line. ****
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RESPONSE --> (5,2) (10,4) m=(4-2)/(10-5)=2/5 y-2=(2/5)(X-5) y-2=2/5x-2 y=2/5x It appears to be a good fit from the chosen points, but the problem in the book is different and I could only see the first two points from the query program. confidence assessment: 3
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06:20:51 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I chose the points (5,2) and (10,4) The slope between these points is slope = rise / run = (4-2)/(10-5) = 2/5 so the equation is y-4 = 2/5(x - 10), which we solve for y to get y = 2/5 x. INTRUCTOR COMMENT: This fits the first two data points, but these are not appropriate points to select. The data set curves, with increasing slope as we move to the right. You need to sketch the best-fit line, as best you can see it, and pick two points on that line. The best-fit line is not likely to pass through any of the data points, and you should never use data points to determine the equation of the best-fit line. Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that comes as close as possible on the average to the points. Extend the line slightly beyond the data set. Estimate the y coordinates of the x = 1 and x = 20 points of this line. Find the equation of the straight line through these points. The coordinates of your points should be reasonably close to (1, 5.5) and (20, 30), though because it's a little difficult to judge exactly where the line should be you are unlikely to obtain these exact results. The equation will be reasonably close to y = .8 x - 3. **
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RESPONSE --> ok self critique assessment: 3
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06:33:20 2.5.18. Incomes 15k, 20k, ..., 70k, loan amts 40.6, 54.1, 67.7, 81.2, 94.8, 108.3, 121.9, 135.4, 149, 162.5, 176.1, 189.6 k.
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RESPONSE --> m=(94800-40600)/(35000-15000) m=54200/20000=1084/400 y-40600=(1084/400)(42000-15000) Y-40600=113820-40650 y-40600=73170 y=113,770 A person with a $42000 income would qualify for a $113,770 loan. confidence assessment: 3
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06:35:52 ERRONEOUS STUDENT SOLUTION WITH INSTRUCTOR COMMENT: Using the points (15,000, 40,600) and (20,000 , 67,700) we obtain slope = rise / run = (67,700 - 40,600) / (20,000 - 15,000) = 271/50 This gives us the equation y - 40,600= (271/50) * (x - 15,000), which we solve for y to obtain y = (271/50) x - 40,700. INSTRUCTOR COMMENT: You followed most of the correct steps to get the equation of the line from your two chosen points. However I think the x = 20,000 value is y = 54,100, not 67,700; the latter corresponds to x = 25,000. So your equation won't be likely to fit the data very well. Another reason that your equation is not likely to be a very good fit is that you used two data points, which is inappropriate; and in addition you used two data points near the beginning of the data list. If you were going to use two data points you would need to use two typical points much further apart. {]In any case to solve this problem you need to sketch the best-fit line, as best you can see it, and pick two points on that line. The best-fit line is not likely to pass through any of the data points, and you should never use data points to determine the equation of the best-fit line. Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that comes as close as possible on the average to the points. Extend the line slightly beyond the data set. Estimate the y coordinates of the x = 10,000 and x = 75,000 points of this line. Find the equation of the straight line through these points. The coordinates of your points should be reasonably close to (5000, 19000) and (75000, 277,000). It's fairly easy to locate this line, which does closely follow the data points, though due to errors in estimating you are unlikely to obtain these exact results. The equation will be reasonably close to y = 2.7 x - 700 . If we let y = 42,000 we can solve for x: 42,000 = 2.7 x - 700 so 2.7 x = 42,700 and x = 42,700 / 2.7 = 15,800 approx.. Your solution will differ slightly due to differences in your estimates of the line and the two points on the line. **
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RESPONSE --> ok self critique assessment: 3
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06:41:37 **** What is the equation of the line of best fit? **** How well does the line fit the scatter diagram of the data? Describe any systematic deviation of the line from the best-fit line. **** What is your interpretation of the slope of this line? **** What loan amount would correspond to annual income of $42,000?
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RESPONSE --> Line of best fit y=(1084/400)x-50 It will fit well slope was 1084/400 loan amount $113,770 confidence assessment: 3
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