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14:18:10 **** query 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 --> the graph on #8 looks like it is nonlinear because it has a slight curve in it from what appears to be points (25,15), (30,23) are the points significantly off the line of best fit
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14:18:15 ** 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
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14:19:09 query 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 --> i have graphed (5,2),(10,4) m=y2-y1/x2-x1 m=(4-2)/(10-5)=2/5 y-y1=m(x-x1) y-2=2/5(x-5) y=2/5x slope of line line of best fit y=ax+b a=.78 b=-3.3 y=.78x+ -3.3 y=.78x-3 slope of line of best fit is .78 this problem is not in the book as #12 in any of chapter 2 or anything similar to it
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14:19:18 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
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14:21:54 query 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 --> the line of best fit is y=2.71x+-66.1 or y=2.71x-66.1 with a slope of 2.71 enter $42000 per year in there as x and you would be able to get a loan for $113,754
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14:24:28 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 --> okay i understood it as the person made 42,000 and wanted to borrow money not that the amount to borrow was 42,000
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14:26:12 **** 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 --> y=2.71x -66.1 line of best fit annual income of $42,000 would be a loan amount of up to $113,754
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nٺԗwT赽 assignment #021 ؾ{]Uӂ͛ξ College Algebra 11-04-2005
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11:25:18 **** query 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> since it varies inversley with the sqrt of x the equation would be y=k/sqrt(x) 4=k/sqrt9 so k=12 thus in all cases y=12/sqrt(x)
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11:25:43 ** The inverse proportionality to the square root gives us y = k / sqrt(x). y = 4 when x = 9 gives us 4 = k / sqrt(9) or 4 = k / 3 so that k = 4 * 3 = 12. The equation is therefore y = 12 / sqrt(x). **
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RESPONSE --> o.k.
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11:33:30 query 2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3.
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RESPONSE --> The equation to find K is z=k(x^3+y^2) 1=k(2^3+3^2) 1=k(17) k=1/17 so the equation would always be z=1/17(x^3+y^2)
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11:36:13 ** The proportionality is z = k (x^3 + y^2). If x = 2, y = 3 and z = 1 we have 1 = k ( 2^3 + 3^2) or 17 k = 1 so that k = 1/17. The proportionality is therefore z = (x^3 + y^2) / 17. **
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RESPONSE --> o.k.
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11:42:44 query 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)
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RESPONSE --> so T=period of time in seconds and L equals length in feet so T=2pi/sqrt(32)(sqrt(L))
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11:43:03 ** The equation is T = k sqrt(L), with k = 2 pi / sqrt(32). So we have T = 2 pi / sqrt(32) * sqrt(L). **
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RESPONSE --> o.k.
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11:44:07 **** What equation relates period and length? ****
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RESPONSE --> T=KL
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12:01:23 query 2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.
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RESPONSE --> R= resistance L= length and D= diameter so R varies direct with L and inversley with D^2 so we have R=KL/D^2 so 1.24=k432/4^2 so 19.84=k(432) 19.84/432=k K=31/675 now we know the constant so if R=1.44 and D=3 we need to find L 1.44=(31/675)*L/3^2 12.96=(31/675)*L 282.194 would be the length
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12:03:22 ** We have R = k * L / D^2. Substituting we obtain 1.24 = k * 432 / 4^2 so that k = 1.24 * 4^2 / 432 = .046 approx. Thus R = .046 * L / D^2. Now if R = 1.44 and d = 3 we find L as follows: First solve the equation for L to get L = R * D^2 / (.046). Then substitute to get L = 1.44 * 3^2 / .046 = 280 approx. The wire should be about 280 ft long. **
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RESPONSE --> I tried to be to percise . but I understand how to solve the equation.
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