course Mth 163 QzW˾CKassignment #007
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00:30:03 STUDENT RESPONSE: I predict that the spring will stretch at a greater rate as the weight is added. In my experience, springs lose their strength as they are stretched, and will not go back to their original shape. INSTRUCTOR COMMENT: ** Within their range of elasticity the graph is very nearly linear. If stretched too far the spring will lose its permanent elastic properties and will then deviate from linearity **
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RESPONSE --> I predict that the spring will stretch at a greater rate as the weight is added. self critique assessment: 3
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00:31:12 ** If you predicted a linear graph then did the actual graph confirm this? If you predicted a curvature did the actual graph confirm this? **
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RESPONSE --> I predicted a linear graph and the curvature did confirm. self critique assessment: 3
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00:31:43 ** Many students find the Linked Outline very helpful. **
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RESPONSE --> Yes I found it helpful self critique assessment: 3
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00:32:07 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It was interesting confidence assessment: 3
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course Mth 163 ǚ}Yy|ډXkTassignment #007 007. Precalculus I 02-10-2008
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00:41:25 `q001. Note that this assignment has 8 questions Sketch a graph of the following (x, y) points: (1,2), (3, 5), (6, 6). Then sketch the straight line which appears to come as close as possible, on the average, to the four points. Your straight line should not actually pass through any of the given points. Describe how your straight line lies in relation to the points. Give the coordinates of the point at which your straight line passes through the y axes, and give the coordinates of the x = 2 and x = 7 points on your straight line. Determine the slope of the straight line between the last two points you gave.
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RESPONSE --> It passes through the y-axis at coordinate (0, 2). At x = 2 and x = 7 the coordinates would be (2, 3) and (7, 8) The slope of the line would be (8-3)/(7-2) which equals 5/5 or 1. confidence assessment: 3
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00:48:34 `q004. The equation y = x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)?
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RESPONSE --> Plugging m = 1 into the first equation we get 3 = 2 * 1 + b, so 3 = 2 + b and b = 3 - 2 = 1. Now the equation y = m x + b becomes y = x + 1. confidence assessment:
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01:02:27 `q005. Using the best-fit equation y =.76 x + 1.79, with the numbers accurate to the nearest .01, how close do the predicted points corresponding to x = 1, 3, and 6 come to the original data points (1,2), (3, 5), and (6, 6)?
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RESPONSE --> For 1, 3, and 6 using the best-fit equation y = .76x + 1.79 we get the following coordinates: (1, 2.55), (3, 4.07), (6, 6.35) These three points on average differ from the the original data points by -.01 confidence assessment: 3
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01:03:51 Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points. The average distance is (.55 + .93 + .35) / 3 = .58 from the points.
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RESPONSE --> I messed up the average distance because I subtracted .93. I thought we should since it was below the given data point. self critique assessment: 3
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01:06:30 The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43. The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51. Thus the best-fit model does give the better result. We won't go into the reasons here why it is desirable to minimize the square of the distance rather than the distance. When doing eyeball estimates, you don't really need to take this subtlety into account. You can simply try to get is close is possible, on the average, to the points.
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RESPONSE --> The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43 The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51 This tells us that the best fit model is the better fit. self critique assessment: 3
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01:13:46 `q007. If the original data points (1,2), (3, 5), and (6, 6) represent the selling price in dollars of a bag of widgets vs. the number of widgets in the bag, then how much is paid for a bag of 3 widgets? How much would you estimate a bag of 7 widgets would cost?
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RESPONSE --> 3 widgets would cost 5 dollars. 7 widgets would cost around $2.80 according to my graph and calculation because the line is decreasing at an increasing rate. confidence assessment: 1
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01:15:28 If we use the best-fit function y =.76 x + 1.79, noting that y represents the cost and x the number of widgets, then the cost of 3 widgets is y = .76 * 3 + 1.79 = 4.05, representing cost of $4.05. The cost of 7 widgets would be y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11.
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RESPONSE --> I didn't use the y = .76 + 1.79 function. I must have misread it because I came up with a whole new funtion using the three coordinates given, but I do fully understand now self critique assessment: 3
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01:20:05 `q008. According to the function y = .8 x + 1.4, how much will a bag of 7 widgets cost? How many widgets would you expect to get for $10?
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RESPONSE --> 7 widgets cost would be calculated as follows: y = .8(7) + 1.4 y = $7.00 If I paid $10 I would calculate the number purchased as follows: 10 = .8x + 1.4 10 - 1.4 = .8x 8.6 = .8x 8.6 / .8 = x = 10.75 widgets confidence assessment: 3
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01:20:41 end program
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RESPONSE --> OK self critique assessment: 3
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