assignment #007 007. Precalculus I 08-12-2008
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17:16:32 `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 --> the line passes through the points (1,2), and (6,6) the line is a bit below the point (3,5). passes through y axis at: (0,1) x=2; (2,3) x=7; (7,8) confidence assessment: 3
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17:16:54 Your straight line should pass above the first and third points and beneath the second. If this is not the case the line can be modified so that it comes closer on the average to all three points. The best possible straight line passes through the y-axis near y = 2. The x = 2 point on the best possible line has a y coordinate of about 3, and the x = 7 point has a y coordinate of about 7. So the best possible straight line contains points with approximate coordinate (2,3) and (5,7). The slope between these two points is rise/run = (7 - 3)/(5 - 2) = 4 / 5 = .8. Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79
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RESPONSE --> self critique assessment: 3
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17:20:19 `q002. Plug coordinates of the x = 2 and x = 7 points into the form y = m x + b to obtain two simultaneous linear equations. Give your two equations. Then solve the equations for m and b and substitute these values into the form y = m x + b. What equation do you get?
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RESPONSE --> confidence assessment: 3
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17:27:36 Plugging the coordinates (2,3) and (7, 6) into the form y = m x + b we obtain the equations 3 = 2 * m + b 5 = 7 * m + b. Subtracting the first equation from the second will eliminate b. We get 4 = 5 * m. Dividing by 5 we get m = 4/5 = .8. Plugging m = .8 into the first equation we get 3 = 2 * .8 + b, so 3 = 1.6 + b and b = 3 - 1.6 = 1.4. Now the equation y = m x + b becomes y = .8 x + 1.4. Note that the actual best-fit line is y = .763 x + 1.79, accurate to three significant figures.
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RESPONSE --> y=mx+b; (2,3) (7,6) 3=2m+b 6=7m+b; subtract 1rst from 2nd. 4=5m m=.8; plug into 1rst equation. 3=2(.8)+b b=1.4 y=.8x+1.4 self critique assessment: 3
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17:31:09 `q003. Using the equation y = .8 x + 1.4, find the coordinates of the x = 1, 3, and 6 points on the graph of the equation.
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RESPONSE --> y=.8(1)+1.4 y=2.2 y=.8(3)+1.4 y=3.8 y=.8(6)+1.4 y=6.2 confidence assessment: 3
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17:31:17 Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2.
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RESPONSE --> self critique assessment:
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17:33:15 `q004. The equation y = .8 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 --> on the 1rst and last set of points the y-value is closer (within .2 points of each other) with the original than (3,5). confidence assessment: 3
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17:41:34 (1, 2.2) differs from (1, 2) by the .2 unit difference between y = 2 and y = 2.2. (3, 3.8) differs from (3, 5) by the 1.2 unit difference between y = 5 and y = 3.8. (6, 6.2) differs from (6, 6) by the .2 unit difference between y = 6 and y = 6.2. {}The average discrepancy is the average of the three discrepancies: ave discrepancy = ( .2 + 1.2 + .2 ) / 3 = .53.
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RESPONSE --> I didnt word it that precisely: (1,2.2) there is a .2 unit difference in the y values (3,3.8) there is a 1.2 unit difference in the y-values (6,6.2) there is again a .2 diff. in the y values self critique assessment:
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17:45:39 `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 --> y=.76(1)+1.79 y=2.55 .55 diff from the original y y=.76(3)+1.79 y=4.07 1.07 diff from the original y y=.76(6)+1.79 y=6.35 .35 diff from orig. value .55+4.07+.35=4.97 confidence assessment: 3
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17:47:13 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 --> again... I forgot to divide .55+4.07+.35=4.97 4.97/3=1.657; ave distance from original points self critique assessment: 3
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17:47:37 `q006. The average distance of the best-fit line to the data points appears to greater than the average distance of the line we obtain by an estimate. In fact, the best-fit line doesn't really minimize the average distance but rather the square of the average distance. The distances for the best-fit model are .55, .93 and .35, while the average distances for our first model are .2, 1.2 and .2. Verify that the average of the square distances is indeed less for the best-fit model.
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RESPONSE --> confidence assessment: 2
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17:49:52 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 --> .55^2=.3025 4.07^2=16.56 .35^2=.1225 .3025+16.56+.1225=16.985 16.985/3=5.66 self critique assessment: 3
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17:51:05 `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 --> confidence assessment: 3
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17:52:07 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 --> y=.76(3)+1.79 y=4.05 y=.76(7)+1.79 y=7.11 self critique assessment: 3
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17:53:36 `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 --> y=.8(7)+1.4 y=7 10=.8x+1.4 8.6=.8x x-10.75 confidence assessment: 3
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17:53:44 Using the model we obtained, y = .8 x + 1.4, we note that the cost is represented by y and the number of widgets by acts. Thus we can find cost of 7 widgets by letting x = 7: cost = y = .8 * 7 + 1.4 = 7. To find the number of widgets you can get for $10, let y = 10. Then the equation becomes 10 = .8 x + 1.4. We easily solve this equation by subtracting 1.4 from both sides than dividing by .8 to obtain x = 10.75. That is, we can buy 10.75 widgets with $10.
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RESPONSE --> self critique assessment:
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IߕяmѦxC_zʿѼ assignment #008 008. Precalculus I 08-12-2008
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19:21:11 `q001. Note that this assignment has 4 questions For the function y = 1.1 x + .8, what are the coordinates of the x = 2 and x = 9 points? What is the rise between these points and what is the run between these points? What therefore is the slope between these points?
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RESPONSE --> y=1.1(2)+.8 y=3 (2,3) y=1.1(9)+.8 y=10.7 (9,10.7) rise= 10.7-3 rise= 7.7 run= 9-2 run=7 slope between 2 points= 1.1 confidence assessment: 3
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19:21:20 Evaluating y = 1.1 x +.8 for x = 2 and x = 9 we obtain y = 3 and y = 10.7. The graph points are therefore (2,3) and (9,10.7). The rise between these points is 10.7 - 3 = 7.7 and the run is 9-2 = 7. Thus the slope is 7.7 / 7 = 1.1.
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RESPONSE --> self critique assessment:
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19:21:48 `q002. For the function y = 1.1 x + .8, what are the coordinates of the x = a point, in terms of the symbol a? What are the coordinates of the x = b point, in terms of the symbol b?
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RESPONSE --> confidence assessment: 3
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19:24:18 If x = a, then y = 1.1 x + .8 gives us y = 1.1 a + .8. If x = b, then y = 1.1 x + .8 gives us y = 1.1 b + .8. Thus the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8).
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RESPONSE --> x=a; y=1.1a+.8 x=b; y=1.1b+.8 (a,1.1a+.8), (b,1.1b+.8) self critique assessment: 3
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19:25:26 `q003. We see that the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8). What therefore is the rise between these two points? What is the run between these two points?
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RESPONSE --> rise= (1.1b+.8)-(1.1a+.8) run= (b-a) confidence assessment: 3
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19:25:33 The rise between the points is the rise from y = 1.1 a + .8 to y = 1.1 b + .8, a rise of rise = (1.1 b + .8) -(1.1 a + .8) = 1.1 b + .8 - 1.1 a - .8 = 1.1 b - 1.1 a. The run is from x = a to x = b, a run of run = b - a.
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RESPONSE --> self critique assessment:
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19:30:23 `q004. We see that the rise between the x = a and x = b points of the graph of y = 1.1x +.8 is 1.1 b + .8 - (1.1 a + .8), while the run is b - a. What therefore is the average slope of the graph between these points? Simplify your answer.
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RESPONSE --> you would obtain the rise/run for each set of points, add those two slopes together and divide by 2 confidence assessment: 3
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19:30:32 The slope is slope = rise / run = (1.1 b - 1.1 a) / (b - a) = 1.1 (b - a) / (b - a) = 1.1. The significance of this series of exercises is that the slope between any two points of the straight line y = 1.1 x + .8 must be 1.1, no matter whether the points are given by numbers (e.g., x = 2 and x = 9) or by symbols (x = a and x = b). Mostly
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RESPONSE --> self critique assessment:
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