course Mth 163 I may be exhausted enough to sleep through Christmas, but I will get this completed. I know I keep saying that, but I will do it-- somehow :) ‡ˆÉÀ”Ÿ¶‹ˆ‚Ÿ€Øû¨ážEy¯Ý
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18:27:06 `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 lies in the middle of the given points. My line crosses the y axis at (0,2) x=2 (2, 3.5) X=7 (7,8) slope= .9 confidence assessment: 3
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18:28:21 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 --> I was close self critique assessment: 3
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18:41:52 `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 --> for coordinates (2,3.5) the equation was as follows: 3.5=m(2) + b 3.5=2m + b for coordinates (7,8) the equation was as follows: 8= m(7) + b 8= 7m + b m= .9 and b= 1.7 therefore: y(x) = .9x + 1.7 confidence assessment: 2
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18:44:58 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 --> ok self critique assessment: 3
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19:00:50 `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 --> (1, 2.2) (3, 3.8) (6, 7.8) confidence assessment: 3
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19:02:04 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 --> well it would help if I could multiply 8x6 correctly ;) self critique assessment: 3
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19:04:40 `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 --> very close, a line drawn between the new points brings the line closer to the two points below the line drawn in question 1 confidence assessment: 3
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19:10:27 (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 see now what you wanted for this question. self critique assessment: 3
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19:27:19 `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 --> using the best fit equation the new coordinates found were: (1, 2.55) this has a difference of .55 (3, 4.07) this has a difference of .93 (6, 6.35) this has a difference of .35 the average difference is (.55+.93 + .35)/3 = .61 confidence assessment: 3
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19:29:58 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 --> The answer I got for the average difference was .61 when i saw the different answer here I redid the math several times and still got an average difference of .61 self critique assessment: 3
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19:37:59 `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 --> average difference of first model was .53 the squares of differences of the best fit model were .3025, .8649, .1225 the average difference of the squares = .43 confidence assessment: 3
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19:41:44 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 --> I only squared the one model I did the math and got .51 for the average of the differences of squares of the first model self critique assessment: 3
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20:10:00 `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 --> would this be the same estimation as earlier for the line? I am unsure if the points are ($, #) or (#,$) if $, # then a bag of 3 widgits would be $2 and 7 widgets would be $7 if #, $ then a bag of 3 widgits would be $5 and 7 widgets would be $8 confidence assessment: 1
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20:13:45 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 --> duh.. Why was I so reluctant to use the math instead of the graph? self critique assessment: 3
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20:19:10 `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 --> according to the given function 7 widgits would cost $7 and for $10 I would expect to receive 11 widgits (10.75 rounded to 11) confidence assessment: 3
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20:19:47 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 --> ok self critique assessment: 3
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