course Mth163 Submitted June 21, 2009 at 2:35 pm. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok If you look back at your x points you have x = 7 giving you the coordinates of (7,7), also showing correct calculations that in your problem 7 – 2 = 5 not 5 – 2 = 5 Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the coordinates ( 2,3) 3 = 2m + b and coordinates ( 7,7) 7 = 7m + b To solve for m & b you subtract the 1st from the 2nd and solve for m: 7 = 7m + b -3 = 2m + b 4 = 5m m = 4/5 0r .8 no solve for b by putting the m into the 1st equation and checking with the 2nd: 3 = 2(.8) + b 3 = 1.6 + b b = 1.4 7 = 7(.8) + 1.4 7 = 5.6 + 1.4 7 = 7 Taking the equation y = mx + b and adding m = .8 & b = 1.4 your new equation would be: y = .8x + 1.4 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x = 1, 3, 6 y = .8(1) + 1.4 y = 2.2 coordinates are (1, 2.2) y = .8(3) + 1.4 y = 3.8 coordinates are ( 3, 3.8) y = .8(6) + 1.4 y = 6.2 coordinates are (6, 6.2) Confidence Assessment: possitive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These points differ from (1, 2) to (1, 2.2) by .2 from (3, 5) to (3, 3.8) by 1.2 and from ( 6, 6) to (6, 6.2) by .2 we add the differences and divide by three to find the average: .2 + 1.2 + .2 / 3 = .53 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = .76(1) + 1.79 y = 2.55 (1, 2.55) compared to (1, 2.2) difference of .55 y = .76(3) + 1.79 y = 4.07 ( 3, 4.07) compared to (3, 5,) difference of .93 y = .76(6) + 1.79 y = 6.35 (6, 6.35) compared to (6, 6) difference of .35 to compare the averages we add these differences together and divide by 3: (.55 + .93 + .35) / 3 = .61 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have looked at my problem and have the same information that you have and still after several rechecks come up with .61 instead of .58 Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .55^2 = .30 .93^2 = .87 .35^2 = .12 The average square distance : (.30 + .87 + .12) / 3 = .43 The square distance from our first model: .2^2 = .04 1.2^2 = 1.44 .2^2 = .04 The average square distance: ( .04 + 1.44 + .04) / 3 = .51 rounded This proves the best- fit model is less Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula y = .76x + 1.79 with x = 3 – the number of widgets: We would receive y = .76(3) + 1.79 y = 4.07 for the cost Then x = 7 the number of widgets: We would get y = .76(7) + 1.79 = y = 7.11 the cost of 7 widgets. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = .8(7) + 1.4 y = $7 the cost of 7 widgets To figure out how many widgets you would get for $10 you would substitute 10 in the y’s place on the original function: 10 = .8x + 1.4 8.6 = .8x x = 10.75 expected widgets for the price of $10 Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. 2nd query questions 007. `query 7 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qquery predictions Sketch your graph representing the predicted height of the low end vs. the weight on the spring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As the weight is added the height gets lower at a faster rate as more weight is applied. The spring cannot hold its shape or strength as it is stretched out with the added weight. Confidence Assessment: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Graph is nearly linear, if stretched too far spring loses spring shape and deviates from being linear. Self-critique Rating: ********************************************* Question: `qcomment on how the actual graph of the data compared with your prediction YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My graph showed a decrease as the weight was applied and was compatible to the predictions Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If you predicted a linear graph then did the actual graph confirm this? If you predicted a curvature did the actual graph confirm this? ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qquery linked outline discuss your experience with the Linked Outline. Did you find it helpful? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It was very helpful with step by step instructions on how to obtain my information. Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Many students find the Linked Outline very helpful. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Again I am reminded how much we can apply our surroundings to mathematics. "