course Mth 163
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20:49:19 `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 --> After graphing the line with the points (1,2) 93,5) (6,6), I drew a line as close to the median of the lines as I could. The points of reference that I can list that will show the line drawn are (0,1.9) (2,3.5) and (5,6). If these coordinates are graphed on the same graph as the ""question"" coordinates, one will see the line in which was drawn. The coordinates that of the point of where the line passes thru the y axis are (0,1.9). The coordinates that would lie where x = 2 are (2,3.5) and the coordinates of where x = 7 are (7,7.7). In order to find the slope of this drawn line between the points (2,3.5) and (7,7.7) one would need to plug the figures into the linear equation of y = mx = b to find that the rise of this line is 4.2 and the run is 5. Using the definition of slope being rise/run, plug in the figures 4.2 / 5 and the slope of this line is .84
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20:51:19 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.
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RESPONSE --> My figures differ a bit from those listed in your answer, but also my points on my drawn line are different also. My slope figured to only two decimal points, therefore I could be no more exact than the .84 listed.
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20:52:57 .763 x + 1.79
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RESPONSE --> I'm not quite sure of what this question is asking, therefore I am going to examine the answer in hopes that the question will make itself available. The question I have listed is .763 x + 1.79
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21:15:22 `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 --> I don't feel my answer is correct, but I'm going to do it anyway. With the 2 coordinates of x listed: x = 2, the equation would be y = 2m + b; x = 7, the equation would be y = 7m + b. To solve these two equations for ""m"" and ""b"" would be: y = 2m = b, subtract 2m from both sides to get y - 2m = b, then plug this solution for ""b"" into the next equation of y = 7m + b, y = ym + y - 2m, y = 5m + y, m = 0; then taking these two solutions and plugging into the original linear equation of y = mx = b, I get, y = (0)x + y - 2m
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21:23:47 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 --> This I understand, in the question stated on the past page, there was only a coordinate for two x coordinates and no y coordinates, therefore my answer looked as greek to you as it did to me. With the coordinates of (2,3) and (7,6). Using the linear equation of y = mx + b you would get the following. 6 = 7m + b - 3 = 2m +b = 3/5 = m = .6 6 = 7(.6) + b 1.8 = b Plugging this back into the linear equation would give you y = .6x + 1.8 This differs from the answer listed since the answer used (7,5) for the coordinates of the second point instead of (7,6) as listed in the question.
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21:26:47 `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 --> By using a table and substituting in the ""x"" into the equation I find that the coodinates of x = 1,3,6 are (1,2.2) (3,3.8) (6,6.2)
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21:26:57 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 --> GREAT
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21:30:20 `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 --> If I'm understanding the question correctly, the average difference is: The x coordinate is perfect and the y coordinate is off an average of -.2667
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21:31:57 (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 --> Correct, I shouldn't have factored in a negative into my equation. When doing it again without using the negative (-1.2), my answer mirrors the one listed.
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21:37:48 `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 --> The first coordinate (1,2.55), the best-fit equation is .45 from the original data points. The second coordinate (3,4.07), the best-fit equation is .93 from the original data points. The third coordinate (6,6.35), the best-fit equation is .35 from the original data points. On average, their deviation from the original data points are .5767
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21:38:14 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 --> same thing I got except that I took the decimal further out than requested by you.
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21:59:30 `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 --> I have the distances for the best-fit model to be .45, .93 and .35 with the average being .58 when squared is .3364 as the distances for the first model are .2, 1.2, .2 where the average is .53 and the square of this is .2809.
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22:04:30 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 --> I did this incorrectly...I took the average distance and squared instead of taking each distance and squaring and then finding the average. First model .2^2 = .04, 1.2^2 = 1.44, .2^2 = .04 .04 + 1.44 + .04 = .51 Best fit model .55 ( I see my mistake now), .55^2 = .30, .93^2 = .86, .35^2 = .12 .3 + .86 + .12 = 1.28 / 3 = .43 So it does show that the best-fit model does give a better result
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22:13:25 `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 --> Making a graph with the coordinates being (1,2), (3,5) and (6,6) and then drawing a line that does not go thru any of the given points, but seems to be an average between the points, the ""y"" coordinate being 7 (in this case the number of widgets, the ""x"" coordinate would be somewhere between 6.5 and 7.0
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22:16:13 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 understand the math, but when the question was ask I was under the understanding that 3 widgets cost $5.00. I understand putting it into the equation and the math that goes along with it
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22:30:48 `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 --> Using the function y = .8x + 1.4, substitute the 7 in for the ""x"" and you would get 7.39 for y therefore 7 widgets would cost 7.39...divide 7.39 by 7 to get the cost of a widget. 7.39 / 7 = 1.06 divided into 10 would give you the number of widgets for $10...9.43
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22:33:01 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 --> I understand and did the first part of the question correctly. I didn't do the second part by way of equation, but I understand now by using the total dollars as the y in the equation.
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