course Mth 163
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11:25:05 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
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RESPONSE --> the graph would be lower by -1 units and it would not go throught the origin like the graph y= x^2
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11:25:13 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
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RESPONSE --> ok
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11:26:47 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. it would be a narrow parabola with all of the coordinates corresponding with each other
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11:26:59 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
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RESPONSE --> ok
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11:27:39 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
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RESPONSE --> that it would be 3 units higher then the orginal y= x^3
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11:28:05 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
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RESPONSE --> ok I misunderstood the function
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11:28:42 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> the graph would be one unit more to the right then the previous unit given
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11:28:49 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
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RESPONSE --> ok
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11:29:26 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. the graph would be three units up
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11:29:38 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
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RESPONSE --> ok
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11:30:25 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> the units would move the graph further from to the right from the x-axis
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11:30:33 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
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RESPONSE --> ok
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11:30:58 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> 3/8 and 1/3
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11:31:20 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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RESPONSE --> ok
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11:31:39 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
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RESPONSE --> I have no idea
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11:32:02 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. ok
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11:34:16 `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds?
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RESPONSE --> y = 2(5^2) + 3 y = 2(25) + 3 y = 53cm and y = 2(9^2) + 3 y = 2(81) + 3 y = 165 cm
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11:35:03 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
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RESPONSE --> I left out the last few steps but I understand the process
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11:36:06 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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RESPONSE --> you would take the points of the depth and clock time and find the slope
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11:36:14 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
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RESPONSE --> ok
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"
course Mth 163
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11:25:05 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
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RESPONSE --> the graph would be lower by -1 units and it would not go throught the origin like the graph y= x^2
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11:25:13 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
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RESPONSE --> ok
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11:26:47 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. it would be a narrow parabola with all of the coordinates corresponding with each other
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11:26:59 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
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RESPONSE --> ok
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11:27:39 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
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RESPONSE --> that it would be 3 units higher then the orginal y= x^3
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11:28:05 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
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RESPONSE --> ok I misunderstood the function
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11:28:42 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> the graph would be one unit more to the right then the previous unit given
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11:28:49 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
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RESPONSE --> ok
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11:29:26 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. the graph would be three units up
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11:29:38 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
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RESPONSE --> ok
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11:30:25 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> the units would move the graph further from to the right from the x-axis
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11:30:33 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
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RESPONSE --> ok
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11:30:58 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> 3/8 and 1/3
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11:31:20 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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RESPONSE --> ok
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11:31:39 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
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RESPONSE --> I have no idea
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11:32:02 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. ok
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11:34:16 `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds?
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RESPONSE --> y = 2(5^2) + 3 y = 2(25) + 3 y = 53cm and y = 2(9^2) + 3 y = 2(81) + 3 y = 165 cm
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11:35:03 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
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RESPONSE --> I left out the last few steps but I understand the process
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11:36:06 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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RESPONSE --> you would take the points of the depth and clock time and find the slope
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11:36:14 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
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RESPONSE --> ok
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}電ml{ Student Name: assignment #007
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11:39:20 `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 comes close to the points but not go through each point
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11:39:30 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 --> ok
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11:39:51 .763 x + 1.79
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RESPONSE --> a type of function
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11:40:25 `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 --> na
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11:40:54 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
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11:43:01 `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 = 9.4 y = 8(3) + 1.4 y = 25.4 y = 8(6) + 1.4 y = 49.4
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11:43:22 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 --> ok
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11:44:03 `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 --> they are alittle bit above the orginal points
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11:44:13 (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 --> ok
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11:46:22 `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 y = .76(3) + 1.79 y = 4.07 y = .76(6) + 1.79 y = 6.35
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11:46:36 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 --> ok
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11:47:27 `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 --> they are less by the fact that they do not go through the dame points in the original
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11:47:35 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 --> ok
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11:48:17 `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 --> you would pay twie as much
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11:48:38 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 --> k
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11:49:50 `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 = $3.80 for 3 you would be able to get10 widgets
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11:49:59 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
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