#$&* course Mth 163 9/19 9
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Given Solution: ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = 2^x f(2) = 2^2 = 4 f(-a) = 2^-a f(x+3) = 2^(x+3) f(x) + 3 = 2^x + 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(-a) = 2^(-a) = 1 / 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot got how to simplify the 2^-a. Could you help me remember that process???? ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.** ONE MORE STUDENT RESPONSE: It is much easier to keep track of what you are doing if you use meaningful names, particularly in multistep procedures. When working with one three different functions, I could call them f(x) = provides the value of the “x” coordinate for any given “y” g(x) = original value of the data for the “x” coordinate for any given y h(x) = the difference between the calculated value for x and the value for x in the data for any given value of y Therefore: f(x) - g(x) = h(x) True, but anything but easy to follow However if I used the designations below, it would be much easier to keep track of what I was doing. Graph(x) = provides the value of the “x” coordinate for any given “y” Data(x) = original value of the data for the “x” coordinate for any given y Resid(x) = the difference between the calculated value for x and the value for x in the data for any given value of y Therefore: Graph(x) - Data(x) = Resid(x) True, and you know at a glance what it is trying to tell you. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: value(0) = 1000 (1.07)^0 = 0 value(2) = 1000 (1.07)^2 = 1000 * 1.14 = 1140 value(t+3) = 1000 (1.07)^(t+3) value (t+3) / value(t) = 1000 (1.07)^(t+3) = 1000 (1.07^t)(1.07^3) value(t+3)/value(t) = 1000(1.07)^(t+3) / 1000(1.07)^t = (1.07)^(t+3) / (1.07)^t = (1.07^t)(1.07^3) / (1.07)^t = 1.07^3 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I simplified value(t+3) more than you did. Am I wrong???? ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This one was a little hard when doing the exercise because of the long words. It's a lot easier to use a, b, c, d, etc. When I worked this out on paper, I abbreviated just to make it easier for me to read and not take up so much room in my notebook. ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My estimate, according to the last problem, x would be around 3-4 or about 3.5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qwhat is your estimate of the value f(7)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My estimate would be around 33 or 33.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(7) would be about 33 or 33.5 and f(9) would be about 27. So the difference would be about 6 or 6.5. confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qwhat is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = 70 and f(x) = 30 x would be about 3 and 8. Difference would be about 5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for each of the following: (Question above was incomplete - assuming you were going to use the same examples from the f(x) nottions and the generalized modeling process.) The temperature at time t = 3. The temperature at time t = 5. The change in temperature between t = 3 and t = 5. The average of the temperatures at t = 3 and t = 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = T(3) y = T(5) y = T(3)-T(5) y = T(3) + T(5) / 2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the 3rd one wrong, I needed to do T(5) - T(3). After reading your notes, I understand my error. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = T(t) = 150 To get the length of time require for the temperature to fall from 80 to 30 you would do T(t) = 80 and T(t) = 30. Then subtract those answers. You would subtract T(t) = 30 minus T(t) = 80. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The answer was given in depth more than mine with an equation added, but the same conclusion. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qquery. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(t) = 34 f(t) = 47 Solve those for a given equation and then subtract those such as f(47) - f(34) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand using abs instead since it was referring to how long instead of clock time. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qBy how much did the depth change between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(34) - (23) = 11 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Did I get this correct???? ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. ** If your solution matches this one, you solved the problem as it was intended and did very well. However, after this solution had been in use for years, a sharp-eyed student noticed that the problem is actually not well-posed. Consider the following: VALID STUDENT OBJECTION (problem is actually not well-posed) Solution shows decrease (indicating direction) rather than measure of change. Question would actually ask for an absolute value - increase and or decrease would not be indicated? INSTRUCTOR RESPONSE: Very good. Technically the question isn't well-posed. The change is the change and not the magnitude of the change, so we're asking for a positive change in depth. The phrasing about the time interval is 'how long ... to change', which implies a positive time interval. The positive change doesn't occur in a positive time interval; a negative time interval doesn't answer the question as phrased. The question would have been well-posed had it asked 'How long does it take for the depth to decrease by 1 cm. ... etc.'. I'm going to leave the phrasing of the question as is, and add this note to the solution. Most students who answer the question correctly will then have an opportunity to consider the idea of a 'well-posed problem'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qOn the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This would be f(34) - f(23)/centimeters confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had it mixed up, but after review, I understand. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used a sample from the simulated data to graph with the following points (4.1, 62.7) (8.2, 57.2) (12.3, 52.1) (16.4, 47.4) (20.5, 42) (24.6, 38.8) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat 3 data point did you use as a basis for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I randomly chose (8.2, 57.2) (16.4, 47.4) (20.5, 42) confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not go by what the student response was. I randomly chose. Should I have chosen like they did????? ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089t^2 - 1.4992t + 98.8544. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qWhat is the average deviation for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I chose x = 4.1, 12.3, and 24.6 and plugged them into the equation y = -.0099t^2 - .9512 + 65.6658 61.7248+ 53.8438+ 17.8658 = 133.4/3 = 44.5 Average deviation is 44.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.** The given points are (0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). If the model is f(t) = .0089t^2 - 1.4992t + 98.8544 then we can create the following table: t y f(t) deviation 0 96 98.544 2.544 10 89 84.442 4.558 20 68 72.12 4.12 30 65 61.578 3.422 40 48 52.816 4.816 50 49 45.834 3.166 60 36 40.632 4.632 70 41 37.21 3.79 For example when t = 30 the data point is (30, 65). The function value is f(30) = 61.578. The deviation between the function value and the data point is | 65 - 61.578 | = 3.422. Note that the function values are calculated to a ridiculous number of significant figures. Since the original data are given only to whole-number values, it would be more appropriate to round the values of f(t) to the nearest whole number, giving us the table t y f(t) deviation 0 96 99 3 10 89 84 5 20 68 72 4 30 65 62 3 40 48 53 5 50 49 46 3 60 36 41 5 70 41 37 4 The average deviation would be the average of the deviations. Adding the deviations up we get 32. Dividing this by 9, the number of data points, we find that the average deviation is about ave dev = 32 / 9 = 3.6, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Did I give enough detail????? ------------------------------------------------ Self-critique Rating: ok
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This was a good insightful exercise. I thought that the curve would be close and it was except for the lower end. It’s neat to see how mathematical equations can be worked out to get precise answers. Math interests me, mostly because it is a challenge, and I love challenges. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first two points were very close, but the point t = 24.6 was off by a significant amount. This made the curve of the graph on the left hand side lower towards the x axis confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This was a good insightful exercise. I thought that the curve would be close and it was except for the lower end. It’s neat to see how mathematical equations can be worked out to get precise answers. Math interests me, mostly because it is a challenge, and I love challenges. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first two points were very close, but the point t = 24.6 was off by a significant amount. This made the curve of the graph on the left hand side lower towards the x axis confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.** Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This was a good insightful exercise. I thought that the curve would be close and it was except for the lower end. It’s neat to see how mathematical equations can be worked out to get precise answers. Math interests me, mostly because it is a challenge, and I love challenges. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!