#$&* course Mth 163 Hello,I apologize for the great delay in submitting the next part of assignment 4. As I already informed you of the death in my family, I feel it necessary to also make you aware of my work schedule over the past two weeks. I have worked 15 days straight now within my ministry internship on top of my family's loss. We have hosted two conferences and many long work days in between. This schedule should ease off for a while now and I should have the time to catch up on my assignments. This explanation is by no means my attempt to come up with excuses as to why the work has not been completed, but more or less an attempt to exhibit my concern about the class. I will diligently work on the next assignment sets and strive to take my major quiz and test by the end of this week/beginning of next week if you see that that is a reasonable goal.
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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: 3 ********************************************* Question: `qWhere f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(2)=2^x 2^(2)=4 f(-a)=2^x 2^(-a) f(x+3)=2^x 2^(x+3) f(x)+3=2^x 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 don’t understand the second problem. Why is it that by flipping the 2 into ½ makes the a positive?
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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: 3 ********************************************* Question: `qWhat were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Value(t) = $1000 (1.07)^t Value(0)= 1000(1.07)^0 Value(0)= $1000 Value(2)= 1000(1.07)^2 Value(2)= 1144.9 Value(t+3)= 1000(1.07)^(t+3) Value(t+3)/(value t)= 1000(1.07)^(t+3)/ 1000(1.07^t) The $1000 will cancel out, giving us: Value (t+3)/(value t)=(1.07)^(t+3)/ (1.07^t) Then the 1.07^t’s cancel out, leaving 1.07^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat did you get for illumination(distance)/illumination(2*distance)? Show your work on this one. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: illumination(distance) = 50 / distance^2 50/distance^2/50/2*distance^2 Flip the fraction: 50/d^2*2d^2/50 The 50s cancel out. 2d^2/d^2 Square the 2 4d^2/d^2 The d^2 cancels out, leaving us with 4. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to graph the function, I placed points at (2,80),(5,40), and (10,25). The graph slopes downward from the first point sharply. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qwhat is your estimate of value of x for which f(x) = 60? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My estimate is around 3.5 for the graph because it is in between the 2 and the 5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: `qwhat is your estimate of the value f(7)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because 7 is in between 5 and 10, my rough estimate of where the y value would be between 30 and 35. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t think about checking the slope to estimate. I looked at my graph and provided the estimate from there very roughly because the scale of the graph is not perfectly accurate. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is your estimate of the difference between f(7) and f(9)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If 7 is between 30 and 35, 9 is going to fall between 25 and 30. That leaves anywhere from 5 to 10 spaces between 7 and 9, which is consistent to the other points on the graph. So, if our estimate point for 7 is 32, which is the closest estimate on my graph, and our estimate point for 9 is 28, that is a difference of 4. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): My estimate points are a little different, leaving me off by a point. Is this an okay estimate?
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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: 3 ********************************************* 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: The temperature at time t = 3. 70 degrees The temperature at time t = 5. 37 degrees The change in temperature between t = 3 and t = 5. 70-37=33 The average of the temperatures at t = 3 and t = 5. 70+37=107/2= 53.5degrees confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the model, all we would do to find the clock time for 150 degrees is to go to 150 and find the time on the graph. To find the length of time between 80 and 30 degrees, we would subtract the two times at 80 and 30 get the number. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK, didn’t give a formula, but I said the same thing as the given formula. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery. use the f(x) notation at every opportunity: For how long was the depth between 34 and 47 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For where f(t)=34 and f(t)=47, we would subtract the two times from one another to find the depth. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qBy how much did the depth change between t = 23 seconds and t = 34 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Model: y=0.0013t^2-0.939t+85.6625 F(23)=64.8 F(34)=55.2 So, the depth is 64.8-55.2=9.6 confidence rating #$&*: 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): My model is very close to the given solution. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: Taken from the same information above, you would simply divide in the opposite direction. So, instead of placing the seconds on top, we put the centimeters on top: 9.6cm/11s= 0.87 centimeters per second confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I graphed all of the given points of (0,96), (10,89), (20,68), (30, 65), (40,48), (50,49), (60, 36), (70,41) and pulled my graph from the first point to the last point on the graph through the middle of all of the given points. confidence rating #$&*: 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: After constructing 2 models that were completely bogus, I read the given solution for the problems below to see that I could choose other points on the graph that were not given in the notes for the example data. So, I chose (5,81), (25,63), (60,34). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat was your function model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=0.0031t^2-0.939t+85.6625 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: `qWhat is the average deviation for your model? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My deviations are as follows: (20,68) 96-76.4=12.6 (30,65) 65-58.6625=9.3 (40,48) 48-50.1825=-2.1825 (50,49) 49-41.9625=7.0375 (60,36) 36-34.0025=1.9973 (70,41) 41-26.3025=14.7 To find the average of my deviations, I added all of them and divided by 6: 12.6+9.3+2.1825+7.0375+1.9973+14.7=47.8173/6=7.9 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): I did not round off my answers at all, so my average would not be set up the same way. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow close is your model to the curve you sketched earlier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Decently close. It is not spot on, but most of the points fit. 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: It is crucial that when you achieve a model that I do it more than once for the best results. Models are lots of work!!! "