course Mth 173 22:14:43`q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.
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RESPONSE --> (0,-4) and (-4/3,0) confidence assessment: 3
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22:15:11 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> correct self critique assessment: 3
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22:15:36 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> no the graph in linear confidence assessment: 3
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22:15:46 The graph forms a straight line with no change in steepness.
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RESPONSE --> true self critique assessment: 3
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22:16:21 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> m=3 confidence assessment: 3
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22:16:30 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> true self critique assessment: 3
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22:20:20 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> the graph is increasing the graph gets steeper as x increases increasing at an increasing rate confidence assessment: 3
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22:20:50 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> true self critique assessment: 3
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22:22:38 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> decreasing as the y values are getting smaller at a decreasing rate confidence assessment: 3
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22:22:49 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> true self critique assessment: 3
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22:25:37 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> increasing at a constant rate given this part of the graph is linear. confidence assessment: 3
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22:26:32 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> I understand. self critique assessment: 3
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22:29:33 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> decreasing at a decreasing rate as x increases steepness is getting lower as the graph is getting flatter confidence assessment: 3
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22:29:57 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> true self critique assessment: 3
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22:30:54 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> increasing at an increasing rate as the car is getting faster and farther away with time confidence assessment: 3
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22:31:13 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> true self critique assessment: 3
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{Ĭr愯x assignment #001 001. Depth vs. Clock Time and Rate of Depth Change 06-05-2007
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18:08:31 `qNote that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> July-December as the growth is approx. $100 per month. confidence assessment: 3
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18:08:53 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> understood self critique assessment: 3
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18:09:49 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> March-July=$75/month July-December=$40/month confidence assessment: 3
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18:10:03 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> true self critique assessment: 3
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18:13:42 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> less quickly between t=40 and 90 than t=10 and 40 confidence assessment: 3
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18:14:06 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> true self critique assessment: 3
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18:15:02 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> they are both problems that use plot data to find and compare slopes in given intervals. confidence assessment: 3
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18:15:31 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> true self critique assessment: 3
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ۤbs竇j|oO¥ assignment #001 赠̀ƶE阎z Calculus I 06-06-2007
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20:44:05 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> The three representitative points are:(20,60), (40,41) and (60,30).
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20:46:34 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> At time=7 temp would be 76.64 At time=19 temp would be 61.16 At time=31 temp would be 48.56
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20:46:41 ** Continue to the next question **
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RESPONSE --> ok
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20:47:27 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> The three points were: (20,60) (40,41) and (60,30).
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20:48:19 ** A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). **
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RESPONSE --> ok
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20:49:20 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> the first equation is composed was 60=400a+20b+c
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20:49:37 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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RESPONSE --> ok
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20:50:15 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> The second equation I composed from the data was: 41=1600a+40b+c
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20:50:29 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok
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20:51:09 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> The third equation I composed from the data was: 30=3600a+60b+c
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20:51:15 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok
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20:54:03 What multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?
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RESPONSE --> Actually I put the equations into matrix fom and used my TI-92 to reduce them down to Row Reduced Echelon Form recieving the following matrix: 1 0 0 1/100 0 1 0 -31/20 0 0 1 87
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20:54:27 ** STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c. By doing this, I obtained my first new equation 3200a + 40b = -30. **
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RESPONSE --> ok
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20:56:37 To get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?
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RESPONSE --> I used the 92's rref command on an augmented matrix
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20:56:41 ** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45**
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RESPONSE --> ok
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20:58:03 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> after viewing the reduced matrix I had the following values for a,b and c respectively: 1/100 -31/20 87
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20:58:08 ** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **
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RESPONSE --> ok
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20:58:31 What equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable?
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RESPONSE --> refer to earlier response
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20:58:35 ** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. **
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RESPONSE --> ok
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20:58:48 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=87
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20:59:28 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok
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21:04:28 What is the resulting quadratic model?
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RESPONSE --> for my model I chose t=time and y=temperature and I recieved the following model: y=(t^2)/100-(31/20)t+87
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21:04:36 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> ok
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21:06:23 What did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times?
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RESPONSE --> predicted temps were 87 72.5 and 60 with respective deviations of 8.0 2.5 and 0.0
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21:06:29 ** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. **
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RESPONSE --> ok
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21:06:50 What was your average deviation?
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RESPONSE --> average deviation for the model was 1.13
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21:07:00 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok
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21:08:07 Is there a pattern to your deviations?
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RESPONSE --> starting with the second listing in the table there is a pattern that emerges for the following entries in the table
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21:08:27 ** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **
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RESPONSE --> ok
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21:08:37 Have you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?
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RESPONSE --> Yes
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21:08:44 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok
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21:12:35 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> No one individual will remember something forever as their brain will not be responsive forever however I will retain this instruction for quite a long time. observe, record, analyze and postulate the data, do some algebra and calculus/linear algebra to find a model and then test the model against real world data
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21:12:50 ** STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!! INSTRUCTOR COMMENT: OK, I'm convinced. **
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RESPONSE --> ok
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21:20:05 Query Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.
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RESPONSE --> I previously called them temp however I meant depth. (95,0) (75,10) (60,20) (49,30) (41,40) (35,50) (30,60) (26,70)
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21:20:29 ** STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems. (5.3, 63.7) (10.6. 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6). **
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RESPONSE --> ok
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22:01:10 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I miss informed you on the previous question my data set for the depth experiment came from the simulated data on the website. I was looking at another data set and transposed headings, the data set was: (62.7,4.1) (57.2,8.2) (52.1,12.3) (47.4,16.4) (42,20.5) (38.8,24.6) the three points were: (62.7,4.1) (52.1,12.3) (42,20.5)
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22:01:19 ** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)**
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RESPONSE --> ok
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22:06:10 Give the first of your three equations.
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RESPONSE --> 3931.29a+62.7b+c=4.1 is the first equation is composed I from the data set
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22:07:18 ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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RESPONSE --> ok
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22:08:09 Give the second of your three equations.
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RESPONSE --> My second equation composed from the data set is: 2714.41a+52.1b+c=12.3
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22:08:12 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok
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22:08:45 Give the third of your three equations.
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RESPONSE --> My third equation is: 1764a+42b+c+20.5
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22:08:54 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok
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22:11:48 Give the first of the equations you got when you eliminated c.
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RESPONSE --> From the three equations I transmuted them into matrix form and used my TI-92 to put them in rref giving me the matrix: 1 0 0 .0019 0 1 0 -.986 0 0 1 58.6473
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22:11:53 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok
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22:14:27 Give the second of the equations you got when you eliminated c.
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RESPONSE --> bypassed with rref command
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22:14:32 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok
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22:16:44 Explain how you solved for one of the variables.
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RESPONSE --> I used a TI-92 and the rref command
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22:16:49 ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **
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RESPONSE --> ok
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22:18:40 What values did you get for a and b?
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RESPONSE --> a=.0019 and b=-.986 based in the information in the data set.
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22:18:45 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok
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22:18:59 What did you then get for c?
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RESPONSE --> c=58.6473
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22:19:04 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok
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22:22:50 What is your function model?
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RESPONSE --> y=depth and t=time y=.0019t^2-.986t+58.6473 is the model for this data set
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22:22:57 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok
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22:24:33 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> at t=15 y=44.2848
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22:24:58 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok
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22:25:32 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> at t=5 y=53.7648
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22:26:23 ** INSTRUCTOR COMMENT: The exercise should have specified a depth. The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation: 68 = .01t^2 - 1.6t + 126 using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. **
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RESPONSE --> ok
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22:40:42 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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RESPONSE --> My data set for grade vs. time is: (2.1,10) (2.5,20) (2.6,30) (3.3,40) (3.5,50) (3.9,60)
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22:41:03 ** STUDENT SOLUTION: Grade vs. percent of assignments reviewed (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). **
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RESPONSE --> ok
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22:42:41 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (2.5,20) (3.3,40) (3.9,60)
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22:42:45 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok
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22:48:29 Give the first of your three equations.
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RESPONSE --> 6.25a+2.5b+c=20
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22:48:36 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> ok
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22:48:55 Give the second of your three equations.
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RESPONSE --> 10.89a+3.3b+c=40
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22:48:59 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> ok
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22:49:23 Give the third of your three equations.
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RESPONSE --> 15.21a+3.9b+c=60
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22:49:28 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> ok
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22:49:55 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Once again I used a 92 and the rref command.
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22:49:58 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> ok
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22:51:03 Give the second of the equations you got when you eliminated c.
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RESPONSE --> TI-92 and rref command
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22:51:08 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok
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22:52:30 Explain how you solved for one of the variables.
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RESPONSE --> TI-92 with the rref command
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22:52:34 ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the variable b. In order to do this, I multiplied the first new equation by 80 and the second new equation by -50. **
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RESPONSE --> ok
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22:53:00 What values did you get for a and b?
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RESPONSE --> a=5.9524 and b=-9.5238
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22:53:04 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> ok
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22:53:13 What did you then get for c?
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RESPONSE --> c=6.6071
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22:53:17 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> ok
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22:55:12 What is your function model?
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RESPONSE --> y=5.9524t^2-9.5238t+6.6071 based on my data set
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22:55:17 y = (0) x^2 + (.01727)x + 1.773
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RESPONSE --> ok
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23:09:40 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> for t=2.0 y=11.3691
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23:09:57 ** The precise solution depends on the model desired average. For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have 3.3 = -.00028 x^2 + .06 x + .5. This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0. We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility. To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range. In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. **
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RESPONSE --> ok
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23:10:34 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> at t=2.9 y=29.0478
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23:10:45 ** Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. **
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RESPONSE --> ok
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23:20:01 How well does your model fit the data (support your answer)?
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RESPONSE --> fairly well given the mean of the deviations is 1.478 on an interval of 10's
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23:20:16 ** You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. **
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RESPONSE --> ok
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23:32:31 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> my data set for illumination vs. distance: (1230,1) (307.5,2) (136.67,3) (76.88,4) (49.2,5) (34.17,6) (25.1,7) (19.22,8) (15.19,9) (12.30,10)
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23:32:41 ** STUDENT SOLUTION: (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)**
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RESPONSE --> ok
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23:45:41 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (1230,1) (49.2,5) (12.3,10) are the three points I chose for the graph
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23:47:41 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> ok
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00:00:43 Give the first of your three equations.
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RESPONSE --> a+b+c=1230 25a+5b+c=49.2 100a+10b+c=12.3
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00:00:51 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> ok
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00:01:06 Give the second of your three equations.
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RESPONSE --> already given
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00:01:09 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> ok
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00:01:15 Give the third of your three equations.
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RESPONSE --> already given
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00:01:18 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> ok
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00:02:33 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Once again I used my TI-92 and the rref command to obtain the matrix 1 0 0 31.98 0 1 0 -487.08 0 0 1 1685.1
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00:02:36 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> ok
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00:02:50 Give the second of the equations you got when you eliminated c.
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RESPONSE --> already given
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00:02:53 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> ok
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00:03:08 Explain how you solved for one of the variables.
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RESPONSE --> using the rref command on a TI-92
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00:03:11 ** STUDENT SOLUTION CONTINUED: I solved for a by eliminating the variable b. I multiplied the first new equation by 4 and the second new equation by -6 **
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RESPONSE --> ok
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00:03:29 What values did you get for a and b?
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RESPONSE --> a=31.98 b=-487.08
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00:03:33 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> ok
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00:03:45 What did you then get for c?
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RESPONSE --> c=1685.1
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00:03:48 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> ok
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00:04:22 What is your function model?
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RESPONSE --> y=31.98t^2-487.08t+1685.1
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00:04:28 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> ok
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01:19:31 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> at distance=10 illumination=12.3
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01:19:43 ** STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun. My illumination prediction was 319.61 w/m^2, obtained by evaluating my function model for x = 1.6. **
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RESPONSE --> ok
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01:21:37 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> at illumination=10 distance=9.98
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01:21:51 ** The precise solution depends on the model and the range of averages. For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations 25=9.4 r^2 - 139 r + 500 and 100 =9.4 r^2 - 139 r + 500 Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data. The solutions which correspond to the data are r = 3.9 when y = 100 and r = 5.4 when y = 25. So when the distance x has range 50% - 69% if the illumination range is 25 to 100. Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **
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RESPONSE --> ok
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