course Mth 163 ƒÜ{žòT„Ï«Ûá`ɬwÀ–™É»Øassignment #004
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20:57:00 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> f(-2)= -8 f(a)= -a f(x-4)= -24x+48 f(x)-4= x^3-4 confidence assessment: 2
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21:05:02 ** 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. **
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RESPONSE --> With f(a) I wasn't sure if I should multiply the a like a 1. I see that all that needs to be done is that I just take the problem and write it as -a^3. I noticed with f(x-4) I needed to work it a little more. I got 2 parts of the problem but I messed up when I was multiplying the varible. I was able to get the same answer for the first and last problem. self critique assessment: 3
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21:06:13 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2)= .25 f(-a)= 1 f(x+3)= 2x+6 f(x)+3= 8 confidence assessment: 2
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21:10:23 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> With the forst problem I was looking at the wrong one but I got the same answer and I see that when you multiply 2 times 2 you get 4 which is the correct answer. I had 2^a but I thought a was the same thing as 1. But if I thought it was I should of gotten 2 but I realize that I don't need to multiply anything because there is a variable. I see that I don't need to multiply out 2^(x+3). Since the problem has x+3 as the exponet there is no need to multiply it out. With the last problem of f(x)+3 I realize that I just need to put it as I see it. There is no need to multiply anything out. When I did it I multiplied it as the 3 being positive which would have given me 8. self critique assessment: 3
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21:13:54 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> When you use meaninful names you are able to say what you are talking about and what you are wanting to know the answer to. This helps you better unstand what is being asked and talked about and you can work the problem. For some people it is easier for them to understand things if it relates to something they are familiar with. confidence assessment: 2
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21:15:25 ** 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.**
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RESPONSE --> I also think that is is helpful to use meaninful names. If you don't think at something for a while and you go back and look at something it is like a light bulb comes on and you are able to remember about it because it realated to something else and it makes you think about it. self critique assessment: 3
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21:18:09 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> value(0)= 0 value(2)= $1144.90 value(t+3)= $1000(1.07)^t+3
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21:24:23 ** 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. **
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RESPONSE --> I see that on value(0) that you don't have to multiply 0 to the $1000. After I took 1.07 and multiplied it by 0 I just multiplied it on out and didn't think that it would do anything else. I realize that it is just like not having the other number there. On the last problem I am a little confused. I see how it is done but it still is confusing. I will problem just need to work it out a few times to completely understand it. I was able to get the same answer for the first and last problem. self critique assessment: 3
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21:42:04 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> Take the distance which is 50 and put it over 2 times the distance of 50. You would have 2*50= 100. 50/100=.50 confidence assessment: 2
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21:47:25 ** 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. **
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RESPONSE --> I see that when you write the equation out you have 50/distance and the you have to multiply 2 times the distance squared and put it over 50. The 50's cancel each other out and you are left woth 2 times the distance squared in parthenesses and put it over the distance squared. You multiply the 2 by 2 and get 4 and then the distance squared cancel each other out. It leaves you with the answer of 4. self critique assessment: 3
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21:48:29 query #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.
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RESPONSE --> I took the points (2,80) (5,40) and (10,25) and placed them on my graph and then connected the points and was able to get a curved line. confidence assessment: 2
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21:49:09 ** 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. **
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RESPONSE --> I used a smooth curve to connect my points. I see how it is done. self critique assessment: 2
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21:50:12 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> To use the point 60 it is between 40 and 80. The x of 40 and 80 is 2 and 5. I figured what would be the middle of 2 and 5 and got 3.5. confidence assessment: 2
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21:51:11 **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**
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RESPONSE --> I realize that I also have to figure in the point (10,25). When I do that is see that is brings 3.5 down a bit anf then I get 3.3. This would give a point of (3.3,60). self critique assessment: 3
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21:59:05 what is your estimate of the value f(7)?
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RESPONSE --> You find the middle between 5 and 10 and that gives you 7. Then you need to try to find where about the middle of 40 and 25 would be. I found 34 for this answer. It would be better to assume that the middle would be somewhere around 32 or 33. confidence assessment: 2
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22:00:18 ** 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. **
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RESPONSE --> I see that you need to use the slope on this one. Since there is a slope of -3 and the graph is decreasing this would make it easier to see that the point would lay somewhere around 32 or 33. self critique assessment: 3
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22:06:01 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> If I use f(7)=32 and deciede to find f(9) I need to try to determine what it would be near. If f(10)=25 and since the graph is decreasind the answe for f(9) would be somehwere between 32 and 25. If I try to find the middle and I see that a good choice would be 27 because it is about 5 in the difference. confidence assessment: 2
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22:07:00 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> I realize that I went about the same way to determine what f(9) would equal. I see that it equals 27. It is done my estimating the middle and the difference of 5 between the two. self critique assessment: 3
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22:08:09 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> Gor 30 the x value would be somewhere around 8 or 9 and the x value would be somewhere around 5 or 6. confidence assessment: 2
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22:09:23 ** 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. **
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RESPONSE --> I see that the x value for 70 is 3. Since the graph is decreasing I realize that I should have had a lower numer. The x value for 30 is about 8 or 9 and you can see that the difference is somewhere around 5 or 6. self critique assessment: 3
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22:11:37 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> t=3 is T(3) t=5 is T(5) t=3 and t=5 you have T(3)-T(5). confidence assessment: 2
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22:13:39 ** 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 **
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RESPONSE --> I realize that I cut my answer short and I should have kept going. I see that that change is somewhere between -30 degrees and 80 degrees is 50 degrees. The change between -90 degrees and 20 degrees is a -70 degrees/ So the change between T(3) and T(5) is T(5)-T(3)/2. self critique assessment: 3
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22:16:51 What 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?
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RESPONSE --> For f(150) you would take T(t)=150 and then takee T(t)=80 and T(t)=30. from 80 to 30 there is a difference of 50. Then you need to take the value of t at T(t)=30 and subtract it from the value of t at T(t)=80. confidence assessment: 2
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22:17:50 ** 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)] **
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RESPONSE --> I see that it is done by take T(t)=80 and T(t)=30 and you would have to subtract the two to find the difference to be able to solve the equation. self critique assessment: 3
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22:19:49 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> You have to take f(t)=34 and use t1 and use f(t)=47 and use t2. You would then have a equation of (t2-t1). You would then be able to use this equation to solve the equation. confidence assessment: 2
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22:20:56 ** 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 ... ' **
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RESPONSE --> I see that you need to solve for t1 and t2. t1 is equal to t=34 and t2 is equal to t=47. Since we don't know which one is grater we have to use absolute value. self critique assessment: 3
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22:22:25 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(34)-f(23) f(t)=.01t^2-1.5t=90 f(34)= 50.6 f(23)= 60.8 so f(34)-f(23)=50.6-60.8=-10.2. confidence assessment: 2
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22:23:48 ** 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. **
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RESPONSE --> I see that you have to solve for each of the f(x). In order to do this you take both f(x) of 34 and 23. You use the model that you have and then solve for each. After you find the answer for both you subtract them from each other to get your answer. self critique assessment: 3
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22:47:27 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(34) - f(23) Changed in t 34s-23s t/change in s which is 11s and then divided ed by f(34-f(23) which gives f(t)=.01t^2-1.5t+90. f(34)=50.6cm 11s f(23)=60.8cm f(34)-f(23)=50.6cm-60.8cm=-10.2cm 11s/f(34-f(23)=11s/(-10.2cm)=-1.08seconds. confidence assessment: 2
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22:50:17 ** 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. **
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RESPONSE --> Inorder to find the change in depth you have to find the difference of f(34)-f(23). This is a difference of 11. You then have to take the model in cm and you would find that at f(34) is 50.6 cm and f(23) is 60.8 cm. You would then take 50.6cm minus 60.8 cm which is -10.2 and then you have to divide 11s by -10.2cm and the depth is decreasing by 1 cm and your answer would be 1.08 seconds. self critique assessment: 3
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22:53:52 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(34)-f(23). The change in t 34s-23s=11s. t=f((34)-f(23) divided by 11s f(t)=.04^2-1.5t+90 f(34)=50.6cm f(23)=60.8cm f(34)-f(23)=50.6cm-60.8cm=-10.2cm f(34)-f(23)/11s=-10.2cm/11s=.92cm/sec t=23sec t=34sec the depth decreased by -92cm in sec on average. confidence assessment: 2
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22:56:52 ** 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. **
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RESPONSE --> To find the change in depth you have to take f(34)-f(23). The difference is 11s. You then have to divide the cm by the seconds to find the average of it. You use the model that you have and find that f(34)=50.6cm and f(23)=60.8cm so then you have 50.6cm-60.8cm and you have -10.2cm. You then take -10.2cm and divide it by 11s to get a -.92cm/sec and that is your average. self critique assessment: 3
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22:58:23 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> I made my graph by using depth vs time and using the following points: (0,96) (10,89) (20,68) (30,65) (40,48) (50,49) (60,36) and (70,41). confidence assessment: 2
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22:59:27 ** 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). **
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RESPONSE --> I see that in order to sketch a graph you have to use the points (0,96) (10,89) (20,68) (30,65) (40,48) (50,49) (60,36) and (70,41). self critique assessment: 3
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22:59:59 What 3 data point did you use as a basis for your model?
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RESPONSE --> The three points that I used was (4,93) (24,68) NS (60,41). confidence assessment:
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23:00:41 ** 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.**
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RESPONSE --> I have to draw a curved line and the best places to use are (4,93) (24,68) and (60,41). self critique assessment: 3
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23:01:55 What was your function model?
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RESPONSE --> My function model is (4,93) (24,68), and (60,41) and the depth(t)=.0089x^2-1.4992x+98.8544. confidence assessment: 2
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23:02:49 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
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RESPONSE --> I see that the function model is found from the 3 points of (4,93) (24,68) and (60,41) and the depth(t)=.0089x^2-1.4992x+98.8544. self critique assessment: 3
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23:03:13 What is the average deviation for your model?
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RESPONSE --> the average deviation for my model is 3.880975. confidence assessment: 2
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23:04:00 ** 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.**
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RESPONSE --> I see that it is helpful to use 2 colums labeled 'model data' and 'deviation'. When you average all the numbers you get 3.880975. self critique assessment: 3
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23:04:25 How close is your model to the curve you sketched earlier?
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RESPONSE --> I noticed that is was really cloose which is really interesting. confidence assessment: 2
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23:04:50 ** 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.**
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RESPONSE --> I saw that it was really close which is good and you can see that you are on the right track. self critique assessment: 3
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23:06:49 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I noticed that the data points that was used on depth vs time model fit a curve drown through it. The thing that sank in for me with this was that you can't always except all the data to be in neat little patterns but it is possible to look for sometype of a pattern in the data. self critique assessment: 3
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23:07:10 ** 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 **
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RESPONSE --> I see and I was also surprized to realize this and found everything to be true. self critique assessment: 3
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