course Mth 271
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14:02:10 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> 80%.....so 3.26
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14:03:17 ** 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 --> Opps. I thought that I just had to plug in 80% in for the t but I needed to plug in in for y.
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14:03:37 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 3.26
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14:04:31 ** 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 --> I think that i did this on right. I pluged in 80 for t and came up with 3.26
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14:05:13 How well does your model fit the data (support your answer)?
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RESPONSE --> It fits it very well. It is off my a few hundreds but it is very close.
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14:05:56 ** 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 --> I only did for 70 and 80 but they were very close
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14:08:28 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (935.1395, 1) (264.4411, 2) (105.1209, 3) (61.0149, 4) (43.0624, 5) (25.9154, 6) (19.9277, 7) (16.2723, 8) (11.2808, 9) (9.4845, 10)
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14:08:48 ** 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 --> This is what I got
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14:10:06 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> I didnt mean to hit that button!...i got (105, 3) (43.06, 5) (16.27, 8)
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14:10:37 Give the first of your three equations.
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RESPONSE --> 105.12=a9+ 3b+ c
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14:10:43 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> ok
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14:11:08 Give the second of your three equations.
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RESPONSE --> 43.06= a 25+ 5b+ c
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14:11:12 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> ok
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14:11:38 Give the third of your three equations.
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RESPONSE --> 16.27= a 64+ 8b+ c
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14:11:48 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> Thats what I got
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14:12:52 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 88.85=a-55 +5b
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14:12:59 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> ok
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14:13:28 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 62.06= a-16+ -2b
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14:13:39 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> ok
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14:14:20 Explain how you solved for one of the variables.
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RESPONSE --> I had to muliply the new second equation by 5 and the first by 2 then the b canceled and I then solved for a
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14:14:26 ** 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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14:14:49 What values did you get for a and b?
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RESPONSE --> a= -2.57 b= -10.47
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14:15:36 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> My values are way off and I am not sure why.....I dont think that they would get off this much my just chooseing different points.
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14:15:49 What did you then get for c?
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RESPONSE --> c= 159.66
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14:15:57 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> ok
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14:16:39 What is your function model?
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RESPONSE --> y= -2.57t^2+ -10.47t + 159.66
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14:17:11 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> I pluged in my numberrs for this and got what I got with my numbers the same way you got with your numbers
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14:17:40 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> 136.33
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14:18:12 ** 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 --> I pluged in 1.6 for t and I got 136.66 w/m^2
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14:18:59 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> y= 25 then 2.95 and when y= 100 then 7.85
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14:19:57 ** 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 --> That is how i worked mine out.
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