course mth 163 ӡV㣀̑Vassignment #001
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13:49:35 Query Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising?
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RESPONSE --> I found it interesting how many real problems math can solve with the right formula. confidence assessment: 3
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13:49:45 ** Continue to the next question **
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RESPONSE --> self critique assessment: 3
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13:50:19 Query Introductory Flow Experiment (no summary needed) Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? Support your conclusion.
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RESPONSE --> it is changing at a slower and slower rate confidence assessment: 3
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13:50:36 ** If you time the water at equal time intervals you should find that the depth changes by less and less with each new interval. If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it. } Either way you would conclude that water flows from the hole at a decreasing rate. The reason is that as the water depth decreases the pressure forcing the water out of the hole decreases. **
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RESPONSE --> self critique assessment: 3
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13:52:02 What does the graph of depth vs. clock time look like? Is it increasing or decreasing? Does the rate of increase or decrease speed up or slow down? Does your graph intercept the y axis? Does it intercept the x axis? How would you describe its overall shape?
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RESPONSE --> it decreases it does not intercept the x or y axis 1/2 parabola confidence assessment: 3
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13:52:29 ** The graph will start on the positive y axis and will decrease at a decreasing rate. The shape of the graph is the left-hand side of a parabola that opens upward. The right-hand half of the parabola does not correspond to the flow. The lef-hand half of the parabola, which corresponds to the flow, gets less and less steep with increasing clock time, matching the fact that that the rate of decrease is slowing. At the instant the containers empties, the water will be at the level of the hole. If depth is measured relative to the hole, then at the instant depth will reach zero. The corresponding graph point will lie on the t axis and will correspond to the vertex of the parabola. **
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RESPONSE --> self critique assessment: 3
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Ծw assignment #002 002. `query2 Precalculus I 08-29-2008 aإdψ|asɘ assignment #002 002. Precalculus I 08-29-2008
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12:51:49 `q001. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. NOTE SOLN IS -1, 10, 100.
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RESPONSE --> 60a + 5b + c = 90 2a + 3b + c = 128 58a + 2b= -38 200a + 10 b + c = 0 60a + 5b + c = 90 140a +5b= -90 (5)(58a + 2b= -38) =290a+10b= -190 (2)(140a +5b= -90) =280a+10b= -180 10a=-10 a=-1 58(-1) + 2b= -38 -58+2b=-38 2b=20 b=10 2(-1) + 3(10) + c = 128 -2+30+c=128 c=100 confidence assessment: 3
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12:52:40 The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become 'new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might have obtain this system, or one equivalent to it, using a slightly different sequence of calculations.
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RESPONSE --> self critique assessment: 2
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12:56:21 `q002. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 , which can be obtained from the system in the preceding problem, by eliminating the easiest variable.
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RESPONSE --> (7)(58 a + 2 b = -38) (2)(198 a + 7 b = -128) 406a+14b= -266 396a+14b= -256 10a= -10 a= -1 confidence assessment: 2
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12:56:34 Neither variable is as easy to eliminate as in the last problem, but the coefficients of b a significantly smaller than those of a. So we will eliminate b. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the program multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128). Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1.
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RESPONSE --> self critique assessment: 3
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12:58:57 `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.
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RESPONSE --> 58 (-1) + 2 b = -38 -58+2b= -38 2b= 20 b=10 198 (-1) + 7 b = -128 -198+7b= -128 7b=70 b=10 confidence assessment: 2
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12:59:04 You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10.
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RESPONSE --> self critique assessment: 3
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13:01:15 `q004. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.
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RESPONSE --> 2a + 3b + c = 128 -2+30+c=128 c=100 60a + 5b + c = 90 -60+50+c=90 c=100 200a + 10 b + c = 0 -200+100+c=0 -100+c=0 c=100 confidence assessment: 3
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13:01:38 Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which was some simple arithmetic gives us c = 100. Substituting these values into the second equation we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will begin obtain an identity. This would completely validate our solution.
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RESPONSE --> self critique assessment: 3
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13:08:32 `q005. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case for the graph you sketched in the preceding assignment, then what equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?
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RESPONSE --> y = a x^2 + b x + c -2=a(1)^2+b(1)+c confidence assessment: 2
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13:08:45 We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2.
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RESPONSE --> self critique assessment: 3
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13:10:52 `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what equations do we get if we substitute the x and y values corresponding to the point (3, 5), then (7, 8) into the form y = a x^2 + b x + c?
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RESPONSE --> y = a x^2 + b x + c 5= a(3)^2+b(3)+C 8= a(7)^2+b(7)+c confidence assessment: 2
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13:11:02 Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 7.
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RESPONSE --> self critique assessment: 3
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13:22:40 `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what system of equations do we get if we substitute the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c? What is the solution of this system?
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RESPONSE --> a+b+c= -2 9a+3b+c=5 49a+7b+c=8 9a+3b+c=5 a+b+c= -2 8a+2b=7 49a+7b+c=8 9a+3b+c=5 40a+4b=3 32a+14b= 28 80a+14b= 6 -48a=22 a= -.458 8a+2b=7 8(-.458)+2b=7 -3.667+2b=7 2b=10.667 b=5.3335 a+b+c= -2 -.458+5.3335+c= -2 c=6.8755 confidence assessment: 3
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13:22:54 The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different system, you should show the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.
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RESPONSE --> self critique assessment: 3
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13:35:26 `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c. What function do you get? What do you get when you substitute x = 1, 3, 5 and 7 into this function?
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RESPONSE --> y=-.458x^2+5.335x+6.8755 -.458(1)^2+5.335(1)+6.8755=11.7525 -.458(3)^2+5.335(3)+6.8755 -4.122+16.005+6.8755=18.7585 -.458(5)^2+5.335(5)+6.8755 -11.45+26.675+6.8755=22.1005 -.458(7)^2+5.335(7)+6.8755 -22.442+37.345+6.8755=21.7785 confidence assessment: 2
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13:38:43 Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values gives us the points (1, -2), (3, 5) and (7, 8) we used to obtain the formula, plus the point (5, 8.33333).
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RESPONSE --> I see what I did I entered the value for c as a pos. instead of neg. i thought things didn't look right. it should be y = - 0.45833 x^2 + 5.33333 x - 6.875 -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2 -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5 -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333 -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8 self critique assessment: 2
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13:38:50 end program
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RESPONSE --> self critique assessment: 2
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zWÔ[ߛ assignment #002 002. `query2 Precalculus I 08-31-2008
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18:04:08 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 --> 0,118 40, 91.96194 80, 73.29324 confidence assessment: 2
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18:04:16 ** Continue to the next question **
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RESPONSE --> self critique assessment: 3
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18:09:10 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> 7, 113.79 19, 106.81033 31, 100.13005 confidence assessment: 2
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18:09:18 ** Continue to the next question **
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RESPONSE --> self critique assessment: 2
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18:11:06 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> 0, 118 60, 81.85297 120, 59.90819 confidence assessment: 3
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18:11:46 ** 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 --> self critique assessment: 1
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18:12:42 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 118= 0a^2+0b+c confidence assessment: 3
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18:13:12 ** 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 --> self critique assessment: 3
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18:14:05 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 81.85297=3600a+60b+c confidence assessment: 3
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18:14:17 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> self critique assessment: 3
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18:14:57 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 59.90819=14400a+120b+c confidence assessment: 3
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18:15:05 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> self critique assessment: 3
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18:17:43 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 --> 118=0a + 0b +c 81.85297=3600a +60b +c 36.14703= -3600a -60b confidence assessment: 3
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18:17:55 ** 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 --> self critique assessment: 3
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18:20:04 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 --> 81.85297=3600a +60b +c 59.90819= 14400a + 120b +c 21.94478= -14040a -60b confidence assessment: 3
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18:20:13 ** 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 --> self critique assessment: 3
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18:22:12 Which variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?
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RESPONSE --> 36.14703= -360a -60b 21.94478= -14040a -60b 14.20225= 13680a a=.00103817617 or aprox. .00104 confidence assessment: 3
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18:22:26 ** 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 --> self critique assessment: 2
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18:24:43 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 --> 36.14703= -360(.00104)- 60b b=-.6086905 or aprox. -.60869 confidence assessment: 3
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18:25:03 ** 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 --> self critique assessment: 3
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18:26:24 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> 118=.00104(0)^2-.60869(0)+c c=118 confidence assessment: 3
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18:26:33 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> self critique assessment: 2
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18:27:10 What is the resulting quadratic model?
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RESPONSE --> y=.00104x^2-.60869+118 confidence assessment: 3
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18:27:23 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> self critique assessment: 2
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18:29:27 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 --> 0, 118 dev. 0 20, 106.2422 dev. 2.3416 40, 95.3164 dev. 3.35446 confidence assessment: 2
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18:29:57 ** 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 --> self critique assessment: 2
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18:31:38 What was your average deviation?
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RESPONSE --> 1.84851625 confidence assessment: 2
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18:31:47 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> self critique assessment: 2
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18:32:08 Is there a pattern to your deviations?
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RESPONSE --> not really confidence assessment: 3
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18:32:16 ** 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 --> self critique assessment: 3
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18:32:41 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 --> i believe so confidence assessment: 3
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18:32:49 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> self critique assessment: 3
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18:33:05 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> i hope so confidence assessment: 3
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18:33:20 ** 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 --> self critique assessment: 3
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18:36:37 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 --> 90, -1.9742 80, 8.8732 70, 20.8646 60, 34 50, 48.2794 40, 63.7028 30, 80.2702 20, 97.9816 10, 116.837 confidence assessment: 2
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18:36:45 ** 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 --> self critique assessment: 2
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18:37:46 What three points on your graph did you use as a basis for your model?
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RESPONSE --> 90, 0 60, 36 20, 100 confidence assessment: 2
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18:37:58 ** 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 --> self critique assessment: 2
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18:38:38 Give the first of your three equations.
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RESPONSE --> 8100a+90b+c=0 confidence assessment: 3
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18:38:44 ** 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 --> self critique assessment: 2
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18:39:16 Give the second of your three equations.
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RESPONSE --> 3600a+60b+c=36 confidence assessment: 2
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18:39:23 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> self critique assessment: 2
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18:39:53 Give the third of your three equations.
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RESPONSE --> 400a+20b+c=100 confidence assessment: 3
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18:40:01 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> self critique assessment: 2
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18:40:44 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 4500a+30b= -36 confidence assessment: 3
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18:40:48 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> self critique assessment: 3
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18:41:15 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 3200a+40b= -64 confidence assessment: 3
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18:41:20 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> self critique assessment: 3
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18:44:05 Explain how you solved for one of the variables.
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RESPONSE --> 4500a+30b= -36)(-40) 3200a+ 40b= -64)( -30) 180,000a+1200b= -1440 96,000a+1200b= -1920 84,000a= 480 a=.00572 confidence assessment: 3
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18:44:11 ** 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 --> self critique assessment: 3
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18:49:25 What values did you get for a and b?
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RESPONSE --> a=.00572 b= -2.058 4500(.00572)+30b= -36 confidence assessment: 2
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18:49:38 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> self critique assessment: 2
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18:49:59 What did you then get for c?
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RESPONSE --> c=136.8364 confidence assessment: 2
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18:50:06 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> self critique assessment: 2
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18:50:56 What is your function model?
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RESPONSE --> y=.00572x^2-2.05714+136.8364 confidence assessment: 1
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18:51:11 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> self critique assessment: 2
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18:53:22 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> 90, -1.9742 80, 8.8732 70, 20.8646 60, 34 50, 48.2794 40, 63.7028 30, 80.2702 20, 97.9816 10, 116.837 confidence assessment: 3
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18:53:27 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> self critique assessment: 1
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18:55:47 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> 14 depth and 72 clock time confidence assessment: 2
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18:56:06 ** 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 --> self critique assessment: 1
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19:00:10 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 --> 0, .96368375 10, 1.36495 20, 1.750933 30, 2.090227 40, 2.388479 50, 2.650655 60, 2.881118 70, 3.083704 80, 3.261786 90. 3.418326 100, 3.53922625 confidence assessment: 3
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19:00:31 ** 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 --> self critique assessment: 1
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19:01:45 What three points on your graph did you use as a basis for your model?
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RESPONSE --> 10, 1.36495 50, 2.650655 100, 3.555931 confidence assessment: 3
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19:01:51 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> self critique assessment: 3
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19:02:57 Give the first of your three equations.
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RESPONSE --> 1.36495= 100a+10b+c 2.650655= 2500a+50b+c 3.555931= 10000a+100b+c confidence assessment: 3
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19:03:07 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> self critique assessment: 1
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19:03:34 Give the second of your three equations.
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RESPONSE --> 2.650655= 2500a+50b+c confidence assessment: 3
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19:03:38 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> self critique assessment: 3
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19:04:00 Give the third of your three equations.
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RESPONSE --> 3.555931= 10000a+100b+c confidence assessment: 3
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19:04:05 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> self critique assessment: 3
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19:04:53 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -1.285705= -2400a -40b confidence assessment: 2
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19:04:58 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> self critique assessment: 2
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19:05:29 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -.905276= -7500a -50b confidence assessment: 2
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19:05:35 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> self critique assessment: 2
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19:07:32 Explain how you solved for one of the variables.
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RESPONSE --> (-1.285705= -2400a -40b)(-50) (-.905276= -7500a -50b)( -40) 28.07421= -180000a a= -.00015968 confidence assessment: 2
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19:07:40 ** 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 --> self critique assessment: 2
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19:08:15 What values did you get for a and b?
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RESPONSE --> a= -.00015968 b= .041723425 confidence assessment: 2
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19:08:21 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> self critique assessment: 2
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19:08:42 What did you then get for c?
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RESPONSE --> c= .96368375 confidence assessment: 2
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19:08:49 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> self critique assessment: 2
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19:09:43 What is your function model?
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RESPONSE --> y= -.00015968x^2+.041723425x+.96368375 confidence assessment: 2
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19:09:52 ** y = -.0000876638 x^2 + (.01727)x + 1.773 **
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RESPONSE --> self critique assessment: 2
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19:11:29 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> 3.0-4.0= 65-100 percent 3,65)-(4,100) confidence assessment: 2
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19:12:30 ** 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 --> i didn't get a 4.0 solution either it was actually up to 3.53922625 self critique assessment: 2
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19:12:59 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 80, 3.27960575 confidence assessment: 2
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19:13:11 ** 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 --> self critique assessment: 2
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19:13:36 How well does your model fit the data (support your answer)?
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RESPONSE --> i think my model fits the data pretty well confidence assessment: 2
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19:19:21 ** 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 --> 0, .96368375 dev. .03782915 10, 1.36495 dev. 0 20, 1.73428025 dev. .01665275 30, 2.0716745 dev. .0185525 40, 2.37713275 dev. .01134625 50, 2.65066 dev. .000005 60, 2.89224125 dev. .01112325 70, 3.1018915 dev. .0181875 80, 3.27960575 dev. .01781975 90. 3.425384 dev. .007058 100, 3.53922625 dev. .01670475 self critique assessment: 2
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19:20:57 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> 1, 1280 2, 320 3, 142.2222 4, 80 5, 51.2 6, 35.55556 7, 26.12245 8, 20 9, 15.80247 10, 12.8 confidence assessment: 3
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19:21:30 ** 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 --> self critique assessment: 1
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19:22:14 What three points on your graph did you use as a basis for your model?
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RESPONSE --> 1,1280 5, 51.2 10, 12.8 confidence assessment: 2
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19:22:20 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> self critique assessment: 2
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19:22:59 Give the first of your three equations.
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RESPONSE --> 1a+1b+c=1280 confidence assessment: 3
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19:23:06 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> self critique assessment: 2
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19:23:26 Give the second of your three equations.
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RESPONSE --> 25a+5b+c=51.2 confidence assessment: 3
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19:23:30 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> self critique assessment: 2
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19:23:51 Give the third of your three equations.
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RESPONSE --> 100a+10b+c=12.8 confidence assessment: 3
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19:23:56 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> self critique assessment: 2
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19:24:32 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -24a -4b=1228.8 confidence assessment: 3
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19:24:37 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> self critique assessment: 2
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19:24:59 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -75a -5b= 38.4 confidence assessment: 3
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19:25:04 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> self critique assessment: 2
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19:26:15 Explain how you solved for one of the variables.
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RESPONSE --> -5( -24a- 4b= 1228.8) -4( -75a -5b=38.4) -5990.4= -180a a=33.28 confidence assessment: 2
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19:26:21 ** 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 --> self critique assessment: 2
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19:26:39 What values did you get for a and b?
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RESPONSE --> a= 33.28 b= -506.88 confidence assessment: 3
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19:26:49 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> self critique assessment: 2
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19:27:03 What did you then get for c?
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RESPONSE --> c=1753.6 confidence assessment: 3
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19:27:09 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> self critique assessment: 2
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19:27:44 What is your function model?
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RESPONSE --> y=33.28x^2-506.88x+1753.6 confidence assessment: 2
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19:27:50 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> self critique assessment: 2
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19:28:18 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> 1.6= aprox. 1027.7888 confidence assessment: 2
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19:28:30 ** 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 --> self critique assessment: 2
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19:30:01 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> my chart had a great variation and acurate solutions could not be given based on my estimates confidence assessment: 1
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19:30:06 ** 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 3.9 - 5.4 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 --> self critique assessment: 1
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i{Nǻ assignment #003 003. `query 3 Precalculus I 08-31-2008
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22:14:12 query graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.
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RESPONSE --> y= 3x^2 opens upwards 0,0 1,3 2,12 -2,12 y=2x^2 opens upwards 0,0 1,2 2,8 -2,8 y=.5x^2 opens upwards 0,0 1,.5 2, 1 -2, 1 y= -.3x^2 opens downwards 0,0 1, -.3 2, -1.2 -2, -1.2 confidence assessment: 2
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22:14:29 ** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up. INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). **
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RESPONSE --> self critique assessment: 0
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22:27:02 query prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1
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RESPONSE --> y = x^2 + 2x + 1 the vertex is -1,0 and the vertex lies on the x axis y = x^2 + 3x + 1 the vertex is -1.5, .25 and the x intercepts are -.381967, 0 and -2.61805, 0 confidence assessment: 2
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22:27:36 ** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1). For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, 1.25) and (-.5, 1.25). **
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RESPONSE --> self critique assessment:
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22:29:00 how did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?
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RESPONSE --> it changed based upon the quadratic formula confidence assessment: 1
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22:29:12 ** The vertices move downward and to the left, but not along a straight line. In fact the vertices lie along a different parabola of their own. **
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RESPONSE --> self critique assessment: 0
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22:30:07 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> They help you get the vertex and the shape around the vertex angled correctly. confidence assessment: 1
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22:30:20 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symetry around the vertex which defines the 'shape' and direction and allows you to extrapolate. INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. **
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RESPONSE --> self critique assessment: 0
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22:30:53 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> b^2-4(a)(c) confidence assessment: 2
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22:31:12 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). **
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RESPONSE --> self critique assessment: 0
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22:33:51 query #4. Questions about vertex between zeros and the shape of the curve connecting vertices: What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?
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RESPONSE --> b/2a is the axcess of symmetry you either add or subtract sqrt(b^2-4(a)(c))/2a to get the zeros confidence assessment: 2
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22:34:01 ** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **
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RESPONSE --> self critique assessment: 0
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22:35:35 What was the shape of the curve connecting the vertices?
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RESPONSE --> the slope tends to become less steep as it comes closer to the vertex confidence assessment: 2
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22:35:52 ** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea. Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **
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RESPONSE --> self critique assessment: 0
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22:35:58 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 0
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ɼz叮Hk{ҥ}J assignment #003 003. Precalculus I 08-31-2008
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21:26:43 `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
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RESPONSE --> y = - 0.45833 x^2 + 5.33333 x - 6.875 -5.33 +- sqrt(5.33^2 -4( -.5)( -6.9)/ 2( -.5) (-5.33 +- sqrt(28.4089-13.75)/-1 -5.33 +- 3.8)/ -1 y = - 0.45833 + 5.33333 - 6.875 y= -2 y = - 0.45833 (3)^2 + 5.33333 (3)- 6.875 y=5 y = - 0.45833 (7)^2 + 5.33333 (7) - 6.875 y=8 confidence assessment: 2
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21:27:01 For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
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RESPONSE --> self critique assessment: 1
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21:27:42 `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
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RESPONSE --> 5.33 confidence assessment: 2
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21:27:53 Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
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RESPONSE --> self critique assessment: 2
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21:32:53 `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
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RESPONSE --> 4.34 7.638731652 (4.34, 7.638731652) confidence assessment: 2
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21:33:35 The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
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RESPONSE --> self critique assessment:
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21:36:49 `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
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RESPONSE --> 5.33333, 8.53248 confidence assessment: 2
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21:38:02 In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024).
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RESPONSE --> self critique assessment: 0
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21:48:22 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
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RESPONSE --> 6.81822049615, 8.18191495933 4.81822049615, 8.18191495933 confidence assessment: 3
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21:48:35 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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RESPONSE --> self critique assessment: 0
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21:56:24 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
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RESPONSE --> vertex= 5,125 to the right= 6, 124 to the left= 4, 124 the parabola crosses the x axis at 0, 16.18 and 0, -6.18 confidence assessment: 3
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21:56:43 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
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RESPONSE --> self critique assessment: 0
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21:56:51 end program
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RESPONSE --> self critique assessment: 0
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