Mth 271
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I need help getting my modeling project to work out. What have I done wrong? I do not understand how to do the deviations. Can you please look at what I have thus far and tell me where I have screwed up.
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002. `Query 2
Question: `qWhat were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
Your solution:
(0,95) (20,60) (40,41)
Confidence Assessment: 3
Given Solution:
`a Continue to the next question **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31?
Your solution:
y = .01 t^2 - 2 t + 100
At clock time 7
y=.01*7^2-2*7+100
y=.01*49-14+100
y=0.49-14+100
y=-13.51=100
Y=86.49
Time=19
Y=.01*19^2-2*19+100
Y=.01*361-38+100
Y=3.61-38+100
Y=65.61
Time=31
Y=.01*31^2-2*31+100
Y=.01*961-62+100
Y=9.61-62+100
Y=47.61
Confidence Assessment: 3
Given Solution:
`a Continue to the next question **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)?
Your solution:
(10,75), (30,49), (50,35)
Confidence Assessment: 3
Given Solution:
`a 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). **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat is the first equation you got when you substituted into the form of a quadratic?
Your solution:
(10,75)
depth = a t^2 + bt + c
75=a(10^2)+10b+c
Confidence Assessment: 3
Given Solution:
`a STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat is the second equation you got when you substituted into the form of a quadratic?
Your solution:
(30,49)
49=a (30^2)+30b+c
Confidence Assessment: 3
Given Solution:
`a STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat is the third equation you got when you substituted into the form of a quadratic?
Your solution:
(50,35)
35=a(50^2)+50b+c
Confidence Assessment: 3
Given Solution:
`a STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat 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?
Your solution:
Subtract
75=a(10^2)+10b+c
49=a (30^2)+30b+c
My new equation is:
26=a(-20^2)-20b
26=400a-20b
Confidence Assessment: 3
Given Solution:
`a 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. **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qTo get the second equation what multiple of which equation did you add to what multiple of which other equation, and what is the resulting equation?
Your solution:
75=a(10^2)+10b+c
35=a(50^2)+50b+c
40= 1600a – 40b
Confidence Assessment: 3
Given Solution:
`a 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**
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhich variable did you eliminate from these two equations, and what was its value?
Your solution:
40=1600a-40b
-2( 26=400a-20b)
I only had to multiply the second equation by another variable for it to cancel out b.
40=1600a-40b
-52= - 800a+40b
-12=2400a
Confidence Assessment: 3
Given Solution:
`a 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. **
Self-critique (if necessary):
ok.
Self-critique Rating: ok
Question: `qWhat 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?
Your solution:
A= -0.005
40=1600*-0.005-40b
40=-8-40b
48/40= b
B=1.2
Confidence Assessment: 3
Given Solution:
`a 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. **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat is the value of c obtained from substituting into one of the original equations?
Your solution:
A= -0.005
B=1.2
75=a(10^2)+10b+c
-0.005(10^2)+10*1.2+c=75
-0.5+12+c=75
11.5+c=75
C=63.5
Confidence Assessment: 3
Given Solution:
`a STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
Self-critique (if necessary): ok
Self-critique Rating:ok
Question: `qWhat is the resulting quadratic model?
Your solution:
depth = a t^2 + bt + c
y = (-0.005) t^2 - (1.2)t + 63.5
Confidence Assessment: 3
Given Solution:
`a STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was
y = (.015) x^2 - (1.95)x + 93. **
Self-critique (if necessary): ok
Self-critique Rating: ok
Question: `qWhat 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?
Your solution:
depth = a t^2 + bt + c
Clock times 95, 60, 41
y = (-0.005) t^2 + (1.2)t + 63.5
First =-0.005*9025+1.2*95+63.5 = -45.125 + 114 +63.5 = 132.375
Deviation
Second= -0.005 * 3600 + 1.2*60+63.5 = -18 + 72 +63.5 = 117.5
Deviation
Third= -0.005 *1681 + 1.2*41+63.5 = -8.405 + 49.2 +63.5 = 104.295
Deviation
I believe 95, 60 and 41 are the depths, not the clock times.
what are the clock times corresponding to these depths?
What does your model give you when you substitute each of these clock times?
By how much do the results of your model differ from the given depths at these three clock times? e.g., by how much does the depth predicted by the model at the first clock time differ from the given depth of 95 cm?
Confidence Assessment:
Given Solution:
`a 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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If you could tell me where I have messed up or how is the best way to complete this exercise and what I have done wrong. As soon as possible I would greatly appreciate it. I don't want to submit something that I know is wrong. And am soo terrified about getting behind. Please help me.
I've inserted a few questions immediately following your calculation of deviations. Insert your answers, and mark them as indicated in the note below. I think this will be fairly easy for you to fix, and should help you straighten this out.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#