question form

#$&*

Phy 201

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.

** Question Form_labelMessages **

submission of cq_1_012

** **

cq_1_012

#$&*

Phy 201

Your 'cq_1_01.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** CQ_1_01.2_labelMessages **

The problem:

Answer the following:

• How accurately do you think you can measure the time between two events using the TIMER program?

answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> (start in the next line):

to the .01 place

#$&*

** **

this is a copy of your response to my submission of cq_1_012. i was wondering if this is all I submitted or if your response got cut off? thank you

@&

Something clearly went wrong. If you have a copy of that document, can you resubmit it? If not, let me know on what day you submitted it and I should be able to locate your original submission.

*@

question form

#$&*

Mth 173

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.

** Question Form_labelMessages **

query 01 correction

** **

query 01

#$&*

course Mth 1713

5/17 2 am

001. `query1

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Question: `qFor the temperature vs. clock time model, what were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(0, 95)

(20, 60)

(40, 41)

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** Continue to the next question **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(7, 85)

(19, 62)

(31, 49)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** Continue to the next question **

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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(20, 60)

(30, 49)

(40, 41)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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 nswers'.

STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps, you should not expect that the numbers given here will be the same as the numbers you obtained when you solved the problem.)

For my quadratic model, I used the three points

(10, 75)

(20, 60)

(60, 30). **

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Self-critique (if necessary): I will use more spread out data points the next time

------------------------------------------------

Self-critique Rating:

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Question: `qWhat is the first equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

60 = a(20)^2 + b(20) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**

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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qWhat is the second equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

49= a(30)^2 + b(20) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is the third equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

41= a(40)^2 + b(40) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **

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Self-critique (if necessary):

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Self-critique Rating:

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I subtracted:

&&&& the second equation : 49= a(30)^2 + b(30) + c

And the first equation: 60 = a(20)^2 + b(20) + c

= -11= a500 + b10

The c’s cross out

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qTo 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&The third equation: 41= a(40)^2 + b(40) + c

Minus

The second equation: 49= a(30)^2 + b(30) + c

-8= a700 + b10

And eliminated c &&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&&I eliminated the b value from these two equations.

I subtracted the equation: -8= a700 + b10

Minus

The equation: -11= a500 + b10

The resulting answer was: 3 = a200

Therefore a = .015 &&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I substituted the value of a into the 2 variable equation:

-11= a500 + b10

-11= (.015)500 + b10

-11 = 7.5 + b10

-18.5 = b10

b = -1.85 &&&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is the value of c obtained from substituting into one of the original equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&& I substituted the values of a = .015 and b= -1.85 into the equation :

60= a(20)^2 + b(20) + c

60= .015(20)^2+ -1.85(20) + c

60 = 6 - 37 + c

60= -31 + c

91 = c

C= 91 &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is the resulting quadratic model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& a = .015, b = -1.85, c = 91

y = .015 x^2 - 1.85 x + c

@&

c = 91, so

y = .015 x^2 - 1.85 x + 91

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was

y = (.015) x^2 - (1.95)x + 93. **

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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&&&&(20):

y= .015 (20)^2 - 1.85 (20) + 91

y= 6 -37 + 91

y= 60

deviation = 0

(30) :

Y= .015 (30)^2 -1.85 (30) + 91

Y= 13.5 - 55.5 + 91

Y= 49

Deviation = 0

(40):

Y= .015 (40)^2 -1.85 (40) + 91

Y= 24 -74 + 91

Y= 41

Deviation = 0 &&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qWhat was your average deviation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& 0 + 0 + 0 / 3 = 0 &&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: My average deviation was .6 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qIs there a pattern to your deviations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& No, there was no deviation &&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

@&

Your deviations would be calculated at these three points. But you would also calculate deviations at the remaining data points. It is unlikely that you would get 0 deviation for any of the other data points.

For example, your data included the point (0, 95). Your model

y = .015 x^2 - 1.85 x + 91

would give you y = 91 when y = 0. The deviation for the first data point is therefore 4.

*@

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Question: `qHave 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Yes I have. I have taken notes and printed off the process as well

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Yes! I have printed off the steps of the modeling process so they can be easily accessed in my notebook when I need a review.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

@&

Your entire process on the first problem was good, and you did a very good job of providing the details. The only flaw is at the very end, where you didn't calculate all the deviations. This would, of course, be very easy to do.

*@

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Question: `qQuery Completion of Model first problem: Completion of model from your data. Give your data in the form of depth vs. clock time ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(2.8, 82.1)

(5.6, 76.8)

(8.4, 72.7)

(11.2, 69.8)

(14, 66.8)

(16.8, 63.7)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(8.4, 72.7)

(11.2, 69.8)

(14, 66.8)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

72.7= a(8.4)^2 + b(8.4) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

69.8= a(11.2)^2 + b(11.2) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

66.8= a(14)^2 + b(14) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

-2.9= a54.88 + b2.8

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

-3= a70.56 + b2.8

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I subtracted

-2.9= a54.88 + b2.8

-3= a70.56 + b2.8

And got .1 = a-15.68

A= -.0064

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A= -.0064 and b= -.912

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

80.82

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y= -.0064a -.912b +80.82

@&

The model you are using is y as a quadratic function of x, so that

y = a x^2 + b x + c.

You substitute your values of a, b and c to get your model, which will be

y = -.0064 x^2 - .912 x + 80.82.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your depth prediction for the given clock time (give clock time also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For time 5.6 seconds the predicted depth was 75.5

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat clock time corresponds to the given depth (give depth also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For given depth 69.8 the clock time 11.2

@&

The problem should have specified a depth which was not equal to one of yoru data points.

You would in any case not have used your knowledge of the data point (11.2, 69.8) to find the time at which depth is 69.8.

How would you do this?

Hint: How would you go about finding the time at which the depth is, say, 60?

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qCompletion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

grade vs. percent of material 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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(20, 2.118034)

(30, 2.369306)

(40, 2.581139)

@&

It isn't a good idea to pick three consecutive points when you can spread the points out. You get a better trend if you do so.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2.118= a(20)^2 + b(20) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2.369 = a(30)^2 + b(30) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

2.581 = a(40)^2 + b (40) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I subtracted the second equation : 2.369 = a(30)^2 + b(30) + c

Minues

The first equation: 2.118 = a(20)^2 + b(20) + c

To get :

.251 = a500 + 10b &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I subtracted the third equation : 2.581 = a (40)^2 + b(40) + c

Minus

The second equation : 2.369 = a(30)^2 + b (30) + c

To find : .212 = a700 + b10 &&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go

9600a + 80b = 1.381966 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I subtracted the two two-variable equations to eliminate b and find a

.212 = a700 + b10

Minus

.251 = a 500 + b10

To find : -.039 = a200

A = -.000195 &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A= - .000195

.251 = a500 +b 10

.251 = (-.000195) (500) + b (10)

B= .03485 &&&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

@&

You largely appear to be following correct procedures in solving these systems. However your result is not consistent with the points you found, which would give you a, b and c values -.00025, .0385, 1.44.

So there are errors in some of the details of your solutions.

Earlier you got the equation

.442= a700 + b10

The right-hand side would have come from subtracting the second equation from the third. The .442 on the left doesn't seem to correspond to any of the possible calculations that would result from the three equations.

Can you repeat your solution for this problem and submit a copy using the Question Form?

*@

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I used the first equation: 2.118 = a(20)^2 + b (20) + c

2.118 = (-.000195) (20)^2 + (.03485) (20) + c

2.118 = -.078 + .697 + c

C= 1.499 &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: c = 1.773. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& Y= -.000195x^2 + .03485x + 1.499 &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** y = -.0000876638 x^2 + (.01727)x + 1.773 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3 - 4 grade range = 43.3% and 57%

??????????? I’m not sure how I got these answers, and when I tried to rework the problem I got an unrealistic answer. I took the quadratic model : y = -.000195x^2 + .03485x + 1.499 and substituted the value of x into the y, then subtracted to get the quadratic form of 0 = -.000196x^2 + -0348x - 1.50. I then tried to solve x using the quadratic formula and got -105.13 or -73. 08. These numbers don’t make sense and I don’t think I am solving this right. Could you help please?????

@&

For grade range 3 - 4, since y represents grade and x represents percent of study time, you would substitute 3 for y and solve for x, then you would substitute 4 for y and solve for x.

Using your model

y = -.000195x^2 + .03485x + 1.499

what equation do you get for y = 3, how do you solve that equation, and what is your solution?

Answer the same question for y = 4.

If you don't know how to solve those equations, send me your best attempt and I'll guide you on the solution.

*@

@&

This requires more detail. You need to show how you got these results. For example you could show the equations you solved and give at least a short indication of how you solved them.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat grade average corresponds to the given percent of review (give grade average also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

80% of classes is grade average of 6.5

@&

Substituting 80 for x in your model

y = -.000195x^2 + .03485x + 1.499

I don't get 6.5. Can you recalculate this and, if you still get 6.5, explain how you got your result?

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

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Question: `qHow well does your model fit the data (support your answer)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I think the model is somewhat off. The predicted results deviate pretty significantly from the actual values

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qillumination vs. distance

Give your data in the form of illumination vs. distance ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your 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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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)**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(3, 105.1209)

(4, 61.01488)

(5, 43.06238)

@&

Again you need to spread your chosen points out to represent a wider range of the data.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED:

(2, 264.4411)

(4, 61.01488)

(8, 16.27232) **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I will spread out data points from now on

------------------------------------------------

Self-critique Rating:

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

105.12 = a(3)^2 + b(3) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

61.01= a(4)^2 + b(4) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

43.06 = a(5)^2 + b(5) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&& I subtracted the second equation: 61.01 = a(4)^2 + b(4) + c

Minus the first equation:

105.12 = a(3)^3 + b(3) +c

To get:

-44.11= a7 + b &&&&&&

@&

To better document your work you would want to indicate which equation was subtracted from which (e.g., first from third).

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&& I subtracted the third equation: 43.06= a (5)^2 + b(5) + c

Minus the second equation : 61.01 = a(4)^2 + b(4) + c

To get:

-17.95= a9 + b

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& I subtracted the two equations to eliminate the variable b and solved from a

-17.95 = a9 + b1

Minus

-44.11 = a7 + b1

To get:

26.16 = 2a

A= 13.08 &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& A= 13.08

-44.11 = (13.08)(7) + b(1)

-44.11 = 91.56 + b

B= -135.67 &&&&&

@&

Your result for b is very far from the correct result, which based on your chosen points is around -135, but it's not possible for me to see what went wrong with the information you've provided.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& 105.12 = (13.08)(3)^2 + (-135.67)(3) + c

105.12 = 117.72 - 407.01 + c

105.12 = -289.29 + c

394.41 = c &&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: c = 588.5691**

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&& Y= 13.08x^2 - 135.67x + 394.41 &&&&&&

@&

The a, b and c values corresponding to your chosen points would be 13, -135, 393.

I can't tell what went amiss in your analysis.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat is your illumination prediction for the given distance (give distance also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

***** Given distance was 1.6

Y= 13.08(1.6)^2 - 135.67 (1.6) + 394.41

Y= 210.8

The illumination prediction is 210.8 W/m^2 &&&&&&&

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat distances correspond to the given illumination range (give illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The given illumination range 25 to 100 W/ m^2

The distances are ?????

?????I don’t understand how to properly solve using the quadratic equation. I found the values of a, b, and c for each 25 and 100, but when I try to solve using the quadratic equation the number under the square root is negative. Could you please show me how you solve and got 4.9 and 5.4??????

@&

I no longer see your model.

In any case I would first ask you to show how you got the negatives under the square root. It is certainly possible that this would happen. A quadratic model has a parabolic graph. A parabola which opens upward has a minumum possible y value, and any y value less than this would lead to a negative under the radical. Similarly a parabola which opens downward has a maximum value, and any y value that exceeds this would again result in a negative under the square root.

You should solve your present model for these illumination levels.

*@

@&

The results for your model would be very approximately 4.9 and 5.4.

These results should have been obtained by solving quadratic equations.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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. **

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!#*&!

@&

You're doing most of the right things, but something is going amiss in your solution of these systems. You have valid equations, and seem to be doing the right things, but you haven't shown enough of the details for me to be sure.

I'm going to ask you to rework the last two models and submit your work using the Question Form or the Submit Work Form (your choice; either is fine).

Check my notes first.

Go ahead and use the same points you have chosen, but do read my note about the advisability of spreading the chosen points out more.

*@

** **

I am resubmitting my work for query 1 per your request. I fixed most things you asked me to but I have a few questions. thank you

** **

self-critique rating

@&

Your models are good. You are doing very well, and you are being very conscientious with your assignments. I believe you're going to do quite well in these courses.

You do have a few things still to fix, and you should take the time to do so. I don't think it will take you long, and over the course of the summer it will save you time and trouble in the long run.

&#Please see my notes and submit a copy of this document with revisions, comments 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.

&#

*@