Assignment 2-Query

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course Mth 163

Question: `qAssignment 2 For 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:

1 (5.3, 63.7)

3 (15.9, 46)

5 (26.5, 32)

Confidence rating #$&* 3

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Given Solution:

** Continue to the next question **

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Question: `qAccording to your graph what would be the temperatures at clock times

7, 19 and 31?

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Your solution:

f(7)=.016(7)^2+-2.019(7)+73.938

f(7)=60.589

f(19)=41.353

f(31)=26.725

Confidence rating #$&* 3

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Given Solution:

** Continue to the next question **

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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:

(5.3, 63.7)

(15.9, 46)

(26.5, 32)

Confidence rating #$&* 3

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

quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a(5.3)^2+5.3b+c=63.7

28.09a+5.3b+c=63.7

Confidence rating #$&* 2

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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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Question: `qWhat is the second equation you got when you substituted into the form

of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a(15.9)^2+15.9b+c=46

252.81a+15.9b+c=46

Confidence rating #$&* 2

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Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my second data

point was 400a + 20b + c = 60 **

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

a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a(26.5)^2+26.5b+c=32

702.25a+26.5b+c=32

Confidence rating #$&* 3

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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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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 used Matrix on a graphing calculator

This is not an acceptable way to solve a system of equations in this course.

If you want to specify the details of the matrix and the process of row reduction, which is equivalent to and based on the process of solving the system by elimination, this would be acceptable.

Confidence rating #$&* 3

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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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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:

Matrix

Confidence rating #$&* 3

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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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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:

Matrix

Confidence rating #$&* 3

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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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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:

Matrix

Confidence rating #$&* 3

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

equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

73.938

Confidence rating #$&* 3

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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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Question: `qWhat is the resulting quadratic model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f(x)=.016x^2+-2.019x+73.938

Confidence rating #$&* 2

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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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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:

f(5.3)=.016(5.3)^2+-2.019(5.3)+73.938

f(5.3)=63.687

f(10.6)=54.334

f(15.9)45.881

Confidence rating #$&* 3

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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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Question: `qWhat was your average deviation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

-8.903

you should document the details of your solution, showing the process you used to arrive at this result

Confidence rating #$&* 3

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Given Solution:

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

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Question: `qIs there a pattern to your deviations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Yeah, it was decreasing at a dcreasing rate/

Confidence rating #$&* 3

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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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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 and yes

Confidence rating #$&* 3

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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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Question: `qHave you memorized the steps of the modeling process, and are you

gonna remember them forever? Convince me.

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Your solution:

Yes, you use the standard form of quadratic equation. Choose 3 points from data site.

Sub for x and y. Solve for a, b, and c.

Confidence rating #$&* 3

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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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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:

(5.3,63.7)

(10.6,54.8)

(15.9,46)

(21.2,37.7)

(26.5,32)

(31.8,26.6)

Confidence rating #$&* 3

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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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3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(5.3, 63.7)

(15.9, 46)

(26.5, 32)

Confidence rating #$&* 3

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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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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

28.09a+5.3b+c=63.7

Confidence rating #$&* 3

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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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ok

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

252.81a+15.9b+c=46

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation

252.81a +15.9b + c = 46 **

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

702.25a+26.5b+c=32

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation

702.25a + 26.5b + c = 32. **

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third

gave me 449.44a + 10.6b = -14. **

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Used matrix

Confidence rating #$&* 3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a=.016

b=-2.019

Confidence rating #$&* 3

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Given Solution:

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

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c=73.937

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

f(x)=.016x^2+-2.019x+73.938

Confidence rating #$&* 3

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Given Solution:

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

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

also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Clock time 46 sec

depth 14.92 cm.

Confidence rating #$&* 3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The given depth 14 cm, clock time 47.78 sec.

You need to document how you found this. Note that the calculator is appropriate only for doing the arithmetic (or checking the results of your analysis).

Confidence rating #$&* 3

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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. **

STUDENT QUESTION

I have done what I could with the completion of flow model page 7 directions when I

hit the sqrt

button on calculator so sqrt -.652 I would get (0,.807465169527) I do not remember

ever doing a problem with 0,.8…. so I

hope I used the correct numbers to solve the rest of quad equation using quad

formula

INSTRUCTOR RESPONSE:

Short answer:

The square root of a negative isn't a real number, so there is no solution to the

equation for your given depth. Your calculator indicated a complex-number solution.

Longer answer:

The square root of a negative number is an imaginary number; the result you got is a

point in the complex-number plane, on the imaginary axis. You might not understand

what that means, but the point is that there is no real-number solution. You can't

square a real number and get a negative, so the square root of a negative isn't a real

number.

What this means is that the equation has no real-number solution. There is no clock

time t for which the depth takes the y value you used in the equation.

In terms of the graph, note that the graph of the quadratic function is a parabola,

which opens upward. So there are y values that lie completely below the parabola. If

you try to solve the quadratic for one of these y values, you won't get a solution.

This sort of thing can certainly happen with a mathematical model. When it does, the

answer is simply that there is no such solution.

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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:

( 0,1)

(10,1.791)

(20,2.118)

(30,2.369)

(40,2.581)

(50,2.768)

(60,2.936)

(70,3.092)

(80,3.236)

(90,3.372)

(100,3.5)

Confidence rating #$&* 3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(10, 1.791), (40, 2.581), (90, 3.372)

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

100a+10b+c=1.791

Confidence rating #$&* 3

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Given Solution:

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

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1600a+40b+c=2.281

Confidence rating #$&* 3

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Given Solution:

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

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

8100a+91b+c=3.372

Confidence rating #$&* 3

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Given Solution:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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Given Solution:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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Given Solution:

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

9600a + 80b = 1.381966 **

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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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. **

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a=-1.317

b=0.033

Confidence rating #$&* 3

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Given Solution:

** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c=1.475

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

-1.317x^2+0.033x+1.475=y

Confidence rating #$&* 3

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

Given Solution:

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

STUDENT QUESTION

Hello! I am working on the Modeling Project #1 still and I am having such issues with

the data sets for the rest of the

worksheet. I keep reading and I see that your doing grade average versus percentage

of assignments, but I am confused on what

it is asking or what method I am supposed to be using. I got the first question, solving

for a, b, and c and I am familiar

with the quadratic forumla, I am just missing something on how to start these next two

problems.

Could you give me a boost to what to do?

INSTRUCTOR RESPONSE

What that boils down to can be summarized by a table.

For example, consider the following

x y

2 20

5 50

12 130

From this table and the form y = a x^2 + b x + c you get the equations

20 = 4 a + 2 b + c

50 = 25 a + 5 b + c

130 = 144 a + 12 b + c

which you can solve by elimination, as you did with the first question.

Now you are given data for grade ave. vs. percent of review.

You could make a table of y vs. x, with y the grade average and x the percent of review.

You could replace the heading 'x' in the first column with the identifier 'percent of

review' and the 'y' in the second column with 'grade ave', so your table would

represent percent of review vs. grade average.

Your table would have several additional rows (one additional row for every 'data

point').

You are instructed to choose three data points, and to base your model on those three

points.

You could for example make a 'shortened table' with just the three points you choose,

very similar to the table given above (but with different numbers).

To get a quadratic model you would again use the form y = a x^2 + b x + c to get three

equations, one for each point.

Solving the equations for a, b and c and plugging those values back into the form y =

a x^2 + b x + c gives you your model.

Let me know if this doesn't help.

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Self-critique (if necessary):

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:

No Solution

You should show the equation(s) needed to answer this question, and the details of the solution process.

Confidence rating #$&* 3

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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. **

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Self-critique (if necessary):

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

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:

No solution

Confidence rating #$&* 3

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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. **

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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:

It didnt fit very well

You need to quantify and document this answer.

Confidence rating #$&* 3

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

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. **

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Self-critique (if necessary):

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)

(4/61.0149)

(5/43.0624)

(6/25.9154)

(7/19.9277)

(8/16.2723)

(9/11.2808)

(10/9.4845)

Confidence rating #$&* 3

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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)**

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Self-critique (if necessary):

ok

Self-critique rating #$&*3

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(1, 935.139)

(5, 43.062)

(10, 9.484)

Confidence rating #$&* 3

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

Given Solution:

** STUDENT SOLUTION CONTINUED:

(2, 264.4411)

(4, 61.01488)

(8, 16.27232) **

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1a+1a+c=935.139

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

25a+5b+c=43.062

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

100a+10b+c=9.484

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Used matrix

Confidence rating #$&* 3

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Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

matrix

Confidence rating #$&* 3

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

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 **

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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=24.034

b=-367.22

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

c=1278.327

Confidence rating #$&* 3

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

Given Solution:

** STUDENT SOLUTION CONTINUED: c = 588.5691**

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Self-critique (if necessary):

Self-critique rating #$&*

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

24.034x^2+-367.22x+1278.327=y

Confidence rating #$&* 3

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

Given Solution:

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

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Self-critique (if necessary):

Self-critique rating #$&*

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

also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When the distance was 1.6 the illumination was 758.30204.

Confidence rating #$&* 3

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

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. **

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Self-critique (if necessary):

Self-critique rating #$&*

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

illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When the illumination was 25 the distance was 5.168.

Confidence rating #$&* 3

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

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. **

You clearly understand the overall process. And it's good that you know how to set the system of equations up as a matrix calculation.

However on this assignment you need to document the steps of your solution. You can choose to give the matrix and the steps you took to reduce it (i.e., which multiple of which row was added to which, what a row was multiplied by, etc.), which is completely equivalent to the elimination process. But simply quoting 'matrix calculation' is not sufficient. That's something that could be done using a calculator with no insight into the elimination process.

You should fill in the details of the process for at least one of these systems, to be sure you know how to document this process. Similar problems will almost certainly appear on at least one test, and you will need to document your solutions there.

You also need to show the equations you are solving when answering questions about the models, and how you solved them. You should insert the details on all those questions. The calculator is not an appropriate tool for solving these equations, though of course it can be used to do the arithmetic.

In general, in this course, work is do be done analytically. The calculator is appropriate only for doing arithmetic, and of course for checking or validating the solutions you have constructed analytically.

&#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.

&#

"

#$&*