qa_007

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course Mth163

Question: `q001. Note that this assignment has 10 questions

Sketch a graph of the following (x, y) points: (1,2), (3, 5), (6, 6). Then sketch the straight line which appears to come as close as possible, on the average, to the three points. Your straight line should not actually pass through any of the given points.

• Describe how your straight line lies in relation to the points.

• Give the coordinates of the point at which your straight line passes through the y axes, and give the coordinates of the x = 2 and x = 7 points on your straight line.

• Determine the slope of the straight line between the last two points you gave.

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Your solution:

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My strait line lies slightly below my other coordinates

(0,.1) (2,2.2)

Rise 2.4-.1/run2=1.15 slope

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confidence rating #$&*: 2

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Given Solution:

Your straight line should pass above the first and third points and beneath the second. If this is not the case the line can be modified so that it comes closer on the average to all three points.

The best possible straight line passes through the y-axis near y = 2. The x = 2 point on the best possible line has a y coordinate of about 3, and the x = 7 point has a y coordinate of about 7. So the best possible straight line contains points with approximate coordinate (2,3) and (7,7).

The slope between these two points is rise/run = (7 - 3)/(7 - 2) = 4/5 = .8.

Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79

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Self-critique (if necessary):

I just completely ran through x=7,

Coordinates are (7,8)

Now I can find average rate

(8-2.2)/(7-2)=1.16

I know this is still off I just wanted to apply my numbers the right way.

Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79, How was this obtained again???

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You are not expected to be able to find the best possible straight line. You're just supposed to come reasonably close, as you did.

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Question: `q002. Plug coordinates of the x = 2 and x = 7 points into the form y = m x + b to obtain two simultaneous linear equations. Give your two equations. Then solve the equations for m and b and substitute these values into the form y = m x + b. What equation do you get?

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Your solution:

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. y=.8(2)+3

. y=4.6

. y=.8(7)+7

. y= 12.6

Or is it y=.8(2)+b

Then y-b=.8(2)

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The form is y = m x + b.

You're asked to use the x and y coordinates of two points.

You aren't asked to plug in a value for b.

You don't appear to have plugged in the y coordinate of this point.

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confidence rating #$&*:

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Given Solution:

Plugging the coordinates (2,3) and (7, 7) into the form y = m x + b we obtain the equations

3 = 2 * m + b

7 = 7 * m + b.

Subtracting the first equation from the second will eliminate b. We get 4 = 5 * m. Dividing by 5 we get m = 4/5 = .8.

Plugging m = .8 into the first equation we get 3 = 2 * .8 + b, so 3 = 1.6 + b and b = 3 - 1.6 = 1.4.

Now the equation y = m x + b becomes y = .8 x + 1.4.

Note that the actual best-fit line is y = .763 x + 1.79, accurate to three significant figures.

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Self-critique (if necessary):

Are we not allowed to use our average rate as slope??? Meaning the m as in y=mx+b???

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Self-critique rating:1

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The point here is to use the x and y coordinates of two points to get the equation of the line, analogous to the way you previously used the coordinates of 3 points to get the equation of the parabola.

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Question: `q003. Using the equation y = .8 x + 1.4, find the coordinates of the x = 1, 3, and 6 points on the graph of the equation.

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Your solution:

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When I plug each value in for x I obtain these y values ( in order)

2.2

3.2

6.2

This graph is not linear it looks exponential ( I counted by ones on x axis and by twos for the y axis)

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confidence rating #$&*: 3

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Given Solution:

Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question: `q004. The equation y = .8 x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)?

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Your solution:

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Between points 2.2 and 2 this is a two point deviation same goes for 6.2 and 6, however once we get to 3.8 and 5 the deviation is 1.2. To find the average of these deviations I will add them up and divide by three. My quotient is .0533(forever 333333’s)

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confidence rating #$&*: 3

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Given Solution:

(1, 2.2) differs from (1, 2) by the .2 unit difference between y = 2 and y = 2.2.

(3, 3.8) differs from (3, 5) by the 1.2 unit difference between y = 5 and y = 3.8.

(6, 6.2) differs from (6, 6) by the .2 unit difference between y = 6 and y = 6.2.

{}The average discrepancy is the average of the three discrepancies:

ave discrepancy = ( .2 + 1.2 + .2 ) / 3 = .53.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question: `q005. Using the best-fit equation y =.76 x + 1.79, with the numbers accurate to the nearest .01, how close do the predicted points corresponding to x = 1, 3, and 6 come to the original data points (1,2), (3, 5), and (6, 6)?

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Your solution:

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y =.76 (1) + 1.79=2.55

y =.76*3 + 1.79=4.07

y =.76 *6 + 1.79=6.35

(1,2.55)(3,4.07)(6,6.35)

Differ from (1,2), (3, 5), and (6, 6)

By an average deviation, which I will obtain by subtracting my new y values by my original y values then dividing by 3. So change in y/3 = .61 This prediction model is off only by a deviation of .61

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confidence rating #$&*: 3

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Given Solution:

Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points.

The average distance is (.55 + .93 + .35) / 3 = .61 from the points.

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Self-critique (if necessary):

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Self-critique rating:OK

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Question: `q006. The average distance of the best-fit line to the data points appears to greater than the average distance of the line we obtain by an estimate. In fact, the best-fit line doesn't really minimize the average distance but rather the square of the average distance. The distances for the best-fit model are .55, .93 and .35, while the average distances for our first model are .2, 1.2 and .2. Verify that the average of the square distances is indeed less for the best-fit model.

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Your solution:

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.55^2+ .93^2 +.35^2=1.2899

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confidence rating #$&*:

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Given Solution:

The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43.

The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51.

Thus the best-fit model does give the better result.

We won't go into the reasons here why it is desirable to minimize the square of the distance rather than the distance. When doing eyeball estimates, you don't really need to take this subtlety into account. You can simply try to get is close is possible, on the average, to the points.

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q007. If the original data points (1,2), (3, 5), and (6, 6) represent the selling price in dollars of a bag of widgets vs. the number of widgets in the bag, then how much is paid for a bag of 3 widgets? How much would you estimate a bag of 7 widgets would cost?

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Your solution:

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$5

1*3=6,3*6=12

(7,12) $12 for 7 widgets

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confidence rating #$&*: 1

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Given Solution:

If we use the best-fit function y =.76 x + 1.79, noting that y represents the cost and x the number of widgets, then the cost of 3 widgets is

y = .76 * 3 + 1.79 = 4.07, representing cost of $4.07.

The cost of 7 widgets would be

y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11.

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Self-critique (if necessary):

Use the best-fit function y =.76 x + 1.79 I thought we were done using that.

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Self-critique rating:3

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Question: `q008. According to the function y = .8 x + 1.4, how much will a bag of 7 widgets cost? How many widgets would you expect to get for $10?

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Your solution:

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y = .8 (7) + 1.4=$7

$10= .8 x + 1.4, you can but 10.75 widgets ( I solved using algebra)

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confidence rating #$&*:

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Given Solution:

Using the model we obtained, y = .8 x + 1.4, we note that the cost is represented by y and the number of widgets by acts. Thus we can find cost of 7 widgets by letting x = 7:

cost = y = .8 * 7 + 1.4 = 7.

To find the number of widgets you can get for $10, let y = 10. Then the equation becomes

10 = .8 x + 1.4.

We easily solve this equation by subtracting 1.4 from both sides than dividing by .8 to obtain x = 10.75. That is, we can buy 10.75 widgets with $10.

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Self-critique (if necessary):

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Self-critique rating:OK

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `q009. Sketch a graph with the points (5, 7), (8, 8.5) (10, 9) and (12, 12). Sketch the straight line you think best fits the data points.

Extend your line until it intercepts both the x and y axes.

What is your best estimate of the slope of your line?

What is your best estimate of the x intercept of your line?

What is your best estimate of the y intercept of your line?

If your graph represents the cost in dollars of a widget vs. the number of widgets sold, then what is the cost of 4 widgits, and how many widgets could you get for $20?

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Your solution:

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What is your best estimate of the slope of your line?

Counting the coordinates I got -.478 for slope

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You should have sketched a line and used two points on that line as a basis for your estimates. The x and y intercepts would have been a good choice for the two points, but any two points, not too close together, would do.

It isn't clear what you mean by 'counting the coordinates'.

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Doing the change in my ys / change in x’s gives me -12.5/-11=1.136

What is your best estimate of the x intercept of your line?

-11

What is your best estimate of the y intercept of your line?

5.5

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Given those intercepts your slope would be .5.

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Clearly, also, the slope of this graph isn't negative.

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If your graph represents the cost in dollars of a widget vs. the number of widgets sold, then what is the cost of 4 widgits, and how many widgets could you get for $20?

y = .8 *4 + 1.4=4.6

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Those numbers aren't relevant to the data of this problem or to the linear-function model that follows from this data.

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$20 = .8 x + 1.4,sloved algebraically I obtain 23.25 so with $20 you can buy 23.25 widgets

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confidence rating #$&*:

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Question: `q010. Plot the three points (1, 2), (2, 3.5) and (3, 4) on a reasonably accurate hand-sketched graph. Sketch a straight line through the first and last points.

What is the distance of each of the three points from the line?

What is the sum of these distances?

What is the sum of the squares of these distances?

Sketch a line 1/4 of a unit higher than the line you drew.

What is the distance of each of the three points from the new line?

What is the sum of these distances?

What is the sum of the squares of these distances?

Which line is closer to the points, on the average?

For which line is the sum of the squares of the distances less?

How far from the line y = x + 7/6 is each of the three points?

What is the sum of these distances?

What is the sum of the squares of these distances?

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Your solution:

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What is the distance of each of the three points from the line?

My two last points touch the line, 0,0,.5 (2, 3.5) this is where it extends out by .5

What is the sum of these distances?

3.5

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Did you have a point that was 3.5 units from your line? If so you didn't position your line very well, but I don't think that's the case.

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If I interpret your previous answer correctly, your distances were 0, 0 and .5. If you add those distances up you don't get 3.5.

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What is the sum of the squares of these distances?

Using d=sqrt[(x1-x2)^2+(y1-y2)^2] I am going to try to apply that to three coordinates expect I am going to cube my values and then cube root everything so it will look like this d=cubert[(x1-x2-x3)^3+(y1-y2-x3)^3] d=cubert[(1-2-3)^3+(2-3.5-4)^3] = -6.13

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You listed your distances. This questions asks you to find the sum of the squares of those distances.

sqrt[(x1-x2)^2+(y1-y2)^2] is an important expression related to distances, but it is not relevant to this question.

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What is the distance of each of the three points from the new line?

From the first point to the last point they are 2 units apart and from the second to the first they more 1.5 units and to the .5 units apart.

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You need to give three distances.

If your line is 1/4 unit higher than before, none the three distances can't vary by more than the corresponding distance from your previous line.

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What is the sum of these distances?

2

What is the sum of the squares of these distances?

d=cubert[(.25-.5-.75)^3+(.5-.875-1)^3]=.125

Which line is closer to the points, on the average?

the middle one?

How far from the line y = x + 7/6 is each of the three points?

(1, 2), (2, 3.5) and (3, 4) using these x values or my new point ones???

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Those are the only three points mentioned so far. You've been asked to draw a couple of lines, but no new points.

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Using original x values {2,3.5,4}

y = 2 + 7/6=19/6=3.17

y = 3.5 + 7/6=4.67

y = 4+ 7/6=31/6=5.17

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Those area the y values on the new line. You need those values, but to answer the question you need to find the distance of each point from the line.

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What is the sum of these distances?

13.01

What is the sum of the squares of these distances?

58.58

confidence rating #$&*:

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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Good work on many of the questions, but you've misinterpreted some questions so I'm going to ask for a revision.

&#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.

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