ISED 160 NOTES ABOUT HOMEWORK, ANNOUNCEMENTS, ETC.
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Assigned on: |
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T 11/10 |
We went over the homework,
and looked at more examples in preparation for the test. Make sure that you
know the required format of the answers in each of the test problem
categories listed above. 1. Show what Ò!Ó means as a
product and how cancellation makes the computation easier. See notes 10/22
and 10/27 above and in class notes for more examples of this. 2. When you state your
answer using C or P, include the correct n and r for the given situation and
do not go further with the calculation. When you are deciding whether order
matters or not, keep in mind that a particular order does not have to be
stated in the problem for it to imply that permutations are needed. When you
ask yourself if Òorder mattersÓ,
it may be the case that the situation implies an order that is not
explicitly stated in words. For example, the number of ways that 3 of the 10
floats in a parade can finish ahead of the others is 10P3. The fact that they
appear in a parade implies that they will appear in an order, because
groups must line up in a long line for a parade. This means that order
matters, in the sense that the parade will look different depending on which
float appears in what order. But also be careful to not inject your own
inferences into a situation about whether order matters. The number of ways
that 3 people can be selected from a group of 10 to receive a prize of a free
raffle ticket is 10C3. Order doesnÕt matter because all 3 are receiving the
same prize. The fact that a prize is awarded does not automatically mean
Òorder mattersÓ, because in this case, the prize is the same for each person:
each has a ticket that has an equal chance to win at the time it is awarded
to them. Do not take it further and assume that after the raffle is drawn, they
will get different prizes—you do not know that this is the case,
because the nature of the raffle has not been told to you in the problem! See
notes from 10/27 (above and from in class, where there were more examples). 3. See notes 10/27 and
10/29 above and in class notes for ample examples of the nCr form required. 4. See notes 11/03 and
11/05 above and in class notes, homework problem #1 from the last homework,
todayÕs quiz, etc. for examples of this more difficult nCr set-up. 5. See notes 10/29 (just before
the hmk assm.) and we looked at more examples in class today. In these, you
will not use nCr notation or decimals, but write the individual probabilities
as fractions of whole numbers using card counts. 6. These must be written as
fractions of whole numbers taken from the table, and not as decimals, so that
you can see where the numbers are coming from. See hmk problem #3 in the last
homework and notes 11/05 and in class today. You will have a table of at
least 2 rows and 2 columns, but no larger than 3 rows or columns (nothing as
large as the table in the 11/05 notes). As this test is fairly
non-calculational and has the capability of providing quick answers on much
of it, it will not require the whole session on Thursday as material such as
on Test #3 did. Please come on time and be prepared to see a short lecture
preparing you for the final segment of the course. We will talk about
graphing of linear functions before you start the test. This should be a
review and will lead to some homework that will be due next Tuesday. |
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Th 11/12 |
Congratulations on
completing about 80% of the semesterÕs material with the above test! Since
the test was shorter than most, I took some time before starting the test to
prepare you for the last segment of the material in the course: visual
representation and analysis of data. I will post the Test#4 grades by code on the ÒtestscoresÓ button from the front page of this website sometime by Saturday afternoon (refresh/reload the page). You will have a good idea of your base grade (without benefit of hmk/quiz portion, which I cannot compute for you until the end of the semester) by dropping the lowest of the first four tests and averaging the other three. With hmk/quiz/participation, your grade will probably go up from there. If you missed 2 tests, you will not pass the course.
The deadline to turn in a withdrawal form is Thursday 11/19. If after you see your Test 4 grade you wish to withdraw, and are eligible to do so, I will sign it as long as you: --fill out the form completely, with your reasons --attach an unofficial copy of your transcript as specified on the form, and --turn it in to BH 239 (or me in person) before the noon on the 11/19 deadline. If you do not obtain my signature by noon on 11/19, do not prepare the forms properly, or are not eligible for a withdrawal, I will have to award you a grade for the semester. I will not approve withdrawals past that point. GRAPHING practice for next weekÕs work! We will start our new segment on graphing next week. You should know how to plot points in a 2-dimensional space (x and y coordinate axes), find slopes of lines given points, and how to write equations of lines in different forms. These are skills that you need to know for our work next week, so if you do not remember your graphing skills from Algebra, maybe check out any Algebra book from the library and take a look at the linear graphing section for yourself.
You must know how to find the slope of a line from the coordinates of two points on the line, where you are also able to estimate the coordinates of those points from a graph. The slope of a line measures the change in y values over the change in x values.
Slope m =(y2 - y1 )/(x2 - x1 )
You must know how to use point-slope form for the equation of a line. The name means what it says: the equation shows both the slope m and a particular point (x1, y1) on the line. The x and y with no subscript of one are variables, holding a spot for the infinite other (x, y) points on the line y - y1 = m ( x - x1 )
You must know how to put a line into slope-intercept form. The name means what it says: when the equation of a line is in this form, you can read off the slope m as the number multiplying x, and the y-coordinate of the y-intercept (0,b) as the number b in the other term:
y = mx + b
This is a good form for the line to enable you to make predictions for new data. For example:
Suppose the points ( 2.3 , 12.7 ) and ( 3.4 , 22 ) lie on a line. The slope of a line measures the change in y values over the change in x values. Slope m =(y2 - y1 )/(x2 - x1 )=(22-12.7)/(3.4-2.3)=9.3/1.1=8.45
We can use either of the two points above to be (x1,y1), but I will use the same one as I did to find the slope ( 2.3, 12.7 ): y - 12.7 = 8.45 ( x - 2.3 )
To take the point-slope form we have in part 2 and put it into slope –intercept form, we solve for y by multiplying 8.45 by both x and -2.3 to remove the parentheses and then adding 12.7 to both sides of the equation so that we may collect like terms. y - 12.7 = 8.45(x) + 8.45(- 2.3 ) using the distributive property to remove parentheses y = 8.45x - 19.44 +12.7 since we can add the same number to both sides of an equation y = 8.45x - 6.74 collecting like terms Now the line is in slope-intercept form, and we know that the slope is still 8.45 but now we also know that the line crosses the y axis at -6.74, that is, the y-intercept is a point that has coordinates of (0, -6.74).
To use this equation to make predictions, suppose that x stands for time in years and y stands for profit in millions of dollars for a company. Using the equation above, how much profit will the company realize in 5 years? This means that we plug in 5 for x and find y = 8.45(5) - 6.74 = 42.25 – 6.74 = 35.51, that is, 35.51 million dollars in profit is projected.
You might want to visit the interactive weblinks below if you want more practice. The coolmath site is geared toward kids, but if you can ignore that, you will find that the lecture notes are well-written, and you have the opportunity to work some example problems if you click on Òcrunch someÓ! IÕve selected a few of the pages that are most relevant, but feel free to go to the home page and look up whatever other lessons you want (such as a visual of rise and run, horizontal and vertical lines, etc.).
Try some problems at the following:
http://www.coolmath.com/algebra/Algebra1/06Lines/07_findingslope2points.htm http://www.coolmath.com/algebra/Algebra1/06Lines/12_findeqpointslope.htm http://www.coolmath.com/algebra/Algebra1/06Lines/13_findeq2points.htm http://www.coolmath.com/algebra/Algebra1/06Lines/14_predicting.htm
Alternately, go check out any Beginning Algebra book from the library and look at the chapter on lines. Whatever the case, take advantage of the FREE knowledge out there! Work on being proficient in the above tasks so that we can more easily work on an activity in class on Tuesday. Be careful about missing class these last few weeks, as things will go quickly towards the end, and Test #5 is a mandatory part of your grade and cannot be dropped! Homework (due Tuesday 11/17): 1. Find the slope of the line that passes thru the points ( 2.5, 256.5 ) and ( 10.7, 42.8). Round calculations to 2 decimal places. 2. Write the equation of the line above using point-slope form with the point ( 2.5, 256.5 ) and solve for y to put it in slope-intercept form. 3. Write the equation of the line above using point-slope form with the point ( 10.7, 42.8) and solve for y to put it in slope-intercept form. (Note that your answers should be close for parts 2 and 3, but rounding may cause a small difference). 4. Graph the points on an informal graph (you do not have to buy graph paper) with equal marks of size 1 on the x axis from 1 to 10 and equal marks of size 50 on the y axis (0, 50, 100, 150, 200, 250), plot the points from part 1, draw a straight line thru them and extend that line to cut the y axis, and see if the y-intercept you saw from parts 2 and 3 in slope-intercept form (the b from y=mx+b) matches about where your line would crosses the y axis. It should be close, but donÕt ask for too much from an informal graph! The work we will do next
week will assume proficiency in the above skills, as we will expand the
knowledge into a new realm: using imperfect linear data to form a scatterplot
and fit a best line to the data. |
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T 11/17 |
We worked on the equations
for the line of best fit for a scatterplot of seemingly linear data. One can
draw what one thinks is the line of best fit thru the data and use points on
that line to write an estimate of the equation of best fit, but one can also
use formulas to find the absolute best line possible. We did each of these
with a set of data as an exercise in class. Here is the idea and the
equations behind finding the line of best fit: Given several data points
(x,y) you fill out the table below (the data points' coordinates are the x
and y values. The symbol means For example, for the data points (2,7) and (4,8) we know that the slope of the line thru them is 1/2 = 0.5 so that y-7 = 0.5(x-2) so that y = 0.5x +6 is the equation of the line thru them, that is, a line with slope 0.5 and y-intercept 6. Now let us use the equations for the line of best fit: Set up a table with the following quantities and sum them up
After having done this with all of the given data points use all of these numbers to plug into the formulas for the line of best fit (which are below)
Using the formulas above, Sxx = 20-(36/2)=2 Sxy = 46-[(6)(15)/2]=46-45=1
The best fit line is then y=0.5x+6 (which matches the equation found at the beginning exactly, because 2 points make a line, not a scatterplot!
Now as another example that does form a scatterplot, for the data pts (1,9), (2,8), (3,6), and (4,3): a. You can graph them and then draw what you think is the line of best fit (no two will be alike, but they will all be close). Then you can pick two points on your line and write the equation of that line. This is your estimate of the best fit line. b. The formulas become more important as the pattern is hard to see, as when the data is more spread out than the example we did in class. We have the following formulas for the actual line of best fit ( that is, the best possible line that can be drawn from the data): Fill out the table and find the actual line of best fit :
Note that n=4 is the number of data points
Using the formulas above, Sxx = 30-(100/4)=30-25=5 Sxy =
55-[(10)(26)/4]=55-65= -10
The best fit line is then y= -2x+11.5 This is in slope-intercept form. How does it compare to your estimate?
HOMEWORK due Thursday 11/19: Use the same procedure as in todayÕs class to find the line of best fit for the data: (19,55), (23,7), (21,20), (15,123), (16,88), (18,76). a. Graph it informally with a different scale for x and y appropriate to their values (0 to 25 by 5 points at a tie for the x axis, and 0 to 125 by 25 points at a time for the y axis for instance), estimate the best line equation from your graph using two points different from any of the data points then put into slope-intercept form b. use the equations above to find the actual line of best fit (using Sxx, Sxy, etc.) SOME NOTES TO READ (no homework to turn in for this part!): Before the end of class, we spent a few
minutes talking about the differences and similarities of linear and
exponential (one kind of non-linear) functions. Given a table of perfectly
linear or exponential data, we can write the appropriate equation that
describes that data in the form y=mx+b for a line or y= b(a)x for an exponential. The b in both cases is the y-intercept and is
found in the table where x=0. With linear relationships, m is the
slope of the line and is the amount by which we add each time. We add
positive numbers for an increasing line (positive slope) or negative numbers
for a decreasing line (negative slope). For instance, if the slope is 5, then
mx = 5x = x+x+x+x+x. After the sum mx is found, b is also added on to get
y=mx+b. So a linear relationship is one built by repeated addition. The a in the exponential is the amount
by which we multiply each time (ÒaÓ contains the rate of increase or decrease
since a=1+r or 1-r, where r is the rate—we will talk more of this next
time). In y= b(a)x , the (a)x means ÒaÓ multiplied by itself x times. For instance, (2)5 = 2*2*2*2*2 = 32. After you find (a)x and then multiply by the b to get y= b(a)x you can see that an exponential relationship is built by completely
by multiplication. The following are some
tables of data to illustrate what sets of linear and exponential data look
like and how their equations are written:
is a decreasing linear set
of data because you are adding -3 each time, so y=-3x+12. (Verify that the
points in the table lie on this line by plugging the values in and checking them).
is a increasing linear set
of data because you are adding +7 each time, so y=7x+20.
is an increasing exponential
set of data because you are multiplying by 3 each time so y= 12(3)x. (plug the x and y values into the exponential equation above and
see that it gives a true statement). Note the placement of the y-intercepts
of each above and how 3 and 12 are used in each!
is a increasing exponential
set of data because you are multiplying by 1.5 each time so y= 50(1.5)x.
is a decreasing exponential set of data because you are multiplying by 0.9 each time so y= 250(0.9)x.
is a decreasing exponential
set of data because you are multiplying by 0.75 each time so y= 420(0.75)x. |
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Th 11/19 |
We went over the previous finding line of best fit homework, and writing equations for lines and exponentials from tables of perfectly linear or exponential data (as in the above notes), and looked at how we can find the exponential of best fit for a scatterplot by turning it into a line using logarithms. Here is another example, but with data that is not perfectly exponential as it was in class: The following below set of data is best represented with an exponential relationship y= b(a)x but since it is not perfectly exponential, we cannot write the relationship from the table values.
It is difficult to estimate an exponential scatterplot relationship, but it can be turned into a linear relationship by taking the logarithm of the y values. We have equations we can use to find the line of best fit. So we use these equations on the linear logged data, then turn the best fit line we get from them back into an exponential relationship by ÒunloggingÓ the slope and y-intercept of the line.
We make a table of logged y values (x values staying the same) to turn the data into a linear scatterplot. Make these logged values accurate to at least the hundredths place.
Now we can find the best fit line for this (x, logy) linear scatterplot by making a summations table with and use the standard deviation calculations (Sxx, etc.) for finding the best fit line.
Sxx = 26-(64/3)=4.67 Sxy = 18.26-[(8)(7.09)/3]= -0.65
So the best fit line for the logged data is y=-0.14x+2.73
Now to find the best fit exponential for the original data, ÒunlogÓ the slope and y-intercept of the line above: raise 10 to the power of each separately and then write the equation for the exponential of best fit for the original table data (x, y). a=10slope =10-0.14 = 0.72 b=10y-intercept = 102.73= 537.03 So the best fit exponential for the original data is y=537.03(0.72)x.
(Check your answer: does plugging x=1 into your best fit exponential give you something close to the original table value of 398.11? It shouldnÕt be exact because the original data was not perfectly exponential, but it should be in the ballpark! Same for the other two points.). Notice that this is a decreasing exponential relationship. For decreasing exponential relationships, a<1 and for increasing exponential relationships, a>1. For decreasing linear relationships y=mx+b, m is negative and for increasing ones, m is positive. We will talk more about this after break. Homework (due Tuesday 12/01): 1. Given the tables of values for functions, decide if the data are best represented by a linear function or an exponential function. Write the equation for the relationship that you decide.
2. The following table is of an exponential scatterplot (not perfectly exponential, but is best described by an exponential function). Do as in the example above:
a. Take the original (x, y) values in the table and make a new table (x, log y). That is, find log25 to start withÉ
b. Using the values in the (x, logy) table, find the line of best fit using the equations (given again below). Hint: you should get the following summations to plug in:
(notice that you are not summing the original y values to get 595.17—sum the logged y values): c. Convert the slope m and
the y intercept b from the best fit line for the (x, logy) data in part b to the
Òa and bÓ that are the components of the best fit exponential y= b(a)x for the original (x, y) data. This is done by a=10slope and b=10y-intercept .
(This is essentially ÒunloggingÓ the data). Does this equation look like it
describes the data well? d. Use your estimate for
the best fit exponential from part c to estimate the value of y when x is 12. |
