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T 03/17

For those seeking tutoring, the following resources are listed in my syllabus but here are some more details:

LAC in HSS 348: has two stats tutors usually on Mon, Tues, Thurs at times before 4pm. You must sign up with this service and make appointments ahead of time. They have heard of ISED 160 and will tutor for it!

CARP in HSS 344: Evening tutoring only. Several students have told me that they are turning people away when they tell them that they need help in ISED 160. I am trying to find out why. Until I get an answer, tell them that you need tutoring in MATH 124. It is a rough equivalent to our course and they have heard of it and will tutor you in it!

MATH in TH 426: They tell me that they have a stats tutor Chris Bernt on MWF 9am-10am and W 11am -12pm. If you go to him for help, tell him that you want help in MATH 124. He only tutors for Math courses.

Some of the best tutoring, though, comes from meeting with your fellow classmates who are working on the same problems as you are! Start getting to know the people in your class and find out when you can get together to work regularly.

In all cases, get help from these resources as problems come up—don’t wait until tests come around to start reaching out!

Th 04/23

We completed Test #4 today, and with it, about 80% of the material for the semester.

I will post the Test#4 grades by code on the “testscores” button from the front page of this website sometime by this evening. 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 tomorrow, Friday 04/24. If after you see your Test 4 grade you wish to withdraw, and you have not withdrawn from this course twice before, I will sign it as long as you:

1. fill out the form completely, with your reasons

2. attach an unofficial copy of your transcript as specified on the form, and

3. turn it in to BH 239 by noon tomorrow, Friday 04/24.

If you continue past noon Friday in the class, you will have to receive a grade. I will not approve withdrawals past that point if you neglect to observe the deadline.

GRAPHING practice for next week!

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 “crunch some”! I’ve 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 04/28):

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.     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.     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.

4.     4. Note that your answers should be close for parts 2 and 3, but rounding may cause a small difference.

5.     5. Graph the points on an informal graph 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 don’t ask for too much from an informal graph!

T 4/28

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. 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 add them ).

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

=x/n=6/2=3 and `=y/n =15/2=7.5 (where n=2 in this short ex. for the 2 data pts given)

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

=6/2=3

=15/2=7.5

= 1/2 = 0.5

= 7.5-(0.5)(3) =7.5-1.5 =6

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

=10/4=2.5

=26/4=6.5

= -10/5= -2

= 6.5-(-2)(2.5) =6.5+5 =11.5

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 04/30:

Use the same procedure as in today’s 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. The m is the slope of the line and is the amount by which we add each time for an increasing line (positive slope) or subtract each time for a decreasing line (negative slope). The a in the exponential is the amount by which we multiply each time (a contains the rate of increase or decrease since a=1+r or 1-r, where r is the rate).

For instance,

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.

Now some data that is not linear (its graph is not a line, but is exponential):

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.

Th 4/29

We went over finding line of best fit homework, writing equations for lines and exponentials from tables of perfectly linear or exponential data (as in the above notes), and looking 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 “unlogging” 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.

=x/n=8/3=2.67 and `=y/n=7.09/3=2.36

Sxx = 26-(64/3)=4.67

Sxy = 18.26-[(8)(7.09)/3]= -0.65

= -0.65/4.67 = -0.14

= 2.36-(-0.14)( 2.67) =2.73

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, “unlog” 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=10^slope =10^-0.14 = 0.72

b=10^y-intercept = 10^2.73= 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 shouldn’t 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 on Thursday.

Homework (due Tuesday):

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).

b. Using the values in the (x, logy) table, find the line of best fit using the equations (given again below). You should get the following summations to plug in:

x= 10 y= 9.54

xy= 21.86 x² = 30

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 “a and b” that are the components of the best fit exponential y= b(a)^x for the original (x, y) data. This is done by a=10^slope and b=10^y-intercept . (This is essentially “unlogging” the data). Does this equation look like it describes the data well?

T 05/05

We went over the homework and then did some work with tables and word problems. Some practice problems like what we talked about are below, then homework, of course!

Practice exponential scatterplot (in case the last one didn’t quite do it for you!):

The following table is of an exponential scatterplot (not perfectly exponential, but is best described by an exponential function):

Take the original (x, y) values in the table and make a new table (x, log y).

Using the values in the (x, logy) table, find the line of best fit using the equations.

ANSWER: You should get the following summations to plug in from the (x, logy) values:

x= 6 y= 9.39 xy= 13.49 x² =14

The best fit line for (x, logy) is y= - 0.12x+2.54

The best fit exponential for (x, y) is y= 346.74(0.76)^x

Practice rate and word problems:

1. Suppose that a car rental company charges a flat fee of $15 plus a per mile fee of 12 cents.

a. What will be owed to the company for the rental if the customer drives 245 miles?

Answer: y=0.12x+15 so for x=245, y=44.40

b. For what number of miles will the rental bill total come to $70?

Answer: y=0.12x+15 so for y=70, x=458.3

2. Given the following exponential functions, state the initial amount present and the rate of growth or decay (indicate which!) as a percentage.

            a. P(t) = 678 (1.072)^t                                                b. P(t) = 2.07 (0.723)^t

Answer: In part a, the initial value is 678 (the y intercept) and it is growing at the rate of 7.2%.

In part b, the initial value is 2.07 and it is decaying at the rate of 27.7% (0.723=1-0.277).

3. Suppose that for a certain type of bacteria, we start with a population of 1000 and we know that this type grows exponentially at a 49% rate. Write the exponential equation for the number of bacteria present as a function of time in hours. How many bacteria are present after 2 hours? How many after 5 hours?

Answer: y= 1000(1.49)^x which when x=2, y=2220 and when x=5, y=7344.

4. Suppose that a population of 50,000 is growing exponentially at a rate of 4.5% per year. Write an equation for the exponential relationship. How long would it take for the city to double its population by your model?

Answer: y= 50,000(1.045)^x so when y=100,000, we have 100,000= 50,000(1.045)^x which

when simplified by dividing by 50,000 yields 2= (1.045)^x . To solve for x, we take the log of both sides of the equation and the properties of logs give us that x=log 2 / log 1.045 = 15.7 yrs.

5. Suppose that in the previous problem the population of 50,000 is decreasing at the rate of 10% per year. Write the equation for this exponential relationship and tell how long it would take for the city to have 75% of its original population left.

Answer: y= 50,000(1-0.10)^x = y= 50,000(0.90)^x so 75% of the population of 50,000 is y=0.75*50,000=37,500, so we have 37,500= 50,000(0.90)^x which when simplified by dividing by 50,000 yields 0.75= (0.90)^x . To solve for x, we take the log of both sides of the equation and the properties of logs give us that x=log 0.75 / log 0.90 = 2.73 years.

Homework due Thursday:

1. During the period after birth, a male white rat gains exactly 40 grams per week. Assuming that the rat grows at a steady rate, and that he weighed 100 grams at birth, give an equation for his weight after x weeks. (1 pound = 453.59237 grams)

a. What is the rat’s weight according to your model after 5 weeks?

b. When would the rat weigh 500 grams?

c. Would you be willing to use this line to predict (extrapolate) the rat's weight at age 2 years?

2. A town has a population of 12,300 people and the population is decreasing exponentially at the rate of 15.5% per year.

a. How many people will be in the town in 2 years if this trend continues?

b. When will the town have half of its original population?

b. When will the town have one-quarter of its original population?

3. Suppose in problem #2, the population is increasing exponentially at the rate of 15.5% per year. When will the town’s population be 2.5 times its starting population of 12,300?

Th 05/07

Today, we went over the homework and then looked at some issues in writing linear and exponential patterns from harder tables than before.

Schedule (or what’s left of it!):

Tuesday 05/12 : last details/review about the test and possibly other course materials.

Thursday 05/14: Test 5 (last graded activity for the course).

Finals week: You only have to come if you miss Test 5 (see below). You may also come if you need to discuss your grade.

Test 5 is a mandatory part of your grade. It will not be dropped from your average for any reason. Only the lowest of the first 4 tests will be dropped, no matter how you perform on Test 5.

If you take Test 5 on 05/14, you are finished with the course! Test 5 grades will be posted at

http://www.smccd.edu/accounts/callahanp/test160Sp09.html (the testscores link on the main page) by Monday morning 05/19. Final grades will be available at mysfsu no later than midnight 05/28!

Only for students who miss Test 5:

If you miss test #5 (for any reason, including family emergencies), you must email me before noon Friday 5/15 and respond to my follow-up emails that day so that we will both know what to expect the following week (finals week). You will then perform a more difficult comprehensive test during finals week to take the place of test 5 only. If you do not contact me by noon Friday 5/15 or fail to show up for your new test appointment during finals week, you will receive a zero score for test 5.

Practice for more difficult tables of linear and exponential data:

Up until this point, we have looked at tables with easy x values (x=0, x=1, x=2, etc.) so that we could more easily learn to see the linear and exponential relationships. We looked today at some harder tables. Here is a brief description:

For linear relationships, note that if the increments for x are not by 1 as in the previous tables, you must find the slope as the change in y divided by change in x. If you are unsure what this means, pick two points from the table and put them into the slope formula m=(Y2-Y1)/(X2-X1). For example, in the following table, the differences in the y values are constant, but the increments of x are not by 1!

The change in y is +3.2 for each change of +5 in x, so this is linear, but with slope 3.2/5=0.64. Then y = 0.64x + 7. If you did not have the y-intercept in the table, you could use y = mx + b and any other point value to solve for it.

For exponential relationships, if the increments for x are not by 1, you must not just find the quotient of the values in the table, but must find what you multiply to get from the x=0 value to the x=1 value (even if it is not in the table). You must find what you are multiplying by to get from one y-value to the next, but then raise it to the reciprocal of the change in x. In class, we looked at a table like the following:

Ordinarily, you would say that since 84.375/25.000=3.375 and 284.766/84.375=3.375 we are constantly multiplying by the same number and would write y= 25(3.375)^x . But this equation does not work! (check when x=3 and see that you do not get the correct table value). The reason is that we do not have the data for x=1 in the table, yet that is what we need for the equation. We must find what we multiply 25.000 by to get to y value for x=1, which would be the same number we would multiply the y value for x=1 by to get to the y value for x=2, which would be the same number that we would multiply by the y value for x=2 to get to the y value for x=3, which is 84.375. That is, we need to find x such that (25.000)(x)(x)(x)=84.375. This is the same as saying (25.000)(x)³=84.375. Solving for x, we must divide by 25.000 on both sides, to get (x)³=84.375/25.000 or (x)³=3.375. Now you must raise both sides of the equation to the ^1/3 power to solve for x: (x³)^1/3=(3.375)^1/3 so that x=1.5. Now we can write the proper relationship:

y= 25(1.5)^x . Check it out by plugging in x=3 and see that you get back y=84.375.

As another example, if you have the points (0, 12) and (4, 112.56) in a table of exponential data, you find that 112.56/12=9.38 so 12 multiplied by 9.38 gives 112.56. However, 9.38 is not the “a” value in y= b(a)^x because “a” is the amount you multiply by to go one increment in x and we have the amount you must multiply by to go 4 increments in x. Since you must multiply 12 by the number “a” 4 times to get to 112.56, to find a you must raise 9.38 to the ¼ power. That is, (9.38)^1/4=(9.38)^0.25=1.75. Now we have the equation y= 12(1.75)^x which will check out if you plug x=4 into it to see that you get y=112.56.

Problems below for practice will not be collected:

1. Given the table with linear data below, write the linear relationship:

ANSWER: the slope is 3/8=0.375 and the y-intercept is 15, so y=0.375x+15

2. Given the table with exponential data below, write the exponential relationship:

ANSWER: To get from 350 to 86.95 you must multiply by 0.25. But to find the number by which you must multiply to get from x=0 to x=1, you must raise 0.25 to the 1/10 power to get 0.87. Then

y= 350(0.87)^x which can be checked by plugging in x=20 and seeing that you get y=21.6.

3. Write the relationship for the following table:

ANSWER: y= 55.6(1.27)^x

TEST 5 FORMAT:

-- given a linear scatterplot, estimate the line of best fit using point-slope form (formulas for slope and equations of lines will be supplied)

-- given a scatterplot and table values and summations for x, y xy and x² , use the given equations to find the line of best fit (formulas will be supplied).

--estimate an equation for an exponential relationship by using two points on the curve (we did this in class today).

-- turn a line of best fit for logged data (x, logy), into the exponential of best fit by “unlogging” the slope and y-intercept. Then write the equation of the exponential of best fit.

-- given tables of perfectly linear or exponential data, write the linear or exponential equation for each without graphing. Some of the easier ones (with nice x values) and some of the harder ones (as in the notes today) will be given.

-- given word problems that describe linear/exponential relationships, write the appropriate equation and use it to find specified values for follow-up questions by being given an x value to plug in and solve for y, or by being given a y value to plug in and solve for x (as in the previous homework).

T 05/12

In class, we looked at the use of Excel to make pie charts, histograms, etc., and the line of best fit add-on feature from the Data Analysis toolpack on Excel, using the practice exponential scatterplot from the 5/5 notes. We then used that example as a starting place for reviewing for the test as above. There is no more to do but take the test!

Come to Test #5 as scheduled on Thursday 05/14 (format above)!!!

Test 5 is a mandatory part of your grade. It will not be dropped from your average for any reason. Only the lowest of the first 4 tests will be dropped, no matter how you perform on Test 5.

If you take Test 5 on 05/14, you are finished with the course! Test 5 grades will be posted at

http://www.smccd.edu/accounts/callahanp/test160Sp09.html (the testscores link on the main page) by Monday morning 05/19. Final grades will be available at mysfsu no later than midnight 05/28!

Only for students who miss Test 5:

If you miss test #5 (for any reason, including family emergencies), you must email me before noon Friday 5/15 and respond to my follow-up emails that day so that we will both know what to expect the following week (finals week). You will then perform a more difficult comprehensive test during finals week to take the place of test 5 only. If you do not contact me by noon Friday 5/15 or fail to show up for your new test appointment during finals week, you will receive a zero score for test 5.

Oh, and after you finish Test #5,

HAVE A GREAT BREAK!!!