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. |
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T 12/01 |
Class was cancelled today due to illness. Homework due today will now be due on Thursday 12/03 (or if you will not be in class on Thursday, turn in the work to BH239 dept office by the due date). Until then, please take a
look at the following applications of linear and exponential patterns and we
will look at more like them on Thursday. 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. |
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Th 12/03 |
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 didnÕt 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:
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 word problems:
1. Suppose that you have $50 left on an out-of-state gift card that you canÕt seem to find and the company that issued the card plans to deduct $2 a month from its value as a service fee. Write an equation to describe the situation. a. How much will be left on the card after 10 months? Answer: y= -2x+50 so for x=10, y=30 b. How much will be left on the card after 30 months? Answer: y= -2x+50 so for
x=30, y= -10. This is not possible, so you cannot project your model that far
into the future. c. When will the card have
no value? Answer: y= -2x+50 so for y=0, x=25 weeks.
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) = 76 (1.196)t b. P(t) = 550 (0.564)t Answer: In part a, the initial value is 76 (the y intercept) and it is growing at the rate of 19.6%. In part b, the initial value is 550 and it is decaying at the rate of 43.6% (0.564=1-0.436).
3. Suppose that for a certain type of bacteria, we start with a population of 5500 and we know that if it is not supplied with food, it will decay exponentially at a 17.5% rate. Write the exponential equation for the number of bacteria present as a function of time in hours. How many bacteria are present after 5 hours? How many after 10 hours? Answer: y= 5500(0.825)x which when x=5, y=2102 and when x=10, y=803.35.
4. Suppose that a population of 32,000 is growing exponentially at a rate of 5.7% per year. Write an equation for the exponential relationship. How long would it take for the city to triple its population by your model? Answer: y= 32,000(1.057)x so when y=96,000, we have 96,000= 32,000(1.057)x which when simplified by dividing by 32,000 yields 3 = (1.057)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 3 / log 1.057 = 19.82 yrs.
5. Suppose that in the previous problem the population of 32,000 is decreasing at the rate of 3.5% per year. Write the equation for this exponential relationship and tell how long it would take for the city to have 60% of its original population left. Answer: y= 32,000(1-0.035)x = y= 32,000(0.965)x so when 60% of the original population is left, 0.60= (0.965)x . (That is, 60% of the population of 32,000 is y=0.60*32,000=19,200, so we have 19,200= 32,000(0.965)x which when simplified by dividing by 32,000 yields 0.60. But you do not need to go thru all of that!).To solve for x, we take the log of both sides of the equation and the properties of logs give us that x=log 0.60/ log 0.965 = 14.34 years.
Homework due Tuesday:
1. During the period after birth, a male 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 ratÕs weight in pounds 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?
ÒI canÕt believe I ate the whole thing! Anybody got some Alka-Selzer?Ó 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 townÕs population be 2.5 times its starting population of
12,300? Some items we might talk about on
Tuesday: 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. Here are some more difficult situations:
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.
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 ÒaÓ value in y= b(a)x because ÒaÓ 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 ÒaÓ 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:
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:
1. Given a linear
scatterplot, estimate the line of best fit using point-slope form by
picking two points on your line (formulas for slope and equations of lines
will be supplied). You did this on graph paper in class and in homework
assigned on 11/17: Example 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 2. 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). You did this for comparison with estimates in the above
item on the same day (11/17): Example 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). b. use the equations to find the actual line of best fit (using Sxx, Sxy, etc.) 3. (possibly on the test) Estimate
an equation for an exponential relationship by using two points on the curve
(we may look at this in class on Tuesday) 4. Turn a line of best fit
for logged data (x, logy), into the exponential of best fit by ÒunloggingÓ
the slope and y-intercept. Then write the equation of the exponential of
best fit. You did this in notes and for homework on 11/19: Example Notes 11/19: If 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. 5. Given tables of
perfectly linear or exponential data, write the linear or exponential
equation for each without graphing. See notes 11/19, hmk due 12/02, todayÕs
notes, etc.: Example From 11/19 notes:
is a decreasing exponential
set of data because you are multiplying by 0.75 each time so y= 420(0.75)x. Example: Notes above:
Using the y-intercept and another point in the table, (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. 6. 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 above homework). Example Notes 12/03: #4. Suppose that a population of 32,000 is growing exponentially at a rate of 5.7% per year. Write an equation for the exponential relationship. How long would it take for the city to triple its population by your model? Answer: y= 32,000(1.057)x so when y=96,000, we have 96,000= 32,000(1.057)x which when simplified by dividing by 32,000 yields 3 = (1.057)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 3 / log 1.057 = 19.82 yrs. 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 12/10, you are finished with the course! Test 5 grades will be posted at http://www.smccd.edu/accounts/callahanp/test160F09.html
(the testscores link on the main page) by Monday morning 12/14. Final grades
will be available at mysfsu by the deadline. 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 12/11 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 longer and more difficult comprehensive test during finals week to take the place of test 5 only. This option is to help you avoid a zero score on test 5, not to improve your grade past what you earned on the first five tests. If you do not contact me by noon Friday 12/11 or fail to show up for your new test appointment during finals week, you will receive a zero score for test 5. |
