Math Matters: Why Do I Need To Know This?
Bruce Kessler, Department of Mathematics
Western Kentucky University
Episode Five
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Expected value – Average winnings when playing the
lottery
Objective: To illustrate the relevance of the concept of expected value as applied
to any gaming situation, but particularly as applied to buying a lottery ticket.
Welcome to “Math Matters: Why do I need to know this?,” a program where we try
to take some of the mathematics that we talk about in our entry level math courses and
explain to you or show you why that’s good stuff, why you should know that kind of thing in
your everyday life. I’ve got some things to talk to you about today that I’ll think you’ll find
interesting. The first one has to do with expected value and I’m going to talk to you again
about lottery issues. We talked a little bit about the Kentucky lottery last time and talked
about how to calculate some of those probabilities. Now I’d like to go back and calculate some
of the expected values of playing the game and I’ll have to explain to you what I mean by
expected values.
The expected value of some kind of game or buying a ticket or something like that is the
sum of the products of the probability of each outcome and the outcome’s net value. Now
that sounds convoluted, but we’re going to multiply things first and then we’re going to add,
and I’ll just give you a real simple example of this. Let’s say that a local FFA chapter sells
one dollar raffle tickets on a pig that’s worth 200 dollars and they sell a thousand tickets.
What is the expected value then of buying a raffle ticket?
Well, you look at all the different things that could happen: you could win the pig, or you
could lose the pig. The value of winning the pig is actually 200 minus $1. You don’t get your
dollar back when you get the pig. They keep your dollar. And if you lose, you simply lose a
dollar. And the probabilities here are very easy. There’s a one in thousand chance that you
win the pig and a thousand minus one in a thousand chance that you don’t win the pig. So,
we multiply the $199 times the 0.001, and then we add to that what we get when we multiply
− 1 times 0.999. And when you do that arithmetic in your calculator, you get − $0.80.
Alright, what that tells me then is the − $0.80, and it makes sense when you think about
it, because they’re raising a thousand dollars, they’re spending $200 on the pig, so they’re
making $800. So what that really tells us is each of the thousand people who bought a raffle
ticket contributed $0.80 to the club. The negative means that pretty much everybody lost
$0.80. When you average it out over all of them, everybody lost $0.80 on this. One person,
everybody basically lost a dollar, but then one person got the $200 pig back in return so it
averages out to be everyone losing $0.80 of their dollar. (Figure 1)
Now I want to look at the bigger issues of the Kentucky lottery. What are the expected
values of playing the lottery? What are they? Are they negative or are they really negative? I
can tell you they are going to be negative, because this is a money-making deal. The Kentucky
budget receives a lot of money from the lottery and it pays for, supposedly, education. But
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Figure 1, Segment 1
there’s a good question here: can the pot grow big enough to where the expected value is no
longer negative, it’s now positive? And there is a point. We just finished a lottery where we
had three hundred, a pot of $365 million. People will tell you that “Well, once it gets up so
big, that’s when I start playing it.” And so the question is where is that big enough to where
it actually is kind of worth it to play – the expected values are no longer negative. They’re
at least zero, perhaps positive.
So let’s do some calculations here. Let’s start with an easy example: the Cash Ball
lottery. Now, in the state of Kentucky, that is for a $1 play, they’ll pick four numbers one
through 33 in any order, no repeats and then a “Cash Ball,” one through 31. We talked about
how to calculate how many different tickets are possible. You do this little calculation with
the four numbers and then you’ve got 31 different ways of doing the cash ball and then you
can calculate them. I’m not going to get into that today. I actually went to the Kentucky
lottery web page and just downloaded their odds for playing the game. And I’ve got to tell
you, they’re incorrect here. We talked about the difference between odds and probabilities last
time. They’re stating odds, but what they’re giving you here are probabilities. They, I did
the calculations just to check them and these are actually the probabilities. This 1,268,520,
that’s actually the number of different tickets that are possible. So they’re doing, they’re
doing probabilities. That’s okay, we can use that information then to calculate the expected
values. And what we’ll do is we’ll calculate the amount minus one dollar times each one of
these probabilities. (Figure 2)
So, it kind of looks like this. There’s the 200 − 1. That big ugly decimal, that’s what you
get if you divide that one by that 1.2 million blah blah blah. And then do that actually with
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Figure 2, Segment 1
each of the different probabilities. Now the one probability that they don’t give you is the
probability that you don’t win anything, but you can get that from these. You take all these
different probabilities and you add them up and then you subtract that from one. That gives
you the probability that you don’t win anything and then you multiply that times the negative
one, that’s the dollar that you lost and didn’t get back, and that gives you a grand total then
of about − $0.46. So you’ve got winners and losers for the cash ball. But on average, it
averages out everyone loses $0.46 when they buy a ticket, okay? (Figure 3)
Now let’s go to the Powerball. The Powerball’s the one where they just had that $365
million pot and it’s a little more complicated because that pot, whoever gets a winning ticket,
if two people get a winning ticket they have to share that jackpot. The Powerball stuff,
you go through, you pick four numbers one through 55, no repeats, and then you pick a
Powerball and again instead of calculating all of these probabilities, I just went to their web
page and downloaded the odds which again are not odds – they’re probabilities. In fact, they’re
probabilities that have been rounded a little bit. This is more, this one right here they actually
rounded quite a bit. But, all the data is there. They start with a $15 million pot and it grows
over time, it’s a shared jackpot. (Figure 4) So if it’s just simply $15 million, okay, that’s 15
million minus one times that probability and then you subtract all the other different pots,
okay? And then you add all these up and subtract from one. There’s actually about a 97
percent chance that you don’t win anything, you just lose your dollar, and that works out
to an expected value of − $0.70 cents. So with the Powerball, there’s more to lose on that.
(Figure 5)
If it goes up to 365 million, now this is curious. I’m going to recalculate this with 365
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Figure 3, Segment 1
Figure 4, Segment 1
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Figure 5, Segment 1
million. So you go through, it actually grows to $1.70. (Figure 6) Wow! That’s a 170%
return on your income, on your ticket, but the thing that you have to watch for on this is
that splitting of the pot. The more tickets that are sold, the more likely you are to have to
share that prize with someone else and so in this case, I called and got the information, they
sold 665,674,691 tickets. The expected value if you look at the decrease in the pot that you’re
going to end up receiving if you win, it goes to, it still goes to down to negative. It still
goes down to − $0.09. (Figure 7) Now, to their credit, they did actually kind of increase the
secondary pots from $200,000 to $600,000. So that raised the expected value to about two or
three cents. So, they didn’t have to do that, but that was pretty nice of them.
The above calculation that considers the shared jackpot is fairly messy, far beyond the
scope of a general math or college algebra course, and involves using the binomial distribution
and conditional probabilities to calculate the expected value of the shared jackpot given the
number of people playing the lottery. The number of different PowerBall lottery tickets that
can be picked is
55
42
= 3, 478, 761 · 42 = 146, 107, 962,
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1
so, as mentioned earlier,
P (picking the winning lottery ticket) =
1
= p.
146, 107, 962
However, the probability that a winning lottery ticket is picked by someone is different and
is dependent upon the number of lottery tickets sold, in this case, N = 665, 674, 691 tickets.
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Figure 6, Segment 1
Figure 7, Segment 1
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Each of these tickets can be considered as independent trials with the probability mentioned
above, so the probability that k tickets win the jackpot when n ≥ k tickets are sold is given
by a binomial distribution, with probability density function
n k
P (exactly k winners) =
p (1 − p)n−k .
k
Therefore, the probability that someone wins the jackpot in the case where n = N is
P (someone wins jackpot) = 1 − P (exactly 0 win) ≈ 0.9894965.
To calculate the expected value of the jackpot, we actually use the conditional probability
P (exactly k winners | someone wins) =
P (exactly k winners )
,
P (k 6= 0)
assuming n tickets are sold, multiplied by the value of the shared jackpot in that case,
365,000,000
− 1, and summing k from 1 to n. The graph in Figure 7 shows the change in the
k
expected value of the shared jackpot as n goes from 1 to N . The expected value of the shared
$365,000,000 jackpot when n = N is approximately $104,556,526. Using this figure in place
of $365,000,000 reduces the expected value of buying a lottery ticket to roughly − $0.09.
One last note – as of air-time, I was unaware of the fact that the secondary pots that
were raised to $600,000 had to be shared as well. This causes the expected value of buying
a lottery ticket to dip back into the negative range – no big surprise there.
If you want to find the break even point though, it’s pretty simple. Just replace that pot
with “x.” Let me show you what this looks like here, and just, you get this little formula, set
it equal to zero and solve for “x.” It’s not a hard thing to do. Now that’s assuming that you
don’t share the pot. If you do share the pot, it can be as high as $365 million and still have
a negative expected value. (Figure 8)
Alright, that’s a lot of stuff to digest in one quick little presentation, so I’m going to flash
up some summary slides here and let you kind of look that over and then we’ll move on to
the next thing.
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Figure 8, Segment 1
Summary page 1, Segment 1
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Summary page 2, Segment 1
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Finding polynomials through points – Modeling real
data
Objective: To illustrate how to find a degree-n polynomial that interpolates n + 1
points using a simple concept not normally shown to algebra students, LaGrange
polynomials. We show how this idea can be useful by looking at a real-world
modeling application.
We’ve done a lot of probability things and I think that’s needed, something that you need
to know. I want to spend a little bit of time today talking about some algebra ideas. We do
a lot of things in algebra that a kid will look at that and think “Why do I, you know, I’m
never going to be put in this situation, and I’ll never have to fool with this.” And I’d like to
show you today kind of a nice little thing you could do that involves algebra to help you know
how to do some things. The first thing I want to talk to you about is how to find polynomials
through a point, through a series of points, and we’re going to use this to model real life
situations.
One of my little pet peeves, flash up this word problem that I found in an algebra book.
(Figure 1) “Here’s a drug administered orally, the concentration in the blood rises to sharply
to a maximum, modeled by the function” and they give you this horrible degree four polynomial and then they say “What’s the maximum level of medicine in the patient and when does
it occur?” And that’s good stuff, you can then graph this curve on your calculator or later
on you can use calculus to find the maximum of it, but the thing that bothers me about this
is this curve they gave you. When on earth are you ever going to be given a problem where
they give you this formula? I mean, that’s unrealistic to think, “Well they just pop out of the
air.” There’s a way to get these, but they never discuss where they come from. That’s what
I want to do today is show you where that formula comes from or how you can generate that
formula given some data. And that’s going to require us to know this little trick for finding
a polynomial, a degree “n” or less through “n” points.
What you want to do is for each point that you have, you want to create a polynomial that’s
1 at that point’s “x” value, but zero at all the other point “x” values. They’re called LaGrange
polynomials after Joseph Louis LaGrange. He was a French mathematician actually born in
Italy, so Italy kind of claims him as well. And then once you have these things, we’ll take
the right amount of each one of them and the right amount is just whatever that y-coordinate
is for the point so that we go through each of the points and then we expand the polynomial,
multiply, collect like terms and so forth and that gives us the answer. (Figure 2)
So let’s start off easy with two points and I’ll show you how this works. I’d like to build
something that is one here and zero there. So what I do is I’m going to include an x − 4 that
way, so when I plug in the four, that will be zero. And then as far as making it one, well let
me plug in zero and then solve for “c”. And so I just divide by the junk. I know that’s − 4,
but I’m leaving it there because I want to show you the pattern, okay? So the first one is
x−4
, and you can kind of simplify that. The second one you do the same thing. I want now
0−4
x − 0 and the bottom will be 4 − 0, so I’ve kind of swapped out things and that simplifies.
And then to get these things through the point, I want three of the first one and eight of the
second one. And so you can kind of multiply that out, three times this would give me 3 − 34 x
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Figure 1, Segment 2
Figure 2, Segment 2
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and then eight times that would give me 2x. Simplify that a bit more you’ve got 54 x + 3 and
when you graph that, you’ll see that that does go right between those two points. Now that is
a hard way to find the line through two points. The easier way to do that is to you know find
the slope of the line through the points and then use the point slope form of the equation, but
this generalizes, the other won’t. (Figure 3)
Figure 3, Segment 2
Let me kind of bump up the odds here and show you three points and show you how that
looks. If I go to three points, now I want something for the first one that is zero at three and
four, so I throw in an x − 3 and an x − 4. And again, I want it to be one when “x” is zero.
So I plug in a zero, set it equal to one, and solve for “c” and you just divide by the junk,
okay? So you see the pattern here: I’ve got my three, I’ve got my four, that’s my other “x”
values. I’ve got the “x”’s here and then I’ve got the x-value for the first point down here. If
I were to follow that pattern, here’s the second one: the zero and the four and then I plug in
the three. For the third one I put in a zero and a three and I plug in the four. It’s a nice
little pattern to follow and then you can multiply all that mess out. I want three for the first
one, that’s the y-value here, six of the second one and eight of the third one. You do all that
arithmetic, you do all that algebra necessary to clean that up and you get this little formula.
It’s a quadratic, it’s a degree two polynomial and when you plot that, you’ll see that it goes
right through those three points: (0, 3), (3, 6), and (4, 8). (Figure 4)
In fact, it’s even nicer than that because if those three points had been collinear, it would
go ahead and give me the line through those three points. So it, to prove that is a different
deal, but it actually gives you the lowest degree polynomial that will go through those points.
So it’s a neat little trick, LaGrange polynomials. I don’t know why we don’t teach that to be
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Figure 4, Segment 2
honest with you because you’re not doing any calculus, you’re just doing some basic algebra,
kind of multiplying things together. It’s got a beautiful little pattern to it, I think it’s really
cool.
So let me give you a situation. This is something that you could do; you could try this
out yourself. Suppose you’re modeling your trip from Glasgow to Bowling Green and on one
hand, you’re timing yourself, you’ve got a watch timing yourself. And then you’ve got your
trip odometer set to measure the miles. So you leave Glasgow, you start at zero in time,
zero in miles and then when you come and get on the parkway, five minutes later you’ve
traveled two and a half miles. When you eventually get to I-65, you’ve traveled 17, you’ve
been traveling for 17 minutes and you’ve traveled 15 and a half miles. 32 minutes after you
start you get to exit 26 you get off the I-65. You’ve traveled 31 point seven. And then you
reach your destination 42 minutes, your trip odometer say 35.9 miles. So the question would
be, can I find, could I find a nice smooth kind of curve through those points that I could then
use on all this other stuff we’ve been talking about? Finding a maximum, you know, and I’ll
show you some other stuff you can do with it. (Figure 5)
Well, if you take these points and all I’ve done is kind of collected the data and put
them into points. You do this little stunt we talked about where you generate the LaGrange
polynomial for each of the five, okay? And I know it’s messy, but you can see the pattern.
Five, 17, 32, 42, and then down here you divide by putting in the zero, the zero, the zero.
You do that each of the times, okay? (Figure 6) Here are your points. I’ve plotted minutes
versus miles traveled, okay? You do all that arithmetic and then you want to add zero of the
first polynomial, two and a half of the second, 15 and a half of the third, 31 point seven of
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Figure 5, Segment 2
the fourth, 35 point nine of the fifth. I grant you it’s messy, but it’s doable stuff. You know,
you can do this and that gives you this formula which you now plot that, it’s messy looking,
but if you plot that, you’ll see that it goes right through those points. Easy as pie, I think
it’s really kind of cool. And then once you have this, you can extrapolate information. You
don’t just have to do things at data points you collected. You can look at well what is my,
for example, average speed. I’m going to talk to you in just a second about how you calculate
average speed. You know, you can calculate things like that between the points. (Figure 7)
That’s a nice little trick and I know it’s kind of, I did it kind of fast, so I’m going to
flash up a page that has that formula on that. Give you a chance to copy it down and then
as always you can go to the web page, too, and get it later.
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Figure 6, Segment 2
Figure 7, Segment 2
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Summary page, Segment 2
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3
Slope of a line – Calculating average rates of change
Objective: To show the equivalence of the slope of a line through the points
(t1 , f (t1 )) and (t2 , f (t2 )) to the average rate of change of f (t) over the time interval
[t1 , t2 ]. We also foreshadow the development of the derivative as a limiting process
by showing how calculating the slope is inadequate for finding instantaneous rate
of change.
I want to continue on some of these same lines and continue with some algebra with you.
We just talked about how to find that crazy curve. The question would be now that you have
that curve, what can you do with it? Also we go back and one of the things we teach in
our college algebra is how to find the slope of a line. You know, why do I need to know how
to do that? “It’s a nice exercise, I’ll pass the test, whatever it is.” But there’s actually a
nice application for slope and I want to show you that and that kind of leads you into where
the things you can do with slope with algebra and calculating slope and things you can’t do,
okay?
So let’s look at a, let’s look at the same situation, I’m looking at the same kind of model
that I’ve talked about. (Figure 1) But before I get to the curve, let’s talk about how I would
calculate the average speed during stages of the trip and really I just guess I’d look at well
what was my average speed during the trip? Well I can look at the distance and the rate and
the time. The difference between those is distance equals rate times time. Dirt, that’s how I
remember it, d = rt. But if you solve for rate, rate is distance over time, okay? Well, the
distance traveled here was 35.9 miles; the time it took was 42 minutes. So I’ve gone that
many miles in that many minutes so we’d probably rather know miles per hour, so we do a
little conversion, we talked about how to do these conversions, multiplying by one, so that
the minutes cancel, and that works out to 51.3 miles per hour. (Figure 2)
Now what does that have to do with slope? Well, if I asked you the question in a different
way, if I said “find..” and I should mention what slope is, it’s rise over run, change in “y”
over change in “x”. If I were to say to you, “Okay, find the slope of the line through the
beginning point and the end point.” That would be, the “y” would be 35.9 − 0, and then the
“x”, 42 − 0, and that works out to be the fraction there if you want to look at that. And it’s
still miles over minutes which you can convert. But the point is is that slope is exactly the
same as that average speed. And so, if we have some way of describing, if we have points
that relate time to distance traveled, there’s no difference at all between the slope between two
points, and the average speed that we traveled between those two points. So, that’s a good
trick for us, that’s an application of slope. (Figure 2)
Now what I’d like to do is be a little more general. You say “That’s great, I quess, but
what about the intervals in between there? I mean how fast was I going on the interstate?”
That’d be a good question to ask. “What was my average speed before I left Glasgow?,” these
kinds of things. Well, to get that, you’ve got to have that curve between the points. And
we’ve talked about how to do that. What you have to do is, you have to get a continuous
model of the situation and I’m going to do, this is actually the same problem we worked
previous where we got the LaGrange polynomials for each of these points, added them up.
I’m going to do one thing, since I want my answers in miles per hour, I’m going to convert
these points, I’m going to change the times, which are currently in minutes, to the times in
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Figure 1, Segment 3
Figure 2, Segment 3
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hours. The easiest thing to do is you divide by 60 each time. And then also, when I ask
the questions, if you know the interval I want the average rate over is in minutes, I have to
convert those into hours as well. (Figure 3)
Figure 3, Segment 3
So you get the LaGrange polynomials for each of these five points, you take zero of the
first one, two and a half the second, blah, blah, blah, multiply it out and you get this formula
which I will go ahead and flash up here and show you what it looks like. It does hit each of
those points exactly and it gives me a nice smooth curve through there, okay? (Figure 4)
Now, what if I want to know my average speed during this interval of the trip, between
30 minutes into the trip and 42 minutes into the trip? And if it were a certain leg of the
journey, I would have to look back and estimate: “Okay, when did I get on the interstate and
when did I get off the interstate?”, that kind of thing. Well, what you do is, you find the slope
of the line. Through those two points on your continuous now curve, on your polynomial
curve that we just graphed. So I take the 40 minutes, and I convert that into an hour, that’s
actually two thirds of an hour, plug that into “y”, I’ll get that answer. Then, 30 minutes,
I have to convert that into an hour, that’s half an hour, right, 30 divided by 60. You take
these, this gives me then the change in “y”. Be consistent with these numbers and get the
change in “x”. This one comes first, so it needs to come first down here. And what I would
suggest you do actually if you have a graphing calculator is just get that in your graphing
calculator and then just use that to evaluate their function values. That’s tedious raising
those things to powers. And it works out to be 33 and about a third miles per hour. And
what that is, here’s a half, okay there’s one point right there, two thirds would be 6.6, so
that’s somewhere right in there and so this miles per hour represents the slope, if you will,
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Figure 4, Segment 3
of that line through the two points. (Figure 4)
Now, if I wanted to shorten that interval, “How fast was I going? I got a ticket 30
minutes into the trip. How fast was I going between 30 and 31, somewhere between 30 and
31 minutes into the trip? How fast was I going? What was my average speed during that
, you have to convert that into hours
time?” Well, it’s the same calculation. If you take 31
60
now, put it in your formula, minus 30
into
your
formula
and then divide by your change in
60
those “t” values. And then again, I’m just using my calculator to get these messy numbers.
That works out to be about 50.5 miles per hour, okay? And again, you can see the slope,
here’s the line that I found the slope of up here. There’s half, and then the other one would
have been somewhere right there and that’s the slope through those two lines. That’s the line
through those two points and we’ve calculated the slope of that line, so it’s a nice use for
slope. (Figure 5)
Now here’s one that I can’t solve very well. And this is really one we worry about the
most. This is going to illustrate one of the weaknesses of algebra. What we’d really like to
know is, what was our speed 30 minutes into the trip? You know, add a point, how fast was
I going right then? “Now when I got the ticket, how fast was I going?” Well, you can try
that. If you plug that in, you’re plugging in “y” at 12 minus “y” at 12 and then 21 minus 21 and
that’s trouble, that’s where things fall apart because what I have here is a zero on the top and
a zero on the bottom. You know from probably repeated attempts to do so that dividing by
zero is bad, it doesn’t give you an answer and so we don’t know what that is. I have actually
drawn the line. You can find that line and you can find the slope of that line which would
tell you your instantaneous rate of speed, but what it, but to do that, you have to use a trick
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Figure 5, Segment 3
from calculus. Algebra won’t pull it off because we are dividing by zero. In calculus, you learn
this little trick for dancing around the zero business in the denominator. In fact, you almost
work out a way to where you work out the zero in the top and the bottom and you get rid of
them and then you can get the answer. (Figure 6)
By the way, using calculus, we can calculate the derivative of
y = f (t) = 112.933t4 − 334.09t3 + 233.512t2 + 12.7953t
to be
f 0 (t) = 451.732t3 − 1002.27t2 + 467.024t + 12.7953,
and so, the instantaneous velocity at t = 12 is f 0 12 = 52.2063 miles per hour.
So, to answer tough questions like that, you’ve got to keep going. Go ahead, and my advice
to you is to find the nearest calculus teacher and ask them about derivatives and they’ll tell
you how to do that. I’ll flash up the work that I did there and the summary and then we’ll
come back and wrap things up.
Closing
We’ve talked about some high math today. We’ve done serious calculations and I know
we’ve done them pretty quick, so I would encourage you if you’re able to go back and watch
the reruns of the program, please do so and you can maybe catch them there. You can also
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Figure 6, Segment 3
Summary page, Segment 3
22
download computer files we have from our web page, which we’ll flash up that address here
at the end, you can pause them and copy these things down.
But I hope you’ve found the things we’ve talked about today useful. If you have other
ideas that you’d like to hear me talk about, applications of mathematics, or something that
you’ve learned and you don’t quite understand why you need to know that, I’m the guy to
ask about that. And we’ve got a place on the web page that you contact me and we also have
a web address, an email address that you can write to. With that, I am done. Thanks for
your attention and we’ll see you next time.
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