Tuesday, May 22, 2007

Friday's Chapter 15 Test Topics

Here’s a list of topics for Friday’s test. Can you believe it’s our last one?

Precalculus Chapter 15 Test Topics:
Identify the degree, number of real and complex zeros and the leading coefficient of a polynomial from its graph
Sketch the graph of a rational function.
Identify transformations of functions.
Classify discontinuities.
Simplify rational functions.
Determine the zeros, the sum of the zeros, the product of the zeros, the sum of the pairwise product of the zeros, and a possible equation from the graph of a cubic function.
Sketch the graph of a given polynomial.
Identify the zeros of the polynomial.
Factor the polynomial.
Prove that a quadratic has no real zeros.
Show that a value is a zero of a polynomial.
Find zeros of a polynomial.
Discuss the implications of nonreal zeros.
Determine an average rate of change.
Provide a formula for an average rate of change.
Determine an instantaneous rate of change.


13 questions on the non-calculator portion, 11 on the calculator portion. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different.
- Johann Wolfgang von Goethe

Is this me?

Here’s a problem to consider:
Driving problem: Hezzy Tate drives through an intersection. At time t = 2 sec she crosses the stripe at the beginning of the intersection. She slows down a bit, but does not stop, and then speeds up again. Hezzy is good at mathematics, and she figures that her displacement, , in feet, from the first stripe is given by
Use synthetic substitution to show that is a zero of d(t).
Use the results of the synthetic substitution and the quadratic formula to find the other two zeros of d(t).
How do the zeros of confirm the fact that Hezzy does not stop and go back across the stripe?
What is Hezzy’s average velocity from t = 3 to 3.01 sec?
Write the equation for the rational algebraic function equal to Hezzy’s average velocity from 3 sec to t sec.
By appropriate simplification of the fraction in Problem 16, calculate Hezzy’s instantaneous velocity at time t = 3 .

15-5: Instantaneous Rate of Change of a Function: The Derivative

Chapter 15 Section 5:

Review:

So, this is the last section of the last chapter of the year! I think that it is appropriate that I am the last person to post a blog seeing as I was the first person as well.

This section is our introduction to Calculus. Calculus is all about the rate of change and through this section we are figuring out how to predict it.

Now something to remember:
Instantaneous rate of change = Derivative = Slope of a tangent line
for the point x sub 0

Definitions:

Derivative= instantaneous rate, velocity
--derivative of the time-height function


Average rate of change of r(x) of function f(x) on an interval starting at x=c is the change in the y-value of the function divided by the corresponding change in the y-value of the function divided by the corresponding change in the x-value.

Instantaneous rate of change of f(x) at x=c is called the derivative and is denoted f '(x) "f prime of x." It is equal to the limit of the average rate as x approaches c.


The value of the derivative of f(x) at x=c equals the slope of the tangent line to the graph of f at x=c.



Example Problem:
Example 1:
Given: f(x)= x^3 + 2x + 3 what is the instantaneous rate of change of f(x) at x=1?

Remember, first you plug in 1 to f(x), so you have 1^3 + 2(1)+3 = 6

f(x)= f(x)-6/(x-1) = ((x^3 + 2x + 3)-6)/(x-1)

Now, use synthetic substitution


1 (pretend it is in the box thing) ___ 1 __ 0 __ 2__-3
______________________________________1 __ 1 __ 3
________________________________------------------------
________________________________1___1___ 3 __0


(sorry if all of the marks above confused you, but when I tried to publish the post, the computer automatically pushed all of the numbers to the left and it just looked like a jumble of numbers)

Yay! We got zero! which means that it is removable~

so:

((x-1)(x^2 + x + 3))/(x-1) the x-1 cancel out so you are left with

x^2 + x + 3. If you plug in 1 for x you get 1^2 + 1 + 3 = 5

So 5 = the Instantaneous rate of change = derivative = slope!

Now, use this information to create the equation for the line. From the information in the problem, you know the slope (= the Instantaneous rate of change = derivative) which is 5 and you also have a point, (1, 6) and as Mr. French said, it is always better to use point slope form because it is so easy!
So, y - 6 = 5 (x - 2).
Example 2:
You are at Six Flags and you are riding a roller coaster. On your favorite ride, there is a dip that has the equation y= x^4 + 5 where the x-axis is the ground and the y-axis cuts the dip in half. You are such a good Precalculus student that you want to use the knowledge that you learned in class to calculate your average velocity between 1 and 1.5 seconds after you reach the minimum of the dip.
Given this information, you know only your x values, 1 and 1.5. To find the y-values, plug the information into the equation. When you plug in 1, you get 6 and when you plug in 1.5, you get 10.0625. Using this information, you can calculate the average velocity. Your points are (1, 6) and (1.5, 10.0625). (10.0625 - 6)/ (1.5-1) = (4.0625/.5) = 8.125 = the average velocity!

For some extra help visit:

http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorials/frames2_1.html
It is actually pretty cool! If you click on "previous tutorial" and "next tutorial" in the left hand corner there is more information about what this section covers.





This is a picture of me, Juan, and Alex. Juan Alderete de la Pena is the base player from my favorite band, The Mars Volta. I think Henry may have told this story on a previous blog, but I will tell you again anyway. So, ok, we were going to the Detour Music Festival in LA and we took the metro there and when were were in Union Station, we saw him and were like "Juan" and he turned around. It was amazing. We took the metro with Juan and walked down the streets of LA with him looking for the entrance. I couldn't believe that night; it was awesome! Anyway, we became buds and it was pretty amazing.

Reminder:

Oh yeah!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! No more blogs!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Monday, May 21, 2007

15.4 - Rational Functions: Discontinuities, Limits, Partial Fractions

Hey everybody!

to update from Anna's post: two
and a half weeks until summer! yay, the greatest moment ever!

down to business...when you divide a polynomial by a linear factor, the result usually resembles a quadratic function:

f(x) = (x^3 - 5x^2 + 8x - 6)/(x-3)

when you use synthetic substitution:
3} 1 -5 8 -6
3 -6 6
1 -2 2 0 is the answer,

so the net equation is x^2 - 2x + 2. there is a discontinuity at x = 3. this means that the answer is undefined because when you plug 3 in for x, the denominator is equal to 0. you can use this later for finding limits.The dot above the 3 on the x axis is the representation of the discontinuity.

LIMITS! If you want to find the y value that the discontinuity should would be, plug the equation into your calculator. then go to TBLSET, change TBLstart to a few decimal places before the desired number, and change the scale to 0.1, or some very small number. then go to TABLE and find the y values for 2.99 and 3.01. it should be obvious what the missing number is. in this case it is 5.

lim f(x) = 5

you can also find this algebraically by plugging 3 back into the net equation:
x^2 - 2x + 2
3^2 - 2(3) + 2
9-6+2 = 5

You can remove the discontinuity algebraically by simplifying the following:

(x-3)(x^2-2x +2)/(x-3) you can cancel the two (x-3)'s.

there are some equations that have more complex discontinuities that are not removable:

with the equation g(x) =
(x^3 - 5x^2 + 8x - 5)/(x-3) there is an asymptote at x=3.
you can use synthetic substitution to find the factored form:

3} 1 -5 8 -5
3 -6 6
1 -2 2 1 is the answer, so it has a remainder. the net equation is therefore x^2-2x+2+1/(x-3) and the graph...


You can't factor this type of problem so it has an infinite or irremovable discontinuity.
the limit of g:
lim g(x) = does not exist or, in some cases, is infinity

PARTIAL FRACTIONS: polynomial/
polynomial = partial fraction

(9x-7)/(x^2-x-6) = ....

you need to factor the denominator to (x-3)(x+2). you can represent the numerator with variables, A and B.

A/(x-3) + B/(x+2) --> give them equal denominators by multiplying the first value by (x+2) and the second by (x-3).

A(x+2)/(x-3)(x+2) + B(x-3)/
(x-3)(x+2) --> distribute the A and B values.

(Ax +2A +Bx+ 2B)/
(x-3)(x+2) = (9x-7)/(x^2-x-6)

you know there are 9x's and that you have A + B x's, so A+B=9
you also know that... 2A-3B = -7,
so you can use matrices or substitution to solve for A and B. they equal 4 and 5, respectively.

a shorter way to do this is by making the factors in the denominator equal zero...like this:
(9x-7)/(x-3)(x+2)
plug in 3 for x... (27-7)/(3+2) --> 20/5 --> 4
plug in -2 for x... (-18-7)/(-2-3) --> -25/-5 --> 5
plug those values back in to get the partial fraction:

4/(x-3) + 5/(x+2) and that is the answer.

Examples!

What is the partial fraction for (4x-14)/(x^2 + 2x - 8)

first, factor the denominator: (x-2)(x+4).

then, set up two fractions: A/
(x-2) + B/(x+4).
make equal
denominators: A(x+4)/(x-2)(x+4) + B(x-2)/(x-2)(x+4)
distribute: (Ax + 4A +Bx - 2B) /
(x-2)(x+4) = 4x-14
A + B = 4 and 4A - 2B = -14
solve with matrices or substitution.
A = 1, B = 5

or...

this way is better for speed.
(4x-14)
/(x-2)(x+4)
plug in 2 for x to find A value, then plug in -4 to find B.
A = -6/6 --> -1
B = -30/-6 --> 5

the partial fraction is 1/(x-2) + 5/(x+4)

LINKS:

here is some extra help for...
partial fractions: http://www.sosmath.com/algebra/pfrac/pfrac.html
discontinuities: http://www.tpub.com/math2/39.htm
limits: http://en.wikipedia.org/wiki/Limit_(mathematics)

Reminder: Mariclare is next

this is good fun: it's german AND stop motion! two of my fave things!
http://youtube.com/watch?v=4RST7CYjfjo
i don't know what he's saying though. i could be exposing you to crude language without knowing it... :(

i hope all of this helped!

Wednesday, May 16, 2007

Thursday's Quiz Topics

Now that we’re through the worst of the AP season, we can finally get back on track (after the Junior retreat on Friday, anyway)! Here’s a list of topics for Thursday’s quiz:

As a reminder, your Weekly Challenge will be due on Monday, and the test for Chapter 15 will be next Friday.

Precalculus Quiz 15.1-3 Topics
Given a graph of a function:
Determine the degree
Determine the leading coefficient
Determine the number of real and complex (nonreal) zeros
Identify an extreme point and a point of inflection.
Given a cubic function:
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Perform synthetic substitution.
Interpret the results of your synthetic substitution.
Given the factored form of a cubic, determine the zeros of the function.
Given data for a cubic function:
Prove the constant third difference property
Determine the equation for the function algebraically (matrices)
Verify the equation with regression techniques
Analyze the equation
Given partial information about the zeros of a cubic function:
Determine the remaining zeros
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Determine an equation for the function
Analyze a function (word problem!)

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 20 questions on the Non-calculator portion and 20 questions on the Calculator section (all worth 1 point each). If you know your stuff, you should be able to finish quickly. I’ll be around after school on Wednesday (today), and on campus by 7:00 AM on Thursday.

"I never did very well in math - I could never seem to persuade the teacher that I hadn't meant my answers literally."
- Calvin Trillin

Trashing math and country music - ouch!

Wednesday, May 02, 2007

Friday's Quest Topics

Here’s a list of topics for Friday’s test:

Precalculus Chapter 14 Test Topics:
Find terms – arithmetic sequence (determine pattern)
Find terms – geometric sequence (determine pattern)
Definitions of sequences and series
Find terms – arithmetic series (use information)
Determine type of sequence and justify
Determine terms of a sequence – unknown type
Determine terms of a series – unknown type
Explain your method of determination/pattern of series
Determine terms of a sequence (determine pattern)
Determine terms of a series (given pattern)
Determine value of n for given term in a series/sequence
Determine term(s) of a binomial expansion
Word problem – analysis of arithmetic and geometric sequences
Arithmetic series – determine formula, apply formula
Geometric series – determine formula, apply formula
Partial sums of arithmetic and geometric series
Infinite sums of geometric series
Sigma notation

11 questions on the non-calculator portion, 14 on the calculator portion. Since this chapter was so short, each question will be worth 3 points instead of 5, making this a “quest” instead of a test. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

"Knowledge is of no value unless you put it into practice."
- Anton Chekhov


This strategy won't work on me:




Tuesday, May 01, 2007

14.2 Arithmetic, Geometric, and other Sequences

Hey Guys! I hope everyone is coping sorta okay with AP's coming up and everything. In 5 weeks it will be summer!

Ok so lets define some terms first:

  • Sequence is a function where the domain is a set of positive integers. The independent variable is n and the dependent is tn.
  • A Recursion formula specifies tn as a function of the preceding term (tn-1)
  • A Explicit formula of a sequence specifies tn as a function of n.

Arithemetic Sequence is where you add a constant to the previous term to get the next term. The constant added is called the common difference. (This can be adding a negative number)

Geometric Squence is where you multiply the previous term by a constant (the common ratio) to get to the next term. (you can multiply by a fraction which is basically dividing)

Formulas:

It is good to use recursive just to find the next few terms but for finding terms later in the sequence use the explicit formula. Also, a sequence can have a finite or an infinite number of terms depending on whether its domain is finite or infinite.

Arithmetic Linear Function:

  • Explicit: Tn = To + D(n-1)
  • Recursive: Tn = Tn-1 + D

so To is the initial value in the sequence, R is the common difference, Tn is a value in the sequence, and Tn-1 is the previous term.

Geometric Exponential Fution:

  • Explicit: Tn = To * R^(n-1)
  • Recursive: Tn = Tn-1*R

so To is the initial value in the sequence, R is the common ratio, Tn is a value in the sequence, and Tn-1 is the previous term.

Calculator!

  • Make sure you are in sequential mode.
  • go to y=
  • nMin=1 (you will usually want to leave this =1)
  • u(n)= this is where you put your equation (u(n) is the same as t(n))
  • to get u go to 2nd 7 and for the n press the regular x button.
  • for u(nMin)= put the first term of the sequence
  • once you plug in your equation you can go to table set to find values for any term. Go to table set to start at a specific value.
  • you can also graph additional sequences under (v) and (w) (the same way) in case you want to compare the sequences.

Examples!

Given the following problem how would you plug it into the calculator (i had a hard time with this so i am giving it as a example problem.)

You put $1000 into the bank with 6% interest a year. What are the steps for putting this scenario into the calculator.

  1. seq mode
  2. y=
  3. then under nMin=1 (leave this alone)
  4. u(n)= u(n-1)(1.06)
  5. u(nMin)= 1000 (which is the starting value)

Now if the problem asked for a different term (for example the 15th). You can either go to the table or since you have the equation entered you can put in u(the term) on the home screen and press enter and get the answer.

Example Problem 2

For the following problem:

  • Tell what kind of sequence it is
  • Write the next 2 terms
  • Find t100
  • Find the term number of the last given term

2. 2, 4, 8...32
Ok so this is an geometric sequence where the common ratio is 2. So to get to the next term you multiply the previous term by 2. The next two terms are 16 and 32. Now to find t100 you can figure out the explicit formula and either plug in 100 for the n value or you can plug in the formula into y= and go to the table at 100. The explicit formula formula for this sequence is Tn=2(2) ^(n-1) . T100 = 1.267 *10^30. Now to find the term number of the last given term you can either look for the term in the table or you can plug it in for Tn in the formula.
32= 2(2) ^(n-1)


16=(2) ^(n-1)


log 16 =( n-1) log 2


n= the 5th term

Additional Help:

http://home.alltel.net/okrebs/page131.html or http://www.purplemath.com/modules/series3.htm

this site is also kinda cool :http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&ResourceID=340

Reminder: Natalie you are next!

Ok so...our dance teacher told us to watch this clip. It's CRAZY. This woman is balancing and dancing in pointe shoes on this guy's head. Absolutely amazing (and crazy!) You have to watch it for a minute before she dances on his head...

http://www.youtube.com/watch?v=8uaVXmWEN-c

14-3: Series and Partial Sums

Okay, I'm sorry if this is really confusing, but I didn't really get it today in class. (ESPECIALLY not the binomial part)

The difference between a series and a sequence:
A sequence is just a list of numbers like 2,4,6,8
A series is different because it's all being added together in a string like 2+4+6+8.

Mr. Frost taught us that to find the sum of a series, you can pair up the different numbers in the series. So if you had the series 3,7,11,15,19,23,27,31,35,39 you could pair it up like:
3 + 39 = 42
7 + 35 = 42
11 + 31 = 42
15 + 27 = 42
19 + 23 = 42

42 x 5 = 210 (the complete sum for the series)
The equation for the partial sum is

S(n) = (t(1) + t(n))(n/2) -- This is for an ARITHMATIC series

The geometric partial sum equation is:

S(n) = t(1) x (1-r^n)/(1-r)

If the absolute value of r is less than 1, however, the n value will only get bigger and bigger, and this will only make the r^n value in the formula smaller and smaller until it eventually doesn't even matter anymore. So, if the absolute value of r is less than 1, the formula for the geometric partial sum is:

S(n) = t(1)/(1-r)

A geometric series will converge to a limit if the common ratio r satisfies the inequatlity "the absolute value of r is less than 1". If the absolute value of r is greater or equal to 1, the series diverges.

Binomial Series
-coefficients of a binomial series can be calculated recursively using the term before it.

(coefficient of a)(exponent of a)/(term number) = coefficient of next term

-coefficients can also be calculuated algebraically. They are equal to the numbers of combinations of n objects taken r at a time. Like in the probability section:

5C2 = 5!/3!2! = 5x4x3x2x1/(3x2x1)(2x1) = 10, so the term containing b^2 is
5!/3!2!a^3b^2

When expanding a binomial (a + b)^n the term with b^r is
(n!/(n-r)!r!)a^(n-r)b^r = (n/r)a^(n-r)b^r = nCra^(n-r)b^r

Practice Problem:
Using the series 7+12+17+22+27+32, find the 6th partial sum.
7 + 32 = 39
12 + 27 = 39
17 + 22 = 39

39 x 3 = 117 you can do it this way OR you can do it the equation way

S(n) = (t(1) + t(n))(n/2)
S(6) = (7 + 32)(3)
S(6) = (39)(3) - which is the same thing from before!
S(6) = 117

Extra Help: go to http://home.alltel.net/okrebs/page133.html

Reminder: Landry! You're the next BLOGMASTER


umm... i hope this posted. it's all just html in this format. This is a picture of one of Picasso's most famous blue period paintings. Also my favorite of his blue period.

Tuesday, April 24, 2007

Parametric Equations for Moving Objects

Property: Parametric Equations of a Cycloid

x = a(t-sint)
y = a(1-cost)

t is the number of radians the wheel has rolled so far
a is the radius

Objective : Be able to find the parametric equations for the path of a moving object.
Here is a example to explain how to do this:

An airplane is flying to LA from Nevada at a speed of 400 mi/hr. It is at point (20,30) on a Cartesian plane at t = 0 hr. LA is the origin. There are two winds going in different directions blowing. The wind blowing south is moving at a velocity of 90 mi/hr. The wind blowing west is moving at a velocity of 120 mi/hr.
Find the parametric equations for the airplane's path. Use t as hours.

x=20-120t
Since the original point of the plane is (20,30) and the velocity of the force blowing horizontally is 120mi/hr, x = the orignial x value + distance (which is rate x time)
In the same way
y=30-90t
The distance is negative becuase the wind is blowing in the negative directions of x and y.

Predict how long it will take for the plane to be 5 miles north of LA.
First plug 5 in for y.
5=30+90t
t=1.66666667
It would take 1.667 hours to get to this point.


try this site for more help: http://www.assembleme.com/post_2004_07_22_parametric_equations.pdf

i love the yeah yeah yeahs...http://youtube.com/watch?v=1UiNr8T2Mrc. -turn into


Anna you're next.

Monday, April 23, 2007

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 13 Test Topics:
Relationships between polar and Cartesian coordinates
Plot polar points.
Convert polar equation to Cartesian
Transform a polar graph/equation
Parametric equations – linear motion (word problem)
Determine a point on a curve given polar data
Describe the formation/creation of a polar curve
Parametric equations – nonlinear motion (word problems)

I’ve posted the solutions to the review handout on the class website .
See you in class! I’ll be around until 3:00 pm on Wednesday afternoon, and in early on Thursday.

If you have built castles in the air, that is where they should be; now put foundations under them.
- Henry David Thoreau

Sunday, April 22, 2007

13-3: Intersections of Polar Curves

Hi Guys! Sorry it took so long for this post to get online, but I wasn't 100% sure if it was my turn or not. I'm just going to assume it is. As usual, a bland, colorless blog...

The main focus of this chapter is on determining whether or not an intersection of two graphs in polar coordinates is a true intersection.

Here we see a graph with the polar equations

3 + 3 cos (theta)
and
5 sin (2 theta)



There appear to be eight intersections, as shown by the black dots (two are kind of smushed together), but not all are TRUE intersections. In order for an intersection to be true, the two points must meet at the same theta value. In order to find out which values are true, go to the "Mode" function on your calculator, and change the standard "Sequential" setting to "Simul" (which is Latin for "at the same time"- just a little fun fact.) If you visit your graph once again, the two equations will be graphed simultaneously (hence the "simul" setting). Watch closely for the intersections- the true intersections of the graphs will meet at the same point at the same time. After graphing our equations using "simul", we find that only these points are true intersections.



In order to find the coordinates of these points, we must leave polar mode and switch back to "function" mode. Plug in the exact same points in function mode, but using "x" as opposed to "theta". Find the intersections between the two graphs using the "calc" function; these points represent to coordinates on the polar graph.

Here is the graph in function (and radian) mode...



Here's a Problem!

Graph the following equations, and indicate the true intersections. calculate the values.

r= 2 + 8cos(theta)
r= 12 sin (3theta)

Solution

Start by graphing the two equations in polar, simultaneous mode. Make sure your window is big enough to see the intersections. Your graphs should look something like this...



By using the "SIMUL" function, you should end up with about these points as "true" intersections.



Plugging the values into the "function" mode gives you a graph like this-



The values for these points are (17.760, 9.619), (47.267, 7.431), (117.329, -1.673)



Additional Help




Debby's Up Next!




Personalization:
There's a Sketch Comedy group called The Whitest Kids You Know who recently got a show on fuse- these guys are extremely talented and have a lot of creative, original ideas. Here are some of their videos.














The Girl in this video is my cousin Sarah- this was the first sketch of the first episode of the show.



This Video is by a comedy group called "Derrick"

Wednesday, April 18, 2007

Thursday's Quiz

Here’s a list of topics for Thursday’s quiz:

Precalculus Quiz 13.1-3 Topics
Plotting polar coordinates
Multiple representations of polar coordinates
Plotting a curve given a table of polar coordinates
Interpreting/Reading a polar graph
Determining the equation of a polar graph
True/False intersections of polar graphs
Converting a polar equation to Cartesian

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 7 questions on the Non-calculator portion and 9 questions on the Calculator section. I’ll be around after school on Wednesday, and on campus by 8:00 AM on Thursday.

People are like stained-glass windows. They sparkle and shine when the sun is out, but when the darkness sets in, their true beauty is revealed only if there is a light from within.
—PHYSICIAN ELISABETH KÜBLER-ROSS

My job:




Wednesday, April 11, 2007

answers to chapter 10 review questions!

hey everybody!
i didn't put the answers on the review sheet, but here they are.

1. square root of 305.
2. 5 root 10 or square root of 250
3. 4i - 11j -13k
4. square root of 306
5.
6. 23
7. 43i + 29j -49k
8. cos alpha = 7/root 102, cos beta = -7/root 102, cos gamma = 2/root 102
9. 3i + 12j + 3k
10. 8i + 43j + 6k
11. dot product of a and b = 15, so the angle = 84.61
12. -2i - 2j - 2k
i hope these are helpful (and right)
-claire, nat, and allison

Monday, April 09, 2007

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 10 Test Topics:
Unit vectors
Planes – Standard Equation vs. normal vectors and points
Points on (or not on) a plane
Lines in Space – points and unit vectors
Direction Angles
Direction Cosines – properties
Vectors between two points
Angles between vectors
Scalar and vector projections
Cross products

I’ve posted a sample test and review on the class website which I’ll distribute in class tomorrow. Review 10-d will not be as helpful as 10-b, 10-c and 10-e.

See you in class!

Here’s a little something that shows the beauty of mathematics:

Math Art: The Beauty and Symmetry of Mathematics.

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111

9 x 9 + 7 = 88
98! x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

Brilliant, isn't it? And finally, take a look at this symmetry:

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

A unique approach to homework:

Wednesday, March 28, 2007

10.7: Direction Angles and Direction Cosines

Direction Angles and Direction Cosines

OBJECTIVE: given a vector, find its direction angles and direction cosines and vice versa

The graph shows a vector v and its direction angles:




the first three letters of the Greek alphabet alpha, beta, and, gamma are used for the three direction angles. like x, y, and z, the letters come in alphabetical order, corresponding to the three axes.

DEFINITIONS: Direction angles and Direction cosines:



the direction angles of a position vector are:


alpha, from the x-axis to the vector
beta, from the y-axis to the vector
gamma, from the z-axis to the vector
the direction cosines of a position vector are the cosines of the direction angles:


c1 = cos (alpha)
c2 = cos (beta)
c3 = cos (gamma)


PROPERTIES: Direction Cosines:


Pythagorean Property of Direction Cosines:
if alpha, beta, and gamma are the direction angles of a position vector and c1 = cos (alpha), c2 = cos (beta), and c3 = cos (gamma) are the direction cosines, then
cos squared (alpha) + cos squared (beta) + cos squared (gamma) = 1
or
c1 squared + c2 squared + c3 squared = 1


Unit Vector Property of Direction cosines:
vector u = c1 vector i + c2 vector j + c3 vector k is a unit vector in the direction of the given vector.




Example Problem:

find the direction cosines and the direction angles for vector v = 3 vector i + 7 vector j + 5 vector k

Solution:

find the dot products of vector v dot vector i, vector v dot vector j, vector v dot vector k. then use these to find the angles.

vector v dot vector i = (3 vector i + 7 vector j + 5 vector k) (1 vector i + 0 vector j + 0 vector k) = 3, which is the coefficient of vector i in vector v

vector v dot vector i = (3 vector i + 7 vector j + 5 vector k) (0 vector i + 1 vector j + 0 vector k) = 7, which is the coefficient of vector j in vector v

vector v dot vector i = (3 vector i + 7 vector j + 5 vector k) (0 vector i + 0 vector j + 1 vector k) = 5, which is the coefficient of vector k in vector v

absolute value of vector v = square root of 3 squared + 7 sqaured + 5 squared = squared root of 83

abolute value of vector i = abolute value of vector j = abolute value of vector k = 1
vectors i, j, and k are unit vectors

square root of 83(1) cos alpha = 3 ==> cos alpha = 3/square root of 83 ==> alpha = 70.774 degrees

square root of 83(1) cos beta = 7 ==> cos beta = 7/square root of 83 ==> beta = 39.794 degrees

square root of 83(1) cos gamma = 5 ==> cos gamma = 5/square root of 83 ==> gamma = 56.713 degrees






Kaori, you're up next! Good luck!


Who is this country singer????

10-6:Vectors Product of Two Vectors

Cross Products
  • The Cross product of two vectors a x b is perpendicular to the plane containing vectors a and b.
  • The magnitude of a x b- the absolute value of vector a multiplied by the absolute value of vector b multiplied by sin(pheta).
  • The direction of a x b is determined by right hand rule.
Right hand rule- Put the fingers of your right hand so they curl in the shortest direction from the first vector to the second vector. The cross product is in the same direction your thumb points.
  • Thumb points in-rotates from vector b to a and b x a
  • Thumb points out-rotates from vector a to b-a x b
Cross products of the unit coordinate vector

i x i=0 i x j=k j x i=-k

j x j=0 j x k=i k x j=-i

K x K=0 k x i=j I x k=-j

Example of Cross Product

Find vectors a x b
a= 4i+5j+9k and b=11i+5j+10k
a x b= 44 i x i + 20 i x j + 40 i x k
+ 55 j x i+ 25 j x j + 50 j x K
+ 99 k x i + 45 k x j + 90 k x k

2ok-40j
-55 k + 50 i
99 j - 45 i

=5i+59j-35k

In determinants-Example 1

(i J K )
( 4 5 9 )
(11 5 10)

i
(5 9 ) -j (4 9) +K (4 5)
(5 10) (11 10) (11 5)

I(5)(10)-(5)(9) - j(4)(10)-(11)(9)+ k(4)(5)-(11)(5)

=5i+59j-35k

Geometrical cross products and meaning of a x b

a
rea of a parallelogram have a and b as adjacent sides= absolute value (a x b)

area of triangle have a and b as adjacent sides= 1/2 absolute value (a x b)

Example

Find the are of the triangle with vertices's P1(-5, 5, 5), P2(-3, 2, 7), P3(1, 12, 6)

P1P2 x P1P3= -17i + 10 j + 32k

Area= 1/2 (-17i + 10 j + 32k)= 1/2 square root of (-17)^2 + 10^2 + 32^2

= 37.5898


Kaori, your up next good luck

Additional sources-
http://en.wikipedia.org/wiki/Cross_product- This website page gives one great detail on the subject of Cross Products.

Personal touch- Kobe Bryant had a great week in basketball, scoring 50 points or more in 5 games straight.

Here a video of his career, check it out-http://www.youtube.com/watch?v=mGbquGyTW34

Monday, March 26, 2007

Lesson 10-5: Planes In Space

In class today we learned how to find the equation of a plane using a point and perpendicular vector contained on the plane. The equation for a plane in space is:
Ax + By + Cz =D
A, B, and C: coefficients of the components of a normal vector
D: value determined after substituting the coordinates of a given point



First, let’s review the equation using a problem…

The figure below shows a plane in space with a vector , and a point P (4, 6, 8) both contained on the plane. How do we find the equation of the plane?


NOTE: To determine this equation, we have to find an equation of the plane relating x, y, and z where the vector is perpendicular to all points

Step One: Using the given vector equation and point, find a new point contained on the plane.

vector:

point a:
(4, 6, 8)
new point b:
(x, y, z)

Step Two: Now that we have two points, we can create a line connecting the point given to us (4,6,8) to point (x,y,z). As we learned from the mini-quiz, multiplying these two points together gives us a dot product of 0, which is equal to the cos(90°).



Step Three: Next we need to distribute the equation of our vector:

to the equation relating our two points (v).



Step Four: Then, we multiply both sets of equations and set the result equal to zero.



Step Five: With this last equation, we discovered that the coefficients were the same as those of the perpendicular vector equation:



Thus, to find the equation of the plane more efficiently, we can use the vector coefficients and replace them with A, B, and C. To find D, we simply distribute the vector coefficients to the corresponding point.

We distribute the coefficients 12, 3, and 14 to the corresponding x, y, and z values of the ordered pair (4, 6, 8). Thus, 12 x 4, 3 x 6, and 14 x 8. After adding up these numbers, we get 178. Thus, our final equation is:

178 = 12x + 3y + 14z

Helpful Notes:

  • If you are given the equation for a plane and need to find the equation for the perpendicular vector, remember the coefficients are the same! All you have to do is replace x, y, z, with i, j, k !
  • If you are given the equation for a plane and need to find a coordinate for your point on the plane, just plug in the coordinates for x, y, and z, and use basic algebra to solve!

For More Info…
Although this website seems a little more complicated, the animation helps me visualize exactly what the equation and subsequent steps are for solving:
http://college.hmco.com/mathematics/larson/precalculus_limits/1e2/ins_resources/ap.html

Fun Stuff…
This is from my favorite comic site, toothpastefordinner.com. (its my favorite comic for the month):
http://www.toothpastefordinner.com/012207/latest-poll-results-what.gif

Annie, you are up next! Good luck!

Thursday, March 22, 2007

Friday's Quiz Topics

Here’s a list of topics for Friday’s quiz:

Precalculus Quiz 10.1-4 Topics
Draw a three-dimensional vector and its “box”
Unit vectors: i, j, k
Adding vectors visually
Subtracting vectors visually
Dot/Scalar/Inner Product calculations – both formulas
Projection of a vector onto another vector – visual
Calculate vector projection of one vector onto another vector
Magnitude of a vector – calculation
Dot product – calculation
Calculate the angle between two vectors
Scalar projection calculation – two methods
Calculation of a unit vector
Vector projection calculation
Three-dimensional resultant forces
Angle between two three-dimensional vectors
Three-dimensional position vectors
Position vector to a point along a displacement vector


That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 10 questions on the Non-calculator portion and 12 questions on the Calculator section. I’ll be around briefly after school on Thursday, and in early on Friday morning.

I don’t know whether my life has been a success or a failure. But not having any anxiety about becoming one instead of the other, and just taking things as they came a long, I’ve had a lot of extra time to enjoy life.
—COMEDIAN HARPO MARX

Wednesday, March 21, 2007

10-4: Scalar Products and Projections of Vectors

10-4 Scalar Products and Projections of Vectors



The Objective: In a situation where you are given two vectors, find their dot product. Use the result to find the angle between the vectors and the projection of one vector on the other.

Basically this chapter is dealing with multiplying vectors together using various methods and answering various questions.

*An important thing to remember* is that dot products, scalar products, and inner products are all the same thing.

The equation for the DOT PRODUCT of vector a and vector b is vector "a" times vector "b" or = a dot b (both are vectors)

If you translate the vectors so that they are tail-to-tail you find the dot product by multiplying the magnitudes of the vectors and the cosine of the angle between them



For example (vectors "v" and "u"):
















The definition of a dot product is vector "a" times vector "b" equals the magnitude of vector "a" times the magnitude of vector "b" times the cosine of the included angle.



Technique for the computation of dot product:

vector a= X1i + Y1j + Z1k
and
vector b= X2i + Y2j + Z2k
then

a dot b = X1X2 + Y1Y2 + Z1Z2

Maybe it will help to clarify in verbal terms: The dot product of two three-dimensional vectors equals the sum of the respective products of the coefficients for the i, j, and k unit vectors.


Example


Find the dot product of vectors a and b.


vector a= 3i - 4j + 12k
vector b= 2i + 6j - 3k


a dot b = 3(2) + (-4 )( 6) + (12 ) (-3) = -54


To find the measure of the angle theta between vectors x and y you would take:


a dot b = -54



































(7 ) (13) ( cos theta)= -54

cos theta= -54/ (7 ) (13 )

cos^-1(-.5934)= 126.399 degrees (theta)



The other portion of this section deals with Projections of Vectors

The component is called vector projection of vector a on vector b.




An example of this would be light rays shining on two vectors. The rays would be perpendicular to the bottom vector made up of two black arrows (forming a right angle).






Let vector "p" be the vector projection of vector "a" on vector "b".

The formula used for this projection would be:

vector p=

There are two types of scalar projections, acute and obtuse.




Acute: an angle less than ninety degrees (x<90)















Obtuse: an angle greater than ninety degrees (x>90)













Finally, the scalar projection of a on b is



p= magnitude (a) cos theta



and the vector projection of vector a on b is



vector p= (p ) (unit vector "u")



If you need help with any of this material go to http://distance-ed.math.tamu.edu/Math640/chapter2/node4.html it is rather helpful in general.



Take this quiz to find out what kind of gnome you are!



http://quizfarm.com/test.php?q_id=226384





Good luck Celene you are next!






Tuesday, March 20, 2007

Lesson 10-3-Vectors in Space

The Overview of the lesson



This Image on the left is referred to as an Octant.

Octant-The xy-plane, yz-plane, and xz-plane divided into 8 regions.

The region in which all the three variables are positive is called first octant.

The X axis- i variable, the Y axis- j variable, and the z axis- k variable. The variables are used for unit vectors in the x-, y-, and z- direction respectively.





To finding the magnitude is similar to finiding the 2-dimensional magnitude.

ex-
Example of this lesson


Example #1- Find the length of vector P. Vector P=4i+5j+9k

Magnitude (length)- square root of (4^2 + 5^2 + 9^2)

Magnitude=11.045

Example #2- Find the displacement vector of point A(9,3,14) to point B(4,11,5).

One always subtracts from head to tail.

(4i+11j+5k) - (9i+3j+14k)= -5i+8j-9k=Displacement vector from A to B.

Example #3- Find the positon vector to the point 70% of the way from point A(9,3,14) to point B(4,11,5).

Position vector is A+.7(AB)

9i+3j+14k + .7(-5i+3j-9k)= 5.5i+5.1j+7.7k

Additional source

Additional Source that will help is found online on Wikipedia. Type in the search vectors and one will find online assitance.

Allison your next up-good luck

I hope my UCLA Bruins win it all this March in the tornument. This is a picture that examplifies UCLA Basketball.






Monday, March 19, 2007

ARASH YOU'RE NEXT!

Section 10-2: Two Dimensional Vector Practice


Hey kids in this lesson we reviewed our vectors. We learned a bunch of new vocab which I will give you further down. Hope you don't find this too frustrating. Also, I couldn't figure out how to get the arrows above the letters, so if you think there should be an arrow above it, it's because there should be.


Vector Definitions and Properties
Vector Quantity - has both magnitude and direction (eg. - force, velocity, displacement)
Scalar - quantity but no direction (eg. - weight, height)
Vector - directed line segment that represents a vector quantity (v w/ an arrow above it)
Tail - where the vector begins
Head - where the vector ends
Magnitude - (absolute value is used) the length of a vector
absolute value(v), If v = xi = xj, Then absolute value(v) = square root (x² + y²)
Unit Vector - absolute value(u) in the direction of v is a vector that is one unit long in the same direction as v. So u = v/abs(v)* - divide the vector by it's length
Vectors are equal if they have the same magnitude and direction.
The opposite of a vector is the same length in the opposite direction (-v)
Position Vector - v = xi + yj; Starts at the origin and the end point is (x,y)
When adding two vectors the resultant vector falls from the tail of the 1st vector to the tip of the 2nd. (Start of the 1st to the end of the 2nd)
* abs() = absolute value

Example Problem 1
If a = 4i + 8j and b = 5i - 3j
Find a+b => Add the two equations => 9i -5j
Find a-b => Subtract the two equations=> -i + 11j
Find -a => -4i - 8j
Find 2a + 4b => First multiply a by 5 and b by 4 => 8i +16j and 20i - 12 j => then add => 28i +4j
When doing problems like this make sure you understand that the abs(a) + abs(b) does NOT equal the abs(a + b)
Also, when finding the angle for the vector, use the inverse tangent.

Example Problem 2
Given point A (5, 10) and B (8, 20)
Find vector AB (Pointing from A to B)
to do this, look at the given points and subtract. AB = 3i - 10j
Find the position vector of the point 3/4 of the way from A to B.
To do this, it is easier to draw the postion vector to point A, then determine what is 3/4 of AB. Once that is found, add the two vectors.
A = 5i + 10j; 3/4 AB = 3/4(3i - 10j) = (9/4)i - (30/4)j
The answer would then be 5i+10j + (9/4)i - (30/4)j
This equals => 7.25i + 2.5j

Now in honor of March Madness, and USC beating Texas, here is a picture of the Trojan basketball team.

Tuesday, March 13, 2007

Questions and Answers from Chapter 9 Review

Here They Are...


The Questions and Answers for Factorial Feud

1. What are mutually exclusive events?

Events in which the occurrence of one event excludes the possibility that the other will occur.

2. A box of pencils has 35 lead pencils and 240 colored pencils. How many ways can you pull out a colored pencil and then a lead pencil?

8400 ways


3. 6 students must form a line. In how many ways can 6 line up from a group of 20?

27907200 ways

4. What is a permutation?

An arrangement in a definite order of some or all of the elements in a given set.

5. What are complementary events?

Events that complete all possibilities of a random experiment . the probability of any of the complementary events occurring is 1.

6. 20 people enter a competition to make it in America’s Best Barbershop Quartet. If only 4 can be selected, how many possible quartets are there?

4845

7. How many different circular permutations could be made with the letters “MY HOUSE”?

720

8. What is the probability of selecting an A, N, or E from the words “BANANA PEELS”?

7/11 or 63.636%

9. How many ways could 14 friends sit around a round table?

6227020800 ways

10. What is a combination?

A combination of elements in a set is a subset of those elements, without regard to the order in which the elements are arranged.

11. Without using your calculator, determine the value of 12C7

792

12. There are 14 mechanical pencils on a desk. 8 are blue, and six are green. If you choose six of them, what is the probability of 3 being blue, 3 being green? 6 being green?

160/429 or 37.3%, 1/3003

13. What does the following mean: A ∩ B?

The set of outcomes in sets A and B/ the intersection of A and B.

14. What does the following mean: A ∪ B

The union of A and B- the set of all outcomes in A or B.

15. If you are planting 2 flowers, and there is a 70% chance of the first living, and a 4% of the second living, what is the probability of both living? Neither living? The second living but not the first?

2.8%, 28.2%, 1.2%

16. Four fleas are living on a dog’s back. If the dog gets a bath, the first flea has a 20% chance of surving, the second has a 43% chance, the third has a 12% chance, and the fourth has an 87% chance. What is the probability of all surviving? None surviving? Just the third and fourth living? Only the 3rd? only the 4th?

.898%, 5.217%, 4.761%, .711%, 34.911%

17. Two dice are rolled (one pink and one purple). Find the probability that: (do these quickly-you should be able to)
a. The total is 10.
b. The total is 7.
c. The total is 2.
d. The numbers are 2 and 5
e. The pink die shows 2 and the purple die shows 5
f. The pink die shows 2 or the purple die shows 5.
1/12, 1/6, 1/36, 1/18, 1/36, 11/36

18. What is a random experiment?
The act of doing something
-
19. What is a trial?
Each time you do something

20. What is another term for outcome?
Simple event

21. What is the term for all the possible outcomes?
Sample space

22. A card is drawn from a 52 card deck.
a. How many outcomes are in the sample space?
b. How many outcomes in the event that the card is a face card?
c. What is the probability that the card is a face card?
d. What is the probability that the card is red?
e. What is the probability that the card is a king?
f. What is the probability that the card is the king of hearts?
g. What is the probability that the card is in the deck?
52, 12, 12/52 or 3/13, _, 1/13, 1/52, 1

23. 8% of men have a peg leg. Suppose 20 men are selected at random at let P(x) be the probability that x of the men have a peg leg.
a. Find the binomial probability equation for this scenario and find the values for x= 0,1,2,3.
P(x)=
.18869, .32816, .27109, .14143

24. Bobby’s batting average is .3. He comes to bat 5 times during a game.
Calculate the probabilities that he got 0, 1, 2, 3, 4, and 5 hits and what is the mathematical expectation of hits for this game?

.168, .360, .309, .132, .028, .0024
Mathematical expectation is 1.5

Thursday's Test Topics

Here’s a list of topics for Thursday’s test:

Precalculus Chapter 9 Test Topics:
Calculate simple permutations/combinations
Know the difference between permutations and combinations
Calculate Probabilities: And vs. Or (For independent/non-independent and exclusive/overlapping events)
Complementary Probabilities
Binomial Experiments (x successes out of n trials)
Binomial Expansions
Mathematical Expectations

That’s it! The format is as expected – ½ calculator, ½ non-calculator. 10 questions on the Non-calculator portion, 17 on the Calculator portion. I’ll be in my classroom for a short period on Wednesday afternoon and before school on Thursday. I’ll try to be available online later on Wednesday evening. If you have specific questions Wednesday night, email me!

I’ve posted a sample test on the class website which I’ll distribute in class tomorrow.

See you in class!

I like nonsense, it wakes up the brain cells. Fantasy is a necessary ingredient in living, It's a way of looking at life through the wrong end of a telescope. Which is what I do, And that enables you to laugh at life's realities.
- Dr. Seuss


Saturday, March 10, 2007

9-8 Mathematical Expectation

Lesson 9-8: Mathematical Expectations

In this lesson, we will learn about how you can calculate an expected value based on outcomes for each event in a random experiment. Basically, we will put our knowledge of probability to good use.

The example the book gives is of a carnival game of rolling a die.
If you...
roll a 6-- you win 100 points
roll a 2 or 4–- you win 10 points
roll an odd number–- lose 50 points

So in chart form,

Number on die ------Points you win
1 --------------------(-50)
2--------------------(10)
3 --------------------(-50)
4 --------------------(10)
5 --------------------(-50)
6 --------------------(100)

Strategy No. 1: To find the average point winning from the rolling-a-die-game...
You add up all the possible point outcomes and divide by the number of rolls!

(-50+10-50+10-50+100)/(6rolls) It simplifies too..
-30points/6 rollsor -5points/roll


What does it mean???---This answer means that you will on average lose 5 points per roll of the die.

Strategy No. 2

(-50 + 10 + -50 + 10 + -50 + 100) /divided by 6 rolls.
--- (The different outcomes of pts are in numerator, and the number of rolls/trials is in denominator)

(3*(-50) + 2*(10) + 1*(100)) /divided by 6 rolls --- you can "combine like terms"

3/6 * (-50) + 2/6 *(10) + 1/6 * (100) ---combine into individual fractions so that you can see what the probabilities for each outcome is!!!

probability of losing 50 pts is 3/6
probability of winning 10 pts is 2/6
probability of winning 100 pts is 1/6

Strategy No. 3 is the Algebraic Formula:

Mathematical Expectation = (Probability of Event A) * (Outcome of Event A) + (probability of event B) * (outcome of event B) +...

In the die-rolling example:avg winning = [(probability of rolling a 6) *(100 points)] + [(probability of rolling an odd number) * (-50 points) ] + [(probability of rolling a 2 or 4) * (10 points)]

PROBLEM #2:
The 2nd example that the book gives is more difficult because the probability involves "binomial probability distribution" from section 9-7...which means you have to use the equation from pg. 387 of the text book...

P(x) = nCx * a^n-x * b^x

b is the probability that the event occurs
a is the probability that the event does not occur
x is the number of times the event occurs in n repetitions

NOTE: use binomial probability distribution when there are only 2 possible outcomes...like getting a basketball shot---you either get it in or not. Whereas, you wouldn't use this with the die problem because there is an equal probability of rolling each of the 6 numbers on the die.

Basketball Toss Problem: You get 3 tries to shoot basketball hoops.

Number of Baskets------- Points (outcome)
0-------------------------(-50)
1 -------------------------(-45)
2 -------------------------(10)
3 -------------------------(200)

Probability of making the shot = 30% or .3
Probability of not making the shot = 70% or .7

To Solve!!!Now, find the probability of making x baskets...USE binomial probability distribution equation!!!



P(0) = 3C0 * .7^3 * .3^0 = .343 probability of getting 0 shots

P(1) = 3C1 * .7^2 * .3^1 =.441 prob. of getting 1 shot

P(2) = 3C2 * .7^1 * .3^2 = .189 prob. of getting 2 shots

P(3) = 3C3 * .7^0 * .3^3 = .027 prob. of getting 3 shots

(**note: the probabilities of all the outcomes add up to 1)


Now, just plug it into the Mathematical Expectation Equation!
mathematical expectation (E) = (prob. of 1st event * outcome of 1st event) + ...2nd...3rd....

E = .343 * (-50) + .441 * (-45) + .189 * (10) + .027 * (200)

E= about -30 points per game

ALL DONe!

Sites for more information on mathematical expectation:
http://www.wiu.edu/users/mfmk/Math101/Probability/PExpect.html
http://engrwww.usask.ca/classes/GE/210/CE%20225%20Webpage%20files/LECTURES/Lecture%208/Updated%20Lecture%208.pdf


Personalized Part: I love Miró. In my 3rd grade class with Mrs. Schroeder, we learned about Miró and all tried to copy his style...



























Reminder!! Nick, you are NEXT!