Monday, June 23, 2008

So Long ...

And so we begin where we left off ... don't let the sky be your limit. ;-)

I'm so glad we've had this time together,

Just to have a laugh or learn some math,

Seems we've just got started and before you know it,

Comes the time we have to say, "So Long!"

So long everybody, I'll see you in September! Watch this space for pointers to new blogs for each of my classes.

Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)

Monday, June 16, 2008

Class Survey

The exam is over and we did a little survey in class. The results are below; 13 students participated. If you'd like to add another comment on what you see here email me or leave a comment below this post.

Without any further ado, here are the results of our class's survey. Please share your thoughts by commenting (anonymously if you wish) below .....

Classroom Environment
The questions in this section were ranked using this 5 point scale:

Strongly DisagreeDisagreeNeutralAgreeStrongly Agree

The bold numbers after each item are the average ratings given by the entire class.

1. The teacher was enthusiastic about teaching the course. 4.92

2. The teacher made students feel welcome in seeking help in/outside of class. 4.69

3. My interest in math has increased because of this course. 4.23

4. Students were encouraged to ask questions and were given meaningful answers. 4.38

5. The teacher enhanced the class through the use of humour. 4.46

6. Course materials were well understood and explained clearly by the teacher. 4.31

7. Graded materials fairly represented student understanding and effort. 3.85

8. The teacher showed a genuine interest in individual students. 4.38

9. I have learned something that I consider valuable. 3.50

10. The teacher normally came to class well prepared. 4.54

Overall Impression of the Course
The questions in this section were ranked using this 5 point scale:

Very PoorPoorAverageGoodVery Good

1. Compared with other high school courses I have taken, I would say this course was: 4.77

2. Compared with other high school teachers I have had, I would say this teacher is: 4.77

3. As an overall rating, I would say this teacher is: 4.77

Course Characteristics

1. Course difficulty, compared to other high school courses:

Very Easy
Very Difficult

2. Course workload, compared to other high school courses:

Very Easy
Very Difficult

3. Hours per week required outside of class:

0 to 2
2 to 3
3 to 5
5 to 7
over 7

4. Expected grade in the course:


Specific Feedback
[Ed. Note: Numbers in parentheses indicate the number of students, over 1, that gave the same answer.]

What was your best learning experience in this course?

Workshop classes (4)
Explaining our work on the SMARTboard (2)
Great teaching with humour
Understanding math concepts instead of just using formulae (2)
Being comfortable with everyone
Learning the language of mathematics
SMARTboard (2)
Interaction (2)
Learning not to be afraid of making mistakes
Developing Expert Voices project (2)
Getting feedback
This blog
Learning in other ways than with a textbook
Learning to solve problems (2)
Learning innovatively in many ways

What was your worst learning experience in this course?

None (5)
More 1 on 1 help in class
Class kept getting unfocused (3)
Circular Functions Unit
Sometimes had to rush
Not asking questions

What changes would you suggest to improve the way this course is taught?

Fewer SMARTboard technical difficulties
Better scribe posts
Show formulae (how to "plug in values") first then explain or derive them
Mr. K. did an incredible job
I was motivated by Mr. K's enthusiasm
Wanted to hear the "Mr. K. Quote" just before the exam
Study periods for tests
More workshop classes
More group work
More guidance for Developing Expert Voices project
Wanted to hear about The Golden Ratio
Give homework for marks
Too much humour
Make Developing Expert Voices projects simpler and more straight forward
We were occasionally distracted — could have got more done
Course was better than I expected
For once I actually enjoyed math
Stay on task
Don't get off topic (2)
Have an alternative to blogging

It's interesting to compare the items that were considered both the worst and best learning experiences. Also, take a look at the list of worst learning experiences compared to suggestions for next year. Help me do a better job next year by commenting on what you see here ....

Sunday, June 15, 2008

Student Voices Episode 4: Justice, Lawrence, and Richard

First an update on this podcast: While we have received few comments on this or any of our class blogs the number of times the audio files have been downloaded is remarkable ...

Episode 1: Jessie 2440 downloads

Episode 2: Tim_MATH_y 1766 downloads

Episode 3: Chris, Craig, Graeme 1367 downloads

Thanks to all our listeners. We might get one more published during this school year but this may be the last until September. In any case feel free to let us know your thoughts about what you heard; every comment is appreciated.

In this episode of Student Voices Justice, Lawrence, and Richard talk about how they put together their Developing Expert Voices project and what they learned in the process: how they they best learn math, how it can best be taught, and many other incidental things like team work and organizational skills.

They have titled their project with one of my favourite reminders to all my students: Mathematics is the Science of Patterns. If you watch any of the video content they created you'll hear several "in jokes", listen for them. Without any further ado, here is the podcast. A copy of the poster they made for their work is below.

(Download File 12.2Mb, 25 min. 30 sec.)

Photo Credit: Shadow singer by flickr user EugeniusD80

Sunday, June 8, 2008

Judgement Day in Less Than Half a Day

Hi friends,

Provincials are coming. It is TOMORROW!!! But don't fret, it is nothing bad. I believe that the reason why people do bad on the exams is because they are nervous or scared and that there isn't sufficient time. The exams are just longer tests. It isn't much more difficult than a test but a bit more tedious because of the lengthy and copious amounts of queBenofschool here. Seeing that my good ol' buddy m@rk didn't scribe I decided to do it for him. What we did TWO (YES TWO [2]) days ago was study for the provincial exams coming up. Thestions in under 3 hours. All that you have to do is just relax, feel confident in all of the studying you did (Don't be over confident because the will backfire. Just be enough that you are beginning to feel comfortable.) Studying would be the best way to cope with that pre-exam fear. Hopefully everybody asked others for help because listening to a peer is just as good if not better than asking a teacher. The voice of a peer is very valuable. There are many ways to be prepared for the exam and it varies among students. So there isn't a right way but only a wrong way to prepare. It all depends on you and what makes you feel comfortable in situations like tomorrow Provincial Math Exam. My method for studying might not be good for others but it makes me feel comfortable. Hopefully everybody has that comfort zone in their studying.

Remember that time is a major issue for many people. That is why studying helps. Studying helps us all see the path to solving that problem quicker, more elegantly, and most of all more efficiently. The exam may play around with words or problems. They might word it differently, so just take your time in reading the questions carefully. If it helps, read the questions out loud but quietly because when you read with just your eyes your brain doesn't register it like when you read it with sounds being heard.

To end this scribe as well as ending the final scribe I tell you this. Sorry if I can't be poetic like my friend Francis but I can be positive. Good Luck at the exam and remember not to be scared. Just feel comfortable and everything will be fine. Drink lots of water, bring water to the exam, bring pencils (note the plural), erasers (again the plural), and a lucky charm that are magically delicious. You can even bring a real rabbit's foot but not a real rabbit's foot because that is CRUEL!
Mr. K I have one favor to ask you. Could you use that before test line I just said above before the exam. I'm sure everybody will be happy when they hear that.

Don't let the sky be the limit. Surpass the sky. And like what my good friend Andrew says, "Carpe Diem!" Which means seize the day. Hope to see everyone again in Calculus!!!

Good Bye and Good Luck! =)

Thursday, June 5, 2008

Last Class of the Jabbamatheez.

Today was the end of of our classes together. It's been quite fun learning about everything from circular functions to Geometric sequences, especially because of all the skits Mr. K put in.

Well, we started the class by solving a geometric sequence, given the 2nd and 3rd term. We had to find the 8th term given this. Since it was a geometric sequence, we know that there's a common ratio. To find this common ratio, we found the product of the 2 terms. Which was 12/24 = 1/2. We then found where y = 0, if we were to graph this. We found the 1st term by going backwards, which means you would have to multiply the 2nd term (24) by 2. 24 x 2 = 48. Then to find the term of 0, we would double that, which was 96. We then multiplied 96 by (1/2)8 because we need to find the 8th term, and it had a common ratio of 1/2. The answer was 0.375

Last class we found out where the infinite geometric sequence came from, but today we learned that when the common ratio is greater than 1, then it is a diverging series, because it will keep going on and on, thus meaning it is infinity. So when |r| > 1 then r = ∞
When the common ratio is less than 1, it's called a converging series, because it hones in on one value, which is 0. So when |r|<>∞ = 0.

This was pretty much all we learned, so we did some questions after. There was a question about a super ball. At this point Paul pulled out a super ball for Mr. K to use, and it gave the class a little interaction. The ball would start at 200 cm and drop to the ground, it would rebound 3/4 of the distance it fell. We had to then find the total distance travelled by the 4th bounce. This question was easily accomplished using a diagram.
The initial distance being 200 cm, we can multiply by 3/4 to find the distance it bounces back up, then multiply that distance by 2, because it travels that distance twice, bouncing up and falling down, then we multiply the distance from the second bounce by 3/4 (150 x (3/4) = 112.5) to get the distance going up, then multiply by 2 for it going up and down,(150 x (3/4) = 112.5 x 2 = 225 cm) then multiply the distance from the 3rd bounce by 3/4 again (112.5 x (3/4) = 84.375) then multiply that by 2 because it goes up that distance and down hat distance, so it would be (84.375 x 2 = 168.75) add all the values together to get 893.75
You can also use the geometric sequence equation to find this also.
After this we had to find the distance travelled until the ball stopped, so we used the infinite geometric series which was 1200, given all the variables, and you would add the initial value which was 200, and that would be 1400 cm. This was the last question.

That was all we did for our last class. I know, we need something special, because I'm the last scribe and this whole course has been oh so special, because of this reason, I wrote a poem. It's a Kyrielle poem. With 8 syllables each line, and rhyming scheme being aabB ccbB ddbB. Enjoy.

One More Step

Unluckily today's the last,
Of this most enjoyable class.
Though it's not the end for us yet,
As we move on just one more step.

Mr. K, we give you homage,
For expanding on our knowledge.
Never dull, you were fun instead.
As we move on just one more step.

Leaving with a fantastic smile,
As every thing's been so worthwhile.
Now to continue, and to prep.
As we move on just one more step.

-Francis Bowers

er... Scribe List

What Am I even doing this for? There's no more scribing! XD

Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

Stay Phi everyone,

Rence ~ Out

... and I wonder, if you know... what it means, to find your dreams come true...

Today's Slides: June 5

Here they are ...

Wednesday, June 4, 2008

Scribe List

Cycle 6


Hi I'm Justus


Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

I don't know about you guys, but honestly, I'm scared of the exam.

Stay Phi everyone,

Rence ~ Out

Infinite Geometric Series


  • Finding the Sum of Numbers in a Sequence
  • Sigma Notation
  • Infinite Geometric Series Formula

Finding the Sum of Numbers in a Sequence
(a) What is the sum of the integers from 1 to 5000?

Credit to Paul for being our Gauss today.

We imagine the sequence of all the numbers 1 to 5000 in the top row while the bottom row has all the numbers from 5000 to 1.

1, 2, 3, ... 5000
5000, 4999, 4998, ... 1

We find the sum of each column in our table and see that they're all 5001. Refer to slide 4 of June 3's slide for a better understanding of what I'm talking about.

We know that there are 5000 of these terms. We also know that if we want to find the sum of the integers between 1 to 5000, then we would have to multiply the number of terms by the total we have then divide by 2. We divided by 2 because there were two sequences we added.

Example: 5000 * 5001 / 2 = 12 502 500

Another way of solving this problem is to pull out the equation (found on June 3's slide #7) and plugging in the appropriate numbers, which Mr. K has a distaste for using this method because we're not really understanding the concept but just plugging numbers into an equation.

(b) What is the sum of all multiples of 7 between 1 & 5000?

The lowest multiple of 7 between 1 and 5000 is 7.
The highest multiple of 7 between 1 and 5000 is 4998.
There are 714 terms that are multiples of 7 between 1 and 5000.

We then imagine, again, that table, listing all the numbers in the sequence from 1 to 5000 (that are multiples of 7) in the top row and all the numbers in the sequence from 5000 to 1 (that are multiples of 7) in the bottom row, and find the sum of each column.

7, 14, 21, ..., 4998
4998, 4991, 4984, ..., 7

Add the top row and the bottom row, and you find that under each column in our imaginary table is 5005.

We multiplied the number of multiples of 7 between 1 and 5000 by the sum of the first and last terms of the sequence and divided that by 2 because we've counted the sequence twice and we get 1 786 785.

(c) What is the sum of all integers from 1 to 5000 inclusive that are not multiples of 7?

Well, we have the sum of all the integers from 1 to 5000. We have the sum of all the multiples of 7 between 1 and 5000. So if we find the difference of those two numbers, we find the sum of all the integers from 1 to 5000 inclusive that are not multiples of 7.

12 502 500 - 1 786 785 = 10 715 715

Sigma Notation
Left Question:
  • The k = 1 means that you start with 1 and evaluate 3k.
  • The number above the sigma symbol, 4, means to keep evaluating 3k until k = 4, then add all the terms.

Right Question:
  • The k = 0 means that you start with 0 and evaluate 2^k.
  • The number above the sigma symbol, 3, means to keep evaluating 2^k until k = 3, then add all the terms.

Then Mr. K says to find another sigma notation that is equivalent to the two questions on the slide. Of course, "mathematics is the science of patterns" is the common phrase that Mr. K always says, and yes, there're patterns here.

Also, a quick way of figuring out the answer to the sigma notation in question is to add the first and last term, then multiply by 2. Example: In the left question, [3(1) + 3(4)] * 2 = 30

Infinite Geometric Series Formula

The formula given in the above slide is something that we already derived from the previous class.

To answer the question "why is that the formula?" let's first use an example to help better our understanding of infinite geometric series. We make a square, cut it in half, and shade the half. We take the unshaded region of the square, cut that in half, and shade the half. Then we take another unshaded region of the square, cut that in half, and shade that half, and so on. So, firstly, we got 1/2 of the square shaded, then 1/2 + 1/4 shaded, then 1/2 + 1/4 + 1/8 shaded, then 1/2 + 1/4 + 1/8 + 1/16 shaded, and so on. If we continue this pattern and keep cutting and shading the square in half an infinite number of times, then we would've a shaded square.

1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... = 1

The formula that we derived from the previous class was

Sn = t1*(1-r^infinity) / (1-r)

If we take r and raise it to the power of infinity, that would equal zero.
Example: r = 1/2

1/2 * 1/2 * 1/2 * 1/2 ... = 0

So, in the part of the formula where it says (1-r^infinity), that part of the formula is going to equal (1-0) = 1 anyways, and multiplying a number by 1 won't really have a visible effect. It's more efficient to just not include the (1-r^infinity) into the formula.

Plug in the numbers into the equation, BING! BANG! BOOM! you have the result. Have a nice day.

  • Start on Exercise 47: Infinite Geometric Series. You should have finished all the questions in the exercise book up to 47.
  • Tomorrow, we will finish off the lesson the infinite geometric series.
  • Hopefully, we will be on task and finish the course tomorrow, and do a review on Friday.
  • Next scribe is Francis.

Today's Slides: June 4

Here they are ...

Tuesday, June 3, 2008

Introducing Series

The first thing we did in class today is answer some simple questions. Pretty straight forward, should be easy to do. Even if you weren't in class yesterday. Noo excuses, 'cuz I wasn't in class yesterday.

We also graphed the function 'y = [square root] x', which looks like half a sideways parabola.

Next, we learned about 'young Gauss', who was a pretty clever seven-year-old. His teacher told the class one day to add up all the numbers from 1 to 100. Instead of writing it all out and adding it, he found a pattern.

When the numbers are laid out like above, their sums equal 101. There should be a hundred 101's, hence the reason we multiply 100 and 101. We divide it because you added one hundred twice. That's what Karl Gauss figured out. If you want to know the details, you can go ahead and visit this link:

Using the same idea as above, we found the formula for an arithmetic series.
Sn = (n/2)[2a+(n-1)d]

A series is the sum of numbers in a sequence to a particular term in a sequence (definition can be found on blog). You can't have a series if you don't have a sequence first, or else, what are you adding?

We also found the formula for a geometric series.
Sn = [a(1-r^n)]/(1-r)

Lastly, we touched a little bit on 'Sigma', which is located on the second-last slide (8).

Mmm, yeah, that's it. I'm tired now, I'm sorry. Hahaha, next scribe is Joseph, since he asked.. again.

Scribe List

Cycle 6


Hi I'm Justus


Quote of the Cycle ;

"There's no such thing in the world as absolute reality. Most of what they call real is actually fiction. What you think you see is only as real as your brain tells you it is."

And remember to watch...
... it's epic

Stay Phi everyone,

Rence ~ Out

Today's Slides: June 3

Here they are ...

Monday, June 2, 2008

Geometric Sequences

WARNING: The geometric sequences unit does not have any new material; rather the unit gives a new perspective to what we already know from previous learning.

  • What is a sequence?
  • Recursive definition vs. implicit definition
  • Common difference vs. common ratio
  • Determining any term (the nth term) in an arithmetic sequence vs. determining any term (the nth term) in a geometric sequence

An example of an arithmetic sequence is the first sequence of numbers found on SLIDE 2. Examples of geometric sequences is the second and the third sequence of numbers. Note that arithmetic sequences and geometric sequences aren't the only sequences that exists, such as the Fibonacci sequence, but the main scope of the unit focuses on geometric sequences.

We find the next three terms by determining a rule for each sequence of numbers.
  1. For the first sequence of numbers (4, 7, 10, 13...), we see that 3 is added to any term to get the next term.
  2. For the second sequence of numbers (3, 6, 12, 24...), we see that 2 is multiplied to any term to get the next term.
  3. For the third sequence of numbers (32, 16, 8, 4...), we see that 1/2 is multiplied to any term to get the next term. (Remember, in grade 12, we multiply by 1/2 instead of dividing by 2; it's to make our calculations easier to do.)
  4. For the fourth sequence of numbers (1, 1, 2, 3...), we see that we have to add the previous two terms to get the next term.

Let's take a closer look at the first sequence of numbers (4, 7, 10, 13...).

How did we determine that the rule is to add 3 to any term to get the next term? We find that out by determining that the common difference is 3 (green font). We then find out that if we're given any term, n, then 3n+1 is the equation we use to get what n equals. For example,

3n+1 = 3(1)+1 = 4.

So if the rank (n) is 1, then its term is 4. If the rank is 2, then its term is 7. If the rank is 7, then its term is 22. This is expressed with the equation 3n+1.

The rank represents the term. For example, in this sequence, the rank of the first term is 1, the rank of the second term is 2, the rank of the third term is 3, the rank of the nth term is 3n+1, and so on.

Graph 3n+1, and we find that it is the equation of a line. Also, we can determine the 0th term by plugging 0 into the equation.

3n+1 = 3(0)+1 = 1.

For the zeroth term, the output 1. It is also the y-intercept of the graph.

Term 1 (t1), or in some textbooks it's called a, equals 4. The common difference, d, is 3.

The recursive definition, in this sequence of numbers, is to take a term and add 3 continuously.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3n+1.


Let's take a closer look at the second sequence of numbers (3, 6, 12, 24...).

How did we determine that the rule is to multiply 2 to any term to get the next term? Firstly, there isn't a common difference in this case but a common ratio. The common ratio is 2. We then find out that if we're given any term, n, then (3/2)(2^n) is the equation we use to get what n equals.

tn = (3/2)(2^n) = [3 * (1/2) * 2^n] = [3 * 2^(n-1)]

(Refer to the bottom-right corner of SLIDE 4.)

So if the rank (n) is 1, then its term is 3. Similarly, if the rank is 2, then its term is 6. If the rank is 7, then its term is 192. Etcetera.

Graph [3 * 2^(n-1)], and we find that it is the equation of an exponential function. Also, we can determine the 0th term by plugging 0 into the equation.

tn = [3 * 2^(n-1)]
t(0) = [3 * 2^(0-1] = 3/2

For the zeroth term, the output 3/2. It is also the y-intercept of the graph.

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 2.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3*2^(n-1).

Let's take a closer look at the third sequence of numbers (32, 16, 8, 4...).

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 1/2.

The implicit definition, in this sequence of numbers, is the equation of the sequence.


  • Exercise 45: Geometric Sequences

Next Scribe is Nelsa.

Today's Slides: June 2

Here they are ...

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Saturday, May 31, 2008

Probability: More Practice

Instead of starting a new unit, we worked on Probability. Mostly we went over the practice sheet we were given on Thursday. The questions we reviewed were one, two, six and nine.

Question One:

  • The probability of a randomly chosen car being defective is 1/3. Four cars are chosen randomly in order. Given that at least two cars are defective, what is the probability that the first car is defective?

The first thing we did, was figure out what the question was asking.
P(1st is defective I
at least two is defective)

Next we found the sample space. It's always easier if you think of it as a word, and then asking yourself how many times the letters in that word can be rearranged, like so:
DDGG -- meaning two of the cars are defective
DDDG -- three cars are defective
DDDD -- all four cars are defective
Then you can find the probability that the cars will be chosen that way, and that becomes your sample space, or denominator.

Next we looked at all the possibilities that the first car will be defective using slots, which, if you remember, was from combinatorics.

Put that numerator on top of the denominator that we found earlier, and you have your answer.

Question Two:

  • Two jars contain red and green marbles. Jar I contains 3 red and 2 green marbles. Jar II contains 4 red and 3 green marbles. A jar is picked at random and two marbles are picked out of that jar in order. If it is known that the first marble is red, what is the probability of the second marble being red?
You can start this problem by drawing a tree, and writing down the probability that each possibility will occur - if that makes sense. For example, if you look at the picture above, you'll see that the first branch is 'Jar I' followed by 'Jar II', the probability of choosing either jar is one half. The next branches are 'red' and 'green', and the probability of choosing red in 'Jar I' is 3/5, while the probability of choosing green is 2/5. So on and so forth. Keep in mind that the marbles are not replaced after you take it out, so the sample space is different everytime you draw.

In this question, what we're looking for is the probability of getting a red after drawing a red marble first. That becomes the numerator, and the denominator consists of all the possibilities, as shown above. So you plug in the proper numbers in their proper places, and you have that question solved.

Question Six:

  • Susan sees her friend, Tim, at his locker with a worried look on his face. She asks, "What's wrong?" Tim has to open his locker and change clothes within the next five minutes. However, he has forgotten the combination to his new lock. He knows that the lock requires three different numbers. He also remembers that all of the numbers are odd, and all of the numbers are divisible by seven. It takes 10 seconds to dial a locker combination and 1.5 minutes to change clothes. Is Tim likely to be ready for gym class on time?
I think you'll understand how we solved this problem just by looking at the above picture. The '1 min 30 sec', written in pink, is how long it takes Tim to change.

Question Nine:
  • Three identical boxes each contain two drawers. In one box, each drawer contains a gold coin. In another box, each drawer contains a silver coin. The remaining box has a silver coin in one drawer and a gold coin in the other. One drawer is opened and a gold coin is found. What is the probability that the other drawer in that box also contains a gold coin.
  • Michael claims that the probability is 1/3. Jessica claims it is 1/2. Raymond says the probability is 2/3. Explain how each person may have arrived at their answer. Who is correct? Justify your answer.

The probability of choosing the right box is 1/3, and the probability of choosing either a gold or silver coin is 1/2, since there's two drawers. Obviously if you get the box that only have two gold coins, you're not going to get a silver coin. But I'm sure we're all smart enough to realize that.

The probability of getting a second gold coin is shown by:
[P(IG)/(P(IIIG) + P(IG) + P(IG))] + [P(IG)/(P(IIIG) + P(IG) + P(IG))]

If we plug in the correct numbers, we have:
[(1/6)/((1/6) + (1/6) + (1/6))] + [(1/6)/((1/6) + (1/6) + (1/6))]
[(1/6)/(1/2)] + [(1/6)/(1/2)]
(1/3) + (1/3)

Therefore, Raymond is correct. Michael could've gotten his answer by thinking that there's only one box that contains two gold coins, and the probability of choosing that box is 1/3. Jessica could've simply thought, well there's two drawers, so there's a 1/2 chance of getting a gold coin.

So that's the end of that. There's also questions in the slides (9-18) that we could do for practice. In the afternoon class we had a pre-test that we're going to go over tomorrow. There isn't going to be a test for this unit because we don't have time. Soo.. we're starting a new unit next week.

And that's that. The next scribe will be Joseph because he asked me yesterday. Whoot. Good luck to everyone doing the English Provincial Exams tomorrow! Remember to pace yourself and.. uhh, I forgot what else Mr. V and Mrs. Strecker said.. but.. yes!

Thursday, May 29, 2008

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Reminder: Roxanne, Francis and Elven - Your D.E.V. Project is due FRIDAY, yeah that's right, FRIDAY by midnight.


If you haven't already, just give Mr.K a quick email to have him add you to the D.E.V. Blog. Do it now! 'Cause I know you want to.

On a side note: That dot on the Visitor map is getting awfully large. Makes us seem like a big target =( Lol, just kidding.

Stay Phi everyone,

Rence ~ Out

Workshop Period

Ohhh my goodness. I nearly forgot I was scribe today because of the basketball game XD. (Go lakers!)

So the title above describes todays class. Mr. K was away again but no worries, he'll be back tomorrow...i think. Anyways, we were given a worksheet that we worked on in our little groups for the whole class. I thought the sheet had some challenging question like 1 and 2. I hope we go over those tomorrow. Yeeaah it was a pretty uneventful period other than that.

So I'll get to the point. Next scriibe is NELSA. "Beautiful" right Nelsa?

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

AND NO, Unicorns aren't real. :) Sorry to burst your bubble guys. Lol.

Stay Phi everyone,

Rence ~ Out

Wednesday, May 28, 2008

DEV Work Period!

Well, today we had Ms. Gonzaga come back as our substitute since Mr. K was away today. Surprisingly, and thankfully, there was no quiz or anything (which some of us may have been expecting since we usually get one when there's a sub). But anywho, who cares about that right?! So then, basically, this class was dedicated to working on our DEV projects. The due dates for them are really close that's why. We were also told to work on Exercise 44 as well, and that's it! Same deal for the afternoon class! Okay, that's it!

Aaaannnnd!!! Here's something Jamie is probably expecting (inside-outside joke?!)...she's not scribe by the way, Joyce is, and no I did not mean harm to her by putting her name in red!

Well then, that's it for today!!!! Did I mention the next scribe? Well for those who don't care about reading scribe posts and just like scrolling RIGHT TO THE BOTTOM immediately, the scribe is Joyce.

Tuesday, May 27, 2008

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Today's Slides: May 27

Here they are ...

Combinations of Permutations and Probability and Combinations O_o

Why hello there!
Benofschool here. I never scribed for a while, I was beginning to think that you guys forgot about me =(. Well anyways time for a scribe. Mr.K was later than usual today but he wasn't late for school which was good. Today's class was a workshop. We were broke up into groups like every other workshop class. Unfortunately the slides aren't up. Questions were put up and we were off...

Question 1
We had to find the probability of getting an ace or a diamond from a deck of cards. Fairly simple. Francis went up to answer it. To get the answer we just add the probabilities of getting an Ace and a Diamond and subtract the probability of getting the Ace of Diamonds. We add because it is an "or" question and we subtract that single card because it was counted twice when we calculated the probability of getting an Ace or Diamond. So the answer was 4/13.

Question 2
The second question involved a venn-diagram and venn skills from grade 11 logic in pre-calculus. The question was what is the probability of picking a person who doesn't like Dr. Pepper or the diet version if 7 people liked Dr.Pepper, 11 people liked the diet version, and 3 people liked both. So first we have to create a two circle venn-diagram. Rule is start with the inside and work outwards. So we put in the middle which means that 3 people prefer both drinks. Then lets work with on side. But remember from grade 11 we have to subtract the number of objects in the center from the separate values. If we look at the first question, we didn't count the Ace of Diamonds because the probability of getting an Ace or a Diamond is not mutually exclusive as well as the Dr. Pepper situation. If one thing occurs, the other probability can still happen. So that means we have to subtract the number of people that liked both drinks from the separate preferences. We get the sum of those numbers (4 +3+8=15) and subtract from the total number of students in the class (25-15=10). That will be the numerator in the probability and the sample space would be the total number of students. So the answer is 2/5.

Question 3
The 3rd question is about picking dresses. There are 15 dresses: 6 green, 5 blue and 4 yellow dresses. We wanted to know what is the probability of getting exactly 2 green dresses from picking 6. This question involves both combinations and probability. So 6 choose 2 because they are indistinguishable. multiplied by 9 choose 4 for the remainder of the dresses divided by 15 choose 6 which is the ways that we can choose 6 dresses out of 15. So we get an answer of 37.8%.
Question 4
Another combination question. A couple has 4 children that are about to be born. What is the probability of getting at least 2 girls. So what we do is find the probability of getting 2 girls first then 3 girls and finally all girls. Then we just find the sum of those values and we get the answer. The answer is:

That were all of the questions we did in today's class. Remember DEV due dates are arriving in a future now. Check the calendar on the right side bar to check for your due dates. Tomorrow's double will be used as a DEV work day. So bring the stuff you want to bring to work on your DEVs. That is all I have to say about today's class. The next scribe will be kristina. Good Night and see you all tomorrow!!!

Costly Test

So yeah, the reason why I'm up so late was because I just got back from the Glow in the Dark concert... and I've completely lost my voice. Anyways, I found this really cool story so I thought I'd share it and hope that you get a laugh out of it too.

A professor was giving a big test one day to his students.

He handed out all of the tests and went back to his desk to wait. Once the test was over the students all handed the tests back in. The professor noticed that one of the students had attached a$100 bill to his test with a note saying, "A dollar per point."

The next class the professor handed the tests back out.

This student got his test back and $64 change.

Monday, May 26, 2008


and no ...
- being cagey does NOT mean you're inside a cage alot ...

So I was chosen for being the scribe for today. Let me tell you guys something. REVENGE GETS YOU NOWHERE. That's why I'm always going to steal Paul's seat from now on. Thanks for choosing me as scribe Paul.

- SLIDE 2, 3
- Identifying events / dice diagram

- SLIDE 5, 4
- Probabilities involving "AND" and "OR"
- Testing for independence

- SLIDE 6, 7, 8, 9
- A test for cancer

- SLIDE 10
- A test for industrial disease

- identifying events / dice diagram


- and so we started the class finishing off what we had last done in class on Friday.
- a list of events which the class had to determine whether it was dependent or independent. Mutually exclusive or not mutually exclusive. I'll take two examples.

b) one card - a red card or a king - is randomly drawn from a deck of cards.
- because it is only ONE event we know that it must be independent
- a red card can also be a king. Therefore it is not mutually exclusive

c) A class president and a class treasurer are randomly selected from a group of 16 students.
- because we can only have ONE president and ONE class treasurer, the event must be dependent. One student cannot be the president and class treasurer. Well in this case it can't. Therefore it is mutually exclusive



I had actually found this diagram on another blog of Mr. K's class.

This was done to answer the question:
e) Rolling two dice and getting an even sum or a double
- It is independent because the first roll does not affect the chances of the probability for the next roll.
- As you can see on the diagram, you are able to roll a double AND get an even sum. There fore it is not mutually exclusive.

- Probabilities involving "AND" or "OR"
- Testing for independence


here is the slide ...


We compared the result of the probability of getting the flu shot and getting the flu. (0.10)
To the result of the probability of the seniors getting the flu (0.15)

There fore the event is dependent if you get the flu shot or not.

- A test for cancer


- The given question, information and possibilities.

0.5% of 1,000,000 = 1,000,000 x 0.005 = 5000
To find who does not have cancer you would subtract 5000 from 1,000,000
98% of the time, the test will be positive.
The amount of people who have cancer and will test positive would be ..
5000 x .98 = 4900
The amount of people who have cancer and will test negative would be ..
5000 x .02 = 100
982% of the time, the test will be negative.
The amount of people who don't have cancer and will test positive would be ..
995,000 x .02 = 19,900
The amount of people who don't have cancer and will test negative would be ..
995,000 x .98 = 975100
Adding it up the amount of people who tested positive for cancer
4,900 + 19,900 = 24,800
When only 4,900 actually do have cancer.

The probability that a person who actually does have cancer and tests positive is the amount of people who do have cancer divided by the total amount.

4,900 / 24,800 = 19.75%

Here are the tree diagrams to see how this all works out.

- A test for industrial disease


The probability of a person who tested positive and actually does have the industrial disease is much similar to what was done in SECTION C.

The probability of people with industrial disease and tested positive.
The probability of people with industrial disease and tested positive,
The probability of people without industrial disease and tested positive.

^ does that make sense?!

In other words, and in this case ..

0.0099 / 0.0099 + 0.0099 = 0.5%

because there is still a great chance that even if you do NOT have cancer, the result may still test POSITIVE

- not intentionally trying to scare any of you, or myself.
- not intentionally trying to offend anyone who is reading this.
- It's just a joke! Humor in OUR math class is NO laugh matter.

That's it, That's all ladies and 'gents! Have yourself a great day. Go out there and commit random acts of kindness! Cheers!

The next scribe will be ...benofschool

Today's Slides: May 26

Here they are ...

Sunday, May 25, 2008

Types of Probability and Exclusivity

So, this is a double scribe post, which covers the lessons we did on May 22 and May 23 (Thursday and Friday respectively).

So lets get started, shall we? First off is our lesson from May 22...


So last Thursday, we talked about the two types of probability. There are dependent and independent probabilities, and thats what we'll explore here.

Slide Two:
The first thing we did was make a tree, which we've all done before. This tree displays all the possible outcomes of flipping two coins. Easy right? We know probability of each outcome is 1/4 because there are four possible outcomes, and each one is one of those four.

Slide Three:
Moving forward, we have "an entirely different question that is not at all similar to the one we
just did." Please note the quotation marks. Because, this question is very similar to the one we just did. Instead of heads and tails, there are reds and blues. But this is similar but not the same. Why? Because unlike our previous probability question, this question deals with a dependent probability, whereas our previous one dealt with an independent probability. Remember, this lesson was supposed to be about dependent and independent probabilities?

But Paul, you say, what the heck are dependent and independent probabilities anyway? Im confused!

Well, to put it simply, independent probabilities are probabilities that are affected by the steps before them. Why don't you think about it in context, like this:

If you flip a coin, and then you flip another entirely different coin, does the chance of you getting tails on the second coin change because you flipped the previous coin?

Answer? Of course not, excluding far-out possibilities such as your hand got so tired flipping the previous coin that you put less energy into flipping your second coin and gave it 0.001% better chance of landing on tails. This is an example of an independent probability, where both probabilities are, well, they're independent of each other. Makes sense, right?

However, with our second question, our probabilities are dependent. Why? Because when you choose a marble (wearing a blindfold, earmuffs, and nose plugs so you cannot possibly see, hear or smell which marble you're picking), you also
remove that marble, thus changing the probability of you getting the same colour marble when you draw a second one. This is all displayed in the slide two tree, where if you drew a red marble and removed that, your chance of drawing a red marble again the second time is 2/5, whereas if you drew a blue marble the first time, you have a 3/5 chance of getting a red marble when you draw a second time. See? They're different, which is exactly why this is a dependent variable.

And that is basically the gist of this entire lesson, so I'll just summarize the rest of the slides...

Slide Four:
Okay, something changed in this question, and I'll give you a cookie if you can spot it.

Alright, found it? Here, have a cookie. You clearly noticed that our question changed in that the marble is no longer simply discarded (or "thrown out the window" as Mr. K would say [hey, I rhymed]). Instead of removing a marble when we draw it, we just put it back. Because we put the marble back, our probability doesnt change, and it becomes independent. Now that its independent, all you'd have to do is change those R's to H's and B's to T's, and you've got your original heads and tails question! Magic right?

Which brings up another point Mr. K mentioned. A lot of the time, your question boils down to stuff much like the heads and tails question, just wrapped up in a pretty little. Remember that (say it to yourself five times, or something), and you'll find things will be a lot easier.

Slides Five & Six:
Just a rehash of what I already wrote, basically definitions of what dependent and independent variables are (in simpler, less rambling and rant-y [is that a word?] terms).

Slide Seven:
A few questions, dealing with figuring out whether or not the question is describing an independent or dependent probability. Simple, right? I'll explain why quickly.

a) Independent because the outcome of the coin toss doesn't affect the outcome of the dice roll (again, this is factoring out such insane possibilities and how your possibility might be changed if you did both at once, or something along those lines).

b) Dependent because when you draw the first card, you remove it from the deck and change the amount of cards in the deck. So when you draw your second card, your deck is one card smaller than it would be had you not removed the first card, and you can no longer draw that particular card. Thus, the first outcome affects the second outcome.

c) Independent because now your deck is static and the probabilities always remain the same (excluding time travel and such nonsense).

Slide Eight:
Slide eight takes our second slide's question one step further by adding another coin flip. Nothing terribly new here, it just makes the chances of getting a particular combination smaller. I won't bother explaining this in great detail, but I'd like to point out that the final probability of an outcome (say, HHH, or flipping three heads in a row) is equal to the product of its previous steps probabilities (in other words, 1/2 * 1/2 * 1/2 = 1/8, not a coincidence), where each coin flip had a 1/2 chance of landing Heads.

Slide Nine:
Now to follow our trend of making small changes to our previous questions, we take our old question and put a fresh splash of paint on it and volia, we get something entirely new. Fortunately for me, all I have to do is take our old solution, paste it on to our new slide and say Boy = H, Girl = T. But wait, our question says, whats the probability of mom and pop getting exactly two girls and one boy. So what do we do? We find all the answers that have two girls (T's) and one boy (H's). Since there are three possible outcomes out of 8 that have two T's and one H, we determine that the chance of getting two girls and one boy exactly is 3/8.

Slide Ten:
Ignore this, this was my poor and not at all though out attempt at solving the question, and it is completely wrong.

Slide Eleven:
Our final slide, with the correct solution, courtesy of Kristina. Another variation on our girl/boy question, except with a different amount of steps and different conditions. So for the first draw, if we draw a blue marble first, our bag with 3 red and 3 blue becomes a bag with 4 red and 2 blue, and vice versa. So when you draw a second marble, you either have a 1/6 or 2/6 chance of drawing a blue marble the second time depending on the colour of the first marble you drew.

And that concludes our lesson on the Types of Probabilities.


And without delay, I give your my scribe post for our May 23rd (Friday) class on Mutual Exclusive Events.

Slide One:
Contains a fox that is entirely unrelated to probability. However, I would like to dub him (or her) the Probability Fox, just for fun.

Slide Two:
A simple slide with a simple question. If you have 56 listed and 144 unlisted phone numbers, you have a total of 200 phone numbers. So if you want the probability of choosing a listed phone number from those 200 total numbers, you have a 56/200 chance, or a 28/100 or 7/25, or .28, or 28% chance.

Slide Three:
Okay, so we have a horse named Gallant Fox (makes total sense, right?) and another horse named Nashau. Gallant Fox runs a race and has a 2/5 chance of winning. Nashau runs an entirely separate race and has a 1/3 chance of winning. What is the probability that:

a) Both Nashau and Gallant Fox win their respective races. That's easy, we simply multiply their probabilities together and get 2/15. Remember earlier how our final probability was the product of all the probabilities of the steps before it? This is just like that, where our first step is Gallant Fox wins (2/5) or he loses (3/5), and then our second step is Nashau wins (1/3) or he loses (2/3). Since we want the path that has both Nashau and Gallant Fox winning, we want to multiply 1/3 by 2/5, which gives us our final probability for that outcome.

b) So this is just a in reverse, and instead of taking the winning path, we take the losing path. So 2/3 * 3/5 = 6/15.

c) Well, there are two ways we could solve this question. We could find the probabilities for all the routes where one of the horses wins their race and then add those together. But thats just long and tedious, and considering what we already know, we have a much simpler solution at our doorstep. The question asks us what the probability is that one of the horses wins, so as long as they both don't lose, we're good right? Wait, don't we already know the probability of them both losing? Well... if we know that, cant we just take all the possibilities and remove the ones we don't want to get the ones we do want? Something like this... 15/15 is our total possibilities, but 6/15 of them are ones that have both our horses losing their races. So we just subtract 6/15 from 15/15 and... our answer is 9/15.

Slide Four:
This slide was probably the most confusing question we had encountered in a while, and it took a while to figure out. But basically, it goes like this:

Chad wants to meet his girlfriend in either the Library or the Lounge. If he goes to the Lounge, he has a 1/3 chance of meeting her (apparently Chad has poor arranging skills as he still has no idea where they're meeting despite the fact that this has all been pre-arranged, but I digress), but if he goes to the Library, he has a 2/9 chance of meeting her.

a) What is the total probability he'll actually meet up with her, either in the Library or the Lounge. Because its an "or" probability, we add the probabilities together (the Library OR the Lounge), and get 5/9 chance they actually meet.

b) Now this is where it kinda gets confusing. The probability of them not meeting is 2/3 * 7/9? Wait, just were adding just a moment ago, why did we suddenly switch to multiplication? Now we're dealing with an "and" probability, where she is not in the Library AND not in the Lounge. Thus, we get our 14/27 chance they don't meet.

Slide Five:
This slide is a little demonstration on how Chad cannot go to the Lounge AND the Library. Thus, going to the Library or going to the Lounge are mutually exclusive events, which is what this lesson is all about.

Slides Six & Seven:
These slides give you the basic definition of Mututally Exclusive Events, and some examples. These are pretty self-explanatory.

Basically, if you have one event that makes another impossible, then they are mutually exclusive. You cannot turn left and right at the same time, you cannot be facing North and South at the same time. Facing North is mutually exclusive to facing South.

Slide Eight:
So we want the probability of drawing a King or a Spade in a single draw from a pack of 52 playing cards.

Our first event is to draw a King. Our second event is to draw a Spade. Drawing a king has 4/52 chances, and drawing a Spade has 13/52 chances. However, because of the one card that is both a King and a Spade, we must remove that card (1/52). So our formula:

Probability(Event A or Event B) = Probability(Event A) + Probability(Event B) - Probability(Event A and Event B)

Means this:

The probability of Event A or Event B is equal to Event A's probability plus Event B's probability subtract any Probabilities that fulfill both Event A and Event B.

And inputting our data, we get the Probability of drawing either a king or a spade as 16/52.

Slide Nine:
And finally, a couple practice questions. Drag'n Drop Baby!

a) Independent because the marble is returned, mutually exclusive because you cannot draw a red and blue marble at the same time.

b) Independent because there is only one step, not mutually exclusive because it is possible to draw a red king.

c) Dependent because once a president or treasurer is selected, the number of people in the selection pool changes and thus so do the probabilities for the next selection. Mutually exclusive because the president cannot be the treasurer at the same time (they are removed from the selection pool once they have been chosen).

d) Independent because there is only one step, mutually exclusive because there is no card that is a red king and black queen.

e) Independent because the second step does not change the probability of the second, mutually exclusive because its not possible to get an even and odd number at the same time.

And that concludes this super long double buttery extra flavoured scribe post, again posted in the middle of the night and written in the dark. I'll see you all in a few hours. Please feel free to point out any mistakes, I'm fairly sure I made a few.

And because he asked so nicely, my next scribe will not be Thi.



I'll just tell you guys in the morning since nobody will read this until then anyway. Good night, and have a pleasant tomorrow.

P.S. That Google translate thing works, atleast, I think so (I cannot read Japanese and verify this). However, from this I have learned that "Eleven" is "11" in Japanese.

Thursday, May 22, 2008

Today's Slides: May 22

Here they are ...


Once again, it’s been days since I’ve gotten access to the blog. But I still manage to get things done. Conics—the in-depth look at the aspects of a cone. It sounds like I’m writing a feature article on it or something. I sort of enjoyed this unit, it had its moments. It was pretty reminiscent of the days when I was master paper folder in origami club. By the end of emphasizing each shape that evolved from a cone, I simply wondered, “what happened to just making cranes?” I also commend Mr. K for his innovative [yet deadly] ways of disarticulating a cone with a shward.

This may have been a short unit but I really have to admit that I learned a lot. I didn’t know that so many shapes could come from a cone, or two cones to be precise. A “hypercone”, just kidding… The proper term of course would be the double-napped cone that gives way to a hyperbola, a parabola on steroids. Sarcasm, my native tongue.

The class learned each of the formulas and also learned how to graph the formulas. Call me juvenile, but I still have trouble with the grade 12 version of the parabola. I think the 4p always throws me off. I also feel like using the other formula from grade 11 though. But using that idea again should be avoided since thinking would not be lateral. This is why practice makes perfect. I just need to do a little adapting. Otherwise, I’d think I’m ready for the test today. Just keep my hocus pocus my focus, look it’s a locus.

BOB: Conics

WOAAAHH!! I almost forgot to do my BOB again! Sorry this is really really late XD

Okay, this unit was one of my favorite units. I actually enjoyed graphing for this unit and I thought that the concepts were very simple. It was a good break from the Counting unit, which I didn't really like much. I also enjoyed doing all that folding during class, such easy homework! This unit was also full of laughs, we really couldn't seem to stay serious most of the time. As for the parts that I found the hardest for this unit, I would probably have to say that the hyperbola and ellipse were definitely the most difficult, although not by much. I got confused at first with how to find the foci and the asymptotes (hyperbola) but then it all made sense after some explaining. I also remember messing up graphing my vertical hyperbolas since I switched around the transverse and conjugate axis. All in all, tis was a good unit, now I got to get ready for school! Good luck on the test guys :P

Wednesday, May 21, 2008

BOB on Conics

Hi guys this is Richard and here is my bob on the unit of conics.

This unit was not as hard as the other units there were some parts that i got really easy like the ellipse. I probably found that this topic was easiest because i had to scribe for it or it is just the simpler out of the conic shapes. Another thing that was cool was that we got to do a lot of fun folding. That led to the geometry of the conic shapes.

There were also topics that i found difficult for example the parabola. I think that the main reason i don't understand it is because i was not present for that Pre cal class. I almost forgot that the graphing of the parabola is also hard. i also felt the same was that Francis felt on that Pre test. I was also puzzled by the transforming parabola question . Well that's just me ah ha

Richard signing off.

Conics Bob

Ahhh Conics. Too bad Mr.K didn't show us any more of his samurai skills with this metre-stick-katana.

To the point. I will admit, like every section in this course, I had troubles in one way or another. The origami was cool, as it went with the whole Japanese theme Mr.K was trying to capture (TOYOTA-HONDA-MITSUBISHI-MAZDA-YOKOHAMA-SONY-etc...) Like every other type of graphing, I don't like to graph. Granted it does give me a better look at whats happening and I can tell what's happening, but like everything else in life, just because you know how to do it doesn't mean you're supposed to like it. Like graphing.

I liked how parabolas, circles, ellipses and hyperbolas had their distinct way to tell each other apart without having to think hard but rather just look at what's switching to make it graph that way. But, that's probably about it. The math was pretty simple once I took the time to sit down and study it all, like I am tonight.

I didn't like word problems however, how they make it seem like it's supposed to be something it's not, but yeah, I know, I'm just over complicating it.

I will be deep in study for a while because well, exams are closing in surprisingly fast. OH and don't forget about your DEV projects.

Rence ~ Out

BOB on Conics

This unit was quite short, but I'm not celebrating. Personally, I need a bit more time with this unit, only because I don't think we went far enough with transformations that can happen with each conic shape. On the pre-test I saw that one question where the graph of a hyperbola was stretched 4 units, and moved down 1 unit, I was sitting there staring at the page for like 5 minutes, forgot all about those transformations. Plus, I don't think I can remember all those equations such as the equation of a circle, ellipse, hyperbola and the new parabola equation. Hopefully we'll get a formula sheet, or some other type of aid. On top of that, I don't fully understand the total anatomy of all these conic shapes, such as the major, minor, semi-major, semi-minor, conjugate, and transverse axis', and some, I suppose. I still don't exactly know about graphing these guys either. Other than that, I really enjoyed folding the paper, being interactive in a class is always a plus.

Good luck on the test everyone.

Until next time,

Bobbing on Shwards and Things

So, the infamous conic sections unit was completed late last week, and this would be a bob for said unit.

Where to begin? the good? the bad? the ugly? Lets start with the ugly

THE UGLY *dun dun dun*

-GRAPHS! Yes I'm terrible with graphs but I am getting better at them, thanks to a certain "perspective." More on that later though.

The Bad

Not to much here this time, thankfully.

-One of the most difficult parts of this unit is how most of the equations, have another equation which is nearly identical. This made memorizing/remembering which was which difficult at times.
-Going from standard to general form, and remembering which is which was often a pain.

The Good

-Short unit
-Easily understood by used of "visual" tricks (what I meant in the ugly section. By thinking of what the graph of some equation looks like, the question is often made much easier.
-Overall the concepts in this unit, and the difficultly level of the questions was not very demanding.

So yeah, there is it, my small and not very impressive bob post. :P Now I'm back to studying haha.


An Introduction to Probability

Alright young fellers, american idol is almost on so I'll try to finish this before it starts but if I don't..I'm sure I can do both! (:

Slide One
The first thing Mr. K asked us today was, "What is the probability of the sky falling?" (chicken little yay!) Each of us were given about a few minutes to write down our answer and we all shared our answers to each other. *refer to the first slide* Most of us came up with different numbers but some of us had the same answer as others. Most of us just chose a random number, like me, but some of us also took some time to why they choose that number. After this dicussion, he then asked us what the numerator and the denominator represented. The numerator is the number of chances of the sky actually falling while the denominator is the set of all possible outcomes.

Slide Two
The next slide explains the defintion of an inderterminate form. So if we take any number such as 6 and multiply it by one then divide by one your answer would be 6. And if you take the 6 again and multiply it by 1/10 then divide by 1/10, you still get the answer 6. Continuing the pattern, you'd eventually reach zero. So you would multiply 6 by zero then divide by zero. Therfore it is in an inderterminate form. If you recall on slide one, 0/0 and 0/∞ are examples of indeterminates as well as ∞/∞ and ∞/0. 1/0 is not in an indeterminate form, it is however, an undefined value.

Slide Three
The next slide includes various vocabulary words which we should know for this unit as well as different examples to explain the meaning of each words.

Slide Four
Calculating the Probability of Event A
This slide includes the formula P(A) = n(A)/n
P(A) - Denotes the probablity of A
n(A) - number of favourable outcomes (number of accourences of A)
n - sample space (the total number of possible outcomes
Probability can not only be expressed as a fraction but also as a ratio, decimal and a percent. However we will not be looking at it as a ratio in this unit.
IMPORTANT TO REMEMBER: Probability of any event is always a number between 0 or 1. This means that if your answer is bigger than one or it is a negative, then your answer must be wrong.

Slide Five
The fifth slide is also a continuation of the different vocabulary. In this specific slide we dicussed mainly about the complimentary event. The example he asked was, if there was a sudden death single tennis tournament and there were 54 players, how many games would be played if there could only be one winner? The easiest way of coming up with the answer is asking yourself, "how many losers would there be" The answer is 53 because only one player could win. Another example is using a coin or a die. Basically, you're looking for the opposite. So if P(E) = a then P(E1) = 1-a.

Slide Six
In the next slide, we had to determine the sample space when a fair die is rolled once. Since it is possible to roll either 1, 2, 3, 4, 5 or 6, the answer to the question is {1, 2, 3, 4, 5, 6} *note: if you are asked to determine the sample space you must use brace brackets { }*
The next question was to determine the sample space for rolling a six sided die and flipping a coin. You could either use a tree diagram or a chart to find the answer. In this case we first used the tree diagram to find the answer. The first two branches consited of Heads and Tails and each had 6 branches to indicate the numbers on the die. Each number could either get H1 to H6 or T1 to T6 so the answer was 12.

Slide 7
We then had to determine the probability of rolling a 2 when rolling a fair die. The answer is 1/6 because there's only once chance of getting a 2 out of 6 different outcomes. We write this as P(2) = 1/6
In the next question it asked to determine the probability of getting a head and an odd number when rolling a die and flipping a coin. In this case we used the previous chart we made on slide 6 to find the answer. We then found that there are 3 possible chances out of 12 possible outcomes which also redues to 1/4. *note: remember to always reduce your fractions.

I believe I covered most of what we learned and talked about in class, so that means..I'm finished! (: This scribe actually took longer than I clearly expected considering I was running back and forth to my room and my sister's to watch American Idol. But enough of my life..toodles everyone, American Idol isn't over yet!

BTW..Homework is on slides 8-10 and don't forget to study for tomorrow's test! Goodluck and Ciao! OH and.. Next scribe is... "what it do PAUL wall" (: