Friday, June 26, 2009
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!
Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)
Tuesday, June 16, 2009
I dreaded going to class everyday but I THINK I'm gonna miss it. It's weird that its all finally over. All the memories created through out the semester. As classes went by I think my interest in math sparked.
Thank you Mr.K for educating us so well in math, math wasn't so boring with you teaching us. Those interesting little facts and stories you always feed us. It was awesome how you got each student involved in someway. It is a shame that DMCI will be losing a teacher like you. Congratulations on your new job, take care and good luck in the future.
On the other note were you this type of student? I know I was.
New school semester:
At the first week:
At the second week:
Before the mid-term test:
During the mid-term test:
After the mid-term test:
Before the final exam:
Once know the final exam schedule:
7 days before final exam:
6 days before final exam:
5 days before final exam:
4 days before final exam:
3 days before final exam:
2 days before final exam:
1 day before final exam:
A night before final exam:
1 hour before final exam:
During the final exam:
Once walk out from the exam hall:
After the final exam, during the holiday:
Isn't that cute, I just thought I'd share.
Thursday, June 11, 2009
I will truly miss this class. A whole whack of memories have been created in such a short amount of time that we will all surely remember throughout the course of our lives. In all honesty, I thought this class would be boring at first, but man was I wrong. Each day presented a new challenge Even if some of it was review, things were somehow twisted to create a brand new challenge. I swear, after this class, I will never look at life the same way again. As I go home, or even when I'm out running, I always find myself looking at the math around us all of a sudden. It's as if a whole new light has suddenly been shone on the Earth. One time me and Dion even stopped to count the branches on a tree to see if the way they grew followed the Fibonacci Sequence (and to our surprise, they did). A lot has been taught in this class, not only about math, but about life and other subjects as well.
About my progress throughout the class, all I can say is that I did the best that I could and that's all that I wanted to do. The mark had some importance to me as it made me want to work harder to raise it up but overall, I'd say it was the experience that suddenly became important to me.
Now that I've done a few older exams for practice, I think I have found my strengths and weaknesses in this course. I would have to say my biggest strength is Identities. I think it was because it came really easy to me when we learned it and when I saw how it could take such complicated looking things and turn them into 1 term answers, I was amazed. It made me even more interested in the subject. The hardest thing would've had to have been Combinatorics. Who knew counting could be so difficult?! For me, the most difficult part is probably the Binomial Theorem. It was a hard concept for me to grasp right away, but I think I get it now. Hopefully I will have known it enough for the exam.
Am I worried about the exam? Not too much. I've graded myself on the previous exams I've done and I've gotten pretty consistent marks that I'm proud of. Hopefully I do really well on the exam. I will try to aim for that 100%.
I guess I should end this by saying a few things.
Thank you so much to the rest of the pc40sw09 class. You have all helped to make this semester a lot less stressful. Math was kept fun because we weren't afraid to speak up. The heat of "competition" was kept burning by a small handful (you know who you are) and it made things a whole lot more interesting. We shared each other's struggles and celebrated each other's triumphs. We were always there to help each other out where it was needed. We were like one big happy math family. I'll always cherish the memories from this class.
Thank you to Mr. K. Math would be so boring if it weren't for people like you. So passionate and so energetic. It makes us kind of want to go to math class. I guess the coffee would explain all the energy, but passion, that comes from straight the heart. So many great memories were made in this class and a lot of them involve you like
- the sine dance
- Pi Day
- your block of wood
- that "ninja" who came to visit
- your cool tricks on the smart board
- the snapping thing you taught us
- that quadratic equation song
- and a whole lot more!
It's such a shame that we must already say our goodbyes after such a short amount of time, right when I was looking forward to AP Calculus, but it's destiny. Even when you leave, remember that you have left your mark at DMCI and you've made a great impact on the lives of many of the people whom you've taught. For that, we all salute you. Congratulations on your new job and good luck.
And lastly, to my D.E.V. partners Mary and Dion. I know this has been said before, but you two really made this semester a lot less stressful, especially when it came down to making our D.E.V. A potentially (non-intentional) stressful project was made really fun because I had the great honor of working with you two. If it had been anyone else, our project wouldn't be the same. Thank you so much for making this experience everything that it was.
It was a great honor and a pleasure to be in a class filled with such brilliant minds. I wish you all the best in your future studies and that you continue to shine. Remember that we'll always have this blog to keep us connected. Good luck to everyone on our exam tomorrow. Everything we have done in this unit has been leading up to this point. Let's knock this one out of the park!
For one last time, peace out!
Wednesday, June 10, 2009
Since the exam is tomorrow and today is our last day i wanted to make sure i post a day before the exam just like how we would usually blog on test. The only exception is that this is gonna be the biggest test were gonna have.
In the classes I've spent here I've learned quite a lot of things. Besides math of course we were taught some pretty cool stuff too. For all of you guys who are reading i encourage you to take at least 5-10min from your busy schedule to post up our last blog reflecting this semester.
Monday, June 8, 2009
Saturday, June 6, 2009
the correct formulas are now and
If you look the june 5 slides, pg 6 and 7 shows how we get to these formulas (sorry, my print srn doesn't really work)
On friday, the main thing we learned was solving infinite geometric series!...here's the deal...basically, any fraction that is below 1 and over 0, to the power of infinity will be equal to zero...of course it's not REALLY zero, but our calculators don't have the potential to hold a number less than google, and there is no exact value for it since infinity is an idea and not a number...since it's really ultimately infinitely close to zero, we make it equal zero, making our lives easier. Look at the June 5th slides, page 13 for an example.
...to be honest i don't really know what else to add...i'm a little preoccupied thinking finishing up my final project and worrying/studying for the exam...if anyone needs more clarification, just leave a comment within the next week and i'll fix this up some more...
Friday, June 5, 2009
Thursday, June 4, 2009
When solving questions like these, always try to go back to the basics. Probability is the number of favorable outcomes over the possible outcomes. From that, we have something to work with.
Finding the possible outcomes, or Sample Space.
- The question states that "4 men and 4 women" would be chosen. This means that out of the 7 men, 4 will be chosen (7C4), and out of 10 women, 4 will be choose (10C4).
- Since the question said that, "Allen and Bridget will be among these 8 chosen people", we already know that they are part of the possible outcomes. They are represented by the green and blue "1"'s.
- Out of the 7 men, only one has been chosen, (Allen). This leaves us with 6 more men, and 3 more spots for men, 6C3.
- Out of the 10 women, only Bridget has been chosen. This leaves us with 9 more women, and 3 more spots for women, 9C3.
Today's class was focused on:
- Arithmetic and geometric sequences
- Arithmetic and geometric series.
- Sigma Notation
An arithmetic sequence is a sequence (ordered list of numbers), where a fixed number(common difference) is found between two consecutive terms. (negative numbers are added too! )
This means each term is going up or down by the same number.
When wanting to find the nth term in an arithmetic sequence, refer to the equation below. (use Carl Friedrich Gauss's 7year old story to help you remember the equation.)
Geometric sequences are like arithmetic sequence, but instead of adding its multiplying. This means instead of a common difference, there's a common ratio.
When finding the nth term in a geometric sequence, refer to...
Series is defined as, the sum of terms in a sequence. (Sn, where S reads as "sum of" and n would be the rank of the nth term. ex. S4 = sum of the first 4 terms.)
If we were to have a sequence of 1,2,3,4,5,6,7 etc, the series would consist of 1,3,6,10,15. Why? Well the first term is a given, 1. The second term would be 3, because that was the sum of the first and second term from the sequence, (1+2). The third term is the sum of the first, second, and third terms from the sequence. (1+2+3). The orange circled numbers in the above picture are the ranks.
Arithmetic series is the sum of numbers in an arithmetic sequence. This would be helpful if you are asked to find the "sum of integers from 1 to 5000" for example. Where n would be 5000, because there are 5000 terms, a would be 1, because that is the value of the first term, and d would be 1, because that's the common difference. (numbers are going up by one)
What you know:
- From the sequence of multiples of 7 between 1-5000, first term is 1.
- Last term is 4998
- common difference is 7.
- number of terms within that sequence.
Now don't get too carried away with questions a and b. These kind of questions wont be asked on the exam, but you'll need to know the methods of a&b in order to solve c. (c = a question likely asked on the exam).
Logic:( The sum of all integers 1-5000) - (sum of all multiples of 7) = sum of all integers not multiples of 7. This "build up" to a question, is called scaffolding.
Sum of numbers in a geometric sequence.
Is the shorthand way of writing a series, also known as the weird looking "E". Sigma is really confusing, if you don't know how to read it. The n=1 tells you the value of the first term, which is 1. The 4 on top of the sigma is nth number of term to stop at. The (2n-3) is the "rule" or equation you follow.
bye guys! good luck on the exams and your DEVs!
The next scribe is jonno!
Wednesday, June 3, 2009
Okay well apparently today was mostly a review from grade 10 (if you can remember that far back.. good job!)
On the first slide there were four sequences...
4, 7, 10, 13, __, __, __
3, 6, 12, 24, __, __, __
32, 16, 8, 4, __, __, __
1, 1, 2, 3, 5, 8, 13, __, __, __,
Mr. K wanted us to fill in the blanks and explain how we found the missing terms...
Now let's take a closer look and try to identify some patterns that can be applied to any other sequence we might encounter in the future.
4, 7, 10, 13, 16, 19, 22
Note: If you are asked to find the 37th term and you plan on adding 3's, you must add 36 threes.How do we get 3n + 1 ?
Well since we found the y-intercept we can graph this.
By looking at this graph we can see that the slope is 3. So for future reference remember that our slope is the constant. So let's make an equation for this sequence.
Where 3 is the slope and 1 is the y-intercept.
Everytime we have an arithmetic sequence it will be a linear function.
This is the graph of our sequence. Here is how you would make that on your calculator! Hit stat, edit. Under L1 enter your rank (1-7) under L2 enter your values (4, 7, 10, 13, 16, 19, 22) Now hit 2nd stat plot (top left corner of the calculator). Hit enter and make sure plot 1 is on, under type make sure the dots are selected. Then hit graph.
Next we covered some definitions:
Recursive: Repeats again and again.
Implicit Definition: This is the teenage way of saying hi, it's an implied hello. Only clear to people who know what they're looking for.
The Common Difference (d): The number that is repeatedly added to successive terms in an arithmetic sequence.
Common Ratio: The number that isrepeatedly multiplied to successive terms in a geometric sequence.
How to find the nth term in an arithmetic sequence.
Where tn is the nth term
a is the first term
n is the rank of the nth term in the sequence
d is the common difference
How to find the nth term in an geometric sequence.
Where tn is the nth term
a is the first term
n is the rank of the nth term in the sequence
r is the common ratio
Here is the next sequence we looked at, 11, 5, -1, -7...
We were asked to find the 51 term.
Look at the difference between the terms.
Therefore we know our constant is -6. Remember we're not subtracting 6 from the first term to get the second term, we are adding -6! :D
Okay so how to find the 51 term. First let's make a formula.
tn= a+ (n-1)-d
Where n is the term. a is the first value and d is the difference. So we have..
t51= 11+ (51-1)(-6)
Next we looked at the sequence 3, 6, 12, 24, 48, 96, 192
Notice the difference is not constant, so this is not an arithmetic sequence.
To make an equation for this kind of sequence we use the formula, tn= ar^(n-1)
a= the first term r= ratio
The ratio for this sequence is 2 because 6/3=2 and 12/6=2 etc.
So the implicit definition is tn= 3(2)^(n-1)
Next question! 32, 16, 8, 4, 2, 1, 1/2
We multiply by 1/2 to get the next term so 1/2 is our ratio. Now we find the 10 term.
If differences in sequences are different it is NOT arithmetic.
If there are common ratios it is geometric.
And the last slide..
Um next scribe is Mary.
Homework is exercise 44 and 45.
Hi guys! This is Pacifico, the scribe for today!
Today, we delved deeper into the cancer question, and other things; but mainly the cancer question. We also learned a new formula! And we also get to know another way to solve a certain probability question. Things to remember first:
v Don’t forget to accept the invite to the DEV blog!
v GET ENOUGH SLEEP FOR THE EXAM!
v STRESS IMPEDES LEARNING!
v In A|B, | is a shortcut for the phrase, “…given that we already know…”
So, let’s get this scribe started!
For this part, I will only talk more about Q. #4 because Q. #1-3 is a no-brainer for us. Q. #4 starts the semi-complicated things, notably c).
In this slide, it tells us that the probability that person who tests positive for cancer has cancer. Mr. K. used the formula P(C|P) = (4900 ÷ 2400) = 19.76%
P(C|P) is said as: probability of people that have cancer given that they’re already positive.
How did the heck did he get that, you ask?
C = has cancer
H = doesn’t have cancer
P = tested positive
N = tested negative
P(C|P) = P(CP) ÷ [P(CP) + P(HP)] or
Probability of people that have cancer given that they’re already positive is equal to the probability of people that have cancer that is tested positive divided by the sum of the probability of the people that don’t have cancer but is tested positive.
P(C|P) = ( 0.005 × 0.98) ÷ [(0.005 × 0.98) + (0.995 × 0.02)] = 0.1976
So the answer to the question: What is the probability that a person who tests for positive for cancer has cancer is 19.76%
This slide is asking for P(H|N). By reading the question, we know that it’s the same as the last question. So we apply the knowledge that we know already and solve the question!
P(H|N) = P(HN) ÷ [P(HN) + P(CN)]
Answer using this formula!
The next slide is also the same question as the last 2, so I’m pretty sure that you can figure this out on your own (unless this scribe is sucky—which by the way you have to tell me so I can change it!)
Ah… This question…
To find the probability that the couple has at least 2 girls assuming that having a boy or a girl is equally likely events…
We used permutations, from the unit Combinatorics, to find the number of ways of getting 2 girls.
You could use a tree diagram to find the answer, like what is shown below, but that’ll take a long time and it’s going to waste space.
GGBB à 4 “slots” (number of letters) and 2 non-distinguishable object that both have 2 same things.
number of ways to get 2 girls = 4! ÷ (2!2!) = 6
And then, we figure out the probability of getting 2 girls and 2 boys.
P(GGBB) = (½)(½)(½)(½) = 1/16
After that, we multiply the P(GGBB) and the number of ways to get 2 girls.
P(2G) = 6 × (1/16) = 6/16 = 3/8
To compress all that: P(2G) = [4! ÷ (2!2!)] × [(½)(½)(½)(½)] = 3/8
The original question was, “find the probability that the couple has AT LEAST 2 girls.” Therefore, we also calculate the P(3G) and P(4G), which is shown in this slide.
Then to get the final answer, we add P(2G), P(3G) and P(4G) together.
There’s another easier way to find the answer!Use the COMPLIMENT PROBABILITY
P(at least 2 girls) = 1 – [P(1G) + P(0G)]
= 1 – [(1/2)^4 × (4+1)]
= 11/16 <-- answer is the same as adding P(2G), P(3G) and P(4G)
That concludes my scribe post…
Next scribe is… REBECCA! Hurr durr…
Tuesday, June 2, 2009
Monday, June 1, 2009
Today we learned about mutually exclusive events and non-mutual exclusive events. A mutual exclusive event is when "A" occurs, it is impossible for "B" to occur. Basically means you get one or the other, never possible to get them together. It's like the question that Mr K put on about Chad: 1/3 chance he's in the lounge or a 2/9 chance he's in the library. Since i don't think there are 2 Chads' and he doesn't have some sort of magical powers, I'm pretty sure he's either in the lounge, in the library or in neither places. So, that's mutually exclusive events. The diagram that would represent it would be:
This expresses the mutual exclusive events because either one occurs in circle "A" or circle "B".
Then we learned about non-mutual exclusive events. That basically means when "A" occurs there's a possibility that "B" will occur, in layman terms, if one thing happens the other thing COULD also happen. An example would be if you have a deck of cards and you want to draw either a diamond or an ace. There's one card that would fall into the non-mutually exclusive category: the ace of diamonds. So that part of it would be non-mutually exclusive.
This picture represents a non-mutual exclusive event because one can occur in "A", also in "B" or it could occur in the intersection "AB".
The formula to calculate the mutual exclusive events and non-mutual exclusive events are the same, which is:
So basally in a mutual exclusive event you would just add the probability of "A" and the probability of "B" and subtract zero because it's impossible for the two things to occur together. While in a non mutual exclusive event, you have to minus the probability of "AB" because you would account an event twice. Kind of confusing right? Here's an example: the card one: whats the probability of getting a ace or diamonds? Well you would have 4/52 because you have 4 aces in a deck and 52 cards in a deck. For diamonds you would have 13/52 because you have 13 diamonds and 52 cards, BUT you have accounts for the ace of diamonds TWICE. Once when you considered it an ace and once when you considered it a diamond, but they're the same card so you can't do that. That's why you subtract the probability of "AB, which would be 1 since you can only have 1 card which is both an ace and a diamond. So you end up with 16/52.
Uhh, well that's basically it and apparently we're wrapping this unit up tomorrow (if I remember correctly). Soo, yeah good luck with life and next scribe is ... lets' go with Pacifico.
All right, so Conics is one crazy unit. Started out learning about the ancient Japanese art from Samurai Kuro Pat Wa and all the different shapes you can form from cutting a cone. A cone is basically if you took a line and just spun it around. From here you can cut out 4 shapes. You can cut out a circle, an ellipse, a parabola, and a hyperbola. Each one is different.
Some commonalities include:
- Focal Radii
- Locuses (Loci?) of points
Equations to remember~
- (x-h)^2=4p(y-k) (vertical parabola)
- (y-k)^2=4p(x-h) (horizontal parabola)
- p=distance from vertex to directrix or a focal point
- ((x-h)^2/a^2)+ ((y-k)^2/b^2)=1 *horizontal*
- ((y-k)^2/a^2]+[(x-h)^2/b^2)=1 *vertical*
- a=semimajor axis
- b=semiminor axis
- c= distance from center to foci
- ((x-h)^2/a^2)-((y-k)^2/b^2)=1 *horizontal*
- ((y-k)^2/a^2)-((x-h)^2/b^2)=1 *vertical*
- a= transverse axis
- b= conjugate axis
- c= distance to foci
- Have had some trouble in identifying vertical/horizontal in some instances.
- Sometimes get confused between the ellipse and hyperbolic equations.
- Issues with completing the square.
But I have done some looking over things this weekend and, all in all, this un it is not extremely difficult and so hopefully I will do quite well on it. Next period here I come!
~ Pokemon Champion