Saturday, March 25, 2006 

A del.icio.us idea ...

We talked about this in class on Thursday. Students often find more, and better, sites than I do. You're better websurfers than I am. ;-) That got me thinking .....

I spend a lot of time looking for good websites that help us learn in this class. But what if we all spent a little time doing that? What if there was an easy way for us to both save our bookmarks (without cluttering up our favourites list) and share them with the whole class with the click of a single button? And what if we could access those bookmarks not just from home, but from any computer in the world? Hmmm .....

Well, there is an easy way to do that! Instead of saving bookmarks on your home computer sign up for a free account at a site called del.icio.us. You can then access them from any computer in the world. You can easily install a little button/bookmark that allows you to save any webpage you're looking at without interupting your surfing. Now we can all make recommendations to each other learning resources with the click of a single button in our del.icio.us accounts! Tag it using this tag:

am40s

You should also tag each entry with several other words that indicate what it is about; things like: trig graphing stats circles etc. Also, include a brief descriptive clipping from the site (or type one in yourself).

As soon as someone starts saving links I'll add a del.icio.us box to the bottom of the sidebar of our blog. It will show the 10 most recently saved links automatically as you find them. There is will also be a link to the entire archive that you can browse at your leisure.

You can read this tutorial on how to get started with del.icio.us. You might also be interested in watching this screencast that illustrates just how powerful this web tool is.

Remember, this is part of your homework for Spring Break. You must save at least one link for each unit we have studied so far. Try not to post a link that someone else has already found. You can see what's already been posted by looking in our archive.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Friday, March 24, 2006 

scribe

One class today, last one before Spring Break!
We did practice problems on the board:

The weights of babies born at HSC average 8 lbs. 1 oz. (There are 16 oz in 1 lbs.) with a standard deviation of 12 oz.


a) Find the percentage of babies with a birth weight between 7 lbs. and 9 lbs.


To find the mean:

We convert the 1 oz. to lbs. : 1/16=.0625 So the mean is 8.0625 lbs. and we store it in M.

To find the standard deviation:
Again convert to lbs. : 12/16=.75 So the standard deviation is .75 lbs. and we store it in S.

Now we use our calculator:
ShadeNorm(7,9,M,S) = 81.6% of babies are between 7 lbs. and 9 lbs.
*normalcdf(7,9,M,S) does the same thing but just doesn't show the graph*
*It has to be in this order - low, high, mean, standard deviation*


b)Find the weight, w, such that the percentage of babies with a birth weight greater than w is 60%.

Since it's using a percentage, we use InvNorm. This means if we use the inverse of 60%, which is 40% we'll get our answer.
On the Calculator: InvNorm(.4,M,S) = 7.87 lbs.

c)Find the weight, w, such that the percentage of babies with a birth weight less than w is 25%. In this question we also use InvNorm but since it wants 25% and less, we just use .25. On the Calculator: InvNorm(.25,M,S) = 7.55 lbs.


A College aptitude test is scaled so tha
t its scores approximate a normal distribution with a mean of 500 and a standard deviation of 100.

a)Find the probability that a student selected at random will score 800 or more points.

Mr. K showed us another way to figure this out:

On the calculater we adjusted the windows:
Xmin=500-5*100
Xmax=500+5*100

Xscl=100

Ymin=-.005
Ymax=.5

*You have to play with the y min and max to get a good graph*


Then after adjusting the windows we go ShadeNorm(800,1000,500,100) and we get .135%

You also get the same answer if you get the z-score and go ShadeNorm(z,5)


b)Find the score, x, such that 76% of the students have a score: i) less than x ii) more than x


i) less than x

Again using inverse norm.
On calculator: InvNorm(.76) and store in z.
We need to find x, so our formula is :
z * σ + µ
=z*100+500 = 570.6


ii)greater than x
Now we use the inverse of 76% which is 24%.

On the calculator: InvNorm(.24) and store in z

same formula as above,
z * σ + µ = z*100+500 = 429.4

After these problems Mr. K started to show us binompdf.
So our example was fliping 10 coins:
First off you edit your list so in list one it goes from 0 - 10.
Then put up binompdf(10,.5) - 10 is the number of tries, .5 is the probability, and store in L2.
Now look at your list and it'll show you the probability of getting 0 heads, 1 heads, 2 heads, etc. to 10.
To graph this we turn our stat plot on, choose the histogram, xlist: L1, and Freq: L2.
Window settings:
Xmin=0
Xmax=11
Xscl=1
Ymin=-.1
Ymax=.3
Yscl=.1

And that was all the time we had for, he'll explain all the things we can do with this stuff when we come back.

Homework for spring break:
-make a delicious account
-make sure ALL homework assignments are completed, all the way to chapter 3.4
-p.345 review the utlilities about the calculator numbers 25-30

That's it I think, have a nice spring break!

Andrew for the next scribe.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

staistics scribe

Hello everyone! this is rubie...sorry i posted this scribel late...i was working and then i had to run to a friends house and use their computer so yeah please bare with me! Anyways today in class we started off with a quiz on statistics...then Mr. K gave us a couple of questions that we had to work on...here they are:

QUESTIONS: 1) Find the values of z and x :


2) To 3 demical places, find the area under the standard normal curve for each of the following (include a sketch with each answer)


3) The beanstalk club has a minimum height requirement of 5' 10". Women in north america have a mean height 5' 5.5" and the standard deviation of 2.5". what percentage of women are eligable ( include a sketch in your answer)

4)Find the probability that a radomly selected person has an IQ between 108 and 130, if the population is normally distributed with a mean of 100 and standard deviation of 16. include a sketch with your answer.

*I dont think i have the right answers to these questions, so ill leave it at that! and im so sorry that this scribe of mine is really cheap and boring.......hope you guys aren't dissapointed!
For second Period we had a group assignment that we worked on in class.

*DONT FORGET TO PREPARE YOURSELVES FOR THE HOMEWORK QUIZ!* AND AGAIN IM VERY SORRY!




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Wednesday, March 22, 2006 

HOMEWORK

I'm sorry forgot to post it...with my ealier post... ummm HOMEWORK AT Page 123 (1-9)??




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

The Scribe: " Need to catch up?, then read my post =D "


Rubi is not here today, so i volunteered to be the scribe....

If you want to know what happened today CLICK HERE*
AND we took some notes TO PUT IN OUR MATH DICTIONARIES, if you want to see it CLICK HERE*

THE NEXT SCRIBE WILL BE RUBI... if she's not in class again the next scribe will be I HAVE NO IDEA.... i'll just pick the guy named ALLAN.








Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Tuesday, March 21, 2006 

7 8 6

hello every one. i am today's scribe we had two classes today. First class we worked with some calculator functions.
and i will start my scribe with Some calculator functions. And second Class was about international day of elimination for all kind of Racial Discrimination.

Some calculator Functions you should know
Sort A(List Name): Sort A Specific list in Acending order
Access -[stat] + [2]
Sum( List Name): Sum all the values in a list.
Access -[2nd] [stat] [>] [>] [5] - (Math)
Sum(List Name,low, high): Sums all the Values in a list from a specified longer value to a higher value.
Access: Same as above.

1-Var spots(List Name):
Calculates Statistics for the value in a specified list.

1-Var Stats(Values list Name, Frequencies list Name):
Calculates Statistics for values entered in 2 lists as a frequency table.
Access: [stat] [>] [1]
To access (and perhaps store) Statistics with "1-var stats" Command
_
x Acess: [vars] [5] [2]
SX Acess: [Vars] [5] [3]
_
OX Acess: [Vars] [5] [4]
_________________________
Home Work
Mr.Kuropatwa asked us to sign up for a delicious account on the following web.
del.icio.us
I won,t go in details but here is a short guidience on how to sign up.
-As you enter the web there is a box on the right asking you for user name, password and email Id.
After you get signed up the next thing you have to do is to enter the code from the box in the space below it.
Congratulations You are Signed Up
- It is strongly recommended that you should use Fire Fox by Mr.Kuropatwa simply by clicking on the following link and download a free Fire Fox.
phttp://www.mozilla.com/firefox/

LONG TERM HOMEWORK
This Spring you have to Find at least on link for each of the following topics.
1)Metrices
2)Probabilities
3)Statistics

- The purpose of this assignment is to help you understand things that you have problem with
by searching through google etc.and share it with other students in your class by posting it on the post box which is not yet uploaded on our blog.(but it will be uploaded soon)

I hope i provided you with enough information.if u have any questions concerns and bla bla... ask me in class tomorrow.

Rubi* is tomorrows Scribe
PEACE




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

scribe

The scribe will be on soon....and for those of you who have nothing to do i would invite them to the following website to Join Us In The international day of elimination for all kind of Racial Discrimination.

http://www.unesco.org/shs/eng/dayagainstracism.shtml


PEACE




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Monday, March 20, 2006 

DISTRIBUTIONS

Hi sorry for the delay cause i have work tonight but here is my scribe......we did not do much discussion just a bunch of note so here it is......

Type of Distributions
1. Uniform Distribution- date may be discrete or continous. Every outcome in the
experiment is equally likely.
Example: graph the distribution that shows what can happen when a 6-sided die is
trown.
2. A Normal Distributios- Data is continous ( height, weight, time, etc) when
certain experiments are carried out many, many, many times
the probability graph of the data tend to be "bell shaped" this is
known as the "Normal Curve"
3. A Binomial distribution- data is discrete (# of head when ten coins are tossed, #
of spades in a 13 card hand , etc) When a binomial experiment is conducted
many, many, many times a portion of the related histogram approches the
shape of the
normal curve.

The Standard Normal curve - the standard normal curve is used to make comparisons
between different normal distributions. This is doe using Z-scoreswhich
allows us to find a particular value , X , will lie on the standard normal
curve, This is what it meant by "standardizing" the scores for a
partuicular normal distribution.

Properties of a normal distributions

  • Each value of mean and standard deviation determines a different normal distributions
  • all normal distributions are symetrical about the mean
  • 99.7% of all the data lies within the 3rd standard deviation on the mean
  • the area under the curve always equals one
  • the x-axis is an asymptote for the curve

The 68-95-99 rule

Generally speaking, approximately 68% of all the data in a normal distribution lie within the 1st standard deviation of the mean, 95% of all the data lie within 2nd standard deviation of the mean, and 99.7% of all the data lie within the 3rd standard deviation of the mean

Properties of a binomial Distributions

  • (n) is the number of trials
  • (p) is the probability of success
  • (q) is the probability of failure
  • every binomial distribution has exactly 2 outcomes
  • u=pn and standard deviation is equal the square root of npq
  • there is different binomial distribution for each value of p and n
  • P(x) is the probability of x successes in every binomial distribution (zero is equal or less than P(x) and P(x) is less than or equal to one)
  • the sum of all the probabilities, P(n), equals one
  • For a sufficiently large number of trials, (n), any binomial distribution will approach the shape of a normal distribution

Sorry guys i dont have any visual images of the graphs because I try to make it but i can't upload and i dont have time now i feel so sleepy right now so here is all i can do but next time i will make it looks better ........The next scribe will be Abdul....




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

The Last Three Scribes ....

... are simply fantastic!

Shane, Corrie and Steph you three have raised scribing to the level of a new art form! Really.

Shane begins organizing the data in a well spaced table. Corrie moves it forward by using underlining, centering and spacing to make things clear and easy to read. And finally, Steph publishes a well formated table and a beautiful graphic image. When she has trouble drawing the image exactly the way she would like (showing the asymptote) she includes explanatory text.

Ladies and gentlemen; the bar has been raised and the gauntlet has been thrown down. Now the challenge falls to you to raise this art form to yet another new level.

Y'know, you guys should seriously think about writing a textbook. ;-)




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Sunday, March 19, 2006 

Box Up Sunday!

This is a clever little game. You've got to get the small blue box inside the large red box. You can only push a box from the inside. The black boxes, if used cleverly, can help you get the blue box inside the red one. But sometimes they're just in the way. I made it to level 4 pretty quickly, but then it starts getting tough. How far can you go? ;-)

Have fun with this!




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Thursday, March 16, 2006 

Mean and Standard Deviation

Hi everyone. We only had one period of math. Today we went over normal distribution; such that 68% of the values are within 1 standard deviation away from the mean, about 95% of the values are within two standard deviations and about 99.7% are within 3 standard deviations.





That is what the 'curve' should look like. Although, the ends of the curve probably shouldn't be touching the bottom like that..I had a hard time drawing it so I hope you know what I meant to leave out.

There were two questions we did on the board. Here's the first one:
This list shows the lengths in millimeters, of 52 arrowheads.

           16  16  17  17  18  18  18  18  19  20  20  21  21
21 22 22 22 23 23 23 24 24 25 25 25 26
26 26 26 27 27 27 27 27 28 28 28 28 29
30 30 30 30 30 30 31 33 33 34 35 39 40


a) Calculate the mean and standard deviation.
b) Determine the lengths of arrowheads, one st. dev. below and one st. dev. above the mean.
c) How many arrowheads are within one st. dev. of the mean?
d) What percent of the arrowhead are within one st. dev. of the mean length?

Solutions:
a) First of all you put the data given into one list. Then go to [stat], arrow over to [calc], and choose 1: [1-Var Stats], [enter] then 2nd function L1. You can get the mean and standard deviation from that information.
x¯ = 25.4423 σ = 5.6207

b) Now before you calculate one st. dev. below and above the mean, it's best to store those numbers first. To do this you go to [vars], then choose 5:[statistics]. Option 2 represents the mean and option 4 represents standard deviation. To store mean; option 2, store it and set it to [alpha] M. To store standard deviation; option 4, store it and set it to [alpha] S.
One st. dev. below : x¯ - σ = 19.8216
One st. dev. above : x¯ + σ = 31.0631

c) Just count the numbers between 19.8216 and 31.0631.
37

d) 37/52 = .7115
So the percentage is 71.15%

We also repeated (b) and (d) for 2 st. dev. below and above the mean, as well as 3 st. dev. below and above the mean.
2 st. dev:
Below; x¯- 2σ = 14.2008 Above; x¯+ 2σ = 36.6838 %; 50/52 = 96.15%
3 st. dev:
Below; x¯- 3σ = 8.5801 Above; x¯+ 3σ = 42.3046 %; 52/52 = 100%

Here's the second question:
The table shows the weights (in pounds) of newborn infants.

Weight Interval          Mean of Interval          # of Newborn Infants
3.5 - 4.5 4 4
4.5 - 5.5 5 11
5.5 - 6.5 6 1
6.5 - 7.5 7 33
7.5 - 8.5 8 29
8.5 - 9.5 9 17
9.5 - 10.5 10 8
10.5 - 11.5 11 4
Total 125


The same 4 questions are asked. The only difference is you put the Mean of Interval list into L1 and # of Newborn Infants into L2. From there on out, it's basically the same procedure. The only difference I can think of is when you go to [1-Var stats], press enter, L1, then comma and then L2.
So here are the answers:
x¯ = 7.4 σ = 1.5798
1 st. dev:
Below; x¯- σ = 5.8201 Above; x¯ + σ = 8.9798 %; 81/125 = 64.8%
2 st. dev:
Below; x¯- 2σ = 4.2402 Above; x¯+ 2σ = 10.5597 %; 117/125 = 93.6%
3 st. dev:
Below; x¯- 3σ = 2.6603 Above; x¯+ 3σ = 12.1396 %; 125/125 = 100%

That was it for that class =) The homework for this weekend is on page 116, questions 1-9. The scribe for Monday will be: Patrick.









Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Wednesday, March 15, 2006 

Statistics

Hey! Sorry that I only gotten this posted up now. Ah, I know its late already too. So today we had two math classes and in the first math class we started writing our notes on statistics in our dictionaries. So for anyone who missed the day here it is!

Statistics

statistics: the branch of mathematics that deals with collecting, organizing, displaying, and analyzing.

statistic: a number that describes one aspect of a group of data.
EXAMPLE: mean, median, mode, range, standard deviation, etc...

datum: one bit (piece) of information.
data: many bits (pieces) of information.


the "flavors" of data
guantitative data and qualitative date

quantitative data

data that is numeric
(eg. height, weight, time..)

there are two kinds of quantitative date
continuous and discrete

continuous - can be represented using real numbers (eg. height, weight, time, etc..)

discrete - can be represented by using ONLY intergers (eg. # of people, # of cars, # of animals, etc..)

qualitative date

data that is non-numeric
(eg. colours, flavours, etc...)


measures of central tendency

mean: ( A.K.A. 'the arithmetic mean") the symbol of mean is an x with a line RIGHT on top. ( i couldnt find it on character map so this will have to do x¯, sorry! )
- the arithmetic average oof a set of values.

x¯ = Σx/n


where the x¯ with the line RIGHT on top is the mean
where Σx means the sumer of all data (x) in the set (Σ - is called sigma)
where n is the number of data in a set

EXAMPLE: find the average mark this set of 5 quizzes.
48,52,65,45,65.

Σx/n = 48+52+65+45+65/5
=275/5
=55

median ( med) : 1) the middle value in an ordered ( say from smallest to largest) set of data.
2) if there are an even number of data, the median is the average of the middle pair in an ordered set of data.

EXAMPLE: find the median of these quiz scores:
12,10,17,11,15
SOLUTION: 10, 11, 12, 15, 17
12 is the median.

EXAMPLE: find the median of these scores:
12,10,17,11,15,11
SOLUTION: 10,11,11,12,15,17
(11+12)/2 =23/2
=11.5
the median is 11.5

mode (mo): the datum that occures most frequently in a set of data.

EXAMPLE: find the mode in the set of quiz scores:
12,10,17,11,15,11
SOLUTION: the mode is 11 because it occurs more often that any other number in the set.

measures of variability
determine how "spread out" or varied" a set of data is.

range: the difference between the largest and smallest value in a set of data.

EXAMPLE: find the range of ages of people in our class

17,17,18,17,16,17,18,18,18,17,17,18,18,17,18,19,18,19


highest value: 19
lowest value: 16
RANGE: 19-16=3

with teacher MR K.-38 yrs old.
highest value: 38
lowest value: 16
RANGE: 38-16=22

standard deviation: σ represents the "population standard deviation"
S represents the "sample standard deviation"

σ=√Σ(x-x¯)²/n


s=√Σ(x-x¯)²/(n-1)

for technical reason we won't consider in this course, the sample 'variance" √Σ(x-x¯)² is deviation is divided by one less than the total number of data (n).

the "standard deviation" is one measure of the average distance each datum is a set is from the mean.

EXAMPLE: find the standard deviation of ages of people in our class.
a) without including the teacher MR. K.
σ=0.7556
x¯=17.61

b) including MR. K.
σ=4.6117
x¯=18.68

...so yeah thats where we left off in the dictionaries, I think tomorrow we'll be continuing on it though. So hopefully my notes are some kind of help you any of you. Thats what we did in first class. In second class well, we kind of went off topic on some infinitie thing? I was lost on that one so I'm not so sure. But then afterwards Mr. K showed us what a bell shaped curve is...but unfortunally I had forgotten my text book at school so I ain't able to decribe it to you...so maybe the next scribe can fill you guys all in about it or maybe we'll get it in our dictionaries tomorrow, we'll see!

And just another remind if you never got to see the chatbox earlier, tomorrow's scribe will be Steph. Okay! welps, later!!!






Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Tuesday, March 14, 2006 

Averages And Probability

Due to the speed inwhich we were taught this lesson... and the pie distracting me ;) there may be some things I might have missed, so here it goes...

Today we learned some tricks for the TI-83's and it had to do with averages and probability (thus the name) so I'll try to recreate them as best as I can.

First we were given a list of 6 weights of potatoe sacks from 2 different farmers (6 each)

Farmer 1Farmer 2
4940
5160
4845
5255
4735
5365

To enter the numbers press:
[stat]->[enter]-> enter the data in the respective lists -> [2nd]->[mode]

When the numbers have been entered in the lists and all is complete we now can change the information into 1-Variable Statistics by pressing:
[stat]->[>](calc)->[1](1-var Stats)->[2nd]->[1](L1)->[enter]

The screen now displays the following info:
X=50 (average of list 1)
ƩX=300 (sum of all numbers on list)
ƩX2=15028 (all the numbers on the list squared then added up)
SX=2.3664... (sample deviation)
őX=2.16024... (standard deviation)
n=6 (number of numbers on list)
minX=47 (minimum)
Q1=48 (inbetween minimum and middle)
med=50 (middle)
Q3=52 (inbetween middlde and maximum)
maxX=53 (maximum)

Pressing the same keys as before but replacing 2 for 1 like so:
[stat]->[>](calc)->[1](1-var Stats)->[2nd]->[2](L2)->[enter]
will give the results above but for list 2
X=50
ƩX=300
ƩX2=15700
SX=11.83215957
őX=10.8012345
n=6
minX=35
Q1=40
med=50
Q3=60
maxX=65



Another thing we learned was how to put info into a probability graph
we used pennies as an example and thier weights.

f(frequency)Mass(g)
22.7
42.8
342.9
713.0
943.1
743.2
173.3
43.4

With this data now entered into the lists we press:
[stat]->[>]->[1]->[2nd]->[1]->[,]->[2nd]->[2]
This mulitplies the numbers together to create a result we wanted (to tell the probability of a penny of a certian weight happening)

X=3.087
ƩX=926.1
ƩX2=2863.35
SX=0.1223
őX=0.12219
n=300
minX=2.7
Q1=3
med=3.1
Q3=3.2
maxX=3.4

Hitting [stat]->[enter] will bring you to the lists and in the L3 columnat the top we will enter another list this time its the probabilities of the pennies, easily calculated by pressing: [2nd]->[2]->[/](divide)->[3][0][0] at the top of the column and the calc does the rest

To plot this as a probability graph press [2nd]->[Y=] turn on plot 1 change to the 3rd type of graph (bar graph) then change Freq to L3 by pressing [2nd][3]
then press [window] change settings to:
Xmin=2.7
Xmax=3.5
Xscl=0.1
Ymin=-0.1
Ymax=0.35
Yscl=0.05

Pressing [graph] will show the graph in a nice pretty easy to read form

To make it a Frequency graph make Freq L2 instead of L3,
Ymin=-25 and Ymax=100

And the graph will a frequency graph instead

So thats it for all we learned, and the homework was p.108 #1-9

Happy Pi Day and I'll find that coin so everybody should give up :P
Later,
Shane

Scribe Tomorrow: Corrie




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Seek and ye shall find ...



The Coin Hunt has officially begun as of 12:30 this afternoon. The race is on! Who will be the first to find the coin? Will the students find their coin before the teachers find theirs? Who will win the pizza party? Which charity will benefit from this year's hunt?

Check the walls of the building as you walk into school in the morning for hints to figuring out the puzzles.

Happy π Day!!
Have fun with it. ;-)




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Monday, March 13, 2006 

NEXT SCRIBE IS SHANE




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

scribe

the scribe for today is going to look short. statistics was the new title we learned. but it looks like continuation of probability. as we used probability to solve questions. we did a frequency table. but unfortunately i can't upload the file. i did the graph with microsoft excel and copied it to microsoft work. but it can't upload the doc. the second thing we did was a histogram sometimes called frequency distribution. kinds of data quantitative and qualitative. then they devide into decrete( one package) and continuous(various) . decrete bar graph and continuous histrogram. the difference between histogram and bar graph is that bar graph have spaces between trends and the data is decrete while histogram is continuous and no space between trends. i.e after a whole number for example: if the data is between 19.00-19.99 that is one trend after 19.99 the the next which is 20 will be on the next trend. we also learned probability distribution ( binomial distribution). probability distribution the sum always add up to one. binomial experiment there's always two way either success or fail. the other graph is uniform distribution i.e the data is the same in all trends. do #1-7 on page 102, 103 etc............. i am sorry if this notes is complicating. but as mr.k did not gave us any notes for this. i just wrote down what i understood about statistics.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Sunday, March 12, 2006 

Sunday Knight



How far can you go? Play here! ;-)




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Thursday, March 09, 2006 

Last Minute Probability Links

First off, see my earlier post.

Next, a few quizzes (5 questions each, refresh the page for 5 new questions) ....



And in the theatre of mathematics ...




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Blogging On Blogging

So far, probability has been nothing but trouble for me. This unit has been very confusing for me and I hardly understood anything. But I have to agree with Mohamed and say that solving problems using Pascal's Triangle and the fundamental principle of counting have been quite easy as they were simple and easy to understand. Same thing goes for tree diagrams. Compliments seemed difficult and so did the counting exercises stencil. Hopefully I will do well on the test. Well that is all I have to say and now I got to study.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

blogging on blogging

Well my thoughts on this unit that were just finishing, probability is that is it is a very unique unit. I say this because there is no real formulas to it. Well there kinda is but main thing I find you need to know about is understanding the question asked. I'm not sure if any of that made sence but thats my take on the unit. I think I'll do ok on the test, i just gotta stay sharp on the test and not write down any stupid anwsers. Good luck on the test everyone.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Scribe post.. or is it?

Our first class today consisted of a pre-test for our upcoming (TOMORROW) test on probability.

There were 5 questions, the last being a two parter. :O!

1) A combination lock requires the owner to choose three numbers from 1-40 in order to open the lock. If the numbers can be repeated, the probability of an individual correctly guessing the combination to this lock is:

a) 1/40P3 b) 1/40C3 c)1/403 d) 1/3x40

The answer is C. The numbers may repeat, but they have to be in order, otherwise you're not going to get the locker open.

2) For this one, you're supposed to find the probability that a chosen route will pass through point B, when trying to get from point A to C. This is assuming the shortest routes from A to C are travelled.

a) 2/7 b) 3/7 c) 4/7 d) 5/7

The answer is C because there are 20 options to get from A to B. Then, the shortest way to get to C from there is to go directly right. That said, all that has to be done is 20/35, simplified that'd be 4/7.

3) Using the word FOOD, the probability that an arrangement of this word will begin with the two O's if all the letters are used, correct to the nearest hundreth, must be:

4!/(4-2)! = 12. You'd do 2/12 because there are two O's, and you can't tell them apart. Simplified that's 1/6, which in decimal form would be 0.17 (rounded).

4) Events A and B have the following probabilities of occuring; 0.2 = P(A), 0.5 = P(B). If these events are mutually exclusive, the value of P(A or B), correct to the nearest tenth, must be:

P(A or B) = P(A) + P(B) - P(A and B) ... And, because these events are mutually exclusive, P(A and B) = 0. Therefore, .2 + .5 - 0 = .7

5) Rupert has either milk or cocoa to drink for breakfast with either oatmeal or pancakes. If he drinks milk, then the probability that he is having pancakes with milk is 2/4. The probability that he drinks cocoa is 1/5. If he drinks cocoa, the probability of him having pancakes is 6/7.
a) List the sample space of probabilities using a tree diagram or any other method of your choice.

MO
MP
CO
CP

b) Find the probability that Rupert will have oatmeal with cocoa tomorrow morning.

P(oatmeal and cocoa)
P(C) = 1/5
P(O) = 1/7

Multiply 1/5 and 1/7 and voila, you have a 1/35 probability that Rupert will have the oatmeal with cocoa.

We got into groups to work on this after we did it ourselves. Is that even a word, ourselves?

Anywho, the afternoon class consisted of going over that pain of a stencil.

We didn't go over all of the answers. I'm wondering if I should go over them. I can't really explain them, but I understand. Does that make any sense? I'm a bit worried that I'll mislead or explain incorrectly. What the hell, here we go.

3) Seven people reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other?

You do 7 choose 4. That means that 4 will go one way. Then, 3 choose 3. You have three people to choose from left, and you choose all three of them. You mutiply them and get 35. Then you do 7 choose 3, multplied by 4 choose 4. This all equals 70.

5) A party of eight boys and eight girls are going for a picnic. Six of the party can ride in one car, and four in another. The rest must walk. (assume anyone can drive.)

a) In how many ways can the party be distrubuted for the trip?
16 (the party) choose 6 (car one) x 10 (ten people left) choose 4 (second car) x 6 (people left) choose 6 (the suckers that have to walk) = 1 681 680

b) In how many of these will no girl have to walk?
16 choose 6 (first car/girls) x 2 choose 2 (two girls left, guaranteed spots in the second car) x 8 choose 2 (8 people left, two of the boys get to be in the second car) x 6 choose 6 (six boys left, and they all get to walk.. bwahahahahahaHA)

c) Ditto to the last question, except there has to be a boy in each car driving
1 choose 1 (the driver of the car) x 8 choose 5 (5 of the 8 girls in the first car) x 1 choose 1 (the next driver) x 3 choose 3 (3 girls left, and 3 spots left in the second car) x 6 choose 6 (again, the suckers that get to walk)

7) In how many ways can 4 balls be drawn together from a bag containing 5 red balls, 7 white and 3 green? ... 15 choose 4

a) 2 red and 2 white balls?
5 choose 2 x 7 choose 2 .. That means you choose 2 of the 5 red balls, and 2 of the 7 white balls.
b) Just two white balls?
7 choose 2 x 8 choose 2 ... 2 of the 7 balls, and then any 2 of the remaining balls of the other two colours
c) Only white balls?
7 choose 4 ... It's only white balls, right? So you just choose four of them.
d) Balls of only one colour?
5 choose 4 + 7 choose 4 *** It's + because it can be wither 4 of the 5 red balls, or 4 of the 7 white balls. You cannot use the green balls, as there are only 3 of them. Poor green... :( :P
e) No red balls?
10 choose 4 ... You just add the 7 white balls and the 3 green ones

8) Ten points are marked in a plane, no three of them lying in a straight line. How many triangles do they determine?
This one is 10 choose 3.

10) From 7 French books and 9 English books, in how many ways can we select 3 French books and 4 English books and arrange them all on a shelf?
7 choose 3 x 9 choose 4 x *7! ... *Because they're arranged on a shelf .. 7! = 7x6x5.. etc.

After we went through the stencil, we started a review on page 43. We're to do questions 1 through 7.

I hope I explained all of these properly. Tomorrow's scribe is going to be Mr. mohamed (aka Mr. Smartypants)




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

The Mystery Coin Hunt!

π Day is around the corner .. it's five days away! Soon, soon, the hunt will be on!




Somewhere on the property of DMCI a coin will be hidden. Hidden so carefully and cleverly that it cannot be discovered by chance or simply by looking for it. On March 14, π Day, the coin's location will be revealed buried in a series of riddles and puzzles. Until it is discovered the coin's location will remain a mystery....





Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Blogging on Blogging

probability is a branch of math that deals with a change. to me this chapter seemed little different compared to probability in S3. so i had some difficulties. doing pascals triangle and solving problem with fundamental principle of counting were the easy part of probability, but we we started doing problems using the complaiment there was some difficulties. i like doing mutually exclusive problems. that is if one way happens there is no way that another can happen. example if you draw a jack and a 6. the probability is zero. which is sometimes called imposible event. probability is a number between zero and one. i can't tell know if i completely understood this chapter (probability) but i will tell letter when i am done my pre. test. thats all folks. hopefully statics chapter will be more fun than probability.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Wednesday, March 08, 2006 

blogging on blogging

One thing that I didn't understand at first was the diagonal lines that was drawn on the pathway problems. I understood how to figure out how many ways to get from one point to the other but diagonal lines kinda confused me, but once Mr. K. explained it a little bit more, I got it. It shows that the numbers at each end of the diagonal line you add them up to get the number of ways to get the point below, or the one to the left, or the one on the right. This unit was a pretty tough one I think so, so hopefully the test won't be that bad. Gnight!




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Reminder!!!

Im not sure if we're having the PRE-TEST or the real TEST tommorow on probability so if I were you i will study. Its TOMMOROW 1st period. Check the sheets that Mr. K gave us, the sheets that have the answers on the back, and retest yourselves. Good LUCK!!!




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Scribe

Wednesday 8, 2006

Today is the last day we looked at probabilities. We only have 1 period today and Mr. K gave us some notes to write out in our Math dictionary so here it is.

P = Probabilities
A,B = Variables
* = multiplication
/ = division (fraction)

Calculating Probabilities
Using "And" and "OR"


-Given any 2 events, A and B, the probability of A and B occuring is given by:
P(A and B) = P(A) * P(B)

NOTE: If A and B are MUTUALLY EXCLUSIVE (info about Mutually Exclusive event is in Mauxi's scribe) :
P(A and B) = 0

-We calculate the probability of A or B as Follows:
P(A or B) = P(A) + P(B) - P (A and B)

ex. Given a standard deck of cards (52 cards).
J: Draw a jack
H: Draw a heart
6: Draw a 6

One card is drawn, Find:
i. P(J and H) = P(J) * P(H) = 4/52 * 13/52 = 1/52

4/52 = 4 jacks (Spades, heart, clover and diamond)
13/52 = ace H, 2hH, 3H, 4H... king H.
1/52 = jack of heart

4*13 = 52 52*52 = 2704
52/2704 / 52/52 = 1/52 (reduced)

ii. P(J or H) = P(J) + P(H) - P(J and H) = 4/52 + 13/52 - 1/52 = 16/52 (reduced) = 4/13

iii. P(J or 6) = P(J) + P(6) - P(J and 6) = is it possible?
The P(J) (probability of drawing a jack) and P(6) (probability of drawing a 6) are possible but drawing a JACK with 6 are not.
Therefore these are muatully exclusive.

Note: If the event is MUTUALLY EXCLUSIVE the probabality "AND" can't be used.

This is my scribe for today and hope that you got something from this.
The scribe for Friday will be Cait.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Blogging On Blogging

I gotta say probability is not an easy chapter and most of the classes i was absent which make things complicated but i have to do alot of work before the test, however, Mr :K i didn't get it the difference between at least 1 or at least 2 and you said there is a big difference so if you can explain me in class one more time i believe that will be help .




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Scribe

Tomorrow's scribe will be Hesy sorr for being late




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Scribe

Hi guys sorry i'm late.

Here is what we covered second class. Mr:K gave us notes to write in our dictionaries.
These are the notes that were given to us

Complimentary Events
The compliment of an events ,E, is writtin as either E 'or E. The compliment of an events reffers to the cases here ,E, Does Not Occur

Example:- Drawing a heart from a deck of cards.
H'(the compliment)= drawing a card that is not a heart.

Calculation:- complimentary probabilities .

If P(E)=a Then P(E)'1-A

Example:Find the probability of filiping and pennies and gitting at least 1 heads.
Solution:-The compliment of "At least 1 head "is "no heads" or "3tails".
P(3)=1/2*1/2*1/2*1/2=1/8

P(at least one head)
=1*1/8
=8/8-1/8
=7/8
Dependant Events:- 2events are dependant if the outcome from second event depends on what occurs in the 1st event.
Example:-a bag contains 3 Red and 3 Blue Marbles. A marble drawn without placement. What is the probability that the 2nd Marble is Blue?

{It depends what happen in the 1st drawn.

Then Mr: k drew a triangel R =red B = blue.

R 2/5
/ P(RR)=(3/6)(2/5)=1/5
R 3/6 P(RB)=(3/6)(3/5)=3/10
p(BR)=(3/6)(3/5)=3/10
\B 3/5 P(BB)=3/6)(2/5)=1/5
Either 3/10 or 1/5 -it depends on what happened 1st. so here
is easy way to solve P(2nd B0=3/10 add 1/5 = 3+2/10=5/10.
R3/5
/
B3/6
B 2/5


Independent Events :- 2 events are independents if the outcomer from the 1st event has no impact on the outcome of the 2nd event.
Example:-giving the same bag of marbles if a Marble is drawn with replacement them the probability of the second marble being blue is un effected by happened on the first .
R3/6
/
R 3/6
\ B 3/6 They are Unchanged


R 3/6
/
B3/6
\ B 3/6


Mutually Exclusive Events:- two or more events are mutually exclusive if they have no outcome in common.
i.e It is Impossible for both events to occur in the same "sample events", in other words the events have no intersection .

Example:- A=flip coin from A standard deck of card
C=draw A 4
B=roll a die 6 D=draw A head.


P(heads or 3)=P(H)+P(3)
=1/2+1/6
=3/6+1/6
=4/6+1/6


p(4 and H)=P(4)+P(H)-P(4 and H)
=4/52+13/52-1/52=16/52 or you can still reduce 4/13
note P(4 and H)=1/4*13/52=1/52.

Sorry guys i didn't use any pictures because i don't know what is wrong with my software ,but anyway , those who didn't go to the class today i get one thing to remind you , we'll have pre-test on thursday and make sure everybody to do blogging on bloggging because the quicker you do it , the quicker you get feedback so keep that in mind.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Tuesday, March 07, 2006 

blogging

This unit was a bit harder than I thought it would be. All the problems on the Fundamental Principle of Counting are pretty easy, and I actually enjoy it.
Something that I found really cool was Pascal's triangle. It's amazing how many 'hidden' patterns there are in that triangle. I also found it very helpful when solving the pathway problems.
The things that are still a bit foggy are the permutation, but mostly combination questions. This is because the second stencil we recieved was pretty difficult. I hope we go over it a bit more..
Although calculating complimentary events was hard to grasp at first, it made sense in today's class. The thing is, you shouldn't approach a question in just 'one way'..it's like if you work your way backwards, for an answer that you're not looking for, it could help you easily get the answer that you are looking for. That's how I understand the complimentary questions work. Well, I just hope I figure everything out before the test!!




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

blogging on blogging

I gotta say I really thought that matricies was a hard unit until we started probability. I was never good at probability but I am trying really hard to understand it. There are some things that I understand like the factorial problems. The stuff that I am still a little worried about doing are the pick and choosing problems, there were some questions that I could do on the stencil that we got yesterday and others that I couldn't. I understand what we were talking about today in class. I just gotta practice more questions and hope I'll pass the upcoming test.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Probability Review

Hi everyone this is what we did in first class of am. Mr Explained how to use the fundamental principle of counting, formulas that deals with factorial. They seem hard at the beginning, but they get easy as you go along and do them. The question were as follows:

first question:
there are 9 people on a baseball team and they are selecting people to play the different positions.

a) how many ways can the position be filled if there are no restrictions on who plays which position?

answer: 9!= 362880

b) what is the probability that AL is the pitcher if the other positions are selected randomly?

answer: P(AL is the pitcher)
=1*8/9!
=8!/9*8!
=1/9

c) what is the probability that AL, BOB, or CAL are selected to be pitcher if all positions are selected randomly?

answer: 3*8!/9!
=3/9=1/3

d) what is the probability that AL, BOB or CAL are pitcher and DAN or ED are catcher if all positions are selected randomly?

p ( DAN or ED are catcher) = 2*7!/9! = 2/9*8= 1/36




P(d)= 3*2*7!/9!
= 3*2!/9*8!
=1/12
second question:

the probability that GALLANT fox will win the 1st race is 2/5 and that NASHAV will win the 2nd race is 1/3


a) what is the probability that both horses win their races?

P (both win) 2/5 * 1/3= 2/5

b) what is the probability that both horses lose their races?

P(both lose)= 3/5* 2/3= 2/5

c) what is the probability that at least one horse will win its race?


all possible outcomes:
G-W N-L
G-L N-W
G-W N-W
G-L N-L


p (at least one win)= 1-2/5 (p both lose) = 3/5

then mr.K did some extra example where he had to divide the class into boys and girls. he used a new words. that he explained ie. mutual exclusive which means if one way happens there is no way the other can happen.

examples as follows:


P (boy)= 14/20
P ( girl)= 6/20
P 9(blond=1/20

p ( girl or blond) = p ( g) p (bl) = 6/20 + 1/20

p ( boy or blond) = p ( b) + p ( bl)= 14/20 + 1/20

p (A or B) = P (a) + p ( b) - P ( anb) = 15/20 -1/20= 14/20.

do pages 31 and 37 odd #'s 1-9 for homework. mr.k said we going to have p. test on thursday second class.

thats all folks. i am going to post the notes around 11 pm. just because i am at work know, and my break is almost over. sorry guys i know you got to go to bed by 10pm.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Blog on blog action ;)

The main thing I have been having trouble with within this unit is the choose formula. The second stencil we got confused me to no end. There were a few questions that I understood, but aside them, well, just no. I have to say, I am a bit nervous about the upcoming test.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Sunday, March 05, 2006 

Ok everyone im up. heres the blog for friday. we had two classes on friday and in the morning mr. K explained to us how to solve number 11 on the stencil and in the afternoon we wrote in our dictionary.

Here is how mr. K solved number 11.

11* A man and his wife invite 4 other couples to dinner. After the host and hostess have been seated at the ends of the table, in how many ways can the guests be seated so that no man sits beside another man or beside his own wife? (hint: make a diagram of the chairs and table; first find the number of ways of seating the ladies a, b, c, d; for a typical arrangement of the ladies, count the ways in which their husbands, A, B, C, D, can be assigned to the vacant chairs so that A and a B and b, ect. are not adjacent.)

If we had to seat all the couples at the tables it would be easiest to seat the men first. there are 4! ways to seat the men but if we chose one way for now these are the different seating arrangements we would have for the women.

After we seat the men there are 5 different ways to seat the women. mr. K mentioned this answer was a sexist answer because that is only the arrangements for the women so the final answer becomes (4!5)2 = 240.

During the afternoon this is what we put into our dictionary.


Solving Pathway Problems
Method 1

This is Pascal's Triangle

Pascal's triangle turns up in many areas of math and science. It can be used to solve pathway problems as illustrated in these two examples:

Example 1:
How many ways can you go from A to B by the shortest route possible?

There are 70 ways. The diagonals contain the respective entries from Pascal's triangle.

Example 2:

How many different ways can we spell the name Yves?

Solution: Using Pascal's triangle we find the sum of the entries in the bottom row 1 + 3 + 3 + 1 = 8 ways

Method 2

Solving Pathway Problems as Permutations of Non-distinguishable objects:

Find the number of ways to get from A to B using the shortest route possible:

Solution: each route goes right (R) 4 times and down (D) 3 times. Therefore each different arrangement of letters R R R R D D D counts as a different route.

This problem is the same as find the number of "different words" that can be spelled using the letters.

R R R R D D D

# of letters: 7 # of R's: 4 # of D's: 3

7! / 4! 3! = 7 * 6 * 5 * 4 * 3 * 2 * 1 / 3 * 2 * 1 * 4 * 3 * 2 * 1

=7 * 5 / 1

= 35

That is all we wrote in our dictionary for friday and im really sorry this took me a long time to post. hope this is helpful for the people who missed out on friday. Monday's scribe is Muuxi.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Sunday Gridlock



In this game you have to move the blocks (vertical blocks move only vertically and horizontal blocks move only horizontally) out of the way so that the blue block can slide out the "door" on the right. Although this game can sometimes get frustrating there is always a way out. Remember Sisyphus!

So far I've made it to level 6, how far can you get?




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Friday, March 03, 2006 

hey everyone. im sorry its late notice but i just got home from work and i didnt have a computer to use earlier to let anyone know. todays scribe will be posted up tomorrow. its hard for me to think right now. soo tired, but i'll be up early tomorrow to post the scribe. goodnight.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Thursday, March 02, 2006 

It's Coming ...



π
Watch it grow.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

 

Scribe

This is Andrew filling in for Muuxi. I decided that since Muuxi and I are the only contributors left to scribe, and since Muuxi was absent today, I will take this opportunity to be today's scribe.

For those who have been absent for today's class, you people need not worry for all we did during the entire class was work on the Fundamental Principle of Counting stencil worksheet. Reason for that is because we had a substitute teacher and he mentioned that Mr. K. was sick.

Muuxi, if you're reading this, then you can be tomorrow's scribe.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

Wednesday, March 01, 2006 

scribe

hey everyone this is jessica i just found out today that i was scribe and i don't see my name anywhere. but thanks for the heads up khan. so today we had two math periods and for those who didn't go to class today, which shouldn't be too many people cause i think i saw everyone, Mr.K gave us notes to write in our dictionaries. After we were given the notes Mr. K gave us a stencil to work on in groups. I got to say the stencil is pretty difficult so if you missed class better get some help from Mr. K.

These are the notes that were given to us

Combinations:An arrangement of objects where order doesn't matter.

Notation: nCr or (n over r) [sometimes called the "choose" formula]

read as: "n choose r"

formula nCr = n! / (n-r)!r!

n is the number of objects to choose from.r is the number of objects being arranged.

meaning: from a set of n objects, how many different ways can you choose r at a time if the order they are chosen in doesn't matter.

example: how many different lotto 6 / 49 tickets can be sold?

Solution 1: As a permutation of non-distinguishable objects we imagine ourselves saying "yes" to 6 of the numbers from 1 to 49 and "no" to the rest.
ie. we say "no" (n) to 49 - 6 or 43 the number of ways to do this is the same as the number of different ways to spell a 49 letter word made up of 6 y's and 43 n's.

49! / 6!43! = 13 983 816

Solution 2: as a combination

49C6 = 49! / (49 - 6)!6! = 49! / 43!6! = 13 983 816

those were the only notes that were given to us and for the rest of the morning class and afternoon class we were given time to work on our stencils.

there was a question on the stencil that a number of us were asking Mr. K about so he explained to us how to solve it. It was question number 4 and here it is.

4.(a) How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6?
(b) How many of these numbers are even?
(c) How many of these numbers are divisable by 5?

(a) answer is 6 * 6 * 5 * 4 * 3 = 2160
the reason that the first number of the 5 digits has only 6 choices is because you can't include 0 to be the first number. The second digit has 6 choices because we can include the 0 back into the number of choices and there would only be 6 choices left, then 5, 4 and 3.

(b) answer is 360 + 900 = 1260
In this question the 0 is a problem again because it is also an even number so instead we take it out. Now we only have 3 choices for the last digit to be even and since the first digit also can't be 0 the number of choices we have is 5. The second digit is also a5 because we can include the 0 back into the choices the 4, and 3. 5 * 5 * 4 * 3* 3 = 900
We do this again but instead of the last choices being 3 it is 1 because we put 0 as the last digit. then the first digit has 6 choices then 5, 4 and 3. 6 * 5 *4 * 3 * 1 = 360
Now all that is left to do is add those numbers up to get the total number of combinations with the last digit being even.

(c) answer is 360 + 300 = 660
In this last question we are asked how many numbers would be divisable by 5, now we can all identify when a number is divisable by 5 when the last digit is a 5 or 0. Like in the other questions 0 becomes a problem, if the last digit is 0 the number of choices for the first digit is 6 then 5, 4, and 3. 6 * 5 * 4 * 3 * 1 = 360
When the last number we choose is 5 we have to take out the 0 for the first digit and the number of choices becomes 5 and the second digit is also 5 because we know that we can put 0 back into the number of choices then 4 and 3. 5 * 5 * 4 * 3 * 1 = 300
The last thing to do is add those two numbers up to get the number of combinations that will be divisable by 5.

Now that im done I hope that sorta helped anyone who didn't understand what we were doing in class today and those who missed it. I tried the best I could to make this understandable, I actually had to do it twice because my first one got deleted. well thats all. tomorrow's scribe will be muuxi.




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch