scribe . . .

*updated by a little =P*
12/05/06 - Day 5 - Period 5 =)
*just a quick note before you go on . . . All fonts in RED are links =D

Class work:
At the beginning of class we looked over our homework . . the visual thingymajjiger on the yellow sheet . . . Here's a bubbleshare slideshow I made on how to do it :

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I used the way that I did it for the slideshow, but there's other ways you can do it. After we went over it he assigned this for homework. I'm not telling you how to do it because that's . . . CHEATING =P

We had a starter (i think that's what it's called; can't remember =P)on converting equivalents:

• 
  
  17/20 -> 20 - 17 = 3 -> 17:3 -> 17 / 20 = 0.85 -> 0.85 = 85%

• 8.9% -> 8.9 x 100 = 89 -> 89 / 1000 -> 1000 / 89 = 0.89 -> 1000 - 89 = 911 -> 89:911
  

Mr. Harbeck then gave us a definition for percent . . .

Percent -> It is an equivalent that is written as a fraction out of 100.

After that he gave us a paper that looked like this.

You had to use the picture on the right side (homework that was due today that we looked over) to apply it to the squares . I can't describe it well so just look at my paper (the "this" above is a link =P). He told us an easier way for finding half and a quarter of the number; just divide by two, and if you wanna find quarters, divide it in half again. If you wanna find 75%, then find one quarter and subtract the answer from the number you're doing. Same thing for 95%, find 5% first then subtract. It goes on and on and on . . .


Homework:
For homework we had to do the same thing as we did with 300 (put them into the squares). I already got started on it as you can see =P. Then you have to find 1%, 5%, 10%, 30%, 65%, and 75% for 400 and 162 and put them onto the squares sheet. Oh! Don't forget to do the Data Investigations on the yellow paper . . .

Websites:
A really good website I find was this one. It has "basic" stuff on how percentages work and some questions and answers.

BTW:
By the way, I commented on this post =)

Next Scribe:
humdeedumdeedum . . . who to pick for the next scribe ? umm . . . I'll just pick . . .

(I picked you cuz are names rhymed =P)

Commenting

Mr. H, I also made a comment on this person't post here: http://apcalc06.blogspot.com/2006/11/scribe-post-day-52.html
It's also the same post that Angelic commented on, is that okkay?

Math Pretest Equivalents

Math Pretest

Convert the following values so that you can place them on the number line below. Show all your work.

0.175

55%

3:10

15/51

9:3

How could you determine whether the average of these numbers is greater than 10 or less than 10 without actually computing the average. Explain how you decided the average was more than or less than 10.

Angelic's and Melissa's Board Game

The Wrong Turn

Rules: 1^st one to ISLE FINI wins!

Decide on a team.

Team 1 uses a 6 sided dice: get a 6 - move 2 spaces.

get a 3,4,5 - move 1 space

Team 2 uses an 8 sided dice: get an 8 - move 3 spaces

get a 4, 5, 6 - move 2 spaces

You must stop on are 9, NO MATTER WHAT.

Additional Rules are on the board.

*Before you go and say, "Why can't you move on the other spaces,", look below.*

If you land on area (spaces where you move) 8, go back to start:

Team 1: get 5 or 6 - move 2 spaces

get 1, 2, 3, 4 - move 1 space

Team 2: get 7 or 8 - move 3 spaces

get 5 or 6 - move 2 spaces

get 1, 2, 3, 4 - move 1 space

Why is it fair?

It's fair because on area 8. if you land on it that is, you gain a chance to catch up. On an 8 sided dice, there's less of a chance for rolling an 8 than it takes to roll a 6 on a 6 sided dice. In fact, you have a 1/6 chance of rolling a 6 on a 6 sided one and a 1/8 chance of rolling an 8 on an 8 sided dice. Make them have the same denominator and its 6/48 (6 sided) and 8/48 (8 sided). THat's why rolling an 8 gives you one more area movement than rolling a 6. same goes for the rest.

**sorry if it's late. I just completely forgot about it!

Ardia's board game =)

My game is just called "Ardia's board game" (pretty orginal huh? =P).

The game is ment to be played by a minimum of 2 players to a maximum of 5.

To play the game you need board markers (when i played i used bears), the board , and 2 8-faced dice.

The object of the game is to be the first one to get to the finish space.

To move is the most difficult part of the game ..
to move one space you have to roll an odd number from 3-7 (by adding both dice, same for all of them).
to move two spaces you have to roll an even number from 2-8.
to move three spaces you have to roll an odd number from 9-15.
to move four spaces you have to roll and even number from 10-16.

My game is fair because everyone has a chance to move and no one can just win because i put about 4 spaces that said "move back 2 (or 1) spaces" and about 4 that says "move forward 2 (or 3) spaces" and one space that said "roll agian" and one space that is closest to the start that says "go back to the start" and i also put a space that says "miss a turn" so no one can just win they have to have luck on their side =).

My game was played, 10 times i played it 3 times out of ten. Then watched other people play my game . Out of the three times i played the game i only won once! the first time i played i always landed on the "go back to the start" space .. then i started to regret that i put that space there! some people can get so competitive =) haha --just a little humor

***ohh yeah just to tell you in the picture my game wasnt fully colored so yeah .. but it looked cool when it was done .. so feel free to comment =)

ardia=)

Michael and Brandon's Game

Our game is called "The Rich and the Poor". To play, you'll need the board, some markers (coins, bears, etc.), and 4 dice. Two of them should be one colour and the other pair should be another.

The game is meant for 2-4 players. Decide who moves first. Put all markers on "start". Then the player picks either the "poorman's dice" or the "richman's dice" each turn to roll with. The numbers needed and spaces moved differ witht the dice:

Poorman's Dice : need 3, 5, 7, 9, 10, 11, to move ONE space.

Richman's Dice : need 2, 6, 12, to move THREE spaces.

The players then alternate turns, following the steps on any spaces they land on until one player reaches "finish" and they win.

The game is fair because both players have an equal chance of moving, and they can decrease or increase the odds themselves by picking the right dice. There is a ~1/2 or ~1/4 chance of moving.

The game is fun because some strategy comes into play on when to pick the Richman's dice and when to pick the Poorman's dice. For example, if one space is in front of your marker is "Go back 2 spaces" then you want to use the Richman's dice to avoid it.

-apuya

My Game

My game is called "Jeffopoly." This is how you play, you first spin the spinner. If the pointer lands on blue you move 3 spaces forward and if you land on white you mave 2 spaces forword. There are spaces that you can "miss a turn," "move spaces," or move back spaces." First person to make it back to the finish wins. This is a fair game because the is a 50/50 chance of getting a blue or whte and either way you can move spaces.

J.A.A

solving # 4

Question of problem 4.



1. If you flip a coin 150 times, about how many times would you expect to get heads?


If you divide 150 into 2 gives you 75/150


2. The letters a, e, i, o, u, and y are vowels. If one letter of the alphabet is chosen at random, what is the probability it is a vowel?


The answer would be 7/26, how, add all the vowels together and, how many the probability is 7/26


3. If you randomly pick a date in April, how many equally likely outcomes are there?


The answer would be 30, because there are 30 days in April.


4. A magician asks you to pick a card, any card, from a standard deck of 52 cards. What is the probability of picking an ace?


The probability of picking a ace would be 4/52 and you can simplify to 2/26and you can do it one more time to be 1/13. So the answer would be 1/13


This is my question if you have 10 pens, 14 pencils, 16 erasers, and 9 rulers. How many outcomes can you have?

Probability Homework !

E-50 #4

mMm dela cruzzz's math homework...

E-49

Marygrace Dela Cruz . . .

19. The probability of winning a raffle if 500 tickets are sold and you buy 5 of them is 5/500 or 1/100. you would only have a 1 percent chance that you would get a winning raffle.

20. The probability that a girl will be chosen from random of 25 students is 15/25.

Every random pick out of 25, 15 times it will be a girl.

21. The probability that a letter is chosen at random in the word MATHEMATICS is 8/26 or 4/13.

Because there are two Ms, two As, and two Ts, there are 8 DIFFERENT letters.

My own question:

What is the probability that a letter chosen at random is in the word . . .

SUPER­CALI­FRAGI­LISTIC­EXPI­ALI­DOCIOUS???

My answer:

Well scince there are ...

- 3 Ss - 3 Ls

- 2 Us - 7 Is

- 2 Ps - 1 F

- 2 Es - 1 G

- 2 Rs - 1 T

- 3 Cs - 1 X

- 3 As - 1 D

- 2 Os

... and only 26 letters in the alphabet,

the probability of a letter being chosen

at random is this word is 15/26.

There are only 15 DIFFERENT letters.

Math Questions

1. 1/5 (red) This is 1/5 because there is only 1 chance out of 5 of getting red

2.1/5 (green) This is 1/5 because there is only 1 chance out of 5 of getting green

3.2/5 (blue or white) This is 2/5 because there is only 2 chances of 5 of getting blue or white
4. 4/5 (not yellow) This is 4/5 because there is only 4 out of 5 chances of not getting yellow

5.4/5 (not red) This is 4/5 because there is only 4 out of 5 chances of not getting red

6. 3/5 (blue, white and yellow) This is 3/5 because there is only 3 out of 5 chances of getting blue white or yellow

Question I created:

Find the Probability of getting a white??

e-50 #2

Angelic M.

October 16,2006

8-41

2. A spinner is shown below for which outcome is NOT equally likely.

(E) If you spin the spinner once, what is the probability that it'll stop on A?
A takes up the same amount of romm that b and c do together. If you divide A in half, there would be 4 parts. Then there are 2 a's so, 2/4 or 1/2.
(A) If you spin the spinner once, what is the probability that it'll stop on B?
Again, A stands for 2 parts, so there are 4 parts all together. B takes up one of the 4 spots so, 1/4.
(T) If you spin the spinner 50 times, about how many times would you expect it to stop on A?
A takes up half of the room. Half of 50 equals 25. 25 is the answer.
(Y) If you spin 80 times, about how many times would you expect to stop on C?
C takes up a quarter of the room. A quarter is the same thing a as a fourth. 80 divided by 4 equals 20. 20 is the answer.
(MY QUESTION) If you spin the spinner 240, how many times would it supposedly stop on B?
240 divided by 4 equals 60 (24 divided by 4 equals six, add a zero at the end, voila!). Again B takes up a fourth of the space, and so forth, the answer is 60!

1)


2) (1) 1/5 (2)1/5 (3) 1/5 (4) 4/5 (5) 4/5 (6) 1/5


3) Find the probability of spinning on white.



Leslie 8-41 Oct. 16, 2006

Math Questions & Answers Page E-50 Part VI (# 2)

A spinner is shown at the right for which each outcome is NOT equally likely.

If you spin the spinner once, what is the probablity that it will stop on A?

The probablity of the spinner landing on A after one spin would be Pa=2/4, or simplified to Pa=1/2, because A takes up half or two quarters of the spinner. (A/AllOtherAreas)

If you spin the spinner once, what is the probablity that it will stop on B?

The probablity of the spinner landing on B after one spin would be Pb=1/4, because B takes up 1 quarter of the spinner. (B/AllOtherAreas)

If you spin the spinner 50 times, about how many times would you expect it to land on A?

You would expect the spinner to stop on A about 25 times, because the probablity of getting A is 1/2. And 25 is half of 50.

If you spin the spinner 80 times, about how many times would you expect it to land on C?

You would expect the spinner to stop on C about 20 times, because the probablity of getting C is 1/4. And 20 is one quarter of 80.

--{*Create-A-Question*}--

Pretend B took up about 40% (2/5) of the spinner, and C took only 10% (1/10) of the spinner.

Then, if you spinned the spinner 20 times, how many times would it land on B?

Or on C?

Or A?

-apuya

Math Questions III: p. E-49

16.
What is the probablility of guessing the correct answer to a multiple choice question if there are 5 choices?
-P 1/5

17.
What is the probability of guessing the correct answer to a true-false question?
-P 1/2

18.
What is the probability that your birthday will fall on Saturday or Sunday?
-P 2/7

19.
What is the probability of winning a raffle if 500 tickets are sold and you buy 5 of them?
P 5/500

20.
A class of 25 students has 15 girls and 10 boys. If one student is chosen at random, what is the probability it is a girl?
P 15/20

21.
There are 26 letters in the alphabet(no, really?). What is the probability that a letter chosen at random is in the word MATHEMATICS?
P 8/26

Math Questions and Answers

Page E-50

4. Solve.

(N) If you flip a coin 150 times, about how many times would you expect to get heads?

You should get 75 times of getting a head... or 1/2 because there is a 50% of chance that you can get a head and another 50% of getting a tail if you flip the coin once. So flipping the coin 150 times will give you an expectation of 75 because 50% of 150 of 75.

(C) The letters a, e, i, o, u, and y are vowels. If one letter of the alphabet is chosen at random, what is the probability it is a vowel?

The answer should be 6/26 or if you simplify it (by the number 2), you should get 3/13 because the alphabet has 26 letters in total and the vowels would be 6 letters out of 26. Also Mr. Harbeck has taught us an easy way of figuring what the answer would be for questions like this.

---------------------------------------

vowels

-----------------

total letters of alphabet

---------------------------------------

6____ 3

----- = -----
26 ___13

---------------------------------------
(K) If you randomly pick a date in April, how many equally likely outcomes are there?

Usually each months should have only 30 or 31 days unless it's February then you have 28 days and if it's a leap year then you get 29 days in February. But we only need to solve for the month of April and the equally likely outcomes should be 30 because April only has 30 days so the number of 30 days is most likely to be the outcomes of April. So if you pick a ramdomly date in April, you should be picking a date between April 1st and April 30th. I have also checked the calendars of the year 2004 and 2005 to make sure that April has only 30 days.

(P) A magician asks you to pick a card, any card, from a standard deck of 52 cards. What is the probability of picking an Ace?

The answer should be a 4 because every number has 4 cards each, a card of hearts, clubs, spades and diamonds. Also, there are 13 numbers (and letters) in a deck. A number 2, 3, 4, 5, 6, 7, 8, 9, and letters such as an Ace, Jack, Queen and a King. So there are 52 cards in a deck, and each number contains 4 cards each, so the probability of getting an Ace is 4/52 or if you simplify it by the number 4, you will get 1/13.
---------------------------------------

number of Aces in deck

-------------------------

number of cards in deck

----------------------------------------------

4______ 1

------- = -------

26 _____13

My own question:

The prize of $500 dollars is only won by winning the 1st bingo game. The other 5 winners will win $20 dollars. If you were playing bingo with the other 299 people and each person only has 1 bingo card, what is the probability of winning $500 ? What is the probability of winning $20? What is the probability of winning $500 and $20?

P.S. Hints are given =)

K.V. 8-41 (October 16, 2006)

Math Post Answers

I. Find each probability if you spin the spinner once. (questions 1-6)

1. P(red)
1/5

2. P(green)
1/5

3. P(blue or white)
2/5

4.P(not yellow)
4/5

5. P(not red)
4/5

6. P (blue or red or yellow)
3/5

P.S. I can't scan because my scanner doesn't work right now and for some reason I can't upload any pictures either so that's why I don't have a picture on this post.

Math Question II

picture is up
Questions and Answers are blow

7P(striped)3/8 8P(shaded)1/8 9P(shaded)1/8 10P(white or shaded)5/8 11P(striped or white)7/8 12P(striped or shaded)4/8-1/2 13P(not striped)5/8 14P(not white)4/8-1/2 15P(striped or white or shaded)8/8

Question 2 on E-49

Find each probability if you choose one card at random.

7. P(striped) = 3/8 = 37.5 %

8. P(white) = 4/8 = 1/2 = 50%

9. P(shaded) = 1/8 = 12.5%

10. P(white or shaded) = 5/8 = 62.5

11. P(striped or white) = 7/8 = 87.5

12. P(striped or shaded) = 4/8 = 1/2 = 50%

13. P(not striped) = 5/8 = 62.5%

14. P(not white) = 4/8 = 1/2 = 50%

15. P(striped or white or shaded) = 1 whole = 100%

If we were to add four more cards to the deck. What will the possibility of drawing a striped, white or shaded at random?

16. P(striped)

17. P(white)

18. P(shaded)

19. P(white or shaded)

20. P(striped or white)

21. P(striped or shaded)

22. P(not striped)

23. P(not white)

24. P(striped or white or shaded)

Measurement of Central Tendency Pretest

3. Find the Value of x in the data.

10, 12, 6, 6, 14, x, 8, 10, 12, 12, 8, 14

First, I arranged the data above in ascending order! Like shown below:

6, 6, 8,8,10,10,12,12,12,14,14, X

A) If the mean is 10. X= 8

Remember mean is also called an average. To get the mean you add up all the numbers in your data set(sum of all DATA). Then divide the sum of all data by the number of pieces of data( numbers) you used.

6+6+8+8+10+10+12+12+12+14+14+X

_____________________________________ = 10

12

6+6+8+8+10+10+12+12+12+14+14 = 112

Remember 10 x 12 is 120 so the sum of all data is 120.

120 - 112 = 8 so the x is 8

6+6+8+8+8+10+10+12+12+12+14+14
__________________________________ = 10
12

There is only one aswer to this question because 120- 112 is 8 and it couldn't be any other number.

B) If the mode is 12. X= 12

6, 6, 8, 8, 10, 10, 12, 12, 12, 12, 14, 14,

* 12 can be the x because if you add a 12, the mode will still be 12 because 12 is the data that appears the most.

There are other answer to this question but you have to make sure the more is 12. Your x could be 1, 2, 3, 4, 5, 7, 9, 11, 12, 13, 15 and so on. Your answer can't be 6, 8, 10 or 14 because they would become the mode since there are 2 datas of 6, 8, 10 and 14 and if you add 1 there would be 3 datas and 12 wouldn't be the mode anymore.

C) If the median is 11.Is there more than one possible answer for each question?

x = 14

6, 6, 8, 8, 10, 10, 12, 12, 12, 12, 14, 14, 14

10 and 12 are the middle 10 +12 = 22 / 2 = 11

There could be other answers to this question but your answer has to be higher than 12 because if you put a lower number the median won't be 11. Your answer could be a 12, 13, 14 and so on but make sure it is higher than 12.

If I made a mistake please kindly tell me so I could change it thank you!!

THYRZA MAY

8-41