The Adventure Continues ...

Our adventures in blogging continue....

Watch for new blogs going live February 5, 2007 ...

• Applied Math 40S (Winter '07) (Grade 12)


• Pre-Cal 40S (Winter '07) (Grade 12)


• AP Calculus AB continues ... (Grade 12)

So Long ...

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

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

Just to have a laugh or learn some math,

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

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

So long everybody! Watch this space for pointers to new blogs for each of my classes.

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

Podcapsule

Last Thursday morning the students in this class wrote their final exam. That afternoon we wrapped up our class and recorded this podcast. It's called a podcapsule because, like a time capsule, it'll stay here as a permanent record of the students reflections and recommendations to themselves on how to guarantee their success in their next math class.

Although the class is over and the podcast is done you can keep it going as long as you like ...

First listen to the podcast (6.3 Mb, 13 minutes 11 seconds), then, if you like, you can keep it alive by adding an audio comment or text comment on this post.

Exam Rehearsal for Identities and Counting

Hey people! It's JessiccaI_198 as scribe today. Today we had two exam rehearsals; one for identities in the morning and the other one was for counting in the afternoon. We talked a bit about the arc sine and about functions, but I didn't really get many notes on that. But here are the questions and solutions for the Exam Rehearsal Identities:

1) The expression (sinx + cosx)^2 - 1 is equivalent to:

a) sin2x

b) cos2x

c) tan^2x

d) 0

The answer is a) sin2x.

Solution:

sin^2x + 2sinxcosx + cos^2x - 1

2sinxcosx

sin2x

2) If sinA = 3/5 and sinB = 2/3, and A and B are acute angles, what is the value of cos(A - B)?

a) -2/3

b) (4√5 - 6)/15

c) (4√5 + 2)/5

d) (4√5 + 6)/15

The answer is B

Solution:

cos(A - B) = cosAcosB + sinAsinB

= (4/5)(√5/3) + (4/5)(√5/3)

= (4√5 +6)/15

3) Find all values of x in the interval 0 < x < cos2x =" -3sinx">

The answer is x = 7π/6, 11π/6

Solution:

cos2x = -3sinx - 1

1 - 2sin^2x = -3sinx - 1

0 = 2sin^2x - 3sinx - 1

= (2sinx + 1) (sinx - 2)

sinx = -1/2 sinx = 2

2 is rejected because in the unit circle, no value can be bigger than 1.

x = 7π/6, 11π/6

4) If sinΘ = √5/3, what is the value of cos2Θ?

The answer is -1/9

Solution:

cos2Θ = 1 - 2sin^2Θ

= 1 - 2(5/9)

= 1 - 10/9

= 9/9 - 10/9

= -1/9

5) Prove the identity: sinΘ/(1-cosΘ) + sinΘ/(1+cosΘ) = 2cscΘ

Here are the problems from this afternoons class on counting:

1) The number of different arrangements of 3 boys and 4 girls in a row, if the girls must stand together, is represented by:

a) 4!x4!

b) 3!x4!

c) 4!x4!x2!

d) 3!x4!x2!

The answer is a.

2) The students in a music department have practised 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does not matter?



a) 25


b) 35


c) 150


d) 330

The answer is c.

Solution:

6C4 . 5C3

= 150

3) All telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. In this system, the number of different are codes possible was:

The answer is 144 different ways.

Solution:

8C1 . 2C1 . 9C1 = 144 different ways

8C1 - there are eight numbers in the first digit - 2, 3, 4, 5, 6, 7, 8, and 9.

2C1 - two number in the second digit - 0 and 1

9C1 - all the numbers, excluding 0







4) A paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house A to house B without backtracking is:






Solution:
There are 60 different trails






5. a) How many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?



b) How many different ways can 7 chairs be arranges in a row if 2 of the chair are blue, 3 are yellow, 1 is red and 1 is green? (Assume that all of the chairs are identical except for colour.)



c) How many different ways can the chairs in (b) above be arranged in a circle?

5. Answers:
a) 35
b) 420
c) 60

Solution:
5a) 7C3 = 35
- since order doesn't matter. 7 is for the total number of chairs, and 3 is for the number of chairs to be chosen.

b) 7! / 3!2! = 420
- 7! represents the total number of arrangements, and 3! 2! represent the non-distinguishable objects.

c) 6! / 3!2! = 60
- one of the chairs is a reference point, and the rest are seated in a circle. 3!2! are the non-distinguishable objects.

Autograph that provincial exam with EXCELLENCE!!

scribe post: Exam Rehearsal Probability

Hello guys!!!!!
Our topics in mathematics just ended. For us it is a relief but we must always put in mind that learning in a never ending process so there are still maths stuffs out there waiting for us. Anyway, this morning we did a rehearsal examination in probability. Mr. K gave us 30 minutes to work on it then after that we had a group discussion where we talked about our answers and did some brainstorming. Afterwards, Mr. K discussed with to us the right answers with their respective solutions. The questions given are:

(1) A certain soccer player has scored on 82% of his penalty kicks throughout his career. Given this information, the probabality that he will score on exactly 4 of his next 5 penalty kicks, correct to the nearest hundredth, is

(a) 0.80
(b) 0.66
(c) 0.41
(d) 0.08

The answer is (c) 0.41 because the solution for this problem is




(2) If P(A) = 3/4 and P(A and B)= 1/2, where A and B are dependant events, then P(BA) equals:

(a) 1/4
(b) 3/8
(c) 2/3
(d) 5/4

The answer is (c) 2/3. The problem indicated that A and B occured in dependant events, which means that both have to happen.

Solution:

P(A) * P(B) = 1/2

3/4 * P(B) = 1/2

P(B) = 1/2 * 4/3

P(B) = 2/3

(3) Peter places 5 equal-sized tiles in a cloth bag. Each tile has a letter on it. The letters are P, E, T, E and R. The probability that Peter selects the 5 tiles, one at a time, in order such that they spell PETER, correct to the nearest hundredth, is

The answer is 0.02.
>


(4) A child 2 quarters, 2 dimes, and 3 nickels in his pocket, but he does not understand the value of any of the coins. He puts 35cents worth of candy on the counter at a store and randomly selects two coins from his pocket. The probability that the two coins he selects will have a total at least as high as the value of the candy is:




(5) It is known that 53% of graduating are boys. Three are chosen at random. Given that at least two of the three grads are boys, determine the probability that all three of the grads are boys. ( Answer accurate to at least 4 decimal places. )

Well that's all folks.....
The next scribe is JESSICA!!!!!!!!
I already told her that she is next scribe...
Anyway guys, good luck to all...
May the good news be ours after the provincial exam....
God bless us all!!!!!!!!!

Sequences & Series Notes

Here are the notes to add to your math dictionaries for this last unit on Sequences and Series. The notes must be copied into your dictionaries by hand.

Here they are: Page 1 of 3, Page 2 of 3, Page 3 of 3.