Sequences

Hey guys im not quite finished because im a little confused on how to make this scribe. But i think its half of what we had learned and im going to fix it when i get home from tommorow.

BOB

The periodic test is tommorow and i think im ready for it. This unit is the 2nd easiest unit, matrices being 1st. This unit was not too hard and not too easy to understand and learn. I dont think i would forget the word "DBAC" during the test since i think its a funny word, so that wont be a worry. Well anyways.. lets pass this test shall we?? =)

Personal Finance

Today we started off with problems on personal finance. It was a little confusing because the P/Y were changing meanings when i got there. P/Y means Payments Per Year. But there are still some problems with it that we have to understand.





So these are some problems we did in class:





We had problems with this problem but all we had to do is find the FV. Today we had problems on stating whether the P/Y was to be 2 because i thought that Payments per year meant how many times have you added money in a year. Here im not sure if we add any money at all.


Same here. We got this question right even though we did not do the correct way of doing it. I put 1 payment per year because were trying to find how long will 12000 turn to 15000 PER YEAR.





So what we really did today was just to help everybody understand that Payment Per Year is how many times money is being added every time it is compounded.

The next scribe will be... my dear friend TENNYSON

BOB

It's a hard unit but today i think i understood what i was being taught. There are some problems i still dont know how to do but hopefully the test questions should be one's that i actually can do. I think that the unit is hard because there are so many steps on doing it that id probably wont remember how to do them. welps i hope i pass this test. *crosses fingers*

Statistics

Today we refreshed our memorys by doing some problems on probability.

The probability that a student owns a cd player is 3/5. If eight students are selected at random, what is the probability?

a) exactly 4 of them own a cd player

To do this we have to use "Binompdf." So if we want to do this on the calculator you press:

• 2nd, DISTR
• find Binompdf or if you have a TI-83 press 0
• Then we use the formula - Binompdf(# of Trials, Probability of success, # of success)

So it should look like this:

Binompdf(8, 3/5, 4)

You hit enter and your get .2322 or 23.22% and thats your probability.

We also learned how to Calculate probabilitys using normal approxamate of binomial distribution.

Example

A basketball player is successful on her freethrows 75% of the tim. Determine the probability that she makes at least 7/10 free throws.

So using normal approxamate of binomial distribution we have to know that:

• N = Number of trials = 10
• P = Probability of success = .75
• # of Success = 7

So we have to know thes three formulas to do this:

M = N*P

Failure = N(1-P)

S.T = Standard Deviation = S.D = Square root (M(1-P)

Z score = Z = (# of success - M)/S.D

Now we put all the info in and find the Z Score

we get the S.D of -0.3693. So now we use shadenorm to find the probabilty. So we go Shadenorm(-0.3651 , 5) and get the probability of 64.25%

Next scribe is Cris J.

Using Factorial Notations and Permutations

Today we went over on how to use factorial notations and permutations. So if anyone else missed the class here it is..

Factorial Notations
We learned that the symbol "!" can be used to multiply a number down to 1. So if we want to multiply 6 * 5 * 4 * 3 * 2 * 1 = 720, we can simply use "!" to make it easier to multiply. So it would look like this: 6!.

Examples for using factorial notations
So this is what we do:

So in the calculator it should look like this:

10!/(7!3!)*(1/2)^7*(1/2)^3

(and make sure that you put brackets in or it will make no sense at all)

When we equal it, it becomes 0.117 which is 11.7%.

Using Permutations
We use permutations to see how many ways can object be arranged and to do that we use the formula:

nPr = n!/(n-r)!

n is the number of objects available to be arranged.
r is the number of objects that are being arranged.

Examples using permutations

If we had 10 students in our class and we wanted to find how many ways can we change seating plans, there would be 3628800 ways to do it. This is how i know:

But what if there were 5 extra seats? We would do that same thing we did but we'd switch the "n's" to 15 so theres 15 vacant seats. So it would look like this: