Anything and everything goes in here... within reason.
Wed Apr 13, 2005 3:06 pm
ok, i'm having some stats issues. despite being forced to take a stats module, it seems the bloke didnt actually teach us anything useful!
i need to compare some values to establish if they are significantly different, but they're means of big samples (smallest sample size is 339) and the samples are all different sizes. i *think* i need to use an un-matched t-test according to
this website.
i've searched on the internet of how to do a t-test, and cant find anything that makes any sence... help! how do i do a t-test? and is it the right test to be using?
Thu Apr 14, 2005 5:59 am
I've got an assignment to do for religion class and it's a survey involving Christian people and their views on certain issues.
Am I allowed to post that here? I just thought because it's about religion, I might not be able to post it.
Oh and if I'm not allowed, does anyone know a site where I can put itup. Like a site where people do surveys online or something?
Thanks.
Thu Apr 14, 2005 6:11 pm
You might try this one:
http://www.christianforums.com
I tend to go there to stir up trouble when work gets slow
Thu Apr 14, 2005 6:14 pm
I thought that was why you came here... *sniffle*
Are you cheating on us with another forum?
Thu Apr 14, 2005 9:32 pm
I wouldn't call it "cheating," per se...I'd...uhm...well, you see....there's this thing...and...hey! Look! A well-placed diversion!
Sun Apr 17, 2005 12:35 am
does anyone know what period in art history/art movement/style Claude-Joseph Vernet painted in?
especially this painting-
Sun Apr 17, 2005 5:39 pm
Soda wrote:does anyone know what period in art history/art movement/style Claude-Joseph Vernet painted in?
especially this painting-

Try this website:
http://www.gegoux.com/JosephVernet/
Thu May 12, 2005 7:23 pm
Hey,
Recently I've been working on trigonometric functions in regards to calculus. I was going okay, what with the quotient rule and such, but after I had differentiated the basic sin, cos, tan and their reciprocals, cosec, sec and cotan. But then when I tried to go in two directions, I failed, so basically I am asking a couple of quetions here:
1. How does one
integrate sin, sec etc. x?
2. How does one differentiate sin
-1 etc. x?
I don't want the answers; I like working it out by myself, but any pointers would really be appreciated. Cheers
Thu May 12, 2005 8:34 pm
We did this just two weeks ago!

I'll try to help you without giving you the answers.
1. You know how to differentiate the 6 basic trig functions, so to integrate sine and cosine, just go backwards.
a. [integral] sin(x) dx --> -cos(x) + C
b. [integral] cos(x) dx --> sin(x) + C
c. [integral] tan(x) dx --> change it to sin(x)/cos(x) and then use substitution. I think I set u = cos(x), so du = sin(x) dx.
d. [integral] cot(x) dx --> again, change cotangent to cos(x)/sin(x) and use substitution, setting u = sin(x).
e. [integral] sec(x) dx --> this one is incredibly strange, because you have to do the multiply by 1 thing. Multiply it by [sec(x) + tan(x)]/[sec(x) + tan(x)] and then simplify. Then use substitution, setting u = sec(x) + tan(x).
f. [integral] csc(x) dx --> do the same as above, except instead of [sec(x) + tan(x)]/[sec(x) + tan(x)], multiply it by [csc(x) - cot(x)]/[csc(x) - cot(x)]. Simplify, and then substitute.
Here is a scan of the notes I took a couple weeks ago. I hope it isn't too messy.
2. Do you mean sin^(-1) or arcsin(x)/sine inverse?
Sat May 14, 2005 10:04 pm
Vk: I'm assuming Matt meant arcsin(x), and not [sin(x)]^-1. If it were the latter, he would have simply asked for the derivative of csc(x).
All the inverse trig function derivatives can be calculated by implicit differentiation.
Let y = arcsin(x).
Therefore, sin(y) = x, (-π/2 ≤ y ≤ π/2).
Differentiating this with respect to x, we get: cos(y)*dy/dx = 1.
Solving for dy/dx, we get: dy/dx = 1/cos(y).
Now we need to invoke the pythagorean trig identity, and solve it for cos(y):
[sin(y)]^2 + [cos(y)]^2 = 1
[cos(y)]^2 = 1 - [sin(y)]^2
cos(y) = √(1-[sin(y)]^2) or -√(1-[sin(y)]^2)
However, because cos(y) is always positive on the interval [-π/2 ≤ y ≤ π/2], the possibility of having to use the negative root does not apply and we can ignore it.
So plug √(1-[sin(y)]^2) = cos(y) into dy/dx = 1/cos(y) to get:
dy/dx = 1/√(1-[sin(y)]^2).
However, because we are differentiating with respect to x, we want our answer in terms of x, if possible. Note that sin(y) = x. Therefore, [sin(y)]^2 = x^2.
dy/dx = 1/√(1-x^2)
The other inverse trig derivatives (arccos, arctan, etc) can be solved similarly using implicit differentiation.
Tue May 17, 2005 10:23 am
After reading this I'm so excited for trig and calc... I just cant wait... *twitch*
Tue May 17, 2005 10:51 am
Right. I have this stuff that was due like 3 weeks ago, which I did, but then forgot how I solved it, and lost my sheets. So... help would be greatly appreciated, and I'll love you all forever. It's probability, by the way ^.^
1) Two bowls each contain 8 pieces of fruit. In bown A there are five oranges and three apples; in bowl B there is one orange and seven apples.
-a- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples.
-b- For each bowl, find the probability that two pieces of fruit chosen at random will both be apples, when the first piece of fruit is replaced before the second is chosen.
-c- One bowl is chosen at random and from it both pieces of fruit are chosen at random without replacement. If both pieces of fruit are apples, find the probability that A was the chosen bowl.
-d- One bowl is chosen at random, and from it two apples are chosen at random, the first apple being replaced before the second is chosen. If both pieces of fruit are apples, find the probability that bowl A was the bowl chosen.
I've solved -a- and -b-, but am totally stumped on -c- and -d-.
2) Three students Dudley, Ted and Michael, have equal claim for an award. They decide to determine the winner by each tossing a coin, and the person whose coin falls unlike the other two will be the winner. If all the coins fall alike thye will toss again.
-a- Determine the probability that Dudley wins on the first toss.
-b- Given that there is a winner on the first toss, what is the probability that it is Dudley?
-c- What is the probability that the winner is not decided..
(i) In the first two tosses?
(ii) In the first n tosses?
-d- GIiven that no winner is decided in the first n tosses, what is the probability that Dudley wins on the next toss?
I've solved -b-, but am stumpd on a, c, and d.
Yeah... it's kinda long, but if you can help out, I'd appreciate it tons.
Tue May 17, 2005 10:59 am
Is it permutation/combination?
Anyway, How do you differentiate e to the power of x to the power of (3-x)?
We have not learnt how to deferentiate x with x as the power
Tue May 17, 2005 11:34 am
Alex wrote:2) Three students Dudley, Ted and Michael, have equal claim for an award. They decide to determine the winner by each tossing a coin, and the person whose coin falls unlike the other two will be the winner. If all the coins fall alike thye will toss again.
-a- Determine the probability that Dudley wins on the first toss.
Do the combinations (order: Dudley, Otherguy, Otherguy2):
HHH (reroll, counts as "lose" (because of the "on the first toss" condition))
HHT (lose)
HTH (lose)
HTT (win)
THH (win)
THT (lose)
TTH (lose)
TTT (reroll, lose)
2/8
-b- Given that there is a winner on the first toss, what is the probability that it is Dudley?
This just says: The result is neither TTT or HHH. Therefore, the possible combinations are:
HHT (lose)
HTH (lose)
HTT (win)
THH (win)
THT (lose)
TTH (lose)
2/6 = 1/3
-c- What is the probability that the winner is not decided..
(i) In the first two tosses?
What's the probability of getting a reroll twice?
2/8 * 2/8 = 4 / 64
(ii) In the first n tosses?
What's the probabliy of getting a reroll n times?
(have 2/8 chance for a reroll on every roll)
(2/8)^n
-d- GIiven that no winner is decided in the first n tosses, what is the probability that Dudley wins on the next toss?
Same as -a-. Probability has no memory.
Yeah... it's kinda long, but if you can help out, I'd appreciate it tons.
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