Recent content by kentm

I understand the issue with my thinking now. The movie never really explains the algorithm. Ben simply says, "before my chances wer 33.3%, and now they are 66.7%. When I looked up the solution, I found explanations that act as though the 2/3 combined probability from doors 2 and 3 transfer...

Edit: Apparantly I screwed up the quote tag. It actually looks fine to me, but w/e. [SIZE="1"]

I guess, I am just saying that you could use the same statistics to show that it would be best to stay with your first choice. It's the exact same reasoning as far as I can tell. Each door has a probability of 1/3 so together (door 1 and 3) there is a probability 2/3 that one of those doors...

The movie "21" went over a statistics problem in an attempt to make the main character look smart. I'm not sure though that the Solution was correct, even though it does seem so. The problem is this: A game show host gives you a selection of 3 doors, two of which enclose goats, and the...

If you are just trying to be prepared for you college courses, your first calculus course will probably go over what you've already learned: Calculus I: Limits, Derivatives, Applications of Derivatives, Anti derivatives/Definite Integrals. Calculus II: Applications of Integrals, Integration...

The washer method is pretty simple, though it's not the best way to do this problem. The area of a washer is \pi(r_1^{}^2-r_1^{}^2) where r1 is the radius of the outer circle, and r2 is the radius of the inner circle. Which gives us a standard function for the volume of solids where you are...

Nothing happens to that limit as x changes, maybe that's why your calculator comes up with something different. \[ \lim_{\theta \to 0} \frac{\cos\theta \tan\theta}{\theta}\]

I am also taking Calc 2, and we just so happen to be working on the same thing. You are correct about using the Shell method for this. You use the shell method for rotation around a vertical line. This is the formula used: V=\int_a^b 2\pi(Radius)(Height)\,dx. In this case your shell...

I forgot to bring down the 36, so the answer is correct: 3^2\pi=\frac{36\pi}{4} plugging in the numbers to the equation: Area=\pi r^2, we end up with a true statement: 9\pi=9\pi Oh well, it was fun anyways.

Haha, I don't know where that came from. So: \frac{1}{2}\int_0^{\frac{\pi}{2}} 1+cos2x\,dx=\frac{1}{2}\left[x+\frac{1}{2}sin2x\right]_0^{\frac{\pi}{2}} It's good to know this stuff, but the way this ended up working out doesn't seem to work. I'm assuming that I don't know what pi is, hence...

Cool, that actually makes a bit of sense. If you told me what the substitution was I could have done it, I think. Where that substitution came from, you're right, is beyond me. It must take a little practice. Now, after having learned something I see that: \begin{math}A=36...

I'm taking calculus 2, but this isn't for school. We haven't covered trig substitution. A key here is that I'm not interested in the area of the circle. I'm only interested in the area of the circle, and the radius of the circle because they are needed for this...

I was just starting to get comfortable with integrals in the easy situations, and then with substitution in some of the more difficult cases. I thought it might be kind of fun to try and calculate pi with the area under the semi circle: \begin{math}y=\sqrt{9-x^2}\end{math} I figured the...

Okay, now I've got another question: What if I want to have negative bounds like this: \int_-3^3? pdflatex doesn't seem to like that very well.

That's exactly what I needed... Thanks.