Recent content by iceblits

maybe all of mathematical calculation CAN be reduced to addition (or at least the broader idea of the operation)...but that's like saying that chess can be reduced to a series of hand movements, the flute a series of finger movements. Although both of these things are true, the idea is too...

Homework Statement Let α(t) be a regular, parametrized curve in the xy plane viewed as a subset of ℝ^3. Let p be a fixed point not on the curve. Let u be a fixed vector. Let θ(t) be the angle that α(t)-p makes with the direction u. Prove that: θ'(t)=||α'(t) X (α(t)-p)||/(||(α(t)-p)||)^2...

Oh my gosh I can't believe I even posted this question haha!..its a constant of course

Suppose x(t) is a curve in ℝ^2 satisfying x*x'=0 where * is the dot product. Show that x(t) is a circle. The hint says find the derivative of ||x(t)||^2 which is zero and doesn't tell me much. I was hoping for x*x= r, r a constant.

So its the same...!

Oh so is it really something as simple as: 4/52+13/52-1/52 lol?

So it should be just 13/52?

Using this method do i get: (4*51!+13*52*51!-52*1*50!)/(52!)?

Last card i would say... 1-(51!/(52!))? Edit: that's not right

Oh! So it should be a factorial..i forgot about the ways to arrange the rest of the cards

is it.. The first card is an ace in: 4*51 cases The second card is a club in: 52*13 cases The second card an ace of clubs: 52*1

Oh are you saying: AUB=A+B-A(and)B implies AUB=(4/52)+(13/51)-(the answer from part b)?

Oh I see!.. So would the and question then be: P(ace first)*P(clubs on second draw|probability of ace on first draw)=4/52*12/51 edit: just realized you are referring to the (b)...I think I'm still stuck on (a)...I will think about what you said

I think that is my second case. AUB=A+B-A(And)B right? so are you saying then that a.) should be p(ace)+p(clubs)-p(ace of clubs) so (4/52)+(13/51)-(1/52)? I guess I am just confused about what to do with the drawing of two cards..because drawing one card decreases the number

Homework Statement You draw 2 cards from a standard deck of cards without replacement. a.) what is the probability that the first card is an ace OR the second card is a clubs. b.) what is the probability that the first card is an ace AND the second card is a clubs Homework Equations...