Recent content by kakab00

If I knew the easier method I wouldn't be here asking :shy: Maybe I calculated something wrongly up there? Or maybe the school system just sucks :-p

hehe but the answers are still wrong I don't really get your method either, because I haven't learned them yet. I think the school wouldn't have set something they hadn't taught yet, so I believe there's still another (easier( way ) of doing, maybe some other members might know? Thanks...

The answers are still wrong though, I used a) 4 (round off to one significant figure) b) 3.0 + - 0.2 (this was the value I calculated from the equation you gave me)

I have no idea how to do it.. upon further thought, even without finding the derivative for a vs s, the answer already seems to be wrong from the 1st and 2nd part of the equation, square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square] would give you a value of around 4 which is still...

it's not easy for me XD for the 2nd part I tried using your method: so delta a = square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square + (0.1 x 50 x 2)square)] = root(150) which would give you a value of 12.2, much bigger than a=3 calculated. doesn't make sense, or did I do something wrongly?

I don't know the exact answers, I only know whether the answers would be right or wrong because we have to key in them inside a system after we found them. I've not learned partial derivatives yet though, so I think there should be another easier way to do it using delta/fractional uncertainty?

I don't get your equations, wouldn't the delta u and delta u or other symbols cancel each other out?

Homework Statement http://img487.imageshack.us/img487/5702/untitledna1.png Homework Equations v square = u square + 2as The Attempt at a Solution a) Let v square - u square be x delta x/x = 2(delta v/v) + 2(delta u/u) delta x/300 = 0.03 delta x = 9 b) i) a = (v square - u...

the other ways to get an ace would be 1/27 ??(the one that was drawn from the first deck)

3/51 x 1/27 is the chance that you would pick one of the other aces from the half deck

probability of drawing 'other cards' would be 48/51 x 1/27? why would you still need this anyway, I thought the answer would be just 3/51 X 1/27 + 1/27(the ace that was drawn from the first deck)

sadly I don't really quite know how , I'm pretty weak in this chapter 3/52 X 1/27?

So you have to use conditional probility? Assuming A is the situation where the 2nd half-deck has only 1 ace B is the situation where the 2nd half-deck has 2 aces C is the situation where the 2nd half-deck has 3 aces I is the situation where the ace that was picked was choosen D is the...

Homework Statement A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns out to be an ace. The ace is then placed in the 2nd half –deck. The half is then shuffled and a card is drawn from it. Compute the probability...

cool I got it, thanks for all of your help :!)