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Yes to your second sentence.

I have set up my account so that the newest posts in a thread are at the top of a page - which means I need to go to the end of the thread to see the original question. I think most people have the reverse. Instead, could it be set up so that the original question is highlighted at the top of...

Centripetal force is NEVER directed outwards. It is the force that makes the object travel in a circle. It is always directed towards the centre (which is basically what the word 'centripetal' means), and for uniform circular motion that is the only net force. You should be calling the...

It is hard to interpret what you are asking. Do you mean 'radial' instead of 'horizontal'? Are you referring only to a horizontal circle? Because in a non-horizontal circle, 'horizontal resultant force' doesn't mean much. Are you referring only to uniform motion, or are you considering the...

I NEVER get logged out. I can leave it all day and I will still be logged on at the end of the day. I can also close down my browser and shut down my computer, and I will be logged on next time I return. Not sure what I am doing differently to you though.

Either I am misinterpreting the question, or the answer to question 14a is not correct. If n=2, k=2, you are making ORDERED arrangements of two distinct digits, so the answer should be 10P2. But your formula gives 10C2. (The question talks of 'combinations', yet it says that order is...

Re: HSC 2015 4U Marathon Only the limited version involving integer powers.

I am going to guess that your first line is an intermediate stage in your working for a different question. If so, could you post the actual question - there might be an easier way to do it.

Re: HSC 2015 3U Marathon But in (1), you have only jumped three at a time (or multipled of three at a time). How do you test out a more random selection?

Re: HSC 2015 3U Marathon If I blindly accept the following two statements, then the calculation itself is fine. But they are not obvious to me. (1) Similarly, the game cannot be won if an isolated card is created on the 3rd, 4th, 5th, ..., 11th turn. (2) This is because any subsequent...

Re: HSC 2015 3U Marathon I've found the source for your question, and I don't understand their solution. Do you?

Re: 2015 permutation X2 marathon The only prisoner's dilemma I'm aware of is whether or not he should pick up the soap.

Re: HSC 2015 3U Marathon Same question with a twist: 20 guests are placed by the same process around a round table. What is the minimum number of seats required at the table so that the probability is less than 5% that when Alice sits down both seats adjacent to her are occupied? Answer...

There are LOTS of people who could do really well at Ext 2 maths, but who drop maths altogether because their other subjects are even better. They typically get very good ATARS.

Re: HSC 2015 3U Marathon The answer is 1/3 regardless of the number of people. Rewind the seating process so that you see people standing and leaving the table. You need only focus on Alice and the two seats to her side. There is one chance in three that Alice will be the first of those three...

The first line is simply adding up all the Ak's. The second line is the result of using part (a), as directed by the question.

Re: 2015 permutation X2 marathon Aren't all picks "not looking"? It is hard to pick randomly if you look.

Re: 2015 permutation X2 marathon Something that is interesting about the Monty Hall problem (assuming the host always reveals a goat, so the probabilities are 1/3 and 2/3): If someone walks onto the set after the host has opened a door (so he doesn't know the player's original choice) and bets...

Re: 2015 permutation X2 marathon AND ..... you KNOW the host's strategy of only ever revealing a goat.

Re: 2015 permutation X2 marathon If the distribution wasn't even in one variable, you would have to integrate the distribution itself. If it was uneven in both variables, you would need a double integral. That is my understanding anyway - correct me if I'm wrong.