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No news here. I've always used a table to do it, but actually do it in columns rather than rows. x ... f(x) ... W ... P

These are official, courtesy of the Australian Mathematics Trust.

As far as I'm aware, there's no "security period" as such. It's supposed to be done on the same day at the same time - or not at all. After that it's basically public domain. So best of luck for Monday! I'm sure there will be answers up by Monday afternoon.

No brainer. Maths. As far as I'm aware, NSW is a state of the Commonwealth of Australia, and no part, in any way, of the United States of America. As soon as the yanks learn to speak and spell properly the world will be a better and safer place.

Re: Australian Mathematics Competition 19 = [5(3 – √3)]/2 22 = 12(5√3 + 9)

Re: How did people go in the maths comp today? I thought it was easier than that. I looked at f1(x) = x. It's a quadratic with solutions of -2 and 3. Then I looked at f2(x) = x. It's a quadratic with solutions of -2 and 3. Then I looked at f3(x) = x. It's a quadratic with solutions of -2 and...

This is just a compound interest question over 4 years. A = P(1 + r)n = 28000 x (1 + 0.09)4 = 28000 x (1.09)4 = 39524.28508 .: Estimated population at start of 2001 is about 39500. This is a future value question - you can tell its an annuity because Adriana saves the same amount each...

I'd do both. Have a brief set of notes for all the topics, summarised to 3-4 pages for each topic. An explanation, a formula, an example. But then really focus your time on doing lots of questions.

Actually, it's possible to get z-scores higher than 3, but it's just very unlikely. Remember that only 99.7% of scores are within 3 standard deviations from the mean, so there are still others with higher z-scores.

Practice! Seriously though, 1. FV formula - use when question talks about stuff like what will the investment grow to after 20 years, or in 20 years time, or something like that. 2. little PV formula - it's just the reverse of the compound interest formula - use when the question says something...

I think you may have found the SAMPLE deviation (sn-1) instead of STANDARD deviation (sn). I get the following stats: Maths (mean = 77.3, standard deviation = 8.3) English (mean = 74.9, standard deviation = 11.6) Then I usually draw up a quick number line to see if that will solve the problem...

Since the total side length is 12 metres, and sides are bent up from both ends, the length and width of the box produced will be (12 – 2x) metres, and the height is x metres. Now the length and width must give a positive number, so: 12 – 2x > 0 –2x > – 12 x < 6 Or, in other words, x can't be...

As previously stated, a tree diagram is most useful here. It's a very visual means of making sure every possibility is counted. With lots of practice, a tree diagram isn't vital. The student is either chosen first, or chosen second. P(first) = 1/5 P(second) = 4/5 x 1/4 = 1/5 .: P(chosen)...

Patrice is born in a particular month. The probability that Henry was born in a different month is 11/12. The probability that Joel was born in a different month to Patrice AND Henry is 10/12. Probability that all three born in a different month = 1 x 11/12 x 10/12 = 55/72

Remember that these are overall statistics, rather than a physical count, so when using a tree, take care NOT to reduce the fractions. For this I'd just use decimals rather than fractions. (a) P(both pass) = 0.9 x 0.6 = 0.54 = 54% = 27/50 (b) P(Carl pass, Sarah fail) = 0.6 x 0.1 = 0.06...

Can't be proved. More information required. You could always show a counter example which disproves it.

That's what I came up with too. I'm just wondering if there's supposed to be a simpler factorial notation that magically appears. It's stumped one of the 3u teachers at our school (and me).

The only suggestion I'd make is to perhaps write the word "Domain" rather than just "D", although it's probably no big deal. I prefer: Domain: {x: x > 0, x ≠ 4} It's terribly old fashioned and quaint, but it's just habit for me now.

Can anyone do anything with this? Simplify: n x (n – 1) x (n – 2) x ... x k x ... x 3 x 2 x 1 ----------------------------------------------------- (n – k + 1) x (n – k) x ... x 3 x 2 x 1 Thanks...

• Introduce the variables. • Write down an equation connecting them. • Obtain an expression for Q, the quantity you want to maximise or minimise. • Express Q in terms of one variable using the equation from step 2. • Differentiate and apply tests.