Recent content by Kyrix

I justed did my maths extension 1 exam today and i was wondering about atar so... decided to play around with this atar calculator http://www.atarcalculator.com.au/ When I inputted 93 HSC mark it says I get 99.95 atar Is this plausble? If so, what advantage would there be obaining say 99 HSC...

Is it possible to arrange a lesson somewhere closer to me, like the interwest regions?

I'm looking for a good Chemistry and Physics Private Tutor to help boost my Chem and Physics Marks. As of now Im ranked 2nd in Chem and 25th in Physics, in a rank 13 school. But I feel like I'm not 100% understanding the course. Id really prefer a tutor in the innerwest region - canterbury...

I think it should work for 3 objects, but not too sure and i dont know why because for two objects say ..X...Y.. In the total number of arrangements X can either be infront of Y or Y can be infront of X So number of ways of half = no of arrangements / 2 For three Objects i usually do be safe...

I think questions 6 and 9 were done wrongly. For 6, it says in how many ways it can be arranged, which means it is unordered so combinations cant be used. So number of ways of placing particular women on bow side = 4x3x2 Number of ways of placing women on stroke = 4x3 Rest of women = 3! Total...

Could you direct me to a particular video? All I could find were people teaching the basics

Combinations definately harder for me =/

Hmm yes Youre probly right. Im very weak at combinations, sorry for any confusion I could of caused =/

Wow thanks. I sort of get it now. Combinations always makes my head spin =/

Isnt this assuming that the maximum height is at the fence? but if you assume its max height you get the answer so... Imo, this is a bad question. The projectile in my experience are generally harder than this question (if they told you it was max height)

(i) Well, i dont think it works like that. If you place one person at table A and then place one person at table B, as place markers, then the rest of the 6 people can be sitted in 6! ways Place person at table A = 1 Place person at table B = 1 Left over people have a arrangement = 6x5x4x3x2x1...

Ok.. since no one is answering =/ Do you think questions in the hsc will get this hard? Because imo the hsc past paper combinations are relatively easy compared to this

Not sure how to obtain the answer to this probability question in the cambridge text book. A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining (a) one pair (answer: 352/833) (b) two pairs...