4U Maths Question Thread

[RuiAce]

Hi all,

Just a general question if anyone knows.
If we copy down a diagram (e.g. circle geometry) and add a new element (e.g. a new point, or line), do we still need to write down what was added?

Thanks

I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.

But if you're in a rush for time you might get away without doing it.

[lsong]

I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.

But if you're in a rush for time you might get away without doing it.

Noted, Thanks Rui

[uusunny]

How would I do this question?
A funfair game has the following set up: a player may toss two balls into any of k chutes and if both balls are returned via a single chute, the player wins. What is the probability that both balls enter via different chutes, but are returned via the same chute, where that chute is neither of the chutes by which any ball entered?

The answer is (k^2 - 3k +2)/ k^3

Thank you!

[RuiAce]

How would I do this question?
A funfair game has the following set up: a player may toss two balls into any of k chutes and if both balls are returned via a single chute, the player wins. What is the probability that both balls enter via different chutes, but are returned via the same chute, where that chute is neither of the chutes by which any ball entered?

The answer is (k^2 - 3k +2)/ k^3

Thank you!

[armtistic]

Hey,
This implicit differentiation is simple but I didn't know that  dy/dx being a constant meant that the original relation represents two parallel lines.
So I was wondering if this is just something you have to know or if there is some explanation for how they concluded that.

[RuiAce]

Hey,
This implicit differentiation is simple but I didn't know that  dy/dx being a constant meant that the original relation represents two parallel lines.
So I was wondering if this is just something you have to know or if there is some explanation for how they concluded that.

(where by extension, that constant is in fact the gradient of the line itself.)
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The full, proper reason why it's not a coincidence how we have a pair of parallel lines somewhat relies on a more formal understanding of either conics and/or graphs that the HSC doesn't teach. A pair of parallel lines is really one of the "degenerate" cases of the conic sections
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[johnk21]

Can someone please help me with this probability q. Thanks

[Opengangs]

Can someone please help me with this probability q. Thanks

a) Total no. of outcomes is given by 9P3 = 504
Now, if it succeeds 400, the first number simply has to be greater than 3.
Thus, final probability = (6P1 * 8 * 7)/504 = 2/3

b) Again, total no. of outcomes is 9P3.
Now, if we were to randomly choose three numbers, there's only one way to arrange them.
So, the total number of arrangements we'll get is: 1 + 1 + 1 + 1 + 1 + ... + 1, since in each selection, there's only one way in arranging them in descending order.

The fact that there are 9C3 ways of arranging them is revealed when we count the fact that (1, 2, 3) arranged would be the same as (1, 3, 2) arranged, which is the same as (3, 2, 1) arranged.

This brings us to conclude that there is a probability of 9C3/9P3 = 1/6

[RuiAce]

b) Again, total probability is 9P3.
Now, if we were to randomly choose three numbers, there's only one way to do so.
So, the total number of arrangements we'll get is: 1 + 1 + 1 + 1 + 1 + ... + 1, since in each arrangement, there's only one way in arranging them in descending order.

The fact that there are 9C3 ways of arranging them is revealed when we count the fact that (1, 2, 3) arranged would be the same as (1, 3, 2) arranged, which is the same as (3, 2, 1) arranged.

This brings us to conclude that there is a probability of 9C3/9P3 = 1/6

Remark: Total outcomes is different to total probability. The sum of all probabilities is 1.

Actually, \( \binom{9}3 \) is the number of ways they can be chosen. Once the three numbers have been chosen, there is only one way for them to be arranged in descending order. We must not get mixed up between what a "selection" and an "arrangement" is in our reasoning.

[Shlomo314]

Hey Atar team,

Could someone give me a hand on 2016 ext 2 multi choice for 7,9, and 10 please?

Cheers

[Natasha.97]

Hey Atar team,

Could someone give me a hand on 2016 ext 2 multi choice for 7,9, and 10 please?

Cheers

Hi!
Rui has uploaded his explanations here :)

[Shlomo314]

Hi,
I was just view the 2016 solutions that RuiAce post and Q16 says I need to update 3rd party hosting. How do I do that

Cheers

[RuiAce]

Hi,
I was just view the 2016 solutions that RuiAce post and Q16 says I need to update 3rd party hosting. How do I do that

Cheers

At some point I had reached the limit for third party hosting with photobucket. I'll extract the images and put them on another image hosting site instead.

[beau77bro]

i dont get the jump from the equals to the inequality sign. 2011 HSC q8ci btw. but i only need the working explained not the question. it is basically the triangle inequality thing - if so could u explain both how it works and how the hell we were meant to see that


and how do we know its a Geometric Progression?!?!? because what does M represent - it doesn't really say. so do all the coefficients equal M? that doesn't make sense doe it?

[RuiAce]

i dont get the jump from the equals to the inequality sign. 2011 HSC q8ci btw. but i only need the working explained not the question. it is basically the triangle inequality thing - if so could u explain both how it works and how the hell we were meant to see that

I gave a brief mention to this in my book. It's quite unfair and I couldn't get it out at the time even though it makes sense now.



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As for how the hell you were meant to see it?

Short answer: You don't. Unless you saw the next step