Vincezor's Methods questions

[Vincezor]

Hey there,

apologies for starting a new thread - I thought it would be better than hijacking someone else's thread or posting in thushan's help thread. This is because as we are reaching the business end of the methods course, I realise that I will probably have a lot of questions I need help with. Thus, it would be beneficial to create my own thread to do so...

Anyway, first question:

[Solved]1) Solve the inequality:


Yes, it seems really simple, yet I am not sure on how to tackle to problem. Any help is appreciated!

Question 1 Solution Here

[|ll|lll|]

x+3 = 2x-4 or x+3 = -(2x-4) or -(x+3) = - (2x-4) or - (x+3) = 2x-4

Effectively, you'll realise that you will get two repeated solutions for four equations, but you have to show all four though.

[b^3]

There are three situations here for the mods. When they both have the original function which is postive, when they both have the original function being negative, and when one of the original function is positive and the other negative.
When they are both either positive or negative, they result in the same working.
i.e. and
so

When one is positive and one is negative
or


So the intersection of these two domains will be
I hope that makes sense.
EDIT: beaten

[luffy]

Hey there,

apologies for starting a new thread - I thought it would be better than hijacking someone else's thread or posting in thushan's help thread. This is because as we are reaching the business end of the methods course, I realise that I will probably have a lot of questions I need help with. Thus, it would be beneficial to create my own thread to do so...

Anyway, first question:

Solve the inequality:


Yes, it seems really simple, yet I am not sure on how to tackle to problem. Any help is appreciated!

I think it'd be best if you simply drew the two graphs first. Drawing graphs for inequalities is essential to minimizing those silly mistakes in my opinion. After you have drawn the two graphs (i.e. y= |x+3| and y = |2x-4|) on the same set of axes, then find the intersections and the set of solutions will become obvious. You should end up with .

EDIT: Beaten as well

x+3 = 2x-4 or x+3 = -(2x-4) or -(x+3) = - (2x-4) or - (x+3) = 2x-4

Effectively, you'll realise that you will get two repeated solutions for four equations, but you have to show all four though.

I think you sort of skipped the "inequalities" part of the question. Also, are you sure you have to show all four solutions? I was never taught to do all 4 in methods last year, nor do I think VCAA would take marks off for not doing so. However, I could be wrong.

[paulsterio]

I agree with "luffy" - I personally always draw graphs with inequalities, you can find out the intersection points algebraically then graph them, with a graph, should be easy to see what you're doing

[Vincezor]

Thanks guys.

Another question:

[Solved]2) In a binomial experiment with 6 trials . Find the variance of this distribution.

My attempt:







Now , where n=6





is this correct? The answer says but I have no idea why!

I know why! I can't count! T_T

[Andiio]

Thanks guys.

Another question:

2) In a binomial experiment with 6 trials . Find the variance of this distribution.

My attempt:







Now , where n=6





is this correct? The answer says but I have no idea why!

(27/400)/20 = 27/20?

[Vincezor]

(27/400)/20 = 27/20?

Argh fml.. it's probably time to head off to sleep me thinks

haha, so embarassing! cheers Andiio for pointing it out! I'll make a correction hahahaha

[paulsterio]

LOL! It's funny cause I notice it now, but when I was reading through your solution, Vincezor, I didn't even pick up that arithmetic error!

[Vincezor]

[Solved]3) Katia and Mikki play a game in which a fair six-sided die is thrown five times. Katia will receive $1 from Mikki if there is an odd number of sixes, and Mikki will receive $x from Katia if there is an even number of sizes. Find x for the game to be fair. (Count zero as even.)

Man, I have no idea where to start with this :/ When it says that the game is 'fair', does that mean they both equally have the same probability of winning $1 (or losing $1)?

Question 3 Solution Here

[luffy]

First, you need to find the probability of each player winning the game.
Let Z~Bi(n=6, p =1/6)

Pr(Mikki winning $1) = Pr(Z = 1) + Pr(Z = 3) + Pr(Z = 5)
= 0.4342 approx
Pr(Katia winning $x) = Pr(Z = 0) + Pr(Z = 2) + Pr(Z = 4)
= 0.5658 (you could have just done 1 - Pr(Mikki wins))

Now, let X represent this system, denoting the amount won by Mikki.
A game being "fair" implies an expected value of $0 for both players. i.e. neither of them will win or lose money in the long-term.

Thus,
E(X) = 0.4342(1) - x(0.5658) = 0
Therefore, x = 0.4342/0.5658
=0.7673
Since, x is a dollar amount, it would be: $0.77 (or $0.75 realistically).

Hope I didn't make any mistakes.

[Vincezor]

Ahhh that makes sense. I guess essentially I was wondering what 'fair game' meant - I remember seeing this near the start of probability, and I guess I forgot about it after not being exposed to these types of questions for a while.

Thanks again

[paulsterio]

you're way too fast luffy! leave me some questions!

[Vincezor]

SPOILER: If you haven't done Insight 2011, shield your eyes!


4)

I honestly don't understand this question, and the answers confuse the s**t out of me

Doesn't the random "k" value dictate whether he will make it or not? for instance as k increases, he will be able to make it over the obstacle. So we have to find the lowest "K value" (even though it can any number) so that the general equation passes (10,1).
 
So changing the k value essentially just translates the graph up and down? (I think anyway :S) I mean dilation! T_T

So now I think what its asking is how much you can translate the graph to the right so that It wont cross the graph of the obstacle ,


Haha as I'm typing this, I'm confusing my self even more, so maybe disregard my "reasoning"

[paulsterio]

I never got my head around that question either, I think Luffy might be the one to ask in regards to something like this

But no, this is a dilation, so changing the value of k will dilate the graph. I remember getting the right answer, even though I didn't fully understand what I was doing

I just substituted (10,1) into y = 9/10(20-x^2/k^2) and solved for k, but then considering it says "passes" i remember it being an inequality