VCE Methods Question Thread!

[Phy124]

Or Let y = u^3, u = ln(2-x) dy/du = 3u^2, du/dx = -1/(2-x) etc.

[stolenclay]

With questions like

Given that

How do you deal with the fraction bit?

I suppose you could let be a function such that .

That would let you evaluate both integrals in terms of .

[BasicAcid]

I suppose you could let be a function such that .

That would let you evaluate both integrals in terms of .

How'd you go from g(x/3) to 3G(x/3)?
Which antidiff rule is that?

[Zealous]

How'd you go from g(x/3) to 3G(x/3)?
Which antidiff rule is that?

When you integrate you divide by the coefficient of x. So the one third becomes a 3.  (Think about most of the integration rules).

[BasicAcid]

When you integrate you divide by the coefficient of x. So the one third becomes a 3.  (Think about most of the integration rules).

But how do you know 1/3 is the coefficient of x in this case?
I think I'm not understanding the notation (I never use g(x/2) and all of that) haha, it looks really efficient though.

[Damoz.G]

But how do you know 1/3 is the coefficient of x in this case?
I think I'm not understanding the notation (I never use g(x/2) and all of that) haha, it looks really efficient though.

Because x/3 is the same as 1*(x/3)                                        When you expand 1*(x/3), you get x/3. True?

Think about this: What if it was 2x/3. What is the coefficient? It's 2/3.

Do you get it now?

[Alwin]

How'd you go from g(x/3) to 3G(x/3)?
Which antidiff rule is that?

From heaiyuo's working:
I suppose you could let be a function such that .

This implies, by the chain rule, that:

Note that many of you spesh kids will recognise that this is a long winded way of substitution, but using methods methods (excuse the pun) to explain it.


Assuming that you followed the rest of heaiyuo's working, we get up to the point:

This is simply because he "took out" the 3 using the rule derived above.

Hopefully you get the rest of his working It's quite elegant

Spoiler

I suppose you could let be a function such that .

That would let you evaluate both integrals in terms of .

[SocialRhubarb]

I think technically primitive functions aren't on the course either, but they are in the essentials book, so I guess that's close enough.

[shadows]

I've got 2 questions

Find probability that a person who is short, is taller 165cm. (Short person is someone who is shorter than 170cm)
This is a normal distribution question.. Apparently it is conditional. I just thought that you had to find the probability between 160 to 170.

The probabiltity that X is greater than 5 is equal to 5/13, when X is less than y. (this is conditional as well D:

How can you distinguish a conditional question?  I find it pretty difficult, lol especially when they don't give clue words like given.

2nd question.

When a question says express to the nearest mm.

will 10.1cm be suffice? or 101mm?

[TrueTears]

I've got 2 questions

Find probability that a person who is short, is taller 165cm. (Short person is someone who is shorter than 170cm)
This is a normal distribution question.. Apparently it is conditional. I just thought that you had to find the probability between 160 to 170.

The probabiltity that X is greater than 5 is equal to 5/13, when X is less than y. (this is conditional as well D:

How can you distinguish a conditional question?  I find it pretty difficult, lol especially when they don't give clue words like given.

2nd question.

When a question says express to the nearest mm.

will 10.1cm be suffice? or 101mm?

The sample space consists of all people shorter than 170cm.

Let X be the random variable that denotes height.

Then we have P(X > 165|X<170).

[San]

This is the Trial exam 2009 Heffernan exam 11 question part b
100x-pi(x^2)-2(x^2)+pi(x^2)/2
become 100x-pi(x^2)/2-2(x^2)
please explain

From a particular event, two events A and B are such Pr(A)=0.3 and Pr(B)=0.4
If A and B are mutually exclusive events find Pr(A/B)
I dont understand, what the question is asking please explain

[lzxnl]

This is the Trial exam 2009 Heffernan exam 11 question part b
100x-pi(x^2)-2(x^2)+pi(x^2)/2
become 100x-pi(x^2)/2-2(x^2)
please explain

From a particular event, two events A and B are such Pr(A)=0.3 and Pr(B)=0.4
If A and B are mutually exclusive events find Pr(A/B)
I dont understand, what the question is asking please explain

For the top expression, if you remove 100x-2x^2 from consideration, you're left with -pi(x^2)+pi(x^2)/2 which is equal to -pi(x^2)/2. That's what the bottom expression has replaced it with.

As for the bottom expression, Pr(A|B) is the probability that A happens given B occurs. Mutually exclusive means if A happens, B cannot happen and vice-versa. As an example, the events that you will take the train to school on a particular Monday and that you will walk to school on that Monday are mutually exclusive; you can take the train, you can walk to school, and you can do neither of them, but if you do one of them, you cannot do the other. So here, Pr(A|B) is just 0 because if one of A or B occurs, the other cannot occur.

[BasicAcid]

From a particular event, two events A and B are such Pr(A)=0.3 and Pr(B)=0.4
If A and B are mutually exclusive events find Pr(A/B)
I dont understand, what the question is asking please explain

Easiest way is just to apply the formula.

Pr(A|B) = Pr(A intersect B) / Pr(B) - From the formula sheet.

But as A and B are mutually exclusive, Pr(A intersect B) = 0
(That's what mutually exclusive means, the intersection between two items are 0)

Therefore 0/0.4 = 0

[TrueTears]

(That's what mutually exclusive means, the intersection between two items are 0)

just a tiny correction, intersection between two m.e events is the null set, the probability of the intersection is 0

[rhinwarr]

Given x=3.0 is an approximate solution to the equation , use the approximation formula to find a better approximation of the solution.