VCE Methods Question Thread!

[minhalgill]

Recall the formula for conditional probability: Pr(E | F) = Pr(E & F)/Pr(F).

The idea behind independent events is that the occurrence of one event does not affect the probability of the other event. Hence, if E and F are independent events, then Pr(E | F) = Pr(E) (that is, the probability of E occurring is not affected by the occurrence of F).

Hence, if E and F are independent, the formula for conditional probability gives: Pr(E | F) = Pr(E) = Pr(E & F)/Pr(F). And then rearranging gives Pr(E)Pr(F) = Pr(E & F).

i understand that for Pr (E) and Pr(F), but i was asking about Pr(E') and Pr(F')

[S_R_K]

i understand that for Pr (E) and Pr(F), but i was asking about Pr(E') and Pr(F')

It's true for any events. Hence it's true for E' and F'.

[minhalgill]

It's true for any events. Hence it's true for E' and F'.

so youre saying that if Pr(E) and Pr(F) are indepent, Pr(E') and Pr(F') will be independant as well?

thanks for your help

[S_R_K]

so youre saying that if Pr(E) and Pr(F) are indepent, Pr(E') and Pr(F') will be independant as well?

thanks for your help

I wasn't saying that, but that is also true.

What I was saying was that since Pr(E & F) = Pr(E)Pr(F) is true for any independent events E, F, it must also be true when E = A' and F = B'. You can just substitute those events into the formula for independent events. (It would also be true if E and F were some crazy compound events like E = ((A & B) or C') and F = ((A' or C) & B'), although you're probably never going to see a question like this.)

As for the second point, yes, it turns out that if E and F are independent, then so are E' and F'. (In fact, so are E' and F; and so are E and F'). Intuitively, it makes sense: if the chance that E occurs isn't affected by F happening, then the chance that E doesn't occur won't be affected by F not happening.

[minhalgill]

I wasn't saying that, but that is also true.

What I was saying was that since Pr(E & F) = Pr(E)Pr(F) is true for any independent events E, F, it must also be true when E = A' and F = B'. You can just substitute those events into the formula for independent events. (It would also be true if E and F were some crazy compound events like E = ((A & B) or C') and F = ((A' or C) & B'), although you're probably never going to see a question like this.)

As for the second point, yes, it turns out that if E and F are independent, then so are E' and F'. (In fact, so are E' and F; and so are E and F'). Intuitively, it makes sense: if the chance that E occurs isn't affected by F happening, then the chance that E doesn't occur won't be affected by F not happening.

woah, that cleared it all up, thanks!!!

[Unsplash]

im having trouble with part (c), drawing the graph, how would we find the endpoints of the graph? (this is paper 1 btw)

Endpoints occur at the start and end of the domain, so sub in x=0 and x=12.

[minhalgill]

how would we do part (c)

[S_R_K]

how would we do part (c)

The range of 3sin(2x), where 0 ≤ x ≤ pi is [–3, 3]. This set of numbers forms the domain of 1 – x^2. So to find the range of the composite function, just find the range of 1 – x^2 over the domain –3 ≤ x ≤ 3.

In general, to find the range of g(f(x)), first find the range of f(x), then find the range of g(x), where domain of g(x) = range of f(x).

[minhalgill]

does anyone know how to do 6?
thanks in advance

[DBA-144]

does anyone know how to do 6?
thanks in advance

 

I believe that this is right but not sure.

First factor is a repeated factor. Hence there is 1 solution there, it is turning point too.
2nd factor has 2 solutions  when u solve that equation.
3 has one dolution.
Therefore, 1+2+1=4. Hence, D

Edit, its wrong, actually C. Sorry, didnt see plus b.

[sailinginwater]

How do you do mcq 20 of the 2014 methods exam 2?

[S_R_K]

does anyone know how to do 6?
thanks in advance

The correct answer is C, two distinct solutions.

If that factorised polynomial is equal to zero, then by the null factor law, there could be a solution when each factor is equal to zero. Hence, either (x - a)^2 = 0, or x^2 + b = 0 or x^3 + c = 0.

If (x - a)^2 = 0, then there is exactly one real solution.
If x^2 + b = 0, then there are no real solutions (since b > 0).
If x^3 + c = 0, then there is exactly one real solution.

How do you do mcq 20 of the 2014 methods exam 2?

The average value of a function f(x) over the interval [a, b] is given by:
.

Which is equivalent to finding the area underneath the graph, between x = a and x = b, then dividing by the length of the interval [a, b].
For this question, the area underneath the graph can be found by breaking up the region into a rectangle and a triangle. Use the given coordinates to work out the needed dimensions of each shape.

[sailinginwater]

2007 methods cas exam 1 question 7

f(x) = xcos(3x), it is known that f’(x) = cos(3x)-3xsin(3x)

use this fact to find an antiderivative of xsin(3x)

cos(3x)-3xsin(3x)dx = xcos(3x)

-3xsin(3x)dx = xcos(3x)-(sin(3x))/3

xsin(3x)dx = ((xcos(3x)-(sin(3x))/3))/-3

is my working out and answer correct?

Is it a general rule that if we’re taking something to the other side, ie. add or subtract, we take the integral along with the term, but if it’s dividing or multiplying, we don’t take the integral, but just the term itself?

Thanks

[sailinginwater]

Is my working out correct?
It's question 9 on the 2010 methods exam 1

[sailinginwater]

Methods exam 1 2017 question 6
Is my working out in part p aligned with "hence"?