VCE Methods Question Thread!

[kensan]

Hey guys, I'm guessing this is a pretty easy question, but I don't know how to do it.

It says:

For f(x) = x^3 + 2x, the average rate of change with respect to x over the interval [1,5] is:

Answer being 33


Also:

The average value of the function f(x) = e^(2x)cos(3x) for 0<=x<=PI is closest to:

with the answer being -26.3

Explanation on this form of rate of change would be great.
Cheeers

With the second one you are using the formula   You use this when is says 'average value''

[Shenz0r]

This may be strange... but can you look through your theory book BEFORE reading time? (i.e. when you go sit down in the exam room and 15 minutes reading hasn't started)

And Calculators - are we allowed to use that as well during this time?

1. No, don't. You can when reading time has commenced, though.

2. No, I wouldn't risk it.

[Homer]

with exam 2 2006 q.3b, why haven't they included

[supermanflyaway]

I'm really stuck on  c)iii and I don't understand that first line of the solutions
I put the VCAA answers next to the corresponding question to save you guys some time

Oops I probably should include that f(x)= x^3e^-2x
(Image removed from quote.)

bump this

[monkeywantsabanana]

For a 2 mark probability question that requires calculator (e.g. 3a, 3b 2009 Exam 2), what are the two marks allocated to if we aren't allowed to write down "technology syntax"?

Would Pr (X<68.5)= 0.9332 give us the whole two marks?

[BubbleWrapMan]

Say we have a tangent at . Then the gradient of the tangent is .

Also, , so the tangent passes through .

Using the above, the equation of the tangent is , and it passes through .

So,



Any tangent that passes through the origin has or

For a 2 mark probability question that requires calculator (e.g. 3a, 3b 2009 Exam 2), what are the two marks allocated to if we aren't allowed to write down "technology syntax"?

Would Pr (X<68.5)= 0.9332 give us the whole two marks?

Before that just write , with the actual values of mu and sigma

with exam 2 2006 q.3b, why haven't they included

That would make negative

[e^1]

I have trouble with the question (see attachment).

If you can do it, then know Pr(X > 30) = 0.05

[BubbleWrapMan]

Before I get ahead of myself, do you have the answer?

[e^1]

Before I get ahead of myself, do you have the answer?

Yes , that's the part where I don't understand it.

[BubbleWrapMan]

Well you are reduced to two situations when choosing a tomato - it's either more or less than 30 g. This implies we can apply the binomial distribution, with p = 0.05 and n = 10.

We want Pr(X ≥ 2) > 0.95, which is the same as 1 - Pr(X=1) - Pr(X=0) > 0.95. Then you use the binomial formula to solve the inequality, and round your answer UP.

[e^1]

Well you are reduced to two situations when choosing a tomato - it's either more or less than 30 g. This implies we can apply the binomial distribution, with p = 0.05 and n = 10.

We want Pr(X ≥ 2) > 0.95, which is the same as 1 - Pr(X=1) - Pr(X=0) > 0.95. Then you use the binomial formula to solve the inequality, and round your answer UP.

Yeah I get that, but I don't get the line . Shouldn't it be ?

[BubbleWrapMan]

Ergh what the hell, yes, you're right.

[e^1]

Ergh what the hell, yes, you're right.

Thank you!
I have one more question (attachment).

Note:   

[BubbleWrapMan]

the max is at x = 20/3, so evaluate db at x = 20/3 - 2 or x = 20/3 + 2

[chestiepop]

solve for y