VCE Methods Question Thread!

[pugs]

wats x and wats multiply sign, i am blind.
Uhh try to put multiply sign in the middle of variables x and n.

whoops sorry i should have made it clearer --> there aren't any x's in the equation, so they're multiply signs

and i've tried with the multiply sign and it still doesn't work

[Tau]

hi guys, i'm not sure if anyone else has experienced this, but i'm having trouble using the solve function on my calculator with equations involving n & log_e(x)

i've been trying to use the solve function on my calculator to find the answer to
(0.8 )ⁿ + n x 0.8⁽ⁿ⁻¹⁾ × 0.2 < 0.05
^^ that is supposed to be 0 point 8 btw, pls ignore the gap! idk how to pull up the maths writing tool

ideally, it would give me a decimal answer, but i'm receiving 5ⁿ - 5 x (n + 4) x 4ⁿ > 0 instead

is there any way to fix this/is there something i'm doing wrong? i really don't want this to happen during my sac/exams haha

thanks!!

For some complex inequalities the CAS does not solve properly. You can try using the ‘nSolve’ function to numerically solve the inequality.

[pugs]

For some complex inequalities the CAS does not solve properly. You can try using the ‘nSolve’ function to numerically solve the inequality.

ah i just tried this and it came up with 'error: argument error' :/

[redpanda83]

ah i just tried this and it came up with 'error: argument error' :/

use = sign instead of <
.............
UPDATE
...............
So i just graphed it see why solve command wasnt working.
As you can see the the equation actually has an asymptote (y=0) on the latter part. That could be the reason why the calulator wasnt able to do it normally.
so you will have two answers. n<-3.91654 and n>21.7744

[pugs]

use = sign instead of <
.............
UPDATE
...............
So i just graphed it see why solve command wasnt working.
As you can see the the equation actually has an asymptote (y=0) on the latter part. That could be the reason why the calulator wasnt able to do it normally.
so you will have two answers. n<-3.91654 and n>21.7744

oooo thank you so much! and thank you so much to Tau as well!!

yea i tried the = sign with the normal solve and it worked as well, it probably needs some sort of domain for it to work with the < sign¿ idk lol

but thank you heaps! and the visual reference with the graph made it even clearer

[redpanda83]

oooo thank you so much! and thank you so much to Tau as well!!

yea i tried the = sign with the normal solve and it worked as well, it probably needs some sort of domain for it to work with the < sign¿ idk lol

but thank you heaps! and the visual reference with the graph made it even clearer

Np.
Take this as a rule of thumb, graph/visualise the problem and it just gets easier .

[peachxmh]

Hello, can someone please explain why the mean of the continuous random variable with the probability density function f(x)=1/x², x≥1 does not exist? Thank uuuu

[Tau]

Hello, can someone please explain why the mean of the continuous random variable with the probability density function f(x)=1/x², x≥1 does not exist? Thank uuuu

Well in this case, we are looking for the (improper) integral from 1 to infinity of  . The problem is that this integral doesn’t converge, in that there is no finite value we can meaningfully assign a value to. Thus, the mean does not exist.

[radiant roses]

Hi, I am having trouble with this mcq from the 2015 methods exam 2.
The answer is D but I am not sure how to get to that.
I know that the average value is 0, so that means 1/(p+2) x the integral of f(x) from -2 to p is 0. I also know that the negative integral from -2 to 0 = 25/8. I don't know what to do from there. It would be great if you could help!

[Tau]

Hi, I am having trouble with this mcq from the 2015 methods exam 2.
The answer is D but I am not sure how to get to that.
I know that the average value is 0, so that means 1/(p+2) x the integral of f(x) from -2 to p is 0. I also know that the negative integral from -2 to 0 = 25/8. I don't know what to do from there. It would be great if you could help!

The general formula for average value is  .

Now we know that . In other words, the area under the graph from 0 to p must be the same as the area under the graph from -2 to 0. Now, this area is just a triangle, with base and height p, so . Which is answer D, as given. We discard the negative solution, as p is greater than zero.

[Jackson.Sprigg]

I'm getting 385 for this question but tbh I have no idea of my own logic so please help

How many students would the organisers for a year 12 end-of-year function need to survey to establish the preffered venue from a list of 2 at an error margin of 5% and a confidence level of 95%?

The part that I think I'm missing is what the list of 2 means.

I've done it like so:

Error margin = z*sqrt[(p(1-p))/n]                Where p is p hat

Then I subbed Error margin for 0.05, z=1.96 and p=05

Then solved for n and got 385 (rounded up)

Their answer is 271 so I believe my error is with what p is equal to

Thanks for any help! 

[AlphaZero]

I'm getting 385 for this question but tbh I have no idea of my own logic so please help

How many students would the organisers for a year 12 end-of-year function need to survey to establish the preffered venue from a list of 2 at an error margin of 5% and a confidence level of 95%?

The part that I think I'm missing is what the list of 2 means.

I've done it like so:

Error margin = z*sqrt[(p(1-p))/n]                Where p is p hat

Then I subbed Error margin for 0.05, z=1.96 and p=05

Then solved for n and got 385 (rounded up)

Their answer is 271 so I believe my error is with what p is equal to

Thanks for any help! 

The question as it's presented at the moment has several problems with it. To answer it, you need an estimate of proportion of students who prefer one venue of the other. It's not correct to assume that equal proportions of students prefer each venue. We simply don't have enough information.

Was the only part of the question, or is there more to it?

Further it's not actually feasible to obtain a margin of error of exactly 5%. The question should ask for the minimum number of students that should be surveyed so that the margin of error is no more than 5%. Only then does rounding up become justified.

[Jackson.Sprigg]

Ha I hate it when they do this. Thanks for confirming my suspicions. That's all there is to the question. I got it from the Nelson VCE Mathematical Methods Unit 4 book my teacher gave me and this whole section has been dodgy. The rest of the book has questions from previous VCAA exams but for the last chapter they've been trying to make their own questions as there mustn't be many available and for at least half of them I can see how they've sorta solved them but they keep assuming things that shouldn't be assumed and wording questions in really dumb ambiguous ways.

Thanks Alpha, you legend

[milanander]

Sorry for the dumb question, but how relevant are kinematics? My school skipped that chapter of the textbook completely and we never had any SAC questions on kinematics. I couldn't find much questions on this topic in past VCAA exams either, however it's on the study design.

Thanks.

[S_R_K]

Sorry for the dumb question, but how relevant are kinematics? My school skipped that chapter of the textbook completely and we never had any SAC questions on kinematics. I couldn't find much questions on this topic in past VCAA exams either, however it's on the study design.

Thanks.

Just to be clear about what's on the study design: in Unit 2 you are required to know the application of differential calculus to rectilinear motion, with position/velocity/acceleration given as functions of time. Units 3 & 4 does not explicitly mention applications of differential calculus to kinematics, but Unit 2 is presumed knowledge. In addition Units 3 & 4 mentions the application of integration to finding the distance travelled using a speed-time graph.

This is pretty minimal, and to cover your bases you should do a few questions to consolidate your understanding – even if your school hasn't assigned any questions. I think the major textbooks also treat it pretty briefly. But you are correct, it is hardly assessed.