VCE General & Further Maths Question Thread!

[AngelWings]

I dont get why standard deviation doesnt change, i just accept that it just doesnt.
The spread will not move if the points are directly changed. That's because it is according to the position of the mean,
Not to sure about that, is that because the ratio of the other points to the mean stay the same?
Not to sure what sd is :/
Thank you so much!

It takes a lot more explaining than the time I have now, but I'd be happy to come back at the end of next week to discuss this if you wish. I currently am on exams (2 down, 3 more to go) and right now on a study break, so I'll come back to this soon.
However, you can kind of say that the ratio remains the same. I guess that's one way you can think of it.

I'm back after a wild 36 hours the day before yesterday (had to deal with some Semester 2 stuff yesterday). I've attached a PDF document below for you guys to read. You'll need Adobe Reader to view it. If anyone has any troubles understanding it, feel free to PM me. 

[danfer073]

Please can you help me work out part C

Mod edit: Merged with Further Maths Q&A

[Jackie Chan]

(Image removed from quote.)
Please can you help me work out part C

Mod edit: Merged with Further Maths Q&A

Hi!
For part c, you need to use the rule for simple interest investments/loans:
Vn = V0 + n x r/100 x V0

So with this in mind, the question asks for the interest rate that company A needs to offer if after 5 years, both investments are worth $6722.75 (answer from part b, which is the amount Meghan will get if she invests her money with company B)

So:
solve (6722.75= 5000 + 5 x r/100 x 5000,r) on your CAS to get 6.891 = 6.9%

[TheCommando]

http://imgur.com/a/P0CCe
Hi, for question 321 B wht do i draw 2x on tthe graph
And how do i do ei and ii

[pmmenotes]

Hey guys my bound reference is two books sticky taped together do you thinks thats alright my teacher said he didnt know.

[brady_price]

Hey guys my bound reference is two books sticky taped together do you thinks thats alright my teacher said he didnt know.

Hey!

Depends what you mean by sticky taped. You are allowed to combine books for your bound reference, but they must adhere to the VCAA guidelines. See if your school (try the library perhaps?) has a heavy duty stapler, and get them to staple the spines of the books together (this can be done, trust me, my school did it for me), and then use some intense heavy duty tape to cover the spines. Basically, the books have to "become one", the guidelines specify that there has to be a "single spine", which basically means you can heavy duty tape along the spines and you're fine, so long as the books cannot come apart.

Here's the link to the guidelines (scroll down to "Mathematics examinations: use of bound references")
http://www.vcaa.vic.edu.au/Pages/vce/exams/authorisedmaterials.aspx#bound

Good luck!

[Shadowxo]

http://imgur.com/a/P0CCe
Hi, for question 321 B wht do i draw 2x on tthe graph
And how do i do ei and ii

Hi, sorry this reply is so late. I don't do further but I'll see if I can help
So for 321.b), the number of dogs clipped is at least 2* number of dogs washed
So y ≥ 2*x
You'd have to draw the line y=2x and since it's a greater than or equal sign, it'd be the area above this line.

Profit = 40x + 30y
e)(i) I'm not sure if they expect you to work it out mathematically as it's only worth 1 mark per part. Profit from one dog washed is more than a clipped dog. This means one x is worth more than one y. Also, x and y have to be whole numbers. So, I believe you could either use the graph or guess-and-check. As washing a dog is both faster and gives more money, you want the largest x value. Using the graph (from VCAA answers) the largest x value within the bounds is 2, and for an x value of 2 the largest y value is 6.
Alternatively, you can start with the maximum number of dogs clipped which is 8, and decrease number of clips by one and increase number of washes by one (they take similar time to complete, so doing this just a few times is easy). 2 washes and 6 clips is the highest number of washes possible, and 6 clips is the highest number of clips with 2 washes. (clips ≥ 2*washes)
This question's more interpreting from the graph rather than solving it.
e)(ii) Just sub in these values.

Sorry my answer's not more concise (and a bit late) but hope this helps anyway

[Melody_Zhou66]

Hello! I'm new to this page:

My question is what are the recommended modules to complete for the end of year exam this year?

Any tips for summary books/bound references?

[Aaron]

My question is what are the recommended modules to complete for the end of year exam this year?

Complete the ones you have done as a cohort at your school. In my opinion it is not worth it to try and be adventurous and try a module you have not had extensive experience with.

Any tips for summary books/bound references?

Don't make it too lengthy, don't bring in a textbook. You have 90 minutes per exam.... and when you consider the amount of questions, it isn't much time per. You have to know *exactly* where to find certain points if you have to use your reference at any point.

[DanicaG17]

Hey... just a question that kinda need help on...
The rectangular prism shown has a volume of 12.8cm3. A second rectangular prism is made with half length, 4 times height and 2 times width. the volume of the second prism is???
Anyone able to help???

[Shadowxo]

Hey... just a question that kinda need help on...
The rectangular prism shown has a volume of 12.8cm3. A second rectangular prism is made with half length, 4 times height and 2 times width. the volume of the second prism is???
Anyone able to help???

So we know the volume of a rectangular prism is L*W*H
Let V_i = volume of the initial prism, L_i be the length of the initial prism etc. V2 is the volume of the second prism
V_i=L_i*W_i*H_i
V2 = 1/2 *L_i*2*W_i *4*H_i* = 4*L_i*W_i*H_i = 4V_i
So it's just 4*12.8

If you want to do it quicker just divide by 2, multiply by 4 and multiply by 2 (earlier typo).

[DanicaG17]

So we know the volume of a rectangular prism is L*W*H
Let V_i = volume of the initial prism, L_i be the length of the initial prism etc. V2 is the volume of the second prism
V_i=L_i*W_i*H_i
V2 = 1/2 *L_i*2*W_i *4*H_i* = 4*L_i*W_i*H_i = 4V_i
So it's just 4*12.8

If you want to do it quicker just divide by 2, multiply by 4 and multiply by 4.

So we know the volume of a rectangular prism is L*W*H
Let V_i = volume of the initial prism, L_i be the length of the initial prism etc. V2 is the volume of the second prism
V_i=L_i*W_i*H_i
V2 = 1/2 *L_i*2*W_i *4*H_i* = 4*L_i*W_i*H_i = 4V_i
So it's just 4*12.8

If you want to do it quicker just divide by 2, multiply by 4 and multiply by 4.

Thanks heaps!!

[PopcornTime]

Hey,

Would appreciate it if someone could help me answer Question 4 d) from the following attachment. I have attached the whole extended response question for context.

Thanks!

[addictwithatextbook]

Hello, this is my first ever post on AtarNotes I don't know if this question has been asked yet, but in

• 2016 Further Maths Exam 2 - http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2016/2016furmath2-w.pdf
• Module 2: Networks and Decision Mathematics
• Question 3c

Can somebody tell me how activity H has a float time of 2 days, as it says in the examination report? I can only see it having a float time of 1 day.

Thanks in advance.

[plato]

The critical path is AEIJ and is 15 days long.
The path CGJM is 14 days and so each of these activities has a float time of one.
The path CHM is 13 days which means H must have a float of two because C and M are known to have a float of one from above.