VCE General & Further Maths Question Thread!

[AngelWings]

thanks for that!! is their anything else you could add please?

thanks

Recall that triangles have 180 degrees' worth of internal angles.
There is already one common angle at the top.
The final two angles are thus: 180 - top angle.
If they were not parallel, the four remaining angles could be anything, such as in the ways seen in the picture attached, and still be accurate. (I exaggerated some, by the way. I also apologise for my lack of ruler usage and decent to poor drawing skills.) You wouldn't be sure that they're similar, as they would no longer be guaranteed to be corresponding angles. (It won't satisfy similarity rules like SAS or AAA.)

Despite this, it is not always going to be the case; in other cases, it could be an alternate angle pair, co-interior pair or vertically opposite pair. You must memorise these for the exam.

Reminder that nearly all diagrams are not to scale and perhaps not even to proportion, so don't heavily rely on the diagram being accurate to the nth degree, unless necessary, such as in contour maps.

If you are unsure about corresponding angles or parallel lines and the related angles, reread your theory notes or textbook.

[darklight]

Is it possible to calculate outliers using this formula (where o is potential outliers)
o < mean - 2*standard deviations and o > mean + 2 standard deviations?
I saw this in an Insight paper and just wanted to confirm it

[keltingmeith]

Is it possible to calculate outliers using this formula (where o is potential outliers)
o < mean - 2*standard deviations and o > mean + 2 standard deviations?
I saw this in an Insight paper and just wanted to confirm it

This is a method that I personally don't like for several reasons, and VCAA don't mention it in the study design (although, they don't talk much about outliers). However, it is listed in textbooks, and it is something that you certainly could do, so I'd just up to use the normal IQR method wherever possible, but then use the 2s method if given no other option.

[Billion]

Does anyone have a table or know where to find a good summary of the best way to display numerical and categorical data?

Thanks.

[darklight]

Hi guys, apparently the answer is E… I don't get why?

Also does correspond mean the same thing as correlate in the context of explaining the relationship between x and y?

[Zealous]

Hi guys, apparently the answer is E… I don't get why?

Also does correspond mean the same thing as correlate in the context of explaining the relationship between x and y?

Well, they've given you a residual plot graph. For a linear correlation between to variables, 'x' and 'y', the residuals should be evenly spaced above and below the x-axis. That is, the linear regression line y=mx+c overestimates some 'y' values and also underestimates some 'y' values, leaving residuals.

In the graph they've given, the residuals aren't evenly spaced and there appears to be a pattern in the graph - almost like a quadratic. Because of this pattern, we can't confirm that there is a linear relationship between the variables.

[LiquidPaperz]

just a general question. when doing a two way frequency table do you do percentages down a column or across? what does each tell us, say the variables were males and females and overweight , underweight

[darklight]

Well, they've given you a residual plot graph. For a linear correlation between to variables, 'x' and 'y', the residuals should be evenly spaced above and below the x-axis. That is, the linear regression line y=mx+c overestimates some 'y' values and also underestimates some 'y' values, leaving residuals.

In the graph they've given, the residuals aren't evenly spaced and there appears to be a pattern in the graph - almost like a quadratic. Because of this pattern, we can't confirm that there is a linear relationship between the variables.

Oh man, so silly of me! Thanks

[LiquidPaperz]

can someone help me with these. It's further but i dont find many people reply to questions on that, but this seems semi methods related. Say i've been given a question like this, what would be the steps to solve it?

[LiquidPaperz]

so the equation of the graph given is y = x^2/5

then do i go trial and error with the given graphs A-E? so the answer is D this is because when i sub the values in they give me the right y value. e.g. the equation is y = x^2/5 so if i look at D its, y= (5)^2/5 = 5 and the point is 5,5 which has been satisfied. is this the best approach?

[Zealous]

so the equation of the graph given is y = x^2/5

then do i go trial and error with the given graphs A-E? so the answer is D this is because when i sub the values in they give me the right y value. e.g. the equation is y = x^2/5 so if i look at D its, y= (5)^2/5 = 5 and the point is 5,5 which has been satisfied. is this the best approach?

Yep, that approach is fine. So convert the graph into an equation, then see which of the options fits the equation. Though, it's highly unlikely you'll get a question like this on a further exam.

[LiquidPaperz]

it was in the 2012 one http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012furmath1-w.pdf

[LiquidPaperz]

best to worst / easiest to hardest further exams we should be doing??

I get the general vibe heffernan is on par with VCAA, MAV a bit harder, Engage is under VCAA, kilbaha dont bother, NEAP only if you have time, Insight if you want etc?

[demand&supply]

Can someone please tell me how to do this question??? (VCCA 2013 Exam 1, Module 2)

Question 8
There are four telecommunications towers in a city. The towers are called Grey Tower, Black Tower, Silver
Tower and White Tower.
Grey Tower is 10 km due west of Black Tower.
Silver Tower is 10 km from Grey Tower on a bearing of 300°.
White Tower is 10 km due north of Silver Tower.
Correct to the nearest degree, the bearing of Black Tower from White Tower is
A. 051°
B. 129°
C. 141°
D. 309°
E. 321°

[LiquidPaperz]

Can someone please tell me how to do this question??? (VCCA 2013 Exam 1, Module 2)

Question 8
There are four telecommunications towers in a city. The towers are called Grey Tower, Black Tower, Silver
Tower and White Tower.
Grey Tower is 10 km due west of Black Tower.
Silver Tower is 10 km from Grey Tower on a bearing of 300°.
White Tower is 10 km due north of Silver Tower.
Correct to the nearest degree, the bearing of Black Tower from White Tower is
A. 051°
B. 129°
C. 141°
D. 309°
E. 321°

Here you go mate, hope it helps!

https://drive.google.com/folderview?id=0B4jDYm8MQMIFMFIxa2ZDdWhrY0U&usp=sharing