VCE Methods Question Thread!

[KANYEWEST]

The impact of cybercrime on individuals and organisations in our society is increasing. A survey of 762 randomly selected elderly Australians found that 290 of them had been victims of some type of cybercrime.
a) Give a 95% confidence interval for the true proportion of all elderly Australians who have been victims of cybercrime, and interpret your interval.
b) Based on your confidence interval, does it appear that at least 50% of elderly Australians have been cybercrime victims? Explain. [

[VanillaRice]

The impact of cybercrime on individuals and organisations in our society is increasing. A survey of 762 randomly selected elderly Australians found that 290 of them had been victims of some type of cybercrime.
a) Give a 95% confidence interval for the true proportion of all elderly Australians who have been victims of cybercrime, and interpret your interval.
b) Based on your confidence interval, does it appear that at least 50% of elderly Australians have been cybercrime victims? Explain. [

Which part of the question are you stuck on? What progress have you made?

[m2121]

I need help with this question! I'm not sure how to work it out. I tried to do it step by step from one transformation to another but I got stuck. Can someone please help and explain it clearly.

[TheBigC]

I need help with this question! I'm not sure how to work it out. I tried to do it step by step from one transformation to another but I got stuck. Can someone please help and explain it clearly.

Okay, we shall perform this transformation step-by-step using mapping:
Firstly, we have:
- Dilation of factor 3 parallel to the x-axis (or from the y-axis)
$$ (x,y) \rightarrow (3x, y) $$
Then,
- Translation 2 units up
$$ (3x, y) \rightarrow (3x, y+2) $$
Then,
- Reflection in the y-axis
$$ (3x, y+2) \rightarrow (-3x, y+2) $$
Continuing, we have:
- Translation 1 unit left
$$ (-3x, y+2) \rightarrow (-3x-1, y+2) $$
Lastly,
- Reflection in the x-axis
$$ (-3x-1, y+2) \rightarrow (-3x-1, -(y+2)) $$
Ultimately, we can summarise this information as:
$$ (x,y) \rightarrow (-3x-1, -y-2) $$
Thus, we can now find our values for x' and y':   
$$ x'=-3x-1 $$
$$ y'=-y-2 $$
Substituting into our original equation gives:
$$y=\frac{1}{x} \rightarrow -y'-2=\frac{1}{(\frac{x'+1}{-3})} $$
Solving for y'
$$ y'=\frac{3}{(x'+1)}-2 $$

[m2121]

More help needed! Please help me with these questions.

[peachxmh]

Hello, can someone explain how I can solve this problem?

Find the equations of the circles which touch the x-axis, have radius 5 and pass through the point (0,8).

I'm not sure how I should use (0,8) in my working, thanks in advance! So far I've figured out that the centre is either on the line y=-5 or y=5

[TheBigC]

Hello, can someone explain how I can solve this problem?

Find the equations of the circles which touch the x-axis, have radius 5 and pass through the point (0,8).

I'm not sure how I should use (0,8) in my working, thanks in advance! So far I've figured out that the centre is either on the line y=-5 or y=5

Question:

Find the equations of the circles which touch the x-axis, have radius of 5 and pass through the point (0,8)

Firstly, we can examine the general equation for a circle:

$$ (x-h)^2+(y-k)^2=r^2 $$

Thus, we know that it must touch the x-axis and have a radius of 5.

Therefore, $$ r=5 $$ and $$ k = 5 $$ yielding:

$$ (x-h)^2+(y-5)^2=25 $$

Now, to solve for h, we use the point (0,8):

$$ (0-h)^2+(8-5)^2=25 $$
$$ h^2+9=25 $$
$$ h^2=16 $$
$$ h=\pm{4} $$

Hence, the equations of the circles are:

$$ (x-4)^2+(y-5)^2=25 $$

and

$$ (x+4)^2+(y-5)^2=25 $$

[Xaki.z]

Hi,
I was wondering what would i need to be averaging in order to get a 30 raw study score in math methods ?
Thanks

[Sine]

Hi,
I was wondering what would i need to be averaging in order to get a 30 raw study score in math methods ?
Thanks

In 2017 you needed roughly
average sac ranking + 20/40 on exam 1  + 42/80 on exam 2
doing better on one section can allow you to drop more marks in other parts.
However last years exam was probably one of the easiest in the past 10-15 years so the amount of marks you can drop this year would be different.

[TheBamboozler]

hey everyone. I have a SAC coming up next week on Monday and there's going to be a question that involves creating a hybrid function that avoids intersecting with two vertical lines. The graph area would be limited in it's range so I can't just make a function that goes over.

It's sort of like there's two hedges and the hybrid function is the pathway that avoids the hedges. Simple question but I was wondering if there was a way to draw on the CAS calculator that allowed me to specify a vertical line to start at e.g. coordinates (7,2) and also limit it's range. On the CAS calculator I'm having trouble doing something similar to what I describe so any help would be hella appreciated yo.

Edit - Forgot to mention but the CAS I'm using is the TI nspire cx one.

[TheBamboozler]

It just occurred to me that I could define the equation then just make it equal 7 and see if it intercepts, rather than drawing a single vertical line.

I feel a bit silly now. Disregard my question lol.

[True Dat]

It just occurred to me that I could define the equation then just make it equal 7 and see if it intercepts, rather than drawing a single vertical line.

I feel a bit silly now. Disregard my question lol.

For the record, if you did want to plot that vertical line you can do it by pressing menu ---> Graph Entry/Edit -----> Relation. Then enter x = 7 to plot your vertical line.
I find using the vertical or horizontal lines useful when I'm trying to work out intervals, for example determining when a waves height is below 1 metre or something like that.

[Yertle the Turtle]

For the record, if you did want to plot that vertical line you can do it by pressing menu ---> Graph Entry/Edit -----> Relation. Then enter x = 7 to plot your vertical line.
I find using the vertical or horizontal lines useful when I'm trying to work out intervals, for example determining when a waves height is below 1 metre or something like that.

On the most recent version of the TI nSpire CAS software there is a vertical line input.
menu --> Graph Entry/Edit --> Templates --> Lines --> Vertical Lines
Hope this helps
Learn your CAS guys, it will really really really help in your life Methods/Further/Spesh course.

[Lear]

Anyone have a formula for finding angle between two lines?

[TheAspiringDoc]

Anyone have a formula for finding angle between two lines?

Yep.
Eg if gradient A is 0.5 and gradient B is 0.333...,angle A is tan-1(0.5) and angle B is tan-1(0.333) (angle relative to the positive x-axis, that is)
The angle between the lines is angle A - angle B
Is tan-1(0.5)-tan-1(0.333)
=8.13 (2dp)