Easiest Paper Ever (1 Viewer)

[tywebb]

Could people please upload the paper so I can have a look?

OK. Here it is.

http://www.boardofstudies.nsw.edu.au/hsc_exams/2016/exams/2016-hsc-maths-general-2.pdf

But there is a picture missing in 28b.

Here is the missing picture: http://community.boredofstudies.org...1477277253-general-2-paper-missingpicture.jpg

 

[BLIT2014]

OK. Here it is.

Thanks mate

 

[Bliske]

Asking to verify the calculation sounds well within the scope of the syllabus. I'm pretty sure I got taught how to do verify it too.

I thought so as well. My teacher assured me that you would never have to calculate the correlation coefficient by calculator as it would have to be given to you, but I learnt how to do it just in case. I'm glad I did.

 

[FootballHooligan]

The financial expectation question...
Was the answer $0?

 

[InteGrand]

The financial expectation question...
Was the answer $0?

Yeah it's $0.

 

[FootballHooligan]

Yeah it's $0.

I'm seriously expecting to get 80+ for this paper
But being ranked 32/50 is not the best

 

[BigBusty]

Did any1 get 8:30am Saturday landing time for Dubai ?

 

[FootballHooligan]

Did any1 get 8:30am Saturday landing time for Dubai ?

That was the answer

 

[BigBusty]

and for the chocolate volume, it was around 192 ?

 

[Mysterious_girl]

If we are talking about question 30C) (The one with the school playground) I agree that we definitely can't use Pythagoras and should use a different formula.

this is how i calculated the triangle value; 1/2 X 45 X 45 X sin100

 

[InteGrand]

and for the chocolate volume, it was around 192 ?

It is 193.1417... (cubic cm), which is roughly 193 to nearest whole number.

 

[FootballHooligan]

I worked it out like this

1) Found the rectangular area surrounding the marshmallows (20 x 15)= 300
2) The depth of the area (volume of the rectangular area) = 300 x 6 (height of two marshmallows) = 1800
3) Space covered by marshmallows = volume of marshmallows (58.9) x 24 = 1413.6
4) Subtract from volume of rectangular area = (1800- 1413.6)= 386.4

Pretty sure I'm wrong

 

[BigBusty]

I think its a rectangle with 15 x 10 (had to draw it on the diagram and then work out the radius distances) and then subtract 6 circles area from that rectangle to find the remaining area and then multiply the area by 6cm to find the volume of the chocolate
I subtracted 6 circles because inside the rectangle there are 4 1/4 of a circle (which equals 1 circle) + 6 semi circles (3 circles) + 2 normal circles. Sorry if im confusing.

 

[BigBusty]

BTW question 19 = A and 21 = D ? does any1 know? and isn't question 25 = B?

 

[InteGrand]

 

[BigBusty]

For the memory speed question did any1 get 61 mbps (to nearest whole number)

 

[InteGrand]

I worked it out like this

1) Found the rectangular area surrounding the marshmallows (20 x 15)= 300
2) The depth of the area (volume of the rectangular area) = 300 x 6 (height of two marshmallows) = 1800
3) Space covered by marshmallows = volume of marshmallows (58.9) x 24 = 1413.6
4) Subtract from volume of rectangular area = (1800- 1413.6)= 386.4

Pretty sure I'm wrong

This method would unfortunately fail to exclude volume from the parts of the rectangle's "edges" where there is no chocolate. These should be excluded (since there's no chocolate there).

 

[whatarethoose]

lmfao fuck my life i got 2:30am friday because youre going 21 hours back in time from 11:30pm

 

[Bliske]

For the memory speed question did any1 get 61 mbps (to nearest whole number)

Yes.

 

[InteGrand]

For the memory speed question did any1 get 61 mbps (to nearest whole number)

Yes.

(Assuming you're talking about Question 30 (b) (ii)) How did people get 61? (Caveat, I don't do General Maths though, so I might have missed something.) Note that 1 byte = 8 bit. So 3072 MB = 3072 megabyte = 3072*8 mb (megabit).

Therefore, the speed in megabit per second required (noting 7 min. = 420 s) is

(3072*8 mb)/(420 s) = 58.51429... mbps, which is 59 to the nearest whole number.