# General Thoughts: Mathematics (1 Viewer)

#### WileyWanimal

##### Member
part i) i just went screw it and put 10/3 + 10. it's wrong 99% sure

I'm curious about question 13 b)ii too im sure i got it wrong
Actually, I think that's right for Q14d) (i).

#### WileyWanimal

##### Member
Surely nexusbrah is trolling??

#### rose_maylie

##### New Member
i ended up rounding to 2 d.p but the q didn't specify, did it?
Me too, although I left them as exact values.

#### WileyWanimal

##### Member
i ended up rounding to 2 d.p but the q didn't specify, did it?
Nope, although since it was length, now I'm thinking I should have calculated them... hmm

#### jj1823748

##### New Member
what would state rank be?

so not 98 ? :\

#### Simile

##### Member
part i) i just went screw it and put 10/3 + 10. it's wrong 99% sure

I'm curious about question 13 b)ii too im sure i got it wrong
That's what I did as well and for the second you just gotta use the limiting sum.

#### photastic

##### Well-Known Member
so not 98 ? :\
We shall see. If that's the 20th highest mark, then you got a state rank

#### jj1823748

##### New Member
misreading that question wasnt my best haha dang

#### photastic

##### Well-Known Member
part i) i just went screw it and put 10/3 + 10. it's wrong 99% sure

I'm curious about question 13 b)ii too im sure i got it wrong
Part i was to find the IMMEDIATE amount, hence, you calculate how much was left from the 1st dose and you add it to the 10mL so it should be 10/3+10 unless we're wrong.

#### chrisdelinet

##### Member
part i) i just went screw it and put 10/3 + 10. it's wrong 99% sure

I'm curious about question 13 b)ii too im sure i got it wrong
you are right for part i) as the question says amount of drugs right after second dose (somewhere on the lines of this) therefore at end of 1st 8hr period there will be 10/3 mL of drugs within patient (1/3 of drugs present at begining of 8hr period remains). therefore right after second dose, which is when 1st 8hr period ends, you have (10 + 10/3)mL of drugs! you should change your response to 'it's right 99% sure' (well thats what i believe) and for 13b) ii), i believe that is the bearings question, i got 333 degrees for that (seeing that most other people have this and including me, i believe that is quite solid evidence that 333 degrees is correct)

#### stampede

##### doin it tuff
superannuation q

A60 = 500(1.003)^60 + 500(1.01)(1.003)^59 + ... + 500(1.01)^58(1.003)^2 + 500(1.01)^59(1.003)

This is a geometric series with the ratio being (1.01/1.003) as each term is changing by this ratio.

n (number of terms) = 60

a (initial value) = 500(1.003)^60

You get A60 = \$44, 404.38

#### BLIT2014

##### The pessimistic optimist.
Moderator
so fking easy band 1 here i come

Are you sure ? Band 1 is very hard to get especially in this subject with such high calibre

#### bio4lyf

##### Member
Alternative Solutions for 15. b) part iii and iv

instead of writting sin D i wrote sin (theta) .....am i going to lose all those marks???

#### HSC_killer

##### Member
with the cut off for a band 6, is the hsc mark counted or is it the raw marks obtained from that exam that will be counted as cutoof of band 6?

#### chrisdelinet

##### Member
instead of writting sin D i wrote sin (theta) .....am i going to lose all those marks???
hmm i believe you'll lose 1, hard to say because you did not specify the angle, depends if bostes wants to be cruel in a fairly easy test.

#### chrisdelinet

##### Member
superannuation q

A60 = 500(1.003)^60 + 500(1.01)(1.003)^59 + ... + 500(1.01)^58(1.003)^2 + 500(1.01)^59(1.003)

This is a geometric series with the ratio being (1.01/1.003) as each term is changing by this ratio.

n (number of terms) = 60

a (initial value) = 500(1.003)^60

You get A60 = \$44, 404.38
sorry to pop your bubble, but the question asked for nearest dollar therefore A60 = \$44, 404

you may think oh yeah pshhhh, but it's a heads up for people doing the exam in future years READ THE BLOODY QUESTION.