• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Extension One Revising Game (3 Viewers)

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
yeh sorry, didnt specify. Its integers only. its only less than 0 when n is between 5 and 6. use induction
 
Last edited:

u-borat

Banned
Joined
Nov 3, 2007
Messages
1,755
Location
Sydney
Gender
Male
HSC
2013
ah right, i differented cos i thought that the lowest point of the graph was >0, and cos its increasing function, i could say that. :p

will try induction noww.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
Another question guys:

A standard pack of 52 cards consists of 13 cards of ecah of the four suits: spades, hearts, clubs and diamonds:

i) In how many ways can six cards be selected without replacement if at least fiv must be of the same suit? (order is not important)

ii) In how many ways can six cards be selected without replacement so that exactly two are spades and four are clubs? (order is NOT improtant)

Come on guys~
 

samwell

Member
Joined
Nov 19, 2007
Messages
400
Gender
Male
HSC
2008
lyounamu said:
Another question guys:

A standard pack of 52 cards consists of 13 cards of ecah of the four suits: spades, hearts, clubs and diamonds:

i) In how many ways can six cards be selected without replacement if at least fiv must be of the same suit? (order is not important)

ii) In how many ways can six cards be selected without replacement so that exactly two are spades and four are clubs? (order is NOT improtant)

Come on guys~
(i) if at least 5 must be from the same suit then the soln is in two cases:
5 cards from same suit selected[13C5*39C1]
+6 cards from same suit[13C6*39C0]

(ii)13C2*13C4*26C0

is this right????? not sure how to treat the 2nd part
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
samwell said:
(i) if at least 5 must be from the same suit then the soln is in two cases:
5 cards from same suit selected[13C5*39C1]
+6 cards from same suit[13C6*39C0]

(ii)13C2*13C4*26C0

is this right????? not sure how to treat the 2nd part
i) 6 cards:

4 x 13C6 + 6864

5 cards:

4 x 13C5 x 39

Add them up, you get 207636

ii)
13C2 x 13C4 = 55770

If anyone can explain better here, please do so because I suck at explaning perm & comb

Another question:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/95MAT3U.PDF
page 10 Q 7 (d)

I don't know this question but I know the answer. So please reply.
 
Last edited:

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
lyounamu said:
Another question guys:

A standard pack of 52 cards consists of 13 cards of ecah of the four suits: spades, hearts, clubs and diamonds:

i) In how many ways can six cards be selected without replacement if at least fiv must be of the same suit? (order is not important)

ii) In how many ways can six cards be selected without replacement so that exactly two are spades and four are clubs? (order is NOT improtant)

Come on guys~
i) 13C6 x 4 is for each suit given that all six cards selected are the same suit. 13C6 is selecting 6 from 13 (one suit) and since there are 4 suits you multiply by four.
13C5 x 39C1 x 4 is for each suit given that five of one suit is selected. 13C5 is selecting 5 of the one suit from the 13 of one suit. 39C1 is choosing 1 from the remaining 3 suits (can't be the same suit as the other five cause that would make is 13C6) which is 13 (number of cards in one suit) multiplied by 3 - remember to multiply this answer by 4 since any suit can have the 5 selections.
Since you can have 6 of one suit OR 5 of one suit and any other card you ADD the values you get for each case.

13C6*4 + 13C5*39C1*4 = value Namu said (I hope).
 
Last edited:

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
ii) 13C2*13C4. 6 selections are already made since taken 2 spades from a possible 13 and 4 clubs from a possible 13. So there's the answer.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
bored of sc said:
i) 13C6 x 4 is for each suit given that all six cards selected are the same suit. 13C6 is selecting 6 from 13 (one suit) and since there are 4 suits you multiply by four.
13C5 x 39C1 x 4 is for each suit given that five of one suit is selected. 13C5 is selecting 5 of the one suit from the 13 of one suit. 39C1 is choosing 1 from the remaining 3 suits (can't be the same suit as the other five cause that would make is 13C6) which is 13 (number of cards in one suit) multiplied by 3 - remember to multiply this answer by 4 since any suit can have the 5 selections.
Since you can have 6 of one suit OR 5 of one suit and any other card you ADD the values you get for each case.

13C6*4 + 13C5*39C1*4 = value Namu said (I hope).

I just read over that and realised it was the most confusing piece of explanation ever written. Sorry about that.
Couldnt' have put it better myself. Well done, bored of sc. Very concisely and well explained.
 

midifile

Na Na Na Na Naa
Joined
Oct 11, 2006
Messages
1,143
Location
Sydney
Gender
Female
HSC
2008
i) Express rt(12)sinx + 2 cosx in the form Acos(x-@)

ii) State the number of solutions that satisfy the equation rt(12)sinx + 2 cosx = 1 in the domain 0<=x<=2pi

iii) Write down the general solution to rt(12)sinx + 2 cosx = 1
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
lyounamu said:
Couldnt' have put it better myself. Well done, bored of sc. Very concisely and well explained.
In case it is incorrect, the answer I got was 207636. You said it was 200772.
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
midifile said:
i) Express rt(12)sinx + 2 cosx in the form Acos(x-@)

ii) State the number of solutions that satisfy the equation rt(12)sinx + 2 cosx = 1 in the domain 0<=x<=2pi

iii) Write down the general solution to rt(12)sinx + 2 cosx = 1
i) A = root (22 + rt(12)2)
= root16
= 4
tan@ = b/a = rt12/2
@ = 60o

rt(12)sinx + 2cosx = 4cos(x-60)

ii) 2 (75.5o and 284.5o)

iii) @ = 360n + 75.5o
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
bored of sc said:
In case it is incorrect, the answer I got was 207636. You said it was 200772.
uh wait. Let me see:

I miscalculated it. Your one is right.
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
Find the sum of all the integers between 1 and 200 that are not multiples of 7.
 

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
sum of all integers from 1 to 200 - sum of all multiples of 7

= (200/2)(1+200) - (28/2)(1+196)
=100(201) - 14(197)
= 17342

ok can someone do my question now.

prove, using induction n^2 - 11n +30 is bigger than or equal to 0 for n is bigger than or equal to 1, where n is integer.
 

QuLiT

Member
Joined
Aug 23, 2007
Messages
84
Gender
Male
HSC
2008
for your question namu:
il try and make it as clear as possible but yaeh not so good with typing

using cartesian equation:
let y = 0 x = R+r

0 = h + (r+R)tan@ - g(r+R)^2/2v^2 cos@
you use v^2 from part c then there for
0> h+ (r+R)tan@ -((r+R)^2 tan @ )/R

then after this you factorise the tan@ and rearrange and you should get your answer if your still having troubles tell me and il scan my working
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
QuLiT said:
for your question namu:
il try and make it as clear as possible but yaeh not so good with typing

using cartesian equation:
let y = 0 x = R+r

0 = h + (r+R)tan@ - g(r+R)^2/2v^2 cos@
you use v^2 from part c then there for
0> h+ (r+R)tan@ -((r+R)^2 tan @ )/R

then after this you factorise the tan@ and rearrange and you should get your answer if your still having troubles tell me and il scan my working
Thank you so much. I will have a look at that more closely
 

QuLiT

Member
Joined
Aug 23, 2007
Messages
84
Gender
Male
HSC
2008
sorry i should have said factorise (r+R)tan@ instead of just tan@
 

Users Who Are Viewing This Thread (Users: 0, Guests: 3)

Top