Extension One Revising Game (1 Viewer)

[azureus88]

yeh sorry, didnt specify. Its integers only. its only less than 0 when n is between 5 and 6. use induction

 

[u-borat]

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.

will try induction noww.

 

[lyounamu]

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]

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]

(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.

 

[bored of sc]

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).

 

[bored of sc]

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]

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.

 

[adnan91]

thats from a past hsc oooo

 

[midifile]

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]

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]

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
@ = 60^o

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

ii) 2 (75.5^o and 284.5^o)

iii) @ = 360n + 75.5^o

 

[lyounamu]

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]

uh wait. Let me see:

I miscalculated it. Your one is right.

Yay!

 

[lyounamu]

Anyone up for this question? I posted earlier but no one responded.

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

 

[bored of sc]

Find the sum of all the integers between 1 and 200 that are not multiples of 7.

 

[azureus88]

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]

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]

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]

sorry i should have said factorise (r+R)tan@ instead of just tan@