need to confirm some answers: (1 Viewer)

[jiakan]

Yeh I got 1/5 too for the family one.
Omg cant belive how lucky I am, binomial theorem and probability are my weakest, and binomial did not even appear!
Proability questions were so easy and I just cant belive my luck!

God is on my side now for the HSC!

 

[pires_w]

7 (b) (iii)
0.8 =< m =< 1.2


5 (b)
2/3




4 (a) (all)
0.01
0.285
0.323


2 (d) (ii)
284




others i couldnt remmeber...

 

[Argen1447]

me and the other girl forgot to add a constant when integrating i think.

the constant was -250 lol

 

[fishy89sg]

7 (b) (iii)
0.8 =< m =< 1.2


5 (b)
2/3




4 (a) (all)
0.01
0.285
0.323


2 (d) (ii)
284




others i couldnt remmeber...

everythings the same except ur answer to the question i bolded

 

[Hikari Clover]

回复: Re: need to confirm some answers:

the constant was -250 lol

i got C=0.......

but -250 is correct......:burn:

 

[Argen1447]

pires_w[/B]]4 (a) (all)
i)0.01
ii)0.285
iii)0.323

everythings the same except ur answer to the question i bolded

well i jus checked it 3 times now and i REALLY think its 0.323


(0.9 + 0.1)^20

P(>2) = 1 - P(0) - P(1) - P(2)
= 1 - 20C0x(0.9)^20 - 20C1x(0.9)^19x(0.1) - 20C2x(0.9)^18x(0.1)^2
= 0.323

the only reason i can think of the FINAL answer maybe being wrong is the calculator you're using.. I used to have the "old" semi box one for ages and that was perfect, but i lost it and this new calculator im using (casio fx-82AU) sumtimes doesn't do the calculation as you'd expect, say if you wrote ab/c it may not necessarily do (ab)/c ... well not exactly that but something along those lines.. other than that i think my working was right.

how have you guys getting 0.4 sumthing been getting you're answer? post your solution? because a few of you seem to be getting that one..

 

[d.karunairatnam]

well for 7)b)iv) i had 0.16 for 0.8<m<1.2
but i dont think that cant be right..because in that range for m, m can equal 1, which means theta is pi/4, which is the maximum range... and so after that it reduces again...which mean the maximum distance is m=1 and the minimum would be eitherwhen m is 0.8 or 1.2. which means 0.16 would be wrong

 

[fishy89sg]

well i jus checked it 3 times now and i REALLY think its 0.323


(0.9 + 0.1)^20

P(>2) = 1 - P(0) - P(1) - P(2)
= 1 - 20C0x(0.9)^20 - 20C1x(0.9)^19x(0.1) - 20C2x(0.9)^18x(0.1)^2
= 0.323

the only reason i can think of the FINAL answer maybe being wrong is the calculator you're using.. I used to have the "old" semi box one for ages and that was perfect, but i lost it and this new calculator im using (casio fx-82AU) sumtimes doesn't do the calculation as you'd expect, say if you wrote ab/c it may not necessarily do (ab)/c ... well not exactly that but something along those lines.. other than that i think my working was right.

how have you guys getting 0.4 sumthing been getting you're answer? post your solution? because a few of you seem to be getting that one..

O M G i think i did P(0) = 20C0x(0.1)^20

 

[blink376]

well for 7)b)iv) i had 0.16 for 0.8<m<1.2
but i dont think that cant be right..because in that range for m, m can equal 1, which means theta is pi/4, which is the maximum range... and so after that it reduces again...which mean the maximum distance is m=1 and the minimum would be eitherwhen m is 0.8 or 1.2. which means 0.16 would be wrong

Yea, the second range is 19.67<x<20 because the range is a maximum at x=1, not 19.51<x<19.67, so the width isn't 0.159.

Btw, are there two solutions for each x and y in 5(c)??

 

[sting24]

I worked out the distance the skydiver fell to be 283 just then, but 533 seems so familiar. Hope I didn't accidentally write 533.

Probability of green eyes-
i) 0.01
ii)0.285
iii)0.323

yep ditto

 

[fishy89sg]

Yea, the second range is 19.67<x<20 because the range is a maximum at x=1, not 19.51<x<19.67, so the width isn't 0.159.

Btw, are there two solutions for each x and y in 5(c)??

noooooooooooooooooooooooooo

EDIT: but the question asked for the widths of the two intervals

 

[Steth0scope]

O M G i think i did P(0) = 20C0x(0.1)^20

i got the same answer as u 0.433 or wateva and its because we didn't include p(0) in the final i.e. we just did

P(more than 2) = 1 - P(one) - P(two)

when we shoulda done

P(more than 2) = 1 - P(zero) - P(one) - P(two)

:'(

 

[gaoOO]

noooooooooooooooooooooooooo

EDIT: but the question asked for the widths of the two intervals

look, i think those that subtracted straight off got the right answer.

it's probably 0.16 or whatever. Me, i actually forgot to subtract.

 

[fishy89sg]

OH my god. i got 2 (c) wrong.

 

[fishy89sg]

sorry about the double post but OMG i got 3 (b) (ii) wrong. how embarrassing.

 

[AMorris]

My numerical solutions:

Q1a 16 + 8rt5
b (10,1)
c 4x^3/(1 + x^8)
d 45 degrees = pi/4
e 2

Q2a proof
b i sketch
ii 0 <= y <= 2pi
c a = -7, b = 10
d i 1.4
ii 284

Q3 a pi^2/12
b i x = 4, y = 1
ii sketch
c proofs

Q4 a i 0.01 = 1%
ii 0.285
iii 0.32
b proof
c proof

Q5a i proof
ii ~1.1662
b 0.2 = 20% = 1/5
c x = 1/rt2, y = rt3/2
d proofs

Q6a i proof
ii pi
iii x' = 4cos(2t - pi/6)
iv t = pi/4, 5pi/12, 3pi/4, 11pi/12

b i proof
ii proof
iii ~1.66

Q7 a i proof
ii k = 1/(ne)

b i proof
ii proof, h 7.5
iii 0.8 <= m <= 1.2
iv ranges are 11.39 <= x <= 12.67
and 19.51 <= x <= 20 (because m = 1 produces the max range)
(thats what i put down cos i didnt read the question ) the widths are then 1.28m and 0.49m)

All in all there were quite a few tricky questions but nowhere near as hard as last years in my opinion. People will get 84 I reckon (though the last question did have some tricks in it)

 

[IceOnFire]

Just wondering, for question 4a) i) whether i can write,

nC2 (0.1)^2 (0.9)^(n-2)

Well technically it's not wrong?

 

[ajr2795]

4 a) i)
Answer should be relative to the size of the population, as it is essentially choosing 2 items withoput replacement. Eg. if population were 10 you would have a 0% chance of picking 2 people with green eyes as only 1 (10% * 10) woulod have them.
Think the answer is something along the lines of:
(x-10)/[100(x-1)]
As x-->Inf, P--> 1/100
Or am I way overthinking this?

 

[jah_lu]

the question assumes the population is large enough for that not to occur