-
2008-ASLM&S
HONG KONG EXAM]NATIONS AND ASSESSMENT AUTHORITY
HONG KONG ADVANCED LEVEL EXAMINATION 2OO8
II
'ls
tt'4
MATHEMATICS AND STATISTICS AS-LEVEL
8.30 am -
11.30 am (3 hours)
This paper must be answered in English
l. This paper consists of Section A and Section B.
2. Answer ALL questions in Section A, using the AL(E) answer
book.
3. Answer any FOUR questions in Section B, using the AL(C)
answer book.
4. Unless otherwise specified, all working must be clearly
shown.
5. Unless otherwise specified, numerical answers should be
either exact or given to 4 decimal places.
t42008-AS-M & S-l
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6:&a\*HBB Hmg Kong Public Lihatus
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- 4. A and B are two events. A' and B' are the complementary
events of A and B respectively.l9rSuppose P(A)=i , P(lvB)=i,
PQqIB)=i and P(B)=* ,where 0
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(qruur 09) fl uottcas
-
8. Abiologist studied the population of fruit fly I under
limited food supply. Let r be the number
ofdayssincethebeginningoftheexperimentand NO bethenumberoffruitfly
I attime /. Thebiologist modelled the rate of change of the number
of fruit tly A by
N'(') = 'o ,- ('> o) 'I + he-^'where h and k are positive
constants.
,rl
$tIi+
(a) (i) Express ,r[jl -
r-] u, a linear tunction of r .' LN'(t) 'l
(ii) It is given that the intercepts on the vertical axis and
the horizontal axis ofthe graph ofthe linear function in (i) are
1.5 and 7.6 respectively. Find the values of h ard k.
(4 marks)
(b) Take h = 4.5 , k = 0.2 and assume that N(0) = JQ .
(i) Let v= h+ekt,find I .'dt
Hence, or otherwise, find N(r) .
(iD The population of fruit fly B can be modelled by
M(r)= x(t*L"-k'\*t\"/ -'[- t- ) " 'where 6 is a constant. It is
known that M(20) = N(20) .
(l) Find the value of 6.(2)
Thebiologistclaimsthatthenumberoffruitfly I
willbesmallerthanthatoffruit ;
[Hint: Consider the difference between the rates of change of
the two populations.](1l marks)
2008-AS-M & S-5 l8
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(lll)
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-
10. Assume that the number of visitors arriving at each corurter
in an immigration hall is independentand follows a Plisson
distrlbution with a rneanpl-3.9 visitols per minute. A counter is
classified asbusy if at least 4 visitors arriving at it in one
minute.
":'
(a) Find the probability that a counter is busy in a certain
minute.(3 marks)
(b) An officer checks 4 counters in a certain minute.counter is
found.
Find the probability that at least one busy
(2 marks)
If 10 counters are open, find the probability that more
thanminute.
7 ofthem are busy in a certain
(3 marks)
(d) Suppose l0 counters are open and one of them is randomly
selected. Find the probability thatmore than 7 of them are busy and
the randomly selected counter is not busy in a certainminute.
(3 marks)
(e) The immigration hall is called congested if more than 90% of
the open counters are busy in aminute. Suppose 15 counters in the
hall are open. A senior officer checks the counters in acertain
minute. It is given that more than 7 of the first 10 checked
counters are busy. Findthe probability that the hall is
congested.
(4 marks)
(c)
1
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-
-I
12. offrcials of the Food Safety centre of a city inspect the
imported "choy s*" !y selecting 40samples of "Choy i.,." to. each
lorry and testing for an unregistered insecticide' A lorry
of;Crr'"v Sum" is .iurrin.a as risky if more than 2 samples show
positive results in the test'
Farm,4 supplies"ChoySum"tothecity. Pastdataindicatedthat l%
oftheFarm l "ChoySum"showed positive ,"*ti, in the test. on a
certain day, "choy Sum" supplied by Farm A istransportld by a
number of lorries to the city'
(a) Find the probability that a lorry of "Choy Sum" is
risky'
END OFPAPER
(3 marks)
(b) Find the probability that the 5ft lorry is the first lorry
transportitg rislE "Choy tu-',, marks)
(c) If t lorries of "choy Sum" are-insp_ected, find the least
value of t such that the probability offinding at least o.t. iotry
of risky "Choy Sum" is greater than 0'05 (3 marks)
(d) Farm B also supplies "choy Sum" to the ciry' It is known
that l'5% of the Farm B "choysum,, showed positive resulis in the
test. on a certain day, "choy Sum" supplied by Farm
IandFarmBistransportedbySand12lorriesrespectivelytothecity.
(i) Find the probability that a lorry of "Choy Sum" supplied by
Farm B is rislE'(ii) Find the probability that exactly 2 of these
20 lorries of "choy Sum" are risfty'(iii) It is given that exactly
2 of these 20 lorries of "choy Sum" are rrsty' Find the'
proUaUility that these i lorries transport "Choy Sum" from Farm 'B
' (7 marks)
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-
It
it..)
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0
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t.
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{lil{)'{*)-h
Solution Marks Remarks
l. (a)t,-r)(-r-,')(i) #=,.[])t*t.V]#(*)2 +"
o 3a2 ')= I --x+-I- +"'28
l-o -3lT- 2
"11t-=uIs
.'.a=-3 ana t={(ii) The expansion is valid fo, l! . !
(b) ..+=,*1[l).!(L\'\u' m
"z[:o] s[:oJ/io 843l- s-
1l q 8ooJi _2s2e
800
IM
IA
IM
IA
IA
IA
1A
For binomial expansion-l
on (l+ ax) 2For any two terms correct.pp-l for omitting
'...'
For both
For both correct
For RHS, accept 1.05375
Accept 3.16125 and 3.161
(7)
2. (a) y3 -rry=|
tu' 9 -( ro'* r.) = o
" du I d, ")dv- vdu 3y2
-u
IM
IA
Alternative Solution'tlu=l---
vdu^l--
t lra---v Idy'yt
dy=du
y22y3 +l
IM
IA
(b) 2^Xa=zlnu = x2 ln2I du
- 2xln'udxdu
-rxz .2xrn2dx
1M
1M
1A
Alternative Solution
u =2" - "*2k'2du
-
"x2tn2 .2xln2
d.r
=2"'.2xln2
IMIM
1A
26
-
LZ
VIn
I
I^II902 ---=t6
(v)a-@ ^V)a= (g v ,V)d.uo-iin-to-S-5[itrtu6lV
I^II
YI
NI
YI
(or\ s (or \ l-l--t 'l:(e,lr Ie.l- (g wr)a- (S)a = (fl v
,V)d
!f = , .r.t
t'2-'*9 =oz "' ,l16
(g v V)a- (g)a + (p.)a = (g ny)4 :.
(c)
(q)
9
4(s)a.(s lv)a=@wv)a(e)
,n
zl+s'zult's+... = r
J + tro-aZl -tls + {u1- lz + {q1s'e ,o
'lt?ur tggg'91 : eruu 3uo1 e rege Snrp eqlgo uorle4uesuoc eql
'e'l998'91=
Tsss er+,r.o_rffl rr- fr *ffii,, ={eess'lr
+ tys-azt- [(s + r)ut - Q+ r;u11 E's]?i/
t9S8'9I + \.s-aZt-[(S + t)vt- Q+ rut]g'S - r'e'!998'9t ^,
Zl+ (Zul- Sq)e'S = ,J+Zt-(Sq-Zq)e's=0'.' .
0=r .0-/ ueqq
(6q7 acurs) C+ \.s_aZt-[(S+ t)q-(Z+r)q]g'9=I (s*t 7+r\ l.
tPlno-'z'r.[
'
-
r Je sl ='(S+t Z+/\ ry \o_az.r*[
, _
, .Jr.r=*
(q)
(e)'e
(9)
VI
ntl
vt
hII
YI
WI
,Z- ,tEZV& yrxZ
'*Z- '{E Zi%Z=n - -,(e
Zulxz' ,*Z'-i- =rynp ry _.__r_=_ (c) np
^p ^p
\L)
YI
hII
$lJ?IueusryeIAtruoBnlos
-
Solution Marks Remarks(d) P(A' A B') =l-P(Av B)
=t-a20
For P(,4'n B')*0Follow through
For P(l ') + P (B') * P(A' vFollow through
Alternative Solution IP(A' n B')=P(A')
-P(A' a B)-f,_1)_ft)_ [,_r./_[4.]
Alternative Solution 2P(A' r: B') = P(A') + P(B) - P(A' w
B')
= P(A') + P(B) - P((A r: B)')
=[,-1).(,- 3 )-[,-1 1)t s/1. lo/1.610)
Hence A,
il20
*0and B' are not mutually exclusive.
IMI
Alternative Solution
;,"*++).(,_*)=3
2+P(A'v B') since P(l'u.B') < I
Hence A' and B' are not mutually exclusive.IMI
(7\
5. (a) Let $X be the amount of money spent by a randomly
selected customer.r(x > :oooo)= o.zqz,( , rtoooo_ tt) = o.ro,I
oooo ).'.
p(o . , .3oooo- P) = o.rr*[ 6000 )
...3oooo_t =o.z6000
i'e' # = 25800
(b) The required probability_p0osoo
-
6Z
Surssnu IIe oJe^{ sexeJoslaqel pue ur8rro:og (1-dd)
edeqs rogsa1o1dru,(sz rog
sldecrslut rog
()
1,_/ | |:+,-l : t/y,,Ai
e s t E z t /lo I- ;, t- t- s- s- L- B- trrrrr-l':'r't"
-"'1t' _--..' I Z_=xt: -l' -- -: t=t II
t,II
t' ' (x "=nrt-' lz+e/ l/
xj/(q)
z- = x sl J ol eloldruxse Issrue^ eqlJo uoqenbe erll '.'Z+x
+Z-
-
Solution Remarks
For numerator
Follow through
OR
oR 1.0063
(pp-l) if *
*^not rejected
5
=... * f?G -2x)dx
"8x+2
(c) (i) f (r\ =3x -2x+2f '(x) = (x+2)(3)-(3x-2)(l)(x+2)2
8a(x+2)'
So the equation of Z, is .y - k = f'(h)(x - h)
3h-2 8v-ffi= rn*;(*-n1
(h+2)2 y -(3h-2)(h+2) =916-91,
8x -
(h +2)2 y +3h2 -
4h -
4 = o
(iD (1) If Z1 passes through the origin, then8(o)
-
(ft +z)2 (o) +3h2 -
4h -
4 = o
h=2 or ] (rejectedsince P liesinthefirstquadrant)J
...p=3(2)-2 =1(2) +2SloPe of t, =@)-t-=-2
-8Hence the equation of Z, is y
-l = -2(x -2)i.e.2x+Y-5=0
(2) The x-intercept of Z2 is IISo the required area
-
(r3x-2 *.!(1-z)rurz x*2 2\2 )"J
=fr('-#)*.i(i)u,3
= [3x-8hlx+ ,ii.;
=!*glo2
For
t,
Itl{Iq
ttl
.
III
Io
1A+IA
IM
30
-
I
qSnorql /Y\olloJ
rrrral $I eql Jod
euo reqlrg
'lreJJoJ sr rurelc s,1sr8o1otq eql ecueH'
OZ < I roJ (rN < (r)y,,t os' OZ < t ueqaa, Surseercu sr
(7)p - (l)ltt puu (02)N = (97)61 ecutg
LtEt'st=#
-
Solution Marks Remarksrro I lr*7 ate. (a) (i) Jo +0,=
l0 ' (,F* *zrlt*zs' *zrlt*s2 *zr[t*t.s2.il.'o? )
- 2(4) 40 [' )=
1.305182044=1.3052So the increase oftemperature is about
1.3052"C '
d(t -\(u) -r-Jr +i l=--dt\40 ) qorlt*t2
a'-( t.1u,')= ' 'o,z(+0, )
"+40(t + tt )2>0
Hence it is an over-estimate.
(b) (i) 100(ln16)2 -
630lnxs + 1960 = 968
50(lnxs)2 -3l5lnx6 + 496 = 0
31 t6lrl -r,r =
" l0 5xo *22.1980 or 24.5325
. 2001n x 630(ii) w'(x) =.'.w'(x)
-
cc
lCerlOC(c) 3u
ICaIIOC(c) 8ur
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Jopuruouep JoJ htrl
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roleuruouep JoJ WI
lceuoc urroJ roJ I II
pqeqord Ienuourg roJlcolloc s0se3 JoJ
hIIhil
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n
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pennber eq1
mIF
r'l+hil+hlfirn00960'0=
lGtzts tg v s' o - i rGtztg L9? 9' 0 [ ) + s(6Ez,ts Lsrs' 0)] x
o rGtzt s ts v s' o),$rpqeqord pannbsr eq1 (e)
0l 9t0'0 =
0 * i
* Gtzts tgvs'o - t) uGtzts t9?9'o) ofr +OI
7 " rGtzts Lens' 0 - i sGEZEs L9?e'0) ofc =,Qrpquqord pennber
eq1 (p)
,rrrooor1loo:,Gtztstsvs'0
- r) 8(6zrsrgrs'0) 0lJ +(e&esDv;o - i 6GEZtsLe?s'O oll +
0{6tztsLsv;0) =
&lgqeqord partnber eq1 (c)
(r)VI
VI+WI
(e)YI
htrI+WI
lz)VIhII
gLs6'0 !(eeustgvs'o),Gtztstsvs'o - I) + GtztsLgns'O zGtztsLgrs'o
- r) +
Gtzestgvs'o)GrctsDvs'o - r) + Gtztstgvs'il *&lllqeqord
pertnber eq1
z uo-iiqo-S-s IIEmIIV-
VIhII
8ls6'0 =,Gtzts ts v s' O + (erc* t9 t 9' o - 0 rG tzts ts v9' o)
rtc +
) zGtT,tsLsn;OP + ,(a*tster;o - DGtztsLg?9'o) ]c =,$y11qeqord
pa:rnbar eq1
f -86-i1fr
ios-sIiiEu5itV-
zGtzESLgng'o-
YIWI
8LS6'0 =
,G*tsnvs'o-t)-r= (pelceqc sr Jermoc qlt eql JoUe prmoJ ore
srelunoc ,tsnq ou)4-1=,tlyrquqord pelnber eq1(q)
(e)89rS'0 0
6tZtSL9v9'0 =(ir.izirio) t- T-t-t-( o c-ac6'e o'r'z6't o't-'r6't
ot-'o6't ) ,{Urqeqord pennbar
eq1e)
VI
IAil+htrl
$lJeruau$FBIAIuormlos
-
Solution(a) The requl.g0e-1.8 l.gle-1.8 l.g2e-I.8 l.g3e-1.8
1.g4e-I.8+-+-+-+-0!|2!3!4l
* 0.963593339=
0.9636
Remarks
lM for P(X s 4)lM for Poisson probabilitywith correct 2
For standardization
For any one correct
For all correct
lM for form correct
lM for form correct
lM for any 2 caseslM for denominator using (a)lA for all
correct
(b) oo =r(rri#) =p(Z >2)=0.5-0.47i2=0.0228o. =p( z< 2 -3\[
0 8 )= '(' < -l '25) = 0'5 - 0'3944 = 0' 1056Pt =l- Po - Pz
=l-0.0228-0.1056 = 0.8716
(c) (D The required probability= cl pzpt2 + C? pz2 po= 3(0.
l0s6x0 .87 rc)z + 3(0. 1056)2 (0.0228)x 0.241431455x 0.2414
(ii) The required probability= ctpz2 poz *fi.rrn'po * pro= 6(0.
I 056)2 (0.022q2 + I 2(0. I 056)(0. 87 I 6)2 (0. 0228) + (0.S7 I
6)4x 0.599107436- 0.5991
(d) The required probabilityt's'i.-' * (o.l0s6)2 . r#,
0.241431455) + t'soi.-" (0. ssstoi436)
- l.lg635%ng=
0.0883
IM
1A
IA
34
-
s
) Sursn Joleurruouep loJ htrllceJ.loc urroJ JoJ wl
lsexoc sews II3 JoJ htrIlcerros esec 1 ,(ue JoJ hll
((t)(p) q pepre^\e eq UBJ)rpqeqord lerruourg roJ hll
lcauoc seses JoJ IAII
\L)
Llsg'0 *08rr9r tt0'0
"
&gllqeqord parnber eq1
zLto'0 x08LtgILtl'0 x
o(rct esozzt o) r r(rct as ozzt o - D .?rl [ . (e s t L6n Loo'
il,Gs t 6t too o - r) f cl +l(lrososozzo'o) r,(rctanzzo'o - r)
zlrl[(r9 t L6nLoo'O LGst L6n Loo'o - r),'JJ +Ir(utawzzud
ot(L6l6gozzo'o - r) rfrJI o(tgt L6nLoo'il sGgtrctnco- r) fe] =
firpquqord perrnber eq1
tzzU'0 x168690220'0 =
[,(sr o'o) rr(s I 0'0 - il olc + (s t o'o) ur(st 0'0 - I) .rIJ +
or(s t o'o - D] - r =,firpqeqord pelnber eq1
(rs)
(lr)
(l)(p)
' L sl 1 Jo enle^ lsuel eql ecuaH
ezzztssts.s =yffi!.,96'0q > 9$Z0SZ66',out1
96'0> ft9z0sz66'0
'i::,':"*'::i: _l _:
VIhtrt+I^trI
VI
WI+I^It
VI
(r)YI
NI
hII
Fs0'0 < (t9et6rt00'0)r-t(tgtL6vLoo'o- r) + "'+
aL6v Lol'O z(t9tL6n Loo'o - i + G9e L6v too'dcsttur too'o__ ! .
,*t* rn ouorlnlos e^rlBrusllv
IIIIIIII
s0' 0 < ?(89 L6v Loo' il + Gw tav n0'0 - r) r-1([9 uev nu
Or-1, + . . .
+
,Gw tev nu o - D rGs e.t6n Loo' il $ c + r- rGs e L6v Loo' o -
:D3.s t L6n Loo' o) f c
IIIII
i r"l
tr00'0 =Gst6vtoo'o) rGstrcvtoo'o - r) =
,$rlrqeqord pernber eq1(q)
sr00'0 !t9tL6tL00'0 *
[.(ro'o)rr(ro'o - i olJ+ (r0'0)6r(10'0 - r),fJ + or(ro'o - r)] -
r =.Z(e)
\Z)
VIhII
()
YI
I^trI+WI
s{J?lueu$lrBIuormlos
,11
-
6t
'lurq ue^IE eqlJo osn e{BIuplnoc peldtua11e oI{1( esoql pu
peldue[B seleplpuec ,{ueu ool loN 'rood
'(f) (q) eleldruoc o1 3utpe3 ,{q pereputq ere.,'t seleplpuBc
atuos 'rleJ
'uorler8alut op 01 uot1rultsqns,(1dde ]ou plnoc seleplpuecJo
requnu V 'llJ
'punoJ eJel* so{elsttu sselorcc euros q8noql'poo8 ,ften
'poo3,ften0s
(z)
(r) (l)(r) (q)
(l) (e)(r) (e)s
'(Z) ttud qcer lou plnoc scueq puu .,ut8ttoq8norql sessed r7,,
Jo uolllpuoc aql esn 01 pelleJ seleprpuec .(ue61 'rood
'an-lnc e Jo lerruou pue lue8uelaql uoo,^ loq uotlBleJ eql
qll,^A IEIIIlueJ lou ele.!\ sel8plpusc Jo requnu v 'rle{
'pooD
'saloldur.,(se oql pug plnoc seleprpuec lsoIAI 'poo8 ,{ren
0ql
L9
(r)
(l) (c)
(q)
(e) t
IBJoUAC uI eCueuuoJJod0lr) ,Qpepdo4r0qunNuorlsen)
(suorlsanb 9 Jo lno t Jo ooroqc y) g uoqcag
',(russeceu lou sI qclqa ',(lyectueqceu: uoIlBIAopptpuuls aql
elelnclec ol palJl euros q8noqlle uotlsenb sql elpuuq plnoc
seleppuec ,(ueyq 'poog9
'pelnber sr ftrlrqeqord luolllpuoc e leql oJE,^ B lou ele,t\
soleplpuec oluos 'rIE{s
'sluela .,1uepuedeput,, Jo leql qlIA pesnJuoc pue sluala
((ellsnlcxe,(11eqnru,,;o uorlrugop eqlJo elns Jou ere,^^
SelBpIpLIEo ,(ueu areqrrl (p) Ued ldecxe'poo3,ftenv
',(pedo.rd ,( Jo lltutl eqlJo uollelou IEcIItuaqlBu aql lueserd
lou ppoc salsplpuc eulos 'poogc
ur uollcuru cruqlueSol ol{l Jo esn eql ql! ^ JeIIIIUBJ lou
oJe,^'uoItrBIlueroJJIp
seleprpuec oluos 'poocZ
eql
JOJ
(e) ur uorsuedxa eqt ,(1dde 1ou
Jo senlu^ 3o e8ue: eql lno llo,{\
'pelnber uolleulxorddeplnoc solprpuec ,(uery '(q) ued ur rood
reqld
'prlA sr uorsuedxe lelr.uourq eql qclq,^tou plnoc soteprpuec
euog '(t)(e) ued ur pooE ,ften
I
IEJeueD uI ocusuuoJlodJequnNuoqsen|
(,tos1nduo3) y uo;1ceg
aJuEruJoJJod 6solBplpuBJ
-
QuestionNumber
Popularity(%) Performance in General
e (a) (i)(iD
(b) (i) (ii)(iii)
(iv)
5l Very.good.
Poor. Many candidates were not aware that the second derivative
of the givend3 x d,2,
equatron ,r *,
rather than fI .
Good.
that dY should be found and somedtPoor. Many candidates did not
realisefailed to apply chain rule to find dI .
dt
very poor. Most candidates could not interpret their own
mathematical f,rndingsand hence failed to make use of the results
to make judgement.
l0(a)
(b)
(c)
(d)
(e)
97 Very good.
Very good.
Good.
Poor. Many candidates overlooked that a joint probability should
be considered.Fair. Many candidates were able to handle conditional
probabilities but somewere careless.
11(a)
(b)
(c) (i) (ii)
(d)
83 Very good.
Good.
Fair. some candidates did not do the counting right and missed
some of theeligible events.
Poor. Many candidates had difficulty in identi$ing the joint
probabilitiesrequired for the numerator of the conditional
probability.
40
-
tv
'8ur>1urqt c1leure1s.ds e,r.ordrut pue s1decuocgo
Surpuelsropun reueq ur dloq ue+o ucc uotluluese:d reelc pue
sloqru(s IEcIleuoI{}Btu Jo esn redot4 't
'Je/v\suE Ieuu eql ro3 ,(curncceSoeer8ep pelnbar er{l saerqce 01
Japro ur sdels alerpeuretut aq1 ur seceld lelutcep sJoIU esn plnoqs
sepplpueJ 'Z
'ruelqord e o^los ol lro,$u^lo rreqt Jo sllnssr ornJo esn e1ur
ppoc feql 1sql it\oqs q pallq lnq sllDls rleql polsrlsuouop
seleplpuec,{ueyrtr 'senbruqcal pcrltuer{1cru oql pqqeq Eunreeru eql
01 suolluraplsuoc orour a,rt8 ppot{s s4eplpue3 'l
:suorl"pueuuocor pue slueuluoc Isjeuac
JoJeqr.unueqtr
'slua.{e eql ?uqunoc ut .{1yncg3tp p"q ssleplpuec euos 'JIed
'slue^e lue^eleJEurlunoc ur ,Qpcgglp peJslunocue so1eplpuc euos
'poog
'serllenbeur Euttpueq q fUIHs ssel ars^{ saPplpLIBJ 'JIBg
'pooE &sn
'poo8 {:enZ9
(lrXu)
(r)(p)
(c)
(q)
(e)zt
IBJau3C ur ecuuuuoJJed(%)f,rue1ndod
JeqlunNuo4sen$