# Mathematics Grade 11 Paper 1 Eastern Cape November 2018 Memo

> Grade 11 Mathematics Paper 1 memorandum from the Eastern Cape November Exam 2018, 19 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-p1-grade-11-nov-2018-memo-eng-afr/ae87434a-55d1-4ba1-9c17-3d73e4eb2adf
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2018
- Exam period: Eastern Cape November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 19

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## About this document

Mathematics · Grade 11 · Eastern Cape November Exam · 2018. Memorandum, 19 pages. Read online or download the PDF.

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### Page 1

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gen

Doan oad basic education

)
Department:
QY HI Basic Education
y REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE/

NASIONALE
SENIOR SERTIFIKAAT

GRADE/GRAAD 11 |

“a *
" MATHEMATICS P1/WISKUNDE V1 .
/ a
M NOVEMBER 2018 :
i. u
iy u
1 MARKING GUIDELINES/NASIENRIGLYNE :
Oo tate stafatetatete ete gtete etete ete a PTREECCOCOT Ere meeans?

MARKS/PUNTE: 150

These marking guidelines consist of 18 pages.
Hierdie nasienriglyne bestaan uit 18 bladsye.

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Mathematics/P1/Wiskunde V1
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

NOTE:
If a candidate answered a question TWICE, mark only the FIRST attempt.

If a candidate crossed out an answer and did not redo it, mark the crossed-out answer.
Consistent accuracy applies to ALL aspects of the marking marking guidelines.
Assuming values/answers in order to solve a problem is unacceptable.

LET WEL:
As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.

As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na.
Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyne van toepassing.

Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los.

QUESTION/VRAAG 1

DBE/November 2018

LL x(2x +1) =0

1
=0 ork =-—
x or/of x

(2)

1.1.2) 5x? +2x-6=0

x ~b+,/b’ —4ac

2(5)
_ 5+ V124

10
x=0,91 or/of x= -1,31

2
_2@t oy ~46)-6)

1.1.3 2x? -2> 3x

2x? -3x-2 20
(2x +1(x—2) 20

xs -5 or/of x22

Y substitution into correct formula/
vervanging in korrekte formule

v answer/antw.
v answer/antw.

= G)
¥ std form/stand. vorm
¥ factors or using formula/
Saktore of gebruik formule
Y0xs-3 orlof x22
(4)

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Mathematics/P1/Wiskunde VI 3 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne
2x +5
Let J2x+5 =k
p-3=-2
‘ 2k
42 ke 3-0 ¥ changing to quadratic/
( ket 3) ( k- 1)= 0 verander na kwadraties
k=—-3 otlof k=1 ¥ factors or using formula/
Saktore of gebruik formule
V¥k=3 orlofk=1
y2x+5 =-3 2x+5 =1
no solution orlof 2x+5=1 .
“no solution/ geen oplossing
2x=—4
x=-2
“square both sides/
kwadreer beide kante
¥Y yx=-2
OR/OF
Jix+5 -—3—=-2 (6)
2x +5
(lax +5 ) -3=-2,2x +5
(lax +5 } +2(/2x +5 }-3-0
(/2e +5 +3\ (2x +5 -1)<0
—= v changing to quadratic/
af =- 2. =
px +5 3 or v r+5 =I verander na kwadraties
V2x+5 =-3 2x45 =1 ¥ factors/fakt.
or/of iosin he
no solution x4+5=1 v 2x +5 =-3or/of\2x +5 =1
2x=-4 .
2 Y¥ no solution/ geen oplossing
x=
¥ square both sides/
kwadreer beide kante
OR/OF ¥ x=-2

(6)

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Mathematics/P 1/Wiskunde V1 4

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

. _
V2x+5—- =-2
v2x+5
2x+5-3=-2V2x+5
2x+2=-2V2x+5 2
2 f
(x +2) = (2 2x+ 5) ¥ Multiplying by/ Vermenigy. met
4x? 48x44 =4(2x45) is
4x? + 8x44 =8x +20 oars!
4x? -16=0
x’ -4=0 ¥ square both sides/
(x+2)(x-2)=0 kwadreer beide kante
2 or lof x#2 % std form/ vorm
¥ factors/fakt.
¥ x#2
¥Y x=-2
(6)
1.2 y+x=2 and/en = x? +3xy+8=0
2 y=2-x
x? +3x(2—-x)+8=0 VY y=2-x
x? + 6x-3x" +8 =0 ¥ substitution/verv.
-2x? +6x+8=0
x°~3x-4=0
(x-4)(x+1)=0 ¥ std form/stand. vorm
x=4orx=-1
¥ 5 4 orfof * 3 ~C) ¥ factors or using formula/
y= y= Jaktore of gebruik formule
¥ both x-values/wrdes
¥ both y-values/wrdes
OR/OF (6)
yrxu=2 and =x? 4+3xy+8=0
“ x=2-y V x=2-y
Q _ yy +3(2-y)y+8=0 ¥ substitution/verv.
4-Ayt+y? +6y—3y? +8=0
—2y? +2y+12=0 ¥ std form/stand. vorm
vy -y-6=0
(y-3y+2)=0 ¥ factors or using formula/
y=3ory=-2 faktore of gebruik formule
x=2-3 or x=2-(-2) both y-values/wrdes
x=-l or x=4 ¥ both x-values/wrdes
©

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Mathematics/P1/Wiskunde V1 5

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

13 4+,/16—4m(—m+5)
eee ee
2m
For non -real roots/Vir nie-reéle wortels: 16—4m(-m+5)<0

16+4m? —20m<0

m=” —5m+4<0

¥16—4m(-—m+5) <0 |

¥ factors or using formula/

(m—4\m—1)<0 Saktore of gebruik formule
l<m<4 VVleme4
(4)
14 | —x? 44x+412
=-1(x? —4x-12
c x-12) v-1(x? -4x—12)
=o vata a1 ¥ ~1(x? 4x 4+4-4-12)
=-1(x-2)' +16 Y¥-1(x—2) +16
The maximum value of/Die maksimum waarde van
—x? +4x+12 is 16
max value of /—x? +4x+12 is4 ¥ V16=4
(4)
OR/OF
OR/OF
Vx? +4x412
max when y=
2a
-4 ¥ subst/verv.
~ 2-1)
=2 “x-value/waarde
-_(oP
max valuey ©) +4(2)4+12 ¥y-value/waarde
The maximum value of/Die maksimum waarde van
~x? +4x412 is 16
.. max value of \)—x* +4x+12 is 4 ¥ Vi6 =4
(4)
[29] |

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Mathematics/P1/Wiskunde V1

DBE/November 2018

CAPS/KABV — Grade/Graad 1 — Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2
2.1 233 3 91
or?
2824-3242!
~ 2%97 v separate bases/aparte basisse
2*(2°-3.2
= rm 72 ¥ common factor/gemene fakt.
a ¥ (23 -3.2)
=-6
-8 _
i
4
__ 47 ¥ answer/antw.
> (4)
2.2.1 3
16x ? =2
3]
x75 g ¥ isolating/isoleer x
2
3)\3 2
: 2 ) = Oe ) 3 v raising both sides by/verhef
albei kante met — 2
x4 3
v answer/antw.
(3)
OR/OF
3
2-16x? =0
-3
2=2' x? ¥ prime base/priembasis
3
27 =x? Y exp law/eksp. wet
x=2°>
x=4 Yanswer/antw.
GB)

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Mathematics/P1/Wiskunde V1
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

DBE/November 2018

2.2.2 | 4*+8=9.2"
| (2? -9.2* +8=0
2* ~9.2* +8=0 ¥ standard form/stand. vorm
(2x -8\2* -1}=0
2*=8 or 2*=1 ¥2*=8 or 2°=1
27 =23 2% =2°
x=3 x=0 x=3
v x=0
OR/OF (4)
4° +8 =9.2*
(2? -9.2" +8 =0
2* -9.27 +8=0
Let 2*=k
kh? -9.k +8=0 ¥ standard form/stand. vorm
(k-8\Xk-1)=0
k=8 or k=l
2° =8 2*=2° ¥2*=8 or 2*=1
x_ 93 _
2*=2 x=0 v 3
x=3 v x=0
(4)
2.23 | [9 =243 =Y a(x
V3) =(3
2 5 ( ) ) ¥ exp form/eksp. vorm
3* =3 32 35
20 5 OR/OF 2=5x ¥ equating the exp/gelykst van eks
x 2
2 x= 5 v answer/aniw.
x= (3)
5
5
23 | Vp*-9' x(p+a)?
a
(p-ap
| 3 v difference of 2 s
—q\pt+@)xlp+qp quares
= Mo~aho+a)xlo+ap verskil van 2 kwadrate
(p-ap:
v
: ay (p + a x (p + q) exponent law/eksponentwet
oa
15
=(p+q)22
3 Y answer/antw.
=(p+q
(p+a) (3)

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Mathematics/P1/Wiskunde V1 8

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

OR/OF

5
vp’ -4 x(p+q)?

(p-a:

_ |@-aDeraXrtay
“y a)

—4

= \(p+4q)°

v difference of 2 squares
verskil van 2 kwadrate

3
= (p + a) ¥ exponent law/eksponentwet
v answer/antw.
(3)
[17]
QUESTION/VRAAG 3
3.1.1 7 2 -
<5 5
T,=an+b ¥ ~
=-5n+12
(2)
3.1.2 | Ty =-5(20)+12 Y substitution/verv.
=-88 ¥ answet/aniw.
(2)
3.1.3 | —5n+12=-138 ¥ substitution/very.
—5n=-150
" ¥ answer/antw.
n=30 (2)
30" term(T,, )
3.2 6 2x+1 3x-3
\ 4 _~ /
2x-5 x-4 v 2x—5 and/en
x-4
2x-5=x-
* x4 ¥ equating/verg.
x=l ¥ answer/antw.
G3)
[9]

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Mathematics/P 1/Wiskunde V1 9

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienrighne

QUESTION/VRAAG 4
4 p 11 22
oN a ON a oN ™ a
p-4 ll-p q-l 22-q
-~2p 415 q+p-2 332g
41

1" difference//*® verskil:
p-4; ll-p; q-ll; 22-q¢
2" difference/2® verskil:
—2p+153q+ p—22; 33-2q¢

—2p+15=1
—2p=-14
p=7
33-2q=1 q+ p-22=1
-2g=-32  orlof = g+7-22=1
g=16 q=16
OR/ OF
—2p+15=1
-2p=-14
p=7
4 7 11
3 4 5
‘ee ee
1 1

Y p—4;11- ps q-11; 22-4
Y ~2p4153q+ p—22: 33-2q

¥—-2p+15=1
and/en 33—2q=1

3)

v p-4;1l-p
VY -2p+15=1

¥ subst. of p in pattern to
find q/ vervang p in
patroon om q te vind

(3)

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Mathematics/P 1/Wiskunde VI DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne
4.2
2a=1
aati Vaal
2 2
3at+b=3
{5} r6=3
vb=2
b=3 2
2
at+b+c=4
Lib yecg
2 2 ¥c=2
c=2 i 3
1 3 VT, ==n?+=n+2
T,==n’ +in+2 2 2
2 2 (4)
OR/OF
T,=c=2 ¥c=2
2a=1
awl
2 Vai
T, =an’ +bn+e 2
4S" +b(I)+2
5-2 Vp=5
2 2
14,3
T=5" +5nt2 VT, 21+ 3n42
2 2
(4)

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Mathematics/P1/Wiskunde V1 11
CAPS/KABV — Grade/Graad 11 -— Marking Guidelines/Nasienriglyne

DBE/November 2018

43 7, =232 Vin? s3nt2 =232
1,3 2" 3
<n +—n+2 =232
2 2
se +5n -230=0 ” standard form/std. vorm
n +3n-460=0 ¥ factors/subst quad. eq.
(n+23\n—20)=0 [faktore/ verv kwadr. verg
n=—23 orn=20 selecting/kies n=20
(4)
44

1? +3 nsoeh (arty +3 (n41)42=1227
2 2 2 2

Le s3ns2t h(n? 42n41e 3 ne 342 =1227
2 2 2 2 #2

Ly tng 2edntans ty 3ng342=1227
2 2 2 22 2

n? +4n+6=1227

n’ +4n-1221=0

(n+37\n-33)=0

n#-37 n=33

1 3 1 3
Ty —Ts, = 564) +5 G4)+ 2-(40) +5 63)+ 2]

Ty, —Ty, = 631-596
Ly, — Ty = 35

¥ subst into/very, in
T, +T,., =1227

¥ expansion/ontw.

¥ standard form/std. vorm

v value off/wrde van n

v answer/antw.

(G)

[16] |

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Mathematics/P 1/Wiskunde V1

12 DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5
5.1 | x=3 ¥x=3
Vy=2
ye
(2)
5.2 4
O= Par es ¥ subst./verv. y= 0
4
-2= 73 ¥ simplification/vereenv.
—AUx-3)=4
-2x+6=4
x=1 v answer/antw.
Q)
OR/OF
(1:0) ¥ ¥ ¥ answer/antw
3)
53) 4
ye 03° ¥subst/verv. x =0
2
=3 ¥ answer/antw.
(2)
OR/OF
¥ ¥ answer/antw
3
5.4
For/Vir f
¥ asymptotes/asimptote
¥ shape/vorm
vx- and y- int. /afsnit
For/Vir g
¥ x-int/afsnit
¥ y-int /afsnit
(5)

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Mathematics/P 1/Wiskunde V1 13

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne

8 ——+2=x+2 V4 gaxs2
x= x-
Aa x42-2
x-3
4
=,
x-3
x(x-3)= 4 ¥ std vorm/stand. vorm
x’ ~3x-4=0 ¥ factors/faktore
(x—4)(x+1)= 0 ¥ answers/aniw.
x=4 or x=-1 (4)
5.6 | -l<x<3 vv answer/antwoord
(2)
5.7 | y=xte
2=(3)+e
-l=c
“.y=x-1 Is an axis of symmetry off simmetrie-as van f
v Olv4 +3; +2)
[v4 +3;4 +2)= 0(5:4) v5
v4
OR/ OF G)
x-l= 4 7) ¥ equating / vergelyk
x-3
oe
x-3
(x-3) =4
x’? -6x+5=0
(x-5)(x-1)=0
= = v5
x=5 or x=1 VA
y=5-1=4 @)
O(5;4) [21]

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Mathematics/P 1/Wiskunde V1 14 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 6
6.1 ya-4 ¥ answer/antwoord
Q)
6.2 D(3;25) v3
v 25
2)
6.3 ys25 ¥ answer/antwoord
or
ye(—0;25] (1)
64 | £(0)=-(x-3) +25
v i =
=-(0- 37 425 substitute/verv, x=0
=16 E(0;16) Vf (0) =16
1 O41
2(0)= {3} -4
O75 ¥g(0)=-3
=-3 B(0;-3)
v)
EB =16—(~3) = 19units/eenhede answealarwWoard @)
6.5 x>3 or/of xE(3;0) ¥ ¥ answer/antwoord
Accept/Aanvaar x>3 or/of xe[3;00) (2)

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Mathematics/P 1/Wiskunde V1

15

DBE/November 2018

CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienrighyne

6.6 | yu
O= {>| -4
2
| xl
4= {>|
2
atl
2= (3}
2

2-2"
l=-x-1
x=-2
A(-2;0)
B(O;-3)

OR/ OF

~(x-3)' +25=0
(x-3)' =25

x-3=5 or x-3=-5
x=8 x=-2

A(-2;0)
B(O;-3)

y27V

Ave gradient/Gemid gradiént =—*—+

% — %

_ —3-0
0+2
=

“2

Hy

_ 3-0

042
-3

~ 2

Ave gradient/Gemid gradiént = aM

¥ substitution/verv,

¥ equating exponent/gelykst. eksp.

v answer/antwoord

¥ subst. into correct formula
/very, in formule

¥ answer/antwoord
¥ substitution/very.
¥ factors or using formula/

faktore of gebruik formule

¥ answer/antwoord

¥ subst. into correct formula
/verv. in formule

v answer/antwoord

(G)

(5)

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Mathematics/P1/Wiskunde V1 16

CAPS/KABV -- Grade/Graad 11 — Marking Guidelines/Nasienriglyne

DBE/November 2018

67 | ae)=-el0)
x+l
=- {5 4
2
wa yt
2
Range/Waardeversameling:y <4 or y€(—»;4) ¥ y<4 or ye(-00:4)
(2)
6.8 | Turning point/ draaipunt ; (3:27) v3
¥27
(2)
6.9 F(x =-(x—3) +25
f@)= 3) VY —x? +6x416
-x°+6x+16=2x+k ¥ equating/vergelyk
-x’ +4x+16-k=0
tangent has one point of intersection thus two equal roots/
raaklyn het een snypunt dus twee gelyke wortels
v —~4(- -k)=
A=(4)' -4(-1)(16-&) =0 (4f -4-1l6- &)=0
16+64-4k =0
80 =4k
k=20 vanswer/antw.
(4)
[23]

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Mathematics/P 1/Wiskunde V1 17 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 7
TA - \m
L
lt+ig = [: + ‘us Y formula/form.
| 011s)” 0,15
Mig =[1+ = yj — ls
12 12
O15)"
i, =|1l+-—| -1
‘f ( 12
i =12,13% ¥ answer/antw. (3)
72 | A=P(f-i)" ¥ formula/form.
v. aie
=4700(1-0,18 )* substitution/verv.
=R2124,97 ¥ answer/antw.
Q)
7.3.1 | A=P(1+i)" Y formula/form.
2x4
=20 ooo + 1 ¥ substitution/verv.
4
=R23 068,12 ¥ answer/antw.
(3)
73.2) A =P(1 +i)"
0.078 2x12
= 23068,12{1+ 2078) 0078 sno
= R26 949,12
R26 949,12—R2500
=R24.449,12 ¥ A(after 2 years) — R2 500
A =P(1 +i)"
0. 078 3x12
=24 449.12{11 3 } Vn=36
=R30871,61 ¥ answer/antw.
(4)
OR/OF
0,078
1x8 1293 vi =—— and n= 60
A =23 vos. + or -2500{1+ | bs
° ° vv ~2500{ 14.9078
=R30871,48 12
| v answer/antw.
(4)
[13]

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Mathematics/P 1/Wiskunde V1 18 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 8
81 Given/Gegee: P(G) = 0,25
. Let x be the total number of balls
P(G)= g_1 y S.1
x 4 x 4
x =32
n(S) = 32 a)
OR/OF
Let x be the number of yellow balls
+. x+14 be the total number of balls
8 1
P(G)= ==
(6) x+14 4 V gol
x+14=32 x+14 4
n(S)=32 (1)
8.2 R (RR)
R G (RG)
$ Y (®/Y)
R (G;R
£ (GR) ¥18 (number of yellow balls/
ES < aantal geel baile)
G G (G;G)
Ss v branches/takke
3D
GY ayy
18 y ¢ ) ¥ probabilities/waarskynlikhede
2 R (y;R)
¥ outcomes/uitkomste
Y G (¥:6)
y (Y¥;Y)
(4)
83 | P(G,G)+P(R,R)+P(Y, Y) “ED
8 7) (6 5) (1817 | a2 >
=) —x—|+] —x— [+] —x— v( 65
32 31 32 31 32 31 32 31
49 v (2)
= 32 31
Io4 ¥answer/ antw
(4)
[9]

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### Page 19

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Mathematics/P1/Wiskunde V1 19 DBE/November 2018
CAPS/KABV — Grade/Graad 11 — Marking Guidelines/Nasienrigiyne

QUESTION/VRAAG 9
9.1 PCV) x P(M) | 32,
= x = = 0,095 Lo.
P(V and/en M)= cap 08 130
P(V and/en M) # P( V) x P(M) v P(-V)xP(M)=0,095

The events are not independent/Die gebeurtenisse is nie onafhanklik | Y PCV and/en M)=0,08

¥ conclusion/gevolgtr.
OR/OF ()
vy 22
P(V)xP(F) 150
32 83 v 83
=A x5 = 0,118 —
150” 150 "150 «)
P(-V)x P(F) =0,118
P(V andF)= ap 70133
P(VandF) # P( V) x P(F) YPC and F) = 0,133
v .
The events are not independent/Die gebeurtenisse is nie onafhanklik vonclusion/gevolgt 5)
92.1 | P(Aand/enB)=0,12 40 vy P(A and B)# 0
¥ conclusion/gevolgtr.
Events are not mutually exclusive/ Gebeurtenisse nie onderling (2)
uitsluitend nie
9.2.2 | P(Aor B)=P(A)+P(B)— P(A and B)
0,83 = P(A)+ 4P(A)- 0,12 , formal formule
0,95 = 5P(A) substitution/very.
P(A)= 0,19 v P(A)
P(B)= 4(0,19) = 0,76 vP(B)
|
(4)
9.2.3 | P(notA)=1-P(A) v P(not A)=1-P(A)
=1-0,19
=0,81 v answer/antw.
2)
[13]

TOTAL/TOTAAL: 150

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