MATHS P1 GR 12 MEMO SEPT 2024 English+Afrikaans hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2024 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 22 pages./ Hierdie nasienriglyn bestaan uit 22 bladsye. Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) NOTE/LET WEL: • If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging. • Consistent accuracy applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn. • If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is. • The mark for substitution is awarded for substitution into the correct formula. • Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1 1.1.1 (2x − 4)( x −1) = 0 x = 2 or/of x = 1  x = 2  x =1 (2) 1.1.2 2 x 2 − 3( x + 2) = 4 2 x 2 − 3x − 6 − 4 = 0  standard form / standaardvorm 2 x 2 − 3x − 10 = 0  substitution / vervanging −(−3)  (−3) 2 − 4(2)(−10) x= 2(2) x = 3,11 OR/OF x = −1,61  x = 3,11or/of  x = −1,61 (4) 1.1.3 x 2 + 4 x − 21  0 ( x + 7)( x − 3)  0  factors / faktore c.v ' s : x {−7;3} OR / OF −7  x  3  answer / antwoord (Accuracy / Akkuraatheid) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 − x −1 = 3 − 2x 2x − 3 = x −1  squaring both sides/ kwadreer beide kante (2 x − 3) 2 = x − 1 4 x 2 − 12 x + 9 = x − 1 4 x 2 − 13 x + 10 = 0  standard form / standaardvorm (4 x − 5)( x − 2) = 0  factors / formula 5 faktore / formule x or/of x = 2 4  answers with selection x=2 antwoorde met keuse (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) 1.2 2 x = 1 − y...................(1) xy − x 2 + y 2 = 5.........(2) y = 1 − 2 x..................(3)  y = 1 − 2x Subst/Vervang (3) into/in (2) x(1 − 2 x) − x 2 + (1 − 2 x) 2 = 5 x − 2x2 − x2 + 1 − 4x + 4x2 − 5 = 0  substitution / vervanging x 2 − 3x − 4 = 0 ( x − 4)( x + 1) = 0  standard form / standaardvorm x = 4 or/of x = −1  factors / faktore For/Vir x = 4 :  x-values / x-waardes y = 1 − 2(4)  y-values / y-waardes y = −7 For/Vir x = −1 : y = 1 − 2(−1) y =3 OR / OF OR/OF y = 1 − 2 x..................(1) xy − x 2 + y 2 = 5.........(2) 1 y 1 y x = − ..................(3)  x= − 2 2 2 2 Subst/Vervang (3) into/in (2) 2 1 y 1 y y  −  −  −  + y2 = 5  substitution / vervanging 2 2 2 2 y y2  1 y y2   standard form / standaardvorm − −  − +  + y2 = 5 2 2 4 2 4   factors / faktore y y2 1 y y2  y-values / y-waardes − − + − + y2 = 5 2 2 4 2 4 y2 21 + y− =0  x-values / x-waardes 4 4 y + 4 x − 21 = 0 2 ( y + 7)( y − 3) = 0 y = −7 or/of y = 3 For/Vir y = −7 : x=4 For/Vir y = 3 : x = −1 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 5 1.3 f ( x) = x 2 + 3x  f (− x) = x 2 − 3x 1 ✓ f (− x) & t (2k ) 2 x = [t ( x)] 2 t ( x) = 4 x 2  t (2k ) = 4(2k ) 2 t (2k ) f (− x) + =0 ✓ simplification / vereenvoudiging 4 4(2k ) 2 x 2 − 3x + =0 4 x 2 − 3 x + 4k 2 = 0 For equal roots / Vir gelyke wortels,  = 0 b 2 − 4ac = 0 (−3) 2 − 4(1)(4k 2 ) = 0 ✓ subst. in / vervang in:  = 0 9 − 16k 2 = 0 ✓ method of solving for k (3 − 4k )(3 + 4k ) = 0 metode vir oplos van k 3 3 ✓ k-values / k-waardes k= or/of k = − 4 4 (5) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 2/VRAAG 2 2.1 −5 −4 −1 4 1 3 5 2 2 2.1.1 2a = 2 a = 1 a + b + c = −5 ✓ a =1 3a + b = 1 c = −5 − 1 + 2 ✓ b = −2 ✓ c = −4 b = 1 − 3(1)  c = −4  b = −2 Tn = n 2 − 2n − 4 ✓ Tn = n − 2n − 4 2 (4) 2.1.2 Tn = n − 2n − 4 2 T35 = (35)2 − 2(35) − 4 ✓ answer / antwoord = 1 151 (1) 2.1.3 Tn = 1 + (n − 1)(2) Tn = 2n − 1 ✓ Tn = 2n − 1 Tn +1 = 2(n + 1) − 1 Tn +1 = 2n + 1 Tn  Tn +1 = 1155 ✓ (2n −1)(2n +1) = 1155 (2n − 1)(2n + 1) = 1155 4n 2 − 1 = 1155 ✓ standard form / standaardvorm 4n = 1156 2 n 2 = 289 ✓ n = 17 and/en n + 1 = 18 n = 17  n = 17, n  T17 and/en T18 will give a product of 1155 (4) sal ’n produk van 1 155 gee OR/OF OR/OF Tn = 2n − 1 ✓ Tn = 2n − 1 Tn −1 = 2n − 3 (2n − 1)(2n − 3) = 1155 ✓ (2n −1)(2n − 3) = 1155 4n 2 − 8n − 1152 = 0 n 2 − 2n − 288 = 0 ✓ standard form / standaardvorm (n − 18)(n + 16) = 0 n = 18 or/of n  −16 , n ∈ N n = 18 and/en n −1 = 17 ✓ n = 18 and/en n − 1 = 17 T17 and/en T18 will give a product of 1155 sal ’n produk van 1 155 gee (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 7 2.2 TP = 430 d =5 TP = a + ( p − 1)d 430 = 60 + ( p − 1)(5) ✓ equating / gelyk stel 430 = 60 + 5 p − 5 ✓ simplification / vereenvoudiging 5 p = 375 p = 75 T75 = 430 ✓ answer / antwoord (3) 2.3 a + (a + d ) + (a + 2d ) = 30 3a + 3d = 30 a + d = 10  a = 10 − d ......................(1) ✓eq(1) and / en eq (2) a (a + d )(a + 2d ) = 510.....(2) Subst./Vervang (1) into/in (2) (10 − d )(10 − d + d )(10 − d + 2d ) = 510 ✓ substitution / vervanging 10(10 − d )(10 + d ) = 510 (10 − d )(10 + d ) = 51 100 − d 2 − 51 = 0 49 − d 2 = 0 ✓ simplification / vereenvoudiging d 2 = 49 d = 7 d = 7 ✓ value of d / waarde van d  a = 10 − 7 = 3 ✓ value of a / waarde van a (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 3/VRAAG 3 3.1.1 2 2 ; 3 9  answer / antwoord (1) 3.1.2 a S = 1− r 2  substitution / vervanging = 1 − 13 =3  answer / antwoord (2) m 3.2 ∑ 8(2)k-1 = 131 040 k=3 32 + 64 + 128 + . . .  value of a / waarde van a r=2 𝑎(1 − 𝑟 𝑛 ) 𝑠𝑛 = 1−𝑟  substitution / vervanging n 32(2 – 1) 131 040 = 2–1 ∴ 2n – 1 = 4 095  simplification / vereenvoudiging 2n = 4 096 2n = 212 OR/OF n = log2 ( 4 096) ⇒ n = 12  value of n / waarde van n n=m−3+1 12 = m − 2  answer / antwoord m = 14 (5) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 4/VRAAG 4 4.1 x = −5 ✓ equation of V.A / vergelyking van V.A ✓ equation of H.A / vergelyking van H.A y = −2 (2) 4.2 −1 0= −2 ✓ y=0 x+5 −1 2= x+5 2( x + 5) = −1 2 x = −11 −11 x= = −5,5 ✓ answer / antwoord 2  11   − ; 0   2  (2) 4.3 −1 y= −2 x+5 −1 y= −2 ✓x=0 0+5 11 ✓ answer / antwoord y = − = −2, 2 5  11    0; −  (2)  5 4.4 ✓ both asymptotes / beide asimptote ✓ x-intercept / x-afsnit and/of y-intercept / y-afsnit ✓ shape / vorm (3) 4.5 y = −( x + 5) − 2 ✓ method / metode y = −x − 7 ✓ answer / antwoord OR/OF y = −x + c Note: Neem kennis Subst./Vervang (−5; −2) ✓✓Answer only – Full marks −2 = −(−5) + c Slegs antwoord - Volpunte −7 = c (2)  y = −x − 7 [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 5/VRAAG 5 5.1 −b x= 2a −( −5) ✓ substitution / vervanging x= 2(1) 5 ✓ equation / vergelyking x = = 2,5 2 OR / OF OR / OF f ( x) = x − 5 x + 6 2 f / ( x) = 2 x − 5 2x − 5 = 0 ✓ f / ( x) = 0 2x = 5 5  x = = 2,5 ✓ equation / vergelyking 2 (2) 5.2 f ( x) = g ( x) x2 − 5x + 6 = x + 1 ✓ f ( x) = g ( x) x2 − 6 x + 5 = 0 ✓ standard form / standaardvorm ( x − 5)( x − 1) = 0 x = 5 or/ of x = 1 ✓ x-values / x-waardes g (5) = 5 + 1 = 6 B (1;2) and/en C (5;6) ✓ y-values / y-waardes g (1) = 1 + 1 = 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 11 5.3 h = x + 1 − ( x 2 − 5 x + 6) ✓ g ( x) − f ( x) h( x ) = x + 1 − x 2 + 5 x − 6 = − x2 + 6 x − 5 ✓ h(x) −6 x= 2(−1) h / ( x) = −2 x + 6 =3 OR/OF 0 = −2 x + 6 ✓ x=3 x = 3 h(3) = −(3) 2 + 6(3) − 5 = 4 ✓ Max. height /  Max. height is 4 units. Maks. hoogte Maks hoogte is 4 eenhede. (4) 5.4 5 5 5 2 1 Min. of f : f   =   − 5   + 6 = − 2 2 2 4 1 9 Min. of/van t ( x) = − − 2 = − 4 4  9   y  − ;   ; y  R  4  ✓✓ Range of t ( x) / OR / OF Terrein van t(x) 9  y  − ; yR (2) 4 5.5 2 x3 ✓ ✓ 2 x3 (accuracy / akkuraatheid) (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 6/VRAAG 6 6.1 f ( x) = − log c x 1 = − logc ( 12 ) ✓ subst. of / vervanging van ( 12 ; 12 ) 2 c =2 c=4 ✓ value of c / waarde van c g ( x) = d x 2 = d ( 12 ) 1 2 2 d =2 ✓ value of d / waarde van d (3) 6.2.1 g : y = 2x 2 g −1 : x = 2 y 2 ✓ swopping x and y 1 omruil van x en y g −1 : y 2 = x 2 1 g −1 : y =  x 2 1 y = x 2 ✓ answer / antwoord (2) 6.2.2 f ( x) = − log 4 x h( x) = log 4 x ✓ h( x) = log 4 x ✓ answer / antwoord h −1 ( x) : y = 4 x (2) 6.2.3 x ✓ answer / antwoord (1) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 7/VRAAG 7 7.1 A = P(1 − i ) n ✓n=6 ✓ substitution / vervanging = 180000(1 − 0,13)6 = R78052, 72 ✓ answer / antwoord (3) 7.2 x[(1 + i ) n − 1] Fv = ✓ n = 120 and/en i 8% 8 i= or / of  0, 08 120   0, 08 60  12 1200 900 1 +  − 1 1300  1 +  − 1 ✓ n = 60 in F  12    12   = + 0, 08 0, 08 ✓ substitution into Fv / 12 12 vervanging in Fv = 164 651, 4317 + 95519,91312 = R 260171,34 ✓ answer / antwoord (5) 7.3.1 x (1 + i ) − 1 n ✓ correct substitution into OB = P(1 + i ) n −  A formula / korrekte i vervanging in A formule  0,13 75  75 9958,39  1 +  − 1 ✓ correct substitution into  0,13     12   OB = 850000 1 +  − FV formula / korrekte  12  0,13 vervanging in FV formule 12 = R 763 890,54 ✓ answer / antwoord OR/OF OR/OF x 1 − (1 + i )  −n OB =   i ✓ n = 165   0,13 −165  9958,39 1 − 1 +   ✓ substitution into the   12   OB = correct formula / 0,13 vervanging in korrekte 12 formule = R 763 889,86 ✓ answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) 7.3.2 4 Missed instalments / onbetaalde paaiemente OB T75 T76 T77 T78 T79 T80 For Outstanding Balance / Vir Uitstaande Balans = R 763 890,54 : ✓ substitution into the A = P(1 + i ) n correct A formula / 4 vervanging in korrekte  0,13  A formule A = 763890,54 1 +   12  A = R 797 534,2651 ✓ accumulated amount / opgeboude bedrag x 1 − (1 + 1) − n  P= i   0,13  −161  ✓ substitution into the x 1 − 1 +   correct formula /   12   797534, 2651 = vervanging in korrekte 0,13 formule 12 0,13 ✓ n = −161 797534, 2651 x= 12   0,13  −161  1 − 1 +     12   x = R 10 490,96 ✓ answer / antwoord  Adjusted instalment is R10 490,96 (5) OR / OF OR / OF For Outstanding Balance / Vir Uitstaande Balans = R 763 889,86 : A = P(1 + i ) n ✓ substitution into the 4 correct A formula /  0,13  vervanging in korrekte A = 763889,86 1 +   12  A formule = R 797 533,5551 ✓ accumulated   0,13  −161  amount/opgeboude bedrag x 1 − 1 +     12   797533,56 = ✓ substitution into the 0,13 correct formula 12 vervanging in korrekte 0,13 formule 797533,56  x= 12   0,13 −161  1 − 1 +   ✓ n = −161   12   x = R 10 490,95 ✓ answer / antwoord (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 8/VRAAG 8 8.1 f ( x) = x 2 − 3 f ( x + h) − f ( x ) f ( x) = lim h →0 h ( x + h) 2 − 3 − ( x 2 − 3) ✓ substitution into the formula = lim h →0 h vervanging in die formule x + 2 xh − h 2 − 3 − ( x 2 − 3) 2 = lim h →0 h x + 2 xh − h − 3 − x 2 + 3 2 2 = lim h →0 h 2 xh − h 2 ✓ simplification / vereenvoudiging = lim h →0 h h(2 x − h) = lim ✓ factorisation / faktorisering h →0 h = lim 2 x − h h →0 ✓ answer / antwoord f ( x) = 2 x (4) 8.2.1 dy ✓ −6x = − 6x + 7 dx ✓ 7 (2) 8.2.2  x − 5x 3 2  Dx  3 − x  x   x3 5 x 2 1  = Dx  3 − 3 − x 2  1 x x  ✓ x2  1  = Dx 1 − 5 x −1 − x 2  ✓ 1 − 5x−1   1 − 12 ✓ 0 & 5x −2 −2 = 0 + 5x − x (zero does not have to be seen) 2 (hoef nie nul te sien nie) 5 1 1 −1 = 2− ✓ − x 2 x 2 x 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) 8.3 h( x ) = − x 3 − 3 x 2 + 1 g ( x) = h / ( x) g ( x) = −3x 2 − 6 x ✓ g ( x) = −3x 2 − 6 x Max of g(x) will occur at g / ( x) = 0 Maks van g(x) sal wees by g / ( x) = 0 g / ( x) = −6 x − 6 = 0 ✓ x = −1  x = −1 g (−1) = −3(−1)2 − 6(−1) g (−1) = −3 + 6 = 3 ✓ answer / antwoord  largest value  maximum = 3 grootste waarde  maksimum = 3 (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 17 QUESTION 9/VRAAG 9 9.1.1 f ( x) = −3 x 3 + mx 2 + nx f / ( x) = −9 x 2 + 2mx + n 0 = −9(1) 2 + 2m(1) + n 0 = −9 + 2m + n n = 9 − 2m...................(1) ✓ equation 1 / vergelyking 1 f ( x) = −9 x + 2mx + n / 2 0 = −9(3) 2 + 2m(3) + n 81 = 6m + n 81 − 6m = n................(2) ✓ equation 2 / vergelyking 2 81 − 6m = 9 − 2m ✓ equating (method) / 81 − 9 = 6m − 2m gelykstel (metode) 72 = 4m  m = 18 ✓ solve for m / oplos vir m n = 9 − 2(18)  n = −27 ✓ substituting value of m / vervanging van waarde van m  f ( x) = −3 x3 + 18 x 2 − 27 x (5) 9.1.2 f (a) – corresponding y-value when x = a , while explanation of/verduideliking van f / (a ) – gradient/derivative/rate of change of f ✓ f ( a) when x = a f (a) – is die ooreenstemmende y-waarde wanneer ✓ f / (a) x = a, terwyl / f (a ) – stel voor die gradiënt/afgeleide/ veranderingskoers van f wanneer x = a (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 18 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) 9.1.3 Point of inflection/Buigpunt/infleksiepunt x + xB x= A 2 1+ 3 x= =2 2 OR / OF ✓ method / metode f ( x) = −9 x + 36 x − 27 / 2 f / / ( x) = −18 x + 36 ✓x=2 f / / ( x) = 0 −18 x + 36 = 0 18 x = 36 x = 2  f (2) = −3(2)3 + 18(2) 2 − 27(2) = −6 ✓ f (2)  (2; − 6) Gradient of / Gradiënt van g ( x) : f / (2) = −9(2)2 + 36(2) − 27 = 9 ✓ f (2) Gradient of / Gradiënt van h( x) : mg  mh = −1 1  mh = − 9 1 1 ✓ h( x ) = − x  h( x ) = − x 9 9 (5) 9.1.4 f ( x)  0 when f ( x) is concave up // x − value for point of inflection is 2 x  2 ✓✓ answer / antwoord f / / ( x)  0 wanneer f ( x) konkaaf op is x-waarde vir buigpunt/infleksiepunt is 2 x  2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 19 9.2 • a=20 ✓ shape / vorm • (−3;0) (0;0) (3;0) ( x-intercepts) (x-afsnitte) ✓ intercepts on the graph/ afsnitte op die grafiek • x-values of stationary points: x-waardes van stasionêre punte ✓ x- values for stationary points x-waardes vir stasionêre punte (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 20 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 10/VRAAG 10 10.1 S (t ) = −3t 2 + 30t ✓ substitution / vervanging S (3) = −3(3) 2 + 30(3) ✓ answer / antwoord = 63 scripts / skrifte (2) 10.2 S (t ) = −6t + 30 / ✓ S (t ) For maximum number of scripts, S / (t ) = 0 / Vir maksimum aantal skrifte, S / (t ) = 0 −6t + 30 = 0 ✓ S (t ) = 0 6t = 30 t = 5 (Day5 / Dag 5) ✓t = 5 (3) 10.3 No / Nee ✓ No / Nee D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 27 48 63 72 75 72 63 48 27 0 ✓ explanation / verduideliking Sum/Som = 495 scripts/skrifte (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/SEPTEMBER 2024) MATHEMATICS P1/WISKUNDE V1 21 QUESTION 11/VRAAG 11 11.1.1 x = 1 − (0,11 + 0,19 + 0, 41) x = 0, 29 ✓ value of x / waarde van x ✓ answer / antwoord P(A) = 0, 29 + 0,11 = 0, 4 (2) 11.1.2 P (A or/of not/nie B) = 0, 29 + 0,11 + 0, 41 = 0,81 ✓✓ answer / antwoord (2) 11.2.1 a=4 ✓ answer / antwoord (1) 14 7 11.2.2 = ✓ answer / antwoord (1) 30 15 7 11.2.3 P (winning a game) = 30 15 1 P (playing at home) = = 30 2 P (winning a game)  P (playing at home) 7 1 =  30 2 7 ✓ P (winning a game)  = = 0,12 60 P(playing at home) 3 P (winning a game and playing at home) = = 0,10 ✓ P(winning a game and 30  events are not independent, since playing at home) P (winning a game and playing at home)  P (winning a game)  P (playing at home) ✓ conclusion 7 P (wen ʼn wedstryd) = 30 15 1 P (speel tuiswedstryd) = = 30 2 P (wen wedstryd)  P (speel tuiswedstryd) 7 1 =  30 2 ✓ P (wen wedstryd)  7 P(speel tuiswedstryd) = = 0,12 60 3 ✓ P(wen wedstryd en P (wen wedstryd en tuis wedstryd) = = 0,10 speel tuiswedstryd) 30  gebeurtenisse is nie onafhanklik nie, omdat P (wen wedstryd en speel tuiswedstryd)  ✓ Gevolgtrekking P (wen wedstryd)  P (speel tuiswedstryd) (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 22 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2024) QUESTION 12/VRAAG 12 12.1 21×20×10×9×19×18 ✓✓ answer / antwoord =12 927 600 codes/kodes (2) 12.2 DIGITS / SYFERS: 4 2 or/of 3 1 4 1 4 1 6 2 6 2 8 4 8 4 9 9 LETTERS BEFORE G / LETTERS VOOR G: A; B; C; D; E and/en F Out of 6 letters remove A and E (vowels) = 4 letters Van die 6 letters verwyder A en E (klinkers) = 4 letters COMBINED / KOMBINASIE : ✓ 4 in n( A) n( A) = (4  20  4  2 19 18) + (4  20  3119 18) ✓ 4  2 and/en 3 1 in = 218 880 + 82 080 n( A) = 300 960 n( A) P ( A) = n( S ) 300960 = ✓ dividing by / deel deur 12927600 12927600 22 = 945 ✓ answer / antwoord  0, 02 (4) [6] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.