Mathematics Grade 12 Paper 1 Eastern Cape Mock 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2016 MATHEMATICS P1/WISKUNDE V1 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 16 pages./ Hierdie memorandum bestaan uit 16 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) NOTE:  If a candidate answered a question TWICE, mark the FIRST attempt ONLY.  Consistent accuracy applies in ALL aspects of the memorandum.  If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.  The mark for substitution is awarded for substitution into the correct formula. LET OP:  Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging.  Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum.  Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is.  Die punt vir substitusie word vir substitusie in die korrekte formule toegeken. QUESTION 1/VRAAG 1 1.1.1 𝑥 2 − 4𝑥 − 12 = 0  standard form/standaard (𝑥 − 6)(𝑥 + 2) = 0 vorm 𝑥 = 6 or/of 𝑥 = −2  𝑥 = 6 (CA applies)  𝑥 = −2 (CA applies) (3) 1.1.2 3𝑥 2 + 2𝑥 − 6 = 0 Penalise 1 mark for  substitution/substitusie incorrect rounding off. −(2)±√(2)2 −4(3)(−6) 𝑥= Penaliseer 1 punt vir 2(3) verkeerde afronding. 𝑥= 6 −2±√76  𝑥 = −1,79  𝑥 = 1,12 𝑥 = −1,79 or/of 𝑥 = 1,12 (3) 1.1.3 2 27−𝑥 3𝑥 −1 = 2 3  3−3𝑥−1 3𝑥 −1 = 3−3𝑥−1 ∴ 𝑥 2 − 1 = −3𝑥 − 1  equating exponents/ 𝑥 2 + 3𝑥 = 0 gelykstelling van eksponente 𝑥(𝑥 + 3) = 0  factors/faktore 𝑥 = 0 or/of 𝑥 = −3  both x-values/beide x- waardes OR/OF 2 27−𝑥 3𝑥 −1 = 3 2 3𝑥 −1 . 3 = 27−𝑥 2  𝑥 2 − 1 + 1 = −3𝑥 3𝑥 −1+1 = 3−3𝑥  equating exponents/ ∴ 𝑥 2 = −3𝑥 gelykstelling van eksponente 𝑥 2 + 3𝑥 = 0  factors/faktore 𝑥(𝑥 + 3) = 0  both x-values/beide x- 𝑥 = 0 or/ of 𝑥 = −3 waardes. (4) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 3 1.2.1 1 1+𝑥 =0 𝑥+1  𝑥 = −1 =0  𝑥=0 𝑥 𝑥 = −1 or/of 𝑥 = 0 (2) 1.2.2 1 𝑥− 𝑥 1 =1 1+ 𝑥 1 𝑥−𝑥 =1+𝑥 1  manipulation of equation/ manipulasie van vergelyking 𝑥2 − 1 = 𝑥 + 1  standardform/standaard 𝑥2 − 𝑥 − 2 = 0 vorm (𝑥 + 1)(𝑥 − 2) = 0  factors/ faktore 𝑥 = −1 or/of 𝑥 = 2  both x-values/beide x- 𝑥 = 2 only/alleenlik waardes  choosing 𝑥 = 2/keuse van 𝑥=2 (5) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 2/VRAAG 2 2.1 𝑥 − 𝑦 = 3 ; 𝑥𝑦 = 28 𝑥 − 𝑦 = 3 … … … … … … … . . (1) 𝑥𝑦 = 28 … … … … … … … (2) From/vanaf (1) 𝑥 = 𝑦 + 3  𝑥 =𝑦+3 Substitute in (2)/vervang in (2)  substitute in (2)/vervanging 𝑦(𝑦 + 3) = 28 in (2) 2 𝑦 + 3𝑦 − 28 = 0  standard form/standaardvorm (𝑦 − 4)(𝑦 + 7) = 0  factors/faktore 𝑦 = 4 or / of 𝑦 = −7  y-values/y-waardes  x-values/x-waardes 𝑥 = 7 or/of 𝑥 = −4 OR/OF From/vanaf (1) 𝑦 = 𝑥 − 3 Substitute in (2)/vervang in (2)  𝑦 =𝑥−3 𝑥(𝑥 − 3) = 28  substitute in (2)/vervanging 2 𝑥 − 3𝑥 − 28 = 0 in (2) (𝑥 + 4)(𝑥 − 7) = 0  standard form/standaardvorm 𝑥 = −4 or / of 𝑥 = 7  factors/faktore  x-values/x-waardes 𝑦 = −7 or / of 𝑦 = 4  y-values/y-waardes (6) 2.2 𝑥 2 ≤ 4 + 3𝑥 ; 𝑥>0 2 𝑥 − 3𝑥 − 4 ≤ 0 (𝑥 + 1)(𝑥 − 4) ≤ 0  standard form/standaardvorm  factors/faktore + - +  solution/oplossing −1 ≤ 𝑥 ≤ 4 -1 4  final answer/finale antwoord 0≤𝑥≤4 Solution/Oplossing −1 ≤ 𝑥 ≤ 4 But/maar 𝑥 > 0 0<𝑥≤4 (4) [10] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 5 QUESTION 3/VRAAG 3 3.1 1 3 7 13 2 4 6 2 2 𝑇𝑛 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐 2𝑎 = 2  𝑎=1 𝑎=1 3𝑎 + 𝑏 = 2 3+𝑏 =2  𝑏 = −1 𝑏 = −1 𝑎+𝑏+𝑐 =1  𝑐=1 1−1+𝑐 =1 𝑐=1  𝑇𝑛 = 𝑛2 − 𝑛 + 1 𝑇𝑛 = 𝑛2 − 𝑛 + 1 Row/Ry 80 Term 1  6321 𝑇80 = 802 − 80 + 1 𝑇80 = 6321 (5) 3.2 Row 80/Ry 80 6321 6323 6325 6327 ... 𝑛 𝑆𝑛 = [2(𝑎) + (𝑛 − 1)𝑑]  𝑛 = 80 2 80  𝑑=2 𝑆80 = 2 [2(6321) + (80 − 1)(2)] Row 80/Ry 80  sub into correct formula/ 𝑆80 = 512000 vervang in korrekte formule  answer/antwoord OR/OF Row/Ry 80 Term 80 (4) 𝑇80 = 6321 + (79 × 2)  calculating term 80 of row 𝑇80 = 6479 80/bepaling van term 80 𝑛 𝑆𝑛 = 2 [𝑎 + 𝑙] van ry 80 80  6479 𝑆80 = 2 [6321 + 6479] Row 80/Ry 80  sub into correct formula/ 𝑆80 = 512000 vervang in korrekte formule  answer/antwoord (4) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) OR/OF 1 5 11 19 4 6 8 2 2 2𝑎 = 2 𝑎=1  𝑇𝑛 = 𝑛2 + 𝑛 − 1 3𝑎 + 𝑏 = 4  𝑇80 = 6479 3+𝑏 =4 𝑏=1 𝑎+𝑏+𝑐 =1  sub into formula/sub in 1+1+𝑐 =1 korrekte formule 𝑐 = −1  answer/antwoord 𝑇𝑛 = 𝑛2 + 𝑛 − 1 𝑇𝑛 = 𝑛2 + 𝑛 − 1 𝑇80 = 802 + 80 − 1 𝑇80 = 6479 𝑛 𝑆𝑛 = 2 [𝑎 + 𝑙] 80 𝑆80 = 2 [6321 + 6479] Row 80/Ry 80 𝑆80 = 512000 (4) [9] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 7 QUESTION 4/VRAAG 4 4.1.1 𝑇10 = 𝑆10 − 𝑆9  setting up of equation/ 𝑇10 = 10(11)(12) − 9(10)(11) opstel van vergelyking 𝑇10 = 330  substitution/vervanging  answer/antwoord (3) 4.2 𝑝 ; 3𝑝 ; 5𝑝 ; … … … … … 𝑑 = 2𝑝  first three terms/eerste drie 𝑛 𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑] terme 𝑝  𝑑 = 2𝑝 𝑆𝑝 = 2 [2𝑝 + (𝑝 − 1)2𝑝] 𝑝  substitution/vervanging 𝑆𝑝 = 2 (2𝑝 + 2𝑝2 − 2𝑝)  answer/antwoord 3 (4) 𝑆𝑝 = 𝑝 OR/OF  𝑎=𝑝 𝑎=𝑝  𝑙 = 2𝑝2 − 𝑝 𝑙 = 2𝑝2 − 𝑝 𝑛  substitution/vervanging 𝑆𝑛 = 2 [𝑎 + 𝑙] 𝑝  answer/antwoord 𝑆𝑝 = 2 [𝑝 + 2𝑝2 − 𝑝] 𝑆𝑝 = 𝑝3 (4) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 5/VRAAG 5 5.1 𝑟= 𝑥+2  ratio/verhouding 𝑥 Answer only/antwoord alleenlik 2  answer/antwoord (𝑥+2)2 2 𝑇3 = 𝑥 (2) 5.2 𝑎 𝑆∞ = 1−𝑟 𝑥  substitution/vervanging −8 = (𝑥+2)  simplification/ 1− -1 mark for (±) 𝑥 -1 punt vir (±) vereenvoudiging 𝑥2  𝑥 2 = 16 −8 = 𝑥−𝑥−2  both answers/beide antwoorde 𝑥 2 = 16 𝑥 = ±4 (4) [6] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 6/VRAAG 6 6.1 𝐴 = 𝑃(1 − 𝑖)𝑛 15  𝑖 = 100 and/en 𝑛 = 5 15 5 𝐴 = 635000 (1 − 100)  sub into correct formula/ 𝐴 = 281 752,87 vervanging in korrekte formule  answer/antwoord (3) 6.2.1 𝑥[1−(1+𝑖)−𝑛 ] 16.75  𝑖 = 1200 𝑃𝑣 = 𝑖 16.75 −48  𝑛 = −48 𝑥[1−(1+ ) ]  sub into correct formula/ 1200 50000 = 16.75 vervanging in korrekte formule 1200  answer/antwoord 𝑥 = 𝑅 1 436,29 (4) 6.2.2 𝑥[1 − (1 + 𝑖)−𝑛 ]  𝑛 = −18 𝑃𝑣 = 16,75 𝑖  𝑖 = 1200 16.75 −18  substitution/substitusie 1436.29[1−(1+ ) ] 1200 𝑃𝑣 = 16.75 1200  answer/antwoord 𝑃𝑣 = 𝑅22 721,97704  rounding/afronding 𝑃𝑣 = 𝑅22 722 (5) OR/OF Outstanding balance/Uitstaande balans (OB)  𝑛 = 30 16.75 30 16,75 16.75 30 1436.29[(1+ 1200 ) −1]  𝑖 = 1200 OB = 50000 (1 + 1200 ) −[ 16.75 ]  sub into both formulae/ 1200 vervang in beide formules OB = 𝑅 22722,14  answer/antwoord OB = 𝑅22722  rounding/afronding (5) 6.3 𝐴 = 𝑃(1 + 𝑖)𝑛 𝐴 = 2𝑥 and/en 𝑃 = 𝑥  𝐴 = 2𝑥 and/en 𝑃 = 𝑥 14.75 𝑛  sub into correct formula/ 2𝑥 = 𝑥 (1 + 100 ) vervanging in korrekte formule  using of logs/gebruik van log 2 𝑛= 14.75 logaritmes log(1+ ) 100 𝑛 = 5.04 𝑦𝑒𝑎𝑟𝑠/𝑗𝑎𝑟𝑒  answer/antwoord (4) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 7/VRAAG 7 7.1.1 𝑥=0  answer/antwoord (1) 7.1.2 𝑥 > −2 ; 𝑥≠0  𝑥 > −2 𝑥≠0 (2) 7.1.3 𝑦 = −4  answer/antwoord (1) 7.1.4 𝑦 = 𝑏𝑥 − 4  sub of/van −4 5 = 𝑏2 − 4  sub of point (2;5)/ 𝑏2 = 9 vervanging van punt (2;5) 𝑏 = ±3  𝑏 = ±3 𝑦 = 3𝑥 − 4  answer with correct 𝑏 value/ antwoord met korrekte 𝑏 waarde (4) 7.1.5 𝑥 = −2  𝑥 = −2 𝑦 = −1  𝑦 = −1 (2) 𝑎 7.1.6 𝑦 = 𝑥+2 − 1  sub of asymptotes/ 𝑎 vervanging van asimptote −3 = 0+2 − 1  sub of point/vervanging van 𝑎 = −4 punt (0;-3) −4 𝑦 = 𝑥+2 − 1  𝑎 = −4 (3) 7.1.7 𝑦 =𝑥+2−1 𝒚=𝒙+𝟏  𝑦 =𝑥+1 𝑦 = −(𝑥 + 2) − 1  𝑦 = −(𝑥 + 2) − 1 𝒚 = −𝒙 − 𝟑  𝑦 = −𝑥 − 3 (3) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 11 7.2.1 𝑦 = log 1 𝑥 2  swopping of x and y / 𝑓 −1 ∶ 𝑥 = log 1 𝑦 2 omruiling van x en y 1 𝑥  answer/antwoord 𝑦 = (2) OR/OF 𝑦 = 2−𝑥 (2) 7.2.2 (−2; 4)  Shape/vorm (0; 1)  y-intercept/y-afsnit −1 𝑓  any other correct point/ enige ander korrekte punt (3) 7.2.3 1 −𝑥 𝑔(𝑥) = (2)  Answer/antwoord OR/OF 𝑔(𝑥) = 2𝑥  Answer/antwoord (2) 7.2.4 𝑥>1  𝑥>1 (2) [25] Copyright reserved Please turn over

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 8/VRAAG 8 8.1 𝑥 = −3  𝑥 = −3 (1) 8.2 𝑦 = 𝑎(𝑥 + 3)2 − 5  sub of turning point 4 = 𝑎(9) − 5 (−3; 5)/substitusie van 9𝑎 = 9 draaipunt (−3; 5) 𝑎=1  sub of (0; 4)/ vervanging van 𝑦 = 𝑥 2 + 6𝑥 + 9 − 5 (0; 4) 𝑦 = 𝑥 2 + 6𝑥 + 4  simplification/ vereenvoudiging 𝑎 = 1 and/en 𝑏 = 6 (3) 8.3 ∆= 𝑏 2 − 4𝑎𝑐  ∆= 20 ∆= 36 − 4(1)(4)  irrational/irrasionaal ∆= 20  unequal/ongelyk 𝑅𝑜𝑜𝑡𝑠 𝑎𝑟𝑒 𝐼𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑈𝑛𝑒𝑞𝑢𝑎𝑙 / 𝑊𝑜𝑟𝑡𝑒𝑙𝑠 𝑖𝑠 𝐼𝑟𝑟𝑎𝑠𝑖𝑜𝑛𝑎𝑎𝑙 𝑒𝑛 𝑜𝑛𝑔𝑒𝑙𝑦𝑘 (3) 8.4 𝑔(𝑥) = 2𝑥 𝑥 2 + 6𝑥 + 4 = 2𝑥  𝑔(𝑥) = 2𝑥 𝑥 2 + 4𝑥 + 4 = 0  equating equations/ (𝑥 + 2)2 = 0 gelykstelling van vergelykings 𝑥 = −2  𝑥 = −2 𝑔(−2) = −4  𝑦 = −4 Point/punt (−2; −4) OR/OF 𝑓(𝑥) = 𝑥 2 + 6𝑥 + 4 𝑓 ′ (𝑥) = 2𝑥 + 6 and/en 𝑚 = 2  derivative/afgeleide 2𝑥 + 6 = 2 𝑓 ′ (𝑥) = 2𝑥 + 6 2𝑥 = −4  equating to gradient of g/ 𝑥 = −2 gelykstelling aan gradiënt 𝑦 = −4 van g.  x-value/x-waarde Point/punt (−2; −4)  y-value/y-waarde (4) [11] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 9/VRAAG 9 9.1 𝑓(𝑥) = 3𝑥 2 − 1 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim  formula/formule ℎ→0 ℎ 3(𝑥 + ℎ)2 − 1 − (3𝑥 2 − 1)  substitution of/substitusie van = lim (𝑥 + ℎ) ℎ→0 ℎ 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) − 1 − 3𝑥 2 + 1 = lim  ℎ→0 ℎ simplification/vereenvoudiging 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 1 − 3𝑥 2 + 1 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 1 − 3𝑥 2 + 1 = lim ℎ→0 ℎ  6𝑥ℎ 6𝑥ℎ = lim = lim ℎ→0 ℎ ℎ→0 ℎ Answer ONLY: 0 marks = 6𝑥 SLEGS antwoord: 0 punte  answer/antwoord Penalise 1 mark for incorrect use of formula. Must show 𝑓 ′ (𝑥). Penaliseer 1 punt vir verkeerde gebruik van formule. Moet 𝑓′(𝑥) toon. (5) 9.2.1 𝑦 = 5𝑥 2 + √𝑥 1 𝑦 = 5𝑥 2 + 𝑥 2 Penalise 1 mark for 1 incorrect notation.  𝑥2 1 Penaliseer 1 punt vir 𝑑𝑦 1 = 10𝑥 + 2 𝑥 −2 verkeerde notasie. 1 1 𝑑𝑥  10𝑥 2 𝑥 −2 (3) 9.2.2 6𝑥−4 𝐷𝑥 [ ] 3𝑥 6𝑥 4  2 𝐷𝑥 [ − ] 4 3𝑥 3𝑥 4 −1  − 3 𝑥 −1 𝐷𝑥 [2 − 𝑥 ]  answer/antwoord 3 4 4 (3) = 𝑥 −2 or/of 3 3𝑥 2 9.2.3 𝑚 = 𝑠 ′ (𝑡) = 3𝑡 2  derivative/afgeleide 𝑡2 ≥ 0  3𝑡 2 ≥ 0 3𝑡 2 ≥ 0 ∴ no value of t will make 𝑠 ′ (𝑡) negative. ∴ geen waarde van t sal 𝑠 ′ (𝑡) negatief maak nie. (2) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 10/VRAAG 10 10.1 𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 8𝑥 + 12 (𝑥 − 2)(𝑥 2 + 𝑥 − 6) = 0  (𝑥 − 2) (𝑥 2 + 𝑥 − 6) (𝑥 − 2)(𝑥 − 2)(𝑥 + 3) = 0  (𝑥 − 2)(𝑥 + 3)(𝑥 − 2) 𝑥 = 2 or/of 𝑥 = 2 or/of 𝑥 = −3 𝐴(−3; 0)  coordinates of 𝐴 (−3; 0)/ koördinate van 𝐴 (−3; 0) OR/OF 𝑥+3  𝑥 2 − 4𝑥 + 4 𝑥 2 − 4𝑥 + 4 𝑥 3 − 𝑥 2 − 8𝑥 + 12  𝑥 + 3 𝑥 3 − 4𝑥 2 + 4𝑥  coordinates of 𝐴 (−3; 0)/ koördinate van 𝐴 (−3; 0) 3𝑥 2 − 12𝑥 + 12 3𝑥 2 − 12𝑥 + 12 𝑓(𝑥) = (𝑥 2 − 4𝑥 + 4)(𝑥 + 3) 𝐴(−3; 0) (4) 10.2 𝑓 ′ (𝑥) = 3𝑥 2 − 2𝑥 − 8 = 0  𝑓 ′ (𝑥) 𝑓 ′ (𝑥) = 0 (3𝑥 + 4)(𝑥 − 2) = 0  factors/faktore −4 𝑥 = 3 or/of 𝑥 = 2  correct 𝑥 value/korrekte 𝑥 4 4 4 3 4 2 4 waarde 𝑥 = − 3 𝑓 (− 3) = (− 3) − (− 3) − 8 (− 3) + 12 −4 500 500 B ( 3 ; 27 )  𝑦 = 27 (5) 10.3 𝑓 ′′ (𝑥) = 6𝑥 − 2  𝑓 ′′ (𝑥) = 6𝑥 − 2 6𝑥 − 2 = 0 1 1 𝑥=3  𝑥=3 OR/OF (2) 4 − +2 3  finding 𝑥 value of midpoint/ 𝑥= bepaal van 𝑥 waarde van 2 1 𝑥=3 middelpunt 1  𝑥=3 (2) 10.4 4 4 𝑥 < − 3 or/of 𝑥 > 2 𝑥 < − 3  𝑥>2 (2) 10.5 𝑦 = 𝑘 ;𝑘 < 0 Only one Real Root/Net een reële wortel  answer/antwoord (2) [15] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2016) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 11/VRAAG 11 1 1 11.1 𝐷(0) = 3 + 2 (0)2 − 4 (0)3  𝐷(0) = 3 𝑚 𝐷(0) = 3 𝑚 (1) 3 ′ 11.2 𝐷 ′ (𝑡) = 𝑡 − 4 𝑡2  𝐷 (𝑡) 3 𝐷′ (3) = 3 − 4 (3)2 27 = 3− 4  𝐷′ (3) 15 15 = − 4 m/h / m/u  − 4 or/of −3.75 (3) 11.3 Decreasing/vemindering  decreasing/vermindering (1) 11.4 𝐷′ (𝑡) = 0  𝐷′ (𝑡) = 0 3 𝑡 − 4 𝑡2 = 0 4𝑡 − 3𝑡 2 = 0  factors/faktore 𝑡(4 − 3𝑡) = 0 4  𝑡 − 𝑣𝑎𝑙𝑢𝑒𝑠 / 𝑡 − 𝑤𝑎𝑎𝑟𝑑𝑒𝑠 𝑡 = 0 or/of 𝑡 = 3 4 = 1ℎ20𝑚𝑖𝑛  answer/antwoord 3 Time: at 08h00 or 9h20 / Tyd: 08h00 of 9h20 (4) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com 16 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2016) QUESTION 12/VRAAG 12 12.1 12.1.1 P(A′ ) = 1 − P(A)  P(A′ ) = 1 − P(A) = 1 − 0,35  answer/antwoord = 0,65 (2) 12.1.2 P(AandB) = 0  answer/antwoord P(AenB) = 0 (1) 12.1.3 P(A or\of B) = 0,35 + 0,52  𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) = 0,87  answer/antwoord (2) 12.2 12.2.1 6! = 720  6! or/of 720 (1) 12.2.2 4!  4! = 24  24 (2) 12.2.3  2! 2!.5! 240 1  5! = 720 = 3 𝑶𝑹/𝑶𝑭 0,333 6!  6!  answer/antwoord (4) [12] TOTAL/TOTAAL: 150 Kopiereg voorbehou Blaai om asseblief