Mathematics Grade 12 Paper 1 Western Cape Mock 2016 Memo

SEPTEMBER 2016 MATHEMATICS PAPER 1 / WISKUNDE VRAESTEL 1 MEMORANDUM NOTE: • If a candidate answers a question TWICE, mark only the first one. • Consistent accuracy applies in ALL aspects of the marking memorandum. LET WEL: • Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk slegs die eerste poging. • Volgehoue akkuraatheid is geurgans op ALLE aspekte van die memorandum van toepassing. QUESTION/ VRAAG 1 (23) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 1.1.1 𝑥(5𝑥 + 2) = 0 2 ✓✓Each root/ Elke wortel 𝑥 = 0 or/ of 𝑥 = − 5 (2) 1.1.2 2 2𝑥 − 3𝑥 − 4 = 0 ✓ Std form/ vorm −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 −(−3) ± √(−3)2 − 4(2)(−4) 𝑥= 2(2) 3 ± √41 ✓Substitution/ substitusie = 4 𝑥 = −0,85 or/ of 𝑥 = 2,35 ✓✓Each root/ elke wortel 𝑥 = −0,85 or/ of 𝑥 = 2,35 (-1, if incorrect rounding) (4) 2 1.1.3 𝑥 −𝑥−6 ≥ 6 𝑥 2 − 𝑥 − 12 ≥ 0 ✓std form/ vorm (𝑥 + 3)(𝑥 − 4) ≥ 0 ✓factors/ faktore 0 0 + −3 − 4 + ✓ both critical values/ beide kritieke waardes 𝑥 < −3 or/ of 𝑥 > 4 ✓notation / notasie (4) 1.2 𝑥 = 𝑦 + 3 …………….(1) ✓x subject/ onderwerp Subst (1) into (2): 𝑥 2 + 𝑥𝑦 − 2𝑦 2 = 0 ………(2) ✓subst in (2) ∴ (𝑦 + 3)2 + y(y + 3) − 2𝑦 2 = 0 ∴ 𝑦 2 + 6𝑦 + 9 + 𝑦(𝑦 + 3) − 2𝑦 2 = 0 ✓simplify / vereenvoudig ∴ 𝑦 2 + 6𝑦 + 9 + 𝑦 2 + 3𝑦 − 2𝑦 2 = 0 ∴ 9𝑦 + 9 = 0 ✓value of 𝑦 / Waarde van 𝑦 ∴ 𝑦 = −1 ∴ 𝑥=2 ✓value of 𝑥/ Waarde van x (5) OR/OF 1

𝑦 = 𝑥 − 3 …………….(1) ✓y subject/ onderwerp Subst (1) into (2): 𝑥 2 + 𝑥𝑦 − 2𝑦 2 = 0 ………(2) ✓subst in (2) ∴ 𝑥 2 + 𝑥(𝑥 − 3) − 2(𝑥 − 3)2 = 0 ∴ 𝑥 2 + 𝑥 2 − 3𝑥 − 2𝑥 2 + 12𝑥 − 18 = 0 ✓simplify / vereenvoudig ∴ 9𝑥 − 18 = 0 ✓value of 𝑥/ Waarde van x (5) ∴ 𝑥=2 ∴ 𝑦 = −1 ✓value of 𝑦 / Waarde van y 1.3 −2 ± √13 − 2𝑘 𝑥= 3 ∴ ∆= 13 − 2𝑘 ≥ 0 ✓∆ ≥ 0 ∴ 2𝑘 ≤ 13 13 ✓value/ waarde 𝑘 ≤ 13 ∴𝑘≤ 2 2 ✓ deriving k = 6 / afleiding 𝑘 = 6 ∴If 𝑘 = 6, ∆ is a perfect square and roots rational/ As 𝑘 = 6 Answer only/ Slegs antwoord = FULL sal ∆ ‘n volkome vierkant wees en wortels Rasionaal (3) MARKS/VOLPUNTE 1.4 If graphs intersect/ as grafieke sny: f(x) = g(x)  𝑥 2 − 2𝑥 − 3 = 2𝑥 + 𝑎  𝑥 2 − 4𝑥 − 3 − 𝑎 = 0 ✓ equating f and g/ stel f en g gelyk If graphs do not intersect/ As grafieke nie sny nie:  < 0 ✓<0 (−4)2 − 4(1)(−3 − 𝑎) < 0 ✓ Subst. ∴ 16 + 12 + 4𝑎 < 0 ✓ Simplify/ Vereenv ∴ 4𝑎 < −28 (5) ∴ 𝑎 < −7 ✓ answer/ antw. [23] QUESTION/ VRAAG 2 (15) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 2.1.1 1 1 3 ; 3; 2 ✓Answer/ Antw. 2 2 (1) 2.1.2 1 Arithmetic sequence/ Rekenkundige ry; d = − or /of ✓ AS/ RR; 2 1 1 1 ✓d=− (2) is subtracted each time/ 2 word elke keer afgetrek 2 2 2.1.3 1 1 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 ; a = 3 2 ; d = − 2 ; Tn = - 44,5 ; n =? ✓ Subst into/ in Tn 1 1 1 ∴ −44 2 = 3 2 + (𝑛 − 1) (− 2) ✓ Tn = - 44,5 1 1 ∴ 4 − 𝑛 = −44 2 2 1 97 ∴− 𝑛=− 2 2 ∴ 𝑛 = 97 ✓Answer/ Antw (3) 2

2.2.1 18 + 6 + 2+… 1 1 1 1 GS with 𝑟 = 3 / MR met 𝑟 = 3 ✓ GS with 𝑟 = / MR met 𝑟 = 3 3 (2) Thus no negative term / Dus geen term negatief. ✓ no negative term / geen term negatief 1 2.2.2 𝑇10 = 𝑎𝑟 𝑛−1 ; a = 18; r = ; n = 10 3 ✓ Subst into/ in Tn 1 9 = 18 (3) = 2187 2 ✓Answer/ Antw (2) 2.2.3 1 𝑆∞ − 𝑆10 ; 𝑎 = 18; 𝑟 = ; 𝑛 = 10 3 ✓ Correct sum formulae 𝑎 𝑎(1 − 𝑟 𝑛 ) = − ✓✓ Subst into sum formulae/ Subst in 1−𝑟 1−𝑟 1 1 10 somformules∶ 𝑎 = 18; 𝑟 = ; 𝑛 = 10 18 18 [1 − ( ) ] 3 3 = 1− 1 1−3 1−3 ✓ Simplify/ Vereenv 1 10 = 27 − 27 [1 − ( ) ] 3 (5) = 4,57 ×10 −4 ✓ answer/ antw [15] QUESTION/ VRAAG 3 (11) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 3.1 −3 − 5; −7 − ⋯ − 65 ; AS/RR: a = -3; n = 32; d = -2 ; l = - 65 ✓ n = 32. 𝑛 𝑛 𝑆𝑛 = 2 [2𝑎 + (𝑛 − 1)𝑑] OR/OF 𝑆𝑛 = 2 [𝑎 + 𝑙] ✓ Subst in sum formula/ Subst in 32 32 𝑆32 = 2 [2(−3) + (32 − 1)(−2)] 𝑆32 = 2 [(−3) + (−65)] somformule 32 ✓ answer/ antw (3) = 2 [−6 − 62] = -1088 3.2.1 6 ; 5 + 𝑥 ; −6 ; 6𝑥 ✓ 1st diff/ verskille 𝑥 + 5 − 6; −6 − (5 + 𝑥); 6𝑥 − (−6) ✓ 2nd diff/verskille 𝑥 − 1; −𝑥 − 11; 6𝑥 + 6 ; 1st diff/ verskil ✓ 2nd diff/verskille = −𝑥 − 11 − (𝑥 − 1) = 6𝑥 + 6 − (−𝑥 − 11) ; 2nd diff/verskil ∴ −2𝑥 − 10 = 7𝑥 + 17 ✓ Simplify/ Vereenv. (4) ∴ 𝑥 = −3 3.2.2 Sequence/ Ry: 𝟔 ; 𝟐 ; −𝟔 ; −𝟏𝟖; … −4 ; −8 ; −12 1st diff/ verskil −4 ; −4 2nd diff/ verskil ✓ a = -2 2𝑎 = −4  a = -2 ✓b=2 3𝑎 + 𝑏 = −4 3(−2) + 𝑏 = −4  b = 2 𝑎 + 𝑏 + 𝑐 = 6  −2 + 2 + 𝑐 = 6  c = 6 ✓c = 6 ∴ 𝑇𝑛 = −2𝑛2 + 2𝑛 + 6 ✓Equation/ Vergelyking (4) [11] 3

QUESTION/ VRAAG 4 (18) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 4.1 OC = 6 units/ eenhede ✓ OC = 6 A; B: x-intercepts/ afsnitte: Let/ Stel y = 0 ✓ Let/ Stel y = 0 ∴ −2𝑥 2 − 4𝑥 + 6 = 0 ∴ 𝑥 2 + 2𝑥 − 3 = 0 ∴ (𝑥 + 3)(𝑥 − 1) = 0 ✓ Factors/ Faktore ✓ Both x-values/ Beide x-waardes ∴ 𝑥 = −3 OR/OF 𝑥 = 1 A(−3; 0) and/ en B(1; 0) AB = 4 units/ eenhede ✓ answer/ antw (5) 4.2 𝑏 −4 𝑥 = − 2𝑎 = − [2(−2)] = −1 ✓Subst (2) ✓ 𝑥 = −1 OR/OF OR/OF f ′ (𝑥) = −4𝑥 − 4 = 0  𝑥 = −1 ✓Derivative/ Afgeleide = 0 ✓ 𝑥 = −1 4.3 Subst. 𝑥 = −1 in f(x) ∴ ST = −2(−1)2 − 4(−1) + 6 ✓ Subst. x = -1 = 8 units/ eenhede ✓ answer/ antw (2) 4.4 𝑦 −𝑦 0−6 ✓ Subst. in m mAC = 𝑥2 − 𝑥1 = −3−0 = 2 2 1 ✓ answer/ antw (2) 4.5 mg = 2 ; // lines/ lyne ✓ mg = 2 but/ maar mg = f ′ (𝑥) = −4𝑥 − 4 = 2 ✓ mg = f ′ (𝑥) 3 ✓ −4𝑥 − 4 = 2 ∴ 𝑥 = −2 3 ✓ 𝑥=− 3 2 3 1 1 2 ∴ 𝑦 = −2 (− 2) − 4 (− 2) + 6 = 7 2 ✓𝑦=7 2 3 1 D(− 2 ; 7 2) (5) OR/OF Let eq. of tangent be/ Laat vgl. van raakl: y = mx + k mg = 2 ; // lines/ lyne  y = 2x + k ✓ mg = 2 at D: f(x) = g(x) ∴ −2𝑥 2 − 4𝑥 + 6 = 2𝑥 + 𝑘 ∴ −2𝑥 2 − 6𝑥 + 6 − 𝑘 = 0……….(1) ✓ eqn/ vgl: −2𝑥 2 − 6𝑥 + 6 − 𝑘 = 0 For tangent/ Vir raaklyn:  = 0 (−6)2 − 4(−2)(6 − 𝑘) = 0 36 + 48 − 8𝑘 = 0  𝑘 = 10,5 ✓ 𝑘 = 10,5 ∴ −2𝑥 2 − 6𝑥 + 6 − 10,5 = 0 ……Subst k in (1) ∴ 2𝑥 2 + 6𝑥 + 4,5 = 0 or/ of 4𝑥 2 + 12𝑥 + 9 = 0 3 3 ✓ 𝑥=− ∴ (2𝑥 + 3)2 = 0  𝑥 = − 2 2 3 1  y = 2(− 2) + 10,5 = 7 2 ✓𝑦=7 1 2 3 1 D(− 2 ; 7 2) 4.6 𝑎 = −1 ; the axis of symmetry/ die simmetrie-as ✓✓ 𝑎 = −1 OR/OF : 𝑓(𝑎 + 𝑡) = 𝑓(𝑎 − 𝑡) ∴ −2(𝑎 + 𝑡)2 − 4(𝑎 + 𝑡) + 6 = −2(𝑎 − 𝑡)2 − 4(𝑎 − 𝑡) + 6 ∴ −2𝑎2 − 4𝑎𝑡 − 2𝑡 2 − 4𝑎 − 4𝑡 + 6 = −2𝑎2 + 4𝑎𝑡 − 2𝑡 2 − 4𝑎 + 4𝑡 + 6 ∴ 8𝑎𝑡 + 8𝑡 = 0 ✓ Subst. ∴ 8𝑡(𝑎 + 1) = 0 ✓ 𝑎 = −1 (2) ∴ 𝑡 = 𝑜 or/𝑜𝑓 𝒂 = −𝟏 4

[18] QUESTION/ VRAAG 5 (10) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 5.1 𝑥=1 ✓ answer/ antw (1) 5.2 x -int/afsn: Let/ Stel y = 0 ✓𝑦=0 2+𝑥 ✓ 𝑥 = −2 𝑥−1 =0  𝑥 = −2 A(−2; 0) (2) 5.3 y -int/afsn: Let/ Stel x = 0 2+0 0−1 =𝑦  𝑦 = −2  B(0; −2) ✓ B(0; −2) 1 Area/ Opp AOB = 2 AO ×OB ✓ Subst. in Area formula 1 = 2 (2)(2) = 2 units 2/ 𝑒𝑒𝑛ℎ2 ✓ answer/ antw (3) 5.4 2+𝑥 𝑥−1+3 𝑓(𝑥) = = 2+𝑥 𝑥−1+3 𝑥−1 𝑥−1 ✓ = 𝑥−1 3 𝑥−1 𝑥−1 = 𝑥−1 + 𝑥−1 𝑥−1 3 3 ✓ Simplify to/ Vereenv. tot∶ + (2) = +1 𝑥−1 𝑥−1 𝑥−1 5.5 (3; 1) ✓ 3 (CA from/ vanaf 5.2 - shift 2 units to the right/ skuif 2 eenh regs) (2) ✓1 [10] QUESTION/ VRAAG 6 (8) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 6.1.1 𝑦 > −1; 𝑦 ∈ R ✓ ✓𝑦 > −1; 𝑦 ∈ R (2) 6.1.2 𝑔(𝑥) = 2𝑥 ✓ 𝑔(𝑥) = 2𝑥 ∴ 𝑔−1 : 𝑦 = log2 𝑥 ✓𝑦 = log2 𝑥 (2) 2 2 6.2.1 𝑘(𝑥) = 3𝑥 ; 𝑥 ≤ 0 ✓ 𝑘(𝑥) = 3𝑥 ✓𝑥≤0 (2) 6.2.2 (0; 0) OR/OF origin/ oorsprong ✓✓ Answer/ Antw (2) [8] 5

QUESTION/ VRAAG 7 (14) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 7.1 𝐴 = 𝑃(1 − 𝑖)𝑛 ; P = R128 000; A = R45 500; n = 6; i = ? ✓ Subst. of P = 128 000 into depreciation 45 500 = 128 000(1 - i)6 formula ✓ Subst. of 45 500 and/ en n = 6 6 45 500 ∴√ =1−𝑖 128 000 ✓ i the subject/ die onderwerp 45 500 6 ∴𝑖 =1− √ 128 000 ✓ Answer/ Antw ∴ 𝑖 = 0,15834 = 15,83% p.a/ p.j (4) 7.2.1 𝐴 = 𝑃(1 + 𝑖)𝑛 ; A = ?; P = 250 000; n = 12y = 48 quarters; 9,5 9,5 i = 9,5% p.a = per quarter/ per kwartaal ✓ Subst i = = 0,02375 400 400 9,5 ✓ Subst n = 48 in correct formula ∴ 𝐴 = 250 000(1 + 400)48 = R 771 343,67 ✓ Answer/ Antw (3) 7.2.2 Home loan/ Huislening = 2 920 000 - 771 343,67 ✓ Answer/ Antw = R2 148 656, 33 (1) 7.2.3 𝑥[1−(1+𝑖)−𝑛 ] 𝑃𝑣 = 𝑖 ; Pv = 2 148 656, 33 ; i = 10,3% p.a 10,3 = p/m ; n = 20 y = 240 m 1200 10,3 −240 𝑥 [1−(1+ ) ]  2 148 656, 33 = 1200 10,3 ✓ Subst into correct formula 10,3 1200 ✓ Subst i =  x = R 21 163, 87 1200 ✓ Subst n = 240 OR/OF 𝑥[(1 + 𝑖)𝑛 − 1] ✓Answer/ Antw 𝐹𝑣 = 𝑖 10,3 240 10,3 240 𝑥 [(1+ ) −1] 2 148 656, 33 (1 + 1200 ) = 10,3 1200 1200  x = R 21 163, 87 (4) 7.2.4 Interest/ Rente = Amount paid/ bedrag betaal- original loan/ lening = 21 163,87 × 240 - 2 148 656,33 ✓Calculation/ berekening = R 2 930 672, 47 ✓Answer/ Antw (2) [14] 6

QUESTION/ VRAAG 8 (15) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 8.1 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim ℎ→0 ℎ 2 2 − ✓Subst 𝑓(𝑥 + ℎ) and/ en −𝑓(𝑥) 𝑥+ℎ 𝑥 = lim ℎ→0 ℎ 2𝑥−2(𝑥+ℎ) 𝑥(𝑥+ℎ) ✓Common denominator/ Gemene noemer = lim ℎ→0 ℎ 2𝑥−2𝑥−2ℎ 𝑥2 +𝑥ℎ = lim ℎ ℎ→0 −2ℎ 1 = lim × ℎ→0 𝑥(𝑥+ℎ) ℎ −2 ✓Simplify/ Vereenv = lim 2 ℎ→0 𝑥 +𝑥ℎ −2 = lim 2 ✓Subst. h = 0 ℎ→0 𝑥 +𝑥(0) −2 𝑓 ′ (𝑥) = ✓ Answer/ Antw (5) 𝑥2 4 ✓ Exponential form/ eksponensiële vorm 8.2.1 ℎ(𝑥) = 2𝑥 3 − 𝑥 + 3√𝑥 1 ✓✓✓ Answer / Antw ∴ ℎ(𝑥) = 2𝑥 3 − 4𝑥 −1 + 3𝑥 2 3 −1 (-1 for notation in question / -1 vir ∴ ℎ′ (𝑥) = 6𝑥 2 + 4𝑥 −2 + 𝑥 2 notasie in vraag) (4) 2 8.2.2 8𝑥 3 − 27 𝐷𝑥 ( ) 2𝑥 − 3 (2𝑥 − 3)(4𝑥 2 + 6𝑥 + 9) = 𝐷𝑥 ( ) 2𝑥 − 3 ✓Simplify fraction/ Vereenv breuk = 𝐷𝑥 (4𝑥 2 + 6𝑥 + 9) = 8𝑥 + 6 (3) ✓✓ Ans. / Antw 8.3 mtan = 𝑓 ′ (𝑥) = −3𝑥 2 − 2 ✓ mtan = 𝑓 ′ (𝑥) = −3𝑥 2 − 2 = −(3𝑥 2 + 2) ✓ −(3𝑥 2 + 2) Which is always/ wat altyd < 0 ✓ Explain/ Verduidelik (3) [15] 7

QUESTION/ VRAAG 9 (15) 𝒚 𝑩 𝑨 𝒙 𝑪 𝑫 # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 9.1 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 𝑥 − 1 ∴ (𝑥 − 1)(𝑥 2 + 2𝑥 + 1) = 0 ✓(x - 1) ∴ (𝑥 − 1)(𝑥 + 1)(𝑥 + 1) = 0 ✓(𝑥 2 + 2𝑥 + 1) ∴ 𝑥 = 1 or/𝑜𝑓 𝑥 = −1 ✓(𝑥 + 1)(𝑥 + 1) ∴ AB = 2 units / eenhede ✓Both roots/ Beide wortels (5) ✓𝐴𝐵 = 2 9.2 𝑓 ′ (𝑥) = 0 ✓Derivative/ Afgeleide = 0 ∴ 3𝑥 2 + 2𝑥 − 1 = 0 ∴ (3𝑥 − 1)(𝑥 + 1) = 0 1 ✓ Values of 𝑥/ Waardes van 𝑥 ∴ 𝑥 = 3 or / of 𝑥 = −1 32 ∴𝑦=− 27 1 32 ∴ 𝐷( ; − ) ✓Coordinates of D/ Koördinate van D (3) 3 27 9.3 𝑓"(𝑥) = 0 ✓ Linking second derivative with ∴ 6𝑥 + 2 = 0 concavity / Verbind tweede afgeleide met 1 konkawiteit ∴ 𝑥 = −3 ℎ"(𝑥) < 0 ℎ"(𝑥) > 0 ✓ 6𝑥 + 2 = 0 1 − 3 ✓ explanation / verduideliking 1 The concavity changes at 𝑥 = −3 Die konkawiteit verander by 𝑥 = − 3 1 (3) 9.4.1 𝑥>1 ✓ Answer/ Antw. (1) 9.4.2 1 ✓ 𝑥 < −1 𝑥 < −1 or/of − < 𝑥 < 1 1 3 ✓✓ − < 𝑥 < 1 3 • Notation (-1) / (3) Notasie (-1) [15] 8

QUESTION/ VRAAG 10 (7) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 10.1 𝑡=2 1 1 𝐷(2) = 3 + (2)2 − (2)3 2 4 =3+2−2 =3𝑚 ✓ Answer/Antw. (1) 10.2 3 𝐷 ′ (𝑡) = 𝑡 − 𝑡 2 ✓ derivative / afgeleide 4 ′ (3) 3 𝐷 = 3 − (3)2 4 = −3,75 𝑚/ℎ ✓ Answer/Antwoord (2) 10.3 𝐷 ′ (𝑡) = 0 ✓ 𝐷 ′ (𝑡) =0 3 𝑡 − 𝑡2 = 0 4 ×−4: 3𝑡 2 − 4𝑡 = 0 ✓ factors / faktore 𝑡(3𝑡 − 4) = 0 1 𝑡 = 0 𝑜𝑟/𝑜𝑓 𝑡 = 1 3 = 1h 20 m ✓ both answers / beide antwoorde ∴ 𝑇𝑖𝑚𝑒/𝑇𝑦𝑑 ∶ 10: 20 ✓ Final answer / Finale antw. (4) [7] QUESTION/ VRAAG 11 (14) # SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD DESCRIPTORS/BESKRYWERS Mark/ Punt 11.1.1 147 49 147 or / of 49 or/ of 27,22% 𝑃(female and/∩ green eyes) = = = 27,22% 540 180 540 180  147 49 𝑃(vrou en/∩ groen oë) = = = 27,22% ✓✓ Answer/ Antw 540 180 (2) 11.1.2 For events to be independent: 𝑃(female and/∩ green eyes) = 𝑃(green eyes)×𝑃(female) 330 240 22 ✓ P(green eyes) = 330/540 𝑃(green eyes)×𝑃(female) = × = = 0,27 540 540 81 ✓ P(female) = 240/540 147 49 𝑃(female and/∩ green eyes) = = = 0,27 540 180 ✓ 0,27 ∴ 𝑃(female and/∩ green eyes) = 𝑃(green eyes)×𝑃(female) ✓ P(female and/∩ green eyes) = 0,27 Events are independent and the learner is correct. ✓ Deduction Vir gebeure om onafhanklik te wees: 𝑃(vrou en/∩ groen oë) = 𝑃(groen oë)×𝑃(vrou) 330 240 22 ✓ P(groen oë) = 330/540 𝑃(groen oë)×𝑃(vrou) = × = = 0,27 540 540 81 ✓ P(vrou) = 240/540 147 49 𝑃(vrou en/∩ groen oë) = = = 0,27 540 180 ✓ 0,27 9

𝑃(vrou en/∩ groen oë) = 𝑃(groen oë)×𝑃(vrou) ✓ P(vrou en/ ∩ groen oë) = 0,27 (5) Gebeure is onafhanklik en die leerder is korrek. ✓ Afleiding Alternative/Alternatief: For events to be independent: 𝑃(male and/∩ green eyes) = 𝑃(green eyes)×𝑃(male) ✓ P(green eyes) = 330/540 330 300 55 𝑃(green eyes)×𝑃(male) = × = = 0,34 540 540 162 ✓ P(male) = 300/540 183 61 ✓ 0,34 𝑃(male and/∩ green eyes) = = = 0,34 540 180 ✓ P(male and/∩ green eyes) = 0,34 𝑃(male and/∩ green eyes) = 𝑃(green eyes)×𝑃(male) ✓ Deduction Events are independent and the learner is correct. Vir gebeure om onafhanklik te wees: 𝑃(manlik en/∩ groen oë) = 𝑃(groen oë)×𝑃(manlik) 330 300 55 𝑃(groen oë)×𝑃(manlik) = × = = 0,34 ✓ P(groen oë) = 330/540 540 540 162 ✓ P(manlik) = 300/540 183 61 𝑃(manlik en/∩ groen oë) = = = 0,34 540 180 ✓ 0,34 𝑃(manlik en/∩ groen oë) = 𝑃(groen oë)×𝑃(manlik) ✓ P(manlik en/ ∩ groen oë) = 0,34 ✓ Afleiding (5) Gebeure is onafhanklik en die leerder is korrek. 11.2.1 10! ✓ Numerator/ Teller 𝑛(𝑆) = = 151200 ✓Denominator/ Noemer 2!×2!×3! (3) ✓ Answer / Antw 11.2.2 8! ✓ 8!  𝑛(𝐸) = = 10080 2!×2! 2!×2!  10080 1 ✓10080 ∴ 𝑃(𝐸) = = 10080 151200 15 ✓𝑃(𝐸) = 151200 ✓ Answer / Antw  (4) [15] TOTAL/ TOTAAL: 150 10