NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12

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1 NATIONAL SENI CERTIFICATE NASIONALE SENI SERTIFIKAAT GRADE/GRAAD MATHEMATICS P/WISKUNDE V NOVEMBER 0 MEMANDUM MARKS/PUNTE: 00 This memorandum consists of 4 pages. Hierdie memorandum bestaan uit 4 bladsye.

2 Mathematics P/Wiskunde V DBE/November 0 If a candidate answered a question TWICE, mark only the first attempt. If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out question. Consistent accuracy applies in ALL aspects of the marking memorandum. LET WEL: As 'n kandidaat 'n vraag TWEE keer beantwoord het, merk slegs die eerste poging. As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, merk die deurgehaalde antwoord. Volgehoue akkuraatheid is DEURGAANS in ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG Calories Total fat (in grams) Scatter Plot showing number of calories versus total fat sandwiches (in 40 5 Total Fat (in grams) /Totale vetinhoud (in gram) The English and Afrikaans diagrams are different in size The point (00 ; 4) is from.4. Do not penalise for this point in question.. The learner may have other points due to trying to draw the least squares regression line all 8 points plotted correctly marks for 5 7 points plotted correctly mark for 4 points plotted correctly No marks for point or less plotted correctly. ()

3 Mathematics P/Wiskunde V DBE/November 0. a = 5,6 ( 5, ) a or b b = 0,07 (0, ) b or a Penalty for rounding yˆ = 5,6 + 0, 07x equation in question.. (4). Drawn on the scatter plot graph passing close to (50 ; 9) / (00 ; 6) / (650 ; 40) gradient (graph close to (450 ; 6)) ().4 The sandwich from Pinky s restaurant has 0 grams of fat. In terms of the model, the sandwich with 6 grams of total fat should provide 00 calories. This suggests that the sandwich from Pinky s restaurant is far removed from expected model and is therefore an outlier. Die toebroodjie van Pinky se restaurant het n vetinhoud van 0 gram. In terme van die model, sal n toebroodjie van 6 gram, 00 kalorieë verskaf. Dit veronderstel dat die toebroodjie van Pinky se restaurant ver van die verwagte model verwyder is en is dus n uitskieter. comparison with model (literal or graphical (i.e. plotting the point)) outlier / explain this concept vergelyking met model uitskieter.5 r = 0, 9 (0, ).6 There is a very strong, positive correlation between the number of calories and the total fat in the sandwiches. A sandwich with higher number of calories will have a higher total fat content. Daar is n baie sterk positiewe korrelasie tussen die aantal kalorieë en die totale vetinhoud in die toebroodjies. n Toebroodjie met n hoë aantal kalorieë sal n hoër totale vetinhoud hê. strong / reference to correlation positive / increasing / explain the concept of increasing () () () () [5]

4 Mathematics P/Wiskunde V 4 DBE/November 0 QUESTION/VRAAG. Random Sampling Method Ewekansige Steekproef Metode / Willekeurig / Onwillekeurig/ Lukraak. Although the sample is selected randomly, there exists a chance that this could be a homogenous sample. For example the sample could contain many learners from one grade and few from other grades. The sample could contain many more learners of a particular gender and few from the other. () homogenous sample / the sample will be biased towards one characteristic () Al is die steekproef ewekansig gedoen bestaan die kans wel dat die steekproef homogeen kan wees. Ter voorbeeld, die steekproef kan baie leerders van een graad insluit en min van ander grade. Die steekproef kan ook meer leerders van n bepaalde geslag insluit en min van die ander geslag.. The sampling should have equal representation across various criteria. In this case the criteria will possibly be grade and gender. There should possibly be a selection of 8 boys and 8 girls randomly from each grade for the sample. homogene steekproef equal representation of grade equal representation of gender () Gelyke verteenwoordiging van verskillende kategorieë moet in die steekproef teenwoordig wees. In die geval sal die kategorieë graad en geslag wees. Daar behoort moontlik 8 seuns en 8 meisies van elke graad ewekansig gekies te word vir die steekproef. any explanation that indicates representing the school for equally Ideas to consider: gender; race; grade () [4]

5 Mathematics P/Wiskunde V 5 DBE/November 0 QUESTION/VRAAG. Disagree. The events A and B have an intersection and are therefore not mutually exclusive. P(A and B) 0 Verskil. Die gebeurtenisse A en B sny mekaar en is daarom nie onderling uitsluitend nie. P(A en B) 0 Disagree P(A) + P(B) + P(C) =,07 Since the sum of the probabilities of the events is greater than, these events must have an intersection. Hence A and B are not mutually exclusive events. Verskil P(A) + P(B) + P(C) =,07 Siende dat die som van die waarskynlikhede van al die gebeurtenisse groter as is, moet daar n snyding wees. Vervolgens sal A en B nie onderling uitsluitend wees nie. disagree intersection between events exist () disagree intersection between events exist ().. P(B or C) = P(B) + P(C) = 0, + 0, = 0,6 Answer only: full marks substitution into correct formula () A 0,5 0,5 B 0,65 C 0, P(A and B) = P(A).P(B) = 0,45 0, = 0,5 P(B or C) = 0,5 + 0,65 + 0, = 0,6.. P(A and B) = P(A).P(B) = 0,45 0, = 0,5 P(A or B) = P(A) + P(B) P(A and B) = 0,45 + 0, 0,5 = 0,65 = 0,6 Answer only: / marks substitution into correct formula 0,5 () () [6]

6 Mathematics P/Wiskunde V 6 DBE/November 0 QUESTION/VRAAG 4 4. The word EQUATIONS has 9 letters Number of different 5 letter codes = = 50 Number of different 5 letter codes 9! = 4! = 50 Answer only: / marks ACCEPT: Number of different 5 letter codes (this is with repetition) = 9 5 = consonants 5 vowels multiplication rule in five slots () multiplication rule in five slots () Number of different codes = = 600 Number of different codes = 5 5! = ! () [5] QUESTION/VRAAG 5 5. The heights / frequency of the histogram for the interval 500 x < 700 are incorrect. Die hoogtes / frekwensies van die histogram vir die wydte 500 x < 700 is nie korrek nie. The histogram has not taken into account the width of the classes. The first class is double the width of the other classes. The area of the rectangles for the class 500 x < 700 is double what it should be. Die histogram het nie die wydte van die klasse in ag geneem nie. Die eerste klas is dubbel die wydte van die ander klasse. Die oppervlakte van die reghoeke vir die klas 500 x < 700, is dubbel wat dit behoort te wees. heights / frequency incorrect width not taken into account () () Area of each rectangle is not in the same ratio as the frequencies. Die area van elke reghoek is nie in dieselfde verhouding as die frekwensies nie. 5. The interval 500 x < 700 should have a frequency of 5. Die frekwensie behoort 5 vir die klas 500 x < 700 te wees. area of rectangle not in same ratio as frequencies () frequency should be 5 keep interval the same () [4]

7 Mathematics P/Wiskunde V 7 DBE/November 0 QUESTION/VRAAG 6 6. A/M 670 N/W x x = 40 x + 40 = 40 x = 70 Number said lack of parental support = = P(exactly problems) = = 40 5 = 4 = 0, (0, ) P/O sum of values in Venn diagram = 40 for x (must not be negative) = (6) () () []

8 Mathematics P/Wiskunde V 8 DBE/November 0 QUESTION/VRAAG = 0 + () P(More than 6 point) = % = 68 learners 7. Mathematics test x + σ = = 70 English test x + σ = 55 + ( 7,5) = 6,5 x + σ = 55 + ( 7,5) = 70 Matilda scored better relative to her classmates in the English test as she scored between the st and the nd deviation from the mean. Matilda het beter as die res van haar klasmaats in die Engelse toets gevaar aangesien haar punt tussen die ste en die de afwyking van die gemiddeld was. 4% 4% 4% 4% % % If 6% is used: max mark std deviations from mean. % () 84 % () x + σ = 70 x + σ = 6, 5 () A mark of 67% in Maths lies within standard deviation to the right of the mean. Hence it is approximately at the 76 th percentile of distribution of the Maths marks. A mark of 67% in English lies between and standard deviations to the right of the mean. Hence it is approximately at the 95 th percentile of distribution of the English marks. Matilda scored better relative to her classmates in the English test as she has outperformed more learners in English than in Maths. n Punt van 67% in Wiskunde lê binne standaardafwyking aan die regterkant van die gemiddeld. Vervolgens is dit ongeveer by die 76 ste persentiel van die verspreiding van die Wisk punte. n Punt van 67% in Engels lê tussen die ste en de standaardafwyking regs van die gemiddeld. Vervolgens is die Engelse punt ongeveer by die 95 ste persentiel van die verspreiding van die Engelse punte. Matilda het relatief tot haar klasmaats beter in die Engelse toets gevaar. Sy het beter gevaar as meer van die leerders in Engels as in Wiskunde. 76 th percentile 95 th percentile () [7]

9 Mathematics P/Wiskunde V 9 DBE/November 0 QUESTION/VRAAG T T T T T 4 0 = 5 = 5 + = + = + = T + S 9 = 5 + = = 7 ( ) 4 = ( ) 4 = ( 4) 4 = 0 9 of the linear pattern [ ( ) + 8( ) ] a = a = + b = 5 b = c = 5 c = 4 Tn = n T0 = = 7 5 By expanding the formula: n 4 5 ( 0) ( 0) 4 T = T = T 4 = 0 T + S 9 of the linear pattern a = d = a = 5 b = c = 4 () (4) (4) 5 ; ; ; 0 ; ; 5 ; 5 ; 7 ; 95 ; ; 50 ; 8 ; 7 ; 55 ; 96 ; 40 ; 87 ; 47 ; 490 ; 546 ; 605 ; 667 ; 7 ; 800 ; 87; 945 ; 0 ; 0 ; 85 ; 7 expansion The 0 th term = 7 Answer only: 4 marks (4) [7]

10 Mathematics P/Wiskunde V 0 DBE/November 0 QUESTION/VRAAG 9 A x B x C F E D 9. AF : FE = : (Prop Th; FB EC) (Eweredigheid St; FB EC) 9. AF = FE AF 8 FE = = = 4 cm AE = cm ED = (BE DC; Prop Th) / (BE DC; Eweredigheid St) AE ED = ED = 6 cm reason FE = 4 cm AE = cm ED = AE () (4) [6]

11 Mathematics P/Wiskunde V DBE/November 0 QUESTION/VRAAG 0 F B G 5 E 4 56 D O C 0. Ê = Ê = 4 H ( s in a semi circle) ( e in n halwe sirkel) 0. C ÊH = Ê + Ê = 66 E Bˆ C = Bˆ + Bˆ = 66 (tan ch th) (raaklyn koord) Ê = 4 s in a semi circle C ÊH = 66 E Bˆ C = 66 tan ch th () () Bˆ = Bˆ = 4 E Bˆ C = 66 (tan ch th) (raaklyn koord) ( s in the same segment) ( e in selfde seg) tan ch th Bˆ = 4 () 0. + Ê + Ê = Ê Fˆ = (tan ch th) / (raaklyn koord) Ĉ = 58 (sum of int s of ) / (som van binne e ) Fˆ = (opp s cyclic quad) / (oorst koordevierhoek) Dˆ = 58 (sum of int s of ) / (som van binne e ) Fˆ = (opp s cyclic quad) / (oorst koordevierhoek) Fˆ = reason Ĉ = 58 sum Fˆ = opp s cyclic quad Dˆ = 58 sum Fˆ = opp s cyclic quad (4) [9]

12 Mathematics P/Wiskunde V DBE/November 0 QUESTION/VRAAG O A B C Construct radii OA and OC. In OAB and OCB i. OB is common ii. OA = OC (radii) iii. O Bˆ A = OBˆ C = 90 (given) OAB OCB (90 HS) AB = BC ( s) Construct radii OA and OC. In OAB and OCB i. OB is common OA = OC (radii) ii. Â = Ĉ ( s opp = radii) iii. O Bˆ A = OBˆ C = 90 (given) OAB OCB (SAA) AB = BC ( s) Construct radii OA and OC. O Bˆ A = OBˆ C = 90 (given) OA = OB + AB (Pythagoras) OC = OB + BC (Pythagoras) OC = OA (radii) OB is common AB = BC AB = BC construction OB common radii O Bˆ A = OBˆ C = HS construction OB common / OA = OC s opp = radii O Bˆ A = OBˆ C = 90 SAA construction OA = OB + AB OC = OB + BC OA = OC / radii AB = BC [5]

13 Mathematics P/Wiskunde V DBE/November 0 QUESTION/VRAAG K x x 4 O E x T x x Y W. Kˆ = x (tan ch th) (raaklyn koord) Ŷ = x ( s in same seg) ( e in selfde seg) Kˆ = x ( s opp = radii) / (= chs subt = s) ( e oork = radiusse) / (= koorde = e) Ŵ = x ( s in same seg) / (= chs subt = s) ( e in selfde seg) / (= koorde = e). Ô + Ô 4 = 80 x (sum of int s of ) / (opp cyclic quad) (som van binne e ) / (oorst koordevierhoek) Tˆ = 90 x ( at circ cent = at circumference) Kˆ = x tan ch th Ŷ = x s in same seg Kˆ = x reason Ŵ = x reason Ô + Ô 4 = 80 reason reason x (8) ()

14 Mathematics P/Wiskunde V 4 DBE/November 0. Ê = 90 (sum of int s of ) / (som van binne e ) KE = ET ( from centre to chord bisects chord) ( van middelpunt tot koord halveer koord) Ê = 90 sum of int s of from centre to chord bisects chord () Kˆ = 90 x (sum of int s of ) In KWE and TEW. Kˆ = Tˆ = 90 x (proven above). Ŵ = Ŵ = x (Proven in.). WE is common KWE TEW ( S) KE = ET.4 In KOE and WTE i. Kˆ = x (proven) = Ŵ ii. Ê = Ê = 90 ( s on str line / sum of int s of ) ( e op reguit lyn / som van binne e ) iii. Ô = Tˆ = 90 x ( rd of ) KOE WTE ( ) Kˆ = 90 x sum of int s of KWE TEW KOE and WTE Kˆ = Ŵ = x = Ê = 90 Ê Ô = Tˆ = 90 x () KE OE = WE TE ( s) KE = TE (proven) KE.TE = OE.WE KE OE.WE = In KOE and KWE i. Kˆ = x (proven) = Ŵ ii. Ê is common iii. Ô = Kˆ ( rd of ) KOE WKE ( ) KE OE = WE TE KE = TE KOE and WTE Kˆ = Ŵ = x Ê is common (6) KE OE = ( s) WE KE KE = OE.WE KE OE = WE KE TOTAL/TOTAAL: 00 (6) [0]

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