NATIONAL SENIOR CERTIFICATE GRADE 0 MATHEMATICS P3 PREPARATORY EXAMINATION 008 NOVEMBER 008 MARKS: 00 TIME: hours This question paper consists of pages and 4 diagram sheets.
Mathematics/P DoE/Preparatory Examination 008 QUESTION / VRAAG. -4; -3; -37; -60 (). T = T = -3 Tk+ = Tk + Tk -; k>. (3) [5] QUESTION / VRAAG. x 505 + 70 + 6+ 838 + 539 + 45 + 34 396 x = = = = 565,86 n 7 7 (). If the reportings of 004 is left out then, As 004 se aanmeldings weggelaat word x 505 + 70 + 6+ 539 + 45 + 34 x = = = 50,5 n 6 The reportings of 004 push the average up radically. If these two averages are compared it will even look as though the number of assaults increased. The opposite is actually true. The number of assaults decreased a lot. Die 004 aanmeldings stoot die gemiddeld radikaal op. As die twee gemiddeldes vergelyk word mag dit selfs lyk asof aanrandingsake toegeneem het. Die teendeel is egter waar. Aanmeldings het baie afgeneem. ()
Mathematics/P 3 DoE/Preparatory Examination 008.3 No, it is not valid to make such a statement on the ground of the statistics of only one police station because Nee, dit is nie geldig om so ʼn stelling te maak op grond van die statistiek van een polisiestasie nie want (i) (ii) (iii) one police station does not represent the total population of S.A. A minimum of 0% of the police stations should be taken from all areas. een polisiestasie is nie verteenwoordigend van Suid-Afrika se populasie nie. Minstens 0% van polisiestasies wat uit alle gebiede verteenwoordig word sou gebruik moes word. the way in which this certain police station was selected is unknown. It is possible that it was not selected on an equal chance. Die wyse waarop die polisiestasie gekies is is onbekend. Dis moontlik dat dit nie ewekansig gedoen is nie. The size of the sample is definitely too small to make a prediction for the total population of S.A. Die grootte van die steekproef is beslis te laag om ʼn voorspelling vir die totale populasie van S.A. te maak. (5).4. Graph A is decreasing 000 to 007 Graph B is increasing 007 to 000 Grafiek A is stygend 000 tot 007 Grafiek B is dalend 007 tot 000.4. It must look if crime increases and not decreases. Dit moet voorkom of misdaad toeneem en nie afneem nie. []
Mathematics/P 4 DoE/Preparatory Examination 008 QUESTION / VRAAG 3 To calculate the standard deviation x = 7,95 Om die standaardafwyking te bereken Hours x Frequency (f) (x - ) (x - ) (x - ) f 0-7,95 3,0 644,405 6 -,95 4,805 73,63 0 3-7,95 63,05 453,6575 5 60 -,95 8,705 5,5 30 77,05 4,05 33,595 35 38 7,05 49,705 888,695 4 8 4,05 97,405 579, σ = ( x x) n f 8 5,35 σ = = 36,93 = 6, 08 0 (4)
Mathematics/P 5 DoE/Preparatory Examination 008 3. 3.3 on graph / op grafiek x = 7,95 σ = 6,08 x + σ = 34,03 x + σ = 40, on graph / op grafiek x + 3σ = 46,9 x σ =,87 on graph / op grafiek x σ = 5,79 x 3σ = 9,7 on graph / op grafiek (4) 3.4 All the hours between,87 and 34,03 will lay within one standard deviation either side of the mean. 60 + 70 00 = 6,7% pupils fall in this range. To have a normal 0 distribution it must be 68,6%. The data does not fit the normal distribution. Alle leerders wat tussen,87 en 34,03 ure TV kyk val binne die eerste 60 + 70 00 standaardafwyking = 6,7% 0 Vir ʼn normaalverspreiding moet die persentasie 68,6% wees. Die data volg nie die normaalverdeling nie. [3]
Mathematics/P 6 DoE/Preparatory Examination 008 QUESTION / VRAAG 4 4. Strong linear positive correlation Sterk lineêre positiewe korrelasie 4. linear equation of regression lineêre regressie vergelyking: y = a + bx (langste metode) n = 0 + 40 + 60 + 80 + 00 + 0 + 40 x = = 80,0 7 6 + 6 + 30 + 48 + 70 + 80 + 0 y = = 5,43 7 b = ( x x)( y y) ( x x) 9 600 = = 0,8574 0,86 00 x y ( x x) ( x x) ( y y) ( x x) ( y y) 0 6-60 3 600-45,43 75,8 40 6-40 600-35,43 47, 60 30-0 400 -,43 48,6 80 48 0 0-3,43 0 00 70 0 400 8,57 37,4 0 80 40 600 8,57 4,8 40 0 60 3 600 58,57 3 54, y ˆ = a + bx y = a + bx a = y bx = 5,43 (0,86)(80) = -7,37 (4) y = -7,37 + 0,86x With calculator: a = -7,4 Met sakrekenaar: b = 0,86 (4) y = -7,4 + 0,86x
Mathematics/P 7 DoE/Preparatory Examination 008 4.3 4.4 x = 80 y = 5,43 x x y y xy 0 400 6 36 0 40 600 6 56 640 60 3 600 30 900 800 80 6 400 48 304 3 840 00 0 000 70 4 900 7 000 0 4 400 80 6 400 9 600 40 9 600 0 00 5 400 r = 0,99 (3) (calculator) / (sakrekenaar) (3) r = x. xy y = 38 400 56 000. 6 896 = 0,989448 0,99 4.5 There is a very high correlation. The faster you travel the further it will take the vehicle to come to a stand still. ʼn Baie hoë graad van korrelasie. Hoe vinniger gery word, hoe langer neem dit om tot stilstand te kom. () []
Mathematics/P 8 DoE/Preparatory Examination 008 QUESTION / VRAAG 5 5. Probability / Waarskynlikheid = 0,9 + 0, 0,4 = 0,6 n(a or B) = n(a) + n(b) n(a B) = 60% (3) 5. 5.. 5 x 4 = 0 (3) 5.. 5 x 4 x 3 = 60 () 5.3 5.3. Any number on any place Enige nommer op enige plek =! = 479 00 () 4.3.3 8 en kan op! =. = maniere gerangskik word. Total number of combinations of numbers. Totale aantal kombinasies van getalle.! =.0.9.8... = 39 96 800! x = 79 833 600 (3) 5.3.3 4P = 4! = 4.3 =! () 5.4 C correct / C korrek W wrong / W verkeerd Second answer correct / Tweede antwoord korrek = (0,6.0,7) + (0,4.0,4) = 0,4 + 0,6 = 0,6 = 0,58 (4) [9] 60
Mathematics/P 9 DoE/Preparatory Examination 008 GEOMETRY / MEETKUNDE QUESTION /VRAAG 6 6. Rˆ = 65 (hoek tussen raaklyn en koord) / (tan-chord) size 6. Rˆ 3 + Rˆ 4 = 90 35 + Rˆ = 90 Rˆ 4 4 = 55 (radius perpendicular to tangent) (radius loodreg op raaklyn) reason size () 6.3 Ŝ 3 = Qˆ + Qˆ (Ext. < of cyc.quad.) (buite statement hoek van koordevierh) Rˆ 8 + Rˆ 4 = Qˆ + Qˆ (tan-chord) (hoek tussen reason raaklyn en koord) Ŝ3 = Rˆ 3 + Rˆ 4 = 90 or () or reason Ŝ + Ŝ = 90 (angle in ½ ) size (hoek in ½ ) Ŝ 3 = 90 [5] QUESTION / VRAAG 7 7. Equal / Gelyk 7. 7.. Bˆ = Dˆ (alt.l s ADII BC) (verw. binne hoeke ADII BC) reason 7.. Ê = Bˆ (L s in same seg) (hoeke in dies. segm.) reason Ê = Bˆ conclusion en Dˆ = Ê3 (L s in same seg) (hoeke in dies. statement segm.) Ê = Ê 3 (3) 7..3 In Δ AEF en ΔCED Ê 3 = Ê (proven) (reeds bewys) statement  = Ĉ (L s in same seg) (hoeke in dies. statement segm.) Fˆ = Dˆ + Dˆ + Dˆ 3 (rem. L s) (3 e hoek van reason driehoek) Δ AEF III ΔCED (AAA) (HHH) (3)
Mathematics/P 0 DoE/Preparatory Examination 008 7..4 EF = EG (given) (gegee) Fˆ 3 = Ĝ (L s opposite equal sides) (hoeke teenoor gelyke sye.) Ĝ = Ĉ (Corr. L s AD II BC) (ooreenk. hoeke AD II BC) Fˆ 3 = Ĉ FBCG is a Cyc quad. (ext L equal to opp. L) (buite hoek van verh.) QUESTION / VRAAG 8 reason statement reason 8. 6x + 5x + 5x = 80 (L s in Δ 80 ) statement 36x = 80 reason (3) [] x = 5 calculation (3) 8. ΔPQR: PS = SQ (given) (gegee) statement PT = TR (given) (gegee) statement ST II QR (midpt. theorem) (lyn uit midpt) reason Tˆ = Rˆ (corr. L s) (ooreenk. Le ) = 5x conclusion = 5(5) = 5 calculation (5) 8.3 ΔPQN PS = SQ (given) (gegee) statement SM II QN (given) (gegee) statement PM = MN (midpt. theorem) conclusion (lyn uit midpt van sy e sy) PM = = MN answer OR ΔPQN SM II QN (given) (gegee) statement PS = SQ (given) (gegee) statement PS PM = = (line 3 rd ratio side Δ) SQ MN answer (lyn 3 e sy Δ) (4)
Mathematics/P DoE/Preparatory Examination 008 8.4 MN = 6 (given) (gegee) PM 6 ratio = MN 6 ΔPQM SO II QM (proven ST II QR) (bewys ST II QR) statement PS = SQ (given) statement PO = OM (midpt. theorem) (lyn uit midpt e sy) answer but PM = 6 PO = OM = (6) PO = 3 (4) [6] QUESTION 9 9. In Δ ABC and Δ AED: Â = Â (common) (gemeenskap) Ê = 90 (angles on a straigtline) (aangr. e op rgtln.) = Bˆ (given) (gegee) Dˆ = Ĉ (3 rd angle in D s) (3e in D) statement reason statement reason ΔABC ΔAED (A,A, A) (4) 9. AB AC = (D s similar) (D e gelykvormig) AE AD reason 9.3 AE ½ = (given) (gegee) AB AD = AC statement answer () 9.4 7 similar triangles 7 gelykvormige driehoeke answer [8] 40 TOTAAL: 00