Advanced Programme Maths P2 Memo 2018 hlayiso.com

Downloaded from hlayiso.com GRADE 12 EXAMINATION NOVEMBER 2018 ADVANCED PROGRAMME MATHEMATICS: PAPER II MARKING GUIDELINES Time: 2 hours 200 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 11 MODULE 2 STATISTICS QUESTION 1 1.1 (a) Binomial  12   12  Zero or one   ( 0,057 ) ( 0,943 ) +   ( 0,057 )( 0,943 ) 0 12 11 0 1 = 0,8531. (b) The remaining 16 not ADD (0,943)16 = 0,3910 1.2 (a) Without replacement hypergeometric (b) 20 (c) 7 (d) 8 (e) 2 (f) 7–k 8 – (7 – k) = k + 1 QUESTION 2 2.1 (a) Let X be the random variable "weight of babies" P ( X > 2,8 ) 2,8 − 3,2 P(z > 0,85 P ( z > −0,4706 ) 0,05 + 0,1808 = 0,6808 Approximately 953 120 babies IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 11 (b) 0,1 0,4 z = −1,28 −1,28 = ( X − 3,2 ) 0,85 = 2,112 kg 2.2 (a) 61 kg 9 (b) 61 + z × =63 8 16 1,77 = z = 1,77 9 0,461 × 2 92% confident QUESTION 3 H0 : µ x =µ y H1 : µ x > µ y 7,2 − 8,1 Test statistic z = = –1,54 ( 2,85 ) + 4 2 35 38 1 tail test 0,05 -1,645 Not enough evidence to reject the null hypothesis in favour of the claim at the 5% significance level. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 11 QUESTION 4 = 4.1 y ∑y 1 = 59 1 1910 n 6 n n = 12 12 × 26270 − 161× 1910 4.2 b= = 4,8464 12 × 2293 − (161) 2 y= a + bx 955  161  = a + 4,8464   ∴ a = 94,1441 6  12  =y 94,1441 + 4,8464 x 4.3 Strong, positive correlation 4.4 No – too far out of the range (extrapolation) QUESTION 5 60 ∫a ( x − 30 ) dx 1 2 5.1 = 30 60 a 3  3 ( x − 30 )  =1   30 a ( 30 ) = 1 3 3 3 1 =a = ( 30 ) 9 000 3 m  1 3 1 5.2  27000 ( x − 30 )  =   30 2 1 1 ( m − 30 ) = 3 27000 2 (m – 30)3 = 1 3500 m m = 54 minutes IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 11 QUESTION 6 6.1 P ( A )= x P ( B ) = y P ( B′ )= 1 − y P (A ∩ B)= xy A B x –xy xy y – xy P ( A ) × P ( B′ ) =x (1 − y ) P ( A ∩ B′ ) =x − xy =x (1 − y ) Events A and B' are independent  16   9   7  6.2   −    4 = 347  5  05  11  9   5   11  10   6   11  9   6  6.3       +       +       13 860 =  2  45  1  4 6  2 36 Total for Module 2: 100 marks IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 11 MODULE 3 FINANCE AND MODELLING QUESTION 1 1 1.1 (a) 5 640 × = 4 904,35 1,15 1,15 − 1,14 (b) × 100 = 0,00877 … = 0,88% 1,14 x x (1 + i ) ∴ i = 0,0468 per month 24 1.2 = 3 x x (1 + 0,0468 ) ∴ n = 15,1423 ≈ 15 months n = 2 1.3 A3 B2 C1 D5 QUESTION 2  0,088  2.1 500 000   = 3 666,67 interest > payments  12    0,088   −96 x 1 −  1 +    12   2.2 500 000 = x = 7 273,33 0,088 12  0,088  95   1 + −  12  7 300 1  95  0,088   2.3 500 000  1 +  – = 3 576,4053  12  0,088 12  0,088  3 576,4053  1 + 12  = 3 602,63  OR  0,088   0,088  95   1 +  7 300  1 +  − 1  12   12  96  0,088   500 000  1 +  –  12  0,088 12 = 1 008 318,445 – 1 004 715,812 = 3 602,63 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 11 OR −96   0,08   −95  0,08   −  +   + y 1 + 12  7 300 1 1   12    500 000 – 0,08 12 ∴y = 3 602,63 QUESTION 3 72 24 72 8  0,08  2  0,08  0,08  1  0,1  x .  1+ . . 1+ + x .  1+ . . 1+  12  3  12   12  3  4  = 20 702,50 1,9169x = 20 702,50 X = 10 800 QUESTION 4 4.1 Logistic Model, presence of carrying capacity 1 4.2 V= (50) = 25 2 ∆P 4.3 The model has regression equation = −0,0025P + r . P r = – Km = – 50 .(– 0,0025) = 0,125 4.4 T n + 1 = T n + 0,13 .T n (1 – T n /50), T O = 10 t = 11 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 11 QUESTION 5 5.1 (a) number of eagles born per annum (b) efficacy rate at which eagles turn prey into offspring (c) f .b.Dn .En = 15 f (6 000) = 15 f = 0,0025 5.2 a = 0,5 × 1,5 × 3 × 0,67 a = 1,51 5.3 6 000 = b.(12 000)(30) b = 0,016 667 for dassie equilibrium, E n + 1 = E n 0,1 = 0,003 × 0,016 667 × D D = 1 999,96 ≈ 2 000 OR 6 000 = b.(12 000)(30) b = 0,016 667 c 0,1 D= = D = 1 999,96 ≈ 2 000 fb 0,003 × 0,016 667 QUESTION 6 6.1 (a) T 4 = 75,77 T 5 = 84,55 T 6 = 91,122 (b) 64 3 = 110,8 sq units 6.2 195 = p.114 + q.60 and 114 = p.60 + q.24 p = 2,5 and q = – 1,5 T n = 5/2.T n – 1 – 3/2.T n – 2 T 1 = 24, T 2 = 60 Total for Module 3: 100 marks IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 11 MODULE 4 MATRICES AND GRAPH THEORY QUESTION 1 3 6    3 6 2 −2   −2 −1  11 22  1.1 PQ =  .    0 −1 4 6   0 5   −40 21    −7 0  1.2 3x + 2y = 11 x – 2z = 0 6y + 4z = 5 3 2 0 11 1 0 –2 0 0 6 4 5 3 2 0 11 0 2 6 11 R1 – 3.R2 0 6 4 5 3 2 0 11 0 2 6 11 0 0 14 28 3.R2 – R3 1 z = 2, y=– , x=4 2 1.3 (a) 3 (b) 0 (c) t IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 11 QUESTION 2 2.1 reflection across y = x 2.2 k=3 1 1 2.3 C= R and R = S so factor is 4 4  −3 0   −0,5 0   1,5 0  2.4     =   0 1  0 −0,5   0 −0,5  OR  3 0   1 0   0,5 0   1,5 0     =   0 1   0 −1  0 0,5   0 −0,5   cosA −sinA   5   4,025  2.5     =   sinA cosA   −2   3,578  5cos A + 2.sin A = 4,025 and – 2.cos A + 5sin A = 3,578 cos A = 0,4472 and sin A = 0,8944 A = 63,44° A = 63,44° QUESTION 3 3.1 det = 25  25 0 0 −1 −4 −10    3.2 0 −10 0 −2 −8 −10  0 −9 10   0 25 4  25 0 0 −1 −4 10    3.3 0 25 0 5 20 −25   −4 −10  0 0 25 9  −1 −4 10  1   Inverse = 25  5 20 −25   −4 9 −10    IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com GRADE 12 EXAMINATION: ADVANCED PROGRAMME MATHEMATICS – PAPER II – MARKING GUIDELINES Page 11 of 11 QUESTION 4 4.1 one vertex has an odd degree 4.2 yes; there is one pair of odd vertices 4.3 8 edges 4.4 19 × 2 = 4 × 6 + 2 × 4 + 1 × 1 + e e=5 QUESTION 5 5.1 TR 3 TV 3 TS 4 SU 3 RQ 5 RW 5 QP 6 length = 29 5.2 RT 3 TV 3 VW 7 WRQ 10 QP 6 PS 7 SU 3 UTR 9 U/B = 48 5.3 37 is the largest Lower Bound  and 41 is the smallest Upper Bound 5.4 R Q P U S T V W R = 41 QUESTION 6 6.1 3 6.2 e = 2n – 3 6.3 4 Steiner Vertices 9 edges Connectivity Total for Module 4: 100 marks IEB Copyright © 2018