NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

1 NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P FEBRUARY/MARCH 00 MEMORANDUM MARKS: 00 This memorandum consists of 8 pages.

2 Mathematics/P DoE/Feb. March 00 QUESTION. ; 6 s. + 7 developing sequence + 7 T n+ Tn T n+ Tn + 7, T (n ) OR T + 7, T (n ) OR n T n n T n+ T n + 0., T (n ) n OR T 0., T (n ) n T n + developing sequence n T 0. n+ T n + T [5] QUESTION. Yes. All three graphs represent the annual profits for the same company (005 R60 million; 006 R00 million and 007 R80 million). There are, however, differences in the way the information is presented the scale on the vertical axis has been changed in graph and the order of the years reversed in graph.. In graph, the impression created is that the annual profit is levelling off or shows a slight increase year on year. In graph, the impression created is that the annual profit is decreasing.. Graph. This graph shows a substantial increase in annual profits year on year. yes profits for the same company but presented differently. graph - annual profit is levelling off graph - decreasing explanation [6] QUESTION. 9 minutes. The standard deviation is 8 minutes. m 9 + (8) 55 n 9 (8) 5. 0 learners represent 6% of total number 0 00 Total number The library assistant should be employed for one hour each afternoon. There is a small percentage (< %) of learners who spend more than more than hour in the library. for m for n 0 6% () one hour justification [7]

3 Mathematics/P DoE/Feb. March 00 QUESTION 4 4. P(A or B) 0, + 0,5 0,8 4. Since A and B are independent P(A or B) P(A) + P(B) P(A and B) 0, + 0,5 0,5 0,65 addition P(A and B) 0,5 0, + 0,5 0,5 [5] QUESTION 5 5. R C x 00 x 70 x 6 6 and 0 6 (inside B only), 00 x and 70 x 6 (outside) (4) 6 6 B x + x x x 40 x 4 Number of learners playing rugby and cricket 0. P(play basketball only) ,08 44 P(does not play cricket) 40 0,600 P(plays at least sports) set up equation x method ,9 []

4 Mathematics/P 4 DoE/Feb. March 00 QUESTION 6 6. Number of ways in which performances take place : 7! Since first and last performance are fixed, the number of different ways performances can be arranged in 5 cities 5! The different ways the coastal cities tours can take place 4! 4 Total number of ways the itinerary can be arranged 4! 4! multiplication rule 5 cities multiplication rule coastal cities 4! 4! 4! (4) [9]

5 Mathematics/P 5 DoE/Feb. March 00 QUESTION 7 7. & 7. Mean Maximum Temperature (degree Celsius) Scatter Plot of North Latitude vs Mean Maximum Temperature for April North Latitude 7. plotting points 7. gradient correct x-intercept 7. a 9, 94 (9, ) b 0,5 ( 0, ) Equation of regression line yˆ 9,94 0, 5x 7.4 The y-intercept represents the mean maximum temperature for April at the equator. 7.5 Mean maximum temperature for April in Madrid 9,94 0,5(40) 9,4 C a-value b-value equation substitution 7.6 r 0, 9 ( 0,9905 ) answer 7.7 The value of r is close to and suggests that there is a very strong relationship between distance from the equator and the mean maximum temperature for April. The further one moves away from the equator, the colder it gets. (4) () very strong and further away from the equator, the colder it gets () [5]

6 Mathematics/P 6 DoE/Feb. March 00 QUESTION 8 8. Equal to reflex Ô ( s round a point) L Mˆ N reflex Ô ( circ centre circumference) 60 L Mˆ N Nˆ 5 (base angles LM MN) Kˆ 5 (angles in same segment) reflex Ô 60 reason L Mˆ N 0 5 reason () [7] QUESTION 9 9. Is equal to the angle subtended by the chord in the alternate segment 9.. Â x (tangent chord theorem) Â x (vertically opp. angles) 5 Pˆ x (tangent chord theorem) reason reason () 9.. PT TA Pˆ Â Â Pˆ Â 6 Â Rˆ 6 Rˆ (vertical opp angles) (tangent chord theorom) APTR is a cyclic quadrilateral (tangents drawn from same point) (angles opp equal sides) ; PT TA (converse : ext angle of cycl.quad.) (5) equal angles reason (5) []

7 Mathematics/P 7 DoE/Feb. March 00 QUESTION 0 0. OC OB (radii) Hence AE BE (midpoint theorem) OR OC OB conclusion and reason C ÂB 90 (diameter subtends right angle) O ÊB CÂB 90 (corresponding angles AC//OE) AE BE (line drawn from centre, perpend. to chord or midpoint theorem) In Δ AED and ΔCEB AÊD CÊB Dˆ Bˆ Â Ĉ ΔAED /// ΔCEB AE CE DE BE AE.BE DE.CE but AE BE AE DE.CE (vertically opp angles) (angles in same segment) (angles in same segment) (equi - angular) (deduction) (proven) AE.BE DE.CE But AE.BE EF.CE DE.CE EF.CE DE EF E is the midpoint of DF OR AE DE.CE AE.BE EF.CE AE EF.CE EF.CE DE.CE EF DE E is the midpoint of DF OÊB CÂB 90 conclusion and reason statement AE DE CE BE AE BE AE.BE DE.CE AE.BE EF.CE DE.CE EF.CE AE.BE EF.CE AE EF.CE EF.CE DE.CE o [0]

8 Mathematics/P 8 DoE/Feb. March 00 QUESTION. In Δ BDA and Δ CDB BDA CDB 90 B C A B Δ BDA /// Δ CDB. AD : DC : CD 5 0 BD CD But AD BD BD AD.CD BD BD 50 (both x ) (remaining angles) (equiangular). AB ( 50) + (5) (Theorem of Pythagoras) AB 75 Ê AE AB AE ABˆ C BC // DE 75 AD AC AE 5 (proportion theorem) 5 5 BDA C DB B C A B CD BD AD CD BD BD using Pythagoras 90 o BC // DE AE AD AB AC (6) [] TOTAL : 00