Mathematical Literacy P2 Nov 2016 Memo Eng hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICAL LITERACY P2 NOVEMBER 2016 FINAL MARKING GUIDELINE MARKS: 150 Symbol Explanation M Method MA Method with accuracy CA Consistent accuracy A Accuracy C Conversion S Simplification RT/RG/RD Reading from a table/graph/map/diagram SF Correct substitution in a formula O Opinion/reason/deduction/example P Penalty, e.g. for no units, incorrect rounding off, etc. R Rounding off NP No penalty for rounding AO Answer only full marks J Justification This memorandum consists of 19 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 2 DBE/November 2016 NSC – Memorandum QUESTION 1 [36 MARKS] Ques Solution Explanation T&L P 1.1.1 11 A 2A numerator L2 P (even number date) = 22 A 1A denominator = 12 or 0,5 or 50% AO (3) F 1.1.2 • Quality of bank services / security / perks. O L4 OR • Proximity or accessibility of the bank. O 2O reason OR • Marketing/advertising appeal O OR • Loyalty to bank O OR O • Religious reasons / Economical reasons Any other suitable reason (2) 1.1.3 2014 Fee = R3,50 + 1,1% × R1 000 SF 1SF substituting R1000 F = R14,50 CA 1CA 2014 fee L2  R15,50  1SF correct values % change =  −1 × 100% SF  R14,50   R1,00  =  × 100%  R14,50  CA = 6, 8965517… 1CA simplification A ≈ 6,9% R 1R rounding OR OR SF 1SF correct values  R15,50  1SF substituting R1000 % change =  − 1 × 100%  R3,50 + 0,011 × R1 000  SF  R15,50  = −1 × 100% 1CA 2014 fee  R14,50 CA = 6,8965517… CA 1CA simplification A ≈ 6,9% R 1R rounding (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 3 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L 1.1.4 Withdrawal fee R15 000 at Bank X F SF L4 = R3,95 + 0,013 × R15 000 1SF substituting = R198,95 CA Fees for 4 withdrawals 1CA weekly charges = R198,95 × 4 = R795,80 CA 1CA fees for 4 withdrawals Withdrawal fee for R15 000 at Bank Y = R4,00 + R15 000 × 1,15% = R176,50 CA 1CA charges Fees for 4 withdrawals = 4 × R176,50 = R706,00 CA 1CA fees for 4 withdrawals Difference in fees = R795,80 – R706,00 = R89,80 CA 1CA difference It is NOT VALID. O 1O conclusion OR OR Withdrawal fee R15 000 at Bank X MA 1MA substituting = R3,95 + 0,013 × R15 000 = R198,95 CA 1CA weekly charges Withdrawal fee for R15 000 at Bank Y = R4,00 + R15 000 × 1,15% = R176,50 1CA charges CA CA 1CA difference Difference in fees = R198,95 – R176,50 = R22,45 M 1M fees for 4 Saving on 4 withdrawals = R22,45× 4 = R89,80 withdrawals CA 1CA October It is NOT VALID O charges 1O conclusion OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 4 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L Bank X: Fee per R1 000 = R3,95 + R1,30 ÷ 100 × 1 000 MA 1MA substituting = R16,95 CA Withdrawal fee for R15 000 = R16,95 × 15 1CA weekly charges = R254,25 1M fees for 4 For 4 withdrawals : R254,25 × 4 M withdrawals = R1 017 Bank Y: Withdrawal fee for 4 times R15 000 1CA charges = R15,50 × 4 × 15 CA = R930 CA 1CA October charges Difference in fees = R1 017 – R930 = R87 CA 1CA difference It is NOT VALID 1O conclusion (Max of 6 marks for a total withdrawal of R60 000 .) (7) 1.1.5 Wage for 4 full weeks = R2 142,85 × 4 A F = R8 571,40 1A 4 weeks wage L2 R2 142,85 Wage for 2 days = ×2 M 1M divide by 5 5 M 1M multiply by2 = R857,14 Total wage = R8 571,40 + R857,14 1CA total wage = R9 428,54 CA OR OR R2 142,85 R2142,85 × 4 Average day wage = OR 1M divide by 5 5 M 20 A 1A daily wage = R428,57 Total wage for October = 22 × R428,57 M 1M multiply by 22 = R9 428,54 CA 1CA total wage OR OR 2 M 2 days of a five day week = of a week 1M divide by 5 5 Total number of weeks = 4 52 A OR 4,4 1A number of weeks Total wage for October = 4 52 × R2142,85 M 1M multiply by weekly = R9 428,54 CA wage 1CA total wage OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 5 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L M 1M multiplying 52 A 1A 52 weeks in year Monthly wage = R2 142,85 × 12 MA 1MA dividing by 12 = R9 285,68 CA 1CA total wage (4) 1.2.1 • More small/local companies may have entered the D market O L4 O 2O factor with reason • The increased use of smartphones, laptops and tablets O • Locally produced no need to import. 2O factor with reason • Cost of transport increased O • Economical reasons / factors O • Maritime piracy / security O • Other means of transport used O O • Durability - demand for new computers became less Or any other valid factors with reasons (4) 1.2.2 Q1 of 2012: D MA 1MA adding correct L2 (15,7 + 11,7 + 10,1 + 9 + 5,4 ) million values CA 1CA total shipment in = 51,9 million or 51 900 000 2012 Q1 of 2013: = ( 12 + 11,7 + 9 + 6,2 + 4,4 ) million MA 1MA total shipment in = 43,3 million or 43 300 000 2013 Difference between 2013 and 2012 1CA difference in CA million = 51,9 mil – 43,3 mil = 8,6 million or 8 600 000 OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 6 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L Differences (in millions) for A = 15,7 – 12,0 = 3,7 2A differences in B = 11,7 – 11,7 = 0 A millions C = 10,1 – 9,0 = 1,1 D = 9,0 – 6,2 = 2,8 A E = 5,4 – 4,4 = 1 1M adding all M differences Total difference = (3,7 + 1,1 + 2,8 + 1) million 1CA total difference in million = 8,6 million CA Penalty if million omitted (4) RT D 1.2.3 __ M 1RT correct values 12 000 000 15 700 000 % change A = × 100 % 1M calculating % L4 15 700 000 change = – 23,56687898% CA 1CA % change RT 1RT correct values __ 6 200 000 9 000 000 1M calculating % % change D = × 100 % M 9 000 000 change = – 31,11111111% CA 1CA % change The statement is NOT VALID. O 1O conclusion OR OR Percentage of 2012 shipped in 2013: RT 1RT correct values 12,0 By A: × 100% 15,7 = 76,43% A 1A percentage ∴ Percentage decrease = 100% – 76,43% = 23,57% M 1M % change RT 1RT correct values 6,2 By D: × 100% 9 A 1A percentage = 68,89% M 1M % change ∴ Percentage decrease = 100% – 68,89% = 31,11% O 1O conclusion D shows the greatest decrease, the statement is NOT VALID NP (7) [36] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 7 DBE/November 2016 NSC – Memorandum QUESTION 2 [47 MARKS] Ques Solution Explanation T&L A F 2.1.1 Amount × 109,7% = R218,9 billion 1A correct value and L2 (a) R 218,9 billion % Total amount spent = 109,7% M 1M dividing by = R199 544 211 500 CA 109,7% or R199,54 billion or R1,9954 × 1011 1CA total amount NP (3) A F 2.1.1 It is more appropriate to round to one decimal place. 1A statement L4 (b) If a rand value in billions is rounded off to a whole number, the amount that is added or lost is hundreds of millions of rands. O 2O explanation OR (Note: More A appropriate can be It is not appropriate to round to off to a whole number since it implied in the has a big financial implication O statement) (3) A A F 2.1.2 International: 43% of R 218,9 billion = R94,127 billion 1A percentage L3 1A amount Number of visitors = 14,3 million or 14 300 000 \ R94 127 000 000 C 1C conversion Average spent per visitor = 1MA average 14 300 000 MA 1CA value = R6 582,31 CA 1O conclusion This is NOT correct. O OR OR A A 1A percentage International: 43% × R 218,9 billion = R94,127 billion 1A amount C 1C conversion R94,127 × 1 000 million Average spent per visitor = 14,3 million MA 1MA average = R6 582,31 CA 1CA value This is NOT correct. O 1O conclusion OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 8 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L Amount spent by the International visitors MA 1MA multiplying = R6 580 × 14,3 million 1A amount A C = R94 094 million = R94,094 billion 1C conversion But spent by international tourists is 1A percentage A A 1A amount 43% × R 218,9 billion = R94,127 billion The amount was NOT CORRECT O 1O conclusion (6) A A F 2.1.3 Air transport and road transport 1A for each item L2 (2) F 2.1.4 Payment of tourism levy O L4 2O example OR O Purchase of souvenirs OR O Entrance fees to tourist attractions OR O Any other suitable example (2) 2.1.5 Growth in 2014 = 2,9% × R103,6 billion M 1M multiplying = R3,0044 billion M 1M adding GDP contribution (2014) = (R3,0044 + R103,6) billion 1CA amount in 2014 = R106,6044 billion CA Growth in 2015 = 2,9% × R106,6044 billion = R3,0915276 billion GDP contribution (2015) = (R3,0915276 + R106,6044) billion = R109,6959276 billion CA 1CA amount in 2015 Growth in 2016 = 2,9% × R109,6959276 billion = R3,1811819 billion GDP contribution (2016) = (R3,1811819 + R109,6959276) bil. 1CA amount in 2016 = R112,8771095 billion CA R 1R correct rounding = R112 877 million or R112 877 000 000 or R112,877 billion OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 9 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L A M F 2.1.5 GDP contribution (2014) = 102,9% × R103,6 billion 1M multiplying L3 = 106,6044 billion CA 1A 102,9% 1CA amount in 2014 GDP contribution 2015 = 102,9% × R106,6044 billion = 109,6959276 billion CA 1CA amount in 2015 GDP contribution 2016 = 102,9% × R109,6959276 billion = R112,8771095 billion. CA 1CA amount in 2016 = R112 877 million R 1R correct rounding or R112 877 000 000 OR GDP contribution 2016 A A 1M multiplying M 2A 102,9% = R103,6 billion × 102,9% × 102,9% × 102,9% = R112,8771095 billion. CA CA amount in 2016 = R112,877 billion or R112 877 million C 1C conversion or R112 877 000 000 R 1R correct rounding (6) 2.2.1 D (a) RT 3RT correct L2 Stopover times = 5 + 20 + 5 + 2 + 8 + 2 + 2 + 2 + 23 + stopover times M 1M adding 26 + 3 + 17 + 3 + 14 + 3 + 3 stopover times CA = 138 minutes or 2 hrs and 18 minutes 1CA total stopover time or 2,3 hours Stopover times: One or two errors only 1 mark penalty, Three or four errors 2 mark penalty AO (5) 2.2.1 CA CA From Q2.2.1 (a) D 2 and 3 minutes (b) 2CA modal time L2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 10 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L CA From Q2.2.1(a) 2.2.1 Actual train travel time: M (c) RT 1RT start and end time L3 13:24 (day2) to 17:30 (day1) – stopover time CA 1CA 19 hours 54 min = 19 hr 54 min – 2 hr 18 min M 1M subtracting stopover time = 17 hr 36 min = 17, 6 hr C 1C conversion D=S×T SF 1SF substitution 992 km = S × 17hr 36 min 992 km S 1S changing subject of S = formula 17,6 hour = 56,36 km/h CA 1CA simplification OR OR RT CA 1RT start and end time Total time = 24 hours – 17h30 + 13h24 = 19hr 54 min 1CA 19 hours 54 min M 1M subtracting 19hr 54 min – 2 hrs 18 min = 17 hrs 36 min = 17,6 hr stopover time C 1C conversion D=S×T SF 1SF substitution 992 km = S × 17,6 hr 992 km 1S changing subject of S= S 17,6 hour formula ≈ 56 km/h CA 1CA simplification OR OR From 17:30 to 00:00 = 6 hrs 30 min RT 1RT start and end From 00:00 to 13:24 = 13hrs 24 min times CA Time of journey = 19 hrs and 54 minutes M 1CA trip time Travel time = 19 hr 54 min – 2 hr 18 min 1M subtracting = 17 hr 36 min stopover time D=S×T SF 992 km = S × 17,6 hr 1SF substitution 1S changing subject of 992 km S formula Average Speed = 17,6 hour C 1C conversion = 56,36 km/h CA 1CA simplification NP (7) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 11 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L Fin 2.2.2 Forward trip in January: L3 1MA two adult price Parents = 2 × R560 = R1 120 MA MA 1MA discounted price Father = R560 – R560 × 25% OR R560 × 75% for over 55 yrs = R420 CA 1CA father's fare Children's fare = R560 × 80% = R448 MA 1MA children fare Two children = 2 × R448 = R896 CA 1CA total children's fare CA Total fare for family: R1 120 + R420 + R896 = R2 436 1CA Jan total fares Return trip in February: A 1A adults Feb fare Parents fare = 2 × R490 = R980 Father = R490 minus R490 × 25% or R490 × 75% = R367,50 A 1A senior citizen fare Two children = 2 × (R490 – R490 × 50%) = R490 A 1A children Feb fare Total fare for return trip = R980 + R490 + R367,50 1CA total Feb trip's fare = R1 837,50 CA 1CA total trip fare Total cost for both trips = R2 436 + R1 837,50 (Note: Max of 6 marks CA if only one trip is = R4 273,50 calculated ; Max of 9 marks for using the same fare for both trip) OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 12 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L MA MA 1MA adding correct Father's fare = (R560 + R490) × 75% values M 1MA 75 % 1M % calculation = R787,50 CA 1CA simplification Parents' fare = 2 ×( R560 + 490) MA 1MA adding and = R2 100 CA multiplying 1CA simplification 1MA 80% MA MA 1MA 50% A Children's fare = (R560 × 80% + R490 × 50%) × 2 1A correct values = R1 386 CA 1CA simplification Total fare for both trips = R787,50 + R2 100 + R1 386 = R4 273,50 CA 1CA total return trip fare (11) [47] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 13 DBE/November 2016 NSC – Memorandum QUESTION 3 [31 MARKS] Ques Solution Explanation T&L M 3.1.1 Capacity of section C = 5 m × 1, 2 m × 15 m SF 1SF correct values L3 = 90 m3 CA 1CA capacity section C Capacity of section A = 2 m × 12,5 m × 15 m SF 1SF correct values = 375 m3 CA 1CA capacity section A Maximum capacity = 90 m3 + 375 m3 + 300 m3 MA 1MA adding capacities in = 765 m3 m3 OR OR Maximum capacity = Capacity of section (A + B + C) 1SF Correct values for A SF SF = 2 m × 12,5 m × 15 m + 300 m3 + 5 m × 1, 2 m × 15 m 1SF correct values for C CA CA 1CA capacity section A = 375 m3 + 300 m3 + 90 m3 MA 1CA capacity section C 1MA adding capacities in = 765 m3 m3 OR OR Volume = 30 m × 15 m × 2 m SF 1SF volume = 900 m3 CA 1CA volume section A Volume beneath C = 5 m × 15 m × 0,8 m = 60 m3 1SF volume beneath B 1 Volume beneath B = 2 × 12,5 m × 15 m × 0,8 m SF CA 1CA volume beneath B = 75 m3 Maximum capacity = 900 m3 – 60 m3 – 75 m3 1MA subtracting volume = 765 m3 MA in m3 (5) 3.1.2 M M Volume of water = 94% × 765 m3 = 719,1 m3 1M calculating % L3 = 719 100 ℓ C 1C convert to litres 719 100 × 1 = gallons C 1C convert to gal. 3,785 ≈ 189 986,79 gallons CA 1CA simplification OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 14 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L Capacity (in litres) = 765 m3 × 1 000 = 765 000 ℓ C 1C convert to litres 765 000 Capacity( in gallons) = 1C convert to gal. 3,785 C = 202 113,6063 1M calculating % Volume of water = 94% × 202 113,6063M 1CA simplification = 189 986,79 gallons CA NP (4) 3.1.3 In 1 hour 2 350 litres of water will flow. 1MA using flow rate In 1 day: 24 ×2 350 litres MA 1CA water in 1 day = 56 400 litres will flow CA M 1M multiplying In 2 1 2 days amount of water flowing = 2 × 56 400 litres 1 2 = 141 000 litres CA 1CA simplification ∴ Statement is NOT VALID. O 1O conclusion OR OR 135 000 MA Time to fill swimming pool = 1MA finding time taken 2 350/h 1CA time ≈ 57,4468 hours CA 57,4468 hrs = 2 days and 9 h 27 min M 1M splitting calc. hrs Two and a half days = 2 days 12 hours C 1C converting two and a half days ∴ Statement is NOT VALID O 1O conclusion OR OR 135 000 MA 1MA finding time taken Time to fill swimming pool = 2 350/h ≈ 57,4468 hours CA 1CA time MA . Two and a half days = (2 ×24 + 12) hours = 60 hours A 1MA multiply with 24 and add 12 ∴ Statement is NOT VALID O 1A hours 1O conclusion OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 15 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L 135 000 Time to fill swimming pool = MA 3.1.3 2 350/h 1MA finding time taken ≈ 57,4468 hours CA 1CA time MA CA 57,4468 hours ÷ 24 hours/day = 2,3936 1MA dividing by 24 h/d 1CA days NOT VALID O 1O conclusion OR OR 1MA multiplying with 24 MA A h/d M 2 12 days × 24 h/d = 60 hours 1A number of hours L3 MA 1MA multiplying hours Volume of water = 60 hours × 2 350 ℓ/hour with flow rate 1CA simplification = 141 000 ℓ CA This is more than the 135 000 ℓ to be topped up O 1O conclusion The statement is NOT VALID (5) Data 3.2.1 Total = 18 × 15 = 270 MA 1MA multiplying L3 M Difference = 270 – 236 = 34 1M subtracting totals x = 34 ÷ 2 M 1M dividing by 2 = 17 CA 1CA value of x OR OR MA 2 x + 236 1MA adding correct Mean = = 15 18 values 2x = 270 – 236 1M subtracting totals M = 34 1M dividing by 2 34 x= M 2 CA 1CA value of x = 17 OR OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 16 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L M 2 x + 236 2x 1M adding correct Mean = = + 13,1111 18 18 M values 1M mean concept 15 – 13,1111 = 1,8888... 2x 1CA manipulating = 1,8888... CA formula 18 x = 1,888... × 18 ÷ 2 = 17 CA 1CA value of x AO (4) Data 3.2.2 RG Q 1 = 15 and Q 3 = 20 RG 1RG finding Q 1 L3 1RG finding Q 3 IQR = 20 – 15 M 1M subtracting = 5 CA 1CA IQR value AO (4) O D 3.2.3 It is more convenient for them to go in the evening 2O reason L4 OR O During daytime other distractions keep people away. OR (2) Small groups receive individual attention O OR Any other sensible reason O P A 3.2.4 6 1A numerator L2 P (Day Group full attendance) = × 100% 18 A 1A denominator ≈ 33% R 1R whole % AO (3) D 3.2.5 The range of the afternoon group was smaller.O 2O reason L4 The afternoon group has a higher median. O 2O reason The afternoon group has smaller inter-quartile range.O Minimum of the afternoon group is higher. O (Any TWO acceptable reasons) (4) [31] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 17 DBE/November 2016 NSC – Memorandum QUESTION 4 [36 marks] Ques Solution Explanation T&L MA M 4.1.1 0,21875 miles = 385 yards 1MA recognising equal L2 parts 385 Hence, 1 mile = yards MA 1MA correct fraction 0,21875 = 1 760 yards OR OR 1 = 4,571428571 MA 0,21875 1MA conversion factor MA 385 × 4,571428571 = 1760 yards 1MA multiplying 385 with conversion factor (2) MP 4.1.2 Approximately 4,5 miles RG 2RG correct distance. L2 (2) (Accept distances in the range 4,3 miles to 4,7 miles) MP 4.1.3 RG C CA 1RG correct distance L2 700 ft = 700 × 0,3038 m = 212,66 m 1C converting to m 1CA max height (Accept heights in the range 700 ft to 710 ft) NP (3) MP 4.1.4 It is uphill. (steep) O 2O reason L4 OR This runner found it difficult to run uphill. O OR It is easier to run downhill. O (2) A A 4.2.1 6 + 3 or 9 2A number of venues MP L2 [Due to the annexure of Limpopo full marks can be awarded (2) if only 6 is given as the number of venues] MP 4.2.2 Hippo A 2A correct enclosure L2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 18 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L MP A 4.2.3 Zoo is 6 times bigger than the elephant exhibit. 2 A estimation L4 M CA 1M multiplying ∴ 6 × 4 = 24 football fields 1CA solution (Max 2 marks for Also accept 5 or 7 as a correct estimation. number of football fields for estimated areas of 3,4 ANSWER ONLY full marks if 20 to 28 football fields. ,8 or 9.) (4) MP 4.2.4 The distance on the map = 85 mm A 1A measured L4 A distance M Bar scale 20 mm is 200 m 1A measured bar 1M relating to bar to measurement 85 mm M Real distance using the bar scale = × 200 m 1M using the given scale 20 mm 1CA simplification = 850 m CA 1C conversion 1,6 km = 1 600 m C 1O conclusion ∴ The scale is NOT correct. O OR OR A 1A measured bar M 1M relating to bar to Bar scale 20 mm is 200 m measurement 1C conversion 1,6 km = 1 600 m C M 1M using the given scale 1 600 m Calculated map distance = × 20 mm 200 m 1CA simplification = 160 mm CA Measured distance = 85 mm A 1A measured distance ∴ The scale is NOT correct. O 1O conclusion (Accept a range from 82 mm to 87 mm for the distance between streets and 18 mm to 22 mm for the bar scale.) (7) D 4.3.1 Saturday A 2A correct day L2 (2) P 4.3.2 Monday is NOT reflected on the given graph. O 2O reasoning L4 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematical Literacy/P2 19 DBE/November 2016 NSC – Memorandum Ques Solution Explanation T&L D 4.3.3 The number of visitors increase to about 12:00. 2O trend L4 on weekdays and then decrease again till 16:00. O OR The number of visitors on weekends is more than the visitors on weekdays. O 2O trend OR The number of visitors increase to about 13:00 on weekends and then decrease again till 16:00. O Any TWO trends relating time and number of visitors. (4) D 4.3.4 The number indicated by the height of the column on L4 Saturday is a little more than double the height of the mean number for a Tuesday O 2O reason OR 2O reason People work during the week O (4) OR Saturdays they go with their families to the zoo. O OR O Cheaper to go during the weekends OR More activities at the zoo on Saturday. O [36] TOTAL: 150 Copyright reserved