2019 VCAA Further Mathematics Exam 2 Solutions 2 PDF

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http://www.learning-with-meaning.com/ 2019 VCAA Further Mathematics Exam 2 Solutions

Q6a

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SECTION A - Core Data analysis Q1a Day number

SI

Q6b

Summer 0.89

Autumn 1.00

Winter 1.41

Spring 0.70

262 ≈186 mm 1.41

Q1b Recursion and financial modelling

Q7a V 0 = 60 000 , V1 = 0.9 × 60 000 , V 2 = 0.9 2 × 60 000 = 48 600 Q7b 0. 1 = 10% Q7c 0.9n × 60 000 < 20 000 , n > 10.43 .: after 10 years, i.e. sometime th in the 11 year Q1ci Q1 = 12.2

3 Q1cii = 0.20 = 20% 15

Q7d Vn+ 1 = Vn − 0.08V0 where V0 = 60 000 Q8a $3700

Q2ai IQR = 17 − 12 = 5 °C

Q8b 0.0035× 12 = 0.042 = 4.2%

Q2aii 25 −15 = 10°C

Q8c By Finance Solver, find N = 60.02495091 for FV = 0 , set N = 60 to find FV ≈ 92.15 . After 60 months the balance is $92.15

Q2aiii 1

4.2 = 700 dollars 100 × 12

Q2b 50% × 30 = 15 days

Q8d 200 000 ×

Q3 The median minimum daily temperature for the month (the response variable) varies with the month (the explanatory variable). The median decreases with the month.

Q9a By Finance Solver, set FV = −262 332.33 to find PMT ≈ −1704.03 . Each fortnightly repayment is $1704.03 Q9b 1704.03 × 78 −( 350 000 −262332.33) ≈45 246.67 dollars

Q4a humidity 9 am Q4b humidity 3 pm = −1.26 + 0.765 × humidity 9 am Q4c By CAS, r ≈ 0.871

4.8 = 1.004 12 × 100 B0 = 262 332.33 , Bn+1 = 1.004 Bn − 3517.28

Q9c 1 +

Q5a pressure 3 pm increased by 0.8894 hPa for 1 hPa increase in pressure 9 am Q5b pressure 3 pm = 111.4 + 0.8894× 1025 ≈ 1023 Q5c Interpolation Q5d Prediction = 111.4 + 0.8894 × 1013 ≈ 1012 .36 Residual ≈ 1015 − 1012 .36 ≈ 3 hPa Q5ei r = b

sx 4.5477 ≈ 0.8894 × ≈ 0.9657 ≈ 0.966 4.1884 sy

Q5eii r 2 ≈ 0.966 2 ≈ 0.933 = 93. 3% Q5fi Assumption: There is a linear relationship between the atmospheric pressure at 3 pm and the atmospheric pressure at 9 am. Q5fii The residual plot is not random enough to support the assumption. 1 2019 VCAA Further Maths Exam 2 Solutions

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http://www.learning-with-meaning.com/ SECTION B - Modules

Module 2: Networks and decision mathematics

Module 1: Matrices

Q1a Office

Q1a Order 3× 2

Q1bi Hamiltonian cycle

Q1b 50 + 20 + 40 = 110

Q1bii Office --- science laboratories --- mathematics class rooms --gymnasium --- computer rooms --- library --- office

Q1c

50 L = 20 40

Q2a Student Blake Charli Huan Marco

Q1d

3 6 22 19   R = 1 10 7 2   1 3 10 26 T

Q1e e32 represents the number of cars parked for 2 hours in area C.

Q2a Student Anita Imani Jordan Lola

400 = 0.2 = 20% Q2a 2000 Q2b

300  A 780  F  S1 =  300  G   620  W

Sport Tennis Football Basketball Athletics

Sprinting distance (m) 400 200 100 300

Q3a 8 Q3b Start of week 13

Q2c 0. 1 × 600 + 0.2 ×600 = 180 ,

180 = 0.6 = 60% 300

Q3c J Q3d 2 + 4 +7 +1 + 2 + 2 + 5 + 6 = 29 weeks

Q2d 0 1 M = 0  0

Q3e

1 0 0 0 0 0  0 1 0  0 0 1

Activity C D G H K

Q3a Location A , 0.3× 500+ 0.4× 600+ 0.6× 500 + 0.3 × 400 = 810

Reduction (weeks) 0 1 2 1 1

Overall completion time to be reduced by 4 weeks at a minimum cost of 2000 ×1 +2500 ×2 +1000 ×1 + 4000 ×1 = 12 000 dollars (this answer is not required)

Q3b − 210   0   B1 =   210     0  Q3c 810  − 210  600        300   0  300  = + R 1 = V × R 0 + B1 =  310   210  520        580   0  580   600 786    288   , R =  288 VR 1 =  282  2  512      600 644 

2 2019 VCAA Further Maths Exam 2 Solutions

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http://www.learning-with-meaning.com/ Module 3: Geometry and measurement

Module 4: Graphs and relations 2

Q1a Area of the shaded region = 160 × 40 + 12 × 25 = 6700 m 3

Q1b Volume = 6 × 2.4 × 2. 6 = 37.44 m

2 Q1c Total surface area = 2( 6 × 2.4 + 6 × 2.6 + 2.4 ×2.6) =72.48 m

Q1a 60 Q1bi Average rate =

32 − 62 = −6 5

Q1bii 32 − 6 × 3 = 14 Q2a 3

Q1d 8 × 3 × 2 = 48 Q2a Sydney ( 34° S ) is closer to the equator than Magadan (60° N ). Q2b Distance = 6400 ×

π 180

× (34 + 60) ≈ 10500 km

Q2b ≤ 10 Q2c

Q2c 6 am Sydney time = 4 am Perth time Number of hours between 4 am 1 June and 10 am 11 June = 24 × 10 + 6 = 246 Q3ai Distance AC = 25 2 − 15 2 = 20 m

 15  Q3aii ∠ACB = sin −1   ≈ 37°  25 Q3b The cosine rule: Distance RP =

202 + 382 − 2( 20)( 38) cos120° ≈ 51m Q2d 9, 17, 25

Q3c Vertical distance from M to the container 2

2

 2 .4   6  = 4. 4 −   −   ≈3 m  2  2

Q3a 5 x +10 y ≤ 60

2

Q3b

Q3c 4 cars and 4 minibuses, cost = 70 × 4 + 100 × 4 = 680 dollars Q3d 3 cars and 4 minibuses, cost = 70 × 3 + 100 × 4 = 610 dollars Q3e Objective function: Cost C = mx + 100 y Point (6,2) gives maximum cost when m > 100 .: 70 + k = 100 , k = 30

Please inform [email protected] re conceptual and/or mathematical errors

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