Band 4 – Page 42

Functions, 2ADV F2 EQ-Bank 14

List a set of transformations that, when applied in order, would transform `y = x^2` to the graph with equation `y = 1 - 6x - x^2`. (3 marks)

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`text(T1: Translate 3 units in negative)\ xtext(-direction)`

`text(T2: Translate 10 units in negative)\ ytext(-direction)`

`text(T3: Reflect in the)\ xtext(-axis)`

Show Worked Solution

`y = x^2`

`text(Transformation 1:)`

`text(Translate 3 units in negative)\ xtext(-direction)`

`y = (x + 3)^2`

`y = x^2 + 6x + 9`

`text(Transformation 2:)`

`text(Translate 10 units in negative)\ ytext(-direction)`

`y = x^2 + 6x – 1`

`text(Transformation 3:)`

`text(Reflect in the)\ xtext(-axis)`

`y`	`= −(x^2 + 6x – 1)`
	`= 1 – 6x – x^2`

Statistics, STD2 S4 EQ-Bank 4

Ten high school students have their height and the length of their right foot measured.

The results are recorded in the table below.

Using technology, calculate Pearson's correlation coefficient for the data. Give your answer to 3 decimal places. (1 mark)

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Describe the strength of the association between height and length of right foot for these students. (1 mark)

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Using technology, determine the least squares regression line that allows height to be predicted from right foot length. (1 mark)

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`0.941\ \ (text(to 3 d.p.))`
`text(The association is positive and strong.)`
`text(Height) =47.4 + 4.7 xx text(foot length)`

Show Worked Solution

i. `text(By calculator,)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 correlation” .

`r`	`= 0.94095…`
	`= 0.941\ \ (text(to 3 d.p.))`

ii. `text(The association is positive and strong.)`

iii. `x\ text(value ⇒ foot length (independent variables))`

`y\ text(value ⇒ height.)`

`text(By calculator:)`

`text(Height) = 47.4 + 4.7 xx text(foot length)`

Functions, 2ADV F2 EQ-Bank 16

`y = -(x + 2)^4/3` has been produced by three successive transformations: a translation, a dilation and then a reflection.

Describe each transformation and state the equation of the graph after each transformation. (2 marks)

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Sketch the graph. (1 mark)

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`text(See Worked Solutions)`

Show Worked Solution

i. `text(Transformation 1:)`

`text(Translate)\ \ y = x^4\ \ 2\ text(units to the left.)`

`y = x^4 \ => \ y = (x + 2)^4`

`text(Transformation 2:)`

`text(Dilate)\ \ y = (x + 2)^4\ \ text(by a factor of)\ 1/3\ text(from the)\ xtext(-axis)`

`y = (x + 2)^4 \ => \ y = ((x + 2)^4)/3`

`text(Transformation 3:)`

`text(Reflect)\ \ y = ((x + 2)^4)/3\ \ text(in the)\ xtext(-axis).`

`y = ((x + 2)^4)/3 \ => \ y = −(x + 2)^4/3`

ii.

Functions, 2ADV F1 EQ-Bank 8

Jacques is a marine biologist and finds that the mass of a crab is directly proportional to the cube of the diameter of its shell.

If a crab with a shell diameter of 15 cm weighs 680 grams, what will be the diameter of a crab that weighs 1.1 kilograms? Give your answer to 1 decimal place. (2 marks)

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`17.6\ text(cm)`

Show Worked Solution

`M`	`prop d^3`
`M`	`= kd^3`

`text(When)\ \ M=680, \ d=15`

`680`	`=k xx 15^3`
`k`	`=0.201481…`

`text(Find)\ \ d\ \ text(when)\ \ M=1100:`

`1100`	`=0.20148… xx d^3`
`d`	`=root3(1100/(0.20148…))`
	`=17.608…`
	`=17.6\ text{cm (to 1 d.p.)}`

Functions, 2ADV F1 EQ-Bank 7

The current of an electrical circuit, measured in amps `(A)`, varies inversely with its resistance, measured in ohms `(R)`.

When the resistance of a circuit is 28 ohms, the current is 3 amps.

What is the current when the resistance is 8 ohms? (2 marks)

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`10.5`

Show Worked Solution

`A \prop 1/R\ \ =>\ \ A=k/R`

`text(When)\ \ A=3, \ R=28:`

`3`	`=k/28`
`k`	`=84`

`text(Find)\ A\ \text{when}\ R=8:`

`A=84/8=10.5`

Statistics, 2ADV S3 EQ-Bank 2

Let `X` denote a normal random variable with mean 0 and standard deviation 1.

The random variable `X` has the probability density function

`f(x) = 1/sqrt(2pi) e^((−x^2)/2)` where `x ∈ (−∞, ∞)`

The diagram shows the graph of `y = f(x)`.

Complete the table of values for the given function, correct to four decimal places. (1 mark)
Use the trapezoidal rule and 5 function values in the table in part i. to estimate

`int_(−2)^2 f(x)\ dx` (2 marks)

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The weights of Rhodesian ridgebacks are normally distributed with a mean of 48 kilograms and a standard deviation of 6 kilograms.

Using the result from part ii., calculate the probability of a randomly selected Rhodesian ridgeback weighing less than 36 kilograms. (2 marks)

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ii. `0.9369`

iii. `3.155text(%)`

Show Worked Solution

ii.

`int_(−2)^2 f(x)`	`~~ 1/2[0.0540 + 2(0.2420 + 0.3989 + 0.2420) + 0.0540]`
	`~~ 1/2 xx 1.8738`
	`~~ 0.9369`

iii. `mu = 48, sigma = 6`

`ztext(-score (36))`

`= (x – mu)/sigma= (32 – 48)/6=-2`

`text(Shaded area = 93.69%)`

`text(By symmetry,)`

`P(X < 36\ text(kgs))`	`= P(z < −2)`
	`= (100 – 93.69)/2`
	`= 3.155text(%)`

Statistics, 2ADV S2 EQ-Bank 3

The table below lists the average life span (in years) and average sleeping time (in hours/day) of 9 animal species.

Using sleeping time as the independent variable, calculate the least squares regression line. (1 mark)

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A wallaby species sleeps for 4.5 hours, on average, each day.

Use your equation from part i to predict its expected life span, to the nearest year. (1 mark)

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`text(life span) = 42.89 – 2.85 xx text(sleeping time)`
`30\ text(years)`

Show Worked Solution

i. `text(By calculator:)`

COMMENT: Issues here? YouTube has short and excellent help videos – search your calculator model and topic – eg. “fx-82 regression line” .

`text(life span) = 42.89 – 2.85 xx text(sleeping time)`

ii. `text(Predicted life span of wallaby)`

`= 42.89 – 2.85 xx 4.5`

`= 30.06…`

`= 30\ text(years)`

Calculus, 2ADV C4 EQ-Bank 5

The velocity of a particle moving along the `x`-axis at `v` metres per second at `t` seconds, is shown in the graph below.

Initially, the displacement `x` is equal to 12 metres.

Write an equation that describes the displacement, `x`, at time `t` seconds. (2 marks)

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Draw a graph that shows the displacement of the particle, `x` metres from the origin, at a time `t` seconds between `t= 0` and `t = 5`. Label the coordinates of the endpoints of your graph. (2 marks)

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`x = 3t – (3)/(10) t ^2 + 12`

Show Worked Solution

i.	`m_v`	`= -(3)/(5)`
	`v`	`= 3 – (3)/(5) t`

`x`	`= int v \ dt`
	`= int 3 – (3)/(5) t \ dt`
	`= 3t – (3)/(10) t^2 + c`

`text(When) \ \ t = 0, x = 12 \ => \ c = 12`

`:. \ x = 3t – (3)/(10) t ^2 + 12`

ii. `text(When) \ \ t = 5:`

`x = 15 – (3)/(10) xx 25 + 12 = 19.5`

Financial Maths, 2ADV M1 EQ-Bank 1

Ralph opens an annuity account and makes a contribution of $12 000 at the end of each year for 9 years.

For the first 8 years, the interest rate is 4% per annum, compounded annually.

For the 9th year, the interest rate decreases to 3% per annum, compounded annually.

Use the Future Value of an Annuity table to calculate the amount in the account immediately after the 9th contribution is made. (3 marks)

Show Answers Only

`$125\ 885.04`

Show Worked Solution

`text(After 8 years:)`

`text(Total)`	`= 9.214 xx $12\ 000`
	`= $110\ 568`

`text{After 9 years (9th contribution made year end):}`

`text(Total)`	`= 110\ 568 xx 1.03 + 12\ 000`
	`= $125\ 885.04`

Statistics, 2ADV S2 EQ-Bank 2

The table below lists the average body weight (in kilograms) and average brain weight (in grams) of nine animal species.

A least squares regression line is fitted to the data using body weight as the independent variable.

Calculate the equation of the least squares regression line. (1 mark)

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If dingos have an average body weight of 22.3 kilograms, calculate the predicted average brain weight of a dingo using your answer to part i. (1 mark)

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`text(brain weight) = 49.4 + 2.68 xx text(body weight)`
`109\ text(grams)`

Show Worked Solution

i. `text(By calculator:)`

COMMENT: Know this critical calculator skill!.

`text(brain weight) = 49.4 + 2.68 xx text(body weight)`

ii. `text(Predicted brain weight of a dingo)`

`= 49.4 + 2.68 xx 22.3`

`=109.164`

`= 109\ text(grams)`

Statistics, 2ADV S2 EQ-Bank 1

The arm spans (in cm) and heights (in cm) for a group of 13 boys have been measured. The results are displayed in the table below.

The aim is to find a linear equation that allows arm span to be predicted from height.

What will be the independent variable in the equation? (1 mark)

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Assuming a linear association, determine the equation of the least squares regression line that enables arm span to be predicted from height. Write this equation in terms of the variables arm span and height. Give the coefficients correct to two decimal places. (2 marks)

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Using the equation that you have determined in part b., interpret the slope of the least squares regression line in terms of the variables height and arm span. (1 mark)

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`text(Height)`
`text(Arm span)\ = 1.09 xx text(height) – 15.63`
`text(On average, arm span increases by 1.09 cm)`
`text(for each 1 cm increase in height.)`

Show Worked Solution

a. `text(Height)`

COMMENT: Calculator skills for finding the least squares regression line were required in NESA sample exam – know this critical skill well!

b. `text(By calculator,)`

`text(Arm span)\ = 1.09 xx text(height) – 15.63`

c. `text(On average, arm span increases by 1.09 cm)`

`text(for each 1 cm increase in height.)`

Functions, 2ADV F2 EQ-Bank 1

The function `f(x) = |x|` is transformed and the equation of the new function is `y = kf(x + b) + c`.

The graph of the new function is shown below.

What are the values of `k`, `b` and `c`. (2 marks)

Show Answers Only

`k = −1/3, b = 3, c = 2`

Show Worked Solution

`y = |x|`

`text(Translate 3 units left) \ => \ y = |x + 3|`

`text(Reflect in the)\ xtext(-axis) \ => \ y = −|x + 3|`

`text(Dilate by)\ 1/3\ text(from the)\ x text(-axis)`

`=>\ text(Multiply by)\ 1/3 \ => \ y = −1/3|x + 3|`

`text(Translate 2 units up) \ => \ y = −1/3 |x + 3| + 2`

`:. k = −1/3, b = 3, c = 2`

Functions, 2ADV F1 EQ-Bank 6

The graph of `f(x)` is shown below. It has asymptotes at `y = 3` and `x = 2`.

Using interval notation, state the domain and range of `f(x)`. (2 marks)

Show Answers Only

`text(Domain:)\ [−4, 2) ∪ (2, ∞)`

`text(Range:)\ (−∞, ∞)`

Show Worked Solution

`text(Domain:)\ [−4, 2) ∪ (2, ∞)`

`text(Range:)\ (−∞, ∞)`

Statistics, 2ADV S3 EQ-Bank 1

A probability density function can be used to model the lifespan of a termite, `X`, in weeks, is given by

`f(x) = {(k(36 - x^2)),(0):}\ \ \ {:(3 <= x <= 6),(text(otherwise)):}`

Show that the value of `k` is `1/45`. (2 marks)

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Find the cumulative distribution function. (2 marks)

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Find the probability that a termite's lifespan is greater than 5 weeks. (1 mark)

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`text(See Worked Solutions)`
`F(x) = {(0),(1/135 (108t – t^3 – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`
`17/135`

Show Worked Solution

i.	`k int_3^6 36 – x^2\ dx`	`= 1`
	`k[36x – (x^3)/3]_3^6`	`= 1`
	`k[(216 – 72)-(108 – 9)]`	`= 1`
	`45k`	`= 1`
	`k`	`= 1/45`

ii.	`F(t)`	`= int_(-∞)^t f(x)\ dx`
		`= int_3^t f(x)\ dx`
		`= 1/45 int_3^t 36 – x^2\ dx`
		`= 1/45 [36x – (x^3)/3]_3^t`
		`= 1/135[108x – x^3]_3^t`
		`= 1/35[(108t – t^3) – (324 – 27)]`
		`= 1/135(108t – t^3 – 297)`

`:. F(x) = {(0),(1/135 (108t – t^3 – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`

iii.	`P(X > 5)`	`= 1 – F(5)`
		`= 1 – 1/135(108 xx 5 – 5^3 – 297)`
		`= 1 – 118/135`
		`= 17/135`

Calculus, 2ADV C4 EQ-Bank 2

The population, `D`, of Tasmanian Devils in a sanctuary is given by `D(t)`, where `t` is the time in years after the sanctuary was established.

The devil population changes at a rate modelled by the function `(dD)/(dt) = 28 e^(0.35t)`.

Calculate the increase in the number of Tasmanian Devils at the end of the first 8 years. Give your answer correct to three significant figures. (3 marks)

Show Answers Only

`1240 \ text((to 3 sig. fig.))`

Show Worked Solution

`int_0^8 28e^(0.35t)`	`= [28 xx (1)/(0.35) e^(0.35t)]_0^8`
	`= 80(e^(0.35 xx 8) – e°)`
	`= 80(16.44 … – 1)`
	`= 1235.57 …`
	`= 1240 \ text((to 3 sig. fig.))`

Functions, 2ADV F1 EQ-Bank 11

Given the function `f(x) = sqrt(3 - x)` and `g(x) = x^2 - 2`, sketch `y = g(f(x))` over its natural domain. (2 marks)

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Show Worked Solution

`g(x) = x^2 – 2,\ \ f(x) = sqrt(3 – x)`

`g(f(x))`	`= (sqrt(3 – x))^2 – 2`
	`= 3 – x – 2`
	`= 1 – x`

`text(S)text(ince)\ \ f(x) = sqrt(3 – x),`

`=> text(Domain:)\ x <= 3`

Functions, 2ADV F1 EQ-Bank 12

Two archers play a game where each can aim for a large target or a small target.

If an arrow hits the large target it scores `L` points, and if it hits the small target, it scores `S` points.

The results of a game are shown in the table below.

By forming a pair of simultaneous equations, or otherwise, find the value of `L` and `S`. (3 marks)

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`L = 3, S = 7`

Show Worked Solution

`5L + 8S = 71\ \ …\ (1)`

`12L + 5S = 71\ \ …\ (2)`

`text(Multiply (1)) xx 5`

`25L + 40S = 355\ \ …\ (3)`

`text(Multiply (2)) xx 8`

`96L + 40S = 568\ \ …\ (4)`

`text{Subtract (4) − (3)}`

`71L`	`= 213`
`:. L`	`= 3`

`text(Substitute)\ \ L = 3\ \ text(into (1))`

`8S`	`= 56`
`:. S`	`= 7`

Probability, 2ADV S1 EQ-Bank 42

The discrete random variable `X` has the probability distribution shown in the table below.

	`X = x`	`4`	`5`	`6`	`7`	`8`
	`P(x)`	`0.3`	`a`	`0.1`	`0.15`	`0.2`

Find the value of `a`, and hence calculate the the expected value and variance of `X`. (3 marks)

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`2.31`

Show Worked Solution

`0.3 + a + 0.1 + 0.15 + 0.2 = 1`

`=> \ a = 0.25`

`E(X) = ∑ x P(x)`

`qquad X = x`	`4`	`5`	`6`	`7`	`8`
`qquad P(x) qquad`	`0.3`	`0.25`	`0.1`	`0.15`	`0.2`
`qquad x xx P(x) qquad`	`1.2`	`1.25`	`0.6`	`1.05`	`1.6`

`E(X)`	`= 1.2 + 1.25 + 0.6 + 1.05 + 1.6`
	`= 5.7`

`text(Var)(X)`	`= E(X^2) – [E(X)]^2`
	`= (4^2 xx 0.3) + (5^2 xx 0.25) + (6^2 xx 0.1) + (7^2 xx 0.15) + (8^2 xx 0.2) – 5.7^2`
	`= 2.31`

Statistics, 2ADV S3 EQ-Bank 6 MC

A continuous probability density function graph is drawn below.

Which of the following is the mode?

0.42
0.08
1.5
5

Show Answers Only

`C`

Show Worked Solution

`text(Mode is the most common)\ xtext(-value.)`

`ytext(-axis measures the probability.)`

`text(Highest probability = 0.42)`

`:.\ text(Mode = 1.5)`

`=>\ C`

Calculus, 2ADV C2 EQ-Bank 4 MC

If `f(x)=log_2(x^(2x))`, which expression is equal to `f^(′)(x)`?

`2/(x^(2x)ln2`
`2/ln2 + 2log_2x`
`log_2x+2/ln2`
`2/ln2 xx log_2(x^(2x-1))`

Show Answers Only

`B`

Show Worked Solution

`f(x)`	`=log_2(x^(2x))`
	`=2x log_2x`
	`=(2x lnx)/ln2`

`f^(′)(x)`	`=1/ln2 (2x*1/x + 2lnx)`
	`=2/ln2 + (2lnx)/ln2`
	`=2/ln2 + 2log_2x`

`=> B`

NETWORKS, FUR2-NHT 2019 VCAA 3

The zoo’s management requests quotes for parts of the new building works.

Four businesses each submit quotes for four different tasks.

Each business will be offered only one task.

The quoted cost, in $100 000, of providing the work is shown in Table 1 below.

The zoo’s management wants to complete the new building works at minimum cost.

The Hungarian algorithm is used to determine the allocation of tasks to businesses.

The first step of the Hungarian algorithm involves row reduction; that is, subtracting the smallest element in each row of Table 1 from each of the elements in that row.

The result of the first step is shown in Table 2 below.

The second step of the Hungarian algorithm involves column reduction; that is, subtracting the smallest element in each column of Table 2 from each of the elements in that column.

The results of the second step of the Hungarian algorithm are shown in Table 3 below. The values of Task 1 are given as `A, B, C` and `D`.

Write down the values of `A, B, C` and `D`. (1 mark)

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The next step of the Hungarian algorithm involves covering all the zero elements with horizontal or vertical lines. The minimum number of lines required to cover the zeros is three.

Draw these three lines on Table 3 above. (1 mark)

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An allocation for minimum cost is not yet possible.

When all steps of the Hungarian algorithm are complete, a bipartite graph can show the allocation for minimum cost.

Complete the bipartite graph below to show this allocation for minimum cost. (1 mark)

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Business 4 has changed its quote for the construction of the pathways. The new cost is $1 000 000. The overall minimum cost of the building works is now reduced by reallocating the tasks.

How much is this reduction? (1 mark)

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`A = 2, B = 1, C = 1, D = 0`
`text(See Worked Solutions)`
`text(See Worked Solutions)`
`text(See Worked Solutions)`

Show Worked Solution

a. `A = 2, \ B = 1, \ C = 1, \ D = 0`

c. `text(After next step:)`

`text(Allocation): `

d. `text{Hungarian Algorithm table (complete):}`

`text(Allocation): B2 -> T3,\ B1 -> T4,\ B3 -> T2,\ B4 -> T1`

`text(C)text(ost) = 4 + 7 + 8 + 10 = 29`

`text(Previous cost) = 5 + 10 + 8 + 8 = 31`

`:.\ text(Reduction)`	`= 2 xx 100\ 000`
	`= $200\ 000`

GEOMETRY, FUR1-NHT 2019 VCAA 6 MC

A cake in the shape of three cylindrical sections is shown in the diagram below.

Each section of the cake has a height of 8 cm, as shown in the diagram.

The middle section of the cake, B, has twice the volume of the top section of the cake, A.

The bottom section of the cake, C, has twice the volume of the middle section of the cake, B.

The volume of the top section of the cake, A, is 900 cm³.

The diameter of the bottom section of the cake, C, in centimetres, is closest to

Show Answers Only

`C`

Show Worked Solution

`text(Volume) \ C`	`= 900 xx 4`
	`= 3600 \ text(cm)^3`

`pi xx r_C^2 xx h`	`= 3600`
`r_C^2`	`= (3600)/(8 pi)`
`r_C^2`	`= 11.968 \ …`

`:. \ text(Diameter of)\ C ≈ 24\ text(cm)`

`=> \ C`

GEOMETRY, FUR1-NHT 2019 VCAA 5 MC

The cities of Lima and Washington, DC have the same longitude of 77° W.

The shortest great circle distance between Lima and Washington, DC is 5697 km.

Assume that the radius of Earth is 6400 km.

Lima has a latitude of 12° S and is located due south of Washington, DC.

What is the latitude of Washington, DC?

39° N
51° S
51° N
63° N
65° S

Show Answers Only

`A`

Show Worked Solution

`text(Arc Distance) \ WL`	`= 5697`
`({theta + 12}/{360}) xx 2 xx pi xx 6400`	`= 5697`
`theta + 12`	`= (5697 xx 360)/(2 pi xx 6400)`
	`= 51°`
`:. \ theta`	`= 39°`

`=> \ A`

GEOMETRY, FUR1-NHT 2019 VCAA 3 MC

A waterfall in a national park is 4 km east of a camp site.

A lookout tower is 4 km south of the waterfall.

The bearing of the camp site from the lookout tower is

045°
090°
135°
300°
315°

Show Answers Only

`E`

Show Worked Solution

`text(Bearing of) \ C \ text(from) \ T`	`= 360 – 45`
	`= 315°`

`=> \ E`

Networks, STD2 N3 2019 FUR2-N 2

The construction of the new reptile exhibit is a project involving nine activities, `A` to `I`.

The directed network below shows these activities and their completion times in weeks.

Which activities have more than one immediate predecessor? (1 mark)

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Write down the critical path for this project. (1 mark)

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What is the latest start time, in weeks, for activity `B`? (1 mark)

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`D, G and I`
`text(See Worked Solutions)`
`2\ text(weeks)`

Show Worked Solution

a. `D, G and I`

b. `text(Scanning forwards and backwards:)`

`text(Critical Path:)\ ACDFGI`

c.	`text{LST (activity}\ B text{)}`	`= 7 – 5`
		`= 2\ text(weeks)`

NETWORKS, FUR2-NHT 2019 VCAA 1

A zoo has an entrance, a cafe and nine animal exhibits: bears `(B)`, elephants `(E)`, giraffes `(G)`, lions `(L)`, monkeys `(M)`, penguins `(P)`, seals `(S)`, tigers `(T)` and zebras `(Z)`.

The edges on the graph below represent the paths between the entrance, the cafe and the animal exhibits. The numbers on each edge represent the length, in metres, along that path. Visitors to the zoo can use only these paths to travel around the zoo.

What is the shortest distance, in metres, between the entrance and the seal exhibit `(S)`? (1 mark)

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Freddy is a visitor to the zoo. He wishes to visit the cafe and each animal exhibit just once, starting and ending at the entrance.
i. What is the mathematical term used to describe this route? (1 mark)

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ii. Draw one possible route that Freddy may take on the graph below. (1 mark)

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A reptile exhibit `(R)` will be added to the zoo.

A new path of length 20 m will be built between the reptile exhibit `(R)` and the giraffe exhibit `(G)`.

A second new path, of length 35 m, will be built between the reptile exhibit `(R)` and the cafe.

Complete the graph below with the new reptile exhibit and the two new paths added. Label the new vertex `R` and write the distances on the new edges. (1 mark)

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The new paths reduce the minimum distance that visitors have to walk between the giraffe exhibit `(G)` and the cafe.

By how many metres will these new paths reduce the minimum distance between the giraffe exhibit `(G)` and the cafe? (1 mark)

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Show Answers Only

`45\ text(metres)`
i. `text(Hamilton cycle.)`
ii. `text(See Worked Solutions)`
`text(See Worked Solutions)`
`85\ text(metres)`

Show Worked Solution

a. `45\ text(metres)`

b.i. `text(Hamilton cycle.)`

b.ii.

`text(Possible route:)`

`text(entrance)\ – LGTMCEBSZP\ –\ text(entrance)`

d. `text{Minimum distance (before new exhibit)}`

`= GLTMC`

`= 15 + 35 + 40 + 50`

`= 140\ text(m)`

`:.\ text(Reduction in minimum distance)`

`= 140 – (20 + 35)`

`= 85\ text(m)`

MATRICES, FUR2-NHT 2019 VCAA 4

After 5.00 pm, tourists will start to arrive in Gillen and they will stay overnight.

As a result, the number of people in Gillen will increase and the television viewing habits of the tourists will also be monitored.

Assume that 50 tourists arrive every hour.

It is expected that 80% of arriving tourists will watch only `C_2` during the hour that they arrive.

The remaining 20% of arriving tourists will not watch television during the hour that they arrive.

Let `W_m` be the state matrix that shows the number of people in each category `m` hours after 5.00 pm on this day.

The recurrence relation that models the change in the television viewing habits of this increasing number of people in Gillen `m` hours after 5.00 pm on this day is shown below.

`W_(m + 1) = TW_m + V`

where

`{:(quad qquad qquad qquadqquadqquadquadtext(this hour)),(qquadqquadqquad quad \ C_1 qquad quad C_2 qquad \ C_3 quad \ NoTV),(T = [(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)]{:(C_1),(C_2),(C_3),(NoTV):}\ text(next hour,) qquad and qquad W_0 = [(400), (600), (300),(700)]):}`

Write down matrix `V`. (1 mark)

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How many people in Gillen are expected to watch `C_2` at 7.00 pm on this day? (2 marks)

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Show Answers Only

`V=[(0),(40),(0),(10)]`
`666`

Show Worked Solution

a. `V=[(0),(40),(0),(10)]`

b.	`W_1`	`=TW_0+V`
		`=[(quad 0.50, 0.05, 0.10, 0.20 quad),(quad 0.10, 0.60, 0.20, 0.20 quad),(quad 0.25, 0.10, 0.50, 0.10 quad),(quad 0.15, 0.25, 0.20, 0.50 quad)] [(400),(600),(300),(700)]+[(0),(40),(0),(10)]=[(400),(640),(380),(630)]`

	`W_2`	`=TW_1+V=[(396),(666),(417),(621)]`

`:.\ text(666 people are expected to watch)\ C_2\ text(at 7 pm.)`

MATRICES, FUR2-NHT 2019 VCAA 3

The basketball finals will be televised on `C_3` from 12.00 noon until 4.00 pm.

It is expected that 600 Gillen residents will be watching `C_3` at any time from 12.00 noon until 4.00 pm.

The remaining 1400 Gillen residents will not be watching `C_3` from 12.00 noon until 4.00 pm (represented by NotC₃).

`{:(qquadqquadqquadquadtext(this hour)),(qquadqquadqquad \ C_3 quadquad \ NotC_3),(P = [(v, qquad quad w quad),(0.35, quad qquad x quad)]{:(C_3),(NotC_3):}\ text(next hour)):}`

Write down the values of `v, w` and `x` in the boxes provided below. (2 marks)

`v =`

`w =`

`x =`

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Show Answers Only

`v = 0.65, \ w = 0.15, \ x = 0.85`

Show Worked Solution

`v = 1-0.35 = 0.65`

`text(Residents who change from)\ C_3\ text(to Not)C_3`

`= 0.35 xx 600`

`= 210`

`text(S) text(ince 600 residents are watching)\ C_3\ text(at any time)`

`w xx 1400`	`= 210`
`w`	`= 0.15`
`:. x`	`= 1-0.15`
	`= 0.85`

MATRICES, FUR2-NHT 2019 VCAA 2

Three television channels, `C_1, C_2` and `C_3`, will broadcast the International Games in Gillen.

Gillen’s 2000 residents are expected to change television channels from hour to hour as shown in the transition matrix `T` below.

The option for residents not to watch television (NoTV) at that time is also indicated in the transition matrix.

`{:(qquadqquadqquadqquadqquadquadtext(this hour)),(qquadqquadqquad \ C_1 quadquad \ C_2 quadqquad C_3 quad \ NoTV),(T = [(0.50,0.05,0.10,0.20),(0.10,0.60,0.20,0.20),(0.25,0.10,0.50,0.10),(0.15,0.25,0.20,0.50)]{:(C_1),(C_2),(C_3),(NoTV):}text(next hour)):}`

The state matrix `G_0` below lists the number of Gillen residents who are expected to watch the games on each of the channels at the start of a particular day (9.00 am).

Also shown is the number of Gillen residents who are not expected to watch television at that time.

`G_0 = [(100), (400), (100), (1400)]{:(C_1), (C_2), (C_3), (NoTV):}`

Complete the calculation below to show that 835 Gillen residents are not expected to watch television (NoTV) at 10.00 am that day. (1 mark)

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	`xx 100 +`		`xx 400 +`		`xx 100 `
	`+\ \ `		`xx 1400 `	`= 835`

Determine the number of residents expected to watch the games on `C_3` at 11.00 am that day. (1 mark)

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Show Answers Only

`0.15 xx 100 + 0.25 xx 400 + 0.20 xx 100 + 0.50 xx 1400 = 835`
`356`

Show Worked Solution

a. `0.15 xx 100 + 0.25 xx 400 + 0.20 xx 100 + 0.50 xx 1400 = 835`

b. `G_1 text{(11:00 am)} = TG_0 = [(400),(584),(356),(660)]`

`:. 356\ text(residents are expected to watch)\ C_3\ text(at 11:00 am).`

MATRICES, FUR2-NHT 2019 VCAA 1

A total of six residents from two towns will be competing at the International Games.

Matrix `A`, shown below, contains the number of male `(M)` and the number of female `(F)` athletes competing from the towns of Gillen `(G)` and Haldaw `(H)`.

`{:(qquad qquad quad \ M quad F), (A = [(2, 2), (1, 1)]{:(G),(H):}):}`

How many of these athletes are residents of Haldaw? (1 mark)

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Each of the six athletes will compete in one event: table tennis, running or basketball.

Matrices `T` and `R`, shown below, contain the number of male and female athletes from each town who will compete in table tennis and running respectively.

	Table tennis	Running
	`{:(qquad qquad quad \ M quad F), (T = [(0, 1), (1, 0)]{:(G),(H):}):}`	`{:(qquad qquad quad \ M quad F), (R = [(1, 1), (0, 0)]{:(G),(H):}):}`

Matrix `B` contains the number of male and female athletes from each town who will compete in basketball.

Complete matrix `B` below. (1 mark)

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`{:(qquad qquad qquad \ M qquad quad F), (B = [(\ text{___}, text{___}\ ), (\ text{___}, text{___}\)]{:(G),(H):}):}`

Matrix `C` contains the cost of one uniform, in dollars, for each of the three events: table tennis `(T)`, running `(R)` and basketball `(B)`.

`C = [(515), (550), (580)]{:(T), (R), (B):}`

1. For which event will the total cost of uniforms for the athletes be $1030? (1 mark)
  
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2. Write a matrix calculation, that includes matrix `C`, to show that the total cost of uniforms for the event named in part c.i. is contained in the matrix answer of [1030]. (1 mark)
  
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Matrix `V` and matrix `Q` are two new matrices where `V = Q xx C` and:

matrix `Q` is a `4 xx 3` matrix
element `v_11 =` total cost of uniforms for all female athletes from Gillen
element `v_21 =` total cost of uniforms for all female athletes from Haldaw
element `v_31 =` total cost of uniforms for all male athletes from Gillen
element `v_41 =` total cost of uniforms for all male athletes from Haldaw
`C = [(515), (550), (580)]{:(T), (R), (B):}`

Complete matrix `Q` with the missing values. (1 mark)

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`Q = [(1, text{___}, text{___}\ ), (0, 0, 1), (0, 1, 1), (\ text{___}, text{___}, 0)]`

Show Answers Only

`2`
`B = [(1, 0), (0, 1)]`
i. `text(Table tennis)`
ii. `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`
`Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`

Show Worked Solution

a. `2`

b. `B = [(1, 0), (0, 1)]`

c.i. `text(Table tennis)`

c.ii. `[2\ \ \ 0\ \ \ 0] xx [(515), (550), (580)] = [1030]`

d. `Q = [(1, \ 1, \ 0),(0, \ 0, \ 1),(0, \ 1, \ 1),(1, \ 0, \ 0)]`

Calculus, 2ADV C4 2019 MET1-N 3

Evaluate `int_0^1 e^x - e^-x\ dx`. (2 marks)

Show Answers Only

`e + (1)/(e) – 2`

Show Worked Solution

`int_0^1 e^x – e^-x\ dx`	`= [e^x + e^-x]_0^1`
	`= [e + (1)/(e) – (1 + 1)]`
	`= e + (1)/(e) – 2`

Functions, 2ADV F1 2019 MET1-N 2

Let `f(x) = -x^2 + x + 4` and `g(x) = x^2 - 2`.

Find `g(f(3))`. (2 marks)

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Express `f(g(x))` in the form `ax^4 + bx^2 + c`, where `a`, `b` and `c` are non-zero integers. (2 marks)

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Show Answers Only

`2`
`-x^4 + 5x^2 – 2`

Show Worked Solution

a. `f(3)`	`= -3^2 + 3 + 4`
	`= -2`

`g(f(3))`	`= g(-2)`
	`= (-2)^2 – 2`
	`= 2`

b. `f(g(x))`	`= -(x^2 – 2)^2 + (x^2 – 2) + 4`
	`= -(x^4 – 4x^2 + 4) + x^2 + 2`
	`= -x^4 + 5x^2 – 2`

Statistics, MET1-NHT 2018 VCAA 8

Let `overset^p` be the random variable that represents the sample proportions of customers who bring their own shopping bags to a large shopping centre.

From a sample consisting of all customers on a particular day, an approximate 95% confidence interval for the proportion `p` of customers who bring their own shopping bags to this large shopping centre was determined to be `((4853)/(50\ 000) , (5147)/(50\ 000))`.

Find the value of `hatp` that was used to obtain this approximate 95% confidence interval. (1 mark)

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Use the fact that `1.96 = (49)/(25)` to find the size of the sample from which this approximate 95% confidence interval was obtained. (2 marks)

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Show Answers Only

`(1)/(10)`
`40\ 000`

Show Worked Solution

a. `overset^p – z sqrt((overset^p(1 – overset^p))/(n)) = (4853)/(50\ 000) \ \ , \ \ overset^p + z sqrt((overset^p(1 – overset^p))/(n)) = (5147)/(50\ 000)`

`2 overset^p`	`= (4853)/(50\ 000) + (5147)/(50\ 000)`
`:. \ overset^p`	`=1/10`

b. `(1)/(10) + (49)/(25) sqrt(({1)/{10}(1 – {1}/{10}))/(n)`	`= (5147)/(50000)`
`(49)/(25) sqrt((9)/(100n))`	`= (147)/(50000)`
`(49)/(25) * (3)/(10) * (1)/(sqrtn)`	`= (147)/(25 xx 2000)`
`(147)/(sqrtn)`	`= (147)/(200)`
`sqrtn`	`= 200`
`:. \ n`	`= 40\ 000`

Calculus, MET1-NHT 2018 VCAA 7

Let `f : [ 0, (pi)/(2)] → R, \ f(x) = 4 cos(x)` and `g : [0, (pi)/(2)] → R, \ g(x) = 3 sin(x)`.

Sketch the graph of `f` and the graph of `g` on the axes provided below. (2 marks)

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`qquad qquad `
Let `c` be such that `f(c) = g(c)`, where `c∈[0, (pi)/(2)]`

Find the value of `sin(c)` and the value of `cos(c)`. (3 marks)

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Let `A` be the region enclosed by the horizontal axis, the graph of `f` and the graph of `g`.

1. Shade the region `A` on the axes provided in part a. and also label the position of `c` on the horizontal axis. (1 mark)
  
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2. Calculate the area of the region `A`. (3 marks)
  
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Show Answers Only

`sin(c) = (4)/(5), \ cos(c) = (3)/(5)`
i.

ii. `2 \ u^2`

Show Worked Solution

b. `text(At intersection:)`

`4cos(c)`	`= 3sin(c)`
`tan(c)`	`= (4)/(3)`

`sin(c) = (4)/(5)`

`cos(c) = (3)/(5)`

c. i.

ii. `A`	`= int_0^c g(x)\ dx + int_c^((pi)/(2)) f(x)\ dx`
	`= int_0^c 3sin x \ dx + int_c^((pi)/(2)) 4cos \ x \ dx`
	`= 3[-cos x]_0^c + 4[sin x]_c^((pi)/(2))`
	`= 3(-cos(c) + cos \ 0) + 4(sin \ (pi)/(2)-sin(c))`
	`= 3(-(3)/(5) + 1) + 4(1-(4)/(5))`
	`= (6)/(5) + (4)/(5)`
	`= 2 \ \ text(u²)`

Statistics, MET1-NHT 2018 VCAA 6

The discrete random variable `X` has the probability mass function

`text(Pr)(X = x) = {(kx), (k), (0):} qquad {:(x∈{1, 4, 6}), (x = 3), (text(otherwise)):}`

Show that `k = (1)/(12)`. (2 marks)

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Find `text(E)(X)`. (1 mark)

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Evaluate `text(Pr)(X ≥ 3 | X ≥ 2)`. (1 mark)

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Show Answers Only

`text(Proof (See Worked Solution))`
`(14)/(3)`
`1`

Show Worked Solution

a.	`qquad X`	`qquad 1 qquad`	`qquad 3 qquad`	`qquad 4 qquad`	`qquad 6 qquad`
	`qquad Pr(X = x) qquad`	`k`	`k`	`4k`	`6k`

`k + k+4k + 6k`	`= 1`
`12k`	`= 1`
`k`	`= (1)/(12)`

b. `E(X)`	`= ∑ x text(Pr)(X = x)`
	`= k + 3k + 16k + 36k`
	`= 56k`
	`= (56)/(12)`
	`= (14)/(3)`

c. `text(Pr) (X ≥ 3 \| X ≥ 2)`	`= (text(Pr) (X ≥ 3 ∩ X ≥ 2))/(text(Pr) (X ≥ 2))`
	`= (text(Pr) (X ≥ 3))/(text(Pr) (X ≥ 2))`
	`= (k + 4k + 6k)/(k + 4k + 6k)`
	`= 1`

Algebra, MET1-NHT 2018 VCAA 4

Solve `log_3(t) - log_3(t^2 - 4) = -1` for `t`. (3 marks)

Show Answers Only

`4 `

Show Worked Solution

`log_3(t) – log_3(t^2 – 4)`	`= -1`
`log_3 ({t}/{t^2 – 4})`	`= -1`
`(t)/(t^2 – 4)`	`= (1)/(3)`
`t^2 – 4`	`= 3t`
`t^2 – 3t – 4`	`= 0`
`(t – 4)(t+ 1)`	`= 0`

`:. t=4 \ \ \ (t > 0, \ t!= –1)`

Calculus, MET1-NHT 2018 VCAA 3

Evaluate `int_0^1 e^x-e^-x\ dx`. (2 marks)

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Show Answers Only

`e + (1)/(e)-2`

Show Worked Solution

`int_0^1 e^x-e^-x\ dx`	`= [e^x + e^-x]_0^1`
	`= [e + (1)/(e)-(1 + 1)]`
	`= e + (1)/(e)-2`

Functions, MET1-NHT 2018 VCAA 2

Let `f(x) = -x^2 + x + 4` and `g(x) = x^2-2`.

Find `g(f(3))`. (2 marks)

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Show Answers Only

`2`
`-x^4 + 5x^2-2`

Show Worked Solution

a. `f(3)`	`= -3^2 + 3 + 4`
	`= -2`

`g(f(3))`	`= g(-2)`
	`= (-2)^2-2`
	`= 2`

b. `f(g(x))`	`= -(x^2-2)^2 + (x^2-2) + 4`
	`= -(x^4-4x^2 + 4) + x^2 + 2`
	`= -x^4 + 5x^2-2`

CORE, FUR1-NHT 2019 VCAA 20 MC

Marty has been depreciating the value of his car each year using flat rate depreciation.

After three years of ownership, the value of the car was halved due to an accident.

Marty continued to depreciate the value of his car by the same amount each year after the accident.

Which one of the following graphs could show the value of Marty’s car after `n` years, `C_n`?

A.		B.
C.		D.
E.

Show Answers Only

`C`

Show Worked Solution

`text(Flat rate depression)`

`=>\ text{graph decreases in a straight line.}`

`text(Value of the car halves between)\ C_3\ text(and)\ C_4\ \ text(and then)`

`text(continues with the same gradient.)`

`=>\ C`

Vectors, EXT1 V1 EQ-Bank 6

Point `C` lies on `AB` such that `overset(->)(AC) = lambdaoverset(->)(AB)`.

Express `underset~c` in terms of `underset~a` and `underset~b`. (2 marks)

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Hence or otherwise, show that `overset(->)(BC) = (1-lambda)(underset~a-underset~b)`. (1 mark)

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Show Answers Only

`lambdaunderset~b + (1-lambda)underset~a`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

`underset~c = underset~a + overset(->)(AC)`

`overset(->)(AB)`	`= underset~b-underset~a`
`overset(->)(AC)`	`= lambdaoverset(->)(AB)`
	`= lamda(underset~b-underset~a)`

`:.underset~c`	`= underset~a + lambda(underset~b-underset~a)`
	`= lambdaunderset~b + (1-lambda)underset~a`

ii.	`overset(->)(BC)`	`= underset~c-underset~b`
		`= lambdaunderset~b + (1-lambda)underset~a-underset~b`
		`=(1-lambda)underset~a +(lambda-1)underset~b`
		`=(1-lambda)underset~a -(1-lambda)underset~b`
		`= (1-lambda)(underset~a-underset~b)`

Vectors, EXT1 V1 EQ-Bank 5

Using vectors, calculate the acute angle between the line that passes through `A(1, 3)` and `B(2,–6)` and the line that passes through `C(1, 5)` and `D(3,–2)`.

Give your answer correct to one decimal place. (2 marks)

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Show Answers Only

`9.6°`

Show Worked Solution

`underset~a = ((1),(3)), \ underset~b = ((2),(−6)), \ underset~c = ((1),(5)), \ underset~d = ((3),(−2))`

`overset(->)(AB)`	`= underset~b – underset~a = ((2),(−6)) – ((1),(3)) = ((1),(−9))`
`overset(->)(CD)`	`= underset~d – underset~c = ((3),(−2)) – ((1),(5)) = ((2),(−7))`

`costheta`	`= (overset(->)(AB) · overset(->)(CD))/(\|overset(->)(AB)\| · \|overset(->)(CD)\|)`
	`= (2 + 63)/(sqrt82 · sqrt53)`
	`= 0.985…`

`:. theta`	`=cos^(-1) 0.985…`
	`= 9.605…`
	`= 9.6°\ \ (text(to 1 d.p.))`

Vectors, EXT1 V1 EQ-Bank 3

A force described by the vector `underset~F = ((3),(6))` newtons is applied to a line `l` which is parallel to the vector `((4),(3))`.

Find the component of the force `underset~F` in the direction of `l`. (2 marks)

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What is the component of the force `underset~F` in the direction perpendicular to the line? (1 mark)

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Show Answers Only

`((4.8),(3.6))`
`((−1.8),(2.4))`

Show Worked Solution

i. `underset~F = ((3),(6)), \ underset~v = ((4),(3))`

`underset~overset^v = underset~v/(|underset~v|) = underset~v/sqrt(4^2 + 3^2) = 1/5 underset~v=((0.8),(0.6))`

`underset~ F · underset~overset^v`	`= ((3),(6))((0.8),(0.6))`
	`= 3 xx 0.8 + 6 xx 0.6`
	`= 6`

`text(proj)_(underset~v) underset~F`	`= (underset~F · underset~overset^v) underset~overset^v`
	`= 6((0.8),(0.6))`
	`= ((4.8),(3.6))`

ii. `text(Component of)\ underset~F ⊥ l`

`= ((3),(6)) – ((4.8),(3.6))`

`= ((−1.8),(2.4))`

Vectors, EXT1 V1 EQ-Bank 2

In the diagram below, `ROT` is a diameter of the circle with centre `O`.

`S` is a point on the circumference.

Using the properties of vectors `underset~r`, `underset~s` and `underset~t`, show that `angleRST` is a right angle. (2 marks)

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Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`\|underset~r\|`	`= \|underset~s\| = \|underset~t\|\ \ \ text{(radii)}`
`underset~r`	`= −underset~t`

`overset(->)(RS)`	`= underset~s-underset~r`
`overset(->)(TS)`	`= underset~s-underset~t`

`overset(->)(RS) · overset(->)(TS)`	`= (underset~s-underset~r) · (underset~s-underset~t)`
	`= (underset~s-underset~r) · (underset~s + underset~r)\ \ \ \ (text{using}\ \ underset~r = – underset~t)`
	`= underset~s · (underset~s + underset~r)-underset~r · (underset~s + underset~r)`
	`= underset~s · underset~s + underset~s · underset~r-underset~r · underset~s-underset~r · underset~r`
	`= \|underset~s\|^2-\|underset~r\|^2`
	`= 0`

`:. overset(->)(RS) ⊥ overset(->)(TS)`

`:. angleRST\ \ text(is a right angle.)`

CORE, FUR2-NHT 2019 VCAA 7

Tisha plays drums in the same band as Marlon.

She would like to buy a new drum kit and has saved $2500.

Tisha could invest this money in an account that pays interest compounding monthly.

The balance of this investment after `n` months, `T_n` could be determined using the recurrence relation below

`T_0 = 2500, \ \ \ \ T_(n+1) = 1.0036 xx T_n`

Calculate the total interest that would be earned by Tisha's investment in the first five months.

Round your answer to the nearest cent. (2 marks)

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Tisha could invest the $2500 in a different account that pays interest at the rate of 4.08% per annum, compounding monthly. She would make a payment of $150 into this account every month.

Let `V_n` be the value of Tisha's investment after `n` months.

Write down a recurrence relation, in terms of `V_0`, `V_n` and `V_(n + 1)`, that would model the change in the value of this investment. (1 mark)

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Tisha would like to have a balance of $4500, to the nearest dollar, after 12 months.

What annual interest rate would Tisha require?

Round your answer to two decimal places. (1 mark)

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Show Answers Only

`$45.33`
`V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`
`5.87%`

Show Worked Solution

a. `T_1 = 1.0036 xx 2500 = 2509`

`T_2 = 1.0036 xx 2509 = 2518.0324`

`vdots`

`T_5 = 2545.33`

`:. \ text(Total interest) `	`= 2545.33-2500`
	`= $45.33`

b. `text(Monthly interest) = (4.08)/(12) = 0.34%`

`:. \ V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`

c. `text(By TVM Solver:)`

`N`	`= 12`
`I text(%)`	`=?`
`PV`	`=-2500`
`PMT`	`=-150`
`FV`	`=4500`
`text(PY)`	`= text(CY)=12`

`=> I = 5.87%`

CORE, FUR2-NHT 2019 VCAA 6

Marlon plays guitar in a band.

He paid $3264 for a new guitar.

The value of Marlon's guitar will be depreciated by a fixed amount for each concert that he plays.

After 25 concerts, the value of the guitar will have decreased by $200.

What will be the value of Marlon's guitar after 25 concerts? (1 mark)

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Write a calculation that shows that the value of Marlon's guitar will depreciate by $8 per concert. (1 mark)

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The value of Marlon's guitar after `n` concerts, `G_n`, can be determined using a rule.

Complete the rule below by writing the appropriate numbers in the boxes provided. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

`G_n =` – × `n`
The value of the guitar continues to be depreciated by $8 per concert.

After how many concerts will the value of Marlon's guitar first fall below $2500? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`$3064`
`text{Proof (See Worked Solution)}`
`G_n = 3264-8 xx n`
`96 \ text(concerts)`

Show Worked Solution

a. `text(Value)`	`= 3264-200`
	`= $3064`

b. `text(Depreciation per concert)`	`= (200)/(25)`
	`= $8 \ text(per concert)`

c. `G_n = 3264 – 8 xx n`

d. `text(Find) \ n \ text(when) \ \ G_n = 2500:`

`2500`	`= 3264 – 8n`
`n`	`= (3264-2500)/8`
	`= 95.5`

`:. \ text(After) \ 96 \ text(concerts, value first falls below $2500)`

CORE, FUR2-NHT 2019 VCAA 5

A random sample of 12 mammals drawn from a population of 62 types of mammals was categorized according to two variables.

likelihood of attack (1 = low, 2 = medium, 3 = high)

exposure to attack during sleep (1 = low, 2 = medium, 3 = high)

The data is shown in the following table.

Use this data to complete the two-way frequency table below. (1 mark)

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The following two-way frequency table was formed from the data generated when the entire population of 62 types of mammals was similarly categorized.

1. How many of these 62 mammals had both a high likelihood of attack and a high exposure to attack during sleep? (1 mark)
  
  --- 5 WORK AREA LINES (style=lined) ---
2. Of those mammals that had a medium likelihood of attack, what percentage also had a low exposure to attack during sleep? (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
3. Does the information in the table above support the contention that likelihood of attack is associated with exposure to attack during sleep? justify your answer by quoting appropriate percentages. It is sufficient to consider only one category of likelihood of attack when justifying your answer. (2 marks)
  
  --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

1. `15`
2. `text(Percentage) = (2)/(4) xx 100`
  `= 50 %`
3. `text(The data supports the contention that animals with a low likelihood)`
  `text(of attack is associated with low exposure to attack during sleep.)`
  
  `text(- 91%) \ ({31}/{34}) \ text(of animals with low exposure to attack)`
  `text(during sleep, have a low likelihood of attack.)`
  
  `text(- Similarly, 89% of animals with a medium exposure to attack during)`
  `text(sleep have a low likelihood of attack.)`
  
  `text(- 11% of animals with a high exposure to attack during sleep have)`
  `text(a low likelihood of attack)`

Show Worked Solution

b. i. `15`

ii. `text(Percentage)`	`= (2)/(4) xx 100`
	`= 50%`

iii. `text(The data supports the contention that animals with a low likelihood)`

`text(of attack is associated with low exposure to attack during sleep.)`

`text(- 91%) \ ({31}/{34}) \ text(of animals with low exposure to attack)`

`text(during sleep, have a low likelihood of attack.)`

`text(- Similarly, 89% of animals with a medium exposure to attack during)`

`text(sleep have a low likelihood of attack.)`

`text(- 11% of animals with a high exposure to attack during sleep have)`

`text(a low likelihood of attack)`

CORE, FUR2-NHT 2019 VCAA 4

The scatterplot below plots the variable life span, in years, against the variable sleep time, in hours, for a sample of 19 types of mammals.

On the assumption that the association between sleep time and life span is linear, a least squares line is fitted to this data with sleep time as the explanatory variable.

The equation of this least squares line is

life span = 42.1 – 1.90 × sleep time

The coefficient of determination is 0.416

Draw the graph of the least squares line on the scatterplot above. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Describe the linear association between life span and sleep time in terms of strength and direction. (2 marks)

--- 2 WORK AREA LINES (style=lined) ---
Interpret the slope of the least squares line in terms of life span and sleep time. (2 marks)

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Interpret the coefficient of determination in terms of life span and sleep time. (1 mark)

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The life of the mammal with a sleep time of 12 hours is 39.2 years.
Show that, when the least squares line is used to predict the life span of this mammal, the residual is 19.9 years. (2 marks)

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Show Answers Only

`text(Strength is moderate.)`

`text(Direction is negative.)`
`text(The gradient of –1.9 means that life span decreases by)`

`text(1.9 years for each additional hour of sleep time.)`
`text(41.6% of the variation in life span can be explained by the)`

`text(variation in sleep time.)`
`text(Proof(See Worked Solution))`

Show Worked Solution

a. `text{Graph endpoints (0, 42.1) and (18, 7.9)}`

b. `text(Strength is moderate.)`

`text(Direction is negative.)`

c. `text(The gradient of –1.9 means that life span decreases by)`

`text(1.9 years for each additional hour of sleep time.)`

d. `text(41.6% of the variation in life span can be explained by the )`

`text(variation in sleep time.)`

e. `text(Predicted value)`	`= 42.1 – 1.9 xx 12`
	`= 19.3 \ text(years)`

`text(Residual)`	`= text(actual) – text(predicted)`
	`= 39.2 – 19.3`
	`= 19.9 \ text(years)`

CORE, FUR2-NHT 2019 VCAA 3

The life span, in years, and gestation period, in days, for 19 types of mammals are displayed in the table below.

A least squares line that enables life span to be predicted from gestation period is fitted to this data.
Name the explanatory variable in the equation of this least squares line. (1 mark)

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Determine the equation of the least squares line in terms of the variables life span and gestation period.
Round the numbers representing the intercept and slope to three significant figures. (2 marks)

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Write the value of the correlation rounded to three decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(gestation period)`
`text(life span) = 7.58 + 0.101 xx \ text(gestation period)`
`0.904`

Show Worked Solution

a. `text(gestation period)`

b. `text(Input data points into CAS:)`

`text(life span) = 7.58 + 0.101 xx \ text(gestation period)`

c. `r = 0.904 \ text{(by CAS)}`

CORE, FUR2-NHT 2019 VCAA 2

The five-number summary below was determined from the sleep time, in hours, of a sample of 59 types of mammals.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \textbf{Statistic} \rule[-1ex]{0pt}{0pt} & \textbf{Sleep time (hours)} \\
\hline
\rule{0pt}{2.5ex} \text{minimum} \rule[-1ex]{0pt}{0pt} & \text{2.5} \\
\hline
\rule{0pt}{2.5ex} \text{first quartile} \rule[-1ex]{0pt}{0pt} & \text{8.0} \\
\hline
\rule{0pt}{2.5ex} \text{median} \rule[-1ex]{0pt}{0pt} & \text{10.5} \\
\hline
\rule{0pt}{2.5ex} \text{third quartile} \rule[-1ex]{0pt}{0pt} & \text{13.5} \\
\hline
\rule{0pt}{2.5ex} \text{maximum} \rule[-1ex]{0pt}{0pt} & \text{20.0} \\
\hline
\end{array}

Show with calculations, that a boxplot constructed from this five-number summary will not include outliers. (2 marks)

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Construct the boxplot below. (1 mark)

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Show Answers Only

`text(Proof (See Worked Solution))`

Show Worked Solution

a. `IQR = Q_3-Q_1 = 13.5-8.0 = 5.5`

`text(Lower fence)`	`= Q_1-1.5 xx IQR`
	`= 8-1.5 xx 5.5`
	`= -0.25`

`text(Upper fence)`	`= Q_3 + 1.5 xx IQR`
	`= 13.5 + 1.5 xx 5.5`
	`= 21.75`

`text(S) text(ince) \ -0.25 < 2.5 \ text{(minimum value) and} \ 21.75 > 20.0 \ text{(maximum value)}`

`=> \ text(no outliers)`

CORE, FUR2-NHT 2019 VCAA 1

The table below displays the average sleep time, in hours, for a sample of 19 types of mammals.

Which of the two variables, type of mammal or average sleep time, is a nominal variable? (1 mark)

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Determine the mean and standard deviation of the variable average sleep time for this sample of mammals.
Round your answer to one decimal place. (1 mark)

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The average sleep time for a human is eight hours.
What percentage of this sample of mammals has an average sleep time that is less than the average sleep time for a human.
Round your answer to one decimal place. (1 mark)

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The sample is increase in size by adding in the average sleep time of the little brown bat.
Its average sleep time is 19.9 hours.
By how many many hours will the range for average sleep time increase when the average sleep time for the little brown bat is added to the sample? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(type of mammal)`
`text(mean)= 9.2 \ text(hours)`

`sigma = 4.2 \ text(hours)`
`31.6text(%)`
`5.4 \ text(hours)`

Show Worked Solution

a. `text(type of mammal is nominal)`

b. `text(mean)= 9.2 \ text(hours) \ \ text{(by CAS)}`

`sigma = 4.2 \ text(hours) \ \ text{(by CAS)}`

c. `text(Percentage)`	`= (6)/(19) xx 100`
	`= 0.3157 …`
	`= 31.6text(%)`

d. `text(Range increase)`	`= 19.9-14.5`
	`= 5.4 \ text(hours)`

Statistics, EXT1 S1 EQ-Bank 10

Four cards are placed face down on a table. The cards are made up of a Jack, Queen, King and Ace.

A gambler bets that she will choose the Queen in a random pick of one of the cards.

If this process is repeated 7 times, express the gambler's success as a Bernoulli random variable and calculate

the mean. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
the variance. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

`7/4`
`21/16`

Show Worked Solution

i. `text(Let)\ \ X = text(number of Queens chosen)`

`X\ ~\ text(Bin) (7,1/4)`

`E(X)`	`=np`
	`= 7 xx 1/4`
	`=7/4`

ii.	`text(Var)(X)`	`= np(1-p)`
		`=7/4(1-1/4)`
		`= 21/16`

GRAPHS, FUR1-NHT 2019 VCAA 6 MC

In January 2018, an online shop had 260 customer accounts.

In June 2018, the shop had 500 customer accounts.

The graph below shows the number of customer accounts, `a`, with the online shop `n` months after January 2018, for a period of 10 months.

The growth in the number of customer accounts that this graph shows is expected to continue beyond these 10 months, following the same trend.

How many customer accounts can the online shop expect to have at the end of December 2019 `(n = 23)`?

1364
1376
1388
1400
1412

Show Answers Only

`A`

Show Worked Solution

`text(Graph passes through)\ \ (0, 260) \ and \ (5, 500)`

`:. m = (500 – 260)/5 ~~ 48`

`text(Equation:)\ \ a = 260 + 48n`

`text(When)\ \ n = 23,`

`a`	`~~ 260 + 48 xx 23`
	`~~ 1364`

`=> A`

GRAPHS, FUR1-NHT 2019 VCAA 5 MC

The revenue, `R`, in dollars, that a company receives from selling `n` caps is given by the equation

`R = {(25n, qquad n <= 1000),(20n + c, qquad n> 1000):}`

The graph of this revenue equation consists of two straight lines that intersect at the point where `n = 1000`.

What is the value of `c` in this revenue equation?

0
1000
2000
4000
5000

Show Answers Only

`E`

Show Worked Solution

`text(Intersection at)\ \ n = 1000`

`25 xx 1000`	`= 20 xx 1000 + c`
`:. c`	`= 5000`

`=> E`

GRAPHS, FUR1-NHT 2019 VCAA 3 MC

The graph below shows a relationship between `x` and `y`.

The rule that represents this relationship between `x` and `y` is

`y = 2/3 x`
`y = 3/2 x`
`y = 2/3 x^2`
`y = 3/2 x^2`
`y = sqrt(3/2) x`

Show Answers Only

`C`

Show Worked Solution

`text(When)\ \ x^2 = 3,\ \ y = 2`

`text(Consider)\ C:`

`y`	`= 2/3 x^2`
`2`	`= 2/3 xx 3` ✔

`=> C`

MATRICES, FUR1-NHT 2019 VCAA 6 MC

Matrix `W` is a `3 xx 2` matrix.

Matrix `Q` is a matrix such that `Q xx W = W`.

Matrix `Q` could be

A.	`\ [1]`	B.	`\ [(1,0),(0,1)]`	C.	`\ [(1,1),(1,1),(1,1)]`
D.	`\ [(1,0,0),(0,1,0),(0,0,1)]`	E.	`\ [(1,1,1),(1,1,1),(1,1,1)]`

Show Answers Only

`D`

Show Worked Solution

`W = (3 xx 2)`

`[(1,0,0),(0,1,0),(0,0,1)][(e_11,e_12),(e_21,e_22),(e_31,e_32)] = [(e_11,e_12),(e_21,e_22),(e_31,e_32)]`

`=>\ D`

MATRICES, FUR1-NHT 2019 VCAA 4 MC

In a game between two teams, Hillside and Rovers, each team can score points in two ways.

The team may hit a Full or the team may hit a Bit.

More points are scored for hitting a Full than for hitting a Bit.

A team’s total point score is the sum of the points scored from hitting Fulls and Bits.

The table below shows the scores at the end of the game.

Let `f` be the number of points scored by hitting one Full.

Let `b` be the number of points scored by hitting one Bit.

Which one of the following matrix products can be evaluated to find the matrix `[(f),(b)]`?

A.	`\ [(4,8),(5,2)]^(−1) xx [(52),(49)]`	B.	`\ [(4,8),(5,2)] xx [(52),(49)]`	C.	`\ [(52,49)][(4,8),(5,2)]`
D.	`\ [(52,49)][(4,8),(5,2)]^(−1)`	E.	`\ [(4,8,52),(5,2,49)]^(−1)`

Show Answers Only

`A`

Show Worked Solution

`[(4,8),(5,2)][(f),(b)]`	`= [(52),(49)]`
`[(f),(b)]`	`= [(4,8),(5,2)]^(−1)[(52),(49)]`

`=>\ A`

MATRICES, FUR1-NHT 2019 VCAA 2 MC

Four teams, blue (`B`), green (`G`), orange (`O`) and pink (`P`), played each other once in a competition.

There were no draws in this competition.

The results of the competition are shown in the matrix below.

`{:(),(),(text(winner)):}{:(qquadqquad\ text(loser)),((qquadquadB,G,O,P)),({: (B), (G), (O), (P):}[(text(−),1,v,1),(0,text(−),1,1),(0,w,text(−),0),(0,0,x,text(−))]):}`

The letters `v`, `w` and `x` each have a value of 0 or 1.

A 1 in the matrix shows that the team named in that row defeated the team named in that column.

A 0 in the matrix shows that the team named in that row was defeated by the team named in that column.

A dash (–) in the matrix shows that no game was played.

The values of `v`, `w` and `x` are

`v = 0, \ w = 1, \ x = 0`
`v = 0, \ w = 1, \ x = 1`
`v = 1, \ w = 0, \ x = 1`
`v = 1, \ w = 1, \ x = 0`
`v = 1, \ w = 1, \ x = 1`

Show Answers Only

`C`

Show Worked Solution

`v = 1\ \ (B\ text(defeated)\ O)`

`w = 0\ \ (O\ text(defeated)\ G)`

`x = 1\ \ (P\ text(defeated)\ O)`

`=> C`

NETWORKS, FUR1-NHT 2019 VCAA 5 MC

In the graph below, the vertices represent electricity transformer substations.

The numbers on the edges of the graph show the length, in kilometres, of cables that connect these substations.

What is the minimum length of cable, in kilometres, that is necessary to make sure that each substation remains connected to the network?

Show Answers Only

`B`

Show Worked Solution

`text(Minimum length)`	`= 7 + 8 + 10 + 7 + 9 + 8 + 11 + 11`
	`= 71\ text(km)`

`=> B`

NETWORKS, FUR1-NHT 2019 VCAA 4 MC

Which one of the following flow diagrams shows a cut that has a capacity of 19?

A.
B.
C.
D.
E.

Show Answers Only

`E`

Show Worked Solution

`text(Consider each option):`

`text(Capacity of:)`

`text(Cut)\ A`	`= 5 + 2 + 7 + 10 + 7 = 31`
`text(Cut)\ B`	`= 5 + 2 + 7 + 7 = 21`
`text(Cut)\ C`	`= 5 + 2 + 10 = 17`
`text(Cut)\ D`	`= 5 + 2 + 7 = 14`
`text(Cut)\ E`	`= 5 + 7 + 7 = 19`

`=> E`

`f′(theta)`	`= 3 xx 2cos(2 theta) xx sin^2(2 theta)`
	`= 6 cos(2 theta)(1 – cos^2(2 theta))`
	`= 6 cos (2 theta) – 6 cos^3(2 theta)`

`f(theta)`	`= sin^3(2 theta)`
	`= (sin(2theta))^3`