Aussie Maths & Science Teachers: Save your time with SmarterEd
The directed network below shows the sequence of 11 activities that are needed to complete a project.
The time, in weeks, that it takes to complete each activity is also shown.
How many of these activities could be delayed without affecting the minimum completion time of the project?
Parcel deliveries are made between five nearby towns, `P` to `T`.
The roads connecting these five towns are shown on the graph below. The distances, in kilometres, are also shown.
A road inspector will leave from town `P` to check all the roads and return to town `P` when the inspection is complete. He will travel the minimum distance possible.
At the Zenith Post Office all computer systems are to be upgraded.
This project involves 10 activities, `A` to `J`.
The directed network below shows these activities and their completion times, in hours.
Write down the critical path. (1 mark)
Write down these two activities. (1 mark)
Complete the following sentence by filling in the boxes provided. (1 mark)
The extra activity could be represented on the network above by a directed edge from the
| end of activity |
|
to the start of activity |
|
a. `text(Longest path to)\ I:`
`B -> E -> G`
| `:.\ text(EST for)\ \ I` | `= 2 + 3 + 5` |
| `= 10\ text(hours)` |
b. `B – E – G – H – J`
c. `text(Scanning forwards and backwards:)`♦ Mean mark 45%.
`:.\ text(Activity)\ A\ text(and)\ C\ text(have a 2 hour float time.)`
d. `text(end of activity)\ E\ text(to the start of activity)\ J`♦♦ Mean mark 25%.
`text(By inspection of forward and backward scanning:)`
`text(EST of 5 hours is possible after activity)\ E.`
`text(LST of 12 hours after activity)\ E -> text(edge has weight)`
`text(of 1 and connects to)\ J`
The graph below shows the possible number of postal deliveries each day between the Central Mail Depot and the Zenith Post Office.
The unmarked vertices represent other depots in the region.
The weighting of each edge represents the maximum number of deliveries that can be made each day.
Two other cuts are labelled as Cut B and Cut C.
A public library organised 500 of its members into five categories according to the number of books each member borrows each month.
These categories are
J = no books borrowed per month
K = one book borrowed per month
L = two books borrowed per month
M = three books borrowed per month
N = four or more books borrowed per month
The transition matrix, `T`, below shows how the number of books borrowed per month by the members is expected to change from month to month.
`{:(),(),(T=):}{:(qquadqquadqquad\ text(this month)),((qquadJ,quadK,quadL,quadM,quadN)),([(0.1,0.2,0.2,0,0),(0.5,0.2,0.3,0.1,0),(0.3,0.3,0.4,0.1,0.2),(0.1,0.2,0.1,0.6,0.3),(0,0.1,0,0.2,0.5)]):}{:(),(),({:(J),(K),(L),(M),(N):}):}{:(),(),(text(next month)):}`
In the long term, which category is expected to have approximately 96 members each month?
A study of the antelope population in a wildlife park has shown that antelope regularly move between three locations, east (`E`), north (`N`) and west (`W`).
Let `A_n` be the state matrix that shows the population of antelope in each location `n` months after the study began.
The expected population of antelope in each location can be determined by the matrix recurrence rule
`A_(n + 1) = T A_n - D`
where
`{:(),(),(T=):}{:(qquadquadtext(this month)),((qquadE,quadN,quadW)),([(0.4,0.2,0.2),(0.3,0.6,0.3),(0.3,0.2,0.5)]):}{:(),(),({:(E),(N),(W):}):}{:(),(),(text(next month)):}`
and
`D = [(50),(50),(50)]{:(E),(N),(W):}`
The state matrix, `A_3`, below shows the population of antelope three months after the study began.
`A_3 = [(1616),(2800),(2134)]{:(E),(N),(W):}`
The number of antelope in the west (`W`) location two months after the study began, as found in the state matrix `A_2`, is closest to
Beginning in the year 2021, a new company takes over the maintenance of the 2700 km highway with a new contract.
Each year sections of highway must be graded `(G)`, resurfaced `(R)` or sealed `(S)`.
The remaining highway will need no maintenance `(N)` that year.
Let `M_n` be the state matrix that shows the highway maintenance schedule of this company for the `n`th year after 2020.
The maintenance schedule for 2020 is shown in matrix `M_0` below.
For these 2700 km of highway, the matrix recurrence relation shown below can be used to determine the number of kilometres of this highway that will require each type of maintenance from year to year.
`qquad qquad M_(n + 1) = TM_n + B`
where
| `{:(\ \ qquad qquad qquad qquad quad text(this year)),(qquad qquad quad quad G qquad quad R qquad quad S quad quad \ N):}` | ||
| `M_0 = [(500),(400),(300),(1500)]{:(G),(R),(S),(N):}text (,)` | `T = [(0.2,0.1,0.0,0.2),(0.1,0.1,0.0,0.2),(0.2,0.1,0.2,0.1),(0.5, 0.7,0.8,0.5)]{:(G),(R),(S),(N):} \ text(next year,)` | `B = [(k),(20),(10),(-60)]` |
The Hiroads company has a contract to maintain and improve 2700 km of highway.
Each year sections of highway must be graded `(G)`, resurfaced `(R)` or sealed `(S)`.
The remaining highway will need no maintenance `(N)` that year.
Let `S_n` be the state matrix that shows the highway maintenance schedule for the `n`th year after 2018.
The maintenance schedule for 2018 is shown in matrix `S_0` below.
`S_0 = [(700),(400),(200),(1400)]{:(G),(R),(S),(N):}`
The type of maintenance in sections of highway varies from year to year, as shown in the transition matrix `T`, below.
`{:(qquad qquad qquad qquad qquad quad text(this year)),(qquad qquad quad quad G qquad quad R qquad quad S quad quad \ N),(T = [(0.2,0.1,0.0,0.2),(0.1,0.1,0.0,0.2),(0.2,0.1,0.2,0.1),(0.5, 0.7,0.8,0.5)]{:(G),(R),(S),(N):} \ text (next year)):}`
|
|
`× 700 +` |
|
`× 400 +` |
|
`× 200 +` |
|
`× 1400 = 460` |
The state matrix describing the highway maintenance schedule for the nth year after 2018 is given by
`S_(n + 1) = TS_n`
`qquad qquad S_1 = [(460),(text{____}),(text{____}),(1490)]{:(G),(R),(S),(N):}`
Of these kilometres, what percentage is expected to have had no maintenance `(N)` in 2019?
Round your answer to one decimal place. (1 mark)
Round your answer to one decimal place. (1 mark)
The Westhorn Council must prepare roads for expected population changes in each of three locations: main town `(M)`, villages `(V)` and rural areas `(R)`.
The population of each of these locations in 2018 is shown in matrix `P_2018` below.
`P_2018 = [(2100),(1800),(1700)]{:(M),(V),(R):}`
The expected annual change in population in each location is shown in the table below.
Julie has retired from work and has received a superannuation payment of $492 800.
She has two options for investing her money.
Option 1
Julie could invest the $492 800 in a perpetuity. She would then receive $887.04 each fortnight for the rest of her life.
Option 2
Julie could invest the $492 800 in an annuity, instead of a perpetuity.
The annuity earns interest at the rate of 4.32% per annum, compounding monthly.
The balance of Julie’s annuity at the end of the first year of investment would be $480 242.25
Round your answer to the nearest cent. (2 marks)
After three years, Julie withdraws $14 000 from her account to purchase a car for her business.
For tax purposes, she plans to depreciate the value of her car using the reducing balance method.
The value of Julie’s car, in dollars, after `n` years, `C_n`, can be modelled by the recurrence relation shown below
`C_0 = 14\ 000, qquad C_(n + 1) = R xx C_n`
What is the annual rate of depreciation in the value of the car during these three years? (1 mark)
What is the value of the car eight years after it was purchased?
Round your answer to the nearest cent. (2 marks)
Liam cycles, runs, swims and walks for exercise several times a month.
Each time he cycles, Liam covers a distance of `c` kilometres.
Each time he runs, Liam covers a distance of `r` kilometres.
Each time he swims, Liam covers a distance of `s` kilometres.
Each time he walks, Liam covers a distance of `w` kilometres.
The number of times that Liam cycled, ran, swam and walked each month over a four-month period, and the total distance that Liam travelled in each of those months, are shown in the table below.
The matrix that contains the distance each time Liam cycled, ran, swam and walked, `[(c),(r),(s),(w)]`, is
| A. | `[(5),(6),(7),(5)]` | B. | `[(8),(6),(1),(9)]` | C. | `[(8),(6),(7),(9)]` |
| D. | `[(8),(8),(9),(8)]` | E. | `[(4290),(4931),(4623),(4291)]` |
Table 3 shows the yearly average traffic congestion levels in two cities, Melbourne and Sydney, during the period 2008 to 2016. Also shown is a time series plot of the same data.
The time series plot for Melbourne is incomplete.
(Answer on the time series plot above.)
(Answer on the time series plot above.)
Write your answer in the box provided below. (1 mark)
|
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% per year |
The yearly average traffic congestion level data for Melbourne is repeated in Table 4 below.
Round this value to four significant figures. (1 mark)
Write the values of the intercept and the slope of this least squares line in the appropriate boxes provided below.
Round both values to four significant figures. (2 marks)
| congestion level = |
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+ |
|
× year |
Explain why, quoting the values of appropriate statistics. (2 marks)
| a. |  |
| b.i. |  |
b.ii. `text(The least squares line is 1.15% higher each year.)`♦ Mean mark (b)(ii) 36%.
COMMENT: Major problems caused by part (b)(ii). Review!
` :.\ text(Average rate of increase) = 1.15 text(%)`
| b.iii. | `text(Find year when:)` | |
| `43` | `= -2280 + 1.15 xx text(year)` | |
| `text(year)` | `= 2323/1.15` | |
| `= 2020` | ||
c. `-1515`
d. `text(congestion level) = -1515 + 0.7667 xx text(year)`
e. `text(Melbourne congestion level in 2008)`♦♦♦ Mean mark 18%.
`= -1515 + 0.7667 xx 2008`
`= 24.5 text(%)`
`text{In 2008 Sydney has higher congestion (29.2 > 24.5)}`
`text(After 2008, Sydney congestion grows at 1.15% per)`
`text(year and Melbourne grows at 0.7667% per year.)`
`:.\ text(Sydney predicted to always exceed Melbourne.)`
Mariska plans to retire from work 10 years from now.
Her retirement goal is to have a balance of $600 000 in an annuity investment at that time.
The present value of this annuity investment is $265 298.48, on which she earns interest at the rate of 3.24% per annum, compounding monthly.
To make this investment grow faster, Mariska will add a $1000 payment at the end of every month.
Two years from now, she expects the interest rate of this investment to fall to 3.20% per annum, compounding monthly. It is expected to remain at this rate until Mariska retires.
When the interest rate drops, she must increase her monthly payment if she is to reach her retirement goal.
The value of this new monthly payment will be closest to
Five lines of an amortisation table for a reducing balance loan with monthly repayments are shown below.
The interest rate for this loan changed immediately before repayment number 28.
This change in interest rate is best described as
Adam has a home loan with a present value of $175 260.56
The interest rate for Adam’s loan is 3.72% per annum, compounding monthly.
His monthly repayment is $3200.
The loan is to be fully repaid after five years.
Adam knows that the loan cannot be exactly repaid with 60 repayments of $3200.
To solve this problem, Adam will make 59 repayments of $3200. He will then adjust the value of the final repayment so that the loan is fully repaid with the 60th repayment.
The value of the 60th repayment will be closest to
The graph below shows the value, `V_n`, of an asset as it depreciates over a period of five months.
Which one of the following depreciation situations does this graph best represent?
A least squares line is fitted to a set of bivariate data.
Another least squares line is fitted with response and explanatory variables reversed.
Which one of the following statistics will not change in value?
The statistical analysis of a set of bivariate data involving variables `x` and `y` resulted in the information displayed in the table below.
Using this information, the value of the correlation coefficient `r` for this set of bivariate data is closest to
The scatterplot below displays the resting pulse rate, in beats per minute, and the time spent exercising, in hours per week, of 16 students. A least squares line has been fitted to the data.
Part 1
Using this least squares line to model the association between resting pulse rate and time spent exercising, the residual for the student who spent four hours per week exercising is closest to
Part 2
The equation of this least squares line is closest to
Part 3
The coefficient of determination is 0.8339
The correlation coefficient `r` is closest to
Let `underset~r(t) = (1 - sqrt(a)sin(t))underset~i + (1 - 1/b cos(t))underset~j` for `t >= 0` and `a, b ∈ R^−` be the path of a particle moving in the cartesian plane.
The path of the particle will always be a circle if
The construction of The Royal Easter show involves activities `A` to `L`. The diagram shows these activities and their completion times in days.
The company contracted to construct it are given a completion deadline of 31 days.
Calculate the float time of Activity `G`. (3 marks)
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`text(Scanning forwards and backwards:)`
`text(Critical Path is)\ \ A B E J L\ \ text{(29 days)}`
`=>\ text(Critical path is 2 days less than 31 day deadline)`
| `:. text(Float time of Activity)\ G` | `= (22 – 16) + 2` |
| `= 8\ text(days)` |
If `cos(x) = -a` and `cot(x) = b`, where `a, b > 0`, then `text{cosec}(-x)` is equal to
`cos(x) < 0, \ cot(x) > 0 => x\ text(is in 3rd quadrant,)`
`text(or)\ \ x in (pi, (3 pi)/2)(+2k pi, k in ZZ)`♦ Mean mark 49%.
`cot(x) = a/y = b\ \ =>\ \ y = a/b`
`text(cosec)(-x) = 1/(sin(-x)) = -1/(sin(x)) =-text(cosec)(x)`
`text(In 3rd quadrant:)`
`text(cosec)(x) < 0 \ =>\ \ text(cosec)(-x) > 0`
`:. text(cosec)(-x) = 1/y = b/a`
`=> A`
Which one of the following, where `A, B, C` and `D` are non-zero real numbers, is the partial fraction form for the expression
`(2x^2 + 3x + 1)/{(2x + 1)^3 (x^2 - 1)}?`
A. `A/(2x + 1) + B/(x - 1) + C/(x + 1)`
B. `A/(2x + 1) + B/(2x - 1)^2 + C/(2x + 1)^3 + (Dx)/(x^2 - 1)`
C. `A/(2x + 1) + (Bx + C)/(x^2 - 1)`
D. `A/(2x + 1) + B/(2x + 1)^2 + C/(x - 1)`
E. `A/(2x + 1) + (Bx + C)/(2x + 1)^2 + D/(x - 1)`
A game consists of two tokens being drawn at random from a barrel containing 20 tokens. There are 17 red tokens and 3 black tokens. The player keeps the two tokens drawn.
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| i. | |
ii. `P(text(at least one red))`
`= 1 – P(BB)`
`= 1 – 3/20 · 2/19`
`= 187/190`
An investment fund purchases 4500 shares of Bank ABC for a total cost of $274 500 (ignore any transaction costs).
The investment fund is paid a divided of $3.66 per share in the first year.
Island A and island B are both on the equator. Island B is west of island A. The longitude of island A is 5°E and the angle at the centre of Earth (O), between A and B, is 30°.
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| a. | `text{Longitude (island}\ B)` | `= 5-30` |
| `= -25` | ||
| `= 25^@\ text(W)` |
`:.\ text(Island)\ B\ text{is (0°, 25°W).}`
b. `text(Time difference) = 30 xx 4 = 120 \ text(mins)\ =2\ text(hours)`
`text(S)text(ince)\ B\ text(is west of)\ A,`
| `text(Time on island)\ B` | `= 10\ text(am less 2 hours)` |
| `= 8\ text(am)` |
Using `(1 + x)^4(1 + x)^9 = (1 + x)^13`
show that
`\ ^9C_4 + \ ^4C_1\ ^9C_3 + \ ^4C_2\ ^9C_2 + \ ^4C_3\ ^9C_1 + 1 = \ ^13C_4` (2 marks)
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Consider `(dy)/(dx) = 2x^2 + x + 1`, where `y(1) = y_0 = 2`.
Using Euler's method with a step size of 0.1, an approximation to `y(0.8) = y_2` is given by
Let `f(x) = x^3 - mx^2 + 4`, where `m, x ∈ R`.
The gradient of `f` will always be strictly increasing for
Given that `(dy)/(dx) = e^x\ text(arctan)(y)`, the value of `(d^2y)/(dx^2)` at the point `(0,1)` is
The number of distinct roots of the equation `(z^4 - 1)(z^2 + 3iz - 2) = 0`, where `z ∈ C` is
The solutions to `cos(x) > 1/4 text(cosec)(x)` for `x ∈ (0,2pi)\ text(\) {pi}` are given by
`cos(x) > 1/4 text(cosec)(x)`♦ Mean mark 37%.
`4cos(x) > text(cosec)(x)`
`text(Consider:)\ \ 4cos(x) = 1/(sin(x))`
| `4cos(x)sin(x)` | `= 1` |
| `2(2sin(x)cos(x))` | `= 1` |
| `2(sin(2x))` | `= 1` |
| `sin(2x)` | `= 1/2` |
| `2x` | `= pi/6,(5pi)/6,(13pi)/6,(17pi)/6` |
| `:. x` | `= pi/12,(5pi)/12,(13pi)/12,(17pi)/12\ \ \ (x ∈ (0,2pi)\ text(\) {pi})` |
`x ∈ (pi/12,(5pi)/12) ∪ (pi,(13pi)/12) ∪ ((17pi)/12,2pi)`
`=>E`
Let `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.
Find the `x` coordinates of these points. (3 marks)
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| i. | `y` | `= e^(-x^2)` |
| `dy/dx` | `= -2x * e^(-x^2)` | |
| `(d^2y)/(dx^2)` | `= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)` | |
| `= 4x^2 e^(-x^2)\ – 2e^(-x^2)` | ||
| `= 2e^(-x^2) (2x^2\ – 1)` |
`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`
| `2e^(-x^2) (2x^2\ – 1)` | `= 0` |
| `2x^2\ – 1` | `= 0` |
| `x^2` | `= 1/2` |
| `x` | `= +- 1/sqrt2` |
| `text(When)\ \ ` | `x < 1/sqrt2,` | `\ (d^2y)/(dx^2) < 0` |
| `x > 1/sqrt2,` | `\ (d^2y)/(dx^2) > 0` |
`=>\ text(Change of concavity)`
`:.\ text(P.I. at)\ \ x = 1/sqrt2`
| `text(When)\ \ ` | `x < – 1/sqrt2,` | `\ (d^2y)/(dx^2) > 0` |
| `x > – 1/sqrt2,` | `\ (d^2y)/(dx^2) < 0` |
`=>\ text(Change of concavity)`
`:.\ text(P.I. at)\ \ x = – 1/sqrt2`
| ii. | `text(In)\ f(x), text(there are 2 values of)\ y\ text(for)` |
| `text(each value of)\ x.` | |
| `:.\ text(The domain of)\ f(x)\ text(must be restricted)` | |
| `text(for)\ \ f^(-1) (x)\ text(to exist).` |
| iii. | `y = e^(-x^2)` |
`text(Inverse function can be written)`
| `x` | `= e^(-y^2),\ \ \ x >= 0` |
| `lnx` | `= ln e^(-y^2)` |
| `-y^2` | `= lnx` |
| `y^2` | `= -lnx` |
| `=ln(1/x)` | |
| `y` | `= +- sqrt(ln (1/x))` |
`text(Restricting)\ \ x>=0,\ \ =>y>=0`
`:. f^(-1) (x)=sqrt(ln (1/x))`
| iv. | `f(0) = e^0 = 1` |
`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`
`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`
| v. |
|
The diagram below shows a sketch of the graph of `y = f(x)`, where `f(x) = 1/(1 + x^2)` for `x ≥ 0`.
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Let `α` be the `x`-coordinate of `P`. Explain why `α` is a root of the equation `x^3 + x − 1 = 0`. (1 mark)
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| i. |
|
ii. `text(Domain of)\ \ f^(−1)(x)\ text(is)`
`0 < x ≤ 1`
iii. `f(x) = 1/(1 + x^2)`
`text(Inverse: swap)\ \ x↔y`
| `x` | `= 1/(1 + y^2)` |
| `x(1 + y^2)` | `= 1` |
| `1 + y^2` | `= 1/x` |
| `y^2` | `= 1/x − 1` |
| `= (1 − x)/x` | |
| `y` | `= ± sqrt((1 − x)/x)` |
`:.y = sqrt((1 − x)/x), \ \ y >= 0`
iv. `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`
`text(i.e. where)`
| `1/(1 + x^2)` | `= x` |
| `1` | `= x(1 + x^2)` |
| `1` | `= x + x^3` |
| `x^3 + x − 1` | `= 0` |
`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`
`text(it is a root of)\ \ \ x^3 + x − 1 = 0`
Let `g(x) = log_2(x),\ \ x > 0`
Which one of the following equations is true for all positive real values of `x?`
A. `2g (8x) = g (x^2) + 8`
B. `2g (8x) = g (x^2) + 6`
C. `2g (8x) = (g (x) + 8)^2`
D. `2g (8x) = g (2x) + 6`
Let `f(x) = log_e(x)` for `x>0,` and `g (x) = x^2 + 1` for all `x`.
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Let `f(x) = e^x + e^(–x).`
`f(2u)` is equal to
A. `f(u) + f(-u)`
B. `2 f(u)`
C. `(f(u))^2 - 2`
D. `(f(u))^2 + 2`
The function `f(x)` satisfies the functional equation `f (f (x)) = x` for `{x:\ text(all)\ x,\ x!=1}`.
The rule for the function is
A. `f(x) = x + 1`
B. `f(x) = x - 1`
C. `f(x) = (x - 1)/(x + 1)`
D. `f(x) = (x + 1)/(x - 1)`
Let `f(x) = sqrt(x + 1)` for `x>=0`
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i. `text(Sketch of)\ \ f(x):`
`:.\ text(Range:)\ \ y>= 1`
ii. `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`
`text(Domain of)\ \ f(x):\ \ x>=0`
`text(Find domain of)\ g(x)\ text(such that range)\ g(x):\ y>=0`
`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`
`x <= –3\ \ text(and)\ \ x>=-1`
`:. c = -3`
Let `h(x) = 1/(x - 1)` for `-1<h<1`.
Which one of the following statements about `h` is not true?
Use the Trapezoidal rule with five function values to show that
`int_(- pi/3)^(pi/3) sec x\ dx ~~ pi/6 (3 + 4/sqrt3)`. (3 marks)
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| `A` | `~~ h/2 [y_0 + 2(y_1 + y_2 + y_3) + y_5]` |
| `~~ pi/12 [2 + 2(2/sqrt3 + 1 + 2/sqrt3) + 2]` | |
| `~~ pi/12 [6 + 8/sqrt3]` | |
| `~~ pi/6 (3 + 4/sqrt3)\ text(u² … as required)` |
An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for `t >= 6`.
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i. `text(Displacement is reducing when the velocity is negative.)`
`:. t > 5\ \ text(seconds)`
ii. `text(At)\ B,\ text(the displacement) = 7\ text(units)`
`text(Considering displacement from)\ B\ text(to)\ D:`
`text(S)text(ince the area below the graph from)`
`B\ text(to)\ C\ text(equals the area above the)`
`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`
`text(in displacement from)\ B\ text(to)\ D.`
`text(Considering)\ t >= 6`
`text(Time required to return to origin)`
| `t` | `= d/v` |
| `= 7/5` | |
| `= 1.4\ \ text(seconds)` |
`:.\ text(The particle returns to the origin after 7.4 seconds.)`
iii.
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| i. |  |
| `A` | `~~ a/2[1/a + 2(1/(2a)) + 1/(3a)]` |
| `~~ a/2(7/(3a))` | |
| `~~ 7/6` |
ii. `text{Area under the curve}\ \ y=1/x`
`= int_a^(3a) 1/x\ dx`
`= [ln x]_(\ a)^(3a)`
`= ln 3a − ln a`
`= ln\ (3a)/a`
`= ln 3`
`text{Trapezoidal rule in part (i) found the approximate}`
`text(value of the same area.)`
`:. ln 3 ≑ 7/6.`
Two events, `A` and `B`, from a given event space, are such that `P(A) = 1/5` and `P(B) = 1/3`.
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i. `text(Sketch Venn diagram:)`
| `:.P(A′ ∩ B)` | `=P(B) – P(A ∩B)` |
| `=1/3 – 1/8` | |
| `=5/24` |
| ii. |  |
`text(Mutually exclusive means)\ \ P(A ∩ B)=0,`
`:. P(A′ ∩ B) = 1/3`
For events `A` and `B` from a sample space, `P\ (A text(|)B) = 1/5` and `P\ (B text(|)A) = 1/4`. Let `P\ (A nn B) = p`.
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| i. | `P\ (A)` | `=(P\ (A nn B))/(P\ (B text(|) A))` |
| `=p/(1/4)` | ||
| `=4p` |
ii. `text(Consider the Venn diagram:)`
`P\ (A′ nn B′) = 1 – 8p`
iii. `text(Given)\ P(A uu B) = 8p`
`=> 0 < 8p <= 1/5`
`:. 0 < p <= 1/40`
Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is `3/4`, and if the weather is unpleasant, the probability that she will walk Mack is `1/3`.
Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.
Find the probability that Sally walked Mack on at least one of these two mornings. (2 marks)
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| a. | `Ptext{(at least 1 walk)}` | `= 1 – Ptext{(no walk)}` |
| `= 1 – 1/4 xx 2/3` | ||
| `= 5/6` |
b.i. `text(Construct tree diagram:)`
| `P(PW) + P(P′W)` | `= 5/8 xx 3/4 + 3/8 xx 1/3` |
| `= 19/32` |
| b.ii. | `P(P | W)` | `= (P(P ∩ W))/(P(W))` |
| `= (5/8 xx 3/4)/(19/32)` | ||
| `= 15/32 xx 32/19` | ||
| `= 15/19` |
Two events, `A` and `B`, are such that `P(A) = 3/5` and `P(B) = 1/4.`
If `A^{′}` denotes the compliment of `A`, calculate `P (A^{′} nn B)` when
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i. `text(Sketch Venn Diagram)`
| `P (A uu B)` | `= P (A) + P (B)-P (A nn B)` |
| `3/4` | `= 3/5 + 1/4-P (A nn B)` |
| `P (A nn B)` | `= 1/10` |
`:.\ P(A^{′} nn B) = 1/4-1/10 = 3/20`
| ii. |  |
`P (A∩ B)=0\ \ text{(mutually exclusive)},`
`:.\ P (A^{′} nn B) = P (B) = 1/4`
There is a daily flight from Paradise Island to Melbourne. The probability of the flight departing on time, given that there is fine weather on the island, is 0.8, and the probability of the flight departing on time, given that the weather on the island is not fine, is 0.6.
In March the probability of a day being fine is 0.4.
Find the probability that on a particular day in March
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| i. |  |
`Ptext{(FT)}\ +\ Ptext{(F′T)}`
`= 0.4 xx 0.8 + 0.6 xx 0.6`
`= 0.32 + 0.36`
`= 0.68`
ii. `text(Conditional probability:)`
| `P(F\ text(|)\ T)` | `= (P(F ∩ T))/(P(T))` |
| `= 0.32/0.68` |
`:. P(F\ text(|)\ T) = 8/17`
For events `A` and `B` from a sample space, `P(A | B) = 3/4` and `P(B) = 1/3`.
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i. `text(Using Conditional Probability:)`
| `P(A | B)` | `= (P(A ∩ B))/(P(B))` |
| `3/4` | `= (P(A ∩ B))/(1/3)` |
| `:. P(A ∩ B)` | `= 1/4` |
| ii. |  |
| `P(A′ ∩ B)` | `= P(B) – P(A ∩B)` |
| `= 1/3 – 1/4` | |
| `= 1/12` |
iii. `text(If)\ A, B\ text(independent)`
| `P(A ∩ B)` | `= P(A) xx P(B)` |
| `1/4` | `= P(A) xx 1/3` |
| `:. P(A)` | `= 3/4` |
| `P(A ∪ B)` | `= P(A) + P(B) – P(A ∩ B)` |
| `= 3/4 + 1/3 – 1/4` | |
| `:. P(A ∪ B)` | `= 5/6` |
Four identical balls are numbered 1, 2, 3 and 4 and put into a box. A ball is randomly drawn from the box, and not returned to the box. A second ball is then randomly drawn from the box.
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If `X` is a random variable such that `P(X > 5) = a` and `P(X > 8) = b`, then `P(X < 5 | X < 8)` is
A. `a/b`
B. `(a - b)/(1 - b)`
C. `(1 - b)/(1 - a)`
D. `(a - 1)/(b - 1)`
`A` and `B` are events of a sample space.
Given that `P(A | B) = p,\ \ P(B) = p^2` and `P(A) = p^(1/3),\ P(B | A)` is equal to
A. `p^3`
B. `p^(4/3)`
C. `p^(7/3)`
D. `p^(8/3)`
Events `A` and `B` are mutually exclusive events of a sample space with
`P(A) = p and P (B) = q\ \ text(where)\ \ 0 < p < 1 and 0 < q < 1.`
`P (A^{′} nn B^{′})` is equal to
| `P (A^{′} nn B^{′})` | `= 1-p-q` |
| `= 1-(p + q)` |
`=> C`
The probability distribution of a discrete random variable, `X`, is given by the table below.
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ & \ \ \ 1\ \ \ & \ \ \ 2\ \ \ & \ \ \ 3\ \ \ &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.6p^{2} & 0.1 & 1-p & 0.1 \\
\hline
\end{array}
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On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by
\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ & \ \ \ 1\ \ \ & \ \ \ 2\ \ \ & \ \ \ 3\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.2 & 0.2 & 0.5 & 0.1 \\
\hline
\end{array}
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What is the probability that Daniel receives a total of four calls over these two days? (3 marks)
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The shaded region is enclosed by the curve `y = x^3 - 7x` and the line `y = 2x`, as shown in the diagram. The line `y = 2x` meets the curve `y = x^3 - 7x` at `O(0, 0)` and `A(3, 6)`. Do NOT prove this.
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The point `P` is chosen on the curve `y = x^3 − 7x` so that the tangent at `P` is parallel to the line `y = 2x` and the `x`-coordinate of `P` is positive
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| i. | `text(Area)` | `= int_0^3 2x – (x^3 – 7x)\ dx` |
| `= int_0^3 9x – x^3\ dx` | ||
| `= [9/2 x^2 – 1/4 x^4]_0^3` | ||
| `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]` | ||
| `= 81/2 – 81/4` | ||
| `= 81/4\ text(units²)` |
ii. `f(x) = 9x – x^3`
| `text(Area)` | `~~ 1/2[0 + 2(8 + 10) + 0]` |
| `~~ 1/2(36)` | |
| `~~ 18\ text(u²)` |
iii. `y = x^3 – 7x`
`(dy)/(dx) = 3x^2 – 7`
`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`
| `3x^2 – 7` | `= 2` |
| `3x^2` | `= 9` |
| `x^2` | `= 3` |
| `x` | `= sqrt 3 qquad (x > 0)` |
| `y` | `= (sqrt 3)^3 – 7 sqrt 3` |
| `= 3 sqrt 3 – 7 sqrt 3` | |
| `= -4 sqrt 3` |
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`
| iv. |  |
| `text(dist)\ OA` | `= sqrt((3 – 0)^2 + (6 – 0)^2)` |
| `= sqrt 45` | |
| `= 3 sqrt 5` |
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`
`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`
| `_|_\ text(dist)` | `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|` |
| `= (6 sqrt 3)/sqrt 5` | |
| `:.\ text(Area)` | `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5` |
| `= 9 sqrt 3\ text(units²)` |
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| i. | `int_0^(pi/3) cos x\ dx` | `= [sin x]_0^(pi/3)` |
| `= sin\ pi/3-0` | ||
| `= sqrt 3/2` |
| ii. |  |
\begin{array} {|l|c|c|c|}\hline
x & \ \ \ 0\ \ \ & \ \ \ \dfrac{\pi}{6}\ \ \ & \ \ \ \dfrac{\pi}{3}\ \ \ \\ \hline
\text{height} & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline
\text{weight} & 1 & 2 & 1 \\ \hline \end{array}
| `int_0^(pi/3) cos x\ dx` | `~~ 1/2 xx pi/6[1 + 2(sqrt3/2) + 1/2]` |
| `~~ pi/12((3 + 2sqrt3)/2)` | |
| `~~ ((3+2sqrt3)pi)/24` |
♦ Mean mark part (iii) 49%.
| (iii) | `((3+2sqrt3)pi)/24` | `~~ sqrt3/2` |
| `:. pi` | `~~ (24sqrt3)/(2(3+2sqrt3))` | |
| `~~ (12sqrt3)/(3 + 2sqrt3)` |
At a certain location a river is 12 metres wide. At this location the depth of the river, in metres, has been measured at 3 metre intervals. The cross-section is shown below.
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Calculate the approximate volume of water flowing through the cross-section in 10 seconds. (1 mark)
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| i. |  |
| `A` | `~~ 3/2[0.5 + 2(2.3 + 2.9 + 3.8) + 2.1]` |
| `~~ 3/2(20.6)` | |
| `~~ 30.9\ text(m²)` |
♦ Mean mark 49%.
| ii. `text(Distance water flows)` | `= 0.4 xx 10` |
| `= 4 \ text(metres)` |
| `text(Volume flow in 10 seconds)` | `~~ 4 xx 30.9` |
| `~~ 123.6 text(m³)` |
The length of each edge of the cube `ABCDEFGH` is 2 metres. A circle is drawn on the face `ABCD` so that it touches all four edges of the face. The centre of the circle is `O` and the diagonal `AC` meets the circle at `X` and `Y`.
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| i. |
|
`text(S)text(ince)\ \ FA, \ AC\ \ text(and)\ \ FC\ \ text(are all)`
`text(diagonals of sides of a cube,)`
`FA = AC = FC`
`:.ΔFAC\ \ text(is equilateral)`
`:.∠FAC = 60^@`
| ii. |
|
`text(In)\ \ ΔAEF`
| `AF^2` | `= EF^2 + EA^2` |
| `= 2^2 + 2^2` | |
| `= 8` | |
| `AF` | `= sqrt8` |
| `= 2sqrt2` |
`text(In)\ \ ΔAFO`
| `sin\ 60^@` | `= (FO)/(AF)` |
| `sqrt3/2` | `= (FO)/(2sqrt2)` |
| `FO` | `= sqrt3/2 xx 2sqrt2` |
| `= sqrt6\ text(metres … as required.)` |
| iii. |
|
`XY\ \ text(is the diameter of a circle AND the width)`
`text(of the cube.)`
| `:.XY` | `= 2` |
| `:.OX` | `= OY = 1` |
| `tan\ ∠OFX` | `=1 /sqrt6` |
| `∠OFX` | `= 22.207…^@` |
| `:.∠XFY` | `= 2 xx 22.407…` |
| `= 44.415…` | |
| `= 44^@\ text{(nearest degree)}` |
