Band 3

Data Analysis, GEN2 2024 VCAA 3

The Olympic gold medal-winning height for the women's high jump, $\textit{Wgold}$, is often lower than the best height achieved in other international women's high jump competitions in that same year.

The table below lists the Olympic year, $\textit{year}$, the gold medal-winning height, $\textit{Wgold}$, in metres, and the best height achieved in all international women's high jump competitions in that same year, $\textit{Wbest}$, in metres, for each Olympic year from 1972 to 2020.

A scatterplot of $\textit{Wbest}$ versus $\textit{Wgold}$ for this data is also provided.

year	1972	1976	1980	1984	1988	1992	1996	2000	2004	2008	2012	2016	2020
Wgold (m)	1.92	1.93	1.97	2.02	2.03	2.02	2.05	2.01	2.06	2.05	2.05	1.97	2.04
Wbest (m)	1.94	1.96	1.98	2.07	2.07	2.05	2.05	2.02	2.06	2.06	2.05	2.01	2.05

When a least squares line is fitted to the scatterplot, the equation is found to be:

$Wbest =0.300+0.860 \times Wgold$

The correlation coefficient is 0.9318

Name the response variable in this equation. (1 mark)

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

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Determine the value of the coefficient of determination as a percentage. (1 mark)
Round your answer to one decimal place.
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Describe the association between $\textit{Wbest}$ and $\textit{Wgold}$ in terms of strength and direction. (1 mark)

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\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text { strength } \rule[-1ex]{0pt}{0pt} & \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\hline
\rule{0pt}{2.5ex}\text { direction } \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Referring to the equation of the least squares line, interpret the value of the slope in terms of the variables $\textit{Wbest}$ and $\textit{Wgold}$. (1 mark)

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In 1984, the $\textit{Wbest}$ value was 2.07 m for a $\textit{Wgold}$ value of 2.02 m .
Show that when this least squares line is fitted to the scatterplot, the residual value for this point is 0.0328. (2 marks)

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The residual plot obtained when the least squares line was fitted to the data is shown below. The residual value from part f is missing from the residual plot.

1. Complete the residual plot by adding the residual value from part f, drawn as a cross ( X ), to the residual plot above. (1 mark)
  
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2. In part b, a least squares line was fitted to the scatterplot. Does the residual plot from part g justify this? Briefly explain your answer. (1 mark)
  
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In 1964, the gold medal-winning height, $\textit{Wgold}$, was 1.90m . When the least squares line is used to predict $\textit{Wbest}$, it is found to be 1.934 m .
Explain why this prediction is not likely to be reliable. (1 mark)
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Show Answers Only

a. $Wbest$

c. $86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

h. $\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

Show Worked Solution

a. $Wbest$

b. $\text{Using points:}\ (1.90, 1.934)\ \text{and}\ (2.00, 2.02)$

c. $r=0.9318$

$r^2=0.9318^2=0.8682\dots$

$\therefore\ \text{Coefficient of determination} \approx 86.8\%$

d. $\text{Strong, positive}$

e. $Wbest\ \text{will increase, on average by 0.86 metres for every metre of increase in}\ Wgold.$

f.	$Wbest$	$=0.300 +0.86\times 2.02$
		$=2.0372$

$\therefore\ \text{Residual}\ =2.07-2.0372=0.0328$

g.i.

g.ii. $\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}$

h. $\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}$

$\text{and therefore cannot be relied upon.}$

Data Analysis, GEN2 2024 VCAA 2

The boxplot below displays the distribution of all gold medal-winning heights for the women's high jump, $\textit{Wgold}$, in metres, for the 19 Olympic Games held from 1948 to 2020.

Describe the shape of this data distribution. (1 mark)

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For this boxplot, what is the smallest possible number of $\textit{Wgold}$ heights lower than 1.85 m (1 mark)

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i. Using the boxplot, show that the lower fence is 1.565 m and the upper fence is 2.325 m. (1 mark)

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ii. Referring to the boxplot, the lower fence and the upper fence, explain why no outliers exist. (1 mark)

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Show Answers Only

a. $\text{Negatively skewed}$

b. $1$

c.i. $Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19$

$\text{Lower Fence}$	$=Q_1-1.5\times IQR$
	$=1.85-1.5\times 0.19$
	$=1.565$

$\text{Upper Fence}$	$=Q_1+1.5\times IQR$
	$=2.04+1.5\times 0.19$
	$=2.325$

c.ii. $\text{No values exist below the lower fence or above the upper fence.}$

$\therefore\ \text{No outliers exist.}$

Show Worked Solution

a. $\text{Negatively skewed}$

b. $25\%\ \text{of the games had heights}\ <1.85\ \text{m}$

$25\%\times 19= 4.75\rightarrow 4\ \text{games}$

$\therefore\ \text{The smallest number of games is 1}$

c.i. $Q_1=1.85,\ Q_3=2.04,\ IQR=2.04-1.85=0.19$

$\text{Lower Fence}$	$=Q_1-1.5\times IQR$
	$=1.85-1.5\times 0.19$
	$=1.565$

$\text{Upper Fence}$	$=Q_1+1.5\times IQR$
	$=2.04+1.5\times 0.19$
	$=2.325$

c.ii. $\text{No values exist below the lower fence or above the upper fence.}$

$\therefore\ \text{No outliers exist.}$

Data Analysis, GEN2 2024 VCAA 1

Table 1 lists the Olympic year, $\textit{year}$, and the gold medal-winning height for the men's high jump, $\textit{Mgold}$, in metres, for each Olympic Games held from 1928 to 2020. No Olympic Games were held in 1940 or 1944, and the 2020 Olympic Games were held in 2021.

Table 1

\begin{array}{|c|c|}
\hline \quad \textit{year} \quad & \textit{Mgold}(m) \\
\hline 1928 & 1.94 \\
\hline 1932 & 1.97 \\
\hline 1936 & 2.03 \\
\hline 1948 & 1.98 \\
\hline 1952 & 2.04 \\
\hline 1956 & 2.12 \\
\hline 1960 & 2.16 \\
\hline 1964 & 2.18 \\
\hline 1968 & 2.24 \\
\hline 1972 & 2.23 \\
\hline 1976 & 2.25 \\
\hline 1980 & 2.36 \\
\hline 1984 & 2.35 \\
\hline 1988 & 2.38 \\
\hline 1992 & 2.34 \\
\hline 1996 & 2.39 \\
\hline 2000 & 2.35 \\
\hline 2004 & 2.36 \\
\hline 2008 & 2.36 \\
\hline 2012 & 2.33 \\
\hline 2016 & 2.38 \\
\hline 2020 & 2.37 \\
\hline
\end{array}

For the data in Table 1, determine:
1. the maximum $\textit{Mgold}$ in metres (1 mark)
  
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2. the percentage of $\textit{Mgold}$ values greater than 2.25 m. (1 mark)
  
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The mean of these $\textit{Mgold}$ values is 2.23 m , and the standard deviation is 0.15 m .
Calculate the standardised $z$-score for the 2000 $\textit{Mgold}$ of 2.35 m. (1 mark)

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Construct a boxplot for the $\textit{Mgold}$ data in Table 1 on the grid below. (2 marks)

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A least squares line can also be used to model the association between $\textit{Mgold}$ and $\textit{year}$.
Using the data from Table 1, determine the equation of the least squares line for this data set.
Use the template below to write your answer.
Round the values of the intercept and slope to three significant figures. (2 marks)

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The coefficient of determination is 0.857
Interpret the coefficient of determination in terms of $\textit{Mgold}$ and $\textit{year}$. (1 mark)

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Show Answers Only

a.i. $2.39$

a.ii. $50\%$

b. $0.8$

e. $\text{A coefficient of determination of 85.7% shows the variation in the}\ Mgold$

$\text{that is explained by the variation in the }year.$

Show Worked Solution

a.i. $2.39$

a.ii. $\dfrac{11}{22}=50\%$

b.	$z$	$=\dfrac{x-\overline x}{s_x}$
		$=\dfrac{2.35-2.23}{0.15}$
		$=0.8$

c. $Q_2=\dfrac{2.33+2.25}{2}=2.29$

$Q_1=2.12, \ Q_3=2.36$

$\text{Min}\ =1.94, \ \text{Max}\ =2.39$

d. $\text{Using CAS:}$

e. $\text{A coefficient of determination of 85.7% shows the variation in the}\ Mgold$

$\text{that is explained by the variation in the }year.$

BIOLOGY, M8 2024 HSC 11 MC

The data shows the proportion of adults living in Australia who are obese.

Which of the following can be observed from the data?

The proportion of obese adults always increases with age.
There is a greater percentage of men who are obese than women in all age groups.
The proportion of women who are obese increases from 13% at 18–24 to 38% at 65–74.
The proportion of men who are obese increases from 18% at 18–24 to 35% at 45–54, then decreases to 23% at age 85 and over.

Show Answers Only

$C$

Show Worked Solution

→ Options $A, B$ and $D$ can be shown to be incorrect with reference to the table.

$\Rightarrow C$

BIOLOGY, M7 2024 HSC 8 MC

Trypanosomes (Trypanosoma brucei) are protozoans that cause African sleeping sickness in humans. The diagram shows the way that the disease is transmitted to humans.

Which row of the table identifies the pathogen, vector and method of disease transmission to humans?

\begin{align*}
\begin{array}{l}
\ & \\
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|l|l|l|}
\hline
\ \ \textit{Pathogen} & \ \ \textit{Vector} & \ \ \textit{Method of disease} \\
\textit{} & \textit{} & \ \ \textit{transmission} \\
\hline
\ \ \text{Trypanosomes}\ \ & \ \ \text{Tsetse fly}\ \ & \ \ \text{Direct} \\
\hline
\ \ \text{Tsetse fly} & \ \ \text{Cow} & \ \ \text{Direct} \\
\hline
\ \ \text{Trypanosomes} & \ \ \text{Tsetse fly} & \ \ \text{Indirect}\ \ \\
\hline
\ \ \text{Tsetse fly} & \ \ \text{Cow} & \ \ \text{Indirect} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$C$

Show Worked Solution

→ African sleeping sickness is caused by trypanosome parasites (pathogen).

→ The pathogens are transmitted indirectly to humans through the bite of tsetse flies (vector).

$\Rightarrow C$

BIOLOGY, M8 2024 HSC 7 MC

How do stomata maintain water balance in plants?

They close in hot weather to decrease transpiration.
They open in cold weather to decrease transpiration.
They open in hot weather to decrease evaporative cooling.
They close in cold weather to decrease evaporative cooling.

Show Answers Only

$A$

Show Worked Solution

→ Stomata close in hot weather to decrease transpiration.

→ This adaptive response helps prevent excessive water loss during high temperatures and maintain proper water balance.

$\Rightarrow A$

BIOLOGY, M6 2024 HSC 2 MC

Resin produced by spinifex grass has long been used by Aboriginal Peoples. Spinifex resin is currently used to produce medicinal creams.

What is this an example of?

Biotechnology
Selective breeding
Artificial insemination
Genetically modified organisms

Show Answers Only

$A$

Show Worked Solution

→ This is an example of biotechnology because it involves using traditional knowledge to harness a biological resource (spinifex resin) for medicinal purposes.

→ This fits the broad definition of biotechnology as using biological systems or processes to create useful products.

$\Rightarrow A$

BIOLOGY, M7 2024 HSC 1 MC

Which of the following are non-cellular pathogens?

Bacteria
Fungi
Prions
Protozoa

Show Answers Only

$C$

Show Worked Solution

→ Bacteria, fungi, and protozoa are all cellular organisms.

→ Prions are misfolded proteins that can cause disease by inducing normal proteins to misfold, making them non-cellular pathogens.

$\Rightarrow C$

CHEMISTRY, M5 2024 HSC 2 MC

Aboriginal and Torres Strait Islander Peoples have used leaching in flowing water over several days to prepare various foods from plants that can be toxic to humans.

What was the reason for this?

To react with toxins
To dissolve low solubility toxins
To prevent the food from decomposing
To break down compounds that are difficult to digest

Show Answers Only

$B$

Show Worked Solution

→ The toxins form a equilibrium system: $\ce{toxins(s)\rightleftharpoons toxins(aq)}$

→ As the water flows away, the concentration of aqueous toxins decreases which shifts the equilibrium system to the right, causing more low solubility toxins to break down.

$\Rightarrow B$

Networks, GEN1 2024 VCAA 34 MC

Consider the following graph.

A Eulerian trail through this graph could be

ABCDEF
ACBDCFDEF
BACBDCFDEF
BDCABCDFCDEF

Show Answers Only

$C$

Show Worked Solution

$\text{An Eulerian trail has exactly 2 vertices of odd degree.}$

$\text{You must start on one and finish on the other ie B to F.}$

$\rightarrow\ \text{Eliminate A and B}$

$\text{A path cannot be repeated}\ \rightarrow\ \text{Eliminate D}$

$\Rightarrow C$

CHEMISTRY, M7 2024 HSC 1 MC

Which two substances are members of the same homologous series?

Show Answers Only

$A$

Show Worked Solution

→ A homologous series refers to compounds with the same or similar functional group and same general formula.

→ Both substances in $A$ contain a hydroxyl group.

$\Rightarrow A$

Matrices, GEN1 2024 VCAA 25 MC

Matrix $J$ is a $2 \times 3$ matrix.

Matrix $K$ is a $3 \times 1$ matrix.

Matrix $L$ is added to the product $J K$.

The order of matrix $L$ is

$1 \times 3$
$2 \times 1$
$2 \times 3$
$3 \times 2$

Show Answers Only

$B$

Show Worked Solution

$J\ \text{is order}\ 2\times 3\ \text{and}\ K\ \text{is order}\ \3\times 1$

$\therefore\ JK\ \text{is order}\ \2\times 1$

$\Rightarrow B$

Recursion and Finance, GEN1 2024 VCAA 19 MC

Liv bought a new car for $35 000. The value of the car will be depreciated by 18% per annum using the reducing balance method.

A recurrence relation that models the year-to-year value of her car is of the form

$L_0=35\,000, \quad L_{n+1}=k \times L_n$

The value of $k$ is

0.0082
0.18
0.82
1.18

Show Answers Only

$C$

Show Worked Solution

$\text{Reducing balance interest rate}\ =1-0.18=0.82$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 15 MC

The table below shows the total number of cans of soft drink sold each month at a suburban cafe in 2023.

\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Month } \rule[-1ex]{0pt}{0pt}& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\rule{0pt}{2.5ex} \textbf{Cans sold } \rule[-1ex]{0pt}{0pt}& 316 & 321 & 365 & 306 & 254 & 308 & 354 & 357 & 381 & 355 & 365 & 324 \\
\hline
\end{array}

The six-mean smoothed value of the number of cans sold, with centring, for month 5 is closest to

Show Answers Only

$C$

Show Worked Solution

$\text{Six-mean smoothed:}$

$\text{Months:}\ 2-7$	$=\dfrac{321+365+306+254+308+354}{6}=318$
$\text{Months:}\ 3-8$	$=\dfrac{365+306+254+308+354+357}{6}=324$

$\therefore\ \text{Centred smoothed mean}$	$=\dfrac{318+324}{2}$
	$=321$

$C$

Data Analysis, GEN1 2024 VCAA 9-10 MC

The least squares equation for the relationship between the average number of male athletes per competing nation, $males$, and the number of the Summer Olympic Games, $number$, is

$males =67.5-1.27 \times number$

Part A

The summary statistics for the variables number and males are shown in the table below.

The value of Pearson's correlation coefficient, $r$, rounded to three decimal places, is closest to

$-0.569$
$-0.394$
$0.394$
$0.569$

Part B

At which Summer Olympic Games will the predicted average number of $males$ be closest to 25.6 ?

31st
32nd
33rd
34th

Show Answers Only

Part A

$A$

Part B

$C$

Show Worked Solution

Part A

$b$	$=r\dfrac{s_y}{s_x}$
$-1.27$	$=r\times\dfrac{19}{8.51}$
$\therefore\ r$	$=\dfrac{-1.27\times 8.51}{19}$
	$=-0.5688\dots$

$\Rightarrow A$

Part B

$males$	$=67.5-1.27\times number$
$25.6$	$=67.5-1.27\times number$
$number$	$=\dfrac{67.5-25.6}{1.27}$
	$=32.992\dots$
	$\approx 33$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 5 MC

The number of siblings of each member of a class of 24 students was recorded.

The results are displayed in the table below.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad 2 \quad \rule[-1ex]{0pt}{0pt} & \quad 1 \quad & \quad 3 \quad & \quad 2 \quad & \quad 1 \quad & \quad 1 \quad & \quad 1 \quad & \quad 4 \quad & \quad 1 \quad & \quad 1 \quad & \quad 1 \quad & \quad 1 \quad \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 2 & 1 & 2 & 2 & 1 & 3 & 4 & 2 & 2 & 3 & 1 \\
\hline
\end{array}

A boxplot was constructed to display the spread of the data.

Which one of the following statements about this boxplot is correct?

There are no outliers.
The value of the interquartile range (IQR) is 1.5
The value of the median is 1.5
All of the five-number summary values are whole numbers.

Show Answers Only

$C$

Show Worked Solution

$Q_1=1,\ Q_2=1.5,\ Q_3=2\rightarrow\ \text{eliminate D}$

$IQR=2-1=1\rightarrow\ \text{eliminate B}$

$Q_3+1.5\times IQR=2+1.5\times 1\rightarrow 4\ \text{is an outlier}\rightarrow\ \text{eliminate A}$

$Q_2=1.5 \longrightarrow \text{Median }=1.5\rightarrow\ \text{C correct}$

$\Rightarrow C$

Data Analysis, GEN1 2024 VCAA 1 MC

A group of students were asked to name their favourite colour.

The results are displayed in the percentage segmented bar chart below.

The percentage of students who named blue as their favourite colour is closest to

Show Answers Only

$B$

Show Worked Solution

$\%\ \text{Blue}\ =74-56=18\%$

$\Rightarrow B$

Mechanics, EXT2 M1 2024 HSC 5 MC

A particle is moving in simple harmonic motion with period 10 seconds and an amplitude of 8 m . The particle starts at the central point of motion and is initially moving to the left with a speed of $V$ m s$^{-1}$, where $V>0$.

What will be the position and velocity of the particle after 7.5 seconds?

At the central point of motion with a velocity of $V \text{ m s} ^{-1}$
At the central point of motion with a velocity of $-V \text{ m s} ^{-1}$
8 m to the left of the central point of motion with a velocity of $0 \text{ m s} ^{-1}$
8 m to the right of the central point of motion with a velocity of $0 \text{ m s} ^{-1}$

Show Answers Only

$D$

Show Worked Solution

$T=\dfrac{2\pi}{n}=10\ \ \Rightarrow\ \ n=\dfrac{\pi}{5}$

$x$	$=-8 \sin\Big(\dfrac{\pi}{5}t\Big) $
$\dot{x}$	$=-8 \cos\Big(\dfrac{\pi}{5}t\Big) $

$\text{At}\ \ t=7.5:$

$x$	$=-8 \sin\Big(\dfrac{\pi}{2}t\Big)=-8 $
$\dot{x}$	$=-8 \cos\Big(\dfrac{\pi}{2}t\Big)=0 $

$\Rightarrow D$

ENGINEERING, TE 2024 HSC 4 MC

Which row of the table correctly identifies characteristics of analogue and digital communications?

\begin{align*}
\begin{array}{c}
\ & \\
\ & \\
\rule{0pt}{2.5ex}\textbf{A.}\\
\rule{0pt}{2.5ex}\textbf{}\\
\rule{0pt}{2.5ex}\textbf{B.}\\
\rule{0pt}{2.5ex}\textbf{}\\
\rule{0pt}{2.5ex}\textbf{C.}\\
\rule{0pt}{2.5ex}\textbf{}\\
\rule{0pt}{2.5ex}\textbf{D.}\\
\rule{0pt}{2.5ex}\textbf{}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule[-1ex]{0pt}{0pt} \quad \quad \quad \quad \quad Analogue & \rule[-1ex]{0pt}{0pt} \quad \quad \quad \quad \quad Digital \\
\hline
\rule[-1ex]{0pt}{0pt} \text{Continuous (varying) signal} & \rule[-1ex]{0pt}{0pt} \text{Discrete (binary) signal} \\
\rule[-1ex]{0pt}{0pt} \text{Less susceptible to signal degradation} & \rule[-1ex]{0pt}{0pt} \text{More susceptible to signal degradation} \\
\hline
\rule[-1ex]{0pt}{0pt} \text{Continuous (varying) signal} & \rule[-1ex]{0pt}{0pt} \text{Discrete (binary) signal} \\
\rule[-1ex]{0pt}{0pt} \text{More susceptible to signal degradation} & \rule[-1ex]{0pt}{0pt} \text{Less susceptible to signal degradation} \\
\hline
\rule[-1ex]{0pt}{0pt} \text{Discrete (binary) signal} & \rule[-1ex]{0pt}{0pt} \text{Continuous (varying) signal} \\
\rule[-1ex]{0pt}{0pt} \text{Less susceptible to signal degradation} & \rule[-1ex]{0pt}{0pt} \text{More susceptible to signal degradation} \\
\hline
\rule[-1ex]{0pt}{0pt} \text{Discrete (binary) signal} & \rule[-1ex]{0pt}{0pt} \text{Continuous (varying) signal} \\
\rule[-1ex]{0pt}{0pt} \text{More susceptible to signal degradation} & \rule[-1ex]{0pt}{0pt} \text{Less susceptible to signal degradation} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

→ Analog signals are continuous and more susceptible to degradation while digital signals are discrete (binary) and less susceptible to degradation since they can be regenerated and error-corrected during transmission.

$\Rightarrow B$

ENGINEERING, PPT 2024 HSC 3 MC

A simplified image of a bicycle chain drive is shown.

If a cyclist is pedalling at 70 revolutions per minute (RPM), what is the RPM of the driven wheel?

3.2
21.7
226.2
546.0

Show Answers Only

$C$

Show Worked Solution

$\text{Velocity Ratio}\ (VR)\ = \dfrac{\text{output teeth}}{\text{input teeth}} = \dfrac{13}{42}=0.3095$

$\text{Output speed (RPM)}\ = \dfrac{\text{input speed}}{VR} = \dfrac{70}{0.3095}=226.2$

$\Rightarrow C$

ENGINEERING, CS 2024 HSC 1 MC

A common house brick is shown.

Which forming process was used to manufacture the brick?

Forging
Extrusion
Slip casting
Shell moulding

Show Answers Only

$B$

Show Worked Solution

→ Common house bricks are manufactured through extrusion, where clay is forced through a die to create a rectangular column that is then cut into individual brick lengths before drying and firing.

$\Rightarrow B$

Vectors, EXT2 V1 2024 HSC 13a

The point $A$ has position vector $8 \underset{\sim}{i}-6 \underset{\sim}{j}+5 \underset{\sim}{k}$. The line $\ell$ has vector equation

$x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}=t(\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})$.

The point $B$ lies on $\ell$ and has position vector $p \underset{\sim}{i}+p \underset{\sim}{j}+2 p \underset{\sim}{k}$.

Show that $\abs{A B}^2=6 p^2-24 p+125$. (1 mark)

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Hence, or otherwise, determine the shortest distance between the point $A$ and the line $\ell$. (2 marks)

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Show Answers Only

i. $A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)$

$\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)$

	$\abs{AB}^2$	$=(p-8)^2+(p+6)^2+(2 p-5)^2$
		$=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25$
		$=6 p^2-24 p+125$

ii. $\abs{AB}_{\text {min}}=\sqrt{101} \text { units}$

Show Worked Solution

$\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)$

	$\abs{AB}^2$	$=(p-8)^2+(p+6)^2+(2 p-5)^2$
		$=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25$
		$=6 p^2-24 p+125$

ii. $\text{Find shortest distance between \(A$ and $\ell$.}\)

$\Rightarrow \text { Find \(p$ when $\abs{A B}$ is a minimum:}\)

$\text{Minimum occurs when}\ \ p=\dfrac{-b}{2 a}=\dfrac{24}{2 \times 6}=2$

$\therefore \abs{AB}_{\text {min}}=\sqrt{6(2)^2-24(2)+125}=\sqrt{101} \text { units}$

Vectors, EXT2 V1 2024 HSC 12e

The line $\ell$ passes through the points $A(3,5,-4)$ and $B(7,0,2)$.

Find a vector equation of the line $\ell$. (1 mark)

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Determine, giving reasons, whether the point $C(10,5,-2)$ lies on the line $\ell$. (2 marks)

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Show Worked Solution

i. $A(3,5,-4), \quad B(7,0,2)$

$\overrightarrow{A B}=\left(\begin{array}{l}7 \\ 0 \\ 2\end{array}\right)-\left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)=\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)$

$\therefore \text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )$

ii. $\text{If \(C$ lies on the line}, \ \exists \lambda \in \mathbb{R}\ \ \text{such that:}\)

$\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)$

$\text{Find \(\lambda$ for $x$-component: }\)

$10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}$

$\text{Check \(\lambda=\dfrac{7}{4}$ for $y$-component:}\)

$5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}$

$\therefore C\ \text{does not lie on the line.}$

Proof, EXT2 P1 2024 HSC 12d

Explain why there is no integer $n$ such that $(n+1)^{41}-79 n^{40}=2$. (2 marks)

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$\text{Show}\ \ \nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2$

$\text {If \(n$ is even:}\)

$\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2$

$\text {If \(n$ is odd:}\)

$\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2$

$\therefore\ \text {By contradiction}$

$\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2$

Show Worked Solution

$\text{Show}\ \ \nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2$

$\text {If \(n$ is even:}\)

$\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2$

$\text {If \(n$ is odd:}\)

$\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2$

$\therefore\ \text {By contradiction}$

$\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2$

Vectors, EXT2 V1 2024 HSC 12a

The vector $\underset{\sim}{a}$ is $\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$ and the vector $\underset{\sim}{b}$ is $\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$.

Find $\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$. (1 mark)

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Show that $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}$ is perpendicular to $\underset{\sim}{b}$. (2 marks)

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i. $(-1,0,2) $

ii. $\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text {Vectors are perpendicular.}$

Show Worked Solution

i. $\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)$

$\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)$

$ \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0$

$\therefore\ \text{Vectors are perpendicular.}$

Complex Numbers, EXT2 N1 2024 HSC 11e

Write the number $\sqrt{3}+i$ in modulus-argument form. (2 marks)

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Hence, or otherwise, write $(\sqrt{3}+i)^7$ in exact Cartesian form. (2 marks)

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i. $2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)$

ii. $-64 \sqrt{3}-64 i$

Show Worked Solution

i. $z=\sqrt{3}+i$

$|z|=\sqrt{3+1}=2$

$\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}$

$z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)$

ii.	$(\sqrt{3}+i)^7$	$=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)$
		$=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)$
		$=-64 \sqrt{3}-64 i$

Vectors, EXT2 V1 2024 HSC 11c

Find the angle between the two vectors $\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)$ and $\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right)$, giving your answer in radians, correct to 1 decimal place. (2 marks)

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Show Worked Solution

$\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right),\abs{\underset{\sim}{u}}=\sqrt{1+4+4}=3$

$\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right),\abs{\underset{\sim}{v}}=\sqrt{16+16+49}=9$

$\cos \theta=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{u}||\underset{\sim}{v}|}=\dfrac{1 \times 4-2 \times 4-2 \times 7}{3 \times 9}=-\dfrac{2}{3}$

$\theta=\cos ^{-1}\left(-\dfrac{2}{3}\right)=2.30 \ldots=2.3^c \ \ \text{(1 d.p.)}$

Complex Numbers, EXT2 N1 2024 HSC 11b

Let $z=2+3 i$ and $w=1-5 i$.

Find $z+\bar{w}$. (1 mark)

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Find $z^2$. (1 mark)

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i. $z+\bar{w}=3+8 i$

ii. $z^2=-5+12 i$

Show Worked Solution

i. $z=2+3 i$

$w=1-5 i \ \Rightarrow \ \bar{w}=1+5 i$

$z+\bar{w}=2+3 i+1+5 i=3+8 i$

ii.	$z^2$	$=(2+3 i)^{2}$
		$=4+12 i+9 i^2$
		$=-5+12 i$

Calculus, EXT2 C1 2024 HSC 11a

Find $\displaystyle \int x e^x\, d x$ (2 marks)

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$x e^x-e^x+c$

Show Worked Solution

$u=x \quad \ \ u^{\prime}=1$

$v^{\prime}=e^x \quad v=e^x$

	$\displaystyle\int x e^x \,d x$	$=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x$
		$=x e^x- \displaystyle \int e^x \cdot 1\, d x$
		$=x e^x-e^x+c$

Calculus, EXT1 C2 2024 HSC 11e

Differentiate the function $f(x)=\arcsin \left(x^5\right)$. (1 mark)

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$f^{\prime}(x)=\dfrac{5 x^4}{\sqrt{1-x^{10}}}$

Show Worked Solution

$f(x)=\sin ^{-1}\left(x^5\right)$

$f^{\prime}(x)=5 x^4 \times \dfrac{1}{\sqrt{1-\left(x^5\right)^2}}=\dfrac{5 x^4}{\sqrt{1-x^{10}}}$

Calculus, EXT1 C2 2024 HSC 11c

Using the substitution $u=x-1$, find $\displaystyle \int x \sqrt{x-1}\, d x$. (3 marks)

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$\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c$

Show Worked Solution

$\displaystyle \int x \sqrt{x-1}\, d x $


	$u=x-1$	$\Rightarrow \ x=u+1 $
	$\dfrac{d u}{d x}=1$	$\Rightarrow \ d u=d x$

	$\displaystyle\int(u+1) \sqrt{u+1-1}\, d u$	$=\displaystyle{\int}(u+1) \sqrt{u} \, d u$
		$=\displaystyle{\int} u^{\frac{3}{2}}+u^{\frac{1}{2}}\, d u$
		$=\dfrac{2}{5} u^{\frac{5}{2}}+\dfrac{2}{3} u^{\frac{3}{2}}+c$
		$=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c$

Functions, EXT1 F1 2024 HSC 11b

Solve $x^2-8 x-9 \leq 0$. (2 marks)

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$-1 \leqslant x \leqslant 9$

Show Worked Solution

		$x^2-8 x-9 \leqslant 0$
		$(x-9)(x+1) \leqslant 0$

$\therefore -1 \leqslant x \leqslant 9$

Vectors, EXT1 V1 2024 HSC 11a

Consider the vectors $\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}$ and $\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}$.

Find $2 \underset{\sim}{a}-\underset{\sim}{b}$. (1 mark)

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Find $\underset{\sim}{a} \cdot \underset{\sim}{b}$. (1 mark)

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i. $\displaystyle \binom{7}{0}$

ii. $5$

Show Worked Solution

i. $\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}$

$2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}$

ii. $\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5$.

CHEMISTRY, M2 EQ-Bank 2 MC

Which is the correct balanced formula equation for the reaction of potassium with water?

$\ce{K(s) + H2O(l) -> KOH(aq) + H2(g)}$
$\ce{2K(s) + 2H2O(aq) -> 2KOH(aq) + H2(g)}$
$\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}$
$\ce{K(s) + 2H2O(aq) -> KOH(aq) + 2H2(g)}$

Show Answers Only

$C$

Show Worked Solution

→ The correct chemical equation is:

$\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}$

→ The state of water is always liquid. An aqueous solution refers to a solution where a substance has been dissolved into water.

$\Rightarrow C$

CHEMISTRY, M2 EQ-Bank 2

In an experiment, calcium carbonate $\ce{(CaCO3)}$ is heated strongly to produce calcium oxide $\ce{(CaO)}$ and carbon dioxide according to the reaction below:

$\ce{CaCo3(s) -> CaO(s) + CO2(g)}$

A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.

Using the law of conservation of mass, calculate the mass of carbon dioxide gas produced in this reaction. (2 marks)

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Explain how the law of conservation of mass applies to this reaction. (2 marks)

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a. $22.0\ \text{g}$

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

Show Worked Solution

a. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products.

Since mass must be conserved:

$m\ce{(CO2)}$	$=m\ce{(CaCO3)}-m\ce{(CaO)}$
	$= 50.0-28.0$
	$=22.0\ \text{g}$

b. The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

CHEMISTRY, M2 EQ-Bank 1 MC

What numbers are required to correctly balance this equation?

__$\ce{Fe2O3 +}$ __$\ce{CO ->}$ __$\ce{Fe +}$ __$\ce{CO2}$

1, 3, 2, 3
2, 3, 1, 3
1, 1, 2, 1
2, 4, 2, 4

Show Answers Only

$A$

Show Worked Solution

→ Start by balancing the iron $\ce{Fe}$ atoms:

→ There are 2 Fe atoms in $\ce{Fe2O3}$, so we place a 2 in front of $\ce{Fe}$ on the product side.

$\ce{Fe2O3 + CO -> 2Fe + CO2}$

→ Next, balance the oxygen $\ce{O}$ atoms:

→ $\ce{Fe2O3}$ has 3 oxygen atoms, and we need 3 $\ce{CO}$ molecules on the reactant side to balance the 3 $\ce{CO2}$ molecules on the product side.

$\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$

→ Finally, check all atoms to ensure balance.

$\Rightarrow A$

CHEMISTRY, M2 EQ-Bank 1

Balance the following chemical equations:

$\ce{HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$ (1 mark)

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$\ce{C2H6(l) + O2(g) -> H2O(l) + CO2(g)}$ (1 mark)

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$\ce{H3PO4(aq) + CuCO3(aq) -> Cu3(PO4)2(aq) + CO2(g) + H2O(l)}$ (1 mark)

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$\ce{CuSO4(aq) + AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$ (1 mark)

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a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

c. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

d. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

Show Worked Solution

a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

c. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

d. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

CHEMISTRY, M2 EQ-Bank 13

If you have a solution with a concentration of 1.2 mol/L and need 0.6 moles of solute, what volume of solution is required? (1 mark)

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Explain why knowing the volume of a solution is important when preparing specific concentrations in laboratory settings. (2 marks)

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a. $0.5\ \text{L}$

b. The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Show Worked Solution

a. $V=\dfrac{n}{c}=\dfrac{0.6}{1.2}=0.5\ \text{L}$

b. The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

CHEMISTRY, M2 EQ-Bank 12

Calculate the amount (in moles) of a solute in a 2.0 L solution with a concentration of 0.75 mol/L. (1 mark)

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Using your answer from part a, determine the mass of the solute if the solute is $\ce{NaCl}$. (2 marks)

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a. $1.5\ \text{mol}$

b. $87.99\ \text{g}$

Show Worked Solution

a. $n = c \times V = 0.75 \times 2 = 1.5\ \text{mol}$

b. $MM\ce{(NaCl)} = 22.99 + 35.45 = 58.44\ \text{g/mol}$

$m\ce{(NaCl)} = MM \times n = 58.44 \times 1.5 = 87.99\ \text{g}$

CHEMISTRY, M2 EQ-Bank 11 MC

A student prepares a solution of potassium nitrate by dissolving 0.05 kg of $\ce{KNO3}$ in enough water to make 2000 mL of solution. Which of the following correctly calculates the concentration of the solution in mol L$^{-1}$?

$25 \times 10^{-6}\ \text{g L}^{-1}$
$25 \times 10^{-4}\ \text{g L}^{-1}$
$25 \times 10^{-2}\ \text{g L}^{-1}$
$25\ \text{g L}^{-1}$

Show Answers Only

$D$

Show Worked Solution

→ Mass of $\ce{KNO3}$ is 50 g.

→ Volume of solution is 2 L.

$c\ce{(KNO3)} = \dfrac{50}{2} = 25\ \text{g L}^{-1}$

$\Rightarrow D$

Calculus, EXT1 C3 2024 HSC 2 MC

Consider the functions $y=f(x)$ and $y=g(x)$, and the regions shaded in the diagram below.

Which of the following gives the total area of the shaded regions?

$\displaystyle \int_{-4}^4 f(x)-g(x)\,d x$
$\displaystyle \left|\int_{-4}^4 f(x)-g(x)\,d x\right|$
$\displaystyle \int_{-4}^{-3} f(x)-g(x)\,d x+\int_{-3}^{-1} f(x)-g(x)\,d x+\int_{-1}^1 f(x)-g(x)\,d x+\int_1^4 f(x)-g(x)\,d x $
$\displaystyle - \int_{-4}^{-3} f(x)-g(x)\,d x+\int_{-3}^{-1} f(x)-g(x)\,d x-\int_{-1}^1 f(x)-g(x)\,d x+\int_1^4 f(x)-g(x)\,d x$

Show Answers Only

$D$

Show Worked Solution

$\text{Intervals where}\ f(x) \gt g(x)\ \ \Rightarrow\ \text{Positive area values}$

$\text{Intervals where}\ g(x) \gt f(x)\ \ \Rightarrow\ \text{Negative area values}$

$\Rightarrow D$

CHEMISTRY, M2 EQ-Bank 8 MC

A student prepares a standard solution of sodium chloride. They dissolve 5.85 g of sodium chloride $\ce{(NaCl)}$ in enough water to make 1.00 L of solution. Determine the concentration of this solution?

0.100 mol L$^{-1}$
0.500 mol L$^{-1}$
1.00 mol L$^{-1}$
5.85 mol L$^{-1}$

Show Answers Only

$A$

Show Worked Solution

$n\ce{(NaCl)} = \dfrac{5.85}{22.99 + 35.45} = 0.100\ \text{mol}$

$\ce{[NaCl]} = \dfrac{n}{V} = \dfrac{0.100}{1} = 0.100\ \text{mol L}^{-1}$

$\Rightarrow A$

Trigonometry, EXT1 T1 2024 HSC 5 MC

Consider the function $g(x) = 2 \sin^{-1}(3x)$.

Which transformations have been applied to $f(x) = \sin^{-1}(x)$ to obtain $g(x)$?

Vertical dilation by a factor of $\dfrac{1}{2}$ and a horizontal dilation by a factor of $\dfrac{1}{3}$
Vertical dilation by a factor of $\dfrac{1}{2}$ and a horizontal dilation by a factor of 3
Vertical dilation by a factor of 2 and a horizontal dilation by a factor of $\dfrac{1}{3}$
Vertical dilation by a factor of 2 and a horizontal dilation by a factor of 3

Show Answers Only

$C$

Show Worked Solution

$\text{A vertical dilation of factor 2:}$

$f(x) = \sin^{-1}(x)\ \ \rightarrow \ \ f_1(x) = 2\sin^{-1}(x)$

$\text{A horizontal dilation of factor}\ \dfrac{1}{3}:$

$f_1(x) = 2\sin^{-1}(x)\ \ \rightarrow \ \ f_2(x) = 2\sin^{-1}(3x)$

$\Rightarrow C$

Functions, EXT1 F2 2024 HSC 1 MC

The polynomial $x^{3} + 2x^{2}-5x-6$ has zeros $-1, -3$ and $\alpha$.

What is the value of $\alpha$?

$-2$
$2$
$3$
$6$

Show Answers Only

$B$

Show Worked Solution

$\alpha \beta \gamma = -\dfrac{\text{d}}{\text{a}} = 6$

$-1 \times -3 \times \alpha$	$=6$
$\alpha$	$=2$

$\Rightarrow B$

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.

What is the length and width of the cupboard, in metres? (1 mark)

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The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
What is the cost of the tiles for the living area? (4 marks)
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a. $4\ \text{m}\times 0.5\ \text{m}$

b. $$1100$

Show Worked Solution

a. $\text{1 metre = 1000 mm}$

$\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}$

b. $\text{Method 1}$

$\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2$

$\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2$

$\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

$\text{Method 2}$

$\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 $

$\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 $

$\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

CHEMISTRY, M2 EQ-Bank 5 MC

A standard solution is best described as:

A solution prepared to an approximate concentration for general use.
A solution with a precisely known concentration, used in quantitative chemical analysis.
A solution containing only one type of solute molecule.
A solution prepared by dissolving a solid solute in a small volume of solvent.

Show Answers Only

$B$

Show Worked Solution

→ A standard solution is a solution with a precise and known concentration, allowing accurate calculations in quantitative chemical analysis.

→ It is prepared by dissolving an exact amount of a primary standard in a specific volume of solvent.

$\Rightarrow B$

CHEMISTRY, M2 EQ-Bank 7

4.56 g of potassium chloride $\ce{KCl}$ is dissolved in 250 mL of water. What is the concentration of this solution? (2 marks)

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How many grams of calcium chloride $\ce{CaCl2}$ will be needed to make 1.50 L of a 0.250 M solution? (2 marks)

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a. $0.24\ \text{mol L}^{-1}$

b. $41.6\ \text{g}$

Show Worked Solution

a. $MM\ce{(KCl)}= 39.10 + 35.45 = 74.55\ \text{g mol}^{-1}$

$n\ce{(KCl)}= \dfrac{m}{MM} = \dfrac{4.56}{74.55} = 0.061\ \text{mol}$

$c\ce{(KCl)}=\dfrac{n}{V}= \dfrac{0.061}{0.25} = 0.24\ \text{mol L}^{-1}$

b. $n\ce{(CaCl2)} = c \times V = 0.25 \times 1.5 = 0.375\ \text{mol}$

$m\ce{(CaCl2)} = MM \times n = (40.08 + 2(35.45)) \times 0.375 = 41.6\ \text{g}$

Probability, STD1 S2 2024 HSC 17

A wheel is shown with the numbers 0 to 19 marked.

A game is played where the wheel is spun until it stops.

When the wheel stops, a pointer points to the winning number. Each number is equally likely to win.

List all the even numbers on the wheel that are greater than 7. (1 mark)

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What is the probability that the winning number is NOT an even number greater than 7? (2 marks)

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a. $8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18$

b. $0.7$

Show Worked Solution

a. $8\ ,\ 10\ ,\ 12\ ,\ 14\ ,\ 16\ ,\ 18\ \text{(6 numbers)}$

b. $\text{Total numbers = 20}$

$\text{Numbers not even and > 7}\ = 20-6=14\ \text{numbers}$

$P\text{(not even and > 7)}\ =\dfrac{14}{20}=0.7$

Probability, STD1 S2 2024 HSC 1 MC

Mark buys one raffle ticket in a raffle with 1000 tickets.

Which of the following best describes the probability that Mark wins?

Certain
Even chance
Unlikely
Impossible

Show Answers Only

$C$

Show Worked Solution

$\text{P(win)}=\dfrac{1}{1000}\ \ \Rightarrow\ \ \text{Unlikely}$

$\Rightarrow C$

Measurement, STD1 M4 2024 HSC 2 MC

Four people completed the same fitness activity.

The graph shows the heart rate for each person before and after completing the activity.

Which person had the LEAST difference in heart rate?

Show Answers Only

$B$

Show Worked Solution

$\text{Option A: }$	$\text{Jo}$	$=120-80=40$
$\text{Option B: }$	$\text{Kim}$	$=120-100=20\ \checkmark$
$\text{Option C: }$	$\text{Lee}$	$=120-90=30$
$\text{Option D: }$	$\text{Mal}$	$=150-100=50$

$\Rightarrow B$

Complex Numbers, EXT2 N2 2024 HSC 4 MC

A monic polynomial, $f(x)$, of degree 3 with real coefficients has $3$ and $2+i$ as two of its roots.

Which of the following could be $f(x)$ ?

$f(x)=x^3-7 x^2-17 x+15$
$f(x)=x^3-7 x^2+17 x-15$
$f(x)=x^3+7 x^2-17 x+15$
$f(x)=x^3+7 x^2+17 x-15$

Show Answers Only

$B$

Show Worked Solution

$\text{Since coefficients are real, roots are:}\ \ 3, 2+i, 2-i$

\[ \sum \text{roots} = 7 = \dfrac{-b}{1}\ \ \Rightarrow \ b=-7\]

$\text{Product of roots}\ = 3(2+i)(2-i)=15=\dfrac{-d}{1}\ \ \Rightarrow \ d=-15$

$\Rightarrow B$

Proof, EXT2 P1 2024 HSC 3 MC

Consider the statement:

'If a polygon is a square, then it is a rectangle.'

Which of the following is the converse of the statement above?

If a polygon is a rectangle, then it is a square.
If a polygon is a rectangle, then it is not a square.
If a polygon is not a rectangle, then it is not a square.
If a polygon is not a square, then it is not a rectangle.

Show Answers Only

$A$

Show Worked Solution

$\text{Statement:}\ P \Rightarrow \ Q$

$\text{Converse of statement:}\ Q \Rightarrow \ P$

$\Rightarrow A$

Proof, EXT2 P1 2024 HSC 2 MC

Consider the following statement written in the formal language of proof

$\forall \theta \in\biggl(\dfrac{\pi}{2}, \pi\biggr) \exists\ \phi \in\biggl(\pi, \dfrac{3 \pi}{2}\biggr) ; \ \sin \theta=-\cos \phi$.

Which of the following best represents this statement?

There exists a $\theta$ in the second quadrant such that for all $\phi$ in the third quadrant $\sin \theta=-\cos \phi$.
There exists a $\phi$ in the third quadrant such that for all $\theta$ in the second quadrant $\sin \theta=-\cos \phi$.
For all $\theta$ in the second quadrant there exists a $\phi$ in the third quadrant such that $\sin \theta=-\cos \phi$.
For all $\phi$ in the third quadrant there exists a $\theta$ in the second quadrant such that $\sin \theta=-\cos \phi$.

Show Answers Only

$C$

Show Worked Solution

$\Rightarrow C$

Networks, STD1 N1 2024 HSC 15

A network of towns and the distances between them in kilometres is shown.

What is the shortest path from $T$ to $H$ ? (2 marks)

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A truck driver needs to travel from $Y$ to $G$ but knows that the road from $C$ to $G$ is closed.
What is the length of the shortest path the truck driver can take from $Y$ to $G$ after the road closure? (2 marks)

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a. $TYWH$

b. $\text{Length of shortest path}\ (YWHMG) = 89 \text{km}$

Show Worked Solution

a. $TYH=30+38=68, \quad TYWH=30+15+20=65$

$\therefore \text{ Shortest Path is}\ TYWH.$

b. $Y W C M G=15+25+25+25=90$

$YWHMG=15+20+29+25=89$

$\Rightarrow \ \text{All other paths are longer.}$

$\therefore\text{ Length of shortest path = 89 km}$

Calculus, 2ADV C4 2024 HSC 22

The graph of the function $f(x) = \ln(1 + x^{2})$ is shown.

Prove that $f(x)$ is concave up for $-1 < x < 1$. (3 marks)
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A table of function values, correct to 4 decimal places, for some $x$ values is provided.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}

Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram. (2 marks)
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Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer. (1 mark)
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a. $f(x)= \ln(1+x^2)$

$f^{′}(x)=\dfrac{2x}{1+x^2}$

$f^{″}(x)$	$=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}$
	$=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}$
	$=\dfrac{2(1-x^2)}{(1+x^2)^2}$

$\text{Consider domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 $

$(1+x^2)^2 \gt 0\ \ \text{for all}\ x$

$\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) $

$\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) $

b. $\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}$

c. $\text{Trapezoidal rule assumes straight lines join points in the table.}$

$\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}$

Show Worked Solution

a. $f(x)= \ln(1+x^2)$

$f^{′}(x)=\dfrac{2x}{1+x^2}$

$f^{″}(x)$	$=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}$
	$=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}$
	$=\dfrac{2(1-x^2)}{(1+x^2)^2}$

$\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 $

$(1+x^2)^2 \gt 0\ \ \text{for all}\ x$

$\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) $

$\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) $

b. $\text{Total shaded area}$

$\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] $

$\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] $

$\approx 0.538275 $

$\approx 0.5383\ \text{(4 d.p.)}$

c. $\text{Trapezoidal rule assumes straight lines join points in the table.}$

$\text{Since the graph is concave up in the given domain, the trapezoidal rule will overestimate the area.}$

Trigonometry, 2ADV T1 2024 HSC 20

A vertical tower $T C$ is 40 metres high. The point $A$ is due east of the base of the tower $C$. The angle of elevation to the top $T$ of the tower from $A$ is 35°. A second point $B$ is on a different bearing from the tower as shown. The angle of elevation to the top of the tower from $B$ is 30°. The points $A$ and $B$ are 100 metres apart.

Show that distance $A C$ is 57.13 metres, correct to 2 decimal places. (1 mark)

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Find the bearing of $B$ from $C$ to the nearest degree. (3 marks)

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$194^{\circ}$

Show Worked Solution

a. $\text{In}\ \Delta TCA:$

$\tan 35°$	$ =\dfrac{40}{AC}$
$AC$	$ =\dfrac{40}{\tan 35°}$
	$=57.125…$
	$=57.13\ \text{m (2 d.p.)}$

b. $\text{In}\ \Delta TCB:$

$\tan 30°$	$ =\dfrac{40}{BC}$
$BC$	$ =\dfrac{40}{\tan 30°}$
	$=69.28\ \text{m}$

$ \text{Find} \ \angle BCA \ \text{using cosine rule:}$

$\cos \angle B CA$	$ = \dfrac{57.13^2+69.28^2-100^2}{2 \times 57.13 \times 69.28}$
	$= -0.2446 $…
$\angle BCA$	$ = 104.2° $

$\therefore\ \text{Bearing of}\ B\ \text{from}\ C= 90+104=194^{\circ} \text{(nearest degree)} $

L&E, 2ADV E1 2024 HSC 13

The graph shows the populations of two different animals, $W$ and $K$, in a conservation park over time. The $y$-axis is the size of the population and the $x$-axis is the number of years since 1985 .

Population $W$ is modelled by the equation $y=A \times(1.055)^x$.

Population $K$ is modelled by the equation $y=B \times(0.97)^x$.

Complete the table using the information provided. (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

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\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

$\text{Calculations:}$

$\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}$

$\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}$

$\text {Predicted population }(W)=34 \times(1.055)^{50}=494$

Algebra, STD2 A1 2024 HSC 24

Sarah, a 60 kg female, consumes 3 glasses of wine at a family dinner over 2.5 hours.

Note: there are 1.2 standard drinks in one glass of wine.

The blood alcohol content $(BAC)$ for females can be estimated by

$B A C_{\text {female}}=\dfrac{10 N-7.5 H}{5.5 M},$

	where $N$	= number of standard drinks
	$H$	= number of hours drinking
	$M$	= mass in kilograms

Calculate Sarah's $BAC$ at the end of the dinner, correct to 3 decimal places. (2 marks)

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The time it takes a person's BAC to reach zero is given by

$\text {Time}=\dfrac{B A C}{0.015}.$

Calculate the time it takes for Sarah's BAC to return to zero, assuming she stopped drinking after 2.5 hours. Give your answer to the nearest minute. (2 marks)

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a. $BAC=0.052$

b. $\text{3 h 28 m}$

Show Worked Solution

a. $N=3 \times 1.2=3.6, H=2.5, M=60$

	$\therefore B A C$	$=\dfrac{10 \times 3.6-7.5 \times 2.5}{5.5 \times 60}$
		$=0.052 \ \text{(3 d.p.)}$

b.	$T$	$=\dfrac{0.052}{0.015}$
		$=3.466$
		$=3 \text{ h 28 m}$

COMMENT: Use calculator degrees/minutes function for part (b).

Statistics, STD2 S4 2024 HSC 19

A teacher was exploring the relationship between students' marks for an assignment and their marks for a test. The data for five different students are shown on the graph.

The least-squares regression line is also shown.

What is the equation of the least-squares regression line for this dataset? (2 marks)

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Another student, whose marks are not on the graph, scored 5 for the assignment and 12 on the test.
Did this student do better or worse on the test than the regression line predicts? Provide a reason for your answer. (1 mark)

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a. $\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)$

$y \,\text {-intercept}=6$

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4$

$\text{LSRL}\ \ \Rightarrow \ y=1.4x+6$

b. $\text {The student’s mark \((5,12)$ sits below the LSRL.}\)

$\text {Therefore the student did worse than expected.}$

Show Worked Solution

a. $\text {Data points: }(3,11),(5,14),(6,12),(7,15),(9,20)$

$y\,\text {-intercept}=6$

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{20-6}{10-0}=1.4$

$\text{LSRL}\ \ \Rightarrow \ y=1.4x+6$

b. $\text {The student’s mark \((5,12)$ sits below the LSRL.}\)

$\text {Therefore the student did worse than expected.}$

Measurement, STD2 M7 2024 HSC 17

The cost of electricity is 30.13 cents per kWh .

Calculate the cost of using a 650 W air conditioner for 6 hours. (2 marks)

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$\text{Cost} =\$ 1.18$

Show Worked Solution

$\text{Usage}=6 \times 650=3900\, \text{W}=3.9\, \text{kW}$

$\text{Cost} =3.9 \times 30.13=117.507 \,\text{c}=\$ 1.18 \, \text {(nearest cent)}$

f.	\(Wbest\)	\(=0.300 +0.86\times 2.02\)
		\(=2.0372\)

f.	\(Wbest\)	\(=0.300 +0.86\times 2.02\)
		\(=2.0372\)

\(\text{Lower Fence}\)	\(=Q_1-1.5\times IQR\)
	\(=1.85-1.5\times 0.19\)
	\(=1.565\)

\(\text{Upper Fence}\)	\(=Q_1+1.5\times IQR\)
	\(=2.04+1.5\times 0.19\)
	\(=2.325\)

\(\text{Lower Fence}\)	\(=Q_1-1.5\times IQR\)
	\(=1.85-1.5\times 0.19\)
	\(=1.565\)

b.	\(z\)	\(=\dfrac{x-\overline x}{s_x}\)
		\(=\dfrac{2.35-2.23}{0.15}\)
		\(=0.8\)

\(\text{Months:}\ 2-7\)	\(=\dfrac{321+365+306+254+308+354}{6}=318\)
\(\text{Months:}\ 3-8\)	\(=\dfrac{365+306+254+308+354+357}{6}=324\)

\(\therefore\ \text{Centred smoothed mean}\)	\(=\dfrac{318+324}{2}\)
	\(=321\)

\(b\)	\(=r\dfrac{s_y}{s_x}\)
\(-1.27\)	\(=r\times\dfrac{19}{8.51}\)
\(\therefore\ r\)	\(=\dfrac{-1.27\times 8.51}{19}\)
	\(=-0.5688\dots\)