Graphs of Practical Situations

Algebra, STD1 EQ-Bank 25

A scientist is studying a colony of bacteria in a laboratory. At the start of the experiment there are 500 bacteria.

Each hour the number of bacteria is modelled to be 1.25 times the number of the previous hour.

Let $t=0$ be the start of the experiment.

By first completing the table of values below, draw a graph showing how the bacteria population is modelled from $t=0$ to $t=8$ hours. (3 marks)

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$\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & & & & \\ \hline \end{array}$

Using your graph from (a), or otherwise, determine after how many hours the bacteria population first exceeds 2000. (1 mark)

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a. $\text{Table of values:}$

$\text{Number of bacteria} = 500 \times 1.25^t$

$\begin{array}{|l|c|c|c|c|c|} \hline \rule{0pt}{2.5ex}t \ \text{(hours)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 4 \ \ \quad & \ \ \quad 8 \ \ \quad \\ \hline \rule{0pt}{2.5ex}\text{Number of bacteria} \rule[-1ex]{0pt}{0pt}& 500 & \textbf{625} & \textbf{781} & \textbf{1221} & \textbf{2980} \\ \hline \end{array}$

b. $\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}$

Show Worked Solution

a. $\text{Table of values:}$

$\text{Number of bacteria} = 500 \times 1.25^t$

b. $\text{From the graph, the bacteria population reaches 2000 at approximately }\ t \approx 6.25 \ \text{hours.}$

Algebra, STD1 EQ-Bank 22

The population of a certain town on 1 January 2024 was 2000 people.

Each year the population of this town is modelled to be 1.18 times the population of the previous year.

Let $t=0$ be 1 January 2024 .

By first completing the table for the indicated years, draw a graph showing how the population is modelled from 1 January 2024 to 1 January 2034. (4 marks)

\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 & & & & \\
\hline
\end{array}

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\begin{array}{|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}t \ \text{(years since 1 January 2024)} \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \quad \ \ & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 5 \ \ \quad & \ \ \quad 10 \ \ \quad \\
\hline
\rule{0pt}{2.5ex}\text{Population} \rule[-1ex]{0pt}{0pt}& 2000 &2360 &2785 &4576 & 10468 \\
\hline
\end{array}

Show Worked Solution

Algebra, STD1 EQ-Bank 13

A boat is purchased for $15 000. It depreciates in value by $3000 per year.

Let $V$ = Value of the boat in dollars, and $t$ = time in years.

Complete the table of values below that models the relationship between the value of the boat and time in years. (1 mark)

$\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}V & 15\,000 & & 9000 & & 3000 & \\[12pt] \hline \end{array}$

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Using the table of values from (a), neatly graph the value of the boat from 0 to 5 years on the grid below. (2 marks)

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Identify ONE limitation of this linear model. (1 mark)

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a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \ \\ \hline \end{array}

c. $\text{Limitations could include ONE of the following:}$

- $\text{The model predicts the boat has zero value after 5 years, which is}$
  $\text{unrealistic as most boats retain some value.}$
- $\text{Beyond 5 years the model would predict a negative value, which is}$
  $\text{not possible.}$
- $\text{The model assumes a constant rate of depreciation, but in reality}$
  $\text{a boat may depreciate more quickly in early years.}$

Show Worked Solution

a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline t & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline V & 15\ 000 & \textbf{12 000} & 9000 & \textbf{6000} & 3000 & \ \ \ \ \textbf{0}\ \ \ \ \\ \hline \end{array}

c. $\text{Limitations could include ONE of the following:}$

- $\text{The model predicts the boat has zero value after 5 years, which is}$
  $\text{unrealistic as most boats retain some value.}$
- $\text{Beyond 5 years the model would predict a negative value, which is}$
  $\text{not possible.}$
- $\text{The model assumes a constant rate of depreciation, but in reality}$
  $\text{a boat may depreciate more quickly in early years.}$

Algebra, STD1 EQ-Bank 20

Sunny SUPs Pty Ltd charges a $2.50 online booking fee plus $25 per hour for stand-up paddle board hire.

A student claims that Sunny SUPs' income can be modelled using a linear equation.

Is the student correct? Justify your answer. (2 marks)

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$\text{Yes, the student is correct.}$

$\text{Correct answers could include any ONE of the following justifications:}$

- $\text{The income increases by a constant amount of \$25 for each additional}$
  $\text{hour hired.}$
- $\text{The income can be written in the form }\ I=25h+2.50,\text{which is a linear}$
  $\text{equation.}$
- $\text{The graph of income against hours hired would be a straight line.}$
- $\text{The rate of change of income is constant at \$25 per hour.}$

Show Worked Solution

$\text{Yes, the student is correct.}$

$\text{Correct answers could include any ONE of the following justifications:}$

- $\text{The income increases by a constant amount of \$25 for each additional}$
  $\text{hour hired.}$
- $\text{The income can be written in the form }\ I=25h+2.50,\text{which is a linear}$
  $\text{equation.}$
- $\text{The graph of income against hours hired would be a straight line.}$
- $\text{The rate of change of income is constant at \$25 per hour.}$

Algebra, STD1 EQ-Bank 11

A farmer in Western Australia observes that the rodent population on his property is doubling every two weeks.

A student claims that a linear model is NOT appropriate to predict the rodent population over time.

Is the student correct? Justify your answer. (2 marks)

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$\text{Yes, the student is correct.}$

$\text{Correct justification could include ONE of the following:}$

- $\text{The rodent population is doubling every two weeks, which means it}$
  $\text{is increasing by a constant multiplier, not a constant amount — this}$
  $\text{is exponential growth, not linear.}$
- $\text{A linear model increases by the same amount each time period, but}$
  $\text{the rodent population increases by a larger amount each fortnight as}$
  $\text{the population grows.}$
- $\text{The graph of the rodent population over time would be an upsloping}$
  $\text{exponential curve, not a straight line.}$
- $\text{A linear model would significantly underestimate the rodent population}$
  $\text{over time.}$

Show Worked Solution

$\text{Yes, the student is correct.}$

$\text{Correct justification could include ONE of the following:}$

- $\text{The rodent population is doubling every two weeks, which means it}$
  $\text{is increasing by a constant multiplier, not a constant amount — this}$
  $\text{is exponential growth, not linear.}$
- $\text{A linear model increases by the same amount each time period, but}$
  $\text{the rodent population increases by a larger amount each fortnight as}$
  $\text{the population grows.}$
- $\text{The graph of the rodent population over time would be an upsloping}$
  $\text{exponential curve, not a straight line.}$
- $\text{A linear model would significantly underestimate the rodent population}$
  $\text{over time.}$

Algebra, STD1 EQ-Bank 26

Rangers at a nature reserve are monitoring the spread of an invasive weed. At the start of monitoring there are 100 weeds. The number of weeds is growing at a rate of 50% per month.

Let $N$ = number of weeds, and $t$ = time in months.

Complete the table of values below that models the growth of the weeds over an 8 month period. Round all answers to the nearest whole number. (2 marks)

$\begin{array}{|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}N & 100 & & & & \\[12pt] \hline \end{array}$

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Using the table of values from (a), neatly plot the points and join with a smooth curve. (2 marks)

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a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|} \hline \quad t \quad & \quad 0 \quad & \quad 1 \quad & \quad 2 \quad & \quad 4 \quad & \quad 8 \quad \\[6pt] \hline \ N & 100 & 150 & 225 & 506 & 2563 \\[6pt] \hline \end{array}

Show Worked Solution

a. $\text{Table of values:}$

$\text{Algebraic method}$

$\text{Formula: }\ N=100\times1.5^t$

$t=0:\ N=100\times1.5^{0}=100$

$t=1:\ N=100\times1.5^{1}=150$

$t=2:\ N=100\times1.5^{2}=225$

$t=4:\ N=100\times1.5^{4}=506.25\approx506$

$t=8:\ N=100\times1.5^{8}=2562.89\ldots\approx2563$

$\text{Using CASIO calculator with constant multiplier}$

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline {0} & 100= & 100 & 100 \\
\hline {1} & \text{Ans}\times 1.5= & 150 & 150 \\
\hline {2} & = & 225 & 225 \\
\hline {4} & = & 506.25 & 506 \\
\hline {8} & = & 2562.890625 & {2563} \\
\hline \end{array}

Algebra, STD1 EQ-Bank 27

A conservation program is tracking the recovery of a native wildflower species in a national park. At the start of the program there are 200 plants. The number of plants is predicted to grow at a rate of 25% per year.

Let $N$ = number of plants, and $t$ = time in years.

Complete the table of values below that models the growth of the plants. Round all answers to the nearest whole number. (2 marks)
\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

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Using the table of values from (a), neatly plot the points and join with a smooth curve. (2 marks)

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a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad t \quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex}N \rule[-1ex]{0pt}{0pt}& 200 & & & & \\
\hline
\end{array}

Show Worked Solution

a. $\text{Table of values:}$

$\text{Algebraic method}$

$\text{Formula: }\ N=200\times1.25^t$

$t=0:\ N=200\times1.25^{0}=200$

$t=1:\ N=200\times1.25^{1}=250$

$t=2:\ N=200\times1.25^{2}=312.5\approx313$

$t=3:\ N=200\times1.25^{3}=390.625\approx391$

$t=4:\ N=200\times1.25^{4}=488.28\ldots\approx488$

$\text{Using CASIO calculator with constant multiplier}$

\begin{array} {|c|c|c|c|}
\hline t & \text{Input} & \text{Output}\ (N) & \text{Rounded}\ (N) \\
\hline \colorbox{lightblue}{0} & 200= & 200 & \colorbox{lightblue}{200} \\
\hline \colorbox{lightblue}{1} & \text{Ans}\times 1.25= & 250 & \colorbox{lightblue}{250} \\
\hline \colorbox{lightblue}{2} & = & 312.5 & \colorbox{lightblue}{313} \\
\hline \colorbox{lightblue}{3} & = & 390.625 & \colorbox{lightblue}{391} \\
\hline \colorbox{lightblue}{4} & = & 488.28\ldots & \colorbox{lightblue}{488} \\
\hline \end{array}

Algebra, STD1 EQ-Bank 15

A household's monthly water bill consists of a fixed service charge of $45 plus $3 per kilolitre of water used.

Let $C$ = monthly cost in dollars, and $k$ = water usage in kilolitres.

Complete the table of values below that models the relationship between water usage and monthly cost. (1 mark)

$\begin{array}{|c|c|c|c|c|c|c|} \hline \vphantom{\dfrac{1}{1}}\quad k \quad & \quad 0 \quad & \quad 10 \quad & \quad 20 \quad & \quad 30 \quad & \quad 40 \quad & \quad 50 \quad \\[6pt] \hline \vphantom{\dfrac{1}{1}}C & & 75 & 105 & & 165 & 195 \\[12pt] \hline \end{array}$

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Using the table of values from (a), neatly graph the monthly cost for water usage from 0 to 50 kL on the grid below. (1 mark)

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Using your graph from (b), or otherwise, find the monthly cost when the household uses 35 kL of water. (1 mark)

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a. $\text{Table of values}$

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

c. $$120$

Show Worked Solution

a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|c|} \hline k & 0 & 10 & 20 & 30 & 40 & 50 \\ \hline C & \textbf{45} & 75 & 105 & \textbf{135} & 165 & 195 \\ \hline \end{array}

c. $\text{From the graph, when } k=35,\ \ C=\$120$

Algebra, STD1 EQ-Bank 14

GreenCut Lawn Services charges a fixed call-out fee of $25 plus $40 per hour of work.

Let $C$ = total charge in dollars, and $h$ = number of hours worked.

Complete the table of values below that models the relationship between GreenCut's Lawn Services hours worked and total charge. (1 mark)

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\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad 2\quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & & 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 & & 225 \\
\hline
\end{array}
Using the table of values from (a), neatly graph the total charge for work completed from 0 to 5 hours on the grid below. (1 mark)

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Using your graph from (b), or otherwise, find the total charge for 2.5 hours of work. (1 mark)

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A customer has a budget of $160. Using your graph from (b), or otherwise, determine the maximum number of complete hours GreenCut can work within this budget. (1 mark)

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a. $\text{Table of values:}$

\begin{array}{|c|c|c|c|c|c|c|}
\hline
\quad \rule{0pt}{2.5ex}h\quad \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad &\quad 2\quad & \quad 3 \quad & \quad 4 \quad & \quad 5 \quad\\
\hline
\rule{0pt}{2.5ex}C & \textbf{25}& 65 & 105 \rule[-1ex]{0pt}{0pt}& 145 &\textbf{185} & 225 \\
\hline
\end{array}

c. $$125$

d. $3\ \text{hours}$

Show Worked Solution

a. $\text{Table of values:}$

c. $\text{From the graph, when }\ h=2.5, \ C=\$125$

d. $\text{From the graph, \$160 lies between } h=3 \text{ and } h=4.$

$\therefore\ \text{Maximum complete hours} = 3\ \text{hours}$

Algebra, STD1 A3 2024 HSC 10 MC

A cyclist rides a bicycle at a constant speed around a circular track.

Which of the graphs best illustrates the distance of the cyclist from the centre of the track as time varies?

Show Answers Only

$A$

Show Worked Solution

$\text{The bike rider is always a fixed distance from the centre (length of the radius).}$

$\Rightarrow A$

♦ Mean mark 42%.

Algebra, STD1 A3 2023 HSC 4 MC

The diagram shows water in a pool which is in the shape of a triangular prism. The pool is being emptied of water at a constant rate.

Which graph best illustrates the change in depth of water with time?

Show Answers Only

$D$

Show Worked Solution

$\text{Eliminate B and C as both graphs show the depth of water}$

$\text{increasing with time, but the pool is being emptied, so}$

$\text{depth must decrease with time.}$

$\text{Eliminate A as rate of flow out of the tank is not linear.}$

$\Rightarrow D$

♦ Mean mark 49%.

Algebra, STD1 A3 2022 HSC 10 MC

The diagram shows a container, closed at the base. It is to be filled with water at a constant rate.

Which graph best shows the depth of water in the container as time varies?

Show Answers Only

$A$

Show Worked Solution

$\text{The container will fill at a constant rate for the cylindrical}$

$\text{section of the container → a straight line (linear graph).}$

$\text{The container will fill more slowly at a decreasing rate }$

$\text{for the conical section of the container → a curved line }$

$\text{(non-linear graph).}$

$\text{Therefore, }A\ \text{and }B\ \text{are the only options.}$

$\text{It cannot be graph}\ B\ \text{as it shows the depth increasing}$

$\text{at an increasing rate after the straight line section.}$

$\Rightarrow A$

♦♦♦ Mean mark 24%.

Algebra, STD1 A3 2021 HSC 25

The diagram shows a container which consists of a small cylinder on top of a larger
cylinder.

The container is filled with water at a constant rate to the top of the smaller cylinder. It takes 5 minutes to fill the larger cylinder.

Draw a possible graph of the water level in the container against time. (2 marks)

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Show Worked Solution

♦ Mean mark 38%.

Algebra, STD1 A3 2020 HSC 19

Each year the number of fish in a pond is three times that of the year before.

The table shows the number of fish in the pond for four years.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \ & \ \ \ 2021\ \ \ & \ \ \ 2022\ \ \ & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & & & 2700\\
\hline
\end{array}

Complete the table above showing the number of fish in 2021 and 2022. (2 marks)
Plot the points from the table in part (a) on the grid. (2 marks)
Which model is more suitable for this dataset: linear or exponential? Briefly explain your answer. (2 marks)

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\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \ & \ \ \ 2021\ \ \ & \ \ \ 2022\ \ \ & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

c. $\text{The more suitable model is exponential.}$

$\text{A linear dataset would graph a straight line which is}$

$\text{not the case here.}$

$\text{An exponential curve can be used to graph populations that }$

$\text{grow at an increasing rate, such as this example.}$

Show Worked Solution

c. $\text{The more suitable model is exponential.}$

$\text{A linear dataset would graph a straight line which is}$

$\text{not the case here.}$

$\text{An exponential curve can be used to graph populations that }$

$\text{grow at an increasing rate, such as this example.}$

♦♦ Mean mark (c) 31%.

Algebra, STD1 A3 2020 HSC 14

Adam travels on a straight road away from his home. His journey is shown in the distance–time graph.

Describe the journey in the first 4 minutes by referring to change in speed and distance travelled. (2 marks)

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After the 4 minutes shown on the graph. Adam rests for 2 minutes and then returns home by travelling on the same road at a constant speed. Adam is away from home for a total of 10 minutes.

On the grid above, complete the distance–time graph using the information provided. (2 marks)

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a. `text{Speed: Adam increases speed until approximately}\ \ t=2,`

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Show Worked Solution

a. `text{Speed: Adam increases speed until approximately}\ \ t=2,`

♦ Mean mark part (a) 35%.

`text{and then decreases speed until he stops when}\ \ t=4.`

`text{Distance travelled: Adam’s distance from home increases}`

`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`

`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`

`text(distance from home remains the same.)`

Mean mark part (b) 51%.

b. `text(S)text(ince Adam travels home at a constant speed, the graph is)`

`text{a straight line and ends at (10, 0).}`

Algebra, STD1 A3 2019 HSC 23

Five rabbits were introduced onto a farm at the start of 2018. At the start of 2019 there were 10 rabbits on the farm. It is predicted that the number of rabbits on the farm will continue to double each year.

Complete the following table. (1 mark)
Complete the scale on the vertical axis and then plot the data from part (a) on the grid. (2 marks)
Would a linear model or an exponential model better fit this graph? Explain the reason for your answer. (1 mark)

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