// Numbas version: exam_results_page_options {"type": "exam", "timing": {"timeout": {"message": "", "action": "none"}, "timedwarning": {"message": "", "action": "none"}, "allowPause": true}, "percentPass": 0, "navigation": {"showfrontpage": true, "showresultspage": "oncompletion", "preventleave": true, "browse": true, "reverse": true, "allowregen": true, "onleave": {"message": "", "action": "none"}}, "duration": 0, "question_groups": [{"pickingStrategy": "all-ordered", "pickQuestions": 1, "name": "Group", "questions": [{"name": " Using a speed graph to find distance", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}], "metadata": {"description": "

Use a piecewise linear graph of speed against time to find the distance travelled by a car.

\n

Finally, use the total distance travelled to find the average speed.

", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {}, "type": "question", "ungrouped_variables": ["d2farea", "area"], "advice": "

We can use a speed graph to calculate the distance travelled in a given time interval by finding the area under the line between the start and end times.

\n

#### a)

\n

The shape made by the speed curve, the line $x=0$, and the lines $t=4$ and $t=6$ seconds is a rectangle, so we can work out the area of this section by multiplying the width by the height.

\n

The rectangle is $2$ seconds wide, and $\\var{c1}$ ms-1 high.

\n

\\begin{align}
\\text{Area} &= \\text{width} \\times \\text{height}\\\\
&= 2 \\times\\var{c1}\\\\
&=\\simplify{2{c1}}\\text{.}
\\end{align}

\n

So the distance covered in this two second interval is $\\simplify{2{c1}}$ m.

\n

#### b)

\n

The shape made by the line and $x=0$ between $0$ and $2$ seconds forms a right-angled triangle with width $2$ and height $\\var{b1}$.

\n

\\begin{align}
\\text{Area}&=  \\frac{1}{2}\\times \\text{width} \\times \\text{height}\\\\
&= \\frac{1}{2} \\times 2 \\times \\var{b1}\\\\
&=\\var{b1} \\text{.}
\\end{align}

\n

So therefore,the distance covered in this two second interval, and our answer, is $\\simplify{{b1}}$ meters.

\n

#### c)

\n

The shape made by the speed curve and $x=0$ between $2$ and $4$ seconds forms a trapezium. This can be broken down in to a right angle triangle (let's call this $A$) and a rectangle (we'll call this $B$).

\n

Triangle $A$ has width $2$ m and height $\\var{c1}-\\var{b1}$ ms-1.

\n

\\begin{align}
A &=  \\frac{1}{2}\\times \\text{width} \\times \\text{height}\\\\
&= \\frac{1}{2}\\times2 \\times\\ (\\var{c1}-\\var{b1})\\\\
&= \\var{c1}-\\var{b1}\\\\
&=\\simplify{{c1}-{b1}}\\text{.}
\\end{align}

\n

We can work out the area of the rectangle $B$ by multiplying its width, $2$ seconds, by its height, $\\var{b1}$ ms-1:

\n

\\begin{align}
B &=  \\text{width} \\times \\text{height}\\\\
&= 2 \\times(\\var{c1}-\\var{b1})\\\\
&=2 \\times \\simplify{{c1}-{b1}}\\\\
&=\\simplify{2{c1-b1}}\\text{.}
\\end{align}

\n

We can now work out the whole area under the line by adding these two areas together:

\n

\\begin{align}
\\text{Area} &= A + B \\\\
&=\\simplify{{c1}-{b1}} + \\simplify{2{c1-b1}} \\\\
&=\\simplify{2{c1-b1}+{c1}-{b1}} \\text{.}
\\end{align}

\n

The distance covered in this interval is $\\var{2(c1-b1)+c1-b1}$ m.

\n

#### d)

\n

Speed is the distance travelled per unit of time.

\n

\\begin{align}

\n

\n

#### b)

\n

To convert $\\var{km}$ km into miles, we multiply $\\var{km}$ by the conversion rate given: $0.62$.

\n

\\\begin{align} \\var{km}\\times\\frac{5}{8}&= \\var{miles}\\\\ &=\\var{dpformat(miles,0)}\\text{ miles, rounded to the nearest integer.} \\end{align}\

\n

#### c)

\n

The {location} Marathon is $26$ miles long. To convert to km we multiply by the inverse of the conversion rate given in part b):

\n

\$26 \\times \\frac{1}{0.62} = 42\\text{ miles, rounded to the nearest integer.} \$

\n

", "tags": ["taxonomy"], "preamble": {"js": "", "css": ""}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["km_to_miles", "km", "miles", "person", "location"], "statement": "

{person['name']} is training for the {location} Marathon.

Convert from km to metres and miles, and miles to km.

"}, "variablesTest": {"condition": "", "maxRuns": 100}}, {"name": "Converting units of height (feet/inches/cm)", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}], "variable_groups": [], "preamble": {"js": "", "css": ""}, "type": "question", "parts": [{"stepsPenalty": 0, "type": "gapfill", "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "allowFractions": false, "scripts": {}, "minValue": "cm", "maxValue": "cm", "marks": "2", "variableReplacements": []}], "marks": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "prompt": "

What is {person['name']}'s height in centimetres?

\n

", "steps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "allowFractions": false, "scripts": {}, "minValue": "60", "maxValue": "60", "marks": "0.5", "variableReplacements": [], "prompt": "

We have information on how to convert feet to inches and inches to cm, but not feet to cm. We will therefore first convert the height into inches only.

\n

What is $5$ft in inches?

"}, {"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "allowFractions": false, "scripts": {}, "minValue": "60+inches", "maxValue": "60+inches", "marks": "0.5", "variableReplacements": [], "prompt": "

What is {person['name']}'s height in inches?

"}, {"variableReplacementStrategy": "originalfirst", "type": "information", "showCorrectAnswer": true, "marks": 0, "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "prompt": "

We can now use the conversion rate for inches to cm to find {person['name']}'s height in cm.

"}]}, {"variableReplacementStrategy": "originalfirst", "type": "gapfill", "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "mustBeReducedPC": 0, "showFeedbackIcon": true, "precisionType": "dp", "correctAnswerStyle": "plain", "allowFractions": false, "scripts": {}, "strictPrecision": false, "minValue": "cm/100", "maxValue": "cm/100", "marks": 1, "variableReplacements": [], "showPrecisionHint": false, "precision": "2", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"]}], "marks": 0, "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "prompt": "

What is {person['name']}'s height in metres?

\n

\n

\n\n

{person['name']} is $5$ft $\\var{inches}$ inches tall.

\n

#### a)

\n

To find {person['name']}'s height into cm, we first convert it into inches.

\n

We can convert feet into inches by multiplying by $12$. So $5$ feet is

\n

\$5\\times12=60 \\text{ inches.}\$

\n

Therefore $5$ft $\\var{inches}$ inches is

\n

\$60+\\var{inches} = \\var{60+inches}\\text{ inches.} \$

\n

We can now convert this to cm by multiplying by $2.54$.

\n

\$\\var{60+inches}\\times2.54=\\var{cm}\\text{ cm, rounded to the nearest integer.}\$

\n

#### b)

\n

To convert centimetres into metres, we divide by $100$:

\n

\$\\var{cm}\\div100=\\var{dpformat(cm/100,2)} \\text{ metres.}\$

\n

Therefore, {person['name']} is $\\var{dpformat(cm/100,2)}$ metres tall.

\n

\n", "tags": ["taxonomy"], "variables": {"cm": {"templateType": "anything", "description": "", "definition": "precround((inchfeet*2.54),0)", "name": "cm", "group": "Ungrouped variables"}, "inchfeet": {"templateType": "anything", "description": "", "definition": "(5*12)+inches", "name": "inchfeet", "group": "Ungrouped variables"}, "person": {"templateType": "anything", "description": "", "definition": "random_person()", "name": "person", "group": "Ungrouped variables"}, "s": {"templateType": "anything", "description": "", "definition": "if(person['gender']='neutral','','s')", "name": "s", "group": "Ungrouped variables"}, "inches": {"templateType": "anything", "description": "", "definition": "random(4..11)", "name": "inches", "group": "Ungrouped variables"}}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["person", "inches", "inchfeet", "cm", "s"], "statement": "

{person['name']} is $5$ft $\\var{inches}$ inches tall and would like to find {person['pronouns']['their']} height in cm.

\n

{capitalise(person['pronouns']['they'])} find{s} the following unit conversion table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $1$ foot $12$ inches $1$ inch $2.54$ cm $1$ metre $100$ cm

Convert a height given in feet and inches into cm and then metres.

"}, "variablesTest": {"condition": "", "maxRuns": "1000"}}, {"name": "Using compound units - speed", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "variable_groups": [], "preamble": {"js": "", "css": ""}, "type": "question", "parts": [{"variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "variableReplacements": [], "precisionMessage": "

Round your answer to $2$ decimal places.

", "precisionType": "dp", "mustBeReducedPC": 0, "precisionPartialCredit": 0, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "allowFractions": false, "showPrecisionHint": false, "strictPrecision": false, "maxValue": "distance/seconds", "precision": "2", "marks": 1, "scripts": {}, "minValue": "distance/seconds", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"]}], "variableReplacements": [], "showFeedbackIcon": true, "prompt": "

What was the runner's average speed, in metres per second?

\n

[[0]] m/s

\n

"}, {"variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "marks": 0, "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "variableReplacements": [], "precisionMessage": "

Round your answer to $2$ decimal places.

", "precisionType": "dp", "mustBeReducedPC": 0, "precisionPartialCredit": 0, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "allowFractions": false, "showPrecisionHint": false, "strictPrecision": false, "maxValue": "distance/seconds*3.6", "precision": "2", "marks": 1, "scripts": {}, "minValue": "distance/seconds*3.6", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"]}], "variableReplacements": [], "showFeedbackIcon": true, "prompt": "

How fast is this in kilometres per hour?

\n

[[0]] km/h

#### a)

\n

To find the average speed of the runner in meters per second (m/s), we divide the distance covered by the runner (in metres) by the time taken for the runner to run this distance (in seconds).

\n

\\\begin{align} \\text{Average speed} &= \\displaystyle\\frac{\\var{distance}}{\\var{seconds}}\\\\ &= \\var{distance/seconds}\\\\ &= \\var{dpformat(distance/seconds,2)}\\; \\text{m/s} \\; (\\text{rounded to 2 decimal places}). \\end{align} \

\n

#### b)

\n

We can convert the average speed of the runner that we calculated in a) in metres per second to kilometres per hour using the following two equivalences:

\n

\$1\\text{m} = \\displaystyle\\frac{1}{1000}\\text{km},\$

\n

\$1 \\; \\text{second} = \\displaystyle\\frac{1}{60} \\; \\text{minutes} = \\displaystyle\\frac{1}{3600} \\; \\text{hours}. \$

\n

We know from a) that the average speed of the runner in m/s was $\\var{dpformat(distance/seconds,5)}$ m/s ($5$ d.p), so to convert this speed to km/h we first need to convert metres to kilometres,

\n

\$\\var{dpformat(distance/seconds,5)} \\; \\text{m/s} = \\var{dpformat(distance/seconds/1000,5)} \\text{km/s} \\; (5 \\; \\text{d.p})\$

\n

Then we convert seconds to hours,

\n

\$1 \\; \\text{second} = \\displaystyle\\frac{1}{3600} \\; \\text{hours}.\$

\n

Now we have

\n

\$\\var{dpformat(distance/seconds,5)} \\; \\text{m/s} = \\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per \\displaystyle\\frac{1}{3600} hours}.\$

\n

We want a rate per one hour, so we multiply by $3600$ to obtain a measurement in km/h:

\n

\\\begin{align}\\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per \\displaystyle\\frac{1}{3600} hours} &= \\var{siground(distance/seconds/1000,5)*3600} \\; \\text{km}/\\text{h}\\\\&=\\var{dpformat(distance/seconds*3.6, 2)} \\; \\text{km}/\\text{h} \\; (\\text{rounded to 2 decimal places}).\\end{align}\

\n

\$\\var{sigformat(distance/seconds/1000,5)} \\; \\text{kilometres per \\displaystyle\\frac{1}{3600} hours} = \\var{distance/seconds*3.6} \\; \\text{km}/\\text{h}.\$

\n

Note that throughout this calculation we have rounded all figures to $5$ decimal places for convenience; when doing calculations which involve long decimals, you should always input the full figure into your calculator to avoid getting an incorrect answer due to rounding.

", "tags": ["taxonomy"], "variables": {"speed": {"templateType": "anything", "description": "

Athlete's speed, in m/s.

\n

4m/s is about 9 mph, a bit faster than a jog. The current world record is 12m/s.

", "definition": "random(4..8#0)", "name": "speed", "group": "Ungrouped variables"}, "distance": {"templateType": "anything", "description": "

Distance that the runner ran.

", "definition": "random(80,100,150,200)", "name": "distance", "group": "Ungrouped variables"}, "seconds": {"templateType": "anything", "description": "

Time taken to cover the distance, in seconds.

", "definition": "floor(distance/speed)", "name": "seconds", "group": "Ungrouped variables"}}, "rulesets": {}, "variablesTest": {"condition": "", "maxRuns": "100"}, "functions": {}, "ungrouped_variables": ["distance", "seconds", "speed"], "statement": "

An athlete runs $\\var{distance}$ m in $\\var{seconds}$ seconds.

\n

Calculate a speed in m/s given distance and time taken, then convert that to km/hour

"}}, {"name": "Use speed and distance to calculate time ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "metadata": {"description": "

Calculate the time taken for a certain distance to be travelled given the average speed and the distance travelled.

\n

Small, simple question.

", "licence": "Creative Commons Attribution 4.0 International"}, "ungrouped_variables": ["kilometres", "speed"], "type": "question", "rulesets": {}, "variable_groups": [], "statement": "

At a greyhound race, a fake rabbit moves around the inside of the track to motivate the dogs to run.

\n

The track is $\\var{kilometres}$km long and the rabbit moves at a constant speed of $\\var{speed}$m/s.

We are told that the track is $\\var{kilometres}$km long and that the speed of the rabbit is $\\var{speed}$m/s (metres per second).

\n

Most people remember the relationship between speed, distance and time with the formula

\n

\$\\text{Average speed} = \\frac{\\text{Distance travelled}}{\\text{Total time taken}}.\$

\n

We can rearrange the formula for average speed to give us the formula for the time taken:

\n

\$\\text{Total time taken} = \\displaystyle\\frac{\\text{Distance travelled}}{\\text{Average speed}}.\$

\n

Firstly, we must convert the units of the length of the race from kilometres to metres. We know that $1\\text{km} = 1000\\text{m}$, therefore

\n

\\begin{align}
\\var{kilometres}\\text{km} &= (\\var{kilometres} \\times 1000)\\text{m}\\\\
&= \\var{{kilometres}*1000}\\text{m}.
\\end{align}

\n

Therefore, the time taken for the rabbit to finish one lap is

\n

\$\\displaystyle\\frac{\\var{{kilometres}*1000}}{\\var{speed}} = \\var{({kilometres}*1000)/{speed}} \\; \\text{seconds}.\$

\n

\\\begin{align} \\displaystyle\\frac{\\var{{kilometres}*1000}}{\\var{speed}} &= \\var{({kilometres}*1000)/{speed}} \\; \\text{seconds}\\\\ &= \\var{dpformat(({kilometres}*1000)/{speed}, 0)} \\; \\text{seconds} \\; (\\text{to the nearest second}). \\end{align}\

", "parts": [{"scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "variableReplacements": [], "stepsPenalty": 0, "steps": [{"scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "information", "showCorrectAnswer": true, "prompt": "

Speed, distance and time are related by the equation

\n

\$\\text{Average speed} = \\frac{\\text{Distance travelled}}{\\text{Total time taken}}.\$

", "variableReplacements": [], "showFeedbackIcon": true, "marks": 0}], "prompt": "

Calculate the time taken for the rabbit to complete a circuit of the track.

\n

", "gaps": [{"correctAnswerFraction": false, "variableReplacements": [], "precisionPartialCredit": 0, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "marks": "2", "precision": 0, "showFeedbackIcon": true, "precisionType": "dp", "minValue": "({kilometres}*1000)/{speed}", "scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "({kilometres}*1000)/{speed}", "mustBeReduced": false, "allowFractions": false, "strictPrecision": false, "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "showPrecisionHint": false, "precisionMessage": "

"}], "showFeedbackIcon": true, "marks": 0, "showCorrectAnswer": true}], "tags": ["compound measures", "Compound measures", "Compound units", "compound units", "Distance", "distance", "km", "metres per second", "speed", "taxonomy", "Time", "time"], "preamble": {"js": "", "css": ""}, "functions": {}, "variables": {"speed": {"description": "

Speed of the greyhound in metres per second.

", "group": "Ungrouped variables", "definition": "random(14..18)", "name": "speed", "templateType": "anything"}, "kilometres": {"description": "

Kilometres to be ran by the greyhound

", "group": "Ungrouped variables", "definition": "random(0.35..0.49 #0.01)", "name": "kilometres", "templateType": "anything"}}, "variablesTest": {"maxRuns": 100, "condition": ""}}, {"name": "Straight line equation application: measuring sunflower height", "extensions": ["jsxgraph", "random_person"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Vicky Hall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/659/"}, {"name": "Bradley Bush", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1521/"}], "metadata": {"description": "

An applied example of the use of two points on a graph to develop a straight line function, then use the t estimate and predict. MCQ's are also used to develop student understanding of the uses of gradient and intercepts as well as the limitations of prediction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{person['name']} is given a sunflower seedling for {person['pronouns']['their']} 30th birthday (day $0$) and observes its height. {capitalise(person['pronouns']['they'])} make{s} the following observations later that week:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 Observation A B Day $\\var{xa}$ $\\var{xb}$ height (cm) $\\var{ya}$ $\\var{yb}$
\n

{person['name']} plots the 2 points:

\n

{plotPoints()}

", "variables": {"m": {"group": "Ungrouped variables", "name": "m", "description": "

", "templateType": "anything", "definition": "random(2..3)"}, "yb": {"group": "point coordinates", "name": "yb", "description": "

y coordinate of point B

", "templateType": "anything", "definition": "m*xb+c"}, "d": {"group": "Ungrouped variables", "name": "d", "description": "

A number of days after receiving the seedling, on which the height is estimated

", "templateType": "anything", "definition": "random([10,14,20])"}, "ya": {"group": "point coordinates", "name": "ya", "description": "

y coordinate of point A

", "templateType": "anything", "definition": "m*xa+c"}, "c": {"group": "Ungrouped variables", "name": "c", "description": "

The intercept

", "templateType": "anything", "definition": "random(2..4)"}, "s": {"group": "Random person", "name": "s", "description": "

He makes, they make.

", "templateType": "anything", "definition": "if(person['gender']='neutral','','s')"}, "xb": {"group": "point coordinates", "name": "xb", "description": "

x coordinate of point B

", "templateType": "anything", "definition": "xa+1"}, "person": {"group": "Random person", "name": "person", "description": "

A random person

", "templateType": "anything", "definition": "random_person()"}, "xa": {"group": "point coordinates", "name": "xa", "description": "

x coordinate of point a

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#### a)

\n

The gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).

\n

\$m = \\frac{y_2-y_1}{x_2-x_1}=\\frac{\\simplify[!collectNumbers]{{yb}-{ya}}}{\\simplify[!collectNumbers]{{xb}-{xa}}}=\\frac{\\simplify{{yb}-{ya}}}{\\simplify{{xb}-{xa}}}=\\simplify[simplifyFractions,unitDenominator]{({yb-ya})/({xb-xa})}\\text{.}\$

\n

#### b)

\n

Rearranging the equation $y=mx+c$ for $c$ and using point A:

\n

\$c = y_1-mx_1 = \\var{ya}-\\var{m}\\times\\var{xa}=\\simplify{{ya-m*xa}}\\text{.}\$

\n

We then check this against point $B$:

\n

\$y_2 = mx_2 + c = \\simplify[fractionNumbers]{{m}{xb}+{c}}=\\simplify{{m}*{xb}+{c}}\\text{.}\$

\n

#### b)

\n

We now substitute the values for $m$ and $c$ into the equation of a straight line, $y=mx+c$,

\n

\$y=\\simplify[!noLeadingMinus,unitFactor]{{m} x+ {c}}\\text{.}\$

\n

\n

#### c)

\n

The gradient represents the vertical change (height in cm) per unit of the horizontal axis (days): the change in height of the sunflower per day.

\n

#### d)

\n

Substituting $x=\\var{d}$ into the straight line equation, the height $y$ after $\\var{d}$ days is

\n

\\begin{align}
y&=\\simplify{{m}}x+\\var{c}\\\\
&=\\simplify[]{{m}{d}}+\\var{c}\\\\
&=\\var{m*d+c}\\text{cm.}
\\end{align}

\n

#### e)

\n

Substituting $x=\\var{1826}$ into the straight line equation, the height $y$ after $1826$ days is

\n

\\begin{align}
y&=\\simplify{{m}}x+\\var{c}\\\\
&=\\simplify[]{{m}1826} + \\var{c}\\\\
&=\\var{m*1826+c}\\text{cm.}
\\end{align}

\n

Note that this is $\\var{(m*1826+c)/100}$ metres. In 2014, a sunflower of $9.17$ metres was entered into the Guinness World Records as tallest sunflower.

\n

#### f)

\n

Possible reasons that the prediction will not be accurate are:

\n
\n
• from everyday observations and basic biology that a sunflower cannot keep growing forever, but our straight line equation would predict otherwise.
• \n
• the set of observations are limited in their time frame. In particular another relationship could easily fit the 2 data points.
• \n
\n

Invalid reasons that the prediction will not be accurate are:

\n
\n
• There can only be one straight line between two points; only one equation describes this.
• \n
• Based on common sense, we know that sunflowers do grow over time.
• \n
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What is the gradient, $m$, of the straight line between the two points?

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$m =$ [[0]]

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Use the gradient and the coordinates of the two points to find the height of the sunflower when {person['name']} received it.

\n

[[0]] cm.

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Let $y$ be the sunflower height and $x$ the time, in days, since {person['name']} received the sunflower. What is the equation of the straight line between the points?

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$y(x) =$ [[0]]

\n

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The length of time taken in days for the sunflower to grow $1$ cm

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The change in height (in cm) of the sunflower over $1$ day

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The width of the ruler used to measure the sunflower

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All of the above

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{person['name']} uses the straight line equation to predict the future height of the sunflower. What will the height be on day $\\var{d}$?

\n

[[0]] cm

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{person['name']} wonders if {person['pronouns']['they']} can guess what the height of the sunflower will be on {person['pronouns']['their']} 35th birthday. {capitalise(person['pronouns']['they'])} work{s} out that this is day 1826. Using the straight line equation, what would the height be on day 1826?

\n

[[0]] cm

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Sunflower height as a function of time may not have a straight linear relationship.

", "

The observations only span a very limited time range.

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There are multiple straight linear relationships that could be obtained using the same $2$ data points.

", "

Sunflower height never actually increases over time.

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Some questions to do with measures of distance and length.

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