// Numbas version: exam_results_page_options {"name": "TLU Workshop", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", ""], "questions": [{"name": "simplify linear", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Simplify the following algebraic expression:

", "advice": "

Remember to group like terms. Group 'x' terms with 'x' terms and numbers with numbers.

\n

$\\var{a}x+\\var{b}+\\var{c}x$

\n

= $\\var{a}x+\\var{c}x+\\var{b}$

\n

=$\\var{d}x+\\var{b}$

\n

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$\\var{a}x+\\var{b}+\\var{c}x$

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rebelmaths

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

The following table lists five pairs of $x$ and $f$ values.

\n\n\n\n\n\n\n\n\n\n\n\n
$\\mathbf{x}${x1}{x2}{x3}{x4}{x5}
$\\mathbf{f} ${f1}{f2}{f3}{f4}{f5}
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$\\sum x = $  [[0]]

\n

$\\sum f = $ [[1]]

\n

$\\sum fx = $ [[2]]

\n

$\\sum fx^2 =$ [[3]]

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$\\Sigma$ just means 'sum of' or total.

\n

For example:

\n

$\\Sigma x$ means add up the $x$ values

\n

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The managers at a chemical company claim that one of their products contains on average {mu0} ml of caustic material per litre. If there is too much caustic material, the product will be dangerous. A government inspector is brought in to test the product. She randomly selects a sample of {sample_size}, litre-size containers of the product and finds that the mean weight of caustic materials is {sample_mean} ml per litre with a sample standard deviation of {sample_stdev} ml

", "advice": "

In this example we have a single sample of {sample_size} data values. 

\n

\\(H_0:\\) $\\mu$ \\(=\\var{mu0}\\)

\n

\\(H_A:\\) $\\mu > \\var{mu0}$ 

\n

Given a sample of size \\(n\\) recall:

\n

the formula for the sample mean:    \\(\\overline{x}=\\frac{\\sum {x}}{n}=\\var{sample_mean}\\)

\n

the formula for the sample standard deviation:   \\(s=\\sqrt{\\frac{\\sum{(x-\\overline{x})^2}}{n-1}}=\\var{sample_stdev}\\)

\n

the formula for the t-statistic:   \\(t_{stat}=\\frac{\\overline{x}-\\mu}{\\frac{s}{\\sqrt{n}}}=\\frac{\\var{sample_mean}-\\var{mu0}}{\\frac{\\var{sample_stdev}}{\\sqrt{\\var{sample_size}}}}=\\var{test_statistic}\\)

\n

the absolute value of $t_{stat} = |t_{stat}| = |\\var{test_statistic}|=\\var{abs_test}$

\n

For a significance level of 5%:

\n

$t_{{\\alpha},\\var{sample_size}-1}=\\var{critical}$

\n


d)

\n

{Correctc}

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Null Hypothesis

\n

$\\operatorname{H}_0\\;: \\; \\mu=\\;$[[0]]

\n

Alternative Hypothesis

\n

$\\operatorname{H}_A\\;: \\; \\mu > $[[1]]

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Give all your answers correct to two decimal places.

\n

Input the sample mean:   \\(\\bar{x}=\\) [[0]]

\n

Input the sample standard deviation:   \\(s=\\) [[1]]

\n

Should we use the z or t test statistic? [[2]] enter z or t

\n

Enter the value for the test statistic:    [[3]]

\n

Enter the value for the absolute value of the test statistic [[4]]

\n

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Use tables to find the critical value for a significance level of 5%. 

\n

\n

critical value = [[0]]

\n

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Your Decision: [[0]]

\n

\n

\n

 

\n

Conclusion: [[1]]

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Simple Interest, $I=P\\times t\\times r$

\n

$I=\\var{initial}\\times \\var{n}\\times \\var{p}/100$

\n

$I=\\var{interest}$

\n

Adding the interest to the initial amount gives €$\\var{answer1}$.

\n

\n

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An investor puts €$\\var{initial}$ in a savings account that pays $\\var{p}\\%$ simple interest at the end of each year. How much would the investor have after $\\var{n}$ years?

\n

Give your answer to $2$ decimal places.

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Simple interest.

\n

rebelmaths

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The compound interest formula is: $\\ A = P(1+i)^n $

\n

Part (a)

\n

P represents the principal sum invested , so in this example it is €$\\var{P}$.

\n

Part (b)

\n

i represents the rate of compound interest, in this example, the annual interest rate is $\\var{perc}$% so i is $\\frac {\\var{perc}} {100}=\\var{int}$.

\n

Part (c)

\n

n represents the number of compounding periods , so in this example it is $\\var{n}$ years.

\n

Part(d)

\n

The amount in the deposit account after $\\var{n}$ years is denoted by A. Using the compound interest formula:

\n

$A=P(1+i)^n$

\n

$A=\\var{P} \\times(1+\\var{int})^\\var{n}=\\var{A}$

", "rulesets": {}, "parts": [{"prompt": "

What is the value of P?

\n

€[[0]]

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What is the value of i ?

\n

\n

[[0]]

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What is the value of n?

\n

[[0]]

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How much will be in the deposit account after $\\var{n}$ years? Please give your answer to the nearest cent.

\n

€[[0]]

\n

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A lump sum of €$\\var{P}$ is deposited into a savings account, that pays compound interest at a rate of  $\\var{perc}$% per annum for $\\var{n}$ years. If no withdrawals are made from the account, then the amount that the lump sum will have grown to is given by the formula: 

\n

\n

$\\ A = P(1+i)^n $

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Calculate the final amount in a savings account where compound interest is earned annually 

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A gold ball on a steel table - effect of its own weight.

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

The size of the contact and the pressure between the contacting surfaces depends on:

\n
    \n
  1. the materials, and here we assume them to be linear, elastic and isotropic;
  2. \n
  3. the geometry, and here we assume them to be either flat (infinite radius) or having a constant radius;
  4. \n
  5. the applied load, and here we assume the load is applied normal to the contact, i.e., there is no friction.
  6. \n
\n

\n

Materials: A single material property - the Elastic Contact Modulus ($E^*$) - combines the elastic properties of both contacting surfaces into an equivalent stiffness:

\n

\\[{1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2}\\]

\n

where $E_1$ and $E_2$ are the Young's elastic moduli of the two materials, and $\\nu_1$ and $\\nu_2$ are the corresponding Poisson's ratios.

\n

Geometry: The two simplest forms of contact between curved surfaces are:

\n\n

In either case, an equivalent radius $R$ can be defined. If both surfaces are convex, e.g., two balls touching, then:

\n

\\[{1 \\over R} = {1 \\over R_1} + {1 \\over R_2}\\]

\n

where $R_1$ is the radius of Surface 1 and $R_2$ is the radius of Surface 2. If, however, one surface is concave, e.g., a ball in a cup, then:

\n

\\[{1 \\over R} = {1 \\over R_1} - {1 \\over R_2}\\]

\n

where $R_1$ is the radius of (convex) Surface 1 and $R_2$ is the radius of (concave) Surface 2. (The 'cup' radius must be larger than the 'ball' radius.)

\n

Applied Load: This is a little tricky because the same notation, $P$, is used to mean different things:

\n\n

2D Contact: The peak contact pressure, $p_0$, and semi-contact width, $a$, are given by:

\n

\\[p_0 = \\left({P E^* \\over \\pi R}\\right)^{1 \\over 2}\\]

\n

\\[a = \\left({4 P R \\over \\pi E^*}\\right)^{1 \\over 2}\\]

\n

3D Contact: The peak contact pressure, $p_0$, and semi-contact width, $a$, are given by:

\n

\\[p_0 = \\left({6 P {E^*}^2 \\over \\pi^3 R^2}\\right)^{1 \\over 3}\\]

\n

\\[a = \\left({3 P R \\over 4 E^*}\\right)^{1 \\over 3}\\]

", "advice": "\n

${1 \\over R} = {1 \\over R_1} + {1 \\over R_2} = {1 \\over \\var{diawheel} \\div 2} + {1 \\over \\var{diarail} \\div 2}$ [units: mm$^{-1}$]

\n

which can be rearranged to give $R = \\var{siground(R*1000,3)}$mm.

\n\n

${1 \\over E^*} = {1 - \\nu_1^2 \\over E_1} + {1 - \\nu_2^2 \\over E_2} ={1 - 0.3^2 \\over 209 \\times 10^9} + {1 - 0.3^2 \\over 209 \\times 10^9}$

\n

which can be rearranged to give $E^* = \\var{siground(ECM,3)}$GPa.

\n\n

$p_0 = \\left({P E^* \\over \\pi R}\\right)^{1 \\over 2} =\\left({ \\left( \\var{load} \\times 10^3 \\div \\var{width} \\times 10^{-3} \\right) \\times \\var{siground(ECM,3)} \\times 10^9 \\over \\pi \\times \\var{siground(R*1000,3)} \\times 10^{-3}} \\right)^{1 \\over 2} = \\var{siground(p0,3)}$MPa.

\n\n

$a = \\left({4 P R \\over \\pi E^*}\\right)^{1 \\over 2} = \\left({4 \\times\\left( \\var{load} \\times 10^3 \\div \\var{width} \\times 10^{-3} \\right) \\times \\var{siground(R*1000,3)} \\times 10^{-3} \\over \\pi \\times \\var{siground(ECM,3)} \\times 10^9}\\right)^{1 \\over 2} = \\var{siground(scw,3)}$mm.

\n\n

area = $2 \\times \\var{siground(scw,3)}$mm $\\times \\var{width}$mm $= \\var{siground(area,3)}$mm$^2$.

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Semi-contact width. [Units: mm]

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Peak contact pressure. [Units: MPa]

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Applied load. [Units: kN]

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Rail disc diameter. [Units: mm]

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Elastic Contact Modulus. [Units: GPa]

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Track width. [Units: mm]

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Contact area. [Units: mm$^2$]

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Load per unit length. [Units: N/m]

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Equivalent radius. [Units: m]

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Wheel disc diameter. [Units: mm]

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Reference values:

\n\n

In a twin-disc test, a 'wheel' disc (machined from a steel train wheel) with diameter $\\var{diawheel}$ mm runs against a 'rail' disc (machined from a steel railway rail) with diameter $\\var{diarail}$ mm.

\n

The disc surfaces are cylindrical with the axes aligned. The width of the surfaces is $\\var{width}$ mm.

\n

The applied load is $\\var{load}$ kN.

\n

Calculate:

\n\n

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Example of a dilution calculation involving mass concentration and molarity calculations.

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A \"stock\" solution contained $\\var{stock_mass}$g of sodium chloride (NaCl) in $\\var{stock_volume}$L of solution. $\\var{stock_volume_used}$ml of the \"stock\" solution was diluted to $\\var{final_volume}$ml. What is the concentration of the final solution in g/L and M? [Relative atomic masses (Da): Na $\\var{Na}$; Cl $\\var{Cl}$].

\n

[[0]] g/L

\n

[[1]] M

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1) Our \"stock\" solution contains $\\var{stock_mass}$g of sodium chloride (NaCl) in $\\var{stock_volume}$L of solution. Calculate the concentration of the \"stock\" solution in g/L.

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2) Calculate the molecular weight of sodium chloride (NaCl)

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3) Calculate how many moles of sodium chloride there are in $\\var{stock_mass}$g.

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4) Calculate the concentration of the \"stock\" solution in M (mol/L).

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5) Calculate what $\\var{stock_volume_used}$ml is as a volume in litres.

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6) Calculate how many grams of sodium chloride there are in $\\var{stock_volume_used}$ml of the \"stock\" solution using your answers to parts 1 and 5.

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7) Calculate how many moles of sodium chloride there are in $\\var{stock_volume_used}$ml of the \"stock\" solution using your answers to parts 4 and 5.

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8) Calculate what $\\var{final_volume}$ml is as a volume in litres.

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9) Using your answers to parts 6 and 8, calculate the concentration in g/L of the final solution. 

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10) Using your answers to parts 7 and 8, calculate the concentration in M of the final solution. 

", "correctAnswerFraction": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain", "type": "numberentry", "mustBeReduced": false}], "scripts": {}, "type": "gapfill"}], "ungrouped_variables": ["stock_volume_used", "Cl", "Na", "NaCl", "stock_mass_concentration", "final_mass_concentration", "stock_volume", "stock_molarity", "final_volume", "final_molarity", "stock_mass"], "functions": {}, "preamble": {"js": "", "css": ""}, "variable_groups": [{"name": "Unnamed group", "variables": []}], "advice": "

We have been asked to do quite a lot of calculations in this example so let's break it up into several smaller parts. To work out the concentrations in g/L and M after dilution, we need to first work out the concentrations in g/L and M of the \"stock\" solution. Let's start by working out the concentration of the \"stock\" solution in g/L.

\n

\n

The \"stock\" solution contains $\\var{stock_mass}$g of sodium chloride dissolved in $\\var{stock_volume}$L of solution. To work out the concentration in g/L we use the formula

\n

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

\n

Putting in our numbers we find that the concentration of the \"stock\" solution in g/L is

\n

$\\dfrac{\\var{stock_mass}}{\\var{stock_volume}} = \\var{stock_mass / stock_volume} \\text{ g/L}.$

\n

\n

Next, we can work out the concentration of the stock solution in M (mol/L). The formula for this is

\n

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}.$

\n

\n

We can see that we need to work out how many moles of sodium chloride (NaCl) there are in $\\var{stock_mass}$g. The formula for this is

\n

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{molecular weight}} = \\text{number of moles}.$

\n

\n

We can see that to do this, we are going to need to work out the molecular weight of sodium chloride (NaCl). We have been told the atomic weights of each of the atoms in sodium chloride (NaCl) so we can do this by adding together the atomic weights of all the atoms to get the molecular weight which is

\n

$1 \\times \\var{Na} + 1 \\times \\var{Cl} = \\var{NaCl} \\text{ Da}.$

\n

\n

We can now use this to work out how many moles of sodium chloride there are in $\\var{stock_mass}$g using the formula above. Putting in the numbers we find that there are

\n

$\\dfrac{\\var{stock_mass}}{\\var{NaCl}} = \\var{stock_mass / NaCl} \\text{ moles}$

\n

of sodium chloride in $\\var{stock_mass}$g. We can now work out the concentration of the \"stock\" solution in M (mol/L). The formula is 

\n

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}.$

\n

We have $\\var{stock_mass}$g of sodium chloride dissolved in $\\var{stock_volume}$L of liquid and we have calculated that there are $\\var{stock_mass / NaCl}$ moles of sodium chloride in $\\var{stock_mass}$g. Putting these numbers into the formula we find that the concentration of the \"stock\" solution in M is

\n

$\\dfrac{\\var{stock_mass / NaCl}}{\\var{stock_volume}} = \\var{stock_molarity} \\text{ M}.$

\n

\n

Let's recap what we have calculated so far. We have worked out that the concentration of the \"stock\" solution in g/L is 

\n

$\\var{stock_mass_concentration} \\text{ g/L}$

\n

and in M is

\n

$\\var{stock_molarity} \\text{ M}.$

\n

\n

Since we are diluting $\\var{stock_volume_used}$ml of the \"stock\" solution, we need to work out how much sodium chloride there is in $\\var{stock_volume_used}$ml of the solution using the concentrations we have calculated. We need to work this amount out in both grams and moles, let's start with grams. If we rearrange the formula 

\n

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

\n

we find that 

\n

${\\text{volume of liquid (in litres)}} \\times \\text{concentration (in g/L)} = \\text{mass of substance (in grams)}.$

\n

\n

Now, $\\var{stock_volume_used}$ml is equal to 

\n

$\\dfrac{\\var{stock_volume_used}}{1000} = \\var{stock_volume_used / 1000} \\text{ L},$

\n

so in $\\var{stock_volume_used}$ml of \"stock\" solution, there are

\n

$\\var{stock_volume_used / 1000} \\times \\var{stock_mass_concentration} = \\var{(stock_volume_used / 1000) * stock_mass_concentration} \\text{ g}$

\n

of sodium chloride. Similarly, we can work out the number of moles of sodium chloride using the formula

\n

${\\text{volume of liquid (in litres)}} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}.$

\n

Putting in our numbers, we find that in $\\var{stock_volume_used}$ml of \"stock\" solution, there are

\n

$\\var{stock_volume_used / 1000} \\times \\var{stock_molarity} = \\var{(stock_volume_used / 1000) * stock_molarity} \\text{ moles}$

\n

of sodium chloride.

\n

\n

Finally, we can work out the concentration of the diluted solution. We know that $\\var{stock_volume_used}$ml of \"stock\" solution is diluted to $\\var{final_volume}$ml so the final volume of liquid is $\\var{final_volume}$ml. We have worked out that in $\\var{stock_volume_used}$ml of stock solution there are $\\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride and we have not added any more sodium chloride so in our diluted solution we have $\\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride in $\\var{final_volume}$l of liquid. The formula for the concentration in g/L is

\n

$\\dfrac{\\text{mass of substance (in grams)}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in g/L)}.$

\n

$\\var{final_volume}$ml is equal to $\\var{final_volume / 1000}$L and so the concentration of the diluted solution in g/L is

\n

$\\begin{align}\\dfrac{\\var{(stock_volume_used / 1000) * stock_mass_concentration}}{\\var{final_volume / 1000}} & = \\var{final_mass_concentration} \\\\ & = \\var{precround(final_mass_concentration, 3)} \\text{ g/L to 3 d.p.}\\end{align}$

\n

The formula for concentration in M (mol/L) is

\n

$\\dfrac{\\text{number of moles of substance}}{\\text{volume of liquid (in litres)}} = \\text{concentration (in mol/L)}$

\n

so the concentration of the diluted solution in M is

\n

$\\begin{align}\\dfrac{\\var{(stock_volume_used / 1000) * stock_molarity}}{\\var{final_volume / 1000}} & = \\var{final_molarity} \\\\ & = \\var{precround(final_molarity, 3)} \\text{ M to 3 d.p.}\\end{align}$

", "variables": {"stock_mass_concentration": {"description": "

The mass concentration of the \"stock\" solution in g/L.

", "name": "stock_mass_concentration", "definition": "stock_mass / stock_volume", "group": "Ungrouped variables", "templateType": "anything"}, "Cl": {"description": "

atomic weight of chlorine

", "name": "Cl", "definition": "35.453", "group": "Ungrouped variables", "templateType": "anything"}, "final_molarity": {"description": "", "name": "final_molarity", "definition": "stock_molarity * stock_volume_used / final_volume", "group": "Ungrouped variables", "templateType": "anything"}, "Na": {"description": "

atomic weight of sodium

", "name": "Na", "definition": "22.990", "group": "Ungrouped variables", "templateType": "anything"}, "stock_mass": {"description": "

The mass of the substance (in grams) dissolved in the \"stock\" solution

", "name": "stock_mass", "definition": "50 * random(2..8)\n", "group": "Ungrouped variables", "templateType": "anything"}, "final_volume": {"description": "

The volume the sample of \"stock\" solution is diluted to in ml.

", "name": "final_volume", "definition": "50 * random(5..10)", "group": "Ungrouped variables", "templateType": "anything"}, "final_mass_concentration": {"description": "", "name": "final_mass_concentration", "definition": "stock_mass_concentration * stock_volume_used / final_volume", "group": "Ungrouped variables", "templateType": "anything"}, "stock_molarity": {"description": "

The molarity of the \"stock\" solution

", "name": "stock_molarity", "definition": "stock_mass_concentration / NaCl", "group": "Ungrouped variables", "templateType": "anything"}, "stock_volume_used": {"description": "

The volume of the \"stock\" solution used for dilution in ml.

", "name": "stock_volume_used", "definition": "100", "group": "Ungrouped variables", "templateType": "anything"}, "stock_volume": {"description": "

The volume of liquid in the \"stock\" solution in litres.

", "name": "stock_volume", "definition": "0.5 * random(2..5)", "group": "Ungrouped variables", "templateType": "anything"}, "NaCl": {"description": "

molecular weight of sodium chloride

", "name": "NaCl", "definition": "Na + Cl", "group": "Ungrouped variables", "templateType": "anything"}}, "rulesets": {}, "statement": "

Answer the following question. Please give your answers as decimals, not as fractions. Give your answers to 3 decimal places but only round your calculations at the final step, use the exact values in your calculator for intermediate steps.

\n

Clicking on 'Show steps' will provide you with some prompts to break down the question into smaller parts.

\n

If you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.

", "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question"}, {"name": "Julie's copy of SLR and Correlation", "extensions": ["stats", "jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}], "functions": {"regfun2": {"definition": "\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',\n{boundingBox:[-5,maxy,maxx,-5],\n axis:true,\n showNavigation:false,\n grid:false,\n });\n var board = div.board; \n //var n=length(r1);\n //var names = 'ABCDEFGHIJKLMNOP';\nfor (j=0;jFit what you think is the best regression line by draggging the points A and B, shown in the scatter plot below, to form an interactive line.

\n

As you alter your line the values for $a$ and $b$ in the equation $Y = a + bX$ will be shown in red below.

\n

\n

Value of $a$ for interactive line:  

\n

Value of $b$ for interactive line:

\n

The Sum of the Squares of the Errors (SSE) for the interactive line is :

\n

Remember you are trying to make the SSE as small as possible.

\n

{regressline(r1,r2,min(r1)-10,max(r1)+10,min(r2)-10,max(r2)+10,n)}

\n

\n

Click on Show steps if you want to see the minimum value for SSE and the best fitting straight line. There are no marks awarded for this part of the question and you will not lose any marks in subsequent parts by clicking on Show steps.

\n

 

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "

\n

The minimum value of the SSE is: {sumr}

\n

 

\n

{regfun3(r1,r2,max(r1)+10,max(r2)+10,rsquared,sumr,n)}</p>

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

Calculate the equation of the best fitting regression line, $Y = a + bX.$

\n

Determine $a$ and $b$ to 5 decimal places and input them below to 3 decimal places.

\n

$a=\\;$[[0]],      $b=\\;$[[1]] (both to 3 decimal places.)

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "

To find $a$ and $b$ you first find  $\\displaystyle b = \\frac{n\\Sigma xy-\\Sigma x \\Sigma y}{n\\Sigma x^2 -(\\Sigma x)^2}$

\n

Then $\\displaystyle a = \\frac{\\Sigma y - b \\Sigma x}{n}$

\n

Note that $n$ is the number of data points. In this case $\\var{n}$

\n

Now go back and fill in the values for $a$ and $b$.

\n

 

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Calculate the correlation coefficient $r$ to 5 decimal places, then input below to 3 decimal places.

\n


$r=\\;$[[0]],   (to 3 decimal places)

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "

\\[r=\\frac{n\\Sigma xy -\\Sigma x \\Sigma y}{\\sqrt{n\\Sigma x^2-(\\Sigma x)^2}\\sqrt{n\\Sigma y^2-(\\Sigma y)^2}}\\]

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There are  $\\var{n}$ observations of datapoints $(X,Y)$ given in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$X$$\\var{r1[0]}$$\\var{r1[1]}$$\\var{r1[2]}$$\\var{r1[3]}$$\\var{r1[4]}$$\\var{r1[5]}$$\\var{r1[6]}$$\\var{r1[7]}$$\\var{r1[8]}$$\\var{r1[9]}$
$Y$$\\var{r2[0]}$$\\var{r2[1]}$$\\var{r2[2]}$$\\var{r2[3]}$$\\var{r2[4]}$$\\var{r2[5]}$$\\var{r2[6]}$$\\var{r2[7]}$$\\var{r2[8]}$$\\var{r2[9]}$
\n

\n

The scatterplot for this data is shown below.

\n

\n

{regfun2(r1,r2,max(r1)+10,max(r2)+10,rsquared,sumr,n)}

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For a given set of data points:

\n

Use an interactive scatterplot to add best fit line and compare to true regression line via scatterplot solution and given minimum SSE.

\n

Determine values for intercept and slope of least squares regression line.

\n

Determine correlation and r squared value.

\n

Determine prediction and residual at a given X value.

\n

rebelmaths

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Use two points on a line graph to calculate the gradient and $y$-intercept and hence the equation of the straight line running through both points.

\n

The answer box for the third part plots the function which allows the student to check their answer against the graph before submitting.

\n

This particular example has a positive gradient.

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

In this question we will identify the equation of the straight line passing through points  $A=(\\var{xa},\\var{ya})$ and  $B=(\\var{xb},\\var{yb})$ in the form $y = mx + c$.

\n

{plotPoints()}

", "advice": "

We find the equation of a straight line passing through two points by finding the gradient and the $y$-intercept of the line.

\n

a)

\n

We can find the gradient ($m$) using the points $A = (x_1,y_1)=(\\var{xa},\\var{ya})$ and $B = (x_2,y_2)=(\\var{xb},\\var{yb})$.

\n

As the definition of gradient is the ratio of vertical change ($y_2-y_1$) to horizontal change ($x_2-x_1$).
The equation for gradient is,

\n

\\begin{align}
m &= \\frac{y_2-y_1}{x_2-x_1} \\\\[0.5em]
&= \\frac{\\simplify[!collectNumbers]{{yb}-{ya}}}{\\simplify[!collectNumbers]{{xb}-{xa}}} \\\\[0.5em]
&= \\frac{\\simplify[]{{yb}-{ya}}}{\\simplify{{xb}-{xa}}} \\\\[0.5em]
&= \\simplify[simplifyFractions,unitDenominator]{({yb-ya})/({xb-xa})}\\text{.}
\\end{align}

\n

b)

\n

Rearranging the equation $y=mx+c$ and substituting either of the points gives

\n

\\[c = y_1-mx_1 \\quad \\mathrm{or} \\quad c = y_2-mx_2 \\,\\text{.} \\]

\n

We can then also use this equation with the other point's coordinates to check our answer.

\n

Let's use point $A$ first:

\n

\\[
\\begin{align}
c &= y_1-mx_1 \\\\
&= \\var{ya}-\\var[fractionnumbers]{m}\\times\\var{xa} \\\\
& = \\simplify[fractionnumbers]{{ya-m*xa}}\\text{.}
\\end{align}
\\]

\n

We then check this against point $B$:

\n

\\[
\\begin{align}
y_2 &= mx_2 + c \\\\[0.5em]
&= \\simplify[fractionNumbers]{{m}{xb}+{c}} \\\\[0.5em]
&= \\var[fractionnumbers]{m*xb+c}\\text{.}
\\end{align}
\\]

\n

c)

\n

We can now substitute these values for $m$ and $c$ into $y=mx+c$  to get:

\n

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

\n

The green line drawn on the graph represents the above line equation.

\n

{correctPoints()}

", "rulesets": {}, "variables": {"m": {"name": "m", "group": "Ungrouped variables", "definition": "(ya-yb)/(xa-xb)", "description": "", "templateType": "anything"}, "yb": {"name": "yb", "group": "Ungrouped variables", "definition": "ya+random([2,4])", "description": "", "templateType": "anything"}, "xa": {"name": "xa", "group": "Ungrouped variables", "definition": "random(-4..-1)", "description": "", "templateType": "anything"}, "ya": {"name": "ya", "group": "Ungrouped variables", "definition": "random(-4..2)", "description": "", "templateType": "anything"}, "xb": {"name": "xb", "group": "Ungrouped variables", "definition": "xa+random([2,4] except -xa)", "description": "", "templateType": "anything"}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "ya-m*xa", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "\n", "maxRuns": 100}, "ungrouped_variables": ["xa", "xb", "ya", "yb", "m", "c"], "variable_groups": [], "functions": {"correctPoints": {"parameters": [], "type": "html", "language": "javascript", "definition": "//point coordinate variables\nvar xa = Numbas.jme.unwrapValue(scope.variables.xa);\nvar xb = Numbas.jme.unwrapValue(scope.variables.xb);\nvar ya = Numbas.jme.unwrapValue(scope.variables.ya);\nvar yb = Numbas.jme.unwrapValue(scope.variables.yb);\nvar m = Numbas.jme.unwrapValue(scope.variables.m);\nvar c = Numbas.jme.unwrapValue(scope.variables.c);\n\n//make board\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',{boundingBox:[Math.min(-1,xa-2),Math.max(2,yb+2,c+1),Math.max(2,xb+2),Math.min(-1,ya-2,c-1)],grid: true});\nvar board = div.board;\nquestion.board = board;\n\n\n//points (with nice colors)\nvar a = board.create('point',[xa,ya],{name: 'A', size: 7, fillColor: 'blue' , strokeColor: 'lightblue' , highlightFillColor: 'lightblue', highlightStrokeColor: 'yellow', fixed: true, showInfobox: true});\nvar b = board.create('point',[xb,yb],{name: 'B', size: 7, fillColor: 'blue' , strokeColor: 'lightblue' , highlightFillColor: 'lightblue', highlightStrokeColor: 'yellow',fixed: true, showInfobox: true});\n\n\n//ans(was tree) is defined at the end and nscope looks important\n//but they're both variables\n\nvar correct_line = board.create('functiongraph',[function(x){ return m*x+c},-22,22], {strokeColor:\"green\",setLabelText:'mx+c',visible: true, strokeWidth: 4, highlightStrokeColor: 'green'} )\n\n\n\nquestion.signals.on('HTMLAttached',function(e) {\nko.computed(function(){\n//define ans as this \ncorrect_line.updateCurve();\nboard.update();\n});\n });\n\n\nreturn div;"}, "plotPoints": {"parameters": [], "type": "html", "language": "javascript", "definition": "//point coordinate variables\nvar xa = Numbas.jme.unwrapValue(scope.variables.xa);\nvar xb = Numbas.jme.unwrapValue(scope.variables.xb);\nvar ya = Numbas.jme.unwrapValue(scope.variables.ya);\nvar yb = Numbas.jme.unwrapValue(scope.variables.yb);\nvar m = Numbas.jme.unwrapValue(scope.variables.m);\nvar c = Numbas.jme.unwrapValue(scope.variables.c);\n\n//make board\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',{boundingBox:[Math.min(-1,xa-2),Math.max(2,yb+2,c+1),Math.max(2,xb+2),Math.min(-1,ya-2,c-1)],grid: true});\nvar board = div.board;\nquestion.board = board;\n\n//points (with nice colors)\nvar a = board.create('point',[xa,ya],{name: 'A', size: 7, fillColor: 'blue' , strokeColor: 'lightblue' , highlightFillColor: 'lightblue', highlightStrokeColor: 'yellow', fixed: true, showInfobox: true});\nvar b = board.create('point',[xb,yb],{name: 'B', size: 7, fillColor: 'blue' , strokeColor: 'lightblue' , highlightFillColor: 'lightblue', highlightStrokeColor: 'yellow',fixed: true, showInfobox: true});\n\n\n//ans(was tree) is defined at the end and nscope looks important\n//but they're both variables\n var ans;\n var nscope = new Numbas.jme.Scope([scope,{variables:{x:new Numbas.jme.types.TNum(0)}}]);\n//this is the beating heart of whatever plots the function,\n//I've changed this from being curve to functiongraph\n var line = board.create('functiongraph',[function(x){\nif(ans) {\n try {\nnscope.variables.x.value = x;\n var val = Numbas.jme.evaluate(ans,nscope).value;\n return val;\n }\n catch(e) {\nreturn 13;\n }\n}\nelse\n return 13;\n },-12,12]\n , {strokeColor:\"blue\",strokeWidth: 4} );\n \nvar correct_line = board.create('functiongraph',[function(x){ return m*x+c},-22,22], {strokeColor:\"green\",setLabelText:'mx+c',visible: false, strokeWidth: 4, highlightStrokeColor: 'green'} )\n\nquestion.lines = {\n l:line, c:correct_line\n}\n\n question.signals.on('HTMLAttached',function() {\nko.computed(function(){\nvar expr = question.parts[2].gaps[0].display.studentAnswer();\n\n//define ans as this \ntry {\n ans = Numbas.jme.compile(expr,scope);\n}\ncatch(e) {\n ans = null;\n}\nline.updateCurve();\ncorrect_line.updateCurve();\nboard.update();\n});\n });\n\n\nreturn div;"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the gradient, $m$, of the straight line between these two points.

\n

$m=$ [[0]]

\n

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "m", "maxValue": "m", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Use this gradient and the coordinates of the points to calculate the $y$-intercept, $c$.

\n

$c=$ [[0]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "c", "maxValue": "c", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Give the equation of the straight line through these points in the form $y=mx+c$. 

\n

$\\displaystyle y=$ [[0]]

\n

Use the graph to plot your answer and check that it goes through these points.

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [{"variable": "m", "part": "p0", "must_go_first": true}, {"variable": "c", "part": "p1", "must_go_first": true}], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{m}*x+{c}", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": true, "singleLetterVariables": false, "allowUnknownFunctions": false, "implicitFunctionComposition": false, "notallowed": {"strings": ["c", "m"], "showStrings": false, "partialCredit": 0, "message": "

You must input your answer in the form y = mx +c where m and c are numbers.

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Julie's copy of TP6", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "parts": [{"showFeedbackIcon": true, "prompt": "

\n

$y = $ [[0]]

", "variableReplacements": [], "sortAnswers": false, "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "customMarkingAlgorithm": "", "steps": [{"showFeedbackIcon": true, "scripts": {}, "prompt": "

It may be helpful to factor out y. For example: 

\n

\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y}=y(\\simplify{x^{{n1}} + {n2}*x - {n3}})\\]

", "showCorrectAnswer": true, "variableReplacements": [], "customMarkingAlgorithm": "", "marks": 0, "type": "information", "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true}], "type": "gapfill", "stepsPenalty": 0, "gaps": [{"showFeedbackIcon": true, "checkVariableNames": false, "variableReplacements": [], "showCorrectAnswer": true, "vsetRangePoints": 5, "expectedVariableNames": [], "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true, "failureRate": 1, "scripts": {}, "customMarkingAlgorithm": "", "showPreview": true, "vsetRange": [0, 1], "type": "jme", "answer": "{n4}/(x^{{n1}} + {n2}*x-{n3})", "marks": "5", "checkingType": "absdiff", "checkingAccuracy": 0.001}]}], "tags": [], "functions": {}, "advice": "

Here, we first have to collect all terms involving $y$ on the same side. Hence, we get:

\n

\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y} = \\var{n4}\\]

\n

We then spot that $y$ appears exactly once in each term on the left, so factorise:

\n

\\[y(\\simplify{x^{{n1}} + {n2}*x - {n3}}) = \\var{n4}\\]

\n

and simple division gives the answer.

", "rulesets": {}, "ungrouped_variables": ["n1", "n2", "n3", "n4"], "statement": "

Consider the equation:

\n

\\[\\simplify{x^{{n1}}*y + {n2}*y*x} = \\simplify{{n3}*y} + \\var{n4}\\]

\n

Re-arrange this equation to make $y$ the subject.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Another transposition question, which requires (basic) factorisation.

\n

rebelmaths

"}, "variables": {"n1": {"definition": "random(2..6)", "group": "Ungrouped variables", "templateType": "anything", "name": "n1", "description": ""}, "n4": {"definition": "random(1..10)", "group": "Ungrouped variables", "templateType": "anything", "name": "n4", "description": ""}, "n3": {"definition": "random(-7..7 except 0 n2)", "group": "Ungrouped variables", "templateType": "anything", "name": "n3", "description": ""}, "n2": {"definition": "random(-7..7 except 0)", "group": "Ungrouped variables", "templateType": "anything", "name": "n2", "description": ""}}, "type": "question"}, {"name": "MATH6005 T3Q5 (adaptive)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "advice": "", "variable_groups": [], "tags": [], "statement": "

Let $N = \\begin{pmatrix} 1 & 2 & 8\\\\ 1 & 3 & 4\\\\ k & k & 1 \\end{pmatrix}$.

", "metadata": {"licence": "None specified", "description": ""}, "rulesets": {}, "ungrouped_variables": ["N1", "invN1", "A"], "preamble": {"js": "", "css": ""}, "parts": [{"customMarkingAlgorithm": "", "customName": "", "useCustomName": false, "prompt": "

For what values of $k$ does $N$ have an inverse?

\n

$k\\neq $ [[0]]

", "type": "gapfill", "marks": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showFeedbackIcon": true, "adaptiveMarkingPenalty": 0, "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"customMarkingAlgorithm": "malrules:\n [\n [\"-1/12\", \"Some good work done but not quite there yet. Did you get a determinant of $1-12k$? Be careful when you let this $=0$ and solve - watch the signs carefully.\"],\n [\"1/28\", \"You're on the right track but it looks like you have made one important mistake. Look back at your notes to check how to calculate the determinant of a $2 \\\\times 2$ matrix. Pay particular attention to the signs.\"],\n [\"-1/28\", \"You're on the right track but it looks like you have made a couple of small but important mistakes. Look back at your notes to check how to calculate the determinant of a $2 \\\\times 2$ matrix. Pay particular attention to the signs. Also be very careful with signs when solving the determinant $=0$.\"],\n [\"-44/5\", \"Some good work done but not quite there yet. It looks like you have made a couple of small but important mistakes. Look back at your notes to check how to calculate the determinant of both a $2 \\\\times 2$ and a $3 \\\\times 3$ matrix. Pay particular attention to the signs in both cases.\"],\n [\"44/5\", \"Some good work done but not quite there yet. It looks like you have made a couple of small but important mistakes. Look back at your notes to check how to calculate the determinant of both a $2 \\\\times 2$ and a $3 \\\\times 3$ matrix. Pay particular attention to the signs in both cases. Also be very careful with signs when solving the determinant $=0$.\"],\n [\"5/28\", \"You're on the right track but it looks like you have made one important mistake. Look back at your notes to check how to calculate the determinant of a $3 \\\\times 3$ matrix. Pay particular attention to the signs.\"],\n [\"-5/28\", \"You're on the right track but it looks like you have made one important mistake. Look back at your notes to check how to calculate the determinant of a $3 \\\\times 3$ matrix. Pay particular attention to the signs. Also be very careful with signs when solving the determinant $=0$.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))Find $N^{-1}$ if $k=-1$.

\n

$N^{-1} = $ [[0]]

", "type": "gapfill", "marks": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showFeedbackIcon": true, "adaptiveMarkingPenalty": 0, "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"customMarkingAlgorithm": "", "customName": "", "correctAnswerFractions": true, "useCustomName": false, "markPerCell": true, "allowResize": false, "type": "matrix", "marks": "9", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showFeedbackIcon": true, "adaptiveMarkingPenalty": 0, "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "showCorrectAnswer": true, "tolerance": 0, "numColumns": "3", "allowFractions": true, "correctAnswer": "invN1", "numRows": "3"}]}, {"customMarkingAlgorithm": "", "customName": "", "useCustomName": false, "prompt": "

Let $k=-1$ as in part (b). Given the matrix $A = \\begin{pmatrix} 13 & 13 & -13\\\\ 26 & 52 & 26\\\\ 0 & 13 & 65 \\end{pmatrix}$, solve for $M$ in the equation $MN = A$.

\n

$M = $ [[0]]

", "type": "gapfill", "marks": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showFeedbackIcon": true, "adaptiveMarkingPenalty": 0, "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"customMarkingAlgorithm": "", "customName": "", "correctAnswerFractions": false, "useCustomName": false, "markPerCell": true, "allowResize": false, "type": "matrix", "marks": "9", "variableReplacementStrategy": "originalfirst", "variableReplacements": [{"part": "p1g0", "variable": "invN1", "must_go_first": true}], "showFeedbackIcon": true, "adaptiveMarkingPenalty": 0, "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "showCorrectAnswer": true, "tolerance": 0, "numColumns": "3", "allowFractions": true, "correctAnswer": "A*invN1", "numRows": "3"}]}], "functions": {}, "variables": {"N1": {"definition": "matrix([1, 2, 8], [1, 3, 4], [-1, -1, 1])", "templateType": "anything", "name": "N1", "description": "", "group": "Ungrouped variables"}, "invN1": {"definition": "matrix([\n[7/13,-10/13,-16/13],\n[-5/13,9/13,4/13],\n[2/13,-1/13,1/13]\n])", "templateType": "anything", "name": "invN1", "description": "", "group": "Ungrouped variables"}, "A": {"definition": "matrix([13, 13, -13], [26, 52, 26], [0, 13, 65])", "templateType": "anything", "name": "A", "description": "", "group": "Ungrouped variables"}}, "variablesTest": {"condition": "", "maxRuns": 100}}, {"name": "Julie's copy of Sheet 2 Q10 Chain rule question with custom feedback", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "tags": [], "metadata": {"description": "Chain rule question with feedback given for anticipated student errors.", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Let $y=\\sin (\\var{a}x^{\\var{n}}-\\var{b}x)$. What is $\\frac{dy}{dx}$?

", "advice": "", "rulesets": {}, "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2..7)", "description": "

multiple of the power of x

", "templateType": "anything"}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "

power on the x

", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(2..9 except a)", "description": "

multiple of x

", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "n", "b"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

$\\frac{dy}{dx}=$ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "malrules:\n [\n [\"-({a*n}*x^{n-1}-{b})*cos({a}*x^{n}-{b}*x)\", \"Almost there! Double check the rule for differentiating $\\\\sin x$.\"],\n [\"cos({a}*x^{n}-{b}*x)\", \"You have used the correct rule for differentiating $\\\\sin x$. However, since the angle is more complicated than $x$, what extra step do you need to include?\"],\n [\"-cos({a}*x^{n}-{b}*x)\", \"Double check the rule for differentiating $\\\\sin x$. Also, since the angle is more complicated than $x$, what extra step do you need to include?\"],\n [\"cos({a*n}*x^{n-1}-{b})\", \"Remember, the angle in a trigonometric function never changes when you differentiate it!\"],\n [\"-cos({a*n}*x^{n-1}-{b})\", \"Remember, the angle in a trigonometric function never changes when you differentiate it! Also, double check the rule for differentiating $\\\\sin x$.\"],\n [\"cos({a}*x^{n}-{b}*x)+({a*n}*x^{n-1}-{b})*sin\", \"The function $\\\\sin (\\\\var{a}x^{\\\\var{n}} - \\\\var{b}x)$ is not a product, so you do not need the product rule.\"],\n [\"-cos({a}*x^{n}-{b}*x)+({a*n}*x^{n-1}-{b})*sin\", \"The function $\\\\sin (\\\\var{a}x^{\\\\var{n}} - \\\\var{b}x)$ is not a product, so you do not need the product rule.\"],\n [\"cos({a}*x^{n}-{b}*x)+sin({a*n}*x^{n-1}-{b})\", \"The function $\\\\sin (\\\\var{a}x^{\\\\var{n}} - \\\\var{b}x)$ is not a product, so you do not need the product rule.\"],\n [\"-cos({a}*x^{n}-{b}*x)+sin({a*n}*x^{n-1}-{b})\", \"The function $\\\\sin (\\\\var{a}x^{\\\\var{n}} - \\\\var{b}x)$ is not a product, so you do not need the product rule.\"]\n ]\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))