We can see from the graph that the points (2,3) and (9,7) are on the line of best fit.

Using our knowledge from the straight line topic:

$$ \begin{eqnarray} m\ &=&\ \small\frac{y_2-y_1}{x_2-x_1}\\[6pt] &=&\ \small\frac{7-3}{9-2}\\[6pt] &=&\ \small\frac{4}{7} \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (2,3) to find the equation:
$$ \begin{eqnarray} y-b &=& m(x-a) \\[6pt] y-3 &=& \small{\frac{4}{7}}\normalsize (x-2) \\[6pt] 7(y-3) &=& 4(x-2) \\[6pt] 7y-21 &=& 4x-8 \\[6pt] -4x+7y &=& -8+21 \\[6pt] -4x+7y &=& 13 \\[6pt] \end{eqnarray} $$

(a)  We can see from the graph that the points (3,9) and (5,4) are on the line of best fit.

In this question, Q is used instead of \(y\small,\) and it's P instead of \(x\small.\) You can either use P and Q throughout the working or you can use \(x\) and \(y\) in your working and just change them to P and Q at the end. We prefer to use the correct letters throughout, but either way is acceptable to the SQA. Use whichever style you prefer.

$$ \begin{eqnarray} m\ &=&\ \small\frac{Q_2-Q_1}{P_2-P_1}\\[6pt] &=&\ \small\frac{4-9}{5-3}\\[6pt] &=&\ -\small\frac{5}{2} \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (3,9) to find the equation:
$$ \begin{eqnarray} Q-b &=& m(P-a) \\[6pt] Q-9 &=& -\small{\frac{5}{2}}\normalsize (P-3) \\[6pt] 2(Q-9) &=& -5(P-3) \\[6pt] 2Q-18 &=& -5P+15 \\[6pt] 5P+2Q &=& 15+18 \\[6pt] 5P+2Q &=& 33 \\[6pt] \end{eqnarray} $$

(b)  We substitute \(P\) = 4 into the equation:

$$ \begin{eqnarray} 5P+2Q &=& 33 \\[6pt] 5(4)+2Q &=& 33 \\[6pt] 20+2Q &=& 33 \\[6pt] 2Q &=& 33-20 \\[6pt] 2Q &=& 13 \\[6pt] Q &=& \frac{13}{2} \\[6pt] \end{eqnarray} $$

(c)  We substitute \(Q\) = 3 into the equation:

$$ \begin{eqnarray} 5P+2Q &=& 33 \\[6pt] 5P+2(3) &=& 33 \\[6pt] 5P+6 &=& 33 \\[6pt] 5P &=& 33-6 \\[6pt] 5P &=& 27 \\[6pt] P &=& \frac{27}{5} \\[6pt] \end{eqnarray} $$

(a)  Point D is (3,100) and point E is (15,340). First we find the gradient.

$$ \begin{eqnarray} m\ &=&\ \small\frac{W_2-W_1}{A_2-A_1}\\[6pt] &=&\ \small\frac{340-100}{15-3}\\[6pt] &=&\ \small\frac{240}{12}\\[6pt] &=& 20 \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (3,100) to find the equation:
$$ \begin{eqnarray} W-b &=& m(A-a) \\[6pt] W-100 &=& 20(A-3) \\[6pt] W-100 &=& 20A-60 \\[6pt] W &=& 20A-60+100 \\[6pt] W &=& 20A+40 \end{eqnarray} $$

(b)  Note that one year is 12 months, so we substitute A = 12 into the equation:

$$ \begin{eqnarray} W &=& 20A+40 \\[6pt] &=& 20\times 12+40 \\[6pt] &=& 240+40 \\[6pt] &=& 280 \end{eqnarray} $$

So the estimated weight of the one year old calf is 280 kg.

(a)  Point A is (8,14) and point B is (12,20). First we find the gradient.

$$ \begin{eqnarray} m\ &=&\ \small\frac{P_2-P_1}{d_2-d_1}\\[6pt] &=&\ \small\frac{20-14}{12-8}\\[6pt] &=&\ \small\frac{6}{4}\\[6pt] &=& \small\frac{3}{2} \end{eqnarray} $$

Now we use (\(a\),\(b\)) = (8,14) to find the equation:
$$ \begin{eqnarray} P-b &=& m(d-a) \\[6pt] P-14 &=& \small\frac{3}{2}\normalsize (d-8) \\[6pt] 2(P-14) &=& 3(d-8) \\[6pt] 2P-28 &=& 3d-24 \\[6pt] 2P &=& 3d-24+28 \\[6pt] 2P &=& 3d+4 \\[6pt] \end{eqnarray} $$

Note that you could divide this equation through by 2 and give it as \(P=\large\frac{3}{2}\normalsize d+2\) if you wish, but we prefer to avoid fractions whenever possible.

(b)  We just substitute d = 5 into the equation:

$$ \begin{eqnarray} 2P &=& 3d+4 \\[6pt] &=& 3\times 5+4 \\[6pt] &=& 15+4 \\[6pt] &=& 19 \\[6pt] P &=& \small\frac{19}{2} \end{eqnarray} $$

19 ÷ 2 = 9.5 so the 5 mile journey costs £9.50.

(a)  From the scattergraph, we obtain points (1.5,14) and (3.5,8). So we find the gradient.

$$ \begin{eqnarray} m\ &=&\ \small\frac{F_2-F_1}{E_2-E_1}\\[6pt] &=&\ \small\frac{8-14}{3.5-1.5}\\[6pt] &=&\ \small\frac{-6}{2}\\[6pt] &=& -\!3 \end{eqnarray} $$

Now we can use (\(a\),\(b\)) = (1.5,14) to find the equation:
$$ \begin{eqnarray} F-b &=& m(E-a) \\[6pt] F-14 &=& -\!3(E-1.5) \\[6pt] F-14 &=& -\!3E+4.5 \\[6pt] 2F-28 &=& -\!6E+9 \\[6pt] 2F &=& -\!6E+9+28 \\[6pt] 2F &=& -\!6E+37 \\[6pt] \end{eqnarray} $$

Note that you could divide this equation through by 2 and give it as \(F=-3E+18.5\) if you wish, but we prefer whole numbers in equations, whenever possible.

(b)  We just substitute E = 1.1 into the equation:

$$ \begin{eqnarray} 2F &=& -\!6E+37 \\[6pt] &=& -\!6\times 1.1+37 \\[6pt] &=& -\!6.6+37 \\[6pt] &=& 30.4 \\[6pt] F &=& \small\frac{30.4}{2} \\[6pt] &=& 15.2 \end{eqnarray} $$

So Amaar can expect to get 15.2 km per litre.

(a)  From the scattergraph, we obtain points (5,20 000) and (25,50 000). So we find the gradient.

$$ \begin{eqnarray} m\ &=&\ \small\frac{P_2-P_1}{T_2-T_1}\\[6pt] &=&\ \small\frac{50\,000-20\,000}{25-5}\\[6pt] &=&\ \small\frac{30\,000}{20}\\[6pt] &=& 1500 \end{eqnarray} $$

Now we can use (\(a\),\(b\)) = (5,20 000) to find the equation:
$$ \begin{eqnarray} P-20\,000 &=& 1500(T-5) \\[6pt] P-20\,000 &=& 1500T-7500 \\[6pt] P &=& 1500T-7500+20\,000 \\[6pt] P &=& 1500T+12\,500 \end{eqnarray} $$

(b)  We just substitute T = 8 into the equation:

$$ \begin{eqnarray} P &=& 1500T+12\,500 \\[6pt] &=& 1500\times 8+12\,500 \\[6pt] &=& 12,000+12\,500 \\[6pt] &=& 24\,500 \end{eqnarray} $$

So the estimated salary of an employee who has been working for the business for eight years is £24 500.