Write down the independent variable.

Write down the boiling temperature of the liquid.

Jim describes the correlation as very strong. Circle the value below which best represents the correlation coefficient.

$$0.992\quad \quad \quad 0.251\quad \quad \quad 0\quad \quad \quad  - 0.251\quad \quad \quad  - 0.992$$

Jim’s model is $d =  - 2.24t + 105$, for $0 \leqslant t \leqslant 20$. Use his model to predict the decrease in temperature for any 2 minute interval.

$t$     A1     N1

[1 mark]

105     A1     N1

[1 mark]

$- 0.992$     A2     N2

[2 marks]

valid approach     (M1)

eg$\,\,\,\,\,$$\frac{{{\text{d}}d}}{{{\text{d}}t}} =  - 2.24;{\text{ }}2 \times 2.24,{\text{ }}2 \times  - 2.24,{\text{ }}d(2) =  - 2 \times 2.24 \times 105,$

finding $d({t_2}) - d({t_1})$ where ${t_2} = {t_1} + 2$

4.48 (degrees)     A1     N2

Notes:     Award no marks for answers that directly use the table to find the decrease in temperature for 2 minutes eg $\frac{{105 - 98.4}}{2} = 3.3$.

[2 marks]