The table below shows the annual global production of plastics, P, measured in millions of tonnes per year, for six selected years - AQA - A-Level Maths Pure - Question 9 - 2021 - Paper 1

After plotting the points from the previous table on a graph with ( t ) on the x-axis and ( \log_{10} P ) on the y-axis, a line of best fit can be drawn that represents the linear relationship observed.

The gradient of the line from the graph is calculated using:

$k = \frac{y_2 - y_1}{x_2 - x_1}$

Using points (0, 1.875) and (25, 2.414), we find:

$k = \frac{2.414 - 1.875}{25 - 0} = \frac{0.539}{25} \approx 0.02156 \approx 0.02$

Substituting t = 0 into the original model:

$P = A \times 10^{0} = A$

From the table, ( P = 75 ) when ( t = 0 ), thus:

$A = 75$

For the year 2030, ( t = 50 ) (since 2030 - 1980 = 50). Using the model:

$P = 75 \times 10^{0.02 \times 50} = 75 \times 10^{1} = 750 \text{ million tonnes}$

The model may be inappropriate because it assumes a constant growth rate. In reality, external factors such as regulations, technology advancements, and shifts in consumer behavior can significantly alter production rates, making the linear model less reliable in predicting future outcomes.