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

To complete the table, we calculate the log₁₀ P values for each t:

To plot the data:

1. Mark the points (0, 1.88), (5, 1.97), (10, 2.08), (15, 2.19), (20, 2.31), and (25, 2.41) on a graph.
2. Draw a line of best fit that accurately represents the trend of these points.

To find k using the graph, the gradient of the line of best fit is calculated. Assuming the line passes through the points (0, 1.88) and (25, 2.41):

$k = \frac{2.41 - 1.88}{25 - 0} = \frac{0.53}{25} = 0.0212 \approx 0.02$

Reconfirming the previous calculation, if we use the two points from the line for the gradient, we maintain:

$k \approx 0.02$

Using the model:

$P = A \times 10^{0.02t}$ For the year 2030, t = 50: $P = 75 \times 10^{0.02 \times 50} = 75 \times 10^{1} = 750 \text{ million tonnes}$

Setting up the equation:

$8000 = A \times 10^{0.02t}$ Using A = 75,

$8000 = 75 \times 10^{0.02t} \implies 106.67 = 10^{0.02t}$ Taking logarithms:

$ext{log}_{10} (106.67) = 0.02t \implies t \approx 101.4$ Thus, the year is approximately:

$1980 + 101 = 2081$

The model's lack of accounting for environmental impact, government regulations, and advancements in alternative materials may render predictions inaccurate. The global production of plastics may not consistently follow the same trends due to changing societal values and technological advancements.