The number of gannets on an island is assumed to follow this exponential growth model. $$N = 0.45 \times 1.07^x$$ $N$ is the number of gannets, in thousands. $x$ is the number of years after 1$^{st}$ January 2010. (a) Complete the table for $$N = 0.45 \times 1.07^x$$. | $x$ | 0 | 5 | 10 | 15 | 20 | |-----|------|------|------|------|------| | $N$ | 0.45 | 0.63 | 1.00 | 1.25 | 1.74 | (b) Draw the graph of $$N = 0.45 \times 1.07^x$$. (c) Use the graph to find the year when the gannet population is predicted to reach 1000.

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The number of gannets on an island is assumed to follow this exponential growth model - OCR - GCSE Maths - Question 21 - 2018 - Paper 4

Question 21

The number of gannets on an island is assumed to follow this exponential growth model. $$N = 0.45 \times 1.07^x$$ $N$ is the number of gannets, in thousands. $x$ i... show full transcript

Worked Solution & Example Answer:The number of gannets on an island is assumed to follow this exponential growth model - OCR - GCSE Maths - Question 21 - 2018 - Paper 4

Step 1

Complete the table for N = 0.45 × 1.07^x.

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Answer

To complete the table, we will calculate the values of $N$ at each specified $x$ using the formula:

$N = 0.45 \times 1.07^x$

For $x = 0$ : $N = 0.45 \times 1.07^0 = 0.45 \times 1 = 0.45$ Therefore, $N = 0.45$ .
For $x = 5$ : $N = 0.45 \times 1.07^5 \approx 0.63$
For $x = 10$ : $N = 0.45 \times 1.07^{10} \approx 1.00$
For $x = 15$ : $N = 0.45 \times 1.07^{15} \approx 1.25$
For $x = 20$ : $N = 0.45 \times 1.07^{20} \approx 1.74$

Thus, the completed table is:

$x$	0	5	10	15	20
$N$	0.45	0.63	1.00	1.25	1.74

Step 2

Draw the graph of N = 0.45 × 1.07^x.

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Answer

To draw the graph of the function $N = 0.45 \times 1.07^x$ , follow these steps:

Plot the points calculated from the table on a grid:
- (0, 0.45)
- (5, 0.63)
- (10, 1.00)
- (15, 1.25)
- (20, 1.74)
After plotting these points, connect them smoothly to illustrate the exponential growth characteristic. Ensure the curve smoothly approaches the plotted points while reflecting the growth model.
Label the axes appropriately: The horizontal axis for ‘Number of years, $x$ , after 1^{st} January 2010 $’ and the vertical axis for ‘Number of gannets,$ N$, (thousands)’. This will help in visualizing the growth of the gannet population.
Ensure that you maintain a consistent scale on both axes.

Step 3

Use the graph to find the year when the gannet population is predicted to reach 1000.

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Answer

To find when the gannet population reaches 1000 (or $N = 1.00$ in thousands), locate $N = 1.00$ on the vertical axis of the graph. Draw a horizontal line from this point until it intersects the curve.

Drop down vertically to see the corresponding value of $x$ on the horizontal axis. Based on the graph, assume you find that this occurs at approximately $x = 11.5$ . Given that the model starts from 1 $^{st}$ January 2010, you can calculate the year:

$2022 = 2010 + 12 ext{ (rounded to the nearest full year)}$

Therefore, the gannet population is predicted to reach 1000 in the year 2022.

The number of gannets on an island is assumed to follow this exponential growth model - OCR - GCSE Maths - Question 21 - 2018 - Paper 4

Worked Solution & Example Answer:The number of gannets on an island is assumed to follow this exponential growth model - OCR - GCSE Maths - Question 21 - 2018 - Paper 4

Complete the table for N = 0.45 × 1.07^x.

Draw the graph of N = 0.45 × 1.07^x.

Use the graph to find the year when the gannet population is predicted to reach 1000.

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