Find the value of \(a\) and of \(b\).

Use the regression equation to estimate the number of coyotes in the reserve when \(t = 7\).

Let \(f\) be the number of foxes in the reserve after \(t\) years. The number of foxes can be modelled by the equation \(f = \frac{{2000}}{{1 + 99{{\text{e}}^{ - kt}}}}\), where \(k\) is a constant.

Find the number of foxes in the reserve on 1 January 1995.

After five years, there were 64 foxes in the reserve. Find \(k\).

During which year were the number of coyotes the same as the number of foxes?

evidence of setup     (M1)

eg\(\;\;\;\)correct value for \(a\) or \(b\)

\(13.3823\), \(137.482\)

\(a{\rm{ }} = {\rm{ }}13.4\), \(b{\rm{ }} = {\rm{ }}137\)     A1A1     N3

[3 marks]

correct substitution into their regression equation

eg\(\;\;\;13.3823 \times 7 + 137.482\)     (A1)

correct calculation

\(231.158\)     (A1)

\(231\) (coyotes) (must be an integer)     A1     N2

[3 marks]

recognizing \(t = 0\)     (M1)

eg\(\;\;\;f(0)\)

correct substitution into the model

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ - k(0)}}}},{\text{ }}\frac{{2000}}{{100}}\)     (A1)

\(20\) (foxes)     A1     N2

[3 marks]

recognizing \((5,{\text{ }}64)\) satisfies the equation     (M1)

eg\(\;\;\;f(5) = 64\)

correct substitution into the model

eg\(\;\;\;64 = \frac{{2000}}{{1 + 99{{\text{e}}^{ - k(5)}}}},{\text{ }}64(1 + 99\(e\)^{ - 5k}}) = 2000\)     (A1)

\(0.237124\)

\(k =  - \frac{1}{5}\ln \left( {\frac{{11}}{{36}}} \right){\text{ (exact), }}0.237{\text{ }}[0.237,{\text{ }}0.238]\)     A1     N2

[3 marks]

valid approach     (M1)

eg\(\;\;\;c = f\), sketch of graphs

correct working     (A1)

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ - 0.237124t}}}} = 13.382t + 137.482\), sketch of graphs, table of values

\(t = 12.0403\)     (A1)

\(2007\)     A1     N2

Note:     Exception to the FT rule. Award A1FT on their value of \(t\).

[4 marks]

Total [16 marks]