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]