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Sagot :
The completed statement is as follows;
The model of the data is; [tex]\underline{P \approx 3.6 \cdot e^{-2.35 \times 10^{-2} \times t}}[/tex]. The exponential model
suggest that the honeybees colonies have an approximately -2.35 percent
growth rate per year. If this trend continues, the number of honeybee
colonies 40 years after 1989 will be around 1.40625 million. Given that the
correlation coefficient indicates a correlation between the model and the
data, predictions made by the model is approximately correct.
Reasons:
The table of values is presented as follows;
[tex]\begin{tabular}{|c|c|}\underline{Years since 1989} &\underline{Honeybee Colonies (millions)}\\0&3.6\\4&3.1\\8&2.8\\12&2.65\\16&2.6\\20&2.25\end{array}\right][/tex]
The general form of an exponential growth function is; [tex]P = \mathbf{P_0 \cdot e^{k \cdot t}}[/tex]
When t = 0, we have;
[tex]P = P_0 \cdot e^{k \times 0} = 3.6[/tex]
Therefore;
P₀ = 3.6
When t = 20, we have;
[tex]P = 3.6 \cdot e^{k \times 20} = 2.25[/tex]
Solving gives;
k ≈ -2.35 × 10⁻²
The exponential model is therefore;
[tex]\underline{P \approx 3.6 \cdot e^{-2.35 \times 10^{-2} \times t}}[/tex]
The exponential model suggest that the honeybees colonies have an
approximately -2.35 percent growth rate.
After 40 years, we have;
[tex]P \approx 3.6 \cdot e^{-2.35 \times 10^{-2} \times 40} \approx 1.40625[/tex]
If the trend continues, the number of honeybee colonies 40 years after 1989 will be approximately 1.40625 million
Expressing the growth rate as a logarithm, we have;
ln(P) = ln(3.6) - 2.35×10⁻²·t
The correlation between the values is found as follows;
[tex]\displaystyle r_p = \mathbf{\frac{\sum y_1 \cdot y_2}{\sqrt{\sum y_1^2 \cdot \sum y_2^2} }} \approx 0.99939[/tex]
Where;
y₁ = The logarithm of the measured correlation.
Therefore, because the correlation coefficient indicates a correlation
between the model and the data, predictions made by this model is
approximately correct.
Learn more about exponential model here:
https://brainly.com/question/16873037
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