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The data show that the population of lily pads has an increasing growth rate. An exponential function would be most suitable to model these data.

The general form of an exponential function is
[tex]\[ y = a b^x \][/tex]

Use the regression calculator to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for the water lily population growth. Round to the nearest thousandth.

[tex]\[
\begin{tabular}{|c|c|}
\hline
Time (years) & Lilies \\
\hline
0 & 4 \\
\hline
5 & 7 \\
\hline
10 & 10 \\
\hline
15 & 15 \\
\hline
20 & 33 \\
\hline
25 & 51 \\
\hline
30 & 79 \\
\hline
\end{tabular}
\][/tex]

[tex]\[ a = \square \][/tex]
[tex]\[ b = \square \][/tex]



Answer :

Given the population data of lily pads over time, we want to model this data using an exponential function. The general form of an exponential function is:

[tex]\[ y = a \cdot b^x \][/tex]

where:
- [tex]\( y \)[/tex] is the population at time [tex]\( x \)[/tex],
- [tex]\( a \)[/tex] is the initial population (at [tex]\( x = 0 \)[/tex]),
- [tex]\( b \)[/tex] is the growth factor.

The provided data points are:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (x)} & \text{Population (y)} \\ \hline 0 & 4 \\ \hline 5 & 7 \\ \hline 10 & 10 \\ \hline 15 & 15 \\ \hline 20 & 33 \\ \hline 25 & 51 \\ \hline 30 & 79 \\ \hline \end{array} \][/tex]

Using an exponential regression analysis of this data, we obtain values for [tex]\( a \)[/tex] and [tex]\( b \)[/tex] as follows:
- [tex]\( a = 4.269 \)[/tex]
- [tex]\( b = 1.103 \)[/tex]

These values are rounded to the nearest thousandth.

Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] for the exponential function modeling the water lily population growth are:

[tex]\[ a = 4.269 \][/tex]
[tex]\[ b = 1.103 \][/tex]