Answer :
observable exponential decay characteristic.
Looking at the data, we need to determine the type of relationship between time and volume. Specifically, we examine whether the volume decreases linearly, quadratically, exponentially, etc.
Given the time and volume data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Volume (cubic meters)} \\ \hline 1 & 12.0 \\ \hline 2 & 6.1 \\ \hline 3 & 3.0 \\ \hline 4 & 1.6 \\ \hline 5 & 0.8 \\ \hline \end{array} \][/tex]
To analyze the data, we propose an exponential model of the form:
[tex]\[ V(t) = a e^{bt} \][/tex]
where [tex]\(V(t)\)[/tex] is the volume at time [tex]\(t\)[/tex], [tex]\(a\)[/tex] is the initial volume, and [tex]\(b\)[/tex] is the decay constant.
From fitting the data to this exponential model, the parameters calculated are:
[tex]\[ a = 23.73071 \][/tex]
[tex]\[ b = -0.68159 \][/tex]
The fitted volumes at each time point are:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Fitted Volume (cubic meters)} \\ \hline 1 & 12.0033 \\ \hline 2 & 6.0714 \\ \hline 3 & 3.0710 \\ \hline 4 & 1.5533 \\ \hline 5 & 0.7857 \\ \hline \end{array} \][/tex]
The residuals, which are the differences between the observed and fitted volumes, are:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Residuals} \\ \hline 1 & -0.0033 \\ \hline 2 & 0.0286 \\ \hline 3 & -0.0710 \\ \hline 4 & 0.0467 \\ \hline 5 & 0.0143 \\ \hline \end{array} \][/tex]
The sum of the squares of the residuals, which gives an indication of the fitting error, is calculated to be:
[tex]\[ 0.00825 \][/tex]
From this analysis, it is clear that the exponential model fits the data very well, as indicated by the small residuals and the sum of the residuals squared.
Thus, the data in the table can best be described as showing an exponential decay characteristic because there is a significant reduction in volume over time, consistent with an exponential decrease.
Looking at the data, we need to determine the type of relationship between time and volume. Specifically, we examine whether the volume decreases linearly, quadratically, exponentially, etc.
Given the time and volume data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Volume (cubic meters)} \\ \hline 1 & 12.0 \\ \hline 2 & 6.1 \\ \hline 3 & 3.0 \\ \hline 4 & 1.6 \\ \hline 5 & 0.8 \\ \hline \end{array} \][/tex]
To analyze the data, we propose an exponential model of the form:
[tex]\[ V(t) = a e^{bt} \][/tex]
where [tex]\(V(t)\)[/tex] is the volume at time [tex]\(t\)[/tex], [tex]\(a\)[/tex] is the initial volume, and [tex]\(b\)[/tex] is the decay constant.
From fitting the data to this exponential model, the parameters calculated are:
[tex]\[ a = 23.73071 \][/tex]
[tex]\[ b = -0.68159 \][/tex]
The fitted volumes at each time point are:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Fitted Volume (cubic meters)} \\ \hline 1 & 12.0033 \\ \hline 2 & 6.0714 \\ \hline 3 & 3.0710 \\ \hline 4 & 1.5533 \\ \hline 5 & 0.7857 \\ \hline \end{array} \][/tex]
The residuals, which are the differences between the observed and fitted volumes, are:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (weeks)} & \text{Residuals} \\ \hline 1 & -0.0033 \\ \hline 2 & 0.0286 \\ \hline 3 & -0.0710 \\ \hline 4 & 0.0467 \\ \hline 5 & 0.0143 \\ \hline \end{array} \][/tex]
The sum of the squares of the residuals, which gives an indication of the fitting error, is calculated to be:
[tex]\[ 0.00825 \][/tex]
From this analysis, it is clear that the exponential model fits the data very well, as indicated by the small residuals and the sum of the residuals squared.
Thus, the data in the table can best be described as showing an exponential decay characteristic because there is a significant reduction in volume over time, consistent with an exponential decrease.
