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Sagot :
To determine the best prediction for the number of bacterial cells at 16 hours, we'll fit an exponential growth model to the given data and use this model for the prediction.
Firstly, we recognize that an exponential model generally has the form:
[tex]\[ P(t) = a \cdot e^{bt} \][/tex]
where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex],
- [tex]\( a \)[/tex] is the initial amount (when [tex]\( t = 0 \)[/tex]),
- [tex]\( b \)[/tex] is the growth rate.
Given the table, we have the following data points:
[tex]\[ \begin{array}{c|c} \text{Time }(t) & \text{Cell Population }(P(t)) \\ \hline 0 & 125 \\ 2 & 162 \\ 4 & 258 \\ 6 & 374 \\ 8 & 518 \\ 10 & 763 \\ \end{array} \][/tex]
We fit the exponential model [tex]\( P(t) = a \cdot e^{bt} \)[/tex] to this data. Through the process of fitting the data, we determine the parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
From the fitting process, the parameters are found to be:
[tex]\[ a \approx 120.94548094516612 \][/tex]
[tex]\[ b \approx 0.18392823656534055 \][/tex]
Now, we use this model to predict the number of cells at 16 hours.
[tex]\[ P(16) = a \cdot e^{b \cdot 16} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ P(16) \approx 120.94548094516612 \cdot e^{0.18392823656534055 \cdot 16} \][/tex]
Calculating the exponent:
[tex]\[ 0.18392823656534055 \cdot 16 \approx 2.9428517845 \][/tex]
So,
[tex]\[ P(16) \approx 120.94548094516612 \cdot e^{2.9428517845} \][/tex]
[tex]\[ P(16) \approx 120.94548094516612 \cdot 18.9738767328 \][/tex]
[tex]\[ P(16) \approx 2294.3197157559675 \][/tex]
Hence, the best prediction for the number of cells in the colony at 16 hours is approximately:
[tex]\[ \boxed{2294} \][/tex]
So the correct answer is closest to:
D. 2,700
Firstly, we recognize that an exponential model generally has the form:
[tex]\[ P(t) = a \cdot e^{bt} \][/tex]
where:
- [tex]\( P(t) \)[/tex] is the population at time [tex]\( t \)[/tex],
- [tex]\( a \)[/tex] is the initial amount (when [tex]\( t = 0 \)[/tex]),
- [tex]\( b \)[/tex] is the growth rate.
Given the table, we have the following data points:
[tex]\[ \begin{array}{c|c} \text{Time }(t) & \text{Cell Population }(P(t)) \\ \hline 0 & 125 \\ 2 & 162 \\ 4 & 258 \\ 6 & 374 \\ 8 & 518 \\ 10 & 763 \\ \end{array} \][/tex]
We fit the exponential model [tex]\( P(t) = a \cdot e^{bt} \)[/tex] to this data. Through the process of fitting the data, we determine the parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
From the fitting process, the parameters are found to be:
[tex]\[ a \approx 120.94548094516612 \][/tex]
[tex]\[ b \approx 0.18392823656534055 \][/tex]
Now, we use this model to predict the number of cells at 16 hours.
[tex]\[ P(16) = a \cdot e^{b \cdot 16} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ P(16) \approx 120.94548094516612 \cdot e^{0.18392823656534055 \cdot 16} \][/tex]
Calculating the exponent:
[tex]\[ 0.18392823656534055 \cdot 16 \approx 2.9428517845 \][/tex]
So,
[tex]\[ P(16) \approx 120.94548094516612 \cdot e^{2.9428517845} \][/tex]
[tex]\[ P(16) \approx 120.94548094516612 \cdot 18.9738767328 \][/tex]
[tex]\[ P(16) \approx 2294.3197157559675 \][/tex]
Hence, the best prediction for the number of cells in the colony at 16 hours is approximately:
[tex]\[ \boxed{2294} \][/tex]
So the correct answer is closest to:
D. 2,700
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