Solve the problem.
-After introducing an inhibitor into a culture of luminescent bacteria, a scientist monitors the luminosity produced by the culture. Use a graphing utility to fit a logarithmic function to the data. Predict the luminosity
After 20 hours. $\begin{array}{l|ccccccc}
\text { Time, hrs } & 2 & 3 & 4 & 5 & 8 & 10 & 15 \
\hline \text { Luminosity } & 77.4 & 60.8 & 54.5 & 45.8 & 30.0 & 24.3 & 10.5
\end{array}$

A) $y = 98.75 - 32.66 \ln ( x ) , 0.91$
B) $y = 107.55 - 41 \ln ( x ) , - 15.27$
C) $y = 112.97 - 45.97 \ln ( x ) , - 24.74$
D) $y = 100.5 - 32.7 \ln ( x ) , 2.54$

Question

Solve the problem.
-After introducing an inhibitor into a culture of luminescent bacteria, a scientist monitors the luminosity produced by the culture. Use a graphing utility to fit a logarithmic function to the data. Predict the luminosity
After 20 hours. $\begin{array}{l|ccccccc}
\text { Time, hrs } & 2 & 3 & 4 & 5 & 8 & 10 & 15 \
\hline \text { Luminosity } & 77.4 & 60.8 & 54.5 & 45.8 & 30.0 & 24.3 & 10.5
\end{array}$

A) $y = 98.75 - 32.66 \ln ( x ) , 0.91$
B) $y = 107.55 - 41 \ln ( x ) , - 15.27$
C) $y = 112.97 - 45.97 \ln ( x ) , - 24.74$
D) $y = 100.5 - 32.7 \ln ( x ) , 2.54$

Quizplus · Accepted Answer

Given the data, a logarithmic function can be fitted using a graphing utility or statistical software. The correct choice, A, represents a logarithmic function that could be derived from such a process. The function $y = 98.75 - 32.66 \ln(x)$ likely fits the given data points well, and the number following the function, 0.91, could represent the correlation coefficient or another statistic related to the fit quality, but it's not part of the function itself for predicting luminosity.

Solve the Problem A) y=98.75−32.66ln⁡(x),0.91y = 98.75 - 32.66 \ln ( x ) , 0.91y=98.75−32.66ln(x),0.91

Solve the Problem A) $y = 98.75 - 32.66 \ln ( x ) , 0.91$