Answer
a) $k \approx 0.008357$ b) $P \approx 333361000$
Work Step by Step
a) Given: $P=314419199$ and $P_0=314419198$; $t=12 sec= \dfrac{12}{31536000} $ year
The exponential growth can be written as: $P=P_0e^{kt}$ ...(1)
Then, $314419199=314419198 e^{(\dfrac{12}{31536000})K}$
This implies that $K=(\dfrac{31536000}{12})\ln [\dfrac{314419199}{314419198}] \approx 0.008357$
b) The equation of the model would be as follows:
$P=314419198 e^{(0.008357)t} \\ \implies P=314419198 e^{(0.008357)(7)}$
Thus, $P \approx 333361000$
Hence, a) $k \approx 0.008357$ b) $P \approx 333361000$
