Anonymous

Let A = 375 in the equation and solve for t.

Anonymous

A(t) = 298e^(0.033)t. t = ln[A/298]/(0.033). Now, for A = 375, t = [1/(0.033)]ln[375/298] = 6.964622408, at the tail end of 2017. Choice B. pertains.

Anonymous

A=298e^(0.033t) => 375=298e^(0.033t) => t=6.9646.. => Ans. 2017 (B) because 6.9646...years 6<6.9646...<7.

Anonymous

The formula A = 298e^0.033t models the population of a particular city, in thousands, t years after 2011. When will the population of the city reach 375 thousand? 375000 = 298 e^(0.033 t) t ≈ 216.291 B. 2017

Anonymous

I assume you mean A = 298e^(0.033t). The parentheses matter. 298e^(0.033t) = 375 e^(0.033t) = 375/298 0.033t = ln(375/298) t = (1000/33)ln(375/298) ≅ 6.96 years 6.96 years after 2011 occurs during 2018.