91Ó°ÊÓ

To find the number of trackers installed in the year 2000, we need to substitute t = 0 in the given function \(N(t)=0.6 e^{0.17 t}\). We have: $$ N(0) = 0.6 e^{0.17 \cdot 0} $$ Since \(e^0 = 1\), we get: $$ N(0) = 0.6 \cdot 1 = 0.6 $$ So, there were 0.6 million automatic vehicle trackers installed in the year 2000.

To find the number of trackers installed in the year 2005, we need to substitute t = 5 in the given function \(N(t)=0.6 e^{0.17 t}\). We have: $$ N(5) = 0.6 e^{0.17 \cdot 5} $$ Using a calculator, we find the value of \(e^{0.85}\), and then multiply it by 0.6: $$ N(5) = 0.6 \cdot e^{0.85} \approx 0.6 \cdot 2.34 \approx 1.4 $$ So, it was projected that about 1.4 million automatic vehicle trackers would be installed in the year 2005.

To sketch the graph of the function \(N(t) = 0.6e^{0.17t}\), we need to understand the behavior of exponential functions. The given function is of the form \(N(t) = ab^{t}\), where a = 0.6 and b = \(e^{0.17}\). Since b > 1, the graph is an increasing exponential function, which means it will have a horizontal asymptote at N = 0 and it will grow increasingly as t increases. The graph starts at the point (0, 0.6) and approaches infinity as t approaches infinity. To better illustrate this, you can plot the function using graphing software or a graphing calculator, remembering that this function is valid for the domain \(0 \leq t \leq 5\). The increasing nature of the graph's curve will show the growth of the number of automatic vehicle trackers installed in fleet vehicles over time.