Let's solve each part of the question step by step.
### (a) Continuous Rate of Growth
The continuous rate of growth for an exponential function of the form [tex]\( n(t) = n_0 e^{rt} \)[/tex] is given by the exponent [tex]\( r \)[/tex], expressed as a percentage. In this case, the given exponential function is [tex]\( n(t) = 930 e^{0.1t} \)[/tex]. Here, [tex]\( r = 0.1 \)[/tex].
To convert the continuous rate of growth to a percentage, multiply [tex]\( r \)[/tex] by 100:
[tex]\[ \text{Continuous rate of growth} = 0.1 \times 100 = 10.0\% \][/tex]
### Answer:
[tex]\[
\boxed{10.0} \text{ percent}
\][/tex]
### (b) Initial Population
The initial population [tex]\( n(0) \)[/tex] is the value of the function at [tex]\( t = 0 \)[/tex]. This is given by:
[tex]\[ n(0) = 930 e^{0.1 \cdot 0} = 930 e^0 = 930 \][/tex]
### Answer:
[tex]\[
\boxed{930}
\][/tex]
### (c) Population at [tex]\( t = 5 \)[/tex]
To determine the population of bacteria at [tex]\( t = 5 \)[/tex], substitute [tex]\( t = 5 \)[/tex] into the given function:
[tex]\[ n(5) = 930 e^{0.1 \cdot 5} \][/tex]
Compute the exponent first:
[tex]\[ 0.1 \cdot 5 = 0.5 \][/tex]
So,
[tex]\[ n(5) = 930 e^{0.5} \][/tex]
Using the numerical value of [tex]\( e^{0.5} \approx 1.64872 \)[/tex]:
[tex]\[ n(5) = 930 \times 1.64872 \approx 1533.31 \][/tex]
Rounding to the nearest whole number gives:
[tex]\[ n(5) \approx 1533 \][/tex]
### Answer:
[tex]\[
\boxed{1533}
\][/tex]
In summary:
- (a) The continuous rate of growth is [tex]\( \boxed{10.0} \)[/tex] percent.
- (b) The initial population is [tex]\( \boxed{930} \)[/tex].
- (c) The population at [tex]\( t = 5 \)[/tex] is [tex]\( \boxed{1533} \)[/tex] bacteria (rounded to the nearest bacteria).
