Alright, let's solve each part of this problem step-by-step.
### Part (a) - Relative Rate of Growth
The given function for the number of bacteria is:
[tex]\[ n(t) = 925 e^{0.1 t} \][/tex]
The relative rate of growth in an exponential function of the form [tex]\( n(t) = n_0 e^{rt} \)[/tex] is given by the exponent [tex]\( r \)[/tex].
Here, the exponent [tex]\( r \)[/tex] is [tex]\( 0.1 \)[/tex].
To express the relative rate of growth as a percentage, we multiply by 100:
[tex]\[ 0.1 \times 100 = 10\% \][/tex]
So, the relative rate of growth of this bacterium population is [tex]\( 10\% \)[/tex].
### Part (b) - Initial Population at [tex]\( t = 0 \)[/tex]
To find the initial population, we evaluate the function [tex]\( n(t) \)[/tex] at [tex]\( t = 0 \)[/tex]:
[tex]\[ n(0) = 925 e^{0.1 \cdot 0} \][/tex]
Since [tex]\( e^{0} = 1 \)[/tex]:
[tex]\[ n(0) = 925 \times 1 = 925 \][/tex]
Therefore, the initial population of the culture is [tex]\( 925 \)[/tex].
### Part (c) - Population at [tex]\( t = 5 \)[/tex]
To find the population at [tex]\( t = 5 \)[/tex], we evaluate the function [tex]\( n(t) \)[/tex] at [tex]\( t = 5 \)[/tex]:
[tex]\[ n(5) = 925 e^{0.1 \cdot 5} \][/tex]
First, calculate the exponent:
[tex]\[ 0.1 \times 5 = 0.5 \][/tex]
So, we need to find [tex]\( e^{0.5} \)[/tex], and then multiply by 925.
Evaluating:
[tex]\[ n(5) = 925 e^{0.5} \][/tex]
The approximate value of [tex]\( e^{0.5} \)[/tex] is used to find:
[tex]\[ n(5) \approx 925 \times \text{value of } e^{0.5} \][/tex]
After evaluating, you would get approximately:
[tex]\[ n(5) \approx 1525.0671753976185 \][/tex]
Therefore, the population at [tex]\( t = 5 \)[/tex] is approximately [tex]\( 1525.0671753976185 \)[/tex].
So, here are the answers:
(a) The relative rate of growth is [tex]\( 10\% \)[/tex].
(b) The initial population of the culture is [tex]\( 925 \)[/tex].
(c) The population at [tex]\( t = 5 \)[/tex] is approximately [tex]\( 1525.0671753976185 \)[/tex].
