To solve this problem, we first need to clarify the growth model for the bacterial population. The exponential growth function can be expressed in the form:
\[ N(t) = N_0 e^{rt} \]
where:
- \( N_0 \) is the initial population,
- \( r \) is the growth rate (as a decimal),
- \( t \) is the time in the same units as the rate,
- \( e \) is the base of the natural logarithm.
Given that the initial population \( N_0 = 950 \) and the growth rate is 4.75% per day, we convert the percentage into a decimal:
\[ r = \frac{4.75}{100} = 0.0475 \]
Since the growth is continuous, we can rewrite the formula using the exponential growth model:
\[ N(t) = 950 e^{0.0475t} \]
However, to fit this into the options provided, we can express the growth model using another exponential base, usually base 10 or 2. But since the options are not directly based on \( e \), we'll interpret which one fits.
Let's analyze the choices:
Options:
- A: \( N(t) = 950(0. 9525)^t \)
- B: \( N(t) = 950(1.0475)^t \)
- C: \( N(t) = 950(4.75)^t \)
- D: \( N(t) = 950(104.75)^t \)
Given that the growth factor per day is \( 1 + r = 1 + 0.0475 = 1.0475 \), the appropriate function corresponds to:
\[ \text{Option B: } N(t) = 950(1.0475)^t \]
Part A: The correct exponential function is:
B: \( N(t) = 950(1.0475)^t \)
Part B: Determine the bacterial population after 60 hours.
Since the function we've chosen is based on days, we need to convert 60 hours into days:
\[ \text{Time in days} = \frac{60 \text{ hours}}{24 \text{ hours/day}} = 2.5 \text{ days} \]
Now, we can plug \( t = 2.5 \) into the chosen function:
\[ N(2.5) = 950(1.0475)^{2.5} \]
Calculating \( (1.0475)^{2.5} \):
\[ (1.0475)^{2.5} \approx 1.1254 \quad (\text{Using a calculator}) \]
Now calculate \( N(2.5) \):
\[ N(2.5) \approx 950 \times 1.1254 \approx 1064.13 \]
Rounding to the nearest bacterium gives:
\[ N(2.5) \approx 1064 \]
Part B: The bacterial population after 60 hours is approximately:
1064 bacteria.