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Over time, the number of organisms in a population increases exponentially. The table below shows the approximate number of organisms after [tex]\( y \)[/tex] years.

\begin{tabular}{|c|c|}
\hline
[tex]$y$[/tex] years & number of organisms, [tex]$n$[/tex] \\
\hline
1 & 55 \\
\hline
2 & 60 \\
\hline
3 & 67 \\
\hline
4 & 75 \\
\hline
\end{tabular}

The environment in which the organism lives can support at most 600 organisms. Assuming the trend continues, after how many years will the environment no longer be able to support the population?

A. 12
B. 24
C. 61
D. 82


Sagot :

To solve this problem, we need to use the given data points and fit an exponential growth function to model the population growth.

1. Identify the form of the exponential growth function:

The general form of an exponential growth function is:
[tex]\[ n = A \cdot e^{Bt} \][/tex]
where [tex]\( n \)[/tex] is the number of organisms, [tex]\( t \)[/tex] is time in years, [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are constants to be determined, and [tex]\( e \)[/tex] is Euler's number (approximately 2.71828).

2. Use the given data points to find [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:

The data points provided are:
- At [tex]\( y = 1 \)[/tex], [tex]\( n = 55 \)[/tex]
- At [tex]\( y = 2 \)[/tex], [tex]\( n = 60 \)[/tex]
- At [tex]\( y = 3 \)[/tex], [tex]\( n = 67 \)[/tex]
- At [tex]\( y = 4 \)[/tex], [tex]\( n = 75 \)[/tex]

From these points, we can determine that the fitted values for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are:
[tex]\[ A = 49.19349550499532 \][/tex]
[tex]\[ B = 0.10408128420803865 \][/tex]

3. Formulate the exponential equation:

Plugging [tex]\( A \)[/tex] and [tex]\( B \)[/tex] into our exponential model, we get:
[tex]\[ n = 49.19349550499532 \cdot e^{0.10408128420803865t} \][/tex]

4. Determine the time when the number of organisms will exceed the environment's capacity:

We need to find the time [tex]\( t \)[/tex] when the number of organisms, [tex]\( n \)[/tex], reaches or exceeds 600. Set [tex]\( n \)[/tex] to 600 and solve for [tex]\( t \)[/tex]:
[tex]\[ 600 = 49.19349550499532 \cdot e^{0.10408128420803865t} \][/tex]

5. Isolate the exponential term:
[tex]\[ \frac{600}{49.19349550499532} = e^{0.10408128420803865t} \][/tex]
[tex]\[ \frac{600}{49.19349550499532} \approx 12.199 \][/tex]

6. Apply the natural logarithm to both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(12.199) = 0.10408128420803865t \][/tex]
[tex]\[ t = \frac{\ln(12.199)}{0.10408128420803865} \][/tex]

7. Calculate the value of [tex]\( t \)[/tex]:
[tex]\[ t \approx \frac{2.502}{0.10408128420803865} \][/tex]
[tex]\[ t \approx 24.030912614811925 \][/tex]

Therefore, the environment will no longer be able to support the population after approximately 24 years. The correct option is:

24