5 answers
The equation \( f(x) = 1600(1.025)^x \) models the population of the school, where \( x \) represents the number of years since 2015.
To find the population of the school in the year 2015, we need to set \( x = 0 \) (since 2015 is the base year).
Substituting \( x = 0 \) into the equation:
\[ f(0) = 1600(1.025)^0 \]
Since any number raised to the power of 0 is 1:
\[ f(0) = 1600 \cdot 1 = 1600 \]
Therefore, the population of the school in the year 2015 was 1600.
To predict the number of students enrolled in the school in the year 2019, we first need to determine how many years have passed since 2015. The year 2019 is 4 years after 2015, so we set \( x = 4 \) in the equation:
\[ f(x) = 1600(1.025)^x \]
Now substituting \( x = 4 \):
\[ f(4) = 1600(1.025)^4 \]
Calculating \( (1.025)^4 \):
\[ (1.025)^4 \approx 1.104486 \]
Now, multiplying that result by 1600:
\[ f(4) \approx 1600 \times 1.104486 \approx 1767.178 \]
Rounding to the nearest whole number, the predicted number of students enrolled in the school in the year 2019 is approximately 1767.
To predict the number of students enrolled in the school in the year 2023, we first determine how many years have passed since 2015. The year 2023 is 8 years after 2015, so we set \( x = 8 \) in the equation:
\[ f(x) = 1600(1.025)^x \]
Substituting \( x = 8 \):
\[ f(8) = 1600(1.025)^8 \]
Now calculating \( (1.025)^8 \):
\[ (1.025)^8 \approx 1.218402 \]
Now, multiplying that result by 1600:
\[ f(8) \approx 1600 \times 1.218402 \approx 1949.443 \]
Rounding to the nearest whole number, the predicted number of students enrolled in the school in the year 2023 is approximately 1949.
