Question
Belmont is a growing industrial town. Every year, the level of CO2 emissions from the town increases by 10%. If the town produced 330,000 metric tons of CO2 this year, how much will be produced 6 years in the future? If necessary, round your answer to the nearest whole number.
Answers
GPT-4o mini
To calculate the future CO2 emissions in Belmont after 6 years with a 10% annual increase, we can use the formula for exponential growth:
\[
E = P(1 + r)^t
\]
where:
- \( E \) is the future emissions,
- \( P \) is the current emissions,
- \( r \) is the rate of increase (as a decimal),
- \( t \) is the number of years in the future.
In this case:
- \( P = 330,000 \) metric tons,
- \( r = 0.10 \) (which is 10%),
- \( t = 6 \) years.
Now, substituting these values into the formula:
\[
E = 330,000(1 + 0.10)^6
\]
\[
E = 330,000(1.10)^6
\]
First, we need to calculate \( (1.10)^6 \):
\[
(1.10)^6 \approx 1.771561
\]
Now, multiply this by the current emissions:
\[
E \approx 330,000 \times 1.771561
\]
\[
E \approx 584,115.12
\]
Rounding this to the nearest whole number, we find:
\[
E \approx 584,115
\]
Thus, the amount of CO2 emissions that will be produced 6 years in the future will be approximately **584,115 metric tons**.
\[
E = P(1 + r)^t
\]
where:
- \( E \) is the future emissions,
- \( P \) is the current emissions,
- \( r \) is the rate of increase (as a decimal),
- \( t \) is the number of years in the future.
In this case:
- \( P = 330,000 \) metric tons,
- \( r = 0.10 \) (which is 10%),
- \( t = 6 \) years.
Now, substituting these values into the formula:
\[
E = 330,000(1 + 0.10)^6
\]
\[
E = 330,000(1.10)^6
\]
First, we need to calculate \( (1.10)^6 \):
\[
(1.10)^6 \approx 1.771561
\]
Now, multiply this by the current emissions:
\[
E \approx 330,000 \times 1.771561
\]
\[
E \approx 584,115.12
\]
Rounding this to the nearest whole number, we find:
\[
E \approx 584,115
\]
Thus, the amount of CO2 emissions that will be produced 6 years in the future will be approximately **584,115 metric tons**.