Answer :
- Investment after 12 years: $A = 475(1 + 0.085)^{12}
- Frog population after 5 years: $A = 100(1 + 0.22)^5$
- Bacteria after 5 hours: $A = 10(1 + 0.80)^5$
- Bacteria after 1 day: $A = 10(1 + 0.80)^{24}$
- Bacteria after 1 week: $A = 10(1 + 0.80)^{168}$.
- Final Answers: $\boxed{1264.30}$, $\boxed{270}$, $\boxed{189}$, $\boxed{13382588}$, $\boxed{7.69 \times 10^{43}}$
### Explanation
1. Introduction to the Problems
We'll tackle these problems one by one, using the formulas for compound interest and exponential growth. Remember, these formulas help us calculate how an initial amount grows over time with a consistent rate of increase.
2. Calculating the Investment Amount
For the investment problem, we use the formula $A = P(1 + r)^t$, where:
- $A$ is the final amount
- $P$ is the principal amount ($475)
- $r$ is the interest rate (8.5% or 0.085)
- $t$ is the time in years (12)
So, $A = 475(1 + 0.085)^{12}$. The result of this calculation is approximately $1264.30.
3. Calculating the Frog Population
For the frog population problem, we use the same formula $A = P(1 + r)^t$, but now:
- $A$ is the final population
- $P$ is the initial population (100 frogs)
- $r$ is the annual growth rate (22% or 0.22)
- $t$ is the time in years (5)
So, $A = 100(1 + 0.22)^5$. The result of this calculation is approximately $270.27. Since we can't have a fraction of a frog, we'll round to the nearest whole number, which gives us 270 frogs.
4. Calculating the Bacteria Population
For the bacteria problem, we again use the formula $A = P(1 + r)^t$, where:
- $A$ is the final number of bacteria
- $P$ is the initial number of bacteria (10)
- $r$ is the growth rate per hour (80% or 0.80)
- $t$ is the time in hours (5, 24, and 168)
Let's calculate for each time period:
- After 5 hours: $A = 10(1 + 0.80)^5 \approx 188.96$. Rounding gives us 189 bacteria.
- After 1 day (24 hours): $A = 10(1 + 0.80)^{24} \approx 13,382,588.45$. Rounding gives us 13,382,588 bacteria.
- After 1 week (168 hours): $A = 10(1 + 0.80)^{168} \approx 7.69 \times 10^{43}$. This is a massive number!
5. Final Results
In summary:
- After 12 years, the investment will be approximately $1264.30.
- After 5 years, there will be approximately 270 frogs.
- After 5 hours, there will be approximately 189 bacteria.
- After 1 day, there will be approximately 13,382,588 bacteria.
- After 1 week, there will be approximately $7.69 \times 10^{43}$ bacteria.
### Examples
These calculations are useful in various real-life scenarios. For example, understanding compound interest helps in making informed investment decisions. Population growth calculations are essential in ecology for predicting species expansion or decline. Similarly, bacterial growth calculations are crucial in medicine and environmental science for understanding infection rates and managing resources.
- Frog population after 5 years: $A = 100(1 + 0.22)^5$
- Bacteria after 5 hours: $A = 10(1 + 0.80)^5$
- Bacteria after 1 day: $A = 10(1 + 0.80)^{24}$
- Bacteria after 1 week: $A = 10(1 + 0.80)^{168}$.
- Final Answers: $\boxed{1264.30}$, $\boxed{270}$, $\boxed{189}$, $\boxed{13382588}$, $\boxed{7.69 \times 10^{43}}$
### Explanation
1. Introduction to the Problems
We'll tackle these problems one by one, using the formulas for compound interest and exponential growth. Remember, these formulas help us calculate how an initial amount grows over time with a consistent rate of increase.
2. Calculating the Investment Amount
For the investment problem, we use the formula $A = P(1 + r)^t$, where:
- $A$ is the final amount
- $P$ is the principal amount ($475)
- $r$ is the interest rate (8.5% or 0.085)
- $t$ is the time in years (12)
So, $A = 475(1 + 0.085)^{12}$. The result of this calculation is approximately $1264.30.
3. Calculating the Frog Population
For the frog population problem, we use the same formula $A = P(1 + r)^t$, but now:
- $A$ is the final population
- $P$ is the initial population (100 frogs)
- $r$ is the annual growth rate (22% or 0.22)
- $t$ is the time in years (5)
So, $A = 100(1 + 0.22)^5$. The result of this calculation is approximately $270.27. Since we can't have a fraction of a frog, we'll round to the nearest whole number, which gives us 270 frogs.
4. Calculating the Bacteria Population
For the bacteria problem, we again use the formula $A = P(1 + r)^t$, where:
- $A$ is the final number of bacteria
- $P$ is the initial number of bacteria (10)
- $r$ is the growth rate per hour (80% or 0.80)
- $t$ is the time in hours (5, 24, and 168)
Let's calculate for each time period:
- After 5 hours: $A = 10(1 + 0.80)^5 \approx 188.96$. Rounding gives us 189 bacteria.
- After 1 day (24 hours): $A = 10(1 + 0.80)^{24} \approx 13,382,588.45$. Rounding gives us 13,382,588 bacteria.
- After 1 week (168 hours): $A = 10(1 + 0.80)^{168} \approx 7.69 \times 10^{43}$. This is a massive number!
5. Final Results
In summary:
- After 12 years, the investment will be approximately $1264.30.
- After 5 years, there will be approximately 270 frogs.
- After 5 hours, there will be approximately 189 bacteria.
- After 1 day, there will be approximately 13,382,588 bacteria.
- After 1 week, there will be approximately $7.69 \times 10^{43}$ bacteria.
### Examples
These calculations are useful in various real-life scenarios. For example, understanding compound interest helps in making informed investment decisions. Population growth calculations are essential in ecology for predicting species expansion or decline. Similarly, bacterial growth calculations are crucial in medicine and environmental science for understanding infection rates and managing resources.
