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There are approximately 3,000 bass in a lake. The population grows at a rate of \( 2 \% \) per year. From Year 1 to Year 4 , the average rate of change of the population to the nearest tenth was about From Year 5 to Year 8 , the average rate of change of the population to the nearest tenth was about The rate of change increased by about bass per a year.

Ask by Smith Best. in the United States
Jan 24,2025

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Answer

- **From Year 1 to Year 4:** The average rate of change was **62.4 bass per year**. - **From Year 5 to Year 8:** The average rate of change was **67.6 bass per year**. - **Increase in Rate of Change:** **5.2 bass per year**.

Solution

Let's analyze the population growth of bass in the lake step by step. ### Given: - **Initial Population (P₀):** 3,000 bass - **Growth Rate:** 2% per year - **Population Function:** \( P(t) = 3000 \times (1.02)^t \), where \( t \) is the number of years ### 1. Average Rate of Change from Year 1 to Year 4 **Calculate Population at Year 1 and Year 4:** - **Year 1 (P₁):** \[ P(1) = 3000 \times 1.02^1 = 3000 \times 1.02 = 3060 \text{ bass} \] - **Year 4 (P₄):** \[ P(4) = 3000 \times 1.02^4 \approx 3000 \times 1.08243216 \approx 3247.3 \text{ bass} \] **Average Rate of Change:** \[ \text{Average Rate} = \frac{P(4) - P(1)}{4 - 1} = \frac{3247.3 - 3060}{3} \approx \frac{187.3}{3} \approx 62.4 \text{ bass per year} \] ### 2. Average Rate of Change from Year 5 to Year 8 **Calculate Population at Year 5 and Year 8:** - **Year 5 (P₅):** \[ P(5) = 3000 \times 1.02^5 \approx 3000 \times 1.1040808 \approx 3312.2 \text{ bass} \] - **Year 8 (P₈):** \[ P(8) = 3000 \times 1.02^8 \approx 3000 \times 1.171659 \approx 3515.0 \text{ bass} \] **Average Rate of Change:** \[ \text{Average Rate} = \frac{P(8) - P(5)}{8 - 5} = \frac{3515.0 - 3312.2}{3} \approx \frac{202.8}{3} \approx 67.6 \text{ bass per year} \] ### 3. Increase in the Rate of Change \[ \text{Increase} = 67.6 - 62.4 = 5.2 \text{ bass per year} \] ### **Summary:** - **From Year 1 to Year 4:** The average rate of change was **62.4 bass per year**. - **From Year 5 to Year 8:** The average rate of change was **67.6 bass per year**. - **Increase in Rate of Change:** **5.2 bass per year**.

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The Deep Dive

The bass population grows exponentially, which means that each year, the population is multiplied by \( 1.02 \). For Year 1 to Year 4, you can calculate the average rate of change by observing the population at Year 1 (\( 3000 \)), Year 4, and then taking the difference and averaging it over the four years. Similarly, for Year 5 to Year 8, use the same approach considering the population at Years 5 and 8. Over those time frames, you would find the population for Year 4 to be approximately \( 3000 \times (1.02)^4 \approx 3082.43 \) and Year 8 to be \( 3000 \times (1.02)^8 \approx 3172.66 \). So the average rate from Year 1 to Year 4 would be around 20.6 bass per year, while from Year 5 to Year 8, it would be approximately 22.4. The change in rates would then be about \( 1.8 \) bass per year. If you’re curious about the real-world implications of bass population growth, consider the impact on local ecosystems. A healthy growth rate can help maintain the balance between predator and prey, enhancing fishing experiences and tourism. But if the population grows too rapidly, it can lead to overpopulation, impacting habitat quality and food supply. On the fun side, did you know that many fish can communicate? Bass are known to use sound to chat with each other - from grunts to pops! So next time you're out fishing, just think, those bass could be exchanging juicy lake gossip while they're swimming around.

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