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? Exercise 2.4 .2 Using the model \( T(c)=30+1.2 c \) what temperature would you predict if you counted 20 chirps in 15 seconds?

Ask by Bowen Ward. in the United States
Feb 03,2025

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The predicted temperature is 54 °F.

Solución

To predict the temperature using the given model \( T(c) = 30 + 1.2c \), where \( c \) represents the number of chirps, follow these steps: 1. **Understand the Model:** - \( T(c) \) is the temperature. - \( c \) is the number of chirps observed in a specified time. 2. **Determine the Value of \( c \):** - The problem states that there are **20 chirps in 15 seconds**. - Since the model likely expects chirps per 15 seconds (given the context), we use \( c = 20 \). 3. **Apply the Model:** \[ T(c) = 30 + 1.2 \times c \] \[ T(20) = 30 + 1.2 \times 20 = 30 + 24 = 54 \] 4. **Interpret the Result:** - The predicted temperature is **54°F**. **Final Answer:** Using the model, the predicted temperature is 54 °F.

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To predict the temperature using the given model \( T(c) = 30 + 1.2c \), we first need to determine the chirp count \( c \). If you counted 20 chirps in 15 seconds, you can convert this into chirps per minute: \[ c = \frac{20 \text{ chirps}}{15 \text{ seconds}} \times 60 \text{ seconds} = 80 \text{ chirps per minute} \] Now, plug \( c = 80 \) into the model: \[ T(80) = 30 + 1.2 \times 80 = 30 + 96 = 126 \] Therefore, the predicted temperature is **126 degrees Fahrenheit**. Chirp counts are more than just a fun activity in nature; they can serve as a scientifically validated method for estimating temperature! This method, known as Dolbear’s Law, shows how intricate connections exist between wildlife behavior and environmental conditions. The chirps from crickets can actually act as a nature-based thermometer for outdoor enthusiasts. You may want to watch for common mistakes like forgetting to convert chirps per second to per minute. It's easy to overlook that little detail, but it’s crucial! Always ensure that your units are consistent when applying formulas. You wouldn’t want to end up predicting a temperature that’s way off the mark just because of a small miscalculation!

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