To find the linear equation that models temperature $T$ as a function of the number of chirps per minute $N$, we can use the formula for the equation of a line:

$T = mN + b$

Where:

• $T$ is the temperature,
• $N$ is the number of chirps per minute,
• $m$ is the slope of the line,
• $b$ is the y-intercept (the temperature when $N = 0$).

Step 1: Find the slope $m$

The slope of a line is given by the formula:

$m = \frac{\Delta T}{\Delta N} = \frac{T_2 - T_1}{N_2 - N_1}$

From the given dataTo find the linear equation that models temperature $T$ as a function of the number of chirps per minute $N$, we can use the formula for the equation of a line:

$T = mN + b$

Where:

• $T$ is the temperature,
• $N$ is the number of chirps per minute,
• $m$ is the slope of the line,
• $b$ is the y-intercept (the temperature when $N = 0$).

Step 1: Find the slope $m$

The slope of a line is given by the formula:

$m = \frac{\Delta T}{\Delta N} = \frac{T_2 - T_1}{N_2 - N_1}$

From the given

• At $N_1 = 116$ chirps per minute, $T_1 = 75^\circ F$,
• At $N_2 = 175$ chirps per minute, $T_2 = 87^\circ F$.

Now calculate the slope:

$m = \frac{87 - 75}{175 - 116} = \frac{12}{59} \approx 0.203$

Step 2: Find the y-intercept $b$

Using the equation $T = mN + b$, substitute one of the given points (say $N_1 = 116$, $T_1 = 75$) and the slope $m = 0.203$ to solve for $b$:

$75 = 0.203(116) + b$ $75 = 23.548 + b$ $b = 75 - 23.548 = 51.452$

Step 3: Write the final equation

Now that we have the slope $m \approx 0.203$ and the y-intercept $b \approx 51.452$, the linear equation is:

$T = 0.203N + 51.452$

This is the linear model for the temperature $T$ as a function of the number of chirps per minute $N$.

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Would you like further clarification or a deeper explanation? Here are 5 related questions:

1. How would you calculate the temperature if the cricket chirps 150 times per minute?
2. What does the slope of the equation tell us in practical terms?
3. Can this model be applied universally to all cricket species?
4. How would this model change if the temperature were in Celsius?
5. How accurate is this linear approximation for extreme temperatures?

Tip: In many real-world problems, interpreting the slope and intercept can give you insight into the nature of the relationship being modeled!