Question

1. crickets and temperature biologists have observed that the chirping rate of crickets of a certain species is related to temperature, and the relationship appears to be very nearly linear. a cricket produces 120 chirps per minute at 70°f and 168 chirps per minute at 80°f. (a) find the linear equation that relates the temperature t and the number of chirps per minute n. (b) if the crickets are chirping at 150 chirps per minute, estimate the temperature.

Explanation:

Part (a)

Step1: Identify two points

We have two points \((t_1, n_1)=(70, 120)\) and \((t_2, n_2)=(80, 168)\) where \(t\) is temperature and \(n\) is chirps per minute.

Step2: Calculate the slope \(m\)

The slope formula is \(m=\frac{n_2 - n_1}{t_2 - t_1}\).
Substitute the values: \(m=\frac{168 - 120}{80 - 70}=\frac{48}{10}=4.8\).

Step3: Use point - slope form to find the equation

The point - slope form of a line is \(n - n_1=m(t - t_1)\). Let's use the point \((70, 120)\).
\(n - 120 = 4.8(t - 70)\)
Expand the right - hand side: \(n-120 = 4.8t-336\)
Add 120 to both sides: \(n = 4.8t - 216\)
We can also write it as \(n-4.8t=- 216\) or solving for \(t\) (if needed), but the problem asks for the equation relating \(t\) and \(n\), so \(n = 4.8t-216\) (or \(t=\frac{n + 216}{4.8}\))

Part (b)

Step1: Substitute \(n = 150\) into the equation

We use the equation \(n = 4.8t-216\). Substitute \(n = 150\):
\(150=4.8t - 216\)

Step2: Solve for \(t\)

Add 216 to both sides: \(150 + 216=4.8t\)
\(366 = 4.8t\)
Divide both sides by 4.8: \(t=\frac{366}{4.8}=76.25\)

Answer:

s:
(a) The linear equation is \(\boldsymbol{n = 4.8t-216}\) (or equivalent forms like \(t=\frac{n + 216}{4.8}\))
(b) The temperature is \(\boldsymbol{76.25^{\circ}\text{F}}\)