Question

biologists have noticed 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 113 chirps per minute at 70°f and 153 chirps per minute at 80°f.
(a) find a linear equation that models the temperature t as a function of the number of chirps per minute n.
t =
(b) what is the slope of the graph?
10/40
what does it represent?
the slope means that the temperature in fahrenheit degrees increases 10 as rapidly as the number of cricket chirps per minute.
(c) if the crickets are chirping at 190 chirps per minute, estimate the temperature. (round your answer to the nearest whole number.)
90 °f

Explanation:

Part (a)

Step1: Find the slope

We have two points \((N_1, T_1)=(113, 70)\) and \((N_2, T_2)=(153, 80)\). The slope \(m\) of a line is given by \(m=\frac{T_2 - T_1}{N_2 - N_1}\).
\(m=\frac{80 - 70}{153 - 113}=\frac{10}{40}=\frac{1}{4}\)

Step2: Use point - slope form

The point - slope form of a line is \(T - T_1=m(N - N_1)\). Using the point \((113, 70)\) and \(m = \frac{1}{4}\):
\(T-70=\frac{1}{4}(N - 113)\)

Step3: Simplify the equation

\(T-70=\frac{1}{4}N-\frac{113}{4}\)
\(T=\frac{1}{4}N-\frac{113}{4}+70\)
\(T=\frac{1}{4}N-\frac{113}{4}+\frac{280}{4}\)
\(T=\frac{1}{4}N+\frac{- 113 + 280}{4}\)
\(T=\frac{1}{4}N+\frac{167}{4}\) or \(T = 0.25N+41.75\)

Step1: Recall the slope formula

We calculated the slope \(m\) using the formula \(m=\frac{T_2 - T_1}{N_2 - N_1}\) with \((N_1,T_1)=(113,70)\) and \((N_2,T_2)=(153,80)\)
\(m=\frac{80 - 70}{153 - 113}=\frac{10}{40}=\frac{1}{4}=0.25\)

Step2: Interpret the slope

The slope \(m = \frac{1}{4}\) means that for each additional chirp per minute, the temperature increases by \(\frac{1}{4}\) (or \(0.25\)) degrees Fahrenheit. So the slope is \(\frac{1}{4}\) (or \(0.25\)) and it represents that the temperature in Fahrenheit degrees increases \(\frac{1}{4}\) (or \(0.25\)) as rapidly as the number of cricket chirps per minute.

Step1: Use the linear equation

We have the linear equation \(T=\frac{1}{4}N + 41.75\). We are given \(N = 190\).

Step2: Substitute \(N = 190\) into the equation

\(T=\frac{1}{4}(190)+41.75\)
\(T = 47.5+41.75\)
\(T=89.25\approx89\) (rounded to the nearest whole number)

Answer:

\(T=\frac{1}{4}N + 41.75\) (or \(T = 0.25N+41.75\))

Part (b)