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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 $ ^{\circ} F $ and 173 chirps per minute at 80 $ ^{\circ} F $.

(a) Find a linear equation that models the temperature $ T $ as a function of the number of chirps per minute $ N $.

(b) What is the slope of the graph? What does it represent?

(c) If the crickets are chirping at 150 chirps per minute, estimate the temperature.

(a) Using $N$ in place of $x$ and $T$ in place of $y,$ we find the slope to be $\frac{T_{2}-T_{1}}{N_{2}-N_{1}}=\frac{80-70}{173-113}=\frac{10}{60}=\frac{1}{6} .$ So a linear

\[

\text { uation is } T-80=\frac{1}{6}(N-173) \Leftrightarrow \quad T-80=\frac{1}{6} N-\frac{173}{6} \quad \Leftrightarrow \quad T=\frac{1}{6} N+\frac{307}{6}\left[\frac{307}{6}=51.1 \overline{6}\right]

\].

(b) The slope of $\frac{1}{6}$ means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket

chirps per minute, Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of $1^{\circ} \mathrm{F}$.

(c) When $N=150,$ the temperature is given approximately by $T=\frac{1}{6}(150)+\frac{307}{6}=76.1 \overline{6}^{\circ} \mathrm{F} \approx 76^{\circ} \mathrm{F}$.

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Catherine A.

October 27, 2020

Heather Z., thanks this was super helpful.

Catherine A.

October 27, 2020

This will help a lot with my midterm

Arkelly M.

September 10, 2021

how did find 307?

Johns Hopkins University

Campbell University

Baylor University

University of Michigan - Ann Arbor

all right here we have the classic cricket chirping problem, and we're going to model the temperature as a function of the number of chirps. So the temperature will be our Y coordinate, and the number of chirps will be our X coordinate. And we're going to use end for the number of chirps and t for the temperature. And the problem gives us a couple of data points that we could translate into ordered pairs. When the number of chirps per minute is 1 13 it's 70 F. And when the number of chirps per minute is 1 73 it's 80 F. So we can use those two points to figure out the equation of a line. And that will be our linear model. So let's start by finding the slope. So we take y tu minus y one a T minus 70 over X two minus x 11 73 minus 1 13 And that gives us 10 60 ifs. And that reduces to 1/6. Okay, now that we have the slope, let's use our point slope form. Why minus y one equals m times X minus X one and let's use one of our points doesn't really matter which one. I'm going to use the 1st 1 And so we have. Why minus 70 equals 1/6 times X minus 1 13 When we simplify that equation first by distributing the 1/6 why minus 70 equals 16 x minus 113 6th and then we add 70 to both sides. We get why equals 16 x plus 307 6 Now we want to use our variables in and T So let's substitute those in there. So we have t temperature equals 1/6 and number of trips per minute plus 307 6 All right, so there's our model, so that takes care of part A. Okay, let's see what we need to do in part B. So it says, what is the slope? So if we look at our equation, we see the slope is 1/6 and what does it represent? Okay, let's think about the units here. So slope is rise over. Run slope is changing. Why? Over change in X and the units on why were the temperature? It was temperature in degrees Fahrenheit, so that would be degrees over the units on acts That was the number of chirps, the degrees over number of chirps per minute. So what this is telling us is that if it goes up, one degree will have a rise in six chirps per minute. So one degree rise results in six more chirps per minute. Okay? And finally, let's look at the temperature. If the crickets were chirping at 150 chirps per minute, let's use our model and substitute 1 50 for N and find the temperature so the temperature would be 1/6 times 1 50 plus 307 6 And that gives us 76 1/6 about 76.17 degrees Fahrenheit at that rate of tripping.