There is a linear relationship between the number of chirps made by the striped ground cricket and the air temperature. A least-squares fit of some data collected by a biologist gives the model $\stackrel{⏜}{y}=25.2+3.3x$, where $x$ is

the number of chirps per minute and $\stackrel{⏜}{y}$ is the estimated temperature in degrees Fahrenheit. What is the predicted increase in temperature for an increase of $5$ chirps per minute?

$$\begin{gathered} \left(a\right)3.3°F\left(c\right)25.2°F\left(e\right)41.7°F \\ \left(b\right)16.5°F\left(d\right)28.5°F \end{gathered}$$

The model is derived from a least-squares fit of certain data acquired by a biologist. $\stackrel{⏜}{y}=25.2+3.3x$, where $x$is the anticipated temperature in Fahrenheit and $y$is the chirps per minute rate.

A regression line is a straight line that depicts the relationship between an explanatory variable $x$ and a response variable $y$ By entering any value of $x$ into the equation of the line, you may use a regression line to anticipate the value of $y$ for any value of $x$

Here, $x$ is the estimated temperature in degrees Fahrenheit and $y$ is the number of chirps per minute.

According to the definition of slope, for every unit increase in chirps per minute, there will be $3.3$units of temperature change.

Assume there is a $5$-chirp-per-minute rise. We know that a unit increment will result in a temperature increase of $3.3$ units. Then, for every $5$ units of $X$ value increment, there will be $5\times 3.3$ units of $Y$ value increment. That is, at an increase of $5$ chirps per minute, the predicted temperature increase is $$\begin{gathered} 5b=5\times 3.3 \\ =16.5°F \end{gathered}$$. As a result, the expected temperature increase is $16.5°F$