In this problem, study time is the explanatory variable $x$, and test score is the response variable $y$. The standard deviation of $y$ should go in the numerator and the standard deviation of $x$ in the denominator.

Once you write $b = r \cdot \dfrac{s_y}{s_x}$, plug in $r = 0.80$, $s_y = 6.0$, and $s_x = 2.0$. Simplify the fraction first, then multiply so you do not confuse multiplying and dividing by 3.

In Desmos, type the expression 0.80*(6/2) (this is $r \cdot (s_y/s_x)$ with $r=0.80$, $s_y=6$, and $s_x=2$).

Look at the value Desmos returns for 0.80*(6/2); that number is the slope of the regression line predicting test score from study time.

When you predict a response variable $y$ from an explanatory variable $x$ using the least–squares regression line, and you are given the correlation and standard deviations, the slope $b$ is calculated by

where:

• $r$ is the correlation between $x$ and $y$,
• $s_y$ is the standard deviation of the response variable (here, test score),
• $s_x$ is the standard deviation of the explanatory variable (here, study time).

From the table:

• The correlation is $r = 0.80$.
• The standard deviation of test scores (the response, $y$) is $s_y = 6.0$.
• The standard deviation of study times (the explanatory variable, $x$) is $s_x = 2.0$.

Substitute these into the formula:

First simplify the fraction:

Now multiply by the correlation:

So, the slope of the least–squares regression line predicting test score from study time is $2.4$, which corresponds to choice C.