The problem states that we have a relationship between the number of students [tex]\( x \)[/tex] and the number of adult chaperones [tex]\( y \)[/tex] given by the equation of the line of best fit:
[tex]\[ y = 0.123x + 5.397 \][/tex]
We need to find the number of students [tex]\( x \)[/tex] when the number of adult chaperones [tex]\( y \)[/tex] is 20.
To do this, we can substitute [tex]\( y = 20 \)[/tex] into the equation and solve for [tex]\( x \)[/tex]:
[tex]\[ 20 = 0.123x + 5.397 \][/tex]
First, we subtract 5.397 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 20 - 5.397 = 0.123x \][/tex]
[tex]\[ 14.603 = 0.123x \][/tex]
Next, we divide both sides of the equation by 0.123 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{14.603}{0.123} \][/tex]
[tex]\[ x \approx 118.72357723577235 \][/tex]
Thus, the predicted number of students is approximately 119.
Therefore, the correct answer is:
B. 119
