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The director of a museum recorded the number of students and adult chaperones in several school groups visiting the museum. She organized the data in a scatter plot, where [tex]\( x \)[/tex] represents the number of students and [tex]\( y \)[/tex] represents the number of adult chaperones. Then she used a graphing tool to find the equation of the line of best fit:

[tex]\[ y = 0.123x + 5.397 \][/tex]

Based on the line of best fit, approximately how many students are predicted to be in a school group with 20 adult chaperones?

A. 8
B. 119
C. 157
D. 206

Sagot :

GinnyAnsweridn
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
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