5\nselect the correct answer\nthe director of a museum recorded the number of students and adult chaperones…

5\nselect the correct answer\nthe 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 x represents the number of students and y represents the number of adult chaperones. then she used a graphing tool to find the equation of the line of best fit:\ny = 0.123x + 5.397.\nbased on the line of best fit, approximately how many students are predicted to be in a school group with 20 adult chaperones?\na. 8\nb. 119\nc. 157\nd. 206

5\nselect the correct answer\nthe 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 x represents the number of students and y represents the number of adult chaperones. then she used a graphing tool to find the equation of the line of best fit:\ny = 0.123x + 5.397.\nbased on the line of best fit, approximately how many students are predicted to be in a school group with 20 adult chaperones?\na. 8\nb. 119\nc. 157\nd. 206

Answer

Explanation:

Step1: Substitute y - value

We are given $y = 20$ and the equation $y=0.123x + 5.397$. Substitute $y$ into the equation: $20=0.123x + 5.397$.

Step2: Isolate the term with x

Subtract 5.397 from both sides: $20 - 5.397=0.123x$, so $14.603 = 0.123x$.

Step3: Solve for x

Divide both sides by 0.123: $x=\frac{14.603}{0.123}\approx119$.

Answer:

B. 119