select the correct answer\na geologist gathered data about the total shoreline and maximum depth of several…

select the correct answer\na geologist gathered data about the total shoreline and maximum depth of several area lakes and organized the data into the table\n| total shoreline (miles) | 22 | 17 | 10 | 23 | 12 | 35 | 7 |\n| maximum depth (feet) | 101 | 85 | 59 | 113 | 64 | 158 | 33 |\nshe then used a graphing tool to display the data in a scatter - plot, with x representing the total miles of shoreline and y representing the maximum depth. she also used the graphing tool to find the equation of the line of best fit.\ny = 4.26x+10.908\nbased on the line of best fit, what is the approximate maximum depth of a lake that has 31 miles of shoreline?\na. 121 feet\nb. 132 feet\nc. 138 feet\nd. 143 feet

select the correct answer\na geologist gathered data about the total shoreline and maximum depth of several area lakes and organized the data into the table\n| total shoreline (miles) | 22 | 17 | 10 | 23 | 12 | 35 | 7 |\n| maximum depth (feet) | 101 | 85 | 59 | 113 | 64 | 158 | 33 |\nshe then used a graphing tool to display the data in a scatter - plot, with x representing the total miles of shoreline and y representing the maximum depth. she also used the graphing tool to find the equation of the line of best fit.\ny = 4.26x+10.908\nbased on the line of best fit, what is the approximate maximum depth of a lake that has 31 miles of shoreline?\na. 121 feet\nb. 132 feet\nc. 138 feet\nd. 143 feet

Answer

Explanation:

Step1: Identify the values

We are given the equation of the line of best - fit $y = 4.26x+10.908$ and $x = 31$ (miles of shoreline).

Step2: Substitute $x$ into the equation

Substitute $x = 31$ into $y = 4.26x+10.908$. So, $y=4.26\times31 + 10.908$. First, calculate $4.26\times31$: $4.26\times31=(4 + 0.26)\times31=4\times31+0.26\times31=124+8.06 = 132.06$. Then, $y=132.06+10.908=142.968\approx143$.

Answer:

D. 143 feet