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On a particularly windy day, the percentage of target hits for different distances from the target were recorded for several participants at a children's archery camp. The data is shown in the scatter
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On a particularly windy day, the percentage of target hits for different distances from the target were recorded for several participants at a children's archery camp. The data is shown in the scatter plot below.

The equation for the line of best fit is as shown below.
Y = 90 - 2.75x

Use the equation for line of best fit to interpret the slope and y-intercept. Based on the line of best fit, predict the percentage of target hits from a distance of 32 meters?

On a particularly windy day, the percentage of target hits for different distances-example-1
User Ekstrakt
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1 Answer

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To find this out, you will plug in the distance for x, so your equation will now be
y= 90- 2.75(32)
y= 90- 88
y= 2
And since y is a percentage, your answer will be
2% of target hits from a distance of 32 meters.
User AKC
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4.2k points
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