Heather is training for a long-distance run. Her data points listed below represent the days of practice, [tex]$x$[/tex], and the number of miles run, [tex]$y$[/tex].
[tex]$(1,2.5),(2,4.2),(4,5.6),(6,7),(8,8.1),(10,11)$[/tex]
Use the equation to interpolate the value and estimate the distance that she could have run on day 3. Round to the nearest tenth of a mile.
day 3 = [$\square$] miles
Answer (2)
Analyze the given data points for days of practice ( x ) and miles run ( y ).
Perform a linear regression to find the equation y = m x + b , where m ≈ 0.85 and b ≈ 2.01 .
Substitute x = 3 into the equation to estimate the distance: y = 0.85 ( 3 ) + 2.01 = 4.56 .
Round the result to the nearest tenth of a mile: 4.6 miles.
Explanation
Analyzing the Data First, let's analyze the given data. We have the days of practice, x , and the corresponding number of miles run, y . The data points are (1, 2.5), (2, 4.2), (4, 5.6), (6, 7), (8, 8.1), and (10, 11). Our goal is to estimate the distance Heather could have run on day 3 using interpolation. This means we need to find a relationship between x and y and then use that relationship to predict the value of y when x = 3 .
Finding the Linear Regression Equation To find the relationship between x and y , we can perform a linear regression. This will give us an equation of the form y = m x + b , where m is the slope and b is the y-intercept. Using the given data points, we find that the slope m ≈ 0.85 and the y-intercept b ≈ 2.01 . Therefore, the equation is approximately y = 0.85 x + 2.01 .
Estimating the Distance on Day 3 Now that we have the equation, we can substitute x = 3 to estimate the distance Heather could have run on day 3: y = 0.85 ( 3 ) + 2.01 = 2.55 + 2.01 = 4.56 So, the estimated distance is approximately 4.56 miles.
Rounding the Result Finally, we round the result to the nearest tenth of a mile, which gives us 4.6 miles.
Examples
Linear interpolation, like we used here, is helpful in many real-world situations. For example, if you're tracking the growth of a plant over several days, you can use interpolation to estimate its height on a day you didn't measure it. Similarly, in finance, you might use interpolation to estimate the value of an investment at a specific point in time based on its values at other times. This technique helps fill in the gaps in your data and make informed predictions.
To estimate the distance Heather ran on Day 3, we found a linear regression equation from the given data, which was approximately y = 0.85 x + 2.01 . By substituting x = 3 into this equation, we calculated the estimated distance as 4.56 miles, which rounds to 4.6 miles. Therefore, the final answer is that Heather could have run approximately 4.6 miles on Day 3.
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