Select the correct answer.
| x | y |
| --- | ----- |
| 2.5 | 23.5 |
| 4 | ? |
| 6.1 | 68.9 |
| 7.9 | 95.6 |
| 9.6 | 134.4 |
The table lists the values for two parameters, [tex]$x$[/tex] and [tex]$y$[/tex], of an experiment. Using interpolation, what is the approximate value of [tex]$y$[/tex] when [tex]$x=4$[/tex]?
A. 42.4
B. 37.6
C. 45.2
D. 40.1
Answer (2)
Identify the two closest x values to x = 4 , which are x 1 = 2.5 and x 2 = 6.1 , and their corresponding y values, y 1 = 23.5 and y 2 = 68.9 .
Apply the linear interpolation formula: y = y 1 + ( x − x 1 ) × ( x 2 − x 1 ) ( y 2 − y 1 ) .
Substitute the values into the formula: y = 23.5 + ( 4 − 2.5 ) × ( 6.1 − 2.5 ) ( 68.9 − 23.5 ) .
Calculate the result and round to one decimal place: y ≈ 42.4 .
The approximate value of y when x = 4 is 42.4 .
Explanation
Understanding the Problem We are given a table of x and y values and asked to estimate the value of y when x = 4 using linear interpolation. Linear interpolation involves finding the equation of the line between two known points and using that equation to estimate a value between those points.
Identifying Nearest Points First, we need to identify the two closest x values in the table to our target x = 4 . These are x 1 = 2.5 and x 2 = 6.1 . The corresponding y values are y 1 = 23.5 and y 2 = 68.9 .
Stating the Interpolation Formula The formula for linear interpolation is: y = y 1 + ( x − x 1 ) × ( x 2 − x 1 ) ( y 2 − y 1 ) This formula calculates y based on the known values.
Substituting Values Now, we substitute the known values into the formula: y = 23.5 + ( 4 − 2.5 ) × ( 6.1 − 2.5 ) ( 68.9 − 23.5 )
Simplifying the Expression Next, we simplify the expression: y = 23.5 + ( 1.5 ) × ( 3.6 ) ( 45.4 ) y = 23.5 + 1.5 × 12.6111 y = 23.5 + 18.9166 y = 42.4166
Calculating the Final Value Finally, we round the result to one decimal place, as the given values are presented: y ≈ 42.4 Therefore, the approximate value of y when x = 4 is 42.4.
Examples
Linear interpolation is used in many real-world applications, such as estimating population sizes between census years, predicting financial values between known data points, or even smoothing out sensor data in robotics. For example, if you know the temperature at 1 PM and 3 PM, you can use linear interpolation to estimate the temperature at 2 PM, assuming a steady change in temperature. This method provides a quick and reasonable estimate when precise data is unavailable.
Using linear interpolation, the approximate value of y when x = 4 is calculated to be 42.4. This is done by identifying the closest x values and then applying the interpolation formula. The correct answer choice is A. 42.4.
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