For each ($P, V$) pair, type the pressure in the $x$ column and the volume in the $y$-column. Then click "Resize window to fit data." Choose the power regression option. Copy the equation, using three significant figures, to match the data.
$V=\square P^{\wedge} \square$
Answer (2)
To express the power regression for volume ( V ) in terms of pressure ( P ), we identify the coefficient as approximately 51.4 and the exponent as approximately -1.00, leading to the equation V = 51.4 P − 1.00 . This reflects the inverse relationship between pressure and volume. Therefore, the final equation is oxed{V = 51.4P^{-1.00}} .
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Identify the coefficient and exponent from the given power regression equation.
Round the coefficient 51.387 to three significant figures, resulting in 51.4.
Round the exponent -0.999 to three significant figures, resulting in -1.00.
Substitute the rounded values into the equation V = a P b , giving the final equation V = 51.4 P − 1.00 .
Explanation
Understanding the Problem We are given a power regression equation y a pp ro x 51.387 x − 0.999 , where x represents pressure ( P ) and y represents volume ( V ). Our goal is to express this equation in the form V = a P b , rounding the coefficients to three significant figures.
Identifying Coefficients The given equation is y a pp ro x 51.387 x − 0.999 . We need to identify the coefficient a and the exponent b . From the equation, we have aa pp ro x 51.387 and ba pp ro x − 0.999 .
Rounding the Coefficient Now, we round the coefficient a to three significant figures. Since aa pp ro x 51.387 , rounding to three significant figures gives us aa pp ro x 51.4 .
Rounding the Exponent Next, we round the exponent b to three significant figures. Since ba pp ro x − 0.999 , rounding to three significant figures gives us ba pp ro x − 1.00 .
Substituting the Values Finally, we substitute the rounded values of a and b into the equation V = a P b . Thus, we have V = 51.4 P − 1.00 .
Final Answer Therefore, the power regression equation, with coefficients rounded to three significant figures, is V = 51.4 P − 1.00 .
Examples
Power regression is a valuable tool in various fields, including physics and engineering. For instance, in thermodynamics, it helps model the relationship between pressure and volume of a gas, as described by the ideal gas law. Understanding this relationship allows engineers to design efficient engines and predict the behavior of gases under different conditions. Similarly, in fluid dynamics, power regression can model the relationship between flow rate and pressure drop in pipes, aiding in the design of efficient pipeline systems. These models are crucial for optimizing performance and ensuring safety in numerous applications.
