Question

find the exponential regression equation that best fits the data (10,4), (12,20), (13,35), and (16,300). (1 point) ○ y = 0.003(2.04)^x ○ y = 2.04(0.003)^x ○ y = 11.27(1.00)^x ○ y = 1.00(11.27)^x

Explanation:

Step1: Recall the general form of an exponential function

The general form of an exponential function is \(y = ab^{x}\), where \(a\) is the initial value and \(b\) is the base.

Step2: Analyze each option

• For \(y = 0.003(2.04)^{x}\):
• When \(x = 10\), \(y=0.003\times(2.04)^{10}\approx0.003\times120.28 = 0.36

eq4\).

• For \(y = 2.04(0.003)^{x}\):
• As \(x\) increases, since \(0 < 0.003<1\), \(y\) will decrease rapidly. When \(x = 10\), \(y = 2.04\times(0.003)^{10}\) is an extremely small number, not \(4\).
• For \(y = 11.27(1.00)^{x}\):
• Since \(1.00^{x}=1\) for all \(x\), \(y = 11.27\) for all \(x\), which does not match the data points like \((10,4)\).
• For \(y = 1.00(11.27)^{x}\):
• When \(x = 10\), \(y=(11.27)^{10}\) is a very large number, not \(4\).

Answer:

None of the options are correct. If we assume there is a mistake in the problem - setting and we use the formula \(y = ab^{x}\) and substitute the point \((x = 10,y = 4)\) and \((x = 12,y = 20)\):
We have the system \(

$$\begin{cases}4=ab^{10}\\20=ab^{12}\end{cases}$$

\). Dividing the second equation by the first gives \(\frac{ab^{12}}{ab^{10}}=\frac{20}{4}\), so \(b^{2} = 5\), \(b=\sqrt{5}\approx2.24\) and \(a=\frac{4}{b^{10}}\). But if we consider the closest form among the given options, there is an error in the problem's option - set. If we force - fit, none of the given options \(y = 0.003(2.04)^{x}\), \(y = 2.04(0.003)^{x}\), \(y = 11.27(1.00)^{x}\), \(y = 1.00(11.27)^{x}\) correctly model the exponential growth from \((10,4)\) to \((12,20)\) to \((13,35)\) and \((16,300)\).

However, if we assume a wrong - keyed option and use the fact that for an exponential function \(y = ab^{x}\), when \(x\) increases, the function value should increase. The form \(y = 0.003(2.04)^{x}\) has a base \(b = 2.04>1\) (exponential growth), while \(y = 2.04(0.003)^{x}\) has \(0<0.003<1\) (exponential decay), \(y = 11.27(1.00)^{x}\) is a constant function and \(y = 1.00(11.27)^{x}\) has a very large base. If we calculate \(y = 0.003(2.04)^{10}\approx0.003\times120.28 = 0.36\), \(y = 0.003(2.04)^{12}=0.003\times(2.04)^{10}\times(2.04)^{2}\approx0.36\times4.16 = 1.5\), \(y = 0.003(2.04)^{13}\approx1.5\times2.04 = 3.06\), \(y = 0.003(2.04)^{16}\approx0.003\times(2.04)^{10}\times(2.04)^{6}\approx0.36\times70.93=25.54\) (still not \(300\)). But among the given options, \(y = 0.003(2.04)^{x}\) is the only one with exponential growth (base \(b>1\)). So if we have to choose the “best” (even though it is not a perfect fit), it is \(y = 0.003(2.04)^{x}\).