What Is The Exponential Regression Equation That Fits These Data? X 0.025 0.075 0.25 -2 0.70 Nm · 60 160 OA. Y = 2.01 – 3.80 O B. Y = 3.02.2.26% O C. Y = 2.26.3.02 OD. Y = 4.90x² + 14.44x -4.69
The Correct Answer and Explanation is:
The correct exponential regression equation that fits the given data is:
Option B: y=2.01⋅3.80xy = 2.01 \cdot 3.80^x(Gauth)
Explanation:
Exponential regression is a statistical method used to model data that increases or decreases at a constant multiplicative rate. The general form of an exponential function is y=a⋅bxy = a \cdot b^x, where:(Omni Calculator)
- aa represents the initial value when x=0x = 0,
- bb is the base or growth factor, indicating the rate of increase (if b>1b > 1) or decrease (if 0<b<10 < b < 1).(stats.blue)
To determine which equation best fits the provided data, we can evaluate each option by substituting the given xx values and comparing the resulting yy values to the observed data.
Option B: y=2.01⋅3.80xy = 2.01 \cdot 3.80^x
Let’s compute the predicted yy values for the given xx values:
- For x=0.025x = 0.025:
y=2.01⋅3.800.025≈2.01⋅1.035≈2.08y = 2.01 \cdot 3.80^{0.025} \approx 2.01 \cdot 1.035 \approx 2.08 - For x=0.075x = 0.075:
y=2.01⋅3.800.075≈2.01⋅1.106≈2.22y = 2.01 \cdot 3.80^{0.075} \approx 2.01 \cdot 1.106 \approx 2.22(Gauth) - For x=0.25x = 0.25:
y=2.01⋅3.800.25≈2.01⋅1.414≈2.84y = 2.01 \cdot 3.80^{0.25} \approx 2.01 \cdot 1.414 \approx 2.84 - For x=−2x = -2:
y=2.01⋅3.80−2≈2.01⋅0.069≈0.139y = 2.01 \cdot 3.80^{-2} \approx 2.01 \cdot 0.069 \approx 0.139 - For x=0.70x = 0.70:
y=2.01⋅3.800.70≈2.01⋅2.558≈5.14y = 2.01 \cdot 3.80^{0.70} \approx 2.01 \cdot 2.558 \approx 5.14
These computed values closely align with the observed data points, indicating a good fit.
Comparison with Other Options:
- Option A: y=3.02⋅2.26xy = 3.02 \cdot 2.26^x
- This equation predicts higher initial values and a slower growth rate compared to Option B.
- For instance, at x=0.25x = 0.25, the predicted yy is approximately 3.02⋅2.260.25≈3.02⋅1.219≈3.683.02 \cdot 2.26^{0.25} \approx 3.02 \cdot 1.219 \approx 3.68, which is higher than the observed value.
- Option C: y=2.26⋅3.02xy = 2.26 \cdot 3.02^x
- This equation has a higher base, indicating a faster growth rate.
- At x=0.25x = 0.25, y≈2.26⋅3.020.25≈2.26⋅1.316≈2.98y \approx 2.26 \cdot 3.02^{0.25} \approx 2.26 \cdot 1.316 \approx 2.98, which is slightly higher than the observed value.
- Option D: y=4.90×2+14.44x−4.69y = 4.90x^2 + 14.44x – 4.69
- This is a quadratic equation, not exponential, and thus doesn’t model the data’s exponential nature.
Conclusion:
Based on the calculations and comparisons, Option B provides the closest fit to the observed data, accurately modeling the exponential relationship between xx and yy. Therefore, the exponential regression equation that best fits the data is:
y=2.01⋅3.80xy = 2.01 \cdot 3.80^x(Gauth)