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To determine the predicted amount of the unstable element left after 6 years, we use the given least-squares regression line equation:
[tex]\[ \ln (\text{Element}) = 2.305 - 0.101 \times (\text{Time}) \][/tex]
1. Identify the given time:
- Time = 6 years
2. Substitute the time into the regression equation:
[tex]\[ \ln (\text{Element}) = 2.305 - 0.101 \times 6 \][/tex]
3. Calculate the value inside the equation:
[tex]\[ \ln (\text{Element}) = 2.305 - 0.606 \][/tex]
[tex]\[ \ln (\text{Element}) = 1.699 \][/tex]
4. Now, to find the predicted amount of the element, we need to eliminate the natural logarithm by exponentiating both sides of the equation. This means applying the exponential function [tex]\(e\)[/tex] to both sides:
[tex]\[ \text{Element} = e^{1.699} \][/tex]
5. Calculate the value of [tex]\(e^{1.699}\)[/tex]:
[tex]\[ \text{Element} \approx 5.468 \text{ grams} \][/tex]
Therefore, the predicted amount of the unstable element left after 6 years is approximately 5.468 grams.
Among the given options, the correct answer is:
[tex]\[ \boxed{5.468 \text{ grams}} \][/tex]
[tex]\[ \ln (\text{Element}) = 2.305 - 0.101 \times (\text{Time}) \][/tex]
1. Identify the given time:
- Time = 6 years
2. Substitute the time into the regression equation:
[tex]\[ \ln (\text{Element}) = 2.305 - 0.101 \times 6 \][/tex]
3. Calculate the value inside the equation:
[tex]\[ \ln (\text{Element}) = 2.305 - 0.606 \][/tex]
[tex]\[ \ln (\text{Element}) = 1.699 \][/tex]
4. Now, to find the predicted amount of the element, we need to eliminate the natural logarithm by exponentiating both sides of the equation. This means applying the exponential function [tex]\(e\)[/tex] to both sides:
[tex]\[ \text{Element} = e^{1.699} \][/tex]
5. Calculate the value of [tex]\(e^{1.699}\)[/tex]:
[tex]\[ \text{Element} \approx 5.468 \text{ grams} \][/tex]
Therefore, the predicted amount of the unstable element left after 6 years is approximately 5.468 grams.
Among the given options, the correct answer is:
[tex]\[ \boxed{5.468 \text{ grams}} \][/tex]
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