The exponential decay model for the radioactive material can be written as:
\[ N(t) = N_0 \cdot 0.5^{\frac{t}{42}} \]
where:
- \( N(t) \) is the amount of material remaining after time \( t \) days
- \( N_0 = 49 \) kg is the initial amount of material
- \( t = 8 \) days
Plugging in the values, we get:
\[ N(8) = 49 \cdot 0.5^{\frac{8}{42}} \]
\[ N(8) = 49 \cdot 0.5^{0.190} \]
\[ N(8) = 49 \cdot 0.702 \]
\[ N(8) \approx 34.398 \text{ kg} \]
Therefore, approximately 34.398 kg of radioactive material remains after 8 days.
The half-life of a certain radioactive material is 42 days. An initial amount of the material has a mass of 49 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 8 days. Round your answer to the nearest thousandth..
5 answers
34.398 is incorrect
I apologize for the error. Let's recalculate it:
\[ N(8) = 49 \cdot 0.5^{\frac{8}{42}} \]
\[ N(8) = 49 \cdot 0.5^{0.190476} \]
\[ N(8) = 49 \cdot 0.827107 \]
\[ N(8) ≈ 40.503 \text{ kg} \]
Therefore, approximately 40.503 kg of radioactive material remains after 8 days.
\[ N(8) = 49 \cdot 0.5^{\frac{8}{42}} \]
\[ N(8) = 49 \cdot 0.5^{0.190476} \]
\[ N(8) = 49 \cdot 0.827107 \]
\[ N(8) ≈ 40.503 \text{ kg} \]
Therefore, approximately 40.503 kg of radioactive material remains after 8 days.
that is also incorrect
I apologize for the oversight. Let's recalculate it one more time:
\[ N(8) = 49 \cdot 0.5^{\frac{8}{42}} \]
\[ N(8) = 49 \cdot 0.5^{0.190476} \]
\[ N(8) = 49 \cdot 0.881415 \]
\[ N(8) ≈ 43.233 \text{ kg} \]
Therefore, approximately 43.233 kg of radioactive material remains after 8 days.
\[ N(8) = 49 \cdot 0.5^{\frac{8}{42}} \]
\[ N(8) = 49 \cdot 0.5^{0.190476} \]
\[ N(8) = 49 \cdot 0.881415 \]
\[ N(8) ≈ 43.233 \text{ kg} \]
Therefore, approximately 43.233 kg of radioactive material remains after 8 days.
1 answer