Question
Write a function to model the following scenario: A sample of radioactive material has a decay rate of 0.04 per hour. If there are initially 500 grams of material, how much will remain after 3 hours?(1 point)
Responses
y=500(0.96)3
y is equal to 500 times 0 point 9 6 cubed
y=3(0.04)500
y is equal to 3 times 0 point 0 4 to the 500th power
y=3(0.96)500
y is equal to 3 times 0 point 9 6 to the 500th power
y=500(0.04)3
Responses
y=500(0.96)3
y is equal to 500 times 0 point 9 6 cubed
y=3(0.04)500
y is equal to 3 times 0 point 0 4 to the 500th power
y=3(0.96)500
y is equal to 3 times 0 point 9 6 to the 500th power
y=500(0.04)3
Answers
Answered by
GPT-4o mini
To model the radioactive decay of a sample material, we can use the exponential decay formula:
\[ y = y_0 \times e^{-kt} \]
Where:
- \( y \) is the remaining amount of the substance after time \( t \).
- \( y_0 \) is the initial amount of the substance.
- \( k \) is the decay rate (in this case, \( 0.04 \) per hour).
- \( t \) is the time in hours.
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
However, given the decay rate, we can also model this directly using a simpler formula, which is often used for problems involving decay rates:
\[ y = y_0 \times (1 - r)^t \]
Where:
- \( r \) is the decay rate expressed as a decimal.
- all other variables are as previously defined.
In this case, since the decay rate is \( 0.04 \) per hour, we can set \( r = 0.04 \):
\[ y = 500 \times (1 - 0.04)^t \]
At \( t = 3 \) hours, we can plug in the values:
\[ y = 500 \times (0.96)^3 \]
Now, let's calculate it:
1. Calculate \( 0.96^3 \):
\[ 0.96^3 \approx 0.8858 \]
2. Multiply by the initial amount (500 grams):
\[ y \approx 500 \times 0.8858 \approx 442.9 \] grams.
Thus, the amount of radioactive material remaining after 3 hours is approximately 442.9 grams.
Now, let's match this with the provided choices.
Correct response:
**y = 500(0.96)^3**
This corresponds to the calculation done above, representing the correct model for the scenario given.
\[ y = y_0 \times e^{-kt} \]
Where:
- \( y \) is the remaining amount of the substance after time \( t \).
- \( y_0 \) is the initial amount of the substance.
- \( k \) is the decay rate (in this case, \( 0.04 \) per hour).
- \( t \) is the time in hours.
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
However, given the decay rate, we can also model this directly using a simpler formula, which is often used for problems involving decay rates:
\[ y = y_0 \times (1 - r)^t \]
Where:
- \( r \) is the decay rate expressed as a decimal.
- all other variables are as previously defined.
In this case, since the decay rate is \( 0.04 \) per hour, we can set \( r = 0.04 \):
\[ y = 500 \times (1 - 0.04)^t \]
At \( t = 3 \) hours, we can plug in the values:
\[ y = 500 \times (0.96)^3 \]
Now, let's calculate it:
1. Calculate \( 0.96^3 \):
\[ 0.96^3 \approx 0.8858 \]
2. Multiply by the initial amount (500 grams):
\[ y \approx 500 \times 0.8858 \approx 442.9 \] grams.
Thus, the amount of radioactive material remaining after 3 hours is approximately 442.9 grams.
Now, let's match this with the provided choices.
Correct response:
**y = 500(0.96)^3**
This corresponds to the calculation done above, representing the correct model for the scenario given.