Answer :
To understand which statement is false, let's carefully analyze each statement using the given model for the amount of substance remaining, [tex]\( A(t) = 50 \left( 0.5 \right)^{\frac{t}{3}} \)[/tex].
### Step 1: Check Statement (1)
Statement (1): In 20 days, there is no substance remaining.
We apply the model to calculate the amount of substance after 20 days:
[tex]\[ A(20) = 50 \left( 0.5 \right)^{\frac{20}{3}} \][/tex]
Evaluating this expression:
[tex]\[ A(20) \approx 50 \left( 0.5 \right)^{6.67} \approx 0.492 \][/tex]
Since [tex]\( 0.492 \)[/tex] grams is significantly more than zero, statement (1) is false.
### Step 2: Check Statement (2)
Statement (2): After two half-lives, there is 25% of the substance remaining.
The half-life is 3 days. After two half-lives (which is 6 days):
[tex]\[ A(6) = 50 \left( 0.5 \right)^{\frac{6}{3}} = 50 \left( 0.5 \right)^{2} = 50 \cdot 0.25 = 12.5 \][/tex]
We need to confirm if 12.5 grams is 25% of the initial amount:
[tex]\[ 0.25 \times 50 = 12.5 \][/tex]
Thus, statement (2) is true.
### Step 3: Check Statement (3)
Statement (3): The amount of the substance remaining can also be modeled by [tex]\( A(t) = 50 \left( 2 \right)^{\frac{-t}{3}} \)[/tex]
We need to confirm if these two forms are equivalent:
[tex]\[ 50 \left( 2 \right)^{\frac{-t}{3}} = 50 \left( (2^{-1})^{\frac{t}{3}} \right) = 50 \left( 0.5 \right)^{\frac{t}{3}} \][/tex]
Since both forms are indeed equivalent, statement (3) is true.
### Step 4: Check Statement (4)
Statement (4): After one week, there is less than 10 grams of the substance remaining.
One week is 7 days. We apply the model to calculate the amount of substance after 7 days:
[tex]\[ A(7) = 50 \left( 0.5 \right)^{\frac{7}{3}} \approx 50 \left( 0.5 \right)^{2.33} \approx 9.921 \][/tex]
Since [tex]\( 9.921 \)[/tex] grams is less than 10 grams, statement (4) is true.
### Conclusion
Based on the detailed analysis above, we can determine that the false statement is:
Statement (1): In 20 days, there is no substance remaining.
Thus, the false statement is clearly statement (1).
### Step 1: Check Statement (1)
Statement (1): In 20 days, there is no substance remaining.
We apply the model to calculate the amount of substance after 20 days:
[tex]\[ A(20) = 50 \left( 0.5 \right)^{\frac{20}{3}} \][/tex]
Evaluating this expression:
[tex]\[ A(20) \approx 50 \left( 0.5 \right)^{6.67} \approx 0.492 \][/tex]
Since [tex]\( 0.492 \)[/tex] grams is significantly more than zero, statement (1) is false.
### Step 2: Check Statement (2)
Statement (2): After two half-lives, there is 25% of the substance remaining.
The half-life is 3 days. After two half-lives (which is 6 days):
[tex]\[ A(6) = 50 \left( 0.5 \right)^{\frac{6}{3}} = 50 \left( 0.5 \right)^{2} = 50 \cdot 0.25 = 12.5 \][/tex]
We need to confirm if 12.5 grams is 25% of the initial amount:
[tex]\[ 0.25 \times 50 = 12.5 \][/tex]
Thus, statement (2) is true.
### Step 3: Check Statement (3)
Statement (3): The amount of the substance remaining can also be modeled by [tex]\( A(t) = 50 \left( 2 \right)^{\frac{-t}{3}} \)[/tex]
We need to confirm if these two forms are equivalent:
[tex]\[ 50 \left( 2 \right)^{\frac{-t}{3}} = 50 \left( (2^{-1})^{\frac{t}{3}} \right) = 50 \left( 0.5 \right)^{\frac{t}{3}} \][/tex]
Since both forms are indeed equivalent, statement (3) is true.
### Step 4: Check Statement (4)
Statement (4): After one week, there is less than 10 grams of the substance remaining.
One week is 7 days. We apply the model to calculate the amount of substance after 7 days:
[tex]\[ A(7) = 50 \left( 0.5 \right)^{\frac{7}{3}} \approx 50 \left( 0.5 \right)^{2.33} \approx 9.921 \][/tex]
Since [tex]\( 9.921 \)[/tex] grams is less than 10 grams, statement (4) is true.
### Conclusion
Based on the detailed analysis above, we can determine that the false statement is:
Statement (1): In 20 days, there is no substance remaining.
Thus, the false statement is clearly statement (1).