To find out how many feet the model train travels in \(4 \frac{1}{5}\) seconds, we first need to convert \(4 \frac{1}{5}\) to an improper fraction.
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Convert \(4 \frac{1}{5}\) to an improper fraction:
\[ 4 \frac{1}{5} = \frac{4 \times 5 + 1}{5} = \frac{20 + 1}{5} = \frac{21}{5} \]
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Calculate the distance traveled:
The model train travels \(\frac{2}{3}\) foot in one second. Therefore, in \(4 \frac{1}{5}\) seconds, the distance it travels is:
\[ Distance = \text{Speed} \times \text{Time} = \frac{2}{3} \times \frac{21}{5} \]
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Multiply the fractions:
\[ Distance = \frac{2 \times 21}{3 \times 5} = \frac{42}{15} \]
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Simplify \(\frac{42}{15}\):
To simplify \(\frac{42}{15}\):
- Find the greatest common divisor (GCD) of 42 and 15, which is 3.
\[ \frac{42 \div 3}{15 \div 3} = \frac{14}{5} \]
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Convert \(\frac{14}{5}\) to a mixed number (if needed):
Dividing 14 by 5, we get:
\[ 14 \div 5 = 2 \text{ remainder } 4 \] So, \(\frac{14}{5} = 2 \frac{4}{5}\).
Thus, the model train will travel \( \frac{14}{5} \) feet or \( 2 \frac{4}{5} \) feet in \(4 \frac{1}{5}\) seconds.
So the answer is:
A. \( \frac{14}{5} \) feet (or \( 2 \frac{4}{5} \) feet)