Let's calculate the product of the fractions [tex]\( \frac{8}{15} \)[/tex], [tex]\( \frac{6}{5} \)[/tex], and [tex]\( \frac{1}{3} \)[/tex].
1. Multiply the numerators:
[tex]\[
\text{Numerator} = 8 \times 6 \times 1 = 48
\][/tex]
2. Multiply the denominators:
[tex]\[
\text{Denominator} = 15 \times 5 \times 3 = 225
\][/tex]
3. Form the resulting fraction:
[tex]\[
\frac{48}{225}
\][/tex]
4. Simplify the resulting fraction:
To simplify the fraction [tex]\(\frac{48}{225}\)[/tex], we'll find the greatest common divisor (GCD) of 48 and 225.
- Prime factorization of 48:
[tex]\[
48 = 2^4 \times 3
\][/tex]
- Prime factorization of 225:
[tex]\[
225 = 3^2 \times 5^2
\][/tex]
The common factors are evaluated:
- The GCD of 48 and 225 is [tex]\(3\)[/tex] (since only 3 is common to both factorizations).
Now, divide both the numerator and the denominator by their GCD [tex]\( \)[/tex]:
[tex]\[
\frac{48 \div 3}{225 \div 3} = \frac{16}{75}
\][/tex]
Therefore, the product of [tex]\(\frac{8}{15}\)[/tex], [tex]\(\frac{6}{5}\)[/tex], and [tex]\(\frac{1}{3}\)[/tex] simplifies to [tex]\(\frac{16}{75}\)[/tex].
Hence, the correct answer is:
D) [tex]\(\frac{16}{75}\)[/tex].
