Step 1: Use the terminating-decimal criterion.
A rational number \(\dfrac{p}{q}\) in lowest terms has a terminating decimal expansion iff the prime factorization of \(q\) contains only \(2\)'s and \(5\)'s, i.e., \(q=2^m5^n\). Step 2: Check each denominator (in lowest terms).
\(\dfrac{2}{15}\): \(15=3\cdot 5\) (contains \(3\)) \(\Rightarrow\) non-terminating.
\(\dfrac{11}{160}\): \(160=2^5\cdot 5\) (only \(2\) and \(5\)) \(\Rightarrow\) terminating.
\(\dfrac{17}{60}\): \(60=2^2\cdot 3\cdot 5\) (contains \(3\)) \(\Rightarrow\) non-terminating.
\(\dfrac{6}{35}\): \(35=5\cdot 7\) (contains \(7\)) \(\Rightarrow\) non-terminating. Step 3: Conclude.
Only \(\dfrac{11}{160}\) meets the criterion, so it has a terminating decimal expansion.