Ordered relations with mixed strict/non-strict signs: Given H > J = K, K ≥ L, L > T, and T < V, determine which conclusion(s) definitely follow from the statements: I) K > T; II) L ≤ H.

Introduction / Context:
This inequality-chain question asks which conclusions are forced by the given relations. We must respect strict (>) vs non-strict (≥, ≤) comparisons and use transitivity carefully.

Given Data / Assumptions:

• H > J = K
• K ≥ L
• L > T
• T < V (does not affect K, L relative to H for our conclusions)

Concept / Approach:
Convert equalities and non-strict bounds into a single order where possible, then test each conclusion independently.

Step-by-Step Solution:
From J = K and H > J ⇒ H > K. From K ≥ L and L > T ⇒ K ≥ L > T ⇒ K > T (I true). From K ≥ L and H > K ⇒ H > K ≥ L ⇒ H > L ⇒ L ≤ H (II true).

Verification / Alternative check:
Assign values respecting the chain, e.g., T=1, L=2, K=2, H=3, V any >1. Then K > T and L ≤ H both hold.

Why Other Options Are Wrong:
a/b/d/e contradict at least one conclusion; both I and II are compelled.

Common Pitfalls:
Forgetting that K ≥ L with L > T implies K > T; also reading L ≤ H as L <= H which is consistent with H > L.

Final Answer:
If both conclusions I and II follow