A. An inorder traversal lists keys in sorted order.B. Search time is always O(log n), regardless of shape.C. Its height can be n−1.D. Every node must have exactly two children.
A. A primary key uniquely identifies a row.B. A foreign key can enforce referential integrity.C. Every relation must contain duplicate tuples.D. Normalization can reduce update anomalies.
Correct Answer: A|B|D
Explanation:
Relations are sets conceptually; normalization reduces anomalies.
A. Mutual exclusion is one Coffman condition.B. Circular wait is one Coffman condition.C. Preemption is always impossible in every resource system.D. Breaking hold-and-wait can prevent deadlock.
Correct Answer: A|B|D
Explanation:
Deadlock prevention can invalidate a necessary condition.
A. An inorder traversal lists keys in sorted order.B. Search time is always O(log n), regardless of shape.C. Its height can be n−1.D. Every node must have exactly two children.
A. A primary key uniquely identifies a row.B. A foreign key can enforce referential integrity.C. Every relation must contain duplicate tuples.D. Normalization can reduce update anomalies.
Correct Answer: A|B|D
Explanation:
Relations are sets conceptually; normalization reduces anomalies.
A. Mutual exclusion is one Coffman condition.B. Circular wait is one Coffman condition.C. Preemption is always impossible in every resource system.D. Breaking hold-and-wait can prevent deadlock.
Correct Answer: A|B|D
Explanation:
Deadlock prevention can invalidate a necessary condition.
A. An inorder traversal lists keys in sorted order.B. Search time is always O(log n), regardless of shape.C. Its height can be n−1.D. Every node must have exactly two children.
A. A primary key uniquely identifies a row.B. A foreign key can enforce referential integrity.C. Every relation must contain duplicate tuples.D. Normalization can reduce update anomalies.
Correct Answer: A|B|D
Explanation:
Relations are sets conceptually; normalization reduces anomalies.
A. Mutual exclusion is one Coffman condition.B. Circular wait is one Coffman condition.C. Preemption is always impossible in every resource system.D. Breaking hold-and-wait can prevent deadlock.
Correct Answer: A|B|D
Explanation:
Deadlock prevention can invalidate a necessary condition.
A. An inorder traversal lists keys in sorted order.B. Search time is always O(log n), regardless of shape.C. Its height can be n−1.D. Every node must have exactly two children.
A. A primary key uniquely identifies a row.B. A foreign key can enforce referential integrity.C. Every relation must contain duplicate tuples.D. Normalization can reduce update anomalies.
Correct Answer: A|B|D
Explanation:
Relations are sets conceptually; normalization reduces anomalies.
A. Mutual exclusion is one Coffman condition.B. Circular wait is one Coffman condition.C. Preemption is always impossible in every resource system.D. Breaking hold-and-wait can prevent deadlock.
Correct Answer: A|B|D
Explanation:
Deadlock prevention can invalidate a necessary condition.