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NEW QUESTION: 1
What is recommended to enhance backup performance without affecting application performance on a host?
A. use a push agent
B. single-mode fibre
C. hardware data compression
D. software data compression
Answer: C

NEW QUESTION: 2
Which of the following biometric parameters are better suited for authentication use over a long period of time?
A. Voice pattern
B. Retina pattern
C. Iris pattern
D. Signature dynamics
Answer: C
Explanation:
The iris pattern is considered lifelong. Unique features of the iris are: freckles, rings, rifts, pits, striations, fibers, filaments, furrows, vasculature and coronas. Voice, signature and retina patterns are more likely to change over time, thus are not as suitable for authentication over a long period of time without needing re-enrollment. Source: FERREL, Robert G, Questions and Answers for the CISSP Exam, domain 1 (derived from the Information Security Management Handbook, 4th Ed., by Tipton & Krause).

NEW QUESTION: 3
The developer wants to write a criteria query that will return the number of orders made by customer of each county.
Assume that customer is an entity with a unidirectional one-to-many relationship to the Order entity and that Address is an embeddable class, with an attribute country of type String.
Which one of the queries below correctly achieves this?
A. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class); cq.select (cb.count(c.join
(customer_. Orders)) , c.get(customers(0), c.get(customer_.address) . get (Address_'country)); (c.get(Customer_.address). get(address_.country));
B. CriteriaBuilder cb = ...
CriteriaQuery cq = cb.createQueryO;
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
Join<Address, String> country = c.join(Customer,.address) .join(Address cq.multiselect(cq.count(o), country );
cq.groupBy(c.get(Customer.address) - get (Address_ . country) ) ;
C. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Custower> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(cb.count(o));
cq.groupBy(c.qet(Customer__.address) - get(Address_.country)) ;
D. CriteriaBuilder cb> = ...
CriteriaQuery cq = cb.createQuery();
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.multiselect(cb.count(0), c,get(customer_.address.get(address_.country) cq.groupBy (c.get(customer_.address) .get(address_.country))
Answer: D
Explanation:
Explanation/Reference:
Incorrect: Not B, Not C: Use multiselect, not select. Not D: Use one Join, not two.
Use multiselect, count and Group By. multiselect() method is used because we are going to get compound result and not one entity type.
Example:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Object[]> query = cb.createQuery(Object[].class);
Root<Department> d = query.from(Department.class);
Join<Department,Teacher> teachers = d.join("teachers");
query.multiselect(d.get("name"),cb.count(teachers)).groupBy(d.get("name")); Reference: Criteria group by clause