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NEW QUESTION: 1
What factor determines whether a supplier can respond to a negotiation task by directly accessing the contract workspace?
A. Supplier user is registered on the Ariba Network
B. Supplier user is a member of the supplier's legal team
C. Supplier user is a contract manager
D. Supplier user is an e-mail reviewer
Answer: B
NEW QUESTION: 2
You have a Windows Server 2016 failover cluster that contains two servers named Server1 and Server2.
You need to apply patches to Server1.
Which two commands should you run before you apply the patches? To answer, select the appropriate options in the answer area.
Answer:
Explanation:
Explanation
References:
https://support.microsoft.com/en-us/help/174799/how-to-update-windows-server-failover-clusters
https://technet.microsoft.com/en-us/library/ee461003.aspx
NEW QUESTION: 3
You are implementing Microsoft Dynamics CRM. Your company delivers training for customers on various topics. Not all trainers are capable of delivering all training classes.
You need to ensure that each training service is associated with the required resource.
What should you create?
A. Selection rules
B. Capabilities
C. Service activities
D. Resource groups
Answer: A
Explanation:
Ref:http://www.powerobjects.com/blog/2010/07/30/service-scheduling-in-dynamics-crmpart-1/
NEW QUESTION: 4
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.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) ) ;
B. 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));
C. 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))
D. 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)) ;
Answer: C
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