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NEW QUESTION: 1
Dr. Smith wants to see the latest test results first. Dr. Brown wants to see the nurse's notes first. The different user views can be available by use of by a
A. data dictionary.
B. GUI
C. schema.
D. subschema.
Answer: D
NEW QUESTION: 2
A desktop technician is responsible for the OS migration of 300 end users. As part of the migration, the technician must save the end users* data to a temporary shared drive. The data must include any hidden files to ensure 100% of the data is retained. Which of the following commands will achieve this result by default?
A. cp
B. robocopy
C. xcopy
D. copy
Answer: B
NEW QUESTION: 3
Application developer has a multitude of ways to specify to use MTOM when sending a Web services request. Which of the following will enable MTOM on the client side?
A. Service svc =Service.create(serviceName); MtomSample proxy= svc.getPort(portName, MtomSample.class); BindingProvider bp = (BindingProvider) proxy; ((SOAPBinding) bp.getBinding().setMTOMenabled(true);
B. Service svc = Service.create(service Name); MTOMFeature mtom = new MTOMFeature(); MtomSample proxy = svc.getPort(port Name, MtomSample.class, mtom);
C. Service svc = Service.create(Service Name); svc.addPort(portName, SOAPBinding.SOAP11HTTP_MTOM_BINDING, endpointUrl);
D. Service svc = Service.create(service Name); svc.createDispatch(port Name, Source. class, PAYLOAD); dispatch.setMTOMEnabled(true);
E. Service svc = Service.create(serviceName); svc.setMTOMEnabled(true);
Answer: A,B,C
NEW QUESTION: 4
A. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN (SELECT CustomerID, ShippingCountry, RANK() OVER (PARTITION BY CustomerID ORDER BY COUNT(OrderAmount) DESC) AS Rnk FROM Orders GROUP BY CustomerID, ShippingCountry) AS o ON c.CustomerID = o.CustomerID WHERE o.Rnk = 1
B. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN (SELECT CustomerID, ShippingCountry, RANK() OVER (PARTITION BY CustomerID ORDER BY OrderAmount DESC) AS Rnk FROM Orders GROUP BY CustomerID, ShippingCountry) AS o ON c.CustomerID = o.CustomerID WHERE o.Rnk = 1
C. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM (SELECT c.CustomerID, c.CustomerName, o.ShippingCountry,
RANK() OVER (PARTITION BY CustomerID ORDER BY COUNT(o.OrderAmount) ASC) AS Rnk FROM Customer c INNER JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.CustomerID, c.CustomerName, o.ShippingCountry) cs WHERE Rnk = 1
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN (SELECT CustomerID, ShippingCountry, COUNT(OrderAmount) DESC) AS OrderAmount FROM Orders GROUP BY CustomerID, ShippingCountry) AS o ON c.CustomerID = o.CustomerID ORDER BY OrderAmount DESC
Answer: A