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NEW QUESTION: 1
Examine the structure of the EMPLOYEES and DEPARTMENTS tables:

You want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?
A. SELECT last_name, department_name, location_id FROM employees , departments ;
B. SELECT employees.last_name, departments.department_name,
departments.location_id FROM employees e, departments D WHERE e.department_id
= d.department_id;
C. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id =d.department_id;
D. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id =manager_id;
Answer: C
Explanation:
Equijoins are also called simple joins or inner joins. Equijoin involve primary key and foreign key.
Incorrect answer:
A. there is no join
B . invalid syntax
C. does not involve the join in the primary and foreign key
Refer: Introduction to Oracle9i: SQL, Oracle University Study Guide, 4-8

NEW QUESTION: 2
ある会社が、スマート冷蔵庫が中央の場所に温度情報を送信できるようにするソリューションを開発しています。既存のService Busがあります。
ソリューションは、メッセージが処理されるまでメッセージを受信して​​保存する必要があります。名前、価格階層、サブスクリプション、リソースグループ、場所を指定して、Azure Service Busインスタンスを作成します。
構成を完了する必要があります。
どのAzure CLIまたはPowerShellコマンドを実行する必要がありますか？

A. オプションC
B. オプションD
C. オプションB
D. オプションA
Answer: D
Explanation:
Explanation
A service bus instance has already been created (Step 2 below). Next is step 3, Create a Service Bus queue.
Note:
Steps:
Step 1: # Create a resource group
resourceGroupName="myResourceGroup"
az group create --name $resourceGroupName --location eastus
Step 2: # Create a Service Bus messaging namespace with a unique name
namespaceName=myNameSpace$RANDOM
az servicebus namespace create --resource-group $resourceGroupName --name $namespaceName --location eastus Step 3: # Create a Service Bus queue az servicebus queue create --resource-group $resourceGroupName --namespace-name $namespaceName
--name BasicQueue
Step 4: # Get the connection string for the namespace
connectionString=$(az servicebus namespace authorization-rule keys list --resource-group
$resourceGroupName --namespace-name $namespaceName --name RootManageSharedAccessKey --query primaryConnectionString --output tsv) Reference:
https://docs.microsoft.com/en-us/azure/service-bus-messaging/service-bus-quickstart-cli

NEW QUESTION: 3
Given classes defined in two different files:
package util;
public class BitUtils {
public static void process(byte[] b) { /* more code here */ }
}
1. package app;
2.
3. public class SomeApp {
4. public static void main(String[] args) {
5. byte[] bytes = new byte[256];
6. // insert code here
7. }
8. }
What is required at line 6 in class SomeApp to use the process method of BitUtils?
A. import util.BitUtils.*;
process(bytes);
B. BitUtils.process(bytes);
C. process(bytes);
D. SomeApp cannot use methods in BitUtils.
E. util.BitUtils.process(bytes);
Answer: E