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Exam Code: C1000-176

Vendor Name: IBM

Total Practice Questions : 60

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NEW QUESTION: 1
次のどのプラットフォームはElastic Beanstalkでサポートしていますか。2つを選んでください。
A. IBM Websphere
B. .NET
C. Jetty
D. Oracle JBoss
E. Apache Tomcat
Answer: B,E

NEW QUESTION: 2
Given:
class Student {
String course, name, city;
public Student (String name, String course, String city) {
this.course = course; this.name = name; this.city = city;
}
public String toString() {
return course + ":" + name + ":" + city;
}
public String getCourse() {return course;}
public String getName() {return name;}
public String getCity() {return city;}
and the code fragment:
List<Student> stds = Arrays.asList(
new Student ("Jessy", "Java ME", "Chicago"),
new Student ("Helen", "Java EE", "Houston"),
new Student ("Mark", "Java ME", "Chicago"));
stds.stream()
.collect(Collectors.groupingBy(Student::getCourse))
.forEach(src, res) -> System.out.println(res));
What is the result?
A. A compilation error occurs.
B. [Java EE: Helen:Houston][Java ME: Jessy:Chicago, Java ME: Mark:Chicago]
C. [Java ME: Jessy:Chicago, Java ME: Mark:Chicago][Java EE: Helen:Houston]
D. Java EEJava ME
Answer: D

NEW QUESTION: 3
教育機関向けにMicrosoft SQL Serverデータベースを開発します。データベースには、学生のスコアを追跡するために使用されるExamEvaluationおよびProjectEvaluationという名前のテーブルが含まれています。テーブルは、次のコードを使用して作成されます。

学生は次のルールに基づいて評価されます。
*各学生は、最大1つの試験と1つのプロジェクトを完了します。
*学生は少なくとも1つの試験を完了する必要があります。
* 1つの試験は複数のセクションで構成され、すべてのセクションの平均スコアに基づいて学生が評価されます。
*単一のプロジェクトはセクションAとセクションBで構成され、学生はセクションAの60%とセクションBの40%を追加して評価されます。
*いつでも、学生は試験のみ、または試験とプロジェクトの両方を完了している場合があります。
学生の評価の要約レポートを作成します。
何らかの形式の評価を完了した学生の評価結果を、アンダータリングの年ごとにグループ化して返すクエリを作成する必要があります。サンプルの結果セットを次の表に示します。

ソリューションを開発するには、どの5つのTransact-SQLセグメントを使用する必要がありますか?回答するには、適切なTransact-SQLセグメントをTransact-SQLセグメントのリストから回答エリアに移動し、正しい順序で並べます。

Answer:
Explanation:

Explanation:
Box 1: SELECT..
The outer select statement
Box 2: (SELECT ..AVG(SectionScore)
Need the average score: A single exam consists of multiple sections, where the student is evaluated based on the average score of all sections.
Box 3: LEFT OUTER JOIN
There might be no finished Project
Box 4: ON..
Box 5: (SELECT ..SUM
Need the SUM function to handle addition of NULL values.
References:
https://docs.microsoft.com/en-us/sql/t-sql/functions/sum-transact-sql


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