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NEW QUESTION: 1
Given:
public class Main {
public static void main(String... ag) {
Vehicle v = new SportsCar();
System.out.println(v.goes());
Tank t = (Tank) v;
System.out.println(t);
}
}
class Vehicle {
public String goes() {
return "goes ";
};
}
class SportsCar extends Vehicle {
public String goes() {
return "fast ";
};
}
class Tank extends Vehicle {
public String goes() {
return "slow ";
};
}
What is the result?
A. fast fast
B. fast followed by an exception
C. fast goes
D. goes goes
E. fast slow
F. Compilation fails
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Line:Vehicle v = new Sportscar();
causes compilation failure:
error: cannot find symbol
Vehicle v = new Sportscar();
symbol: class Sportscar
location: class VehicleTest

NEW QUESTION: 2
HOTSPOT
You need to implement the SCL configurations to support the planned changes.
Which value should you set for each SCL option? (To answer, configure the appropriate options in the dialog box in the answer area.)

Answer:
Explanation:

NEW QUESTION: 3
開発者は、クエリアクティビティの知識を広げたいと考えています。彼らは、過去30日間にバウンスした電子メールアドレスを、バウンス理由およびいくつかの追加のサブスクライバー固有のデータとともに特定したいと考えています。ただし、彼らが書いたSQLはレコードを返しません。以下はSQLステートメントです。

このSQLステートメントが目的の結果を返すようにするには、どのような更新を行う必要がありますか？
A. Marketing Cloudは、クエリアクティビティでDateAdd関数を許可していません。特定の日付を定義する必要があります。
B. Marketing Cloud_BounceデータビューにEmailAddressが含まれていません。彼らはSubscriberlDに参加する必要があります。
C. MarketingCloudのデータビューは20日分のデータしか保持しません。 DateAdd関数の範囲を狭める必要があります。
D. MarketingCloudはGETDATE関数の使用を許可していません。特定の日付を定義する必要があります。
Answer: B

NEW QUESTION: 4
The SAS data set TEMP has the following distribution of values for variable A:
A Frequency
1 500,000
2 500,000
6 7,000,000
8 3,000 Which one of the following SAS programs requires the least CPU time to be processed?
A. data new;
set temp;
if a = 8 then
b = 'Small ';
else if a in(1, 2) then
b = 'Medium';
else if a = 6 then
b = 'Large';
run;
B. data new;
set temp;
if a in (1, 2) then
b = 'Medium';
else if a = 8 then
b = 'Small';
else if a = 6 then
b = 'Large';
run;
C. data new;
set temp;
if a = 6 then
b = 'Large ';
else if a in (1, 2) then
b = 'Medium';
else if a = 8 then
b = 'Small';
D. data new;
set temp;
if a = 6 then
b = 'Large ';
if a in (1, 2) then
b = 'Small';
run;
Answer: C