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NEW QUESTION: 1
Which of the following is a security limitation of File Transfer Protocol (FTP)?
A. Anonymous access is allowed.
B. FTP uses Transmission Control Protocol (TCP) ports 20 and 21.
C. Passive FTP is not compatible with web browsers.
D. Authentication is not encrypted.
Answer: D

NEW QUESTION: 2
A customer asked for your assistance sizing a MirrorView/A solution. Following EMC best practices, which factors would you consider?
A. RLP LUN size, RLP LUN RAID type, RLP LUN data access pattern, RLP LUN IOPs
B. Source LUN size, Source LUN data access pattern, Source LUN R/W ratio
C. RLP LUN size, RLP LUN disk quantity, RLP LUN data access pattern, RLP LUN R/W ratio
D. Source LUN size, source LUN RAID type, source LUN expected data access pattern, source LUN expected R/W ratio
Answer: B

NEW QUESTION: 3
조직은 다음 요구 사항을 충족해야 하는 애플리케이션을 위해 Amazon DynamoDB 테이블을 설계하고 있습니다.
항목 크기는 40KB입니다
각각 2000/500의 읽기 / 쓰기 비율 유지
읽기 중심이 높고 밀리 초 단위의 지연 시간이 짧은 애플리케이션 Amazon EC2 인스턴스에서 애플리케이션이 실행 됨. DynamoDB 테이블에 대한 액세스는 VPC 내에서 최소한의 애플리케이션 코드 변경을 통해 보안을 유지해야 Write-through 캐시를 사용하여 성능을 향상시켜야 합니다. 이러한 요구 사항을 충족합니까?
A. 10000 RCU / 20000 WCU로 DynamoDB 테이블 크기를 조정하고, 읽기 성능을 위해 Amazon ElastiCache를 구현하고, EC2 인스턴스가 DynamoDB에 액세스하도록 VPC에 NAT 게이트웨이를 설정하고, EC2 인스턴스에서 IAM 역할을 구현하여 DynamoDB 액세스를 보호하십시오 .
B. DynamoDB 테이블의 크기를 20000 RCU / 20000 WCU로 조정하고, 읽기 성능을 위해 DAX (DynamoDB Accelerator)를 구현하고, DynamoDB에 대한 VPC 엔드 포인트를 활용하고, EC2 인스턴스에서 IAM 사용자를 구현하여 DynamoDB 액세스를 보호하십시오.
C. DynamoDB 테이블의 크기를 10000 RCU / 20000 WCU로 조정하고, 읽기 성능을 위해 DAX (DynamoDB Accelerator)를 구현하고, DynamoDB에 VPC 엔드 포인트를 사용하고, EC2 인스턴스에서 IAM 역할을 구현하여 DynamoDB 액세스를 보호하십시오.
D. DynamoDB 테이블의 크기를 20000 RCU / 20000 WCU로 조정하고, 읽기 성능을 위해 Amazon ElastiCache를 구현하고, DynamoDB에 대한 VPC 엔드 포인트를 활용하고, EC2 인스턴스에서 IAM 사용자를 구현하여 DynamoDB 액세스를 보호하십시오.
Answer: C

NEW QUESTION: 4
Which sentence is correct about the code below?
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
class A {
int a;
public:
A(int a) : a(a) {}
int getA() const { return a; }
void setA(int a) { this?>a = a; }
/* Insert Code Here */
};
struct add10 { void operator()(A & a) { a.setA(a.getA() + 10); } };
int main() {
int t[] = { 10, 5, 9, 6, 2, 4, 7, 8, 3, 1 };
vector<A> v1(t, t + 10);
for_each(v1.begin(), v1.end(), add10());
vector<A>::iterator it = find(v1.begin(), v1.end(), A(7));
cout << it?>getA() << endl;
return 0;
}
A. it will compile and print 7
B. adding code:
bool operator !=(const A & b) const {
if (this?>a != b.a) { return true; } return false; }
at Place 1 will allow the program to compile
C. it will not compile
D. it will compile but the program result is unpredictable
Answer: C