Oracle 1Z0-1064-21 Deutsch - 1Z0-1064-21 Vorbereitungsfragen, 1Z0-1064-21 Antworten - Condocubeapp

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NEW QUESTION: 1
What is the initial probability recorded in the risk register based on?
A. Post-response assessment
B. Pre-response assessment
C. Proximity of the risk
D. Stage of the activity
Answer: B

NEW QUESTION: 2
한 회사에서 모든 애플리케이션에 대한 배포 도구로 AWS CloudFormation을 사용하고 있습니다. 개발자는 통합 개발 (IDE)을 호스팅하는 Amazon EC2 인스턴스에 액세스 할 수 있도록 버전 관리를 통해 Amazon S3 버킷 내의 모든 애플리케이션 바이너리 및 템플릿을 스테이징합니다. 개발자는 Amazon S3에서 EC2 인스턴스로 애플리케이션 바이너리를 다운로드하고 변경 한 다음 유닛을 로컬에서 실행 한 후 S3 버킷에 업로드합니다. 개발자는 기존 배포 메커니즘을 개선하고 AWS CodePipeline을 사용하여 Ci / CD를 구현하려고 합니다.
개발자에게는 다음과 같은 요구 사항이 있습니다.
소스 제어에 AWS CodeCommit 사용
단위 테스트 및 보안 검색을 자동화 합니다.
단위 테스트가 실패하면 개발자에게 경고
애플리케이션 기능을 켜고 끄고 Ci / CD의 일부로 배포를 동적으로 사용자 지정합니다.
애플리케이션을 배포하기 전에 리드 개발자에게 승인을 요청하세요.
이러한 요구 사항을 충족하는 솔루션은 무엇입니까?
A. AWS CodeBuild를 사용하여 단위 테스트 및 보안 스캔을 실행합니다. 파이프 라인의 후속 단계에서 Lambda를 사용하여 테스트가 실패 할 때 개발자에게 Amazon SNS 알림을 보냅니다. 다양한 솔루션 기능에 대한 Amplify 플러그인을 작성하고 사용자 프롬프트를 활용하여 기능을 켜고 끕니다. Amazon SES를 사용하면 리드 개발자가 애플리케이션을 승인 할 수 있습니다.
B. AWS CodeDeploy를 사용하여 단위 테스트 및 보안 스캔을 실행합니다. 파이프 라인에서 Amazon CloudWatch 경보를 사용하여 단위 테스트가 실패 할 때 개발자에게 Amazon SNS 알림을 보냅니다. 다양한 솔루션 기능에 Docker 이미지를 사용하고 AWS CLI를 사용하여 기능을 켜고 끕니다. 파이프 라인에서 수동 승인 단계를 사용하여 리드 개발자가 애플리케이션을 승인 할 수 있도록 합니다.
C. AWS CodeBuild를 사용하여 테스트 및 보안 스캔을 실행합니다. Amazon EventBridge 규칙을 사용하여 단위 테스트가 실패 할 때 개발자에게 Amazon SNS 알림을 보냅니다. 다양한 솔루션 기능에 대한 AWS Cloud Developer Kit (AWS CDK) 구성을 작성하고 매니페스트 파일을 사용하여 AWS 애플리케이션에서 켜고 끕니다. 파이프 라인에서 수동 개선 단계를 사용하여 리드 개발자가 애플리케이션을 승인 할 수 있도록 합니다.
D. Jenkins를 사용하여 단위 테스트 및 보안 스캔을 실행합니다. 파이프 라인에서 Amazon EventBridge 규칙을 사용하여 단위 테스트시 개발자에게 Amazon SES 알림을 보냅니다 (실패. 다양한 솔루션 기능 및 매개 변수에 대해 AWS CloudFormation 중첩 스택을 사용하여 기능을 켜고 끕니다. 파이프 라인에서 AWS Lambda를 사용하여 리드를 허용합니다. 개발자는 응용 프로그램을 승인합니다.
Answer: C

NEW QUESTION: 3
Given the code fragment:
public class Test {
static String[][] arr =new String[3][];
private static void doPrint() {
//insert code here
}
public static void main(String[] args) {
String[] class1 = {"A","B","C"};
String[] class2 = {"L","M","N","O"};
String[] class3 = {"I","J"};
arr[0] = class1;
arr[1] = class2;
arr[2] = class3;
Test.doPrint();
}
}
Which code fragment, when inserted at line //insert code here, enables the code to print
COJ?
A. int i = 0;
for (String[] sub: arr) {
int j = sub.length -1;
for (String str: sub) {
System.out.println(str[j]);
i++;
}
}
B. for (int i = 0;i < arr.length-1;i++) {
int j = arr[i].length-1;
System.out.print(arr[i][j]);
i++;
}
C. private static void doPrint() {
for (int i = 0;i < arr.length;i++) {
int j = arr[i].length-1;
System.out.print(arr[i][j]);
}
}
D. int i = 0;
for (String[] sub: arr[][]) {
int j = sub.length;
System.out.print(arr[i][j]);
i++;
}
Answer: C
Explanation:
Incorrect:
not A: The following line causes a compile error:
System.out.println(str[j]);
Not C: Compile erro line:
for (String[] sub: arr[][])
not D: Output: C

NEW QUESTION: 4
You have been tasked with configuring multilayer SwitchC, which has a partial configuration and has been attached to RouterC as shown in the topology diagram.
You need to configure SwitchC so that Hosts H1 and H2 can successfully ping the server S1. Also SwitchC needs to be able to ping server S1.
Due to administrative restrictions and requirements you should not add/delete vlans or create trunk links.
Company policies forbid the use of static or default routing. All routes must be learned via EIGRP 65010 routing protocol.
You do not have access to RouteC. RouterC is correctly configured. No trunking has been configured on RouterC.
Routed interfaces should use the lowest host on a subnet when possible. The following subnets are available to implement this solution:
- 10.10.10.0/24
- 190.200.250.32/27
- 190.200.250.64/27
Hosts H1 and H2 are configured with the correct IP address and default gateway.
SwitchC uses Cisco as the enable password.
Routing must only be enabled for the specific subnets shown in the diagram.
Note: Due to administrative restrictions and requirements you should not add or delete VLANs, changes VLAN port assignments or create trunks. Company policies forbid the use of static or default routing. All routes must be learned via the EIGRP routing protocol.
1Z0-1064-21 Deutsch
1Z0-1064-21 Deutsch
1Z0-1064-21 Deutsch
1Z0-1064-21 Deutsch
Answer:
Explanation:
See the explanation for step by step solution:
Explanation
There are two ways to configure interVLAN routing in this case:
+ Use RouterC as a "router on a stick" and SwitchC as a pure Layer2 switch. Trunking must be established between RouterC and SwitchC.+ Only use SwitchC for interVLAN routing without using RouterC, SwitchC should be configured as a Layer 3 switch (which supports ip routing function as a router). No trunking requires.
The question clearly states "No trunking has been configured on RouterC" so RouterC does not contribute to interVLAN routing of hosts H1 & H2 -> SwitchC must be configured as a Layer 3 switch with SVIs for interVLAN routing.
We should check the default gateways on H1 & H2. Click on H1 and H2 and type the "ipconfig" command to get their default gateways.
C:\>ipconfig
We will get the default gateways as follows:
Host1:+ Default gateway: 190.200.250.33
Host2:+ Default gateway: 190.200.250.65
Now we have enough information to configure SwitchC (notice the EIGRP AS in this case is 650) Note: VLAN2 and VLAN3 were created and gi0/10, gi0/11 interfaces were configured as access ports so we don't need to configure them in this sim.
SwitchC# configure terminalSwitchC(config)# no switchport -> without using this command, the simulator does not let you assign IP address on Gi0/1 interface.SwitchC(config-if)# ip address 10.10.10.2 255.255.255.0 ->RouterC has used IP 10.10.10.1 so this is the lowest usable IP address.SwitchC(config-if)# no shutdown SwitchC(config)# int vlan 2 SwitchC(config-if)# ip address 190.200.250.33 255.255.255.224SwitchC(config-if)# no shutdown SwitchC(config-if)# int vlan 3SwitchC(config-if)# ip address 190.200.250.65 255.255.255.224 SwitchC(config-if)# no shutdownSwitchC(config-if)#exitSwitchC(config)# ip routing (Notice: MLS will not work without this command)SwitchC(config)# router eigrp 650SwitchC(config-router)# network 10.10.10.0
0.0.0.255SwitchC(config-router)# network 190.200.250.32 0.0.0.31SwitchC(config-router)# network
190.200.250.64 0.0.0.31
NOTE: THE ROUTER IS CORRECTLY CONFIGURED, so you will not miss within it in the exam , also don't modify/delete any port just do the above configuration. Also some reports said the "no auto-summary" command can't be used in the simulator, in fact it is not necessary because the network 190.200.0.0/16 is not used anywhere else in this topology.
In order to complete the lab, you should expect the ping to SERVER to succeed from the MLS , and from the PCs as well.
Also make sure you use the correct EIGRP AS number (in the configuration above it is 650 but it will change when you take the exam) but we are not allowed to access RouterC so the only way to find out the EIGRP AS is to look at the exhibit above. If you use wrong AS number, no neighbor relationship is formed between RouterC and SwitchC.
In fact, we are pretty sure instead of using two commands "network 190.200.250.32 0.0.0.31 and "network
190.200.250.64 0.0.0.31 we can use one simple command "network 190.200.0.0 because it is the nature of distance vector routing protocol like EIGRP: only major networks need to be advertised; even without "no auto-summary" command the network still works correctly. But in the exam the sim is just a flash based simulator so we should use two above commands, just for sure. But after finishing the configuration, we can use "show run" command to verify, only the summarized network 190.200.0.0 is shown.

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