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NEW QUESTION: 1
2 차원 배열을 사용하는 응용 프로그램을 개발 중입니다.
다음 코드를 사용하여 배열을 선언합니다.
int [,] numbers = 새로운 int [,]
{
{11, 7, 50, 45, 27},
{18, 35, 47, 24, 12},
{89, 67, 84, 34, 24},
{67, 32, 79, 65, 10}
};
다음으로 표현식 번호 [2, 3]를 사용하여 배열 요소를 참조합니다.
이 식의 반환 값은 무엇입니까?
A. 0
B. 1
C. 2
D. 3
Answer: D

NEW QUESTION: 2
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.1.0.0.0.254
B. 許可10.0.0.0.255.255.255.254
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.0.0.0.1
Answer: A
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.

NEW QUESTION: 3

A. Option E
B. Option A
C. Option B
D. Option D
E. Option C
Answer: B,D,E
Explanation:
Explanation
http://docs.oracle.com/cd/B19306_01/server.102/b14231/schema.htm#CBBBIADA

NEW QUESTION: 4
You enable Automatic Static NAT on an internal host node object with a private IP address of 10.10.10.5, which is NATed into 216.216.216.5. (You use the default settings in Global Properties / NAT.) When you run fw monitor on the R80 Security Gateway and then start a new HTTP connection from host 10.10.10.5 to browse the Internet, at what point in the monitor output will you observe the HTTP SYN-ACK packet translated from 216.216.216.5 back into
10.10.10.5?
A. o=outbound kernel, before the virtual machine
B. O=outbound kernel, after the virtual machine
C. I=inbound kernel, after the virtual machine
D. i=inbound kernel, before the virtual machine
Answer: C