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NEW QUESTION: 1
Given:
class MarksOutOfBoundsException extends IndexOutOfBoundsException { }
public class GradingProcess {
void verify(int marks) throws IndexOutOfBoundsException {
if (marks > 100) {
throw new MarksOutOfBoundsException();
}
if (marks > 50) {
System.out.print("Pass");
} else {
System.out.print("Fail");
}
}
public static void main(String[] args) {
int marks = Integer.parseInt(args[2]);
try {
new GradingProcess().verify(marks));
} catch(Exception e) {
System.out.print(e.getClass());
}
}
}

Java grading process 89 50 104

A. Pass
B. Fail
C. Class IndexOutOfBoundsException
D. Class MarketOutOfBoundsException
E. Class Exception
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
MPLSレイヤー3 VPN構成でトラフィックの負荷分散を可能にするコマンドはどれですか。
A. 最大パスibgp 2
B. マルチパス2
C. 最大パス2
D. マルチパスeibgp 2
Answer: C

NEW QUESTION: 4
A developer uses a parsed JSON string to work with user information as in the block below:

Which two option access the email attributes in the object? Choose 2 answers
A. userInformation. email
B. userInformation ( ''email'' )
C. userInformation. Get (''email'')
D. userInformation(email)
Answer: A,B