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Exam Code: C_SACP_2102

Exam Name: SAP Certified Application Associate - SAP Analytics Cloud: Planning

Certification Provider: SAP

Related Certification: SAP Certified Application Associate - SAP Analytics Cloud: Planning

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NEW QUESTION: 1
与えられた:

コードで次の出力を生成する必要があります。
ジョン
ジョー
ジェーン
出力を生成するために、1行目と2行目にどのコードフラグメントを挿入する必要がありますか?
A. Insert Comparable<Person> on line 1.
Insert
public int compare(Person p1, Person p2) {
return p1.name.compare(p2.name);
}
on line 2.
B. Insert Comparator<Person> on line 1.
Insert
public int compare(Person person) {
return person.name.compare(this.name);
}
on line 2.
C. Insert Comparator<Person> on line 1.
Insert
public int compareTo(Person person) {
return person.name.compareTo(this.name);
}
on line 2.
D. Insert Comparator<Person> on line 1.
Insert
public int compare(Person p1, Person p2) {
return p1.name.compare(p2.name);
}
on line 2.
Answer: C
Explanation:
参照:
https://www.coursehero.com/file/p320ss6/Override-public-int-compareTo-Person-other-Compare- this-objects-name-to-others /

NEW QUESTION: 2
Which two fields in the Classifications application construct Classification hierarchies? (Choose two.)
A. Classification Path
B. Use With
C. Attributes
D. Classification Parent
E. Children
Answer: D,E

NEW QUESTION: 3
When sales and output have passed the break-even point, the contribution per unit, for each unit then sold, becomes:
A. The margin of safety
B. Bigger
C. The profit per unit
D. Smaller
Answer: C

NEW QUESTION: 4
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10.0.0.0.0.0.0.1
B. 許可10.0.0.1.0.0.0.254
C. 許可10.0.0.1.0.0.0.0
D. 許可10.0.0.0.255.255.255.254
Answer: B
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.


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