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NEW QUESTION: 1
ソリューションアーキテクトは、10個の既存のアカウントを持つマルチアカウント構造を設計しています。設計は次の要件を満たしている必要があります。
*すべてのアカウントを1つの組織に統合します。
*マスターアカウントとセカンダリアカウントからAmazon EC2サービスへのフルアクセスを許可します。
*セカンダリアカウントを追加するのに必要な労力を最小限に抑えます。
ソリューションに含めるべきステップの組み合わせはどれですか? (2つ選択してください。)
A. フルEC2アクセスを有効にするサービスコントロールポリシー(SCP)を作成し、ポリシーをOUにアタッチします。
B. マスターアカウントから組織を作成します。各セカンダリアカウントからマスターアカウントに参加要求を送信します。要求を受け入れて、OUを作成します。
C. 完全なEC2アクセスポリシーを作成し、ポリシーを各アカウントのロールにマッピングします。他のすべてのアカウントを信頼して役割を引き受けます。
D. マスターアカウントから組織を作成します。マスターアカウントからセカンダリアカウントに招待状を送信します。招待を受け入れて、OUを作成します。
E. マスターアカウントとセカンダリアカウントの間にVPCピアリング接続を作成します。 VPCピアリング接続の要求を受け入れます。
Answer: A,D
Explanation:
Explanation
There is a concept of Permission Boundary vs Actual IAM Policies That is, we have a concept of "Allow" vs
"Grant". In terms of boundaries, we have the following three boundaries: 1. SCP 2. User/Role boundaries 3.
Session boundaries (ex. AssumeRole ... ) In terms of actual permission granting, we have the following: 1.
Identity Policies 2. Resource Policies
NEW QUESTION: 2
You have enabled the Slow Query Log for a short period.
When you process the Slow Query Log, you receive the following snip of output:
Count: 100 Time=0 .22a (22s) Lock=0.00s (0s) Rows=0.0 (0), root[root] @localhost
CREATE TABLE 't1' (id serial,id0 varchar(N) unique key,intcaoll INT (N)
,intco12 INT(N) ,intco13 INT(N) ,intco14 INT(N) ,intco15 INT(N)
,charcol1 VARVHAR(N) ,charcol2 VARCHAR(N) charcol3 VARCHAR (N)
,charcol4 VARVHAR(N) ,charcol5 VARCHAR(N) charcol6 VARCHAR (N)
,charcol7 VARVHAR(N) ,charcol8 VARCHAR(N) charcol9 VARCHAR (N) .charcol 10 VACHAR
(N) )
Count: 64000 Time-0.02s (1213s) Lock=0.00s (6s) Rows=1.0 (64000), root [root]@ localhost
SELECT intocl1, intco12, intco13, intco14, intco15, intco16,intco17, intco18
,intcol9, intcol10, charcol1, charcol2, charcol3, charcol4, charcol5, charcol6
,charcol7, charcol8, charcol9, charcol10 FROM t1 WHERE id = 's'
Count: 1 Time=0.02s (0s) Lock=0.00s (0s) Rows=1.0 (1) agent [agent] @localhost
SELECT Select_priv, Repl_client_priv, Show_db_priv, Super_priv,
Process_priv FROM mysql.user WHERE CONCAT (user, 's', host) = CURRENT_USER ()
Count: 48000 Time=0.02s (778s) Lock=0.00 (3s) Rows=1.0 (48000), root[root]@localhost
SELECT intocl1,intcol2,intcol3, intcol4, intcol5, charcol1, charcol2, charcol3
,charcol4, charcol5, charcol6, charcol7, charcol8, charcol9, charcol10 FROM t1 WHERE id = 's'
You want to tune the query such that it provides the greatest overall time savings.
Which query will accomplish this?
A. SELECT intcol1, intcol2, intcol3, intcol4, intcol5, intcol6, intcol7, intcol8, intcol9,
Intcol10, intcol11, intcol12, intcol13, intcol14, intcol15, intcol16, intcol17, intcol18, intcol19,
charcol10
FROM t1
WHERE id = 's';
B. CHEATE TABLE 't1' (id serial, id0 varchar (N) unique key, intcol1 INT (N)
,intcol2 INT (N), intcol3 INT(N) ,intcol4 INT(N), intcol5 INT(N), charool1 VARCHAR
(N)
,charcol2 VARCHAR (N), charcol3 VARCHAR(N), charcol4 VARCHAR(N), charcol5 VARCHAR
(N)
,charcol6 VARCHAR (N), charcol7 VARCHAR(N), charcol8 VARCHAR(N), charcol9 VARCHAR
(N)
,charcol10 VARCHAR (N);
C. SELECT intcol1, intcol2, intcol3, intcol4, intcol5, charcol1, charcol2, charcol3, charcol4,
charcol5, charcol6, charcol7, charcol8, charcol9, charcol10
FROM t1
WHERE id = 's';
D. SELECT Select_priv, Repl_client_priv, Show_db_priv, Super_priv, Process_priv
FROM mysql.user
WHERE CONCAT (user,'s', host) = CURRENT_USER();
Answer: B
NEW QUESTION: 3
Ihre Einkaufsabteilung möchte überfällige Bestellungen überwachen. Was sind die Voraussetzungen für die Ausstellung von Liefererinnerungen für eine Bestellung?
Es gibt 3 richtige Antworten auf diese Frage.
A. Legen Sie die Nachrichtenermittlung im Customizing fest und pflegen Sie die Nachrichtendatensätze.
B. Erinnerungsfristen in der Bestellposition pflegen.
C. Pflegen Sie die Einkaufswertschlüssel in der Bestellposition.
D. Geben Sie die Bestellung aus.
E. Pflegen Sie einen Bestätigungssteuerungsschlüssel im Lieferantenstamm.
Answer: A,B,C