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NEW QUESTION: 1
あなたはモバイルアプリケーションを設計する会社で働いています。彼らは、プレーヤーの記録が異なるゲームに割り当てられるサーバーを維持しています。追跡システムは新しく、開発中です。
アプリケーションは、EntityFrameworkを使用してAzureデータベースに接続します。データベースには、PlayerテーブルとGameテーブルがあります。
プレーヤーを追加するとき、コードは新しいプレーヤーレコードを挿入し、既存のゲームレコードと新しいプレーヤーレコードの間に関係を追加する必要があります。
アプリケーションは、正しいgameIdとplayerIdを指定してCreatePlayerWithGameを呼び出し、プロセスを開始します。 （行番号は参照用にのみ含まれています。）
次の各ステートメントについて、ステートメントがtrueの場合は、[はい]を選択します。それ以外の場合は、[いいえ]を選択します。
注：正しい選択はそれぞれ1ポイントの価値があります。
Answer:
Explanation:
Explanation
Many-to-many relationships without an entity class to represent the join table are not yet supported.
However, you can represent a many-to-many relationship by including an entity class for the join table and mapping two separate one-to-many relationships.
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<PostTag>()
HasKey(t => new { t.PostId, t.TagId });
modelBuilder.Entity<PostTag>()
HasOne(pt => pt.Post)
WithMany(p => p.PostTags)
HasForeignKey(pt => pt.PostId);
modelBuilder.Entity<PostTag>()
HasOne(pt => pt.Tag)
WithMany(t => t.PostTags)
HasForeignKey(pt => pt.TagId);
}
}
NEW QUESTION: 2
Which table should be used to determine a confidence interval on the mean when standard deviation is NOT known and the sample size is 10?
A. z
B. F
C. t
D. Chi-Square
Answer: C
NEW QUESTION: 3
Which two pieces of information about a device can be seen from the show version command?
A. Installed DRAM
B. CPU load
C. Locally configured IP addresses
D. local CLI
E. Device model
Answer: A,E
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Feedbacks
Aalk - 2014-05-05 16:45:18
Plato - 2014-05-05 16:45:51
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Eleanore - 2014-09-28 16:36:48