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Aleks04 [339]
3 years ago
12

explain why the database system would not allow a tuple to be inserted into the database through this view

Computers and Technology
1 answer:
Andrej [43]3 years ago
6 0
For the view of Exercise 4.18, explain why the database system would not allow a tuple to be inserted into the database through this view.
For reference
For the database of Figure 4.12, write a query to find the ID of each employee with no manager. Note that an employee may simply have no manager listed or may have a null manager. Write your query using an outer join and then write it again using no outer join at all.
OUTER JOIN
select e.ID from employee e left outer join manages m on e.ID = m.ID
where m.manager_id is null;
NO OUTER
select e.ID from employee e where e.ID not in (select m.ID from manages m) or e.ID in (select m.ID from manages m where m.manager_id is null);
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Interactive sites where users write personal topics and comments to a threadded discussion are called?
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The correct answer would be Forums
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When issued a GFE device, you are required to sign an AUP
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GFE stands for Government Furnished Equipment. While AUP is Acceptable Use Policy (AUP). The AUP  should be read and signed by employees annually.  <span><span> </span></span>The statement that "When issued a GFE device, you are required to sign an AUP" is true. 
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3 years ago
Read 2 more answers
Write an algorithm to find perimeter of circle<br>​
Anna11 [10]

Answer:

Here’s one!

Given [math]R[/math], the radius of the circle.

Let [math]N,D\leftarrow 0[/math]

Repeat until [math]D[/math] is large enough (about 1,000,000)

[math]x,y\leftarrow U[0,1][/math]

If [math]x^2 + y^2\le 1[/math] then [math]N\leftarrow N+1[/math]

[math]D\leftarrow D+1[/math]

[math]P\leftarrow\frac{8NR}{D}[/math]

Return [math]P[/math]

[math]U[0,1][/math] is a uniform random number in the range [math][0,1][/math].

Explanation:

6 0
3 years ago
5.
larisa86 [58]

Answer:

Explanation:

5 is correct answer

3 0
3 years ago
) Consider a router that interconnects four subnets: Subnet 1, Subnet 2, Subnet 3 and Subnet 4. Suppose all of the interfaces in
erik [133]

Answer:

Check the explanation

Explanation:

223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.

The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000

Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)

So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26

Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)

and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25

Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)

So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28

4 0
3 years ago
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