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Snowcat [4.5K]
3 years ago
9

discuss the benefits of normalization, and do you think we should normalize our designs to higher levels than 3NF? Why or why no

t?
Computers and Technology
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

Normalization allows us to design database tables efficently.it reduces data redundancy and avoid data anomalies

Explanation:

Normalization on database can go till 3rd NormalForm.BCNF is always not possible to apply on all tables .so BCNF and Higher Normal forms which is dependent on BCNF is always not possible to apply on all database designs. In general we will go till 3rd normal form.1st normal form illustrates that there should not be any repeating data in a record it should be atomic.2nd normal form illustrates that it should be in 1st normal form and  if any composite key exist then all remaining  values should dependent on that composite key. 3rd normal form illustrates that if should be in 2nd normal form and all keys should dependent directly on primary key.

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The technology Herman Hollerith created in the 1880s began the modern-day data processing function in computers.
ololo11 [35]

Answer:

punched-card machine

Explanation:

We know this as the punched-card machine, and it is also known as the Hollerith cards or the IBM cards. At that time, and through this technology, we used to punch a hole with hand or with the machine, and each hole used to represent the data and the instruction. And these were widely used for inputting the data in the early computers.

6 0
2 years ago
A security engineer is configuring a system that requires the X.509 certificate information to be pasted into a form field in Ba
zavuch27 [327]

Answer:

PFXB

Explanation:

3 0
3 years ago
Which is the correct order for arranging these numbers in terminal-digit filing order? 30 15 39 87 39 64 12 46 33 16 75 24 30 15
Pavel [41]

Answer:

16 75 24

12 46 33

30 15 39

87 39 64

Explanation:

8 0
3 years ago
COO
Sveta_85 [38]

Answer:

t ensures that you understand what they are saying and you are both on the same track

Explanation:

6 0
3 years ago
Consider sending a 1,600-byte datagram into a link that has an mtu of 500 bytes. suppose the original datagram is stamped with t
dsp73

Explanation:

Step one

The maximum size of data field in each fragment = 480

(because there are 20 bytes IP header) Thus the number of required

fragments  =\frac{1600-20}{480} \\\\= \frac{1580}{480} \\\\=3.29\\\\

thus the number of required fragment is 4

Step two

Each fragment will have identification number 291. each fragment except the last one will be of size 500 bytes (including IP header). the offset of the fragments will be 0, 60, 120, 180. each of the first 3 fragments will have

flag = 1; the last fragment will have flag =0

6 0
3 years ago
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