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SSSSS [86.1K]
3 years ago
14

Which of the following words is an anatomy for cautious

Computers and Technology
1 answer:
liraira [26]3 years ago
8 0
Uh, I think you mean antonym.

Anyways, here are some antonyms.

Careless, Certain, Foolish, Inattentive, Incautious
You might be interested in
If you anticipate running a particular query often, you can improve overall performance by saving the query in a special file ca
mestny [16]

If you anticipate running a particular query often, you can improve overall performance by saving the query in a special file called a(n)  stored procedure.

<h3>What is meant by stored procedure?</h3>

An application that uses a relational database management system can access a subroutine known as a stored procedure. These processes are kept in the data dictionary of the database. Access control and data validation are two applications for stored processes.

A stored procedure is a collection of SQL statements that have been given a name and are kept together in a relational database management system (RDBMS) so they may be used and shared by various programs.

If you expect to run a certain query frequently, putting it in a special file known as a(n) stored procedure can enhance overall speed.

To learn more about stored procedure refer to:

brainly.com/question/13692678

#SPJ4

3 0
1 year ago
Pendant Publishing edits multi-volume manuscripts for many authors. For each volume, they want a label that contains the author'
lbvjy [14]

Answer:

Check the explanation

Explanation:

Pseudocode For Reading File:

start

Declarations

InputFIle records.txt

OutputFile result.txt

string authorName

string title

int numOfVol

open InputFile

open OutputFile

input authorName, title, numOfVol from records.txt

while not eof

output authorName

output title

output numOfVol

end while

close records.txt

close results.txt

END

The flowchart for reading the file can be seen below.

3 0
3 years ago
- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -&gt; {D}, {B} -&
Olenka [21]

Answer:

The key of R is {A, B}

Explanation:

A key can be seen as a minimal set of attributes whose closure includes all the attributes in R.

Given that the closure of {A, B}, {A, B}+ = R, one key of R is {A, B} But in this case, it is the only key.

In order for us to to normalize R intuitively into 2NF then 3NF, we have to make use of these approaches;

First thing we do is to identify partial dependencies that violate 2NF. These are attributes that are

functionally dependent on either parts of the key, {A} or {B}, alone.

We can calculate

the closures {A}+ and {B}+ to determine partially dependent attributes:

{A}+ = {A, D, E, I, J}. Hence {A} -> {D, E, I, J} ({A} -> {A} is a trivial dependency)

{B}+ = {B, F, G, H}, hence {A} -> {F, G, H} ({B} -> {B} is a trivial dependency)

To normalize into 2NF, we remove the attributes that are functionally dependent on

part of the key (A or B) from R and place them in separate relations R1 and R2,

along with the part of the key they depend on (A or B), which are copied into each of

these relations but also remains in the original relation, which we call R3 below:

R1 = {A, D, E, I, J}, R2 = {B, F, G, H}, R3 = {A, B, C}

The new keys for R1, R2, R3 are underlined. Next, we look for transitive

dependencies in R1, R2, R3. The relation R1 has the transitive dependency {A} ->

{D} -> {I, J}, so we remove the transitively dependent attributes {I, J} from R1 into a

relation R11 and copy the attribute D they are dependent on into R11. The remaining

attributes are kept in a relation R12. Hence, R1 is decomposed into R11 and R12 as

follows: R11 = {D, I, J}, R12 = {A, D, E} The relation R2 is similarly decomposed into R21 and R22 based on the transitive

dependency {B} -> {F} -> {G, H}:

R2 = {F, G, H}, R2 = {B, F}

The final set of relations in 3NF are {R11, R12, R21, R22, R3}

4 0
3 years ago
In what year did the manager and team depicted in the blockbuster film "Moneyball
mariarad [96]
2006 is the answer and the boston red sox were the first team to win 5 chamships 
3 0
3 years ago
Consider the two computers A and B with the clock cycle times 100 ps and 150 ps respectively for some program. The number of cyc
Kipish [7]

Answer:

Option d) B is 1.33 times faster than A

Given:

Clock time, t_{A} = 100 ps

t_{A} = 150 ps

No. of cycles per instructions,  n_{A} = 2.0

n_{B} = 1.0

Solution:

Let I be the no. of instructions for the program.

CPU clock cycle, f_{A} = 2.0 I

CPU clock cycle, f_{B} = 1.0 I

Now,

CPU time for each can be calculated as:

CPU time, T = CPU clock cycle\times clock time

T_{A} = f_{A}\times t_{A} = 2.0 I\times 100 = 200 I ps

T_{B} = f_{B}\times t_{B} = 1.0 I\times 100 = 150 I ps

Thus B is faster than A

Now,

\frac{Performance of A}{Performance of B} = \frac{T_{A}}{T_{B}}

\frac{Performance of A}{Performance of B} = \frac{200}{150}

Performance of B is 1.33 times that of A

7 0
3 years ago
Read 2 more answers
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