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ipn [44]
3 years ago
14

HELP PLEASE!!!Salman wrote a SMART goal for class. Part of the goal includes the fact that he will complete Ms. Murray's extra-

credit assignment. What part of his SMART goal does this fact represent? measurable timely specific Success​
Computers and Technology
2 answers:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

Hello!

The answer to the question on the test in Edgenuity is Specific

Hope this helps!

kirill115 [55]3 years ago
5 0

Answer:

the answer is specific

Explanation:

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- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -> {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
Using the Sakila database, create a query that displays the film title, description, as well the actor first and last name for a
Alex787 [66]

Answer:

Select title, description , first_name, last_name from film inner join film_actor on film.film_id = film_actor.film_id inner join actor on film_actor.actor_id = actor.actor_id where title LIKE 'zo%';

Explanation:

  • The INNER JOIN keyword selects records that have matching values in both tables.
  • The WHERE clause is used to filter records.
  • The WHERE clause is used to extract only those records that fulfill a specified condition.
7 0
3 years ago
The set of rules for how computers talk to one another is called
lianna [129]
Protocol. Like TCP/IP or HTTP etc
6 0
3 years ago
Steps of booting a computer
Lostsunrise [7]
Steps of Booting
The Startup. It is the first step that involves switching the power ON. ...
BIOS: Power On Self Test. It is an initial test performed by the BIOS. ...
Loading of OS. In this step, the operating system is loaded into the main memory. ...
System Configuration. ...
Loading System Utilities. ...
User Authentication.
8 0
2 years ago
A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on average once every 100 sec, e
Sunny_sXe [5.5K]

Explanation:

Here, in the given statement, maximum flow capacity of the ALOHA channel = 18.4% = 0.184

Then, the stations N share 56 kbps

And, the channel rate = 0.184*56 = 10.304 kbps.  

Then, the outputs of the wach stations is 1000-bit frames/100 sec

\therefore, the output of the bits/ sec by each stations = \frac{1000}{100} bits/sec = 10 bits/sec

So, The output of the N stations is 10 bits/sec on the channel that having 10.304kb/sec

Finally, N = \frac{(10.304 \times 10^{3})}{10} = 1030 stations

3 0
3 years ago
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