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Lady bird [3.3K]
3 years ago
6

A company decides to create static design diagrams for an upcoming project. Which statement correctly describes the attribute of

these types of diagrams?
A.
They describe the behavior of a system.
B.
They include sequence diagrams.
C.
They show collaborations among objects.
D.
They include component diagrams.
Computers and Technology
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

A

Explanation:

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Complete the statement about WYSIWYG editors when you use a WYSIWYG editor you typically specify the continent and blank A.layou
erastovalidia [21]

Answer:

When you use a WYSIWYG editor you typically specify the content and layout while the editor generates the HTML code

Explanation:

Required

Complete the blanks

In WYSIWYG, the user of the application provides contents to the WYSIWYG software and also designs the appearance; The appearance is referred to as the layout.

Throughout the design, the user will not use HTML codes; it is the duty of the WYSIWYG editor to generate HTML code based on the input designs by the user.

6 0
3 years ago
Wendy has to capture a sculpture in the middle of the park at night. She is not able to get the entire sculpture into focus in o
dsp73

Answer:

Focus Stacking

Explanation:

F-stop stacking which is also referred to as image stacking is a powerful technique that improves the quality of an image by stacking images taken at different f-stops in order to improve corner sharpness and overcome blurriness. Once photographs are taken at different f-stops (focus stops), a final complete composite picture is created using only the sharpest portions of the photograph.

6 0
3 years ago
For the given program, how many print statements will execute? public static void printShippingCharge(double weight) { if((weigh
Step2247 [10]

CORRECT QUESTION:

For the given program, how many print statements will execute?

public static void printShippingCharge(double weight) { if((weight > 0.0) && (weight <= 10.0)){ System.out.println(weight * 0.75); }

else if((weight > 10.0) && (weight <= 15.0)) { System.out.println(weight * 0.85); }

else if((weight > 15.0) && (weight <= 20.0)) { System.out.println(weight * 0.95); } }

public static void main(String args[]) {

printShippingCharge(18);

printShippingCharge(6);

printShippingCharge(25); }

Answer:

Two of the print statements will output values

Explanation:

These two calls to the printShippingCharge are the ones that will output values: printShippingCharge(18); and printShippingCharge(6);

The first if statement is true when the value of weight is 6 (weight > 0.0) && (weight <= 10.0)

The Third if statement  is true when weight is 18 (weight > 15.0) && (weight <= 20.0)

NOTE: That I made correction to the variable weight. The question isn't consistent with the variable name. It used weight and itemWeight at different points this will lead to a compiller error

4 0
3 years ago
1.Which of the following class definitions defines a legal abstract class?a. class A { abstract void unfinished() { } }b. class
xxMikexx [17]

Answer:(c) abstract class A { abstract void unfinished(); }

Explanation:

A legal abstract class must have the keyword abstract before the class and an abstract class has abstract functions with the keyword abstract written and a void as the return type.

8 0
4 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
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