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Ray Of Light [21]
2 years ago
9

helppppp asap!!!!!!!!!! Sara has just started using the Internet. She would like to be a little more efficient. In 3–4 sentences

, give Sara some advice about how to be more efficient while using the Web.
Computers and Technology
2 answers:
Artist 52 [7]2 years ago
4 0
Use the web for educational purposes.
Use the web for instructional reasons.
Use the web only when something you find interesting does not make sense to you.
Don’t use the web on a day to day basis.
attashe74 [19]2 years ago
3 0

Answer:

Do not log into any social media.

Do not open youtub.

Do not open online games tabs.

Explanation:

In order to be more efficient when using internet you culd just start by not loggin into any social media while you are working, this can help you have a better performance and less opportunities to get distracted, then you should open youtub, even if it helps you to work with some music, youtub often is a source of distraction, better get a music player or a music streaming program, and do not get into online game pages, this could seriously damage yoru prpductivity.

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Pavel [41]

Answer:

c

Explanation:

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1 year ago
Which option describes wearable technology? A. the incorporation of technology into objects we use regularly B. rugged technolog
vichka [17]

The correct answer is

A.  the incorporation of technology into objects we use regularly

#Platogang

8 0
1 year ago
Following are groups of three​ phrases, which group states three ways to improve your​ concentration?
andreev551 [17]
I believe its <span>c-set up a study​ space, preview your​ materials, create a routine</span>
3 0
1 year ago
Write a function called sum_scores. It should take four arguments, where each argument is the team's score for that quarter. It
saw5 [17]

Answer:

Here is the Python program which has a function sum_scores:

def sum_scores(score1, score2, score3, score4):

   sum = score1 + score2 + score3 + score4

   print(sum)    

sum_scores(14,7,3,0)

Explanation:

  • Method sum_scores takes four arguments, score1, score2, score3, score4.
  • The sum variable adds these four scores and stores the value of their addition.
  • Lastly print statement is used to print the value stored in sum variable which is the value obtained by adding the four scores.
  • Last statement calls the sum_scores method and passes four values to it which are 14,7,3,0
  • The output of the above program is:
  • 24
  • If you want to use return statement instead of print statement you can replace print(sum) with return sum. But in order to display the sum of the scores you can replace sum_scores(14,7,3,0) with print(sum_scores(14,7,3,0))
  • The program along with the output is attached as a screenshot.

5 0
2 years ago
Write a function that implements another stack function, peek. Peek returns the value of the first element on the stack without
devlian [24]

Answer:

See explaination

Explanation:

StackExample.java

public class StackExample<T> {

private final static int DEFAULT_CAPACITY = 100;

private int top;

private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);

/**

* Returns a reference to the element at the top of this stack.

* The element is not removed from the stack.

* atreturn element on top of stack

* atthrows EmptyCollectionException if stack is empty

*/

public T peek() throws EmptyCollectionException

{

if (isEmpty())

throw new EmptyCollectionException("stack");

return stack[top-1];

}

/**

* Returns true if this stack is empty and false otherwise.

* atreturn true if this stack is empty

*/

public boolean isEmpty()

{

return top < 0;

}

}

//please replace "at" with the at symbol

Note:

peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.

3 0
2 years ago
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