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Igoryamba
2 years ago
8

PLEASE HELP POWER POINT:

Computers and Technology
2 answers:
Allushta [10]2 years ago
6 0
18: b format
20: b ctrl + and left mouse button
Nataly [62]2 years ago
5 0

18 is playback, not formatting.

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Semmy [17]
B. Spreadsheets is the answer
3 0
3 years ago
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Which SCSI standard allows for the technique known as “hot swapping”? Ultra SCSI Original SCSI Serial SCSI Fast-Wide SCSI
miss Akunina [59]

Answer:

Serial SCSI

Explanation:

Hot swapping can be defined as a process which typically involves fitting or replacing CD-ROM drive, hard-disk drive, power supply or other peripheral devices while a computer system is powered on. Thus, it allows for the installation or removal of a peripheral device from a computer while power is still being supplied to the computer i.e without having to shutdown the computer.

Serial SCSI is a SCSI standard which allows for the technique known as “hot swapping” because it's a point to point connection that is designed to move data to and from computer storage serially.

6 0
3 years ago
Which of the following documents cannot be created using the Microsoft® Word® application?
denis23 [38]
The correct answer is C. Customer spreadsheet
7 0
3 years ago
Read 2 more answers
Write a Java method to delete a record from a B-tree of order n.
max2010maxim [7]

Answer:

Deleting a record on a B-tree consists of three main events:

- Searching the node where the key to be deleted exists

- Deleting the key

- Balancing the tree if required

Explanation:

q = NULL;

   p = tree;

   while (p) {

       i = nodesearch(p, key);

       q = p;

       if (i < used(p) -1 && key == k(p,i)) {

           found = TRUE;

           position = i;

           break;

       }

       p = son(p,i);

   }

   if (!found)

   else if (subtree(p)) {

         if (used(p) > ((n-1)/2)+1)

         delkey (p, position, key);

       else {

           replace (p, position, fsucc(p));

           p0 r1 p1 r2 p2 r3 ……. pn-1 rn-1 pn

           q = &fsucc(p);

           qpos = index of fsucc;

           if (used(rbrother(p)) > ((n-1)/2)+1)

               replace (q, qpos, sonsucc(q));

           else

               while (q && used(q) < (n-1)/2) {

                   concatenate(q, brother(q));

                   q = father(q);

               }

       }

   }

   else

   delkey(p, position, key);

}

6 0
2 years ago
Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the elemen
Verdich [7]

Answer:

Replace <STUDENT CODE> with

for (i = 0; i < SCORES_SIZE; ++i) {

       if(lowerScores.at(i)<=0){

           lowerScores.at(i) = 0;

       }

       else{

           lowerScores.at(i) = lowerScores.at(i) - 1;

       }  

   }

Explanation:

To do this, we simply iterate through the vector.

For each item in the vector, we run a check if it is less than 1 (i.e. 0 or negative).

If yes, the vector item is set to 0

If otherwise, 1 is subtracted from that vector item

This line iterates through the vector

for (i = 0; i < SCORES_SIZE; ++i) {

This checks if vector item is less than 1

       if(lowerScores.at(i)<1){

If yes, the vector item is set to 0

           lowerScores.at(i) = 0;

       }

       else{

If otherwise, 1 is subtracted from the vector item

           lowerScores.at(i) = lowerScores.at(i) - 1;

       }  

   }

Also, include the following at the beginning of the program:

<em>#include <vector></em>

8 0
3 years ago
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