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Lisa [10]
3 years ago
9

Difference between netbook and PDA

Computers and Technology
1 answer:
Mazyrski [523]3 years ago
6 0
Search Results
Featured snippet from the web
A PDA and a laptop computer share many of the same features, yet they are quite different. A PDA is outstanding for its light weight and portability. The laptop is portable, but is heavier and somewhat more cumbersome. ... A PDA is lightweight - more so than the laptop.
You might be interested in
The ACT science test takes___minutes.<br> A. 45<br> B. 30<br> C. 35<br> D. 60
Masja [62]

Answer:

a

Explanation:

8 0
3 years ago
Read 2 more answers
Write a program that prints the day number of the year, given the date in the form month-day-year. For example, if the input is
vlabodo [156]

Answer:

C++:

C++ Code:

#include <iostream>

#include <string>

using namespace std;

struct date

{

  int d,m,y;

};

int isLeap(int y)

{

  if(y%100==0)

  {

      if(y%400==0)

      return 1;

      return 0;

  }

  if(y%4==0)

  return 1;

  return 0;

}

int day_no(date D)

{

  int m = D.m;

  int y = D.y;

  int d = D.d;

  int i;

  int mn[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

  for(i=0;i<m;i++)

  {

      d += mn[i];

  }

  if(isLeap(y))

  {

      if(m>2)

      d++;

  }

  return d;

}

date get_info(string s)

{

  date D;

  int i,p1,p2,l = s.length();

  for(i=0;i<l;i++)

  {

      if(s[i] == '-')

      {

      p1 = i;

      break ;

      }

  }

  for(i=p1+1;i<l;i++)

  {

      if(s[i] == '-')

      {

      p2 = i;

      break ;

      }

  }

 

  D.m = 0;

  for(i=0;i<p1;i++)

  D.m = (D.m)*10 + (s[i]-'0');

 

  D.d = 0;

  for(i=p1+1;i<p2;i++)

  D.d = (D.d)*10 + (s[i]-'0');

 

  D.y = 0;

  for(i=p2+1;i<l;i++)

  D.y = (D.y)*10 + (s[i]-'0');

 

  return D;

 

}

int main()

{

  string s1 = "4-5-2008";

  string s2 = "12-30-1995";

  string s3 = "6-21-2000";

  string s4 = "1-31-1500";

  string s5 = "7-19-1983";

  string s6 = "2-29-1976";

 

  cout<<"Date\t\tDay no\n\n";

  cout<<s1<<"\t"<<day_no(get_info(s1))<<endl;

  cout<<s2<<"\t"<<day_no(get_info(s2))<<endl;

  cout<<s3<<"\t"<<day_no(get_info(s3))<<endl;

  cout<<s4<<"\t"<<day_no(get_info(s4))<<endl;

  cout<<s5<<"\t"<<day_no(get_info(s5))<<endl;

  cout<<s6<<"\t"<<day_no(get_info(s6))<<endl;

 

 

  return 0;

}

Explanation:

4 0
3 years ago
____ refers to the order in which values are used with operators.
mylen [45]

Answer:

b. Associativity

Explanation:

Associativity specifies the order in which operators are processed vis a vis operands/values in an expression. For example: Consider the expression: a + b + c. Here a + b is evaluated first before adding the result with c as the '+' operator is left associative. The default associativity can also be overridden by the use of parentheses. For example a + (b + c) . In this case b+c will be evaluated first.

5 0
3 years ago
1)
spayn [35]

Answer:

1)

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

  if(userValues[i] == matchValue) {

     numMatches++;  }    }

2)  

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

      scanf("%d", &userGuesses[i]);   }

  for (i = 0; i < NUM_GUESSES; ++i) {

         printf("%d ", userGuesses[i]);    }

3)

sumExtra = 0;

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

     if (testGrades[i] > 100){  

        sumExtra = testGrades[i] - 100 + sumExtra;    }       }

4)

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

    if (i<(NUM_VALS-1))  

   printf( "%d,", hourlyTemp[i]);

    else  

    printf("%d",hourlyTemp[i]); }      

Explanation:

1) This loop works as follows:

1st iteration:

i = 0

As i= 0 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 0 So the statement becomes:

userValues[0] == 2

2 == 2

As the value at 0th index (1st element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 1

Now value of i is incremented to 1 so i=1

2nd iteration:

i = 1

As i= 1 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 1 So the statement becomes:

userValues[1] == 2

2 == 2

As the value at 1st index (2nd element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 2

Now value of i is incremented to 1 so i=2

The same procedure continues at each iteration.

The last iteration is shown below:

5th iteration:

i = 4

As i= 4 and NUM_VALS = 4 This means for condition i<NUM_VALS  is false so the loop breaks

Next the statement: printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches);  executes which displays the value of

numMatches = 3

2)

The first loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which reads the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration each element at i-th index is read using scanf such as element at userGuesses[0], userGuesses[1], userGuesses[2]. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

The second loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which prints the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration, each element at i-th index is printed on output screen using printf such as element at userGuesses[0], userGuesses[1], userGuesses[2] is displayed. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

So if user enters enters 9 5 2, then the output is 9 5 2

3)

The loop works as follows:

At first iteration:

i=0

i<NUM_VALS is true as NUM_VALS = 4 so 0<4. Hence the loop body executes.

if (testGrades[i] > 100 checks if the element at i-th index of testGrades array is greater than 100. As i=0 so this statement becomes:

if (testGrades[0] > 100

As testGrades[0] = 101 so this condition evaluates to true as 101>100

So the statement sumExtra = testGrades[i] - 100 + sumExtra; executes which becomes:

sumExtra = testGrades[0] - 100 + sumExtra

As sumExtra = 0

testGrades[0] = 101

So

sumExtra = 101 - 100 + 0

sumExtra = 1

The same procedure is done at each iteration until the loop breaks. The output is:

sumExtra = 8

4)

The loop works as follows:

At first iteration

i=0

i < NUM_VALS is true as  NUM_VALS = 4 so 0<4 Hence loop body executes.

if (i<(NUM_VALS-1))   checks if i is less than NUM_VALS-1 which is 4-1=3

It is also true as 0<3 Hence the statement in body of i executes

printf( "%d,", hourlyTemp[i]) statement prints the element at i-th index i.e. at 0-th index of hourlyTemp array with a comma (,) in the end. As hourlyTemp[0] = 90; So 90, is printed.

When the above IF condition evaluates to false i.e. when i = 3 then else part executes which prints the hourlyTemp[3] = 95 without comma.

Same procedure happens at each iteration unless value of i exceeds NUM_VAL.

The output is:

90, 92, 94, 95

The programs along with their output are attached.

4 0
4 years ago
a blank search examines the first item to see if it is a match, then the second, and so on. answers: linear, binary, and orderly
Mazyrski [523]

Answer:

The answer is a linear search

Explanation:

3 0
3 years ago
Read 2 more answers
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