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yanalaym [24]
1 year ago
14

Choose the attribute used to provide accessibility by configuring a text alternative that is available to browsers and other use

r agents that do not support graphics. text alt src none of these
Computers and Technology
1 answer:
musickatia [10]1 year ago
8 0

Answer:

alt

Explanation:

The alt attribute is an HTML attribute that is used in an HTML and XHTML documents to indicate alternative text (alt text) that is to be rendered whenever the element to which it is applied can no longer be rendered.

The alt attribute also provides alternative details for an image if a pc user for a number of reasons can’t view it (as a result of either slow connection, an error in the src attributes, or if there’s a screen reader being used by the user).

You might be interested in
Write multiple if statements: If carYear is before 1968, print "Probably has few safety features." (without quotes). If after 19
jekas [21]

Answer:

Following is the statement in the C language :

if(carYear < 1968)

printf("\nProbably has a few safety features.\n");

if(carYear > 1970 && carYear <=1991 )

printf("\nProbably has head rests.\n");

if(carYear > 1991 && carYear <=2002)

printf("\nProbably has anti-lock brakes\n.");

if(carYear > 2002)

printf("\nProbably has airbags.\n");

Explanation:

Following is the description of the statement:

  • In the given question we used if block. The if block is only executed when their condition is true.
  • if(carYear < 1968) In this we check we the value of "carYear" variable is less then 1968 then it prints "Probably has a few safety features." in the console window.
  • if(carYear > 1970 && carYear <=1991) In this we check we the value of "carYear" variable is greater then 1970 and less then 1992 it prints "Probably has head rests" in the console window.
  • if(carYear > 1991 && carYear <=2002 ) In this we check we the value of "carYear" variable is greater then 1991 and less then 2003 it prints "Probably has anti-lock brakes" in the console window.
  • if(carYear > 2002) In this we check we the value of "carYear" variable is greater then 2002 then it prints "Probably has airbags" in the console window.

6 0
2 years ago
Write a copy constructor for carcounter that assigns origcarcounter.carcount to the constructed object's carcount. sample output
Drupady [299]

#include <iostream>

using namespace std;

class CarCounter {

  public:

     CarCounter();

     CarCounter(const CarCounter& origCarCounter);

     void SetCarCount(const int count) {

         carCount = count;

     }

     int GetCarCount() const {

         return carCount;

     }

  private:

     int carCount;

};

CarCounter::CarCounter() {

  carCount = 0;

  return;

}

CarCounter::CarCounter(const CarCounter &p){

carCount = p.carCount;

}

void CountPrinter(CarCounter carCntr) {

  cout << "Cars counted: " << carCntr.GetCarCount();

  return;

}

int main() {

  CarCounter parkingLot;

  parkingLot.SetCarCount(5);

  CountPrinter(parkingLot);

  return 0;

}

Sample output:  

Cars Counted: 5

8 0
2 years ago
Read 2 more answers
Which data type change will require the app builder to perform the additional steps in order to retain existing functionalities?
Readme [11.4K]

Answer:

Option D is the correct option.

Explanation:

The following option is correct because the lead alteration from number to text and the number to text types of data will reconstruct the important number of the app builder to accomplish those steps which is extra to continue to have that functionality which is in the existence. That's why the app builder has to be reconstructed the types of data for the custom fields.

7 0
1 year ago
Produce a list named prime_truths which contains True for prime numbers and False for nonprime numbers in the range [2,100]. We
LenKa [72]

Answer:

  1. def is_prime(n):
  2.    for i in range(2, n):
  3.        if(n % i == 0):
  4.            return False  
  5.    return True  
  6. prime_truths = [is_prime(x) for x in range(2,101)]
  7. print(prime_truths)

Explanation:

The solution code is written in Python 3.

Presume there is a given function is_prime (Line 1 - 5) which will return True if the n is a prime number and return False if n is not prime.

Next, we can use the list comprehension to generate a list of True and False based on the prime status (Line 7). To do so, we use is_prime function as the expression in the comprehension list and use for loop to traverse through the number from 2 to 100. The every loop, one value x will be passed to is_prime and the function will return either true or false and add the result to prime_truth list.

After completion of loop within the comprehension list, we can print the generated prime_truths list (Line 8).

3 0
2 years ago
8.Change the following IP addresses from binary notation to dotted-decimal notation: a.01111111 11110000 01100111 01111101 b.101
Andrews [41]

Answer:

a. 01111111 11110000 01100111 01111101 dotted decimal notation:

(127.240.103.125)

b. 10101111 11000000 11111000 00011101 dotted decimal notation: (175.192.248.29)

c. 11011111 10110000 00011111 01011101 dotted decimal notation:

(223.176.31.93)

d. 11101111 11110111 11000111 00011101 dotted decimal notation:

(239.247.199.29)

a. 208.34.54.12 class is C

b. 238.34.2.1 class is D

c. 242.34.2.8 class is E

d. 129.14.6.8 class is B

a.11110111 11110011 10000111 11011101 class is E

b.10101111 11000000 11110000 00011101 class is B

c.11011111 10110000 00011111 01011101 class is C

d.11101111 11110111 11000111 00011101 class is D

Explanation:

8 a. 01111111 11110000 01100111 01111101

we have to convert this binary notation to dotted decimal notation.

01111111 = 0*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1*4 + 1*2 + 1

           = 64 + 32 + 16 + 8 + 4 + 2 + 1

           = 127

11110000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

               = 1*128 + 1*64 + 1*32 + 16 + 0 + 0 + 0 + 0

               = 128 + 64 + 32 + 16

               = 240

01100111 = 0*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 0 + 0 + 4 + 2 + 1

               = 64 + 32 + 4 + 2 + 1

               = 103

01111101   = 0*2^7 + 1*2^6 + 1*2^5+ 1*2^4 +1*2^3 +1*2^2 + 0*2^1 + 1*2^0

               = 0 + 1*64 + 1*32 + 1*16 + 1*8 + 1* 4 + 0 + 1

               = 64 + 32 + 16 + 8 + 4 + 1

               = 125

So the IP address from binary notation 01111111 11110000 01100111 01111101 to dotted decimal notation is : 127.240.103.125

b) 10101111 11000000 11111000 00011101

10101111 = 1*2^7 + 0*2^6 + 1*2^5 + 0*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

             = 175

11000000 = 1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                 = 192

11111000 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +1*2^3 + 0*2^2 + 0*2^1 + 0*2^0

              = 248

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 10101111 11000000 11111000 00011101  to dotted decimal notation is : 175.192.248.29

c) 11011111 10110000 00011111 01011101

11011111 = 1*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 223

10110000 =  1*2^7 + 0*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 0*2^2 + 0*2^1 + 0*2^0

                = 176

00011111 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 31

01011101 = 0*2^7 + 1*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

              = 93

So the IP address from binary notation 11011111 10110000 00011111 01011101 to dotted decimal notation is :223.176.31.93

d) 11101111 11110111 11000111 00011101

11101111 = 1*2^7 + 1*2^6 + 1*2^5 + 0*2^4 +1*2^3 + 1*2^2 + 1*2^1 + 1*2^0

            = 239

11110111 = 1*2^7 + 1*2^6 + 1*2^5 + 1*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

           = 247

11000111 =  1*2^7 + 1*2^6 + 0*2^5 + 0*2^4 +0*2^3 + 1*2^2 + 1*2^1 + 1*2^0

              = 199

00011101 = 0*2^7 + 0*2^6 + 0*2^5 + 1*2^4 +1*2^3 + 1*2^2 + 0*2^1 + 1*2^0

               = 29

So the IP address from binary notation 11101111 11110111 11000111 00011101 to dotted decimal notation is : 239.247.199.29

9. In order to the find the class check the first byte of the IP address which is first 8 bits and check the corresponding class as follows:                

Class A is from 0 to 127

Class B is from 128 to 191

Class C is from 192 to 223

Class D is from 224 to 239

Class E is from 240 to 255

a. 208.34.54.12

If we see the first byte of the IP address which is 208, it belongs to class C as class C ranges from 192 to 223.

b. 238.34.2.1

If we see the first byte of the IP address which is 238, it belongs to class D as Class D ranges from 224 to 239.

c. 242.34.2.8

If we see the first byte of the IP address which is 242, it belongs to class E as Class E ranges from 240 to 255.

d. 129.14.6.8

If we see the first byte of the IP address which is 129, it belongs to class B as Class B ranges from 128 to 191.

10. In order to find the class of the IP addresses in easy way, start checking bit my bit from the left of the IP address and follow this pattern:

0 = Class A

1 - 0 = Class B

1 - 1 - 0 = Class C

1 - 1 - 1 - 0 = Class D

1 - 1 - 1 - 1 = Class E

a. 11110111 11110011 10000111 11011101

If we see the first four bits of the IP address they are 1111 which matches the pattern of class E given above. So this IP address belongs to class E.

b. 10101111 11000000 11110000 00011101

If we see the first bit is 1, the second bit is 0 which shows that this is class B address as 1 0 = Class B given above.

c. 11011111 10110000 00011111 01011101

The first bit is 1, second bit is 1 and third bit is 0 which shows this address belongs to class C as 110 = Class C given above.

d. 11101111 11110111 11000111 00011101

The first bit is 1, the second bit is also 1 and third bit is also 1 which shows that this address belongs to class D.

3 0
1 year ago
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