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Lyrx [107]
3 years ago
6

Imagery functions as a coding system to help individuals acquire movement patterns. this describes

Computers and Technology
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

Imagery can be used to develop qualities in yourself you'd like to have — it's like emotional body-building — and using a technique called “Evocative Imagery” you can cultivate courage, patience, tolerance, humor, concentration, self-confidence or any other quality you'd like to embody.

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Write a function `has_more_zs` to determine which of two strings contains # more instances of the letter "z". It should take as
nekit [7.7K]

Answer:

// Method's name: has_more_zs

// Parameters are text1 and text2 to hold the two phrases to be tested

public static String has_more_zs(String text1, String text2) {

 // Create and initialize z1 to zero

 // Use z1 to count the number of zs in text1

 int z1 = 0;

 // Create and initialize z2 to zero

 // Use z2 to count the number of zs in text2

 int z2 = 0;

 

 //Create a loop to cycle through the characters in text1

 //Increment z1 by one if the current character is a 'z'

 int i = 0;

 while (i < text1.length()) {

  if (text1.charAt(i) == 'z' || text1.charAt(i) == 'Z') {

   z1 += 1;

  }

  i++;

 }

 //Create a loop to cycle through the characters in text2

 //Increment z2 by one if the current character is a 'z'

 

 i = 0; //Re-initialize i to zero

 

 while (i < text2.length()) {

  if (text2.charAt(i) == 'z' || text2.charAt(i) == 'Z') {

   z2 += 1;

  }

  i++;

 }

 

 

 //Using the values of z1 and z2, return the necessary statements

 if (z1 > z2) {

  return "The phrase '" + text1 + "'" + " has more occurences of z than the phrase " + "'" + text2 + "'";

 }

 else if (z1 < z1) {

  return "The phrase '" + text2 + "'" + " has more occurences of z than the phrase " + "'" + text1 + "'";

 }

 else if (z1 == z2) {

  return "The strings have the same number of z";

 }

 else {

  return "Neither string contains the the letter z";

 }

}

Explanation:

Explanation to answer has been given in the code as comments. Please refer to the comments for the details of the code.

The source code file has been attached to this response and saved as "NumberOfZs.java"

Hope this helps!

Download java
5 0
3 years ago
A determinant is any attribute whose value determines other values within a column.
guajiro [1.7K]

Answer:

true

Explanation:

4 0
3 years ago
Assume that the message M has to be transmitted. Given the generator function G for the CRC scheme, calculate CRC. What will be
ANEK [815]

Let the message be M : 1001 0001   and the generator function is G : 1001

Solution :

CRC sender

            <u>                                     </u>

1001     | 1001  0001  000

           <u>  1001                             </u>

           <u> 0000  0001             </u>

                        1000

            <u>            1001             </u>

                        0001  000

                             <u>    1 001  </u>

                                 0001

Here the generator is 4 bit - 1, so we have to take three 0's which will be replaced by reminder before sending to received--

eg    1001  0001  001    

Now CRC receiver

                <u>                                     </u>

1001         | 1001    0001   001

                <u> 1001                           </u>

                 0000 0001

                       <u>     1001                  </u>

                            1000

                          <u>  1001              </u>

                            0001  001

                    <u>        0001  001      </u>

                                   0000

No error

7 0
3 years ago
Một số được gọi là thác đổ nếu phần tử biểu diễn thập phân của nó nhiều hơn một chữ số đồng thời theo chiều từ trái qua phải chữ
kirza4 [7]
I don’t understand this
3 0
3 years ago
In this problem, we want to compare the computational performance of symmetric and asymmetric algorithms. Assume a fast public-k
MariettaO [177]

Answer:

The AES decryption time is 8 minutes

Explanation:

GIVEN THAT:

The 1 GByte is equal to 230 bytes and equal to 233 bits

The RSA decryption time is calculated as = 233 / (100*230 )

= 83886. 06 seconds

= 23 hours

The RSA decryption time is 23 hours

The AES decryption time is calculated as = 233 / (17 * 230 )

= 481.81 seconds

= 8 minutes .

The AES decryption time is 8 minutes

4 0
4 years ago
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