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Semmy [17]
3 years ago
5

What are the advantages of using digital signals over analog signals? Scientist have found advantages when they convert digital

signals to analog signals. Advantages of using digital signals include greater ____ , reduced ____ (unwanted signals), and an increased ____ for sending information.
Word Bank:
capacity accuracy Aliens noise
blank 1
blank2
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Computers and Technology
2 answers:
Viktor [21]3 years ago
6 0

Answer:

Hope it Helps

Explanation:

Brainliest please

borishaifa [10]3 years ago
6 0

Answer:

Advantages of using digital signals include greater <u>accuracy</u>, reduced <u>noise</u> (unwanted signals), and an increased <u>capacity</u> for sending information.

Explanation:

I hope this helps you!

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Explanation:

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A(n) is the tool that will help you the most when developing the content you will use in your presentation.
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Answer:

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Explanation:

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2 years ago
Write a function listLengthOfAllWords which takes in an array of words (strings), and returns an array of numbers representing t
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Answer:

   public static int[] listLengthOfAllWords(String [] wordArray){

       int[] intArray = new int[wordArray.length];

       for (int i=0; i<intArray.length; i++){

           int lenOfWord = wordArray[i].length();

           intArray[i]=lenOfWord;

       }

       return intArray;

   }

Explanation:

  1. Declare the method to return an array of ints and accept an array of string as a parameter
  2. within the method declare an array of integers with same length as the string array received as a parameter.
  3. Iterate using for loop over the array of string and extract the length of each word using this statement  int lenOfWord = wordArray[i].length();
  4. Assign the length of each word in the String array to the new Integer array with this statement intArray[i]=lenOfWord;
  5. Return the Integer Array

A Complete Java program with a call to the method is given below

<em>import java.util.Arrays;</em>

<em>import java.util.Scanner;</em>

<em>public class ANot {</em>

<em>    public static void main(String[] args) {</em>

<em>       String []wordArray = {"John", "James", "David", "Peter", "Davidson"};</em>

<em>        System.out.println(Arrays.toString(listLengthOfAllWords(wordArray)));</em>

<em>        }</em>

<em>    public static int[] listLengthOfAllWords(String [] wordArray){</em>

<em>        int[] intArray = new int[wordArray.length];</em>

<em>        for (int i=0; i<wordArray.length; i++){</em>

<em>            int lenOfWord = wordArray[i].length();</em>

<em>            intArray[i]=lenOfWord;</em>

<em>        }</em>

<em>        return intArray;</em>

<em>    }</em>

<em>}</em>

This program gives the following array as output: [4, 5, 5, 5, 8]

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3 years ago
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A digital computer has a memory unit with 16 bits per word. The instruction set consists of 72 different operations. All instruc
OLga [1]

Answer:

a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1

Explanation:

a. How many bits are needed for the opcode?

Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72

72 = 64 + 8 = 2⁶ + 8

Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128

So, we require 7 bits for the opcode.

b. How many bits are left for the address part of the instruction?

bits left = bits per word - opcode bit = 16 - 7 = 9 bits

c. What is the maximum allowable size for memory?

Since there are going to be 2⁹ bits to addresses each word and 16 bits  for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.

We convert this to bytes

8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB

d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus

2¹⁶ - 1

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