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laila [671]
3 years ago
11

The color pattern in which marine organisms are light on the bottom and dark on the top of their bodies camouflaging them agains

t the water-air interface is ________.
Computers and Technology
1 answer:
suter [353]3 years ago
4 0

Answer: countershading

Explanation:

This is a type of protective camouflaging used by marine organism whereby they are light at the bottom and dark on top.

They have got receptors in their body which help helps them to camouflage their upper and lower half body using a distinctive colour pattern.

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1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0
3 years ago
Instructions Write a program that allows the user to enter the last names of five candidates in a local election and the number
KengaRu [80]

Here is code in java.

import java.util.*;

class Election

{ //main method

   public static void main(String args[]){

      // create an object of Scanner class to read input

      Scanner s = new Scanner(System.in);

      // string array to store name of candidates

       String name[] = new String[5];

       // int array to store vote count of candidates

       int v_count[] = new int[5];

       double p = 0;

       int i = 0, sum = 0, high =0, win = 0;

for(i = 0; i < 5;i++)

       {

          System.out.print("last name of Candidate " + (i+1) + ":");

          // read name of Candidate

          name[i] = s.next();

          System.out.print(" number of votes received: ");

          // read vote of Candidate

           v_count[i] =s.nextInt();

if(v_count[i] > high)

              {

                  // highest vote

                  high = v_count[i];

                  win = i;

              }

              // total vote count

           sum +=v_count[i];

       }

     // printing the output

      System.out.println("\nCandidate\tVotes Received\t% of TotalVotes\n");

       for(i = 0; i < 5;i++)

       {

           // % of vote of each Candidate

           p =(v_count[i]*100)/sum;

           // print the output

          System.out.println(name[i] + "\t\t" + v_count[i] + "\t\t\t" +p);

       }

       // print the total votes

      System.out.println("\nTotal Votes:\t" + sum);

      // print the Winner of the Election

       System.out.println("Winner of the Election is: " + name[win]);

   }

}

Explanation:

Create a string array "name" to store the name of candidates.Create Integer array "v_count" to store votes os each candidates.Calculate total votes of all candidates and find the highest vote amongst all candidate and print it.Find % of votes received by each candidate.Print these stats.

Output:

last name of Candidate 1:patel

number of votes received: 54

last name of Candidate 2:singh

number of votes received: 76

last name of Candidate 3:roy

number of votes received: 33

last name of Candidate 4:yadav

number of votes received: 98

last name of Candidate 5:sharma

number of votes received: 50

Candidate       Votes Received  % of TotalVotes

patel           54                      17.0

singh           76                      24.0

roy             33                      10.0

yadav           98                      31.0

sharma          50                      16.0

Total Votes:  311

Winner of the Election is: yadav

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Time shifting occurs when
Aleksandr-060686 [28]
Answer: C

Time shifting is when you move from one period in time to another.
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DO NOT JOIN ANY Z O O M MEETINGS FROM THIS PERSON! IS A TRAP Please help me get them banned!!!!!
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Answer:

how do i help?

Explanation:

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An incident response plan should be created be for a software system is released for use.
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The answer is choice “a. True”.
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