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Zielflug [23.3K]
2 years ago
15

1. Suppose the memory address in an embedded system has 12 bits. How many unique memory locations can this 12-bit address access

? How many bits this memory space can hold? What about if memory address has 16 bits?
Computers and Technology
1 answer:
const2013 [10]2 years ago
3 0

Answer:

14 bits memory address - 4096 bits.

16 bits memory address - 16384 bits.

Explanation:

The computer system memory is a hardware component that provides the necessary memory space and location to run software application on the computer system. The memory address location comes in various sizes.

A 14 bit memory can address 2^14 bits which is 4096 bits, since each 14 bit line carries a bit on which line. So for a 16 bit memory address, 2^16 bits location is addressed.

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Answer:

The caps and bold letters makes it seem like his boss is yelling at him.

Explanation:

5 0
3 years ago
Both instructions and data in a digital<br>computer are represented as binary<br>digits. True False​
MrRissso [65]

Answer:

False

Explanation:

insturctions are represented in binary numbers, data are represented in decimal numbers.

6 0
2 years ago
Wireless networks usually use ____ signals to send data.
barxatty [35]
<span>Wireless networks usually use radio (b) signals to send data. </span>
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2 years ago
Read 2 more answers
Suppose that the tuition for a university is $10,000 this year and increases 4% every year. In one year, the tuition will be $10
seropon [69]

Answer:

<em>The programming language is not stated; however, I'll answer using Python programming language (</em><em>Se</em><em>e attachment</em><em> </em><em>for</em><em> </em><em>proper </em><em>for</em><em>mat</em><em>)</em>

tuition = 10000

rate = 0.04

for i in range(1,15):

tuition = tuition + tuition * rate

if i <= 10:

print("Year "+str(i)+" tuition:",end=" ")

print(round(tuition,2))

if i == 14:

print("Tuition 4th year after:",end=" ")

print(round(tuition,2))

Explanation:

<em>The first 2 lines initializes tuition and rate to 10000 and 0.04 respectively</em>

tuition = 10000

rate = 0.04

<em>The next line iterates from year 1 to year 14</em>

for i in range(1,15):

<em>This line calculates the tuition for each year</em>

tuition = tuition + tuition * rate

<em>The next 3 lines prints the tuition for year 1 to year 10</em>

if i <= 10:

print("Year "+str(i)+" tuition:",end=" ")

print(round(tuition,2))

<em>The next 3 lines prints the tuition at the 4th year after year 10 (i.e. year 14)</em>

if i == 14:

print("Tuition 4th year after:",end=" ")

print(round(tuition,2))

7 0
3 years ago
Write a Java program to encrypt and decrypt a phrase using two similar approaches, each insecure by modern standards. The first
xeze [42]

Answer:

See explaination

Explanation:

//CryptoManager.java

public class CryptoManager {

static int LOWER_BOUND=32;

static int UPPER_BOUND=95;

/*This method determines if a string is within the allowable bounds of ASCII

codes according to the LOWER_BOUND and UPPER_BOUND characters. The parameter

plainText is the string to be encrypted. The method returns true if all

characters are within the allowable bounds, false if any character is outside.*/

public static boolean stringInBounds (String plainText)

{

boolean flag=true;

//determines if a string is within the allowable bounds of ASCII

//codes according to the LOWER_BOUND and UPPER_BOUND characters.

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

{

if(!((int)plainText.charAt(i)>=LOWER_BOUND && (int)plainText.charAt(i)<=UPPER_BOUND))

{ //false if any character is outside the bounds

flag=false;

break;

}

}

//returns true if all characters are within the allowable bounds

return flag;

}

/*This method encrypts a string according to the Caesar Cipher. The integer key

specifies an offset and each character in plainText is replaced by the character

the specified distance away from it. The parameter plainText is an uppercase

string to be encrypted. The parameter key is an integer that specifies the

offset of each character. The method returns the encrypted string.*/

public static String encryptCaesar(String plainText, int key)

{

//Wrap around the key, if it is greater than the UPPER_BOUND

key=Wrap_around(key);

//encrypted text

String res="";

//encryption

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

{

res+=Character.toString((char) ((int)plainText.charAt(i)+key));

}

//return result

return res;

}

/* This method decrypts a string according to the Caesar Cipher. The integer

key specifies an offset and each character in encryptedText is replaced by

the character "offset" characters before it. This is the inverse of the

encryptCaesar method. The parameter encryptedText is the encrypted string

to be decrypted, and key is the integer used to encrypt the original text.

The method returns the original plain text string.*/

public static String decryptCaesar(String encryptedText, int key){

//Wrap around the key, if it is greater than the UPPER_BOUND

key=Wrap_around(key);

//decrypted text

String org="";

//encryption

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

{

org+=Character.toString((char) ((int)encryptedText.charAt(i)-key));

}

//return result

return org;

}

public static int Wrap_around(int key)

{

while(key>UPPER_BOUND)

{

key-=(UPPER_BOUND-LOWER_BOUND);

}

return key;

}

/* This method encrypts a string according to the Bellaso Cipher. Each character

in plainText is offset according to the ASCII value of the corresponding

character in bellasoStr, which is repeated to correspond to the length of

plaintext. The method returns the encrypted string.*/

public static String encryptBellaso(String plainText, String bellasoStr)

{

//encrypted text

String res="";

//Adjust length of bellasoStr to plainText

while(bellasoStr.length()<plainText.length())

{

bellasoStr+=bellasoStr.substring(0,(plainText.length()-bellasoStr.length()));

}

//encryption

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

{

char c=(char)Wrap_around((int)plainText.charAt(i)+(int)bellasoStr.charAt(i) );

res+=Character.toString(c);

}

//return result

return res;

}

/*

This method decrypts a string according to the Bellaso Cipher. Each character

in encryptedText is replaced by the character corresponding to the character in

bellasoStr, which is repeated to correspond to the length of plainText. This is

the inverse of the encryptBellaso method. The parameter encryptedText is the

encrypted string to be decrypted, and bellasoStr is the string used to encrypt

the original text. The method returns the original plain text string.*/

public static String decryptBellaso(String encryptedText, String bellasoStr)

{

//decrypted text

String res="";

//Adjust length of bellasoStr to plainText

while(bellasoStr.length()<encryptedText.length())

{

bellasoStr+=bellasoStr.substring(0,(encryptedText.length()-bellasoStr.length()));

}

//decryption

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

{

char c=(char)Wrap_around((int)encryptedText.charAt(i)-(int)bellasoStr.charAt(i) );

res+=Character.toString(c);

}

//return result

return res;

}

}

6 0
2 years ago
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