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yan
2 months ago
6

The following is a string of ASCII characters whose bit patterns have been converted into hexadecimal for compactness: 73 F4 E5

76 E5 4A EF 62 73.
Of the eight bits in each pair of digits, the leftmost is a parity bit. The remaining bits are the ASCII code.

(a) Convert the string to bit form and decode the ASCII.

(b) Determine the parity used: odd or even?
Computers and Technology
1 answer:
Rzqust [1K]2 months ago
5 0

Answer:

a) Transforming each character into its binary equivalent:

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

Likewise

F4:

F stands for 15 and its binary is: 1111

4 in binary: 100

Consequently

F4:  1_111_0100

E5:

E corresponds to 14 and its binary form is: 1110

5 in binary: 101

Therefore

E5:  1_110_0101

76:

7 in binary: 111

6 in binary: 110

Hence

76:  0_111_0110

E5:

E has a binary representation of: 1110

5 in binary: 101

Consequently

E5:  1_110_0101

4A:

4 in binary: 100

A represents 10

A in binary form: 1010

Therefore

4A:  0_100_1010

EF:

E in binary is: 1110

F in binary is: 1111

Thus

EF: 1_110_1111

62:

6 in binary form: 110

2 in binary form: 10

Therefore

62:  0_110_0010

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

for 0_1110011: the decimal equivalent is: 115 which translates to s

for 1_1110100:  the decimal equivalent is: 116 which translates to t

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1110110: the decimal equivalent is:  118 which translates to v

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1001010: the decimal equivalent is:  74 which translates to j

for 1_1101111: the decimal equivalent is: 111 which translates to o

for 0_1100010:  the decimal equivalent is: 98 which translates to b

for 0_1110011: the decimal equivalent is:  115 which translates to s

Thus the decoded sequence is:  stevejobs

b) The parity being utilized is odd.

for 0_1110011:  There are 5 instances of 1s and the parity is 0 indicating it is odd.

for 1_1110100: There are 4 instances of 1s and the parity is 1 indicating it is odd.

Thus, we count the number of 1s and then verify if the parity is odd or even.

Likewise

for 1_1100101:  the parity is odd

for 0_1110110: the parity is odd

for 1_1100101:  the parity is odd

for 0_1001010: the parity is odd

for 1_1101111: the parity is odd

for 0_1100010: the parity is odd

for 0_1110011: the parity is odd

Therefore, the parity being used is odd.

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Answer:

Pseudo CODE

a)

n= Input “Enter 5 integer values”

b)

sum=0.0

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Inside loop sum=n[i]+sum

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Print avg

c)

small=n[0] # assume initial number is the smallest

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Inside loop if n[i]<small # check if current number is smaller than small

Inside if Then small=n[i]

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Inside if then large=n[i]

Print small

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d)

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You are a police officer trying to crack a case. You want to check whether an important file is in the evidence room. Files have
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Answer:

Since RANDY operates randomly, any file within the specified index range will have the recurrence relation as follows:

T(n) = T(n-i) + O(1)

Here, the probability is 1/n, where i can vary between 1 and n. The variable n in T(n) denotes the size of the index range, which will subsequently reduce to (n-i) in the following iteration.

Given that i is probabilistically distributed from 1 to n, the average case time complexity can then be expressed as:

T(n) = \sum_{i=1}^{n}\frac{1}{n}T(n-i) + O(1) = T(n/2)+O(1)

Next, solving T(n) = T(n/2) + O(1)

yields T(n) = O(log n).

Thus, the complexity of this algorithm is O(log n).

It should be noted that this represents the average time complexity due to the algorithm's randomized nature. In the worst-case scenario, the index range may only decrease by 1, resulting in a time complexity of O(n) since the worst-case scenario would be T(n) = T(n-1) + O(1).

3 0
2 months ago
Read 2 more answers
You want to register the domain name ABCcompany.org, but the registration service is not allowing you to do that. What's the mos
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Options :

The domain name is already taken.

Domain names are required to end with ".com".

You do not hold legal ownership of ABC Company.

Domain names must exclusively use lowercase letters.

Answer:

The domain name is already taken.

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6 0
3 months ago
Describe a strategy for avoiding nested conditionals. Give your own example of a nested conditional that can be modified to beco
maria [1035]

Answer:

To prevent nested conditionals, one can utilize logical expressions like the AND operator.

A recommended approach is creating an interface class with a method designed for a shared function. This design approach is known as the strategy design pattern. The conditional statements can be consolidated into this method. Subsequently, each class can implement this interface and invoke that shared method as needed by constructing instances of subclasses and calling the common method for those objects. This illustrates polymorphism.

Explanation:

Nested conditionals occur when if or else if statements are placed within another condition. For instance:

if( condition1) {

//runs when condition1 is true

  if(condition2) {

//runs when both condition1 and condition2 are true

  }  else if(condition3) {

 //when condition1 is true and condition3 is also true

} else {

 //condition1 is true but neither condition2 nor conditions3 are satisfied

}  }

Excessive nested conditionals can complicate the program, rendering it hard to read or comprehend, particularly if there's improper indentation. Debugging also becomes challenging when there are numerous nested statements.

Thus, several strategies can be adopted to eliminate nested conditionals, such as utilizing a switch statement.

For instance, I’ll present an example of the strategies discussed earlier:

Logical Expressions Usage:

A method to avoid nested conditionals is by employing logical expressions with logical operators like the AND operator. The previous nested conditionals can be reframed as:

if(condition1 && condition2){ //only runs when both condition1 and condition2 are true

} else if(condition1 && condition3) {

executes only if both condition1 and condition3 are true

} else if(condition1 ){

//condition1 is satisfied but neither condtion2 nor condtion3 are true  }

This can be further simplified to one condition as:

if(!condition3){

// when  condition1 and condition2 are both satisfied

}

else

// condition3 is met

Now, consider a simple instance dealing with whether to attend school based on certain conditions.

if (temperature< 40)

{

   if (busArrived=="yes")

   {

       if (!sick)

       {

           if (homework=="done")

           {

               printf("Go to school.");

           }

       }                    

   }

}

Here, nested conditionals are evident. This can be restructured into a solitary conditional using the AND logical operator.

if ((temperature <40) && (busArrived=="yes") &&

(!sick) && (homework=="done"))

{    cout<<"Eligible for promotion."; }

The alternate method is utilizing an interface. For example, you can

abstract class Shape{

//declare a method common to all sub classes

  abstract public int area();

// same method that varies by formula of area for different shapes

}

class Triangle extends Shape{

  public int area() {

     // implementation of area code for Triangle

return (width * height / 2);

  }

}

class Rectangle extends Shape{

  public int area() {

     // implementation of area code for Rectangle

    return (width * height)

  }

}

Now, you can readily create Rectangle or Triangle instances and invoke area() for any of those objects, resulting in execution of the appropriate version accordingly.

4 0
2 months ago
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