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Sergeeva-Olga
3 months ago
5

Which of the following statements is true? A. Marking a presentation as final is stronger security than password protection. B.

Marking a presentation as final is useful only when saving in Text Only format. C. Marking a presentation as final is the same level of security as password protection. D. Marking a presentation as final is not as strong security as password protection.
Computers and Technology
2 answers:
8_murik_8 [964]3 months ago
6 0
<span>The correct choice is D. Marking a presentation as final provides weaker security compared to password protection.

Experienced computer users can easily bypass the "mark as final" feature, which is why using passwords is recommended since they offer a much stronger defense. The more complicated the password, the harder it is to crack.</span>
Amiraneli [1K]3 months ago
5 0

The answer is B.

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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
Harlamova29_29 [1022]

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
Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS
Amiraneli [1052]
Your question is missing the section C, so here is the C segment while (k<n v="" k="k+1;" answer:="" while:="" bge="" end="" while="" n="" addi="" sll="" making="" indexable="" add="" lw="" sw="" j="" end:="" explanation:="" the="" mips="" assembly="" code="" reflecting="" c="" segment="" is="" />
4 0
1 month ago
CHALLENGE ACTIVITY 2.1.2: Assigning a sum. Write a statement that assigns total_coins with the sum of nickel_count and dime_coun
Harlamova29_29 [1022]

Answer:

In Python:

total_coins = nickel_count + dime_count

Explanation:

The sums of nickel_count and dime_count are combined and assigned to total_coins.

Cheers.

8 0
2 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
Determine the number of bytes necessary to store an uncompressed binary image of size 4000 × 3000 pixels
maria [1035]

Response: 1,500,000 bytes.

Clarification:

If we assume the image dimensions are 4000 pixels in width and 3000 pixels in height, the total uncompressed image will consist of 4000*3000= 12,000,000 pixels.

In the case of a binary image, each pixel can have only two values, which necessitates one bit for each pixel.

This indicates that we need to accommodate 12,000,000 bits.

Given that 1 byte equals 8 bits.

So, to store an uncompressed binary image sized 4000 x 3000 pixels, 12,000,000/8 bytes is required ⇒ 1,500,000 bytes.

3 0
2 months ago
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