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saul85 [17]
3 years ago
15

A user is attempting to format a 4 TB HDD using NTFS but only has the option to format the disk with a size of approximately 2 T

B.
What is the most likely reason the user is unable to format the disk to its full capacity?
a. The disk is using the MBR partitioning scheme.
b. The disk is using the GUID partitioning scheme.
c. NTFS only supports up to 2.2 TB volume sizes.
d. The user chose a primary partition instead of extended partition.
Computers and Technology
1 answer:
matrenka [14]3 years ago
4 0

Answer:

The correct option is option A.

Explanation:

The MBR ( master boot record) partitioning scheme maximum partition size is 2 TB.

This limit occurs because the locations were the partitions are stored, can be of a maximum of 4 bytes long, and because the disk works in hexadecimal, the max value it can store is all F's. Thus, we have that the maximum partition sector would be 4,294,967,295, and since each sector is limited to 512 bytes, this result in the partition being limited to only 2 TB.

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You are troubleshooting a Windows workstation that is having problems communicating with the host at IP address 1.2.3.4. What ut
PSYCHO15rus [73]

Answer:

Tracert 1.2.3.4 is the correct answer.

Explanation:

Tracert is the command that is used to trace the track of the packet of the Internet Protocol. That's why when the user troubleshoots their workstation and then the problem will occur at the time of conversation to the host's IP. So, that's why he using the following command of the to track route of the IP packets.

4 0
4 years ago
Lindsey also needs to calculate the commissions earned each month. If the company earns $200,000 or more in a month, the commiss
lutik1710 [3]

Answer:

I have created and attached an image with example values for this question.

First we'll learn the IF function syntax:

IF(logical_test, value_if_true, [value_if_false])

Where

logical_test is the decision making comparison

value_if_true is the result to return when TRUE

value_if_false is the result to return when FALSE

We assume that cells B10 to G10 contain the Total Sales for January to June. And cells B17 to G17 contain the Commissions for January to June.

a.

The formula to use in cell B17 is:

<em>=IF(B10>=200000;True;False)</em>

<em />

b.

The formula to use in cell B17 is:

<em>=IF(B10>=200000;B10*0.35;False)</em>

<em />

c.

The formula to use in cell B17 is:

<em>=IF(B10>=200000;B10*0.35;B10*0.27)</em>

<em />

d.

To copy the formula in B17 to C17:G17, you can just copy and paste or select cell B17 and drag from the black square on the bottom right of the cell.

8 0
3 years ago
Read 2 more answers
A sinewave has a period (duration of one cycle) of 645 μs. What is the corresponding frequency of this sinewave, in kHz, express
irga5000 [103]

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: <em>155 kHz.</em>

<u>Given the following data:</u>

  • Period = 645 μs

Note: μs represents microseconds.

<u>Conversion:</u>

1 μs = 1 × 10^{-6} seconds

645 μs = 645 × 10^{-6} seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

Frequency = \frac{1}{Period}

Substituting the value into the formula, we have;

Frequency = \frac{1}{645\; * \;10^{-6}}

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

<u>Conversion:</u>

1 hertz = 0.001 kilohertz

1550.3876 hertz = X kilohertz

Cross-multiplying, we have;

X = 0.001 × 1550.39

X = 155039 kHz

To 3 significant figures;

<em>Frequency = 155 kHz</em>

Find more information: brainly.com/question/23460034

5 0
3 years ago
10. When does teasing become cyberbullying?
Vlad [161]

Answer:

A) When the words, images or videos shift from humorous to hurtful.

Unlike kids who are being bullied, kids who are being teased can influence whether it continues or ends. If they get upset, the teaser usually stops.

4 0
2 years ago
and assuming main memory is initially unloaded, show the page faulting behavior using the following page replacement policies. h
Svet_ta [14]

FIFO

// C++ implementation of FIFO page replacement

// in Operating Systems.

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using FIFO

int pageFaults(int pages[], int n, int capacity)

{

   // To represent set of current pages. We use

   // an unordered_set so that we quickly check

   // if a page is present in set or not

   unordered_set<int> s;

   // To store the pages in FIFO manner

   queue<int> indexes;

   // Start from initial page

   int page_faults = 0;

   for (int i=0; i<n; i++)

   {

       // Check if the set can hold more pages

       if (s.size() < capacity)

       {

           // Insert it into set if not present

           // already which represents page fault

           if (s.find(pages[i])==s.end())

           {

               // Insert the current page into the set

               s.insert(pages[i]);

               // increment page fault

               page_faults++;

               // Push the current page into the queue

               indexes.push(pages[i]);

           }

       }

       // If the set is full then need to perform FIFO

       // i.e. remove the first page of the queue from

       // set and queue both and insert the current page

       else

       {

           // Check if current page is not already

           // present in the set

           if (s.find(pages[i]) == s.end())

           {

               // Store the first page in the

               // queue to be used to find and

               // erase the page from the set

               int val = indexes.front();

               

               // Pop the first page from the queue

               indexes.pop();

               // Remove the indexes page from the set

               s.erase(val);

               // insert the current page in the set

               s.insert(pages[i]);

               // push the current page into

               // the queue

               indexes.push(pages[i]);

               // Increment page faults

               page_faults++;

           }

       }

   }

   return page_faults;

}

// Driver code

int main()

{

   int pages[] = {7, 0, 1, 2, 0, 3, 0, 4,

               2, 3, 0, 3, 2};

   int n = sizeof(pages)/sizeof(pages[0]);

   int capacity = 4;

   cout << pageFaults(pages, n, capacity);

   return 0;

}

LRU

//C++ implementation of above algorithm

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using indexes

int pageFaults(int pages[], int n, int capacity)

{

   // To represent set of current pages. We use

   // an unordered_set so that we quickly check

   // if a page is present in set or not

   unordered_set<int> s;

   // To store least recently used indexes

   // of pages.

   unordered_map<int, int> indexes;

   // Start from initial page

   int page_faults = 0;

   for (int i=0; i<n; i++)

   {

       // Check if the set can hold more pages

       if (s.size() < capacity)

       {

           // Insert it into set if not present

           // already which represents page fault

           if (s.find(pages[i])==s.end())

           {

               s.insert(pages[i]);

               // increment page fault

               page_faults++;

           }

           // Store the recently used index of

           // each page

           indexes[pages[i]] = i;

       }

       // If the set is full then need to perform lru

       // i.e. remove the least recently used page

       // and insert the current page

       else

       {

           // Check if current page is not already

           // present in the set

           if (s.find(pages[i]) == s.end())

           {

               // Find the least recently used pages

               // that is present in the set

               int lru = INT_MAX, val;

               for (auto it=s.begin(); it!=s.end(); it++)

               {

                   if (indexes[*it] < lru)

                   {

                       lru = indexes[*it];

                       val = *it;

                   }

               }

               // Remove the indexes page

               s.erase(val);

               // insert the current page

               s.insert(pages[i]);

               // Increment page faults

               page_faults++;

           }

           // Update the current page index

           indexes[pages[i]] = i;

       }

   }

   return page_faults;

}

// Driver code

int main()

{

   int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

   int n = sizeof(pages)/sizeof(pages[0]);

   int capacity = 4;

   cout << pageFaults(pages, n, capacity);

   return 0;

}

You can learn more about this at:

brainly.com/question/13013958#SPJ4

4 0
1 year ago
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