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Andreas93 [3]
2 years ago
9

For example, consider a file with protection mode 644 (octal) contained in a directory with protection mode 730. How might the f

ile be compromised in this case? What are the limitations?
Computers and Technology
1 answer:
jeka942 years ago
8 0

Answer:

Archivo: 110100100 (propietario, grupo, otro: rw- r-- r--)

Directorio: 111 011 000 (propietario, grupo, otro: rwx -wx ---)

Explanation:

Directory permissions are critical. Since group members can access directory ('x') and can modify directory ('w'), even though they can't list directory (not 'r'), it means that if a group member knows the name of the file, that person can also delete it because deleting a file requires permission to write to the directory: file permissions are irrelevant (although commands like 'rm' warn you when you don't have write permission to the file directory , that's a courtesy, because it doesn't matter to the system call 'unlink ()').

Therefore, a member of your group (or, more precisely, a member of the group to which the directory belongs) can delete the file if you know its name. They can also read the file if they know its name, and can create a file with the same name if the original is already missing. Based on the file permissions, being able to read the file is not a compromise - you would have denied group read access (and public read access) if that mattered.

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What are the set of rules to move data from one computer to another?
ss7ja [257]
I don't believe that there are rules.
Hope I helped,
 Ms. Weasley
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3 years ago
Vẽ sơ đồ lắp đặt gồm 2 cầu chì, 1 ổ điện, 2 cực điều khiển 2 đèn mắc song song
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7 0
2 years ago
Serena, an analyst at an environmental agency, wants to prepare a report using data from the Car Emissions database.
konstantin123 [22]

Answer:

The correct answer is:

Option 1: export the query results from the database to a spreadsheet, then export the graph to a document.

Explanation:

Spreadsheets are used to represent numerical data, perform calculations and display the results numerically or graphically.

When using a database, the query results can be exported to other software using the query web address.

So in order to include the graph in her report, Serena will export the result to spreadsheet and then export the graph to document.

Hence,

The correct answer is:

Option 1: export the query results from the database to a spreadsheet, then export the graph to a document.

5 0
2 years ago
Read 2 more answers
Recall that within the ArrayBoundedQueue the front variable and the rear variable hold the indices of the elements array where t
Citrus2011 [14]

Answer:

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Explanation:

Options are not made available; However, I'll answer the question base on the following assumptions.

Assumptions

Array name = elements

The front and the rear variable hold the current indices elements

Element to enqueue is "element" without the quotes

To enqueue means to add an element to an already existing queue or to a new queue.

But first, the queue needs to be checked if it can accept a new element or not; in other words, if it's full or not

To do this, we check: if rear < the length of the queue

If this statement is true then it means the array can accept a new element; the new element will be stored at elements[rear] and the rear will be icremented by 1 rear by 1

Else if the statement is false, then an overflow is said to have occurred and the element will not be added.

Going by the above explanation, we have

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Additional explanation:

The first line calculates the length of the queue

The if condition on line 2 tests if the array can still accept an element or not

Let's assume this statement is true, then we move to liine 3

Line 3 determines the new position of rear.

Let's assume that n = 6 and current rear is 4.

Line 3 will produce the following result;

rear = (rear + 1) % n;

rear = (4 + 1)% 6

rear = 5%6

rear = 5.

So, the new element will be added at the 5th index

Then line 4 will be equivalent to:

elements[rear] = element;

elements[5] = element;

Meaning that the new element will be enqueued at the 5th index.

Lastly, rear is incremented by 1 to denote the new position where new element can be added.

6 0
3 years ago
In case of rdt 3.0 Stop and Wait, suppose we send a packet of 1KB through 1 Gbps link and RTT=20 msec. Find the sender utilizati
Mamont248 [21]

Answer:

a. Utilization =  0.00039

b. Throughput = 50Kbps

Explanation:

<u>Given Data:</u>

Packet Size = L = 1kb = 8000 bits

Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

RTT = 20 msec

<u>To Find </u>

a. Sender Utilization = ?

b. Throughput = ?

Solution

a. Sender Utilization

<u>As Given </u>

Packet Size =  L = 8000 bits

Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec

Utilization =  Transmission Time / RTT + Transmission Time

                 =  8 micro-sec/ 20 msec + 8 micro-sec

                 =  0.008 sec/ 20.008 sec

Utilization =  0.00039

b. Throughput

<u>As Given </u>

Packet Size = 1kb

RTT = 20ms = 20/100 sec = 0.02 sec

So,

Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps

So, the system has 50 kbps throughput over  1 Gbps Link.

5 0
2 years ago
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