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koban [17]
3 years ago
5

For security reasons a network administrator needs to ensure that local computers cannot ping each other. which settings can acc

omplish this task?
Computers and Technology
1 answer:
liq [111]3 years ago
6 0
<span>The administrator of the network can access the admin settings on the network's primary DNS servers. From there, there are hundreds of various settings and changes to make. There is one, however, that is able to disable pings. This will prevent computers from responding to ping requests. This can help to ensure internal security and reduce internal network traffic.</span>
You might be interested in
A search engine is aprogram to search......<br>​
coldgirl [10]
Yes it is. Why are you asking?
4 0
4 years ago
Zoom Vacuum, a family-owned manufacturer of high-end vacuums, has grown exponentially over the last few years. However, the comp
Sati [7]

Answer:

The best advice for Zoom Vacuum is  to start with a Transaction Processing System(TPS). This system will process all the day to day transactions of the system. It is like a real time system where users engage with the TPS and generate, retrieve and update the data. it would be very helpful in lowering the cost of production and at the same time manufacture and produce standard quality products.

Although, TPS alone would not be sufficient enough. so Management Information System(MIS) would be needed, that can get the data from the TPS and process it to get vital information.

This MIS will bring about information regarding the sales and inventory data about the current and well as previous years.

With the combination of TPS as well as MIS is a minimum requirement for a Zoom Vacuum to become successful in the market.

Explanation:

Solution

When we look at the description of the Zoom Vacuum manufacturing company closely, we see that the business is currently a small scale business which is trying to become a small to mid scale business. This claim is supported by the fact that there is only one manufacturing plant and three warehouses. And here, we also need to have the knowledge of the fact that small business major aim is to keep the cost low and satisfy the customers by making good products. So we need to suggest what information system is best suited for small scale business enterprises.

The best recommendation would be to start with a Transaction Processing System(TPS). This system will process all the day to day transactions of the system. It is like a real time system where users interact with the TPS and generate, retrieve and modify the data. This TPS would be very helpful in lowering the cost of production and at the same time manufacture and produce standard quality products . Also TPS can help in communicating with other vendors.

But TPS along would not be sufficient. Also Management Information System(MIS) would be required that can get the data from the TPS and process it to get vital information. This MIS will help in generating information regarding the sales and inventory data about the current and well as previous years. Also MIS can create many types of graphical reports which help in tactical planning of the enterprise.

So a combination of TPS as well as MIS is a minimum requirement for a Zoom Vacuum to become successful in the market.

7 0
4 years ago
4. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switchednetwork (VoIP). Host A conve
mina [271]

Answer:

The time elapsed is 0.017224 s

Solution:

As per the question:

Analog signal to digital bit stream conversion by Host A =64 kbps

Byte packets obtained by Host A = 56 bytes

Rate of transmission = 2 Mbps

Propagation delay = 10 ms = 0.01 s

Now,

Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.

Time taken to generate and convert all the bits into digital signal is given by;

t = \frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}

t = \frac{56\times 8}{64\times 10^{3}}          (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}

t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s

Now, the time elapse between the bit creation and its decoding is given by:

t + t'  + propagation delay= 0.007 + 2.24\times 10^{- 4} s + 0.01= 0.017224 s

8 0
3 years ago
Which statement best describes embedded material and hyperlinks? If the original file is moved or deleted, embedded material and
never [62]
<span>The second statement is the best description. If the original file is moved, hyperlinks will still take you to that file's page address, but the file will no longer be there to display. And embedded material won't be able to function.</span>
3 0
3 years ago
Half of the integers stored in the array data are positive, and half are negative. Determine the
jonny [76]

Answer:

Check the explanation

Explanation:

Below is the approx assembly code for above `for loop` :-

1). mov ecx, 0

2). loop_start :

3).    cmp ecx, ARRAY_LENGTH

4).    jge loop_end

5).    mv temp_a, array[ecx]

6).    cmp temp_a, 0

7).    branch on nge

8).        mv array[ecx], temp_a*2

9).   add ecx, 1

10).   jmp loop_start

11). loop_end :

Assumptions :-

*ARRAY_LENGTH is register with value 1000000

*temp_a is a register

Frequency of statements :-

1) will be executed one time

3) will be executed 1000000 times

4) will be executed 1000000 times

5) will be executed 1000000 times

6) will be executed 1000000 times

7) `nge` will be executed 1000000 times, branch will be executed 500000 times

8) will be executed 500000 times

9) will be executed 1000000 times

10) will be executed 1000000 times

Cost of statements :-

1) 10 ns

3) 10ns + 10ns + 10ns [for two register accesses and one cmp]

4) 10ns [for jge ]

5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]

6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]

7) 10ns for nge, 10ns for branch

8) 30ns + 110ns + 10ns

10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],

110ns for array[ecx],

10ns for mv

9) 10ns for add, 10ns for `ecx` register access

10) 10ns for jmp

Total time taken = sum of (frequency x cost) of all the statements

1) 10*1

3) 30 * 1000000

4) 10 * 1000000

5) 120 * 1000000

6) 20 * 1000000

7) (10 * 500000) + (10 * 1000000)

8) 150 * 500000

9) 20 * 1000000

10) 10 * 1000000

Sum up all the above costs, you will get the answer.

It will equate to 0.175 seconds

7 0
3 years ago
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