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Alex Ar [27]
3 years ago
5

What is the function of the NOS? Select all that apply.

Computers and Technology
2 answers:
Zigmanuir [339]3 years ago
3 0

Answer:

.network management

Explanation:

pls need brainliest

SIZIF [17.4K]3 years ago
3 0

Answer:

gives access privileges

network management

network security

data protection

Explanation:

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Give an example of a process that may need a flowchart to clarify procedures.<br><br> HELP ASAP!
11111nata11111 [884]

To understand flow of programming logic flow chat technique been used. Once end user starts any learning programming flow chat design been learned and corrected the mistake. After success of flow chat design end user been advice to do programming and tested in computer.

<u>Explanation:</u>

In flow chart design is used for beginner level programming. Some time to make higher level managers to understand this flow chart design is used.

Flow chart has start, procedure, condition, function or procedure and end with stop diagrams.

Flow chart makes end user understand the flow of logics.

Suppose we need to do a process for a procedure, end user has to design with start flow chart symbol and put a procedure flow chart symbol with identification character, then end user has use same identification character and start designing the procedure in the flow chart diagram.

Procedure in flow char design is used to call same procedure number of time, so that repeated coding is avoided.

5 0
3 years ago
What’s the answer for this
attashe74 [19]
I believe the answer is B<span />
3 0
3 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
Pete would like to respond only to users within his organization with an automatic reply. He is configuring the automatic respon
Ahat [919]

Answer:

megan pete

Explanation:lol just  had to

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3 years ago
State 4 &amp; circumstances under which warm<br>booting of a computer may be necessary​
Brums [2.3K]

Answer:

to close an open application

3 0
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