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LekaFEV [45]
3 years ago
14

Host A is sending Host B a large le over a TCP connection. Assume Host B has no data to send Host A. Host B will not send acknow

ledgments to Host A because Host B cannot piggyback the acknowledgments on data.a. Trueb. False
Computers and Technology
1 answer:
mestny [16]3 years ago
3 0

Answer:

False.

Explanation:

Piggybacking is an optimization that is used when both sides have to send data to each

other so that the receiver, instead of sending two packets i.e., an ACK and a data packet, it just sends one.

When the receiver (B) does not have any data to send, it will still send an ACK with the sequence number

field containing the next sequence of data it is supposed to send

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The HTML tag for the smallest heading is​ what​
Gre4nikov [31]

Answer:

The HTML <h1> to <h6> tag is used to define headings in an HTML document. <h1> defines largest heading and <h6> defines smallest heading.

Explanation:

5 0
3 years ago
Plz answer me will mark as brainliest ​
victus00 [196]

Answer:

Application

Explanation:

Application software comes in many forms like apps, and even on computers. The software is most common on computers of all kinds while mobile applications are most common on cellular devices.

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3 years ago
Read 2 more answers
Describe what a layer 3 router is and what are its advantages and drawbacks compared to other network devices.
Oduvanchick [21]

Answer:

Did you mean layer 3 switch? Because a router always operates at layer 3

Explanation:

If the answer is yes, then a layer 3 is a switch that combines the functions of a switch and a router. So it is capable of operate layer 2 and layer 3. Some of its benefits are: Support routing between VLAN, decrease network latency because the packets don’t have to make extra hops to go through a router and reduce security management. But they are really expensive and lack of WAN functionality so they are used mostly for large intranet environments.

3 0
4 years ago
Calculate the weighted grade for each student that appears in the data file. You may calculate the grade based on total earned p
lutik1710 [3]

Answer:

I don't know the answer, sorry

6 0
3 years ago
Use semaphore(s) to solve the following problem. There are three processes: P1, P2, and P3. Each process Pi has a segment of cod
Lisa [10]

Answer:

See explaination

Explanation:

Here we will use two semaphore variables to satisfy our goal

We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way

Code:-

s1,s2=1

P1 P2 P3

P(s1)

P(s2)

C1

V(s2) .

P(s2). .

. C2

V(s1) .

P(s1)

. . C3

V(s2)

Explanation:-

The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.

The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process

The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process

Now in the above code:

1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute

So our first condition gets satisfied

2)If C1 is not executed earlier means:-

a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .

b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked

So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.

4 0
3 years ago
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