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jekas [21]
3 years ago
11

Collisions and Ethernet errors typically occur within the first _____ bytes of an Ethernet frame, which is why fragment-free swi

tching catches most Ethernet errors.
Computers and Technology
1 answer:
Slav-nsk [51]3 years ago
6 0

Answer:

64

Explanation:

These errors occur when two devices on the same Ethernet try to send their data at the same time. this is the reason that these errors occurs in first 64 bytes in the start of transmission data over Ethernet.

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Which two options are negotiated via ncp during the establishment of a ppp connection that will use the ipv4 network layer proto
liberstina [14]
<span>NCP are protocols that are used for establishing PPP sessions. </span>During the establishment of a PPP connection that will use the IPv4 network layer protocol the following options are negotiated: the algorithm to compress TCP and IP headers and the IPv4 address used for routing IP over the PPP link.

6 0
3 years ago
Given a int variable named callsReceived and another int variable named operatorsOnCall write the necessary code to read values
erik [133]

Answer:

The code to this question can be given as:

Code:

int callsReceived,operatorsOnCall;    //define variable  as integer

Scanner ob= new Scanner(System.in);    

//create object of scanner class for user input

System.out.println("Insert the value of callsReceived");  //print message.

callsReceived = ob.nextInt();    //input value.

System.out.println("Insert the value of operatorsOnCall"); //print message.  

operatorsOnCall = ob.nextInt();    //input value.

if (operatorsOnCall == 0)  //check number  

{

System.out.println("INVALID");   //print message.

}

else

{

System.out.println(callsReceived/operatorsOnCall);   //print value.

}

Explanation:

In the above code firstly we define 2 integer variable that name is already given in the question. Then we create the scanner class object for taking user input. Then we print the message for input first and second value from the user. then we use conditional statement. If the second variable that is  operatorsOnCall is equal to 0. So It print INVALID. else it divide the value and print it.

3 0
3 years ago
Which of the following best explains how algorithms that run on a computer can be used to solve problems?
timama [110]

The statement which best explains how algorithms running on a computer can be used to solve problems is; D. Some problems cannot be solved by an algorithm.

<h3>What is an algorithm?</h3>

An algorithm is simply a standard formula or procedures which is made up of a set of finite steps and instructions that must be executed on a computer, so as to proffer solutions to a problem or solve a problem under appropriate conditions.

However, it should be noted that it is not all problems that can be solved by an algorithm, especially because the required parameters and appropriate conditions are not feasible or met.

Read more on algorithm here: brainly.com/question/24793921

7 0
2 years ago
From your computer you are able to establish a telnet connection to a remote host but a traceroute to the host IP address result
Tanya [424]

Mostly firewalls may protect trace route to the host IP address.

<u>Explanation:</u>

When telnet has established the connection to a remote host. Then kindly check the host IP address and their gateway settings.  

If any resident firewall such as antivirus is stopping the ping method or gate way firewall is protecting both ends during trace route. Once the connection is established don’t worry about trace route on host IP  address.

Kindly check where the disconnections happens during the trace route.  in case it happens in between then they're stability on the connection issue which should be solved immediately.

8 0
3 years ago
Create a function that will perform linear interpolation from a set of measured data stored in a list or array. The function sho
wel

Answer:

This question is incomplete, here is the complete question:

Python

a) You should create a function that will perform linear interpolation from a set of measured data. The function should take as input a list of values at which samples were taken, and then another list giving the measurements (you can assume each measurement is a single value) at those values. It should also take in a query value, and should give the best estimate it can of the value at that query. Be sure to handle values that are outside of the range, by extrapolating. You should write a program that allows you to test your function by reading the lists from a file where each line of the file is a pair of numbers separated by spaces: the value where the sample was taken, and the measurement at that value. Your program should ask the user for the name of the file and for a query value. Important: The two lists will correspond to each other: i.e. for the i-th value in the first list, the measurement will be the i-th element of the second list (these are called parallel lists or arrays). But, you should not assume that the input values are in increasing/decreasing order. That is, the values in the first list can be in any random ordering, not necessarily from smallest to largest or largest to smallest. You will have to account for this in your program, and there is more than one way to do so. You should discuss what options you can think of to handle the data arriving in any order like that, and decide what you think the best option for handling it is.

Explanation:

from __future__ import division

from cStringIO import StringIO

import numpy as np

from scipy.interpolate import RectBivariateSpline

np.set_printoptions( 1, threshold=100, edgeitems=10, suppress=True )

   # a file inline, for testing --

myfile = StringIO( """

# T P1 P2 P3 P4

0,   80,100,150,200

75, 400,405,415,430

100, 450,456,467,483

150, 500,507,519,536

200, 550,558,571,589

""" )

   # file -> numpy array --

   # (all rows must have the same number of columns)

TPU = np.loadtxt( myfile, delimiter="," )

P = TPU[0,1:] # top row

T = TPU[ 1:,0] # left col

U = TPU[1:,1:] # 4 x 4, 400 .. 589

print "T:", T

print "P:", P

print "U:", U

interpolator = RectBivariateSpline( T, P, U, kx=1, ky=1 ) # 1 bilinear, 3 spline

   # try some t, p --

for t, p in (

   (75, 80),

   (75, 200),

   (87.5, 90),

   (200, 80),

   (200, 90),

   ):

   u = interpolator( t, p )

   print "t %5.1f p %5.1f -> u %5.1f" % (t, p, u)

8 0
4 years ago
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