Consider a packet of length 1,500 bytes that begins at end system A and travels over three links to a destination end system B. These three links are connected by two packet switches. The propagation speed on all three links is 2.5 × 108 m/s, the capacity of the three links is 10 Mbps, 5 Mbps and 20 Mbps, respectively, the packet switch processing delay is 1 msec, and the lengths of the three links are 5,000 km, 2,500 km, and 1,000 km, respectively. Suppose there is no queuing delay. What is the end-to-end delay? Describe the components in the end-to-end delay and show intermediate steps.

Perform the following tasks: a. create a class named testclass that holds a single private integer field and a public constructor. the only statement in the constructor is one that displays the message “constructing”. write a main()function that instantiates one object of the testclass. save the file as testclass. cpp in the chapter 08 folder. run the program and observe the results. b. write another main()function that instantiates an array of 10 testclass objects. save the file as test class array. c . run this program and observe the results.

Write a function called checkfactor that takes two arrays of positive numbers, firstnumberrow and secondnumberrow. checkfactor checks if the first entry in firstnumberrow is divisible by the first entry in secondnumberrow, and performs the same operation on the next array elements until all entries have been checked. all the numbers are positive and the number of entries in the arrays are the same. the function should return the identified divisible numbers in two row arrays named firstdivisible and seconddivisible. restrictions: branches or loops should not be used. the code must use the internal mod and logical functions. hint: the mod function should be used to determine if two numbers are divisible. ex: for num1 and num2 if mod(num1,num2) is 0, then the two numbers are divisible. this is matlab

Aconsumer would pay an extra they used the rent to own program to buy the computer, rather than using cash. for all of the items, is the cheapest option over the life of the contract. the most expensive overall option is to use purchase the item.