Answer:
Two systems are connected by a router. Both systems and the router have transmission rates of 1,000bps. Each link has a propagation delay of 10ms. Also, it takes router 2ms in order to process the packet (e.g. decide where to forward it). Suppose the first system wants to send a 10,000 bit packet to the second system. How long will it take before receiver system receives the entire packet.
Transmission time for first Router = 10,000 bits / 1000 bps = 10 seconds
Receiving time for seond route r= 10,000 bits / 1000 bps = 10 seconds
Propagation delay = 10ms = .01 seconds x 2 for two delays = .02 seconds
First router 2ms to process = .002 seconds
Add all the times together and we get 20.022 seconds which is the same as or 20 seconds and 22 ms
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Answer:
The correct word for the blank space is: object.
Explanation:
Access Control System (<em>ACS</em>) is a computer structure that allows or restricts users to access to features and information of a server. ACS is used to protect confidential data and to make sure the servers are going to be used by authorized users which diminish malfunctions. One of the basic principles of ACS is letting subjects and objects to interact properly.
Answer:
Following are the output of the given code:
Output:
12 20
Explanation:
Description of the code:
- In the java program code, two integer variable "j and k" is defined, that stores a value, that is "10 and 8", in its respective variable.
- After storing the value it uses the "j and k" variable, in this, it increments the value of j with 2, and in the k variable, it adds the value of j and stores the value in k.
- After incrementing the value, the print method is used that prints the value of "j and k", i.e, "12 and 20".
