Answer:
It will be average (B2:C2)
as it will select all the data between B2 and C2
Answer:
You need exit condition like If, otherwise method will repeat endlessly.
Answer:
The answer is "Fiber Optic Cable".
Explanation:
In the given question options are defined, but in these options, there is some error.
The optical fiber cable is also known as internet wiring, which includes fiber strands inside an isolated housing. It is used to transferring data at a very high speed. This cable is developed for long-range data networking, high-performance connectivity, and multimedia, and other choices are wrong that can be described as follows:
- A modem is used to transmits data over a network, that's why it is wrong.
- Router is used to transport traffic between your computer and the internet, that's why it is wrong.
Loop takes only positive numbers and terminates once it encounters a negative numbers.
Answer and Explanation:
Using javascript:
Var positiveInt= window.prompt("insert positive integer");
While(positiveInt>=0){
Alert("a positive integer");
Var positiveInt= window.prompt("insert positive integer");
}
Or we use the do...while loop
Var positiveInt= window.prompt("insert positive integer");
do{
Var positiveInt= window.prompt("insert positive integer");
}
While (positiveInt>=0);
The above program in javascript checks to see if the input number is negative or positive using the while loop condition and keeps executing with each input until it gets a negative input
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec