Answer:
The answer to this question can be given as:
code:
class CollegeCourse
{
String Dept;
int CourseNumber;
double Credits;
double fee;
}
Explanation:
We all know that class is a collection of data members and member functions. In the above code we use the following syntax for class declaration that can be given as:
Syntax of class declaration :
class class_name
{
data member & member function
}
In the above code firstly we declare a class that is CollegeCourse. In this class, we define a variable that name and datatype is already given in the question that is String Dept, int CourseNumber, double Credits, double fee. In the variables first variable data type is string. It is used for store string value like ('us','xxx','aaa').The second variable datatype is an integer. It is used for store integer value like (1,23,5,56). The third and fourth variable datatype is the same that is double. This data type is used to store the floating-point value.
Explanation:
In an educational setting, standards-based assessment is assessment that relies on the evaluation of student understanding with respect to agreed-upon standards, also known as "outcomes". The standards set the criteria for the successful demonstration of the understanding of a concept or skill -WIKIPEDIA
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
i need help on that too :cc
Answer:
i think its 29
- add all the numbers up and divide by how many numbers there are.
