Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
Answer:
Pseudo CODE
a)
n= Input “Enter 5 integer value”
b)
sum=0.0
For loop with i ranging from 0 - 5
Inside loop sum=n[i]+sum
Outside loop avg= sum/5
Print avg
c)
small=n[0] # assume the first number in the list is smallest
large= n[0] # assume the first number in the list is largest
For loop with i ranging from 0 - 5
Inside loop if n[i]<small #if any another number is smaller than small(variable)
Inside if Then small=n[i]
Inside loop if n[i]>large # if any another number is larger than large(variable)
Inside if then large=n[i]
Print small
Print large
d)
print avg
print small
print large
Answer: A :is concerned with defending users’ freedom of use
C:makes source code available for editing
Explanation:
Answer:
Explanation:
The following code is written in Java. It is hard to fully create the code without the rest of the needed code including the T class and the Measurable interface. Regardless the following code can be implemented if you have that code available.
public static T minmax(ArrayList<T> mylist) {
T min = new T();
T max = new T();
for (int x = 0; x < mylist.size(); x++) {
if (mylist.get(x) > max) {
max = mylist.get(x);
} else if (mylist.get(x) < min) {
min = mylist.get(x);
}
}
return (min, max);
}
