Answer:
b. they are located at the intersections of the network of wires and cables that make up the internet.
Explanation:
Given that Router is a computer networking device that operates by transferring data packets between computer networks, while also ensuring traffic organizing tasks on the Internet.
Hence, Routers put packets in the right order after they arrive at their destination by matching the destination IP address of the packet and one of the addresses in the routing table.
Also, Routers check the address on each packet that arrives and decide which way it should go next by ensuring the data packets are summarized for the outgoing interface recorded in the table entry.
Therefore, the correct answer is option B
<span>The IP address 198.162.12.254/24 is a </span>Unicast address.
Answer:
- #include <iostream>
- using namespace std;
- class myinteger {
-
- private:
- int value;
-
- public:
- myinteger(int x){
- value = x;
- }
-
- int getValue(){
- return value;
- }
-
- };
- int main()
- {
- myinteger obj(4);
- cout<< obj.getValue();
- return 0;
- }
Explanation:
Firstly, we use class keyword to create a class named myinteger (Line 5). Define a private scope data field named value in integer type (Line 7 - 8).
Next, we proceed to define a constructor (Line 11 - 13) and a getter method for value (Line 15 -17) in public scope. The constructor will accept one input x and set it to data field, x. The getter method will return the data field x whenever it is called.
We test our class by creating an object from the class (Line 23) by passing a number of 4 as argument. And when we use the object to call the getValue method, 4 will be printed as output.
Answer:
B. Crashing is not possible unless there are multiple critical tasks.
Answer:
Explanation:
Present value is the value in the present of a sum of money, in contrast to some future value it will have when it has been invested at compound interest.
It can be calculated using future value formula below
A = P(1+r/100)^n
where
A = Future value
P = Present value
r = Rate of interest
n = time period
Present Value = Value at Year 1 + Value at Year 4
Calculating Value at Year 1
A = $5,500
r = 8%
n = 1
From A = P(1+r/100)^n ; Make P the subject of formula
P = A ÷ (1 + r/100)^n
Substitute in values
P = $5,500 ÷ (1 + 8/100)^1
P = $5,500 ÷ (1 + 0.08)
P = $5,500/1.08
P = $5092.5925926
P = $5092.59 ----
Present Value at Year 1 = $5092.59
Calculating Value at Year 4
A = $5,000
r = 8%
n = 1
From A = P(1+r/100)^n ; Make P the subject of formula
P = A ÷ (1 + r/100)^n
Substitute in values
P = $5,000 ÷ (1 + 8/100)⁴
P = $5,000 ÷ (1 + 0.08)⁴
P = $5,000/1.08⁴
P = $3675.149263982267
P = $3675.15 ----
Present Value at Year 4 = $3675.15
Present Value = Value at Year 1 + Value at Year 4
Substitute each value
Present Value = $5092.59 + $3675.15
Present Value = $8,767.74
