Answer:
public static double areaSum(Circle c1, Circle c2){
double c1Radius = c1.getRadius();
double c2Radius = c2.getRadius();
return Math.PI * (Math.pow(c1Radius, 2) + Math.pow(c2Radius, 2));
public static void main(String[] args){
Circle c1 = new Circle(6.0);
Circle c2 = new Circle(8.0);
areaSum(c1,c2);
}
Explanation:
It would say that the person read the message on the picture of the person. Example if you texted your friend at 7.00 and it said that the person read at 7.15. got it now?
Answer:
Post-mortem.
Explanation:
It refer to the discussion or analysis of event like here disaster why it doesn't work like plan based and through whole discussions what we learn so in future avoid such type of issue or mistakes.
Answer:
a) Yes
b) Yes
c) Yes
d) No
e) Yes
f) No
Explanation:
a) All single-bit errors are caught by Cyclic Redundancy Check (CRC) and it produces 100 % of error detection.
b) All double-bit errors for any reasonably long message are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
c) 5 isolated bit errors are not caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit since CRC may not be able to catch all even numbers of isolated bit errors so it is not even.
It produces nearly 100 % of error detection.
d) All even numbers of isolated bit errors may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
e) All burst errors with burst lengths less than or equal to 32 are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
f) A burst error with burst length greater than 32 may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit.
Cyclic Redundancy Check (CRC) does not detect the length of error burst which is greater than or equal to r bits.
Answer:
ella puede usar un cepillo de dientes eléctrico
