Add the following constants in the class definition, before the main method:
private final static float START_COST = 1.5f;
private final static float HIGH_TARIFF = 0.5f;
private final static float LOW_TARIFF = 0.25f;
private final static int FIXED_MINS = 2;
private final static int HIGH_TARIFF_MINS = 10;
Then add this to the main method you already had:
float price = START_COST;
if (x > FIXED_MINS) {
if (x <= HIGH_TARIFF_MINS) {
price += HIGH_TARIFF*(x-FIXED_MINS);
}
else {
price += HIGH_TARIFF*(HIGH_TARIFF_MINS-FIXED_MINS)
+ LOW_TARIFF*(x-HIGH_TARIFF_MINS);
}
}
System.out.printf("A %d minute call costs %.2f", x, price);
Answer:
print("Let's play Silly Sentences!")
print(" ")
name=input("Enter a name: ")
adj1=input("Enter an adjective: ")
adj2=input("Enter an adjective: ")
adv=input("Enter an adverb: ")
fd1=input("Enter a food: ")
fd2=input("Enter another food: ")
noun=input("Enter a noun: ")
place=input("Enter a place: ")
verb=input("Enter a verb: ")
print(" ")
print(name + " was planning a dream vacation to " + place + ".")
print(name + " was especially looking forward to trying the local \ncuisine, including " + adj1 + " " + fd1 + " and " + fd2 + ".")
print(" ")
print(name + " will have to practice the language " + adv + " to \nmake it easier to " + verb + " with people.")
print(" ")
print(name + " has a long list of sights to see, including the\n" + noun + " museum and the " + adj2 + " park.")
Explanation:
Got it right. Might be a longer version, but it worked for me.
Automation. ... It involves taking a machine or software that was taught to do simple repetitive tasks (traditional automation) and teaching it to intuitively adapt or correct its performance based on changing conditions, at speed and scale.
Answer:
code = 010100000001101000101
Explanation:
Steps:
The inequality yields 
, where M = 16. Therefore,
The second step will be to arrange the data bits and check the bits. This will be as follows:
Bit position number Check bits Data Bits
21 10101
20 10100
The bits are checked up to bit position 1
Thus, the code is 010100000001101000101
Answer:-
(10111.001)₂
Explanation:
To convert a decimal number to a binary number we have to constantly divide the decimal number by 2 till the decimal number becomes zero and the binary number is writing the remainders in reverse order of obtaining them on each division.
Hence the binary number is 10111.001
To convert binary to hexa decimal we to make a group 4 binary bits starting from the decimal and moving outwards if the last group is not of 4 then add respective 0's and write the corresponding hexa decimal number.
<u>0001</u> <u>0111</u> . <u>0010</u>
1 7 2
Hence the hexadecimal number is 17.2
