Answer:
The answer to this question can be given as:
Method Definition:
public static int secondHalfLetters(String letters)
{
int count = 0;
for (int i = 0; i < letters.length(); i++)
{
if (Character.toLowerCase(letters.charAt(i)) >= 'n')
{
count++;
}
}
return count;
}
Explanation:
In the above method definition first we declare the method that name is already given in the question that is secondHalfLetters() function. In this function we pass a string parameter that is letters.In the function we declare an integer variable count.Then we declare the for loop. In this loop we declare the conditional statement in the if block we convert character into lowercase and check that value is greater the equal to n. If this condition is true then it count increase by 1. In the lass we return count value.
The domain name<span> part of an email address has to conform to strict guidelines: it must match the requirements for a </span>hostname<span>, consisting of letters, digits, hyphens and dots. </span>
Answer:
dncdcnkjjdjedncd
Explanation:
Not securely held or in position *
1 point
Brink
Lope
Precarious
Cajole
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb