Is this the entire question?
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Answer:
2.29
Step-by-step explanation:
2.95 plus 1.26 is 4.21, and 6.50 minus 4.21 is 2.29.
Answer:
u will not fail i promise
Step-by-step explanation:
and the answer is the first becz we have to see the difference in arthimatic terms
In this case, the answer would be 10(a+b). We are multiplying 10 * a + b. Hope this helps you and satisfies your query. Feel free to ask more questions.