How many bit strings of length 10 do not contain the substring 00? In other words, how many strings of length 10, consisting only of 1 and 0, in which there are no two consecutive zeros? (In response, write down only the number without spaces.)

  144

Step-by-step explanation:

For a bitstring of length n, there are Fibonacci(n+2) strings containing no two consecutive zeros. This can be seen by constructing the strings starting with n=1.

1-bit strings: 1, 0 -- 2 strings not containing consecutive 0s

2-bit strings: 11, 10, 01 -- 3 strings not containing consecutive 0s

Note that we have added 1 to all the 1-bit strings, and added 0 only to the string ending in 1.

3-bit strings: 111, 110, 101, 011, 010 -- 5 strings not containing consecutive 0s

Note that these 5 strings consist of all (3) of the 2-bit strings with 1 appended, and all (1) of the 2-bit strings ending in 1 with 0 appended. The number that now end in 0 is the number previously ending in 1.

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If (x, y) represents the numbers of n-bit strings ending in (0, 1), then the number of (n+1)-bit strings ending in (0, 1) is (y, x+y). That is, the recursive relation is ...

For n=1 to n=10, these pairs are ...

  (1, 1), (1, 2), (2, 3), (3, 5), (5, 8), (8, 13), (13, 21), (21, 34), (34, 55), (55, 89)

The sequence of b[n] values is ...

  2, 3, 5, 8, 13, 21, 34, 55, 89, 144

which are the n=3 to n=12 numbers from the Fibonacci sequence.

That is, there will be Fibonacci(12) = 144 10-bit strings with no consecutive 0s.

SI=(P*R*T)/100

P=2000

R=1.5

T=6

SI=(2000*1.5*6)/100

=(2000*9)/100

=180

Neil will earn interest of 180

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

The solution is given in the image below

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The answer is in the image 

Area of rectangle= L*b

here,

L=12.4 cm

b= 8.8 cm

A= L*b

= 12.4 * 8.8 = 109.12 cm^2

Let the father's age be x and son's be y

10 years before-

Father age=x-10

sons age=y-10

Given,

x-10=10(y-10)

x-10=10y-100

Given present age of father=40

therefore,

x=40

40-10=10y-100

10y-100=30

10y=130

y=130/10

y=13

Therefore present age of son=13years

The solution is given below