The first one I'll go with 2nd one ...But I'm dumb so hope that helps
Answer:
Step-by-step explanation:
A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your
b2n+32n+3 = 2 (2n+2)22 b2n+12n+1
your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:
1. Either placing a 1 in available n places (because you can't place it at the end)
2. or by placing a zero in available n+1 places.
0 ϵ P
x ϵ P → 1x ϵ P , x1 ϵ P
x' ϵ P,x'' ϵ P → xx'x''ϵ P
Three times ten plus seven times one plus three times one-tenth plus five times one-hundredth
The answer is 0.0016 or 0.0017
Answer:
The Answer is: 75.
Step-by-step explanation:
Set up the equation:
45 / 600 = y / 1000
600y = 45 * 1000
600y = 45000
y = 45,000 / 600 = 75.
