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
1 - 1/5 = 5/5 - 1/5 = 4/5
4/5 - 1/5 = 3/5
3/5 - 1/5 = 2/5
2/5 - 1/5 = 1/5
1/5 - 1/5 = 0
We can subtract 1/5 from 1 a total of 5 times, so 1 ÷ 1/5 = 5.
Answer:
y=135
Step-by-step explanation:
To find the solution, we must put 7 in for x and solve, as shown:
The one that is wrong is. C. If all the x’s are different it is a function
3.84/16 = 0.24 per tangerine
