Answer:
Explanation:
From the given information:
in binary is:
![a_{n-1}\times 2^{n-1} + a_{n-2}}\times 2^{n-2}+ ...+a_o](https://tex.z-dn.net/?f=a_%7Bn-1%7D%5Ctimes%202%5E%7Bn-1%7D%20%20%2B%20a_%7Bn-2%7D%7D%5Ctimes%202%5E%7Bn-2%7D%2B%20...%2Ba_o)
So, the largest number posses all
nonzero, however, the smallest number has
all zero.
∴
The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times
i.e.
![(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}](https://tex.z-dn.net/?f=%2811111111...1%29_2%20%3D%20%28%201%20%5Ctimes%202%5E%7Bn-1%7D%20%2B%201%5Ctimes%202%5E%7Bn-2%7D%20%2B%20...%20%2B%201%20%29_%7B10%7D)
![= \dfrac{1(2^n-1)}{2-1}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B1%282%5En-1%29%7D%7B2-1%7D)
![\mathbf{=2^n -1}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D2%5En%20-1%7D)
![(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}](https://tex.z-dn.net/?f=%281000...0%29_2%20%3D%20%281%20%5Ctimes%202%5E%7Bn-1%7D%20%2B%200%20%5Ctimes%202%5E%7Bn-2%7D%20%2B%200%20%5Ctimes%202%5E%7Bn-3%7D%20%2B%20...%20%2B%200%29_%7B10%7D)
![\mathbf {= 2 ^{n-1}}](https://tex.z-dn.net/?f=%5Cmathbf%20%7B%3D%202%20%5E%7Bn-1%7D%7D)
Hence, the smallest value is
and the largest value is ![\mathbf{2^{n}-1}](https://tex.z-dn.net/?f=%5Cmathbf%7B2%5E%7Bn%7D-1%7D)
Resolution is the correct option
Have a great day
Answer:
True
Explanation:
In this pseudocode the programmer defines 3 variables, x=1, y=2 and z=3, then he creates a conditional that states that if any of the previously defined variables has the value of 1 then a True will be displayed, on the other hand, if none of the variables has the value of 1 then a False will be displayed. In this case x=1, so it will display a True.
I'm almost certain it's D. I hope I'm right
Language is the primary code humans use to communicate.