Both of the following are group codes. For each one (i) determine the minimum distance between code words. Using this value determine (ii) the maximum number of errors that can reliably be detected, and (ii) the maximum number of errors that can reliably be corrected.

(a) {(0000000), (1000111), (010 1011), (0011101), (1101100), (101 1010), (0110110), (1110001)}
(b) {(000 000 0000), (100101 1110), (0101101111), (001 0111101), (110011 0001), (101 1100011), (011 101 0010), (111000 1100)}

See attached files

Step-by-step explanation:

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!