I believe the answer would be true
Answer:
The answer is "Option c".
Explanation:
In the given question the device A is connected by 3 wires, contributing all of them, which also includes several connector paths. When all the wires of A are broken down, and if all of this leaves no routes that can be used. Even so, if it is done to E, it's also linked to four different routes. Its solution would've been C because its value will be the MINIMUM.
Open this occurence and open the series is the two options to choose from, from the dialog box that shows when attempting to modify the appointments
