This is the most likely cause of preventing the users from accessing the application server because a VLAN is a logical segmentation of a network that isolates traffic based on certain criteria. If the human resources department and the application server are on different VLANs, they will not be able to communicate with each other unless there is a router or a switch that can route between VLANs. Reference: https://www.cisco.com/c/en/us/support/docs/lan-switching/inter-vlan-routing/41860-howto-L3-intervlanrouting.html

Explanation
These are two methods that can prevent a user from installing an OS on a server if properly configured. A bootloader password is a password that protects the bootloader from unauthorized access or modification. The bootloader is a program that loads the operating system into memory when the system boots up. If a user does not know the bootloader password, they cannot change the boot order or boot from another device such as a CD-ROM or USB drive that contains an OS installation media. A BIOS/UEFI password is a password that protects the BIOS (Basic Input Output System) or UEFI (Unified Extensible Firmware Interface) from unauthorized access or modification. The BIOS or UEFI is a firmware that initializes and configures the hardware components of the system before loading

To calculate the minimum number of disks needed for this application, we need to consider the RAID levels and their disk requirements. RAID 1 requires a minimum of two disks and provides mirroring, which means that data is duplicated on both disks. RAID 5 requires a minimum of three disks and provides striping with parity, which means that data is distributed across all disks with one disk storing parity information for error correction. RAID 5 can tolerate one disk failure without losing data. To create a 10GB RAID 1 array for log files, we need two 10GB disks. To create a 20GB RAID 5 array for data files, we need four 10GB disks (three for data and one for parity). To create a 20GB RAID 5 array for the operating system, we need another four 10GB disks (three for data and one for parity). Therefore, the total number of disks needed is 2 + 4 + 4 = 10. However, since we can use different RAID levels for different partitions on the same disk, we can optimize the disk usage by using only eight disks as follows: Disk 1: 10GB RAID 1 (log files) + 10GB RAID 5 (data files) Disk 2: 10GB RAID 1 (log files) + 10GB RAID 5 (data files) Disk 3: 10GB RAID 5 (data files) + 10GB RAID 5 (OS) Disk 4: 10GB RAID 5 (data files) + 10GB RAID 5 (OS) Disk 5: 10GB RAID 5 (parity for data files) + 10GB RAID 5 (OS) Disk 6: 10GB RAID 5 (OS) + unused space Disk 7: 10GB RAID 5 (parity for OS) + unused space Disk 8: unused space Reference: https://en.wikipedia.org/wiki/Standard_RAID_levels

One thing that stood out to me from a context perspective, C is the only private IP address. Could only assume this was referring to private IP addressing.