Which of the following is true when describing a unique local address?

The IEEE 802.11b standard provides 3 non-overlapping channels.

A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.

Bridges break up collision domains, which allow more bandwidth for users.

The main advantage of a layered model is that it can allow application developers to change aspects of a program in just one layer of the layer model's specifications.

Advantages of using the OSI layered model include, but are not limited to, the following:

It divides the network communication process into smaller and simpler components, thus aiding compo- nent development, design, and troubleshooting;

it allows multiple-vendor development through standardization of network components;

it encourages industry standardization by defining what functions occur at each layer of the model;

it allows various types of network hardware and software to communicate;

Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.

Although answer D almost seems right, it is not; the mask is the mask used on the remote network, not the source network. Since there is no number at the end of the static route, it is using the default administrative distance of 1.

All Cisco routers have a default configuration register setting of 0x2102, which tells the router to load the IOS from flash memory and the configuration from NVRAM.

No DR is assigned on any type of point-to-point link. No DR/BDR is assigned on the NBMA point-to-multipoint due to the hub/spoke topology. DR and BDR are elected on broadcast and non-broadcast multi-access networks. Frame Relay is a non-Broadcast Multi-Access (NBMA) network by default.

To see console messages through your Telnet session, you must enter the terminal monitor command.

A /28 is a 255.255.255.240 mask. Let's count to the ninth subnet (we need to find the broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting at 16 (remember, the question stated that we will not use subnet zero, so we start at 16, not 0), 16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128 and the next subnet is 144, so our broadcast address of the 128 subnet is 143. This makes the host range 129-142. 142 is the last valid host.