Answer:
1st question=no
2nd question=consumer will never be satisfied.Unless the technology is super strong that nothing can stop it
3rd question=almost everything,first they build technology because they want humans life to be easy.
wish it helps
Answer:
The recommended type of trunk for interoperability is an IEEE 802.1Q trunk.
Explanation:
IEEE 802.1Q is an open industry standard and is the most commonly implemented on layer 2 switches of different vendors, assuring interoperability.
Commonly know as <em>dot1q</em>, is the networking standard that supports virtual LANs (VLANs) on an IEEE 802.3 Ethernet network. It specifies the mechanisms for tagging frames with VLAN data and the procedures for handling this data by switches and bridges.
D. Field data because that it a report of what happened over the week that a manager can reflect on for possible changes.
Answer:
A project is successful when it achieves its objectives and meets or exceeds the expectations of the stakeholders. But who are the stakeholders? Stakeholders are individuals who either care about or have a vested interest in your project. They are the people who are actively involved with the work of the project or have something to either gain or lose as a result of the project. When you manage a project to add lanes to a highway, motorists are stakeholders who are positively affected. However, you negatively affect residents who live near the highway during your project (with construction noise) and after your project with far-reaching implications (increased traffic noise and pollution).
Explanation:
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
