Answer:
1. PANs uses Star Topology.
2. LAN uses four topology (Bus, Ring, Star and Tree).
3. MAN uses Star Topology.
4. WAN uses Bus topology.
Explanation:
1. PAN is the personnel area network, in which different personnel devices of a person are connect to each other with the help of the central computer with the help of Bluetooth, WiFi or some other medium. The central computer will work like a hub and all the devices are directly connected to the central PC. It is same as the ring topology where all the devices are connected to the central PC. So we can say that, PANs use star topology.
2. LAN is the local area network that has been established with in the premises of the organization. In this type of network, four typologies involve to complete the network connection. First is star topology, that is used to connect all the devices with the switches. Then Bus topology is used to connect all the switches with the single main cable. Ring topology is involved to connect all the switches with each other. Tree topology is used to connect different block if the organization in the form of branch to connect the central router or switch.
3. MAN is the Metropolitan Area Network which is comprise of different LANs. All the LANs are connected to the Router in the form of Star topology.
4. WAN is the wide area network. In WAN different MANs are connected to the network through single cable. This type of network uses Bus topology.
Windows Movie Maker, iMovie, Final Cut Pro (x), QuickTime?
The answer is 1) JPEG 2) TIFF 3) BMP.
<span>The best option that should be considered when preparing these images to.be used for the web :
</span>1) JPG (jpeg) - Used to compress web images to a small file size without
sacrificing quality
2) TIFF (tiff) - Used for professional print images to save in a uncompressed
file format at high resolution.
3) BMP - Used for either web or print because of the ability to save high
quality images ether compressed or uncompressed.
I think that A could be the correct answer. The others are not as credible as A.
Answer:
Check the explanation
Explanation:
1.
System = 256Byte = 8 bit
Cache = 64B , block size = 16 byte.
A) Direct Mapped Cache:
Block offset = log (Block size) = log 16 = 4bit
Total # of block inside cache = 4.
Therefore index offset = log 4 = 2bit.
Remaining is tag bits.
Therefore tag bits = 8-(4+2) = 8-6 = 2 bits
Tag Index offset Block offset
(2 bits) (2 bits) (4 bits)
Fully associative cache :
In fully associative cache, any of main memory block can be placed anywhere in cache. Therefore index offset =0 bits.
Therefore tag bits = 8-block offset bit= 8-4 =4bits
Tag Block offset
(4 bits) (4 bits)
