WebSep 25, 2011 · 64MB = 67108864 Bytes/4 Bytes = 16777216 words in memory, and each single word can thus be addressed in 24 bits (first word has address 000000000000000000000000 and last has address 111111111111111111111111). Also 2 raised to 24 = 16777216, so 24 bits are needed to address each word in memory. WebSo putting the unknown size of each location aside, and converting 1000000h to decimal i.e. 16,777,216 the answer is either: 16MB (16777216/1024/1024=16) I SUSPECT THIS IS THE ANSWER THE TEACHER IS AFTER 16.777216MB (according to the ISO standards) 16 mebibytes (according to the joint IEC/ISO standard - IEC 80000-13)
IP Addressing and Subnetting Flashcards Quizlet
WebA 2nd level page table contains 16 entries, each entry pointing to a 3rd page table. A 3rd page table contains 256 entries, each entry pointing to a page. The process's address space consists of 16 pages, thus we need 1 third-level page table. Therefore we need 1 entry in a 2nd level page table, and one entry in the first level page table. how to see google maps date
How many IP addresses are in a 16? – KnowledgeBurrow.com
WebJul 24, 2024 · A 16-bit integer can store 216 (or 65,536) distinct values. In an unsigned representation, these values are the integers between 0 and 65,535; using two’s complement, possible values range from −32,768 to 32,767. Hence, a processor with 16-bit memory addresses can directly access 64 KB of byte-addressable memory. WebAddresses Hosts Netmask Amount of a Class C / 30: 4: 2: 255.255.255.252: 1 / 64 / 29: 8: 6: 255.255.255.248: 1 / 32 / 28: 16: 14: 255.255.255.240: 1 / 16 / 27: 32: 30 ... WebJan 9, 2024 · By referring to the above table and finding the row with 16 available addresses, and taking the CIDR and netmask, we find that we will need: Subnet mask of 255.255.255.240 CIDR of 192.168.1.0/28 This network will have: The network address of 192.168.1.0 The broadcast address of 192.168.1.15 14 IP addresses left for hosts to use … how to see google image full hd