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What should I write in address line 1 and address line 2 in us lottery form?
Address 1) House number and street name Address 2) City or town Address 3) State and/or county Address 4) Zip or postal code Address 5) Country [optional]
How many numbers do a ip number have?
An IPv4 address has four groups of digits from 0 to 255 (256 total in one group which is equal 2^8), so it could vary from 0.0.0.0 to 255.255.255.255 Total length of IPv4 address space is 2^32 = 4,294,967,296 total addresses. An IPv6 address has eight groups of four hexadecimal digits from 0 to ffff (65,536 total in each group which is equal 2^16), so it could vary from 0:0:0:0:0:0:0:0 to ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff Total length of IPv6 address space is 2^128 = 340,282,366,920,938,463,463,374,607,431,768,211,456 total addresses, which means it's more than 4,8 * 10^28 of addresses for each of 7 billion people on the planet.
How do you find how many proper divisors 7920 has?
To find the number of divisors of a number, factor it into primes in power format; then the number of factors the number has is the product of the powers of the primes each incremented by 1. To find the proper divisors subtract one from thin number. The prime factorization of 7920 is 2^4 x 3^2 x 5 x 11, so it has (4 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 5 × 3 × 2 × 2 = 60 factors. As this includes the number itself and the proper factors of a number exclude the number, 7920 has 60 - 1 = 59 proper factors (divisors).
How many address lines and data lines are required from 256K 16 memory system?
In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.
A microprocessor has a data bus with 64 lines 32 lines address bus.what will be the Maximum number of bits stored in memory?
2^32 is amount of blocks that address bus could locate. and each blocks is 64bit because data bus has 64 lines. then maximum number of bits stored in memory is (2^32)*64 bit. By: Mohammad Saghafi Email: mohammads1364@yahoo.com
