A 1Mx64-bit memory chip contains <1,048,576> , or <2^20> 64-bit Lwords. You could express this either in terms of MegaBits or MegaBytes. In Megabits, 64Mbits because 1Mx64 = 64Mbits and divide by 8 to get the answer in terms of bytes. You divide by 8 because there are 8 bits in 1 byte. So a 1Mx64-bit memory contains 8 Mbytes, or 64 Mbits. If the width had been 1Mx32-bit, you would have 32Mbits, and 4 Mbytes.
Registers represent the number of memory locations. A 2K memory chip has 2x1024=2048 memory locations. Hence there are 2048 registers in a 2K memory.
A java.util.Date object will take about 32 bytes in memory.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
There are 8 bytes in a double
how many bytes are there in a 64-bit machine? Another Answer: It takes 8 bytes to store a 64 bit number.
To construct a RAM memory system of 2 bytes using 1288 RAM memory chips, first, determine the capacity of a single 1288 chip. Each 1288 chip typically has 128 bits (or 16 bytes). Therefore, to achieve 2 bytes, you would need 2 bytes / 16 bytes per chip = 0.125 chips. Since you cannot use a fraction of a chip, you would need at least 1 chip to construct the RAM memory system of 2 bytes.
To determine how many memory chips are required to make 2048 bytes of memory, first convert bytes to bits: 2048 bytes × 8 bits/byte = 16,384 bits. Next, divide the total bits needed by the size of one memory chip: 16,384 bits ÷ 10,244 bits/chip ≈ 1.60. Since you cannot have a fraction of a chip, you would need 2 memory chips to meet or exceed the requirement.
16
a) To provide a memory capacity of 4096 bytes using 256x8 RAM chips, you need 4096 bytes / 256 bytes per chip = 16 chips. b) Each memory address for the 256 locations in a chip will require 8 bits (since 2^8 = 256). Therefore, each chip will require 8 address lines to select one of the 256 locations.
2147483648 bytes
4 bytes
A kilobyte (or a k) is 1024 bytes, so 16k is 16*1024 bytes or 16384 bytes.
1024 mega bytes
its simple and easy, one megabyte is equal to 1 million bytes so do the math bytes megabytes 1,000,000 = 1 10,000,000 = 10 100,000,000 = 100 250,000,000 = 250 and just check how many chips the memory have and do the math ...had to edit this answer substantially for inaccuracies, I think what he was trying to get at is divide the total DIMM memory by the number of chips on the module(ie: 256/8 = 32 megabytes per chip, or 32,000,000 bytes). Then to convert to bits you would need to multiply by 8 again (8 bits in a byte). Mind you this is for how much memory is on each chip and I don't think that is what this question is asking. I think http://wiki.answers.com/Q/How_many_bits_of_information_does_a_single_cell_of_a_memory_chip_on_a_256MB_PC2700_DDR_module_hold would be a more reliable answer.
Registers represent the number of memory locations. A 2K memory chip has 2x1024=2048 memory locations. Hence there are 2048 registers in a 2K memory.
Depends on how many bytes it has
4,096 In computer memory, "kilo" refers to 2 to the 10th power, or 1024, rather than to 1000.