Answer:
hiii
Explanation:
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Answer:
a. 2^6, or 64 opcodes.
b. 2^5, or 32 registers.
c. 2^16, or 0 to 65536.
d. -32768 to 32768.
Explanation:
a. Following that the opcode is 6 bits, it is generally known that the maximum number of opcodes should be 2^6, or 64 opcodes.
b. Now, since the size of the register field is 5 bits, we know that 2^5 registers can be accessed, or 32 registers.
c. Unsigned immediate operand applies to the plus/minus sign of the number. Since unsigned numbers are always positive, the range is from 0 to 2^16, or 0 to 65536.
d. Considering that the signed operands can be negative, they need a 16'th bit for the sign and 15 bits for the number. This means there are 2 * (2^15) numbers, or 2^16. However, the numbers range from -32768 to 32768.
Answer: Please find below the answer along with explanation.
Explanation:
For a given communication channel (for instance, a LAN segment using Ethernet ) the Bandwidth refers to the theoretical maximum data rate that the channel can support, for instance, 100 Mbps in a 100Base T network.
The throughput, instead, refers to the actual data rate achieved in a given communications channel, taking into account the different channel impairments.
For instance, in a LAN segment that uses the original Ethernet 802.3 standard (CSMA/CD), a frequent occurrence of collisions can take down the actual data rate from the theoretical 100 Mbps to a very lower figure, i.e., 5 Mbps.
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