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:
A domestic air carrier airplane lands at an intermediate airport at 1815Z. The latest time it may depart without a specific authorization from an aircraft dispatcher is <u>1915Z (1 hour)</u>.
Explanation:
Under the domestic operations, an airplane landed on intermediate airport can remain their for not more than one hour so the time would me 1 hour.
Here the time represented by 1815Z and 1915Z is in Zulu Time Zone as depicted from letter "Z". The first two digits represent the hour (0-24) and the next two represent the minutes (0-59).
- Here the landing time is 6:15 pm while departing time is after one hour that is 7:15 pm (1915Z).
i hope it will help you.
Answer: for 5 its b and for 6 i may be wrong but i think its also b
Explanation:
