<span>The answer is Ten. Substituting numbers for letters, available selections are 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. Note that, for example, 112, 121, and 211 are considered as the same selection.
</span>AAA, BBB, CCC, <span>AAB, AAC, </span><span>BBC, </span><span>ABB, ACC, </span><span>CCB & ABC.</span>
The answer is C. Jurassic.
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:
He is working on the outcome
Explanation:
Because this is the outcome if you fail or die and the other objectives are the following
operation: A single objective in a level of the the whole game
obstacles: Trying to stop the player like blocks or walls or traps
objective:The main goal of the game
Answer:
(b) (int)(Math.random() * 101)
Explanation:
Given
--- minimum
--- maximum
Required
Java expression to generate random integer between the above interval
The syntax to do this is:
(int)(Math.random((max - min) + 1) + min)
Substitute the values of max and min
(int)(Math.random((100 - 0) + 1) + 0)
Simplify the expression
(int)(Math.random(100 + 1) + 0)
(int)(Math.random(101) + 0)
(int)(Math.random(101))
Hence, the right option is:
(b) (int)(Math.random() * 101)
