Answer:
Step-by-step explanation:
well i don't know what this means because it says which so that means there has to be a multiply choice. so i cant really help with that
Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
Considering that the addresses of memory locations are specified in hexadecimal.
a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ ) = 65536 memory locations
b) The range of hex addresses in a microcomputer with 4096 memory locations is ; 4095
<u>applying the given data </u>:
a) first step : convert FFFF₁₆ to decimal ( note F₁₆ = 15 decimal )
( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )
= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )
= 61440 + 3840 + 240 + 15 = 65535
∴ the memory locations from 0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations
b) The range of hex addresses with a memory location of 4096
= 0000₁₆ to FFFF₁₆ = 0 to 4096
∴ the range = 4095
Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.
Learn more : brainly.com/question/18993173
