Answer: To detect and correct errors, additional bits are added to the data bits at the time of transmission. The additional bits are called parity bits. They allow detection or correction of errors. The data bits along with the parity bits form a code word.
Explanation:
Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
Answer: a. 0.4667
b. 0.4667 and C 0.0667
Explanation:
Given Data:
N = population size (10)
n = random selection (2)
r = number of observations = 7
Therefore
f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )
When y = 1
f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 2
f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 0
f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )
= 1 / 15
= 0.0667
Answer:
Under no circumstances
Explanation:
I'm not 100% sure why, but I remember hearing that you're not suposed to go over the speed limit no matter what
