Answer:
(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010
(b)
(1 – p) = 0.010
(c)
E(x) = 25000 x 0.010
= 259.2
Explanation has given below.
Step-by-step explanation:
Solution:
(a) Probability that a triplet is decoded.
2 out of three
P = 0.94, n = 3
m= no of correct bits
m bit (3, 0.94)
At p(m≤1) = B (1; 3, 0.94)
= 0.010
(b) Using your answer to part (a),
(1 – p) = 0.010
Error for 1 bit transmission error.
(c) How does your answer to part (a) change if each bit is repeated five times (instead of three?
P( m ≤ 2 )
L = Bit (5, 0.94)
= B (2; 5, 0.94)
= 0.002
(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?
Given:
h = 25000
Bits are switched during transmission between two computers = 6% = 0.06
m = Bit (25000, 0.06)
E(m) = np
= 25000 x 0.06
= 1500
m = Bit (25000, 0.01)
E(m) = 25000 x 0.010
= 259.2
Answer:
4x+40=120
Step-by-step explanation:
Vertical angles are congruent so you would set them equal to each other.
I think that the answer is A because you are being asked for the tens place. Since you are looking to identify the place value, choice B and C are automatically eliminated because each place value increases by multiplying by ten. 20 and 60 are not multiples of 10. To identify the tens place, you would have to multiply by ten. I thought of this question in the sense of scientific notations.
10^3 is 1000
So we have 2.5 x 1000 which is 2500
10^2 is 100
So we have 3.5 x 100 which is 350
Now we add 2500+350 and the answer is 2,850
