Answer:
0.77978
Explanation:
This is a Poisson distribution problem
Poisson distribution formula is given as
P(X = x) = (e^-λ)(λˣ)/x!
λ = mean = 10 tankers per day
x = variable whose probability is required
The probability that more than 7 tankers arrives in a certain day = 1 - (Probability that 7 or less tankers arrive in a certain day)
P(X > 7) = 1 - P(X ≤ 7)
P(X ≤ x) = Σ (e^-λ)(λˣ)/x! (Summation From 0 to x)
P(X ≤ 7) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + + P(X=6) + P(X=7) + P(X=8)
Computing this,
P(X≤7) = 0.22022
P(X > 7 ) = 1 - P(X≤7) = 1 - 0.22022 = 0.77978
Answer: b) 3.47 nj
Explanation:
Given that;
length l = 5m
radius of inner conductor r = 10cm = 0.1m
radius of outer conductor D = 20cm = 0.2m
current I = 100A = 100×10⁻³ = 0.1
medium between conductor in air u₀ = 4π × 10⁻⁷
Energy in a coaxial cable transmission line is
w = u₀ /2π I² en(b/a)
we substitute
L = 4π × 10⁻⁷ /4π ×10⁻²× 5 en (20/10)
L =3.4657 × 10⁻⁹ J
L = 3.4657 nJ ≈ 3.47 nJ
Answer:
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Answer:
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