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Sedbober [7]
3 years ago
15

A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. Duri

ng the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.
Engineering
1 answer:
notsponge [240]3 years ago
3 0

Answer:

LOS = A

Explanation:

Given all the parameters the level of service as seen from the attached graph

is LOS =  A

<u>To determine the LOS from the attached graph </u>

<em>calculate the trial value of Vp</em>

Vp = V / PHF

     = (100 + 150) / 0.95  =  263 pc/h

since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1

next we will calculate the flow rate

flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]

             Fhr  = 1 / 1.035 = 0.966 ≈ 1

next calculate the real value of Vp

Vp = V / ( PHF * N * Fhr * Fp )

     = ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )

Vp ≈ 126 pc/h/In

Next calculate the density

D = Vp /  S  =  126 / ( 45 * 1.61 )  = 1.74 pc/km/In

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a)- True

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If two statements are inconsistent with each other it means that they are not telling the same, if they are not telling the same it means that only one of them COULD be true, but there is a third option where the two statements are wrong and non statement is telling the true...so:

If we have two statements inconsistent with each other, AT LEAST one of the statements is false.

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In a voltage-divider biased pnp transistor, there is no base current, but the base voltage is approximately correct. The most li
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4 0
4 years ago
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have
drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = \lambda e^{-\lambda t}

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [  -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905

(c) The Poisson distribution is presented as follows;

P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!}  = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!}  = 3.915\times 10^{-12}

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}.

4 0
4 years ago
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