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2 years ago

Motor oil is responsible for

Engineering

1 answer:

Lelechka [254]2 years ago

5 0

Answer:

lubricating all moving parts in the engine

Explanation:

like the pistons, pushrods, and the crank

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A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t

Ganezh [65]

Answer:

a) $F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

b) $P(10 < X$

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that $X\sim Exp(\lambda)$

And we know the probability denisty function for x given by:

$f(x) = \lambda e^{-\lambda x} , x\geq 0$

In order to find the cdf we need to do the following integral:

$F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

Part b

Assuming that $X \sim Exp(\lambda =0.1)$ , then the density function is given by:

$f(x) = 0.1 e^{-0.1 x} dx , x\geq 0$

And for this case we want this probability:

$P(10 < X$

And evaluating the integral we got:

$P(10 < X$

4 0

3 years ago

A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo

Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, $e_{min}$ = 0.51

Void ratio in the loosest state, $e_{max}$ = 0.87

Now,

Dry density, $\gamma_d=\frac{\gamma_t}{1+w}$

$=\frac{18}{1+0.05}$

= 17.14 kN/m³

Also,

$\gamma_d=\frac{G\gamma_w}{1+e}$

here, G = Specific gravity = 2.7 for sand

$17.14=\frac{2.7\times9.81}{1+e}$

e = 0.545

Relative density = $\frac{e_{max}-e}{e_{max}-e_{min}}$

= $\frac{0.87-0.545}{0.87-0.51}$

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

S = 0.2477

S = 0.2477 × 100% = 24.77%

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3 years ago

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