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2 years ago

Explain why veracity, value, and visualization can also be said to apply to relational databases as well as Big Data.

Engineering

1 answer:

Hatshy [7]2 years ago

5 0

Answer:

Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...

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A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stre

disa [49]

Answer:

0.234

Explanation:

True stress is ratio of instantaneous load acting on instantaneous cross-sectional area

σ = k × (ε)^n

σ = true stress

ε = true strain

k = strength coefficient

n = strain hardening exponent

ε = ( σ / k) ^1/n

take log of both side

log ε = $\frac{1}{n}$ ( log σ - log k)

n = ( log σ - log k) / log ε

n = (log 578 - log 860) / log 0.20 = 0.247

the new ε = ( 600 / 860)^( 1 / 0.247) = 0.234

6 0

2 years ago

Read 2 more answers

Convert 25 mm into in.

mihalych1998 [28]

Answer:

25 mm = 0.984252 inches

Explanation:

Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:

1 mm = 1/25.4 inches

From the question, we have to convert 25 mm into inches

Thus,

25 mm = (1/25.4)*25 inches

So,

$25 mm=\frac{25}{25.4} inches$

Thus, solving we get:

25 mm = 0.984252 inches

4 0

3 years ago

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

What is the magnetic force on a moving electric charge called

crimeas [40]

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field.

4 0

3 years ago

Read 2 more answers

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

2 years ago