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3 years ago

"explain the difference between a point forecast and an interval forecast"

2 answers:

8 0

In a nutshell, point forecast is used in a more detailed data presentation while interval forecast is used for short data presentation.

Point forecast involves meticolous data presentation which includes graphs which includes and not limited to quadratic formula and the like. It is used to indephtly explain a quadratic system and quadratic system + SV and lin ear system and linear system + SV.

Interval forecast is for shorter presentation. It usually does not involve quadratic formulas and linear formulas, it just shows the inteval growth of the data. To explain it better, interval forecast is like a time lapse.

Arlecino [84]3 years ago

5 0

The distinction between the point forecast is a solitary answer case it will rain at 1 pm. An interim figure depends on vulnerability case it will rain between 12 pm and 2 pm. While the interval forecast is utilized on the grounds that a conjecture is dependable accurately wrong and roughly right.

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0.07 divided by 6.65

Kipish [7]

Answer:

0.01052

Step-by-step explanation:

It would take me a million years to show you how to solve it but its long division

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4 years ago

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Let c be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. find the exact length of c from t

Mandarinka [93]

Parameterize the intersection by setting $x(t)=t$ , so that

$x^2=2y\iff y=\dfrac{x^2}2\implies y(t)=\dfrac{t^2}2$
$3z=xy\iff z=\dfrac{xy}3\implies z(t)=\dfrac{t^3}6$

The length of the path $C$ is then given by the line integral along $C$ ,

$\displaystyle\int_C\mathrm dS$

where $\mathrm dS=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dt}\right)^2}\,\mathrm dt$ . We have

$\dfrac{\mathrm dx}{\mathrm dt}=1$
$\dfrac{\mathrm dy}{\mathrm dt}=t$
$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{t^2}2$

and so the line integral is

$\displaystyle\int_{t=0}^{t=2}\sqrt{1^2+t^2+\dfrac{t^4}4}\,\mathrm dt$

This result is fortuitous, since we can write

$1+t^2+\dfrac{t^4}4=\dfrac14(t^4+4t^2+4)=\dfrac{(t^2+2)^2}4=\left(\dfrac{t^2+2}2\right)^2$

and so the integral reduces to

$\displaystyle\int_{t=0}^{t=2}\frac{t^2+2}2\,\mathrm dt=\dfrac{10}3$

3 0

3 years ago

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. x

miskamm [114]

Answer:

$\frac{784}{15} \pi$

Step-by-step explanation:

According to the given situation, the calculation of volume of the solid is shown below:-

Here we will consider the curves that is

$x = 7y^2, x = 7$

Now, rotating the line for the line x which is equals to 7

$7y^2 = 7\\\\y^2 = 1\\\\ y = \pm1$

So, the inner radio is

7 - 7 = 0

and the outer radius is

$7y^2 - 7\\\\ = 7(y^2 - 1)$

Now, the area of cross section is

$A(y) = \pi(7(y^2 - 1))^2\\\\ = 49\pi(y^4 - 2y^2 + 1)$

The volume is

$V = \int\limits^1_{-1} A(y)dy$

now we will put the values into the above formula

$= \int\limits^1_{-1} 49\pi(y^4 - 2y^2 + 1)dy\\\\ = 49\pi(\frac{y^5}{5} - \frac{2y^3}{3} + y)^{-1}\\\\ = 49\pi(\frac{1}{5} - \frac{2}{3} + 1 + \frac{1}{5} - \frac{2}{3} + 1)\\\\ = 49\pi(2 + \frac{2}{5} - \frac{4}{3} )\\\\ = 49\pi(\frac{30+6-20}{15} )\\\\ = \frac{49\pi}{15} (16)$