5392.7 in scientific notation

Prime factorization is a way of showing which prime numbers (numbers that can only be divided by one and itself) can go into a number. For instance, the prime factorization of 4 would be 2 x 2. Two can only be divided by one and itself, so it is a prime number, and 2 x 2 = 4. For larger numbers, divide by a prime number until the number that is being divided is a prime number. Each divisor will be used as part of the prime factorization. The final dividend will also be used. Add the dividend, order each divisor from least to greatest, and then you're done. Therefore, the prime factorization of 40 would be 2 x 2 x 2 x 5.

3 0

3 years ago

Choose the correct answer below. A. Increasing the level of confidence has no effect on the interval. B. Increasing the level of

murzikaleks [220]

Answer:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

Step-by-step explanation:

Let's assume that we have a parameter of interest $\mu$ who represent for example the true mean for a population. And we can construct a confidence interval in order to estimate this parameter if we know the distribution for the statistic let's say $\bar X$ and for this particular example the confidence interval is given by:

$\bar X \pm ME$

Where ME represent the margin of error for the estimation and this margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

3 0

2 years ago

Find the flux of F=(x^5+y^5+z^5-2x-3y-4z)i+sin(2y)j+4zsin^2(y)k across the surface of the tetrahedron bounded by the coordinate

Tju [1.3M]

Use the divergence theorem.

$\mathbf F(x,y,z)=(x^5+y^5+z^5-2x-3y-4z)\,\mathbf i+\sin2y\,\mathbf j+4z\sin^2y\,\mathbf k$
$\implies(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial(x^5+y^5+z^5-2x-3y-4z)}{\partial x}+\dfrac{\partial(\sin2y)}{\partial y}+\dfrac{\partial(4z\sin^2y)}{\partial z}=5x^4-2+2\cos2y+4\sin^2y$

The flux of $\mathbf F$ across the tetrahedron's surface $S$ is then given by the integral of $\nabla\cdot\mathbf F$ over the interior of the tetrahedron $\mathbf R$ .

$\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\iiint_T\nabla\cdot\mathbf F\,\mathrm dV$
$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=1-x-y}(5x^4-2+2\cos2y+4\sin^2y)\,\mathrm dz\,\mathrm dy\,\mathrm dx$
$=\dfrac1{42}$

4 0

3 years ago

5392.7 in scientific notation​

5392.7 in scientific notation