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Use a computer algebra system to plot the vector field F(x, y, z) = sin(x) cos2(y)i + sin3(y) cos4(z)j + sin5(z) cos6(x)k in the

Olenka [21]

By the divergence theorem,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\iiint_{[0,\pi/2]^3}\mathrm{div}\vec F\cdot\mathrm dV$

We have

$\mathrm{div}\vec F(x,y,z)=\cos x\cos^2y+3\sin^2y\cos y\cos^4z+5\sin^4z\cos z\cos^6x$

Then the volume integral is

$\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}\mathrm{div}\vec F\,\mathrm dx\,\mathrm dy\,\mathrm dz=\boxed{\frac{19\pi^2}{64}}$

7 0

4 years ago

Mia's cat weighs 13 pounds, 7 ounces. about what is the weight in kilograms?

gtnhenbr [62]

There are 0.453592 kilograms in 1 pound.

Multiply 13 by 0.453592 and get 5.8967

Now 0.0283495 kilograms in one ounce.

Multiply 7 by 0.0283495 and get 0.198447.

Now add them and get 6.095147

7 0

3 years ago

Read 2 more answers

An increased number of colleges have been using online resources to research applicants. According to a study from last year, 3

Aneli [31]

Answer:

D. H0: p ≤0.36 H1: p>0.36

$$$(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155$

$Z_p=-0.25516$

D. Reject H0. There is sufficient evidence to support the hypothesis that the proportion of admissions officers who visit an applying students' social networking page has increased in the past year

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the population proportion.

$$$Sample proportion: $\bar P=0.47\\Sample size n=100$\\Significance level \alpha=0.10$\\(Left tail test) Z_{1-\alpha}=Z_{0.90}=-1.28155$\\(Right tail test) Z_{\alpha}=Z_{0.1}=1.28155$\\(Two-tailed test) $Z_{1-\alpha/2}=Z_{0.95}=-1.64485$ and $ Z_{\alpha/2}=Z_{0.05}=1.64485\\\\$

The appropriate hypothesis system for this situation is:

$H_0:\pi_0=0.35\\H_a:\pi_0 > 0.35\\\\$Proportion in the null hypothesis is:\\\pi_0=0.35\\\\$

$$$The test statistic is $Z=\frac{(\bar P-\pi_0)\sqrt{n}}{\sqrt{\pi_0(1-\pi_0)}}\\$The calculated statistic is Z_c=\frac{(0.47-0.35)\sqrt{100}}{\sqrt{0.35(1-0.35)}}=-0.25516\\p-value = P(Z \geq Z_c)=0.01620\\\\$

Since, the calculated statistic $Z_c$ is greater than critical $Z_{\alpha}$ , the null hypothesis should be rejected. There is enough statistical evidence to state that the proportion of admissions officers who visit an applying students' social networking page has increased in the past year.