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Reptile [31]
3 years ago
7

A study of the effects of color on easing anxiety compared anxiety test scores of participants who completed the test printed on

soft yellow paper versus a different group of students who completed the test printed on harsh green paper. The scores for five participants who completed the test printed on the yellow paper were 17, 19, 28, 21, and 18. The scores for four participants who completed the test on the green paper were 20, 26, 17, and 24. Using the .05 significance level, one-tailed (predicting lower anxiety scores for the yellow paper), what should the researcher conclude
Mathematics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

Step-by-step explanation:

To solve this, we would follow these simple steps. We have

unvrs :

The arithmetic mean, x-bar for the yellow paper group (Y) = 20.6

The arithmetic mean, x-bar for the green paper group (G) = 21.75

Recall that, H0: µY = µG

And from the data we have, we can see that

H0: µY< µG

We proceed to say that the

T-Test-statistic = -0.404

Also, the p-value: 0.349

From our calculations, we can see that the p-value > 0.05, and as such, we conclude that we will not reject H0. This is because there is not enough evidence to show that test that is printed on the yellow paper decreases anxiety at a 0.05 significance level.

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First expand the parentheses:-

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x = -37 / 104 =  -0.3558 
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3 years ago
Using the right triangle below. What is the name for side AC ?
irga5000 [103]

Answer:

B) Adjacent

Step-by-step explanation:

The hypotenuse will ALWAYS be designated as the longest side in a right triangle.

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4 0
3 years ago
Evaluate the surface integral:S
rjkz [21]
Assuming S does not include the plane z=0, we can parameterize the region in spherical coordinates using

\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v

Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
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So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
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3 years ago
The solution to this system of equations lies between the x-values -2 and -1.5. At which x-value are the two equations approxima
katovenus [111]
D) -1.8
-1.8
6 0
3 years ago
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