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emmainna [20.7K]
3 years ago
5

Results of a poll stated that 40% of U.S. adults identify as Republican, and you want to test the claim at alpha = 0.05. What le

vel of confidence should you use? (Answer to two decimal places, e.g., 0.75).
Mathematics
1 answer:
Nookie1986 [14]3 years ago
4 0

Answer:

The confidence level is 95% or 0.95.

Step-by-step explanation:

Consider the provided information.

Results of a poll stated that 40% of U.S.

If you want to test the claim at alpha = 0.05.

As we know:

Confidence level = 1 - significance level (\alpha)

Confidence level = 1 - 0.05

Confidence level  = 0.95

Therefore, the confidence level is 95% or 0.95.

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Can someone please help me? I don't understand this problem. How would they be the same???
musickatia [10]
Area of rectangle = 15 x 8
                           =120

area of parallelogram = 8 x 15 
                               =120
so, they have the same area, because the length of a rectangle is the same as the height of a parallelogram and also for the base.
4 0
2 years ago
Find the surface area of the tool box. Round your answer to the nearest tenth and explain your answer. ​Pls I NEED THE ANSWER NO
Andrej [43]

Answer:

807.8 in^2

Step-by-step explanation:

The total area of the box is the sum of the areas of all faces of the box. The top, bottom, front, and back faces are rectangles 18 in long. The end faces each consist of a rectangle and a triangle. We can compute the sum of these like this:

The areas of top, bottom, front, and back add up to be 18 inches wide by the length that is the perimeter of the end: 2·5in +2·8 in + 9.6 in = 35.8 in. That lateral area is ...

(18 in)(35.6 in) = 640.8 in^2

The area of the triangle on each end is equivalent to the area of a rectangle half as high, so we can compute the area of each end as ...

(9.6 in)(8.7 in) = 83.52 in^2

Then the total area is the lateral area plus the area of the two ends:

640.8 in^2 + 2·83.52 in^2 = 807.84 in^2 ≈ 807.8 in^2

7 0
2 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
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Ne4ueva [31]
<h2>... Answer is in the pictures above... </h2><h3>... Hope this will help... </h3>

4 0
2 years ago
The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
alex41 [277]

Answer:

B) w\approx3.12 pounds

Step-by-step explanation:

w=0.9\sqrt{E}

w=0.9\sqrt{12}

w\approx3.12

6 0
2 years ago
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