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Basile [38]
3 years ago
13

A poll of 2,210 likely voters was conducted on the president's performance. approximately what margin of error would the approva

l rating estimate have if the confidence level is 95%? (round your answer to 4 decimal places.) margin of error (b) the poll showed that 48 percent approved the president's performance. construct a 90 percent confidence interval for the true proportion. (round your answers to 4 decimal places.) the 90% confidence interval to (c) would you agree that the percentage of all voters opposed is likely to be 50 percent? no, the confidence interval does not contain .50. yes, the confidence interval contains .50.
Mathematics
1 answer:
I am Lyosha [343]3 years ago
3 0

Answer:

the answer is explained above

Step-by-step explanation:


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I need help with this math worksheet.im not smart at all.
garik1379 [7]

Answer:

1) 11.581

2) 10.78

3) 14.8

4) 3.5

5) 16.28

6) .64

7) 23.3

8) 1.24

9) 6.53

10) 1.59

11) 8.79

12) 27.7919

Step-by-step explanation:

Ur smarter than u think. ^_^

7 0
3 years ago
Read 2 more answers
Please help please please
Liula [17]

Length of shorter piece is 13 ft

Length of longer piece is 61 - 13 => 48 ft

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2
Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

but the integrand is even, so this is really just

\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

Substitute x = 1/2 arccot(u/2), which transforms the integral to

\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du

There are lots of ways to compute this. What I did was to consider the complex contour integral

\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}

and it follows that

\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}

7 0
2 years ago
Help quick please!!
valkas [14]

Answer:

4.5 I hope I was not to late

6 0
2 years ago
Find the lateral surface area of a
balandron [24]

Answer:

15.1 mm²

Step-by-step explanation:

Lateral surface area of a cylinder = 2πrh

(where r is the radius and h is the height)

Given:

  • r = 1.2 mm
  • h = 2 mm

Substitute given values into the formula:

⇒ Lateral surface area = 2π(1.2)(2)

                                      = 4.8π

                                      = 15.1 mm²  (nearest tenth)

5 0
2 years ago
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