-1/2(7z+4)+1/5(5z-15)

It seems to you that fewer than half of people who are registered voters in the City of Madison do in fact vote when there is an

Ad libitum [116K]

Answer:

a) Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

b) The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

c) We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

d) The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

(a) How might a simple random sample have been gathered?

Going to public places like restaurants, parks, theaters, etc in Madison and asking voters.

(b) Construct an 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

You take an SRS of 200 registered voters in the City of Madison, and discover that 122 of them voted in the last non-presidential election. This means that $n = 200, \pi = \frac{122}{200} = 0.61$ .

We want to build an 80% CI, so $\alpha = 0.20$ , z is the value of Z that has a pvalue of $1 - \frac{0.20}{2} = 0.90[tex], so [tex]z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 - 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.5533$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{200}} = 0.61 + 1.645\sqrt{\frac{0.61*0.39}{200}} = 0.6667$

The 80% CI to estimate the true proportion of registered voters in the City of Madison who vote in non-presidential elections is (0.5533, 0.6667).

(c) Interpret the interval you created in part (b).

We are 80% sure that our confidence interval contains the true proportion of registered voters in the City of Madison who vote in non-presidential elections.

(d) Based on your CI, does it seem that fewer than half of registered voters in the City of Madison vote in non-presidential elections? Explain.

The lower limit of the interval is higher than 0.5. This means that it does seem that MORE than half of registered voters in the City of Madison vote in non-presidential elections.

4 0

3 years ago

Hideki buys a skateboard that costs c dollars. After 6.5% sales tax is added, the total cost can be expressed by the expression

natulia [17]

Answer: c(1 + 0.065)

Step-by-step explanation:

3 0

3 years ago

Solve the equation

Alborosie

C. There is no solution.

6(x＋4)＝5(x－5)＋x
6x＋24＝5x－25＋x
6x＋24＝6x－25
24＝－25
The statement is false for any value of x
∴ There is no solution.

Hope my answer helped u :)

7 0

3 years ago

Read 2 more answers

What are the zeros of the polynomial function f(x) = x^3-10x^2+24x

Ludmilka [50]

Replace F(x) with y
y=x^3-10x^2+24x
To find the roots replace y with 0.
0=x^3-10x^2-24x
Rewrite the equation as
x^3-10x^2+24x=0
Factor the left side
x(x-6)(x-4)=0
The solution is a result of
x=0, x-6=0, and x-4=0

The answer is:
x=0, 6, 4

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3 years ago

Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the paraboloid z = 9x2 y2 and the p

dolphi86 [110]

The vector function is, r(t) = $\bold{ < t,2t^2,9t^2+4t^4 > }$