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erastova [34]
3 years ago
10

Please help me now plz I promise I will mark you brainliest

Mathematics
1 answer:
den301095 [7]3 years ago
7 0

Answer:

C. Outside the circle

I hope this helps

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A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he
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Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, <em>MOE</em> = 0.04.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, <em>MOE</em> = 0.02.

The formula for margin of error is:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}

The critical value of <em>z</em> for 95% confidence interval is: z_{\alpha/2}=1.96

Compute the minimum sample size required as follows:

MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

8 0
3 years ago
Please help me with these factoring problems. Please help me solve. I can't do these!!
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Simple...

you have: 

1.) -8x-16  --->>>

Factor out a -8 -->>

-8(x+2)

2.)9d+15db --->

Factor out 3d -->

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3.) 2a-9ax-->>

Factor out -a -->

-1a(9x-2)

Thus, your answer.
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The bus driver made 6 stops on his way into town. No passengers exited the bus. By stop 3, there were 9 passengers on the bus. 
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