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Alexxx [7]
2 years ago
7

Need help and im almost done

Mathematics
1 answer:
Sav [38]2 years ago
6 0

Answer:

1 3

Step-by-step explanation:

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For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co
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Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b)   Margin of error ( E) = $1,000 , n = 216

(c)   Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation \sigma = $7,500

\alpha = 1 - confidence interval = 1 - .95 = .05

Z_{\frac{\alpha}{2}} =  Z_{\frac{.05}{2}} = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

                           E   = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}

n =  54.0225

n = 54 ( approximately)

(b)   Margin of error ( E) = $1,000

          E     = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

         1000   =  Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}

n = 216

(c)   Margin of error ( E) = $500

   E = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

  500 = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}

n = 864

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Answer: -6x+2y=8 and y=3x

Step-by-step explanation:

they must have the same slope to be parallel and when you convert -6x+2y=8 into y=mx+b it is y=3x+4

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Step-by-step explanation:

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Answer:

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