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2 years ago

I need help I have been stuck for an hour

Mathematics

2 answers:

Alla [95]2 years ago

7 0

Answer:

same here, i cant figure it out.

Step-by-step explanation:

Mandarinka [93]2 years ago

4 0

Answer:

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

Rzqust [24]

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

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

7 0

3 years ago

Kate ate 1/4 of her orange. Ben ate 2/4 of his banana. Did Kate and Ben eat 1/4+2\4=3/4? Explain

kozerog [31]

Answer:

yes

Step-by-step explanation:

1 + 2 = 3

You don't add equal denominators. so 3/4

3 0

2 years ago

Open image to see question

Agata [3.3K]

Answer:

Step-by-step explanation:

To start off this problem, we are given that the line AB is equal to the line CD. In addition to that, we are given that the line EF equally intersects line AB and line CD. This provides us proof that angle AGH is equivalent to angle DHG. From this information, we can solve this problem relatively easily.

Lets work with this equation:

$80 = 5x$