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3 years ago

PLEASE HELPPP!!!!

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

8 0

40% of $3,200 is $ 1,280

william payed patio town $1,280 for the materials

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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

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4 years ago

Simplify it plzzzzz and thx

frozen [14]

Answer:

this q is on quizizz

Step-by-step explanation:

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3 years ago

Courtney is a retail store manager and will make $40000 this year. She expects to pay 28% of her income in tax, how much money w

AlexFokin [52]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Find the multiplier:

28/100 = 0.28

1 - 0.28 = 0.72

Multiply the total amount by this multiplier:

40,000 x 0.72 = 28,800

She will make $28,800

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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3 years ago

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