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3 years ago

11

$550 at 4% compounded daily for 13 years

2 answers:

natali 33 [55]3 years ago

6 0

:> HOLA QUE ASE? (人◕ω◕)

vichka [17]3 years ago

5 0

Answer:Sydney’s

Step-by-step explanation:

Rhrurjeuejej

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NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud

SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

$M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }$

The 85% confidence level $z_{\alpha } = 1.44$

$\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}$

$\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}$

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The large sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

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3 years ago