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3 years ago

10

And adopt blinks about 450 times in 30 minutes a 12 year old blinks about 150 times in 15 minutes how many more times does an ad

ult blink in 60 minutes then a 12 year old

1 answer:

zhenek [66]3 years ago

5 0

Answer:

EASY.

Step-by-step explanation:

DWndf liFBWwfbbhfaifBFfbayfWabfwyabfab fBAfByhWE8bu

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Okay, so I checked on my calculator, and 5,6, and 7 are right

Step-by-step explanation:

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4 years ago

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juin [17]

6.5693%
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ch4aika [34]

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3 years ago

. Compute the required sample size given the required confidence in the sample results is 99.74% (Z score of 3). The level of al

Shtirlitz [24]

Answer:

900 sample size

Step-by-step explanation:

To determine the sample size for a proportion, the margin of error formula is used to determine this:

$E=Z_{\frac{\alpha}{2} }*\sqrt{\frac{\hat p \hat q}{{n} }$

$n=\hat p \hat q*(\frac{Z_{\frac{\alpha}{2} }}{E} )^2$

Where p is the proportion, E is the margin of error, n is the sample size, q = 1 - p, $Z_\frac{\alpha }{2}$ is the z score.

Since the proportion is not known, the sample size needed to guarantee the confidence interval and error is at p = 0.5 and q = 1 - p = 1 - 0.5 = 0.5

E = 5% = 0.05, $Z_\frac{\alpha }{2}$ = 3. Hence:

$n=0.5*0.5*(\frac{3}{0.05} )^2\\\\n = 900$

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3 years ago