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3 years ago

9

I have alg 2 midterms tomorrow, does anyone know a good site to study?

2 answers:

attashe74 [19]3 years ago

7 0

Yes the notes you took in class and friends from class as well sites to study from wont help as much and we don't know what part of alg 2 you're on.

Kisachek [45]3 years ago

7 0

Um for me I like to use
- https://www.khanacademy.org/math/<span>algebra2
</span>- www.regentsprep.org/Regents/math/algtrig/math-algtrig.htm
<span>- www.classzone.com › Algebra 2
</span>
Hope this helps (:
❤❤❤Please mark as Brilliant ❤❤❤
- A

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Shtirlitz [24]

Answer:

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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

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gregori [183]

Answer:

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Step-by-step explanation:

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4 years ago

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Vanyuwa [196]

Answer:

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Step-by-step explanation:

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3 years ago

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Dmitry_Shevchenko [17]

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3 years ago

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Gwar [14]

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Hope this helps :D

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