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Aloiza [94]
3 years ago
11

Look at dis my grades bad

Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

Answer:

oof

Step-by-step explanation:

sub to tapl

:L

faust18 [17]3 years ago
6 0

Answer: I don't understand kk and why am I from Brazil ^-^

Step-by-step explanation:

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In the figure below, mLJKM = 105°, mLLKM<br> 56°, and KN bisects LKM. Find m LIKN.
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Step-by-step explanation:

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If 2tanA=3tanB,then prove that tan(A-B)=sin2B/(5-cos2B) ...?
andreev551 [17]
Tan ( A - B ) = ( tan A - tan B ) / ( 1 + tan A tan B )
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8 0
3 years ago
Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a
Vikki [24]

Answer:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

Step-by-step explanation:

Information given:

n=131 represent the sample size

\hat p=0.81 represent the estimated proportion

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.326

And replacing into the confidence interval formula we got:

0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730

0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890

And the confidence interval would be:

0.730 \leq \p \leq 0.890

6 0
3 years ago
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