Mathematics: What is the simplest form of the expression below?

And the 95% confidence interval for the difference of the two proportions is given by:Step-by-step explanation:For this case we have the following info given: number of people that they read at least one book in the last 3 months in 2011 the sample size selected for 2011 number of people that they read at least one book in the last 3 months in 2015 the sample size selected for 2015The estimated proportions people that they read at least one book in the last 3 months for each year are given by:And...

What is the simplest form of the expression below?

Ответ:

janelly66

10.03.2022

Mathematics

Let us simplify each Expression Separately :

✿ $\mathsf{(\frac{cot\theta.cos\theta}{sin\theta}).tan\theta}$

$\mathsf{\implies (\frac{cot\theta.tan\theta.cos\theta}{sin\theta})}$

$\mathsf{We\;know\;that : tan\theta.cot\theta = 1}$

$\mathsf{\implies (\frac{cos\theta}{sin\theta})}$

$\mathsf{We\;know\;that : cot\theta = \frac{cos\theta}{sin\theta} }$

$\mathsf{\implies (\frac{cot\theta.cos\theta}{sin\theta}).tan\theta = cot\theta }$

✿ $\mathsf{(\frac{sin\theta}{cos\theta.tan\theta})}$

$\implies \mathsf{(\frac{\frac{sin\theta}{cos\theta}}{tan\theta})}$

$\mathsf{We\;know\;that : tan\theta = \frac{sin\theta}{cos\theta} }$

$\implies \mathsf{(\frac{tan\theta}{tan\theta})}$

$\implies \mathsf{1}$

$\mathsf{\implies (\frac{sin\theta}{cos\theta.tan\theta}) = 1}$

$\mathsf{The\;Simplest\;form\;of\;the\;Given\;Expression\;is : cot\theta}$

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Ответ:

shanicar33500

01.02.2021

Mathematics

$(0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376$

$(0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024$

And the 95% confidence interval for the difference of the two proportions is given by:

$0.0376 \leq p_1 -p_2 \leq 0.1024$

Step-by-step explanation:

For this case we have the following info given:

X_1 = 948 number of people that they read at least one book in the last 3 months in 2011

n_1 = 1200 the sample size selected for 2011

X_2 = 1080 number of people that they read at least one book in the last 3 months in 2015

n_2 = 1500 the sample size selected for 2015

The estimated proportions people that they read at least one book in the last 3 months for each year are given by:

$\hat p_1 = \frac{948}{1200}= 0.79$

$\hat p_2 = \frac{1080}{1500}= 0.72$

And the confidence interval for the true difference of proportions is given by:

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

The confidence level is 95% so then the significance is 0.05 or 5% and $\alpha/2 =0.025$ and the critical value for this case using the normal standard distribution is:

$z_{\alpha/2}=\pm 1.96$

And replacing into the confidence interval formula we got:

$(0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376$

$(0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024$

And the 95% confidence interval for the difference of the two proportions is given by:

$0.0376 \leq p_1 -p_2 \leq 0.1024$

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What is the simplest form of the expression below?

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