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barxatty [35]
2 years ago
8

Can someone please help me

Mathematics
1 answer:
Marrrta [24]2 years ago
7 0

Based on the knowledge of <em>trigonometric</em> expressions and properties of <em>trigonometric</em> functions, the value of the <em>sine</em> function is equal to - √731 / 30.

<h3>How to find the value of a trigonometric function</h3>

Herein we must make use of <em>trigonometric</em> expressions and properties of <em>trigonometric</em> functions to find the right value. According to trigonometry, both cosine and sine are <em>negative</em> in the <em>third</em> quadrant. Thus, by using the <em>fundamental trigonometric</em> expression (sin² α + cos² α = 1) and substituting all known terms we find that:

\sin \theta = -\sqrt{1 - \cos^{2}\theta}

\sin \theta = - \sqrt{1 - \left(-\frac{13}{30} \right)^{2}}

sin θ ≈ - √731 / 30

Based on the knowledge of <em>trigonometric</em> expressions and properties of <em>trigonometric</em> functions, the value of the <em>sine</em> function is equal to - √731 / 30.

To learn more on trigonometric functions: brainly.com/question/6904750

#SPJ1

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Answer:

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Step-by-step explanation:

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Part a

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case we need to find the sample standard deviation with the following formula:

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The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

Now we just take square root on both sides of the interval and we got:

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

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