Can you send a picture, because this doesn’t make it clear what the options are/what to do
The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area =
<span>
<span>
<span>
209.4395102393
</span>
</span>
</span>
segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412
</span>
</span>
</span>
triangle DEF Area =
<span>
<span>
<span>
173.206
</span>
</span>
</span>
segment area = <span>
<span>
209.4395102393
</span>
-173.206
</span>
segment area =
<span>
<span>
<span>
36.2335102393
</span>
</span>
</span>
segment area =
36.23 m
Source:
http://www.1728.org/circsect.htm
Answer:
[300.202 , 329.798]
Step-by-step explanation:
The 95% confidence interval is given by the interval
![\large [\bar x-t^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x-t%5E%2A%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D%2C%20%5Cbar%20x%2Bt%5E%2A%5Cfrac%7Bs%7D%7B%5Csqrt%20n%7D%5D)
where
<em>is the sample mean </em>
<em>s is the sample standard deviation </em>
<em>n is the sample size (n = 7) </em>
is the 0.05 (5%) upper critical value for the Student's t-distribution with 6 degrees of freedom (sample size -1), which is <em>an approximation to the Normal distribution for small samples (n<30).</em>
Either by using a table or the computer, we find
and our 95% confidence interval is
![\large [315-2.447*\frac{16}{\sqrt{7}}, 315+2.447*\frac{16}{\sqrt{7}}]=\boxed{[300.202,329.798]}](https://tex.z-dn.net/?f=%5Clarge%20%5B315-2.447%2A%5Cfrac%7B16%7D%7B%5Csqrt%7B7%7D%7D%2C%20315%2B2.447%2A%5Cfrac%7B16%7D%7B%5Csqrt%7B7%7D%7D%5D%3D%5Cboxed%7B%5B300.202%2C329.798%5D%7D)
<span>Step 1: Make sure the bottom numbers (the denominators) are the same.
Step 2: Add the top numbers (the numerators), put the answer over the denominator.<span>
Step 3: Simplify the fraction (if needed)</span></span>
