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Nezavi [6.7K]
3 years ago
8

Help please! show work

Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0
Your answers are
A = 35.7°
B = 67.6°
C = 76.7°

cosine law

a^2 = b^2 + c^2 -2bc \cos A \\
-2bc \cos A = a^2 - b^2 - c^2 \\ \\
\cos A = \dfrac{a^2 - b^2 - c^2}{-2bc} \\ \\
A = \cos^{-1}\left[ \dfrac{a^2 - b^2 - c^2}{-2bc} \right] \\ \\
A = \cos^{-1}\left[ \dfrac{12^2 - 19^2 - 20^2}{-2(19)(20)} \right]  \\ \\
A = 35.723697

A = 35.723697
sine law for the rest of the angles

\displaystyle
\frac{\sin B}{b} = \frac{\sin A}{a} \\ \\
\sin B = \frac{b \sin A}{a} \\ \\
B = \sin^{-1} \left[ \frac{b \sin A}{a}  \right] \\ \\
B = \sin^{-1} \left[ \frac{19 \sin 35.723697 }{12}  \right]  \\ \\
B \approx 67.58886795

B = 67.58886795
All angles in triangle sum to 180 so find C with that

A + B + C = 180
C = 180 - A - B
C = 180 - 35.723697 - 67.58886795
C = 76.7°

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Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of \mu and its 90% margin of error.

b. A 95% confidence interval for \mu.

Answer:

a

\= x  = 13.4   .   E = 2.3

b

10.7 <  \mu < 16.1

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 18

  The 90% confidence interval is  (11.1, 15.7)

Generally the point estimate of  \mu is mathematically  evaluated  as

       \= x  = \frac{11.1 + 15.7 }{2}

=>    \= x  = 13.4

Generally the margin of error is mathematically evaluated  as

     E = \frac{15.7 - 11.1}{2 }

=> E = 2.3

  From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the equation for the lower limit of the confidence interval is  

      \= x  -  Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1

=> 13.4   -  0.3877 s  = 11.1

=>  s = 5.932

  From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =  1.96 *  \frac{5.932}{\sqrt{18} }

=>    E =  2.7      

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>    13.4  -  2.7  <  \mu < 13.4  +   2.7

=>    10.7 <  \mu < 16.1

3 0
3 years ago
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