That big triangle is made up of 2 congruent 30-60-90 triangles with each base measuring 4.5. You could use the tangent ratio to get tan60 = x / 4.5. Multiply both sides by 4.5 to get 4.5 tan60 = x. Do that on your calculator in degree mode to get that x = 7.79 or 7.8
Part A:
Significant level:
<span>α = 0.05
Null and alternative hypothesis:
</span><span>h0 : μ = 3 vs h1: μ ≠ 3
Test statistics:
![z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7B%5Cbar%7Bx%7D-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B2.78-3%7D%7B0.9%2F%5Csqrt%7B15%7D%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B-0.22%7D%7B0.2324%7D%20%3D-0.9467)
P-value:
P(-0.9467) = 0.1719
Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438
Conclusion:
Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.
Part B:
The power of the test is given by:
![\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105](https://tex.z-dn.net/?f=%5Cbeta%3D%5Cphi%5Cleft%28Z_%7B0.025%7D%2B%20%5Cfrac%7B3-3.25%7D%7B0.9%2F%5Csqrt%7B15%7D%7D%5Cright%29%20-%5Cphi%5Cleft%28-Z_%7B0.025%7D%2B%20%5Cfrac%7B3-3.25%7D%7B0.9%2F%5Csqrt%7B15%7D%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%5Cleft%281.96%2B%20%5Cfrac%7B-0.25%7D%7B0.2324%7D%20%5Cright%29-%5Cphi%5Cleft%28-1.96%2B%20%5Cfrac%7B-0.25%7D%7B0.2324%7D%20%5Cright%29%3D%5Cphi%280.8842%29-%5Cphi%28-3.0358%29%20%5C%5C%20%20%5C%5C%20%3D0.8117-0.0012%3D0.8105)
Therefore, the power of the test if </span><span>μ = 3.25 is 0.8105.
Part C:
</span>The <span>sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:
![1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898](https://tex.z-dn.net/?f=1-0.9%3D%5Cphi%5Cleft%28Z_%7B0.025%7D%2B%20%5Cfrac%7B3-3.75%7D%7B0.9%2F%5Csqrt%7Bn%7D%7D%5Cright%29%20-%5Cphi%5Cleft%28-Z_%7B0.025%7D%2B%20%5Cfrac%7B3-3.75%7D%7B0.9%2F%5Csqrt%7Bn%7D%7D%5Cright%29%20%5C%5C%20%5C%5C%20%5CRightarrow0.1%3D%5Cphi%5Cleft%281.96%2B%20%5Cfrac%7B-0.75%7D%7B0.9%2F%5Csqrt%7Bn%7D%7D%5Cright%29-%5Cphi%5Cleft%28-1.96%2B%20%5Cfrac%7B-0.75%7D%7B0.9%2F%5Csqrt%7Bn%7D%7D%20%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Cphi%5Cleft%281.96%2B%28-3.2415%29%5Cright%29-%5Cphi%5Cleft%281.96%2B%28-3.2415%29%5Cright%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Cfrac%7B-0.75%7D%7B0.9%2F%5Csqrt%7Bn%7D%7D%3D-3.2415%20%5C%5C%20%5C%5C%20%5CRightarrow%5Cfrac%7B0.9%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B-0.75%7D%7B-3.2415%7D%3D0.2314%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Csqrt%7Bn%7D%3D%5Cfrac%7B0.9%7D%7B0.2314%7D%3D3.8898)
![\Rightarrow n=(3.8898)^2=15.13](https://tex.z-dn.net/?f=%5CRightarrow%20n%3D%283.8898%29%5E2%3D15.13)
Therefore, the </span>s<span>ample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.</span>
If you give english i might be able to help
34 is the original number.
This is because 34 times 2 is 68 and 68 plus 3 is 71.