Answer:
As
, it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
Step-by-step explanation:
1. Relevant data:
![\mu=0.70\\N=40\\\alpha=0.05\\X=0.65\\s=0.20](https://tex.z-dn.net/?f=%5Cmu%3D0.70%5C%5CN%3D40%5C%5C%5Calpha%3D0.05%5C%5CX%3D0.65%5C%5Cs%3D0.20)
2. Hypotesis testing
![H_{0}=\mu=0.70](https://tex.z-dn.net/?f=H_%7B0%7D%3D%5Cmu%3D0.70)
![H_{1} =\mu< 0.70](https://tex.z-dn.net/?f=H_%7B1%7D%20%3D%5Cmu%3C%200.70)
3. Find the rejection area
From the one tail standard normal chart, whe have Z-value for
is 1.56
Then rejection area is left 1.56 in normal curve.
4. Find the test statistic:
![Z=\frac{X-\mu_{0} }{\sigma/\sqrt{n}}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu_%7B0%7D%20%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D)
![Z=\frac{0.65-0.70}{0.20/\sqrt{40}}\\Z=-1.58](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B0.65-0.70%7D%7B0.20%2F%5Csqrt%7B40%7D%7D%5C%5CZ%3D-1.58)
5. Hypotesis Testing
![Z_{\alpha}=1.56\\Z=-1.58](https://tex.z-dn.net/?f=Z_%7B%5Calpha%7D%3D1.56%5C%5CZ%3D-1.58)
![-1.58](https://tex.z-dn.net/?f=-1.58%3C-1.56)
As
, it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.
Think of the greater than sign as an equals sign. Then, you solve for m
Begin by using the cosine rule to find the unknown side.
Where a- unknown side, b is 12, c is 16 and angle in between is 108°
The cosine rule...
a² = b² + c² - 2bc cosA
= 12² + 16² - (2×12×16 cos108°)
= 518.655076
Taking the positive
square root...
a = 22.774
a = 22.8 cm nearest tenth
6 + y × 9 (the x is a times sign btw)