Answer:
Cannot be determined
Step-by-step explanation:
The information given for the triangles are not enough to evaluate them for similarity. How ever they are both equilateral triangles given the angle measures.
Answer:
![\bar x = 260.1615](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20260.1615)
![\sigma = 70.69](https://tex.z-dn.net/?f=%5Csigma%20%3D%2070.69)
The confidence interval of standard deviation is:
to ![103.25](https://tex.z-dn.net/?f=103.25)
Step-by-step explanation:
Given
![n =20](https://tex.z-dn.net/?f=n%20%3D20)
See attachment for the formatted data
Solving (a): The mean
This is calculated as:
![\bar x = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
So, we have:
![\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B242.87%20%2B212.00%20%2B260.93%20%2B284.08%20%2B194.19%20%2B139.16%20%2B260.76%20%2B436.72%20%2B355.36%20%2B.....%2B250.61%7D%7B20%7D)
![\bar x = \frac{5203.23}{20}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B5203.23%7D%7B20%7D)
![\bar x = 260.1615](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20260.1615)
![\bar x = 260.16](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20260.16)
Solving (b): The standard deviation
This is calculated as:
![\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum%28x%20-%20%5Cbar%20x%29%5E2%7D%7Bn-1%7D%7D)
![\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7B%28242.87%20-%20260.1615%29%5E2%20%2B%28212.00-%20260.1615%29%5E2%2B%28260.93-%20260.1615%29%5E2%2B%28284.08-%20260.1615%29%5E2%2B.....%2B%28250.61-%20260.1615%29%5E2%7D%7B20%20-%201%7D%7D)
![\sigma = \sqrt{\frac{94938.80}{19}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7B94938.80%7D%7B19%7D%7D)
![\sigma = \sqrt{4996.78}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B4996.78%7D)
--- approximated
Solving (c): 95% confidence interval of standard deviation
We have:
![c =0.95](https://tex.z-dn.net/?f=c%20%3D0.95)
So:
![\alpha = 1 -c](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-c)
![\alpha = 1 -0.95](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-0.95)
![\alpha = 0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.05)
Calculate the degree of freedom (df)
![df = n -1](https://tex.z-dn.net/?f=df%20%3D%20n%20-1)
![df = 20 -1](https://tex.z-dn.net/?f=df%20%3D%2020%20-1)
![df = 19](https://tex.z-dn.net/?f=df%20%3D%2019)
Determine the critical value at row
and columns
and ![1 -\frac{\alpha}{2}](https://tex.z-dn.net/?f=1%20-%5Cfrac%7B%5Calpha%7D%7B2%7D)
So, we have:
---- at ![\frac{\alpha}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Calpha%7D%7B2%7D)
--- at ![1 -\frac{\alpha}{2}](https://tex.z-dn.net/?f=1%20-%5Cfrac%7B%5Calpha%7D%7B2%7D)
So, the confidence interval of the standard deviation is:
to ![\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }](https://tex.z-dn.net/?f=%5Csigma%20%2A%20%5Csqrt%7B%5Cfrac%7Bn%20-%201%7D%7BX%5E2_%7B1%20-%5Calpha%2F2%7D%20%7D)
to ![70.69 * \sqrt{\frac{20 - 1}{8.907}](https://tex.z-dn.net/?f=70.69%20%2A%20%5Csqrt%7B%5Cfrac%7B20%20-%201%7D%7B8.907%7D)
to ![70.69 * \sqrt{\frac{19}{8.907}](https://tex.z-dn.net/?f=70.69%20%2A%20%5Csqrt%7B%5Cfrac%7B19%7D%7B8.907%7D)
to ![103.25](https://tex.z-dn.net/?f=103.25)
What is the expression? Tell me and I will help you.
Answer: 2 miles
Step-by-step explanation:
Afternoon-x; morning- 1.5x; Evening: 1/2x
Make a equation: 1.5x+x+1/2=6
2.5x+1/2x=6
5/2x+1/2x=6
6x=12
x=2
She would’ve ran 6 miles in an hour.