Write the equation of the line passing through the points (-7,5) and (-5,9):
![y-5= \dfrac{9-5}{-5-(-7)} (x-(-7)) \\ y-5= \dfrac{4}{2}(x+7) \\ y-5=2(x+7)](https://tex.z-dn.net/?f=y-5%3D%20%5Cdfrac%7B9-5%7D%7B-5-%28-7%29%7D%20%28x-%28-7%29%29%20%5C%5C%20y-5%3D%20%5Cdfrac%7B4%7D%7B2%7D%28x%2B7%29%20%5C%5C%20y-5%3D2%28x%2B7%29%20)
.
You also have another two points (-3,13) and (-1,17). Look whether coordinates of these points satisfy the line equation:
1. For (-3,13) you have
![13-5=2(-3+7) \\ \{8=8\}](https://tex.z-dn.net/?f=%2013-5%3D2%28-3%2B7%29%20%5C%5C%20%5C%7B8%3D8%5C%7D)
;
2. For (-1,17) you have
![17-5=2(-1+7) \\ \{12=12\}](https://tex.z-dn.net/?f=%2017-5%3D2%28-1%2B7%29%20%5C%5C%20%5C%7B12%3D12%5C%7D)
.
Conclusion: All four points lie on the line y-5=2(x+7), so <span>the relationship shown by the data is linear.</span>
Answer: Correct choice is D.<span />
27.43 * 4.5 = 123.435
Joe jogged 123.435 meters
Answer: a) 0.84 b) 0.67 c) 1.28
Step-by-step explanation:
Using the standard normal distribution table for z-value , we have
(a) The value of
would result in a 80% one-sided confidence interval : ![z_{(1-0.80)}=z_{0.20}=0.8416\approx0.84](https://tex.z-dn.net/?f=z_%7B%281-0.80%29%7D%3Dz_%7B0.20%7D%3D0.8416%5Capprox0.84)
(b) The value of
would result in a 85% one-sided confidence interval : ![z_{(1-0.85)}=z_{0.25}=0.6744897\approx0.67](https://tex.z-dn.net/?f=z_%7B%281-0.85%29%7D%3Dz_%7B0.25%7D%3D0.6744897%5Capprox0.67)
(c) The value of
would result in a 90% one-sided confidence interval : ![z_{(1-0.90)}=z_{0.10}=1.2815515\approx1.28](https://tex.z-dn.net/?f=z_%7B%281-0.90%29%7D%3Dz_%7B0.10%7D%3D1.2815515%5Capprox1.28)
Answer:,n ,
Step-by-step explanation: