Answer: a. The correlation coefficient of the data is positive.
Step-by-step explanation:
Estimated slope of sample regression line = 
Here , confidence interval : (-0.181, 1.529)
Estimated slope of sample regression line = 
![=\dfrac{1.348}{2}\\\\=0.674\ \ \ \ [\text{ positive}]](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B1.348%7D%7B2%7D%5C%5C%5C%5C%3D0.674%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7B%20positive%7D%5D)
⇒Correlation coefficient(r) must be positive, So a. is true.
But, d. and e. are wrong(0.674 ≠ 0 or 1.348).
We cannot check residuals or its sum from confidence interval of slope of a regression line, so b is wrong.
We cannot say that scatterplot is linear as we cannot determine it from interval, so c. is wrong
So, the correct option : a. The correlation coefficient of the data is positive.
Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) = 
= 
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4
1.8 km >> 1,800 m
1,800 - 740 = 1,060 m
1,060 m >> 1.06 km
The shorter route is 1.06 km.
Answer: 126
Step-by-step explanation:
The triangle translated 3 units right and 4 units down. Hope this helps!
