NOT NECESSARILY would a triangle be equilateral if one of its angles is 60 degrees. To be an equilateral triangle (a triangle in which all 3 sides have the same length), all 3 angles of the triangle would have to be 60°-angles; however, the triangle could be a 30°-60°-90° right triangle in which the side opposite the 30 degree angle is one-half as long as the hypotenuse, and the length of the side opposite the 60 degree angle is √3/2 as long as the hypotenuse. Another of possibly many examples would be a triangle with angles of 60°, 40°, and 80° which has opposite sides of lengths 2, 1.4845 (rounded to 4 decimal places), and 2.2743 (rounded to 4 decimal places), respectively, the last two of which were determined by using the Law of Sines: "In any triangle ABC, having sides of length a, b, and c, the following relationships are true: a/sin A = b/sin B = c/sin C."¹
On average you get like about 12 to 14 miles a gallon
Answer:
The dependent variable is the final grade in the course and is the vriable of interest on this case.
H0: ![\beta_1 = 0](https://tex.z-dn.net/?f=%5Cbeta_1%20%3D%200)
H1: ![\beta_1 \neq 0](https://tex.z-dn.net/?f=%5Cbeta_1%20%5Cneq%200)
And if we reject the null hypothesis we can conclude that we have a significant relationship between the two variables analyzed.
Step-by-step explanation:
On this case w ehave the following linear model:
![Y= 21.839 +0.724 X](https://tex.z-dn.net/?f=Y%3D%2021.839%20%2B0.724%20X)
Where Y represent the final grade in the course and X the student's homework average. For this linear model the slope is given by
and the intercept is ![\beta_0 = 21.839](https://tex.z-dn.net/?f=%5Cbeta_0%20%3D%2021.839)
Which is the dependent variable, and why?
The dependent variable is the final grade in the course and is the vriable of interest on this case.
Based on the material taught in this course, which of the following is the most appropriate alternative hypothesis to use for resolving this question?
Since we conduct a regression the hypothesis of interest are:
H0: ![\beta_1 = 0](https://tex.z-dn.net/?f=%5Cbeta_1%20%3D%200)
H1: ![\beta_1 \neq 0](https://tex.z-dn.net/?f=%5Cbeta_1%20%5Cneq%200)
And if we reject the null hypothesis we can conclude that we have a significant relationship between the two variables analyzed.
Hailey's mixing two different coffee blends. Represent them by x and y (in pounds). Then x + y = 5 lb, and x = 5 - y.How much puree Sum. beans are we talking about here?
0.20x +0.80y = 0.60(5 lb) Mult all 3 terms by 100 to get rid of factions:
20x + 80 y = 300. Substitute 5-y for x:
20(5-y) + 80y = 300 => 100-20y + 80y = 300 => 60y = 200, so y = 20/6 or 10/3 lb den x = 5-10/3, or x =5/3 lb
Use 5/3 lb of the first blend and 10/3 lb of the second blend to come up with 5 lb of a 60% blend.