Answer: the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
Step-by-step explanation:
Given data;
lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min
hence, height = 1 / ( 52.0 - 50.0) = 1 / 2
now the probability that the class length is between 50.8 and 51 min = ?
P( 50.8 < X < 51 ) = base × height
= ( 51 - 50.8) × 1/2
= 0.2 × 0.5
= 0.1 ≈ 10%
therefore the probability that the class length is between 50.8 and 51 min is 0.1 ≈ 10%
The slope-intercept form:
m - slope
b - y-intercept
The formula of a slope:
We have two points (2, 0) and (-2, -4). Substitute:
Therefore we have the equation of a line
Put the coordinates of the point (2, 0) to the equation:
<em>subtract 2 from both sides</em>
Answer: 
<span>tan(15) =
sin(15) / cos(15) =
sin(45 - 30) / cos(45 - 30) =
[ sin(45)cos(30) - sin(30)cos(45) ] / [ cos(45)cos(30) + sin(45)sin(30)]
Since sin(45) = cos(45) = √2/2, you can just factor that out from the top and bottom
[ cos(30) - sin(30) ] / [ cos(30) + sin(30)]
[ √3/2 - 1/2 ] / [ √3/2 + 1/2]
(√3 - 1) / (√3 + 1)
(√3 - 1)^2 / (√3+1)(√3 - 1)
(√3 - 1)^2 / (3 - 1)
(3 - 2√3 +1) / 2
2 - √3
There's also a formula for tan(a-b), but I couldn't remember it off hand.</span>
The answer is d: 16y^6/x^2
