Fun. I prefer Oxymetazoline.
For the control group we have a headache probability of
c = 368/1671 = .220
For the experimental group we have a headache probability of
e = 494/2013 = .245
The observed difference is
d = e - c = .025
The variance of the difference is
s² = c(1-c)/n₁ + e(1-e)/n₂
so the standard deviation is

We get a t statistic on the difference of
t = d/s = .025/.0139 = 1.79
We're interested in the one sided test, P(d > 0). We have enough dfs to assume normality. We look up in the standard normal table
P(z < 1.79) = .96327
so
p = P(z > 1.79) = 1 - .96327 = 0.037 = 3.7%
Answer: That's less that 10% so we have evidence to conclude that headaches are significantly greater in the experimental group.
Answer:
25.899sq.metres
Step-by-step explanation:
Area of this section of wood = Area of the sector
Area of the sector = theta/360 * πr²
r is the radius
theta is the central angle
Substitute
Area of the sector = 85.3/360 * πr²
Area of the sector = 85.3/360 * 3.14(5.90)²
Area of the sector = 85.3/360 * 109.3034
Area of the sector = 9,323.58002/360
Area of the sector = 25.899
Hence the area of the section of wood is 25.899sq.metres
4+x+2-2x=5
-x+6=5
-x+6-6=5-6[tex]
-x/-1 = -1/-1
x+1


notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k
Answer:
If you mean 4x to the power of 3 - 18x to the power of 2 +3x -13 then the answer is false