'll use the binomial approach. We need to calculate the probabilities that 9, 10 or 11
<span>people have brown eyes. The probability that any one person has brown eyes is 0.8, </span>
<span>so the probability that they don't is 1 - 0.8 = 0.2. So the appropriate binomial terms are </span>
<span>(11 C 9)(0.8)^9*(0.2)^2 + (11 C 10)(0.8)^10*(0.2)^1 + (11 C 11)(0.8)^11*(0.2)^0 = </span>
<span>0.2953 + 0.2362 + 0.0859 = 0.6174, or about 61.7 %. Since this is over 50%, it </span>
<span>is more likely than not that 9 of 11 randomly chosen people have brown eyes, at </span>
<span>least in this region. </span>
<span>Note that (n C r) = n!/((n-r)!*r!). So (11 C 9) = 55, (11 C 10) = 11 and (11 C 0) = 1.</span>
Cos (π/2 - x) = sin x = 3/5
tan x = sin x / cos x = 3/5 / 4/5 = 3/4
csc x = 1/sin x = 1 / 3/5 = 5/3
sec x = 1/cos x = 1 / 4/5 = 5/4
cot x = 1/tan x = 1 / 3/4 = 4/3.
Answer: A. 3 ways: k, DE, ED (both DE and ED have line markers over top)
To name a line, we just need two points on the line. We list them in any order because the line extends forever in both directions. Contrast this with a ray where order does matter. The little k is another way to name a line, potentially simplifying things.
Choice B is close, but it mentions ray DE instead of line DE. Choice C is missing line ED. Choice D is a similar story as choice B. These facts allow us to rule out B through D.
