Answer:
Line QT and QS is the geometric mean of RT because QT and QS added and divided by 2 will equal the middle of both sums.
T is the midpoint of segments QT and QS
Answer:
0.99804932311
Step-by-step explanation:
We solve this using binomial probability
Binomial probability formula
= nCx × p^x × q^n - x
= n!/(n - x)! x!
Where n = Number of trials = 25 samples
x = Number of successes = 23
p = probability of success = 99% = 0.99
q = probability of failure = 1 - p
= 1 - 0.99
= 0.01
Hence,
p(at least 23 are properly filled) = p(X ≥ x)
= [25!/(25 - 23)! × 23! × 0.99^23 × 0.01^25 - 23 ]+ [25!/(25 - 24)! × 24! × 0.99^24 × 0.01^25 - 24 ]+ [25!/(25 - 25)! × 23! × 0.99^25 × 0.01^25 - 25]
= [300 × 0.99 ^23 × 0.01^2] + [25 × 0.99^24 × 0.01^1] + [1 × 0.99^25 + 0.01^0]
= 0.0238084285 + 0.1964195352 + 0.7778213594
= 0.99804932311
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
