GH: chord
M: point of tangency
MJ: diameter
J: center
MH: radius
GH: secant
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall that
tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)
so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:
(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall the double angle identity for cosine,
cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
so the 1 in the denominator also vanishes:
(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))
Recall the Pythagorean identity,
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1
which means
sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):
-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))
Cancel the cos²(<em>θ</em>) terms to end up with
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2
First, we get the z-score using formula:
z =( X - μ ) / σ
From the problem, we substitute the given values:
z = (9 - 7.8) / 1.7
z = 0.71
From the z-table (one-tailed), for a z-score of 0.7059, the proportion of numbers that are greater than 9 is 0.76.
29 , 35 , 41
the sequence is going up
in 6s
<span>= 49567/1000
</span>rite the decimal number as a fraction
(over 1)
49.567 = 49.567 / 1
Multiplying by 1 to eliminate 3 decimal places
we multiply top and bottom by 3 10's
Numerator (N)
N = 49.567 × 10 × 10 × 10 = 49567
Denominator (D)
D = 1 × 10 × 10 × 10 = 1000
N / D = 49567 / 1000
Simplifying our fraction
<span>= 49567/1000</span>
