31 m= how many cm? Who got it in there System check the comments
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∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find
where is the length of the altitude originating from vertex O, and so
where is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²
We should first calculate the average number of checks he wrote per day. To do that, divide 169 by 365 (the number of days in a year) and you get (rounded) 0.463. This will be λ in our Poisson distribution. Our formula is
. We want to evaluate this formula for X≥1, so first we must evaluate our case at k=0.
To find P(X≥1), we find 1-P(X<1). Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0). Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.<em />
To find P(X≥1), we find 1-P(X<1). Since the author cannot write a negative number of checks, this means we are finding 1-P(X=0). Therefore we have 1-0.3706=0.6294.
There is a 63% chance that the author will write a check on any given day in the year.<em />