In circle J with m \angle HJK= 92m∠HJK=92 and HJ=9HJ=9 units find area of sector HJK. Round to the nearest hundredth.
1 answer:
Answer:
65.03square units.
Step-by-step explanation:
Find the diagram in the attachment below.
Area of the sector HJK = theta/360° × πr²
theta is the angle subtended by the sector = ∠HJK = 92°
r is the radius = HJ = 9units.
Substituting the given values in the formula:
Area of sector HJK = 92°/360° × π(9)²
Area of sector HJK = 23,399.28/360
Area of sector HJK = 65.03 square units to nearest hundredth
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Answer:
Step-by-step explanation:
We know that between 1 to 10 there are 5 even and 5 odd numbers.
We could get 4 even cards , 4 odd cards or 2 odd and 2 even cards
Let´s check all this combinations
Case 1: When all 4 numbers are even:
We are going to take 4 of the 5 even numbers in the box so we have
Case 2: When all 4 numbers are odd:
We are going to take 4 of the 5 odd numbers in the box, so we have
Case 3: When 2 are even and 2 are odd:
We are giong to take 2 from 5 even and odd cards in the box so we have
Remember that we obtain the probability from
So we have the number of favourable outcomes but we need the Total cases for drawing four cards, so we have that:
We are taking 4 of the 10 cards:
Hence we have that the probability that their sum is even