Answer:
Given that a car is a coupe, there is a 10% chance it is from the previous model year.
Step-by-step explanation:
The consideration "Given that a car is from the previous model year, there is a 10% chance that it is a coupe." wouldn't give much information. Previous year models are 10% Coupe and 25% Sedan.
Same for consideration "There is a 10% chance that the car is from the previous model year." This consideration would be good if only a specific model would be taken.
Consideration "There is a 10% chance that the car is a coupe." makes no sense because, in order for this to be true, 90% of specific models would be a sedan which table clearly shows 25%.
The sector (shaded segment + triangle) makes up 1/3 of the circle (which is evident from the fact that the labeled arc measures 120° and a full circle measures 360°). The circle has radius 96 cm, so its total area is π (96 cm)² = 9216π cm². The area of the sector is then 1/3 • 9216π cm² = 3072π cm².
The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines
b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°) ⇒ b = 96√3 cm
Call b the base of this triangle.
The vertex angle is 120°, so the other two angles have measure θ such that
120° + 2θ = 180°
since the interior angles of any triangle sum to 180°. Solve for θ :
2θ = 60°
θ = 30°
Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find
tan(30°) = h / (b/2) ⇒ h = 48 cm
Then the area of the triangle is
1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²
and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:
3072π cm² - 2304√3 cm² ≈ 5660.3 cm²
6,000,000 + 7,000+ 200 is your expanded form
hope this helps
Try this solution:
1. m∠A=m∠L; m∠B=m∠M and m∠C=m∠N;
2. m∠B=m∠M=35° and m∠C=m∠N=95°;
3. m∠A=m∠L=180°-(m∠B+m∠C)=180-35-95=50°
answer: 50°
-1-4x+7x=1x+4x
-1-4x+7x=1x+4x
-1+-4x+7x=x+4x
(-4x+7x)+(-1)=(x+4x)
3x+-1=5x
3x-1=5x
3x-1-5x=5x-5x
-2x-1=0
-2x-1+1=0+1
-2x=1
-2x/-2=1/2
x=-1/2
