Garrett throws a dart at a circular dartboard. the dartboard has a radius of 15 inches, and the bowls are in the center of the dartboard has a radius of 2 inches. what is the probability that a dart thrown at random within the dartboard will hit the bull’s-eye? roger answer the nearest 10th, if necessary. literature cir garrett throws a dart at a circular dartboard. the dartboard has a radius of 15 inches, and the bulls eye in the center of the dartboard has a radius of 2 inches. what is the probability that a dart thrown at random within the dartboard will hit the bull’s-eye? roger answer the nearest 10th, if necessary. a. 1.8% b. 5.6% c. 8.4% d. 13.3%

First we calculate the area of the entire dartboard which will constitute the total area. The total area is in the form of a circle with radius 15 inches

Total area = area of circle =

Total area =

Total area =

Next, we calculate the area of the bull's eye. Again the bull's eye is another circle whose radius is 2 inches

Bull's eye area =

Bull's eye area =

Probability that a dart thrown at random within the dartboard will hit the bull’s-eye =

Probability that a dart thrown at random within the dartboard will hit the bull’s-eye = 4/225 = 0.01777 = 1.77% = 1.8% (Option A)

my answer is$ 12

step-by-step explanation:

because 15 -2=12

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204