Question

An electronic dartboard has a diameter of 10 inches. The bull’s-eye has a radius of 1.5 inches. If a dart hits the board randomly, what is the chance it hits the bull’s-eye? Check all that apply.

The dartboard area is roughly 78.54 square inches.
The dartboard area is roughly 314.16 square inches.
The bull’s-eye area is roughly 7.07 square inches.
The probability that a random dart strikes the bull’s-eye is roughly 0.09.
The probability that a random dart strikes the bull’s-eye is roughly 0.02.

Answer 1

We first have to calculate the area of both the dartboard and the bull's-eye.

Both are circles, so their area will be calculated the same way: A = π r²

Dartboard, diameter: 10 inches - so radius = 5 inches

A = π 5² = 25 π = 78.54 square inches

Bull's-eye, radius = 1.5 inches

A = π 1.5² = 2.25 π = 7.07 square inches

Ratio between bulls-eye and dartboard: 7.07 / 78.54 = 0.09

So, the following 3 statements are accurate:

The dartboard area is roughly 78.54 square inches.

The bull’s-eye area is roughly 7.07 square inches.

The probability that a random dart strikes the bull’s-eye is roughly 0.09.

Answer 2

Options

a); c); d)

Explanation:

We calculate the area of the target using the formula of the area of a circle

`A = \pi r ^ 2`

Where r is the radius, which is equal to half the diameter.

Then the radius of the dartboard is 10 inches

Your area is:

`A_d = \pi((10)/(2)) ^ 2\\\\A_d = 25\pi\\\\A_d = 78.54\ in ^ 2`

We use the same formula to calculate the area of the porthole.

Where r = 1.5 in

`A_b = \pi(1.5) ^ 2\\\\A_b = 7.07\ in ^ 2.`

The probability that a randomly launched dart hits the bull's eye is:

`P = (A_b)/(A_d)\\\\P = (7.07\ in ^ 2)/(78.54\ in ^ 2)\\\\P = 0.0900`

Finally the answer is:

- The dartboard area is roughly 78.54 square inches.

- The bull’s-eye area is roughly 7.07 square inches.

- The probability that a random dart strikes the bull’s-eye is roughly 0.09.