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shusha [124]
3 years ago
13

The object of the game is to toss a beanbag in the circular hole of a 48-by-24-inch board. If the diameter of the circle is 6 in

ches, what is the geometric probability an object will hit the circle on the board? Assume that the object will hit the board.
Mathematics
1 answer:
larisa [96]3 years ago
6 0

Answer:

Geometric probability of an object hitting a circular hole is 0.0245.

Step-by-step explanation:

We have given,

A board of size 48 by 24 inch. There is a circular hole in the board having diameter 6 inches.

So,

Area of a board = 48 × 24 = 1152 square inches

And area of circular hole = π×r²   {where r = diameter/2  = 6 / 2 = 3 inches}

Area of circular hole = π×3² = 9π = 28.27 square inches

Now, we need to find the geometric probability of an object will hit the circle.

Geometric probability =  Area of circular hole / Area of board

Geometric probability = \frac{28.27}{1152}

Geometric probability = 0.0245

hence geometric probability of an object hitting a circular hole is 0.0245.

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ANSWER

The required equation is:

9 {x}^{2}  - 25{y}^{2}  + 250y  - 85 0=0

EXPLANATION

The given equation is

9 {x}^{2}  - 25 {y}^{2}  = 225

Dividing through by 225 we obtain;

\frac{ {x}^{2} }{25}  -  \frac{ {y}^{2} }{9}  = 1

This is a hyperbola that has it's centre at the origin.

If this hyperbola is translated so that its center is now at (0,5).

Then its equation becomes:

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We now expand to get;

9 {x}^{2}  - 25( {y}^{2} - 10y + 25 )= 225

9 {x}^{2}  - 25{y}^{2}  + 250y  - 6 25 = 225

The equation of the hyperbola in general form is

9 {x}^{2}  - 25{y}^{2}  + 250y  - 85 0=0

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