A game has a rectangular board with an area of 44in2. There is a square hole near the top of the game board in which you must not toss in a bean bag. The square has side lengths of 3in. What is the probability of tossing the bag through the hole?

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Ozzem
9/44 chance of tossing it through the hole.

Answer: Probability of tossing the bag through the hole = [tex]\frac{9}{44}[/tex]

Step-by-step explanation:

Since we have given that

A game has a rectangular board with an area of 44 inĀ².

There is a square hole near the top of the game board in which you must not toss in a bean bag.

Length of Side of a square = 3 in.

So, Area of square of square is given by

[tex]Area=Side\times Side\\\\=3\times 3\\\\=9\ in^2[/tex]

So, Probability of tossing the bag through the hole will be

[tex]\frac{\text{Number of favourable outcome}}{\text{ Total number of outcomes }}\\\\=\frac{9}{44}[/tex]

Hence, Probability of tossing the bag through the hole = [tex]\frac{9}{44}[/tex]

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