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DochEvi [55]
2 years ago
8

A circle with a radius of 10 inches is placed inside a square with a side length of 20 inches. Find the probability that a dart

thrown will land inside the circle.
a. 87.5%
b. 75.8%
c. 57.8%
d. 78.5&
Mathematics
2 answers:
melamori03 [73]2 years ago
7 0

Answer:

The correct answer is last option 78.5%

Step-by-step explanation:

Points to remember

Area of circle = πr²

Where r is the radius of circle

Area of square = a²

Where 'a' is the side length of square

<u>To find the area of circle</u>

Here r = 10 cinches

Area =  πr²

 =  3.14 * 10²

 = 3.14 * 100 = 314

<u>To find the area of square</u>

Here a = 20 inches

Area = a²

 = 20²

 = 400

<u>To find the probability percentage</u>

Probability = area of circle/Area of square

 = (314/400)*100

 78.5 %

kiruha [24]2 years ago
4 0
I think the answer is 78.5 . I mean it should be , if you don’t get it right I’m sorry !
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Answer:

The probability that a randomly selected person  from this sample liked recreational reading or disliked recreational reading is 1.

Step-by-step explanation:

The probability of an event <em>E</em> is the ratio of the favorable number of outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

Mutually exclusive events are those events which cannot occur together. They are also known as disjoint events.

If events <em>A</em> and <em>B</em> are mutually exclusive then:

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The data provided is:

Opinion                      Like academic     Dislike academic      Total

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Liked recreational                   136                      40                      176

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Disliked recreational                16                        8                        24

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Total                                       152                      48                      200

Denote the events as follows:

<em>X</em> = a person liked recreational reading.

<em>Y</em> = a person disliked recreational reading.

Compute the probability that a randomly selected person  from this sample liked recreational reading or disliked recreational reading as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)

                =\frac{176}{200}+\frac{24}{200}-0\\\\=\frac{176+24}{200}\\\\=1

Thus, the probability that a randomly selected person  from this sample liked recreational reading or disliked recreational reading is 1.

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