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MaRussiya [10]
3 years ago
15

The area of a circle is 452.39 ft^2 . Why would the diameter have to be?

Mathematics
2 answers:
serg [7]3 years ago
4 0
I believe it would be 24ft.
MAVERICK [17]3 years ago
3 0
STEP 1:
Find the radius.

Area=πr^2
plug in the known numbers

452.39= (3.14)r^2
divide both sides by 3.14

452.39 ÷ 3.14= r^2

144.07= r^2
take the square root of 144.07 and r^2

12 feet= r


STEP 2:
multiply radius by 2 to equal diameter

Diameter= Radius * 2
Diameter= 12 feet * 2
Diameter= 24 feet


ANSWER: The diameter would be 24 feet.

Hope this helps! :)
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Step-by-step explanation:

Multiply straight across the numerator and do the same for the denominator.

7x1=7 (numerator)

16x2=32 (denominator)

Answer is 7/32

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Step-by-step explanation:

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A professor has learned that three students in his class of 20 will cheat on the final exam. He decides to focus his attention o
balu736 [363]

Answer:

a

P(X \ge 1) = 0.509

b

P(X  \ge 1) = 0.6807

Step-by-step explanation:

From the question we are told that

   The number of students in the class is  N  =  20  (This is the population )

   The number of student that will cheat is  k =  3

   The number of students that he is focused on is  n  =  4

Generally the probability distribution that defines this question is the  Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.

Generally  probability mass function is mathematically represented as

      P(X = x) =  \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}

Here C stands for combination , hence we will be making use of the combination functionality in our calculators  

Generally the that  he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

      P(X \ge 1) =  1 - P(X \le 0)

Here  

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{20 - 3} C_{4- 0}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ ^{3} C_0 *  ^{17} C_{4}}{ ^{20}C_4}

   P(X \le 0) =  \frac{ 1 *  2380}{ 4845}

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Hence

    P(X \ge 1) =  1 - 0.491

     P(X \ge 1) = 0.509

Generally the that  he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

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Here n =  6

So

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    P(X  \ge 1) =1- [  \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]

    P(X  \ge 1) =1- [  \frac{1  *  12376}{38760}]

    P(X  \ge 1) =1- 0.3193

    P(X  \ge 1) = 0.6807

   

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Answer:

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Step-by-step explanation:

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