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dimaraw [331]
3 years ago
13

An above ground swimming pool in the shape of a cylinder has a diameter of 21 feet and a height of 3.5 feet. If the pool is fill

ed with water up to 9 inches from the top of the pool, what is the volume (to the nearest cubic foot) of water in the pool?
Mathematics
2 answers:
andreev551 [17]3 years ago
7 0
The answer is 952 cubic feet
svlad2 [7]3 years ago
3 0

Answer:

952 cubic feet

Step-by-step explanation:

The diameter is 21, so the radius is 10.5. The area of a circle is πr^2, so 10.5^2π, or 110.25π. The pool is filled 9 inches from the top, so it is 3.5 feet - 9 inches high, 9 inches is 0.75 feet, so 2.75 feet filled. Now multiply 110.25π by 2.75 to get 303.1875π and then 952.4916, closer to 952 cubic feet.

Tell me if I'm right

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3 years ago
Find the mean of the data summarized in the given frequency distribution . Compare the computed mean to the actual mean of 57.6
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The mean of the frequency distribution is 52.6° F. The computed mean of 52.6° F is lesser than the actual mean of 57.6° F.

<h3>What is the mean of a distribution?</h3>

The mean of the distribution can be defined as the average value of the distribution, It can be expressed as the total sum of all the observed values divided by the frequency of the distribution.

From the parameters given:

Low-temperature             Frequency

40 - 44                                    2

45 - 49                                    5

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The first thing to do is to determine the class midpoint. The class midpoint is the sum of the class interval divided by 2.

The class midpoint of 40 - 44 is \mathbf{\dfrac{40 + 44}{2} = 42}

The class midpoint of 45 - 49 is \mathbf{\dfrac{45 + 49}{2} = 47}

The class midpoint of 50 - 54 is \mathbf{\dfrac{50 + 54}{2} = 52}

The class midpoint of 55 - 59 is \mathbf{\dfrac{55 + 59}{2} = 57}

The class midpoint of 60 - 64 is \mathbf{\dfrac{60 + 64}{2} = 62}

Now, the table can be represented as:

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40 - 44                                    2                   42

45 - 49                                    5                   47

50 - 54                                    9                   52

55 - 59                                    6                   57

60 - 64                                    3                   62

The mean can now be determined as follows:

\mathbf{\bar x = \dfrac{\sum fx}{\sum f }}

\mathbf{\bar x = \dfrac{(2 \times 42)+ (5 \times 47) + ( 9\times 52) +(6\times 57) + (3 \times 62)}{2 + 5 + 9 + 6 + 3 }}

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