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Mars2501 [29]
3 years ago
13

Area of the circle. Leave your answer in terms of pi.

Mathematics
1 answer:
kramer3 years ago
6 0

Answer:

3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679

Step-by-step explanation:

by google

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Volume: 3x2x4=24 so the volume is 24 Meter3

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Coach Evans recorded the height, in inches of each player on his team. The results are shown.
marysya [2.9K]

Answer:

3

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile (Q_1)

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile (Q_3)

Terms in arranged in ascending order:

57,60,60,61,62,62,63,63,64

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

Q_1 is median of terms 57,60,60,61

Number of terms (n) = 4

Median = \frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th}  }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60

So, Q_1=60

So, 25% of the heights of a team is less than 60 inches

Q_3 is the median of terms 62,63,63,64

Median = \frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th}  }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63

So, Q_3=63

So, 75% of the heights of a team is less than 63 inches

Interquartile range = Q_3-Q_1=63-60=3

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

4 0
3 years ago
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