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3 years ago

8 miles is approximately equal to 6 km. How many km are equal to 28 miles? How many miles are equal to 42 km?

Mathematics

2 answers:

Elan Coil [88]3 years ago

5 0

It is to 6km approximately and im in 6th grade <span />

miv72 [106K]3 years ago

5 0

28 miles = 21 km
56 miles = 42 km
And I'm not sure, but you may have gotten the units mixed up

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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:

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.

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Nadya [2.5K]

<h3>Answer: 30 inches</h3>

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Explanation:

x = unknown width

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Set this equal to the given perimeter (94) and solve for x.

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malfutka [58]

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