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solong [7]
3 years ago
8

christina lost 1/10 of her marbles when she was on the playground. She lost 8/10 of her marbles while walking home. How many mar

bles did she lose altogether?
Mathematics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

9/10

Step-by-step explanation:

She lost 1/10 and 8/10 marbles. When you add them up you get 9/10. There for, Christina lost 9/10 marbles altogether.

Hope this helps :)

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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

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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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