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3 years ago

5+1/2(14b-8)=-6 5+b-4=-6 7b+=-6 7b= b= ANSWERS 7-7-1 1

Mathematics

1 answer:

Lunna [17]3 years ago

3 0

Answer:

7, 1, -7, -1.

Step-by-step explanation:

5+1/2(14b-8) = -6

5 + 7b - 4 = -6

7b + 1 = -6

7b = -7

b = -1

So, your answer will be 7, 1, -7, -1.

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:

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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5+1/2(14b-8)=-6 5+__b-4=-6 7b+__=-6 7b=__ b=__ ANSWERS 7-7-1 1

5+1/2(14b-8)=-6 5+b-4=-6 7b+=-6 7b= b= ANSWERS 7-7-1 1