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2 years ago

11

In the data set below, what is the interquartile range? 7 1 5 5 9 Submit

1 answer:

Marizza181 [45]2 years ago

3 0

Answer:

5

Step-by-step explanation:

The interquartile range is the difference between the upper quartile and the lower quartile.

First find the median.

The median is the middle value of the data set arranged in ascending order

1 5 5 7 9 ← data in ascending order

↑ median

The lower quartile is the middle value of the data to the left of the median. If there is not an exact middle then it the average of the values on either side of the middle.

1 5

↑ lower quartile = $\frac{1+5}{2}$ = 3

The upper quartile is the middle value of the data to the right of the median.

7 9

↑ upper quartile = $\frac{7+9}{2}$ = 8

Thus

interquartile range = 8 - 3 = 5

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Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

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$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

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- Multiply both sides by $m-3$ :

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- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

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