Find the iqr of the data set 0, 0, 1/4, 1/2, 1/2, 5/4, 1, 1, 1, 2, 2

Mathematics

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

0, 0, 1, 1, 1, 2, 2

Step-by-step explanation:

25th Percentile: 0

50th Percentile: 1

75th Percentile: 2

Interquartile Range: 2

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zhenek [66]

Answer: $\bold{y=\dfrac{1\pm \sqrt{4x+13}}{2}}$

<u>Step-by-step explanation:</u>

Inverse is when you swap the x's and y's and solve for y:

$y+\dfrac{13}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2\\\\\\x+\dfrac{13}{4}=\bigg(y-\dfrac{1}{2}\bigg)^2\rightarrow \text{(swapped the x's and y's)}\\\\\\\sqrt{x+\dfrac{13}{4}}=\sqrt{\bigg(y-\dfrac{1}{2}\bigg)^2}\rightarrow \text{(took square root of both sides)}\\\\\\\sqrt{\dfrac{4x+13}{4}}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(created common denominator in radical)}\\\\\\ \dfrac{\sqrt{4x+13}}{2}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(simplified radical)}$

$\pm \dfrac{\sqrt{4x+13}}{2}=y-\dfrac{1}{2}}\rightarrow \text{(divided both sides by}\ \pm )\\\\\\\dfrac{1}{2}\pm \dfrac{\sqrt{4x+13}}{2}=y\quad \rightarrow \text{(added}\ \dfrac{1}{2}\ \text{to both sides)}\\\\\\\dfrac{1\pm \sqrt{4x+13}}{2}=y\quad \rightarrow \text{(combined numerators)}$