Answer:
The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
Step-by-step explanation:
Answer: complex equations has n number of solutions, been n the equation degree. In this case:
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i11,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi11%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i101,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi101%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i191,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi191%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i281,25°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi281%2C25%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i78,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi78%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i168,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi168%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i258,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi258%2C75%C2%B0%7D)
![Z=\frac{\sqrt[8]{2} }{\sqrt[4]{2}} e^{i348,75°}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B%5Csqrt%5B8%5D%7B2%7D%20%7D%7B%5Csqrt%5B4%5D%7B2%7D%7D%20e%5E%7Bi348%2C75%C2%B0%7D)
Step-by-step explanation:
I start with a variable substitution:

Then:

Solving the quadratic equation:


Replacing for the original variable:
![Z=\sqrt[4]{0,5+0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5%2B0%2C5i%7D)
or ![Z=\sqrt[4]{0,5-0,5i}](https://tex.z-dn.net/?f=Z%3D%5Csqrt%5B4%5D%7B0%2C5-0%2C5i%7D)
Remembering that complex numbers can be written as:

Using this:

Solving for the modulus and the angle:
![Z=\left \{ {{\sqrt[4]{\frac{\sqrt{2}}{2} e^{i45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i45}} } \atop {\sqrt[4]{\frac{\sqrt{2}}{2} e^{i-45}} = \sqrt[4]{\frac{\sqrt{2}}{2} } \sqrt[4]{e^{i-45}} }} \right.](https://tex.z-dn.net/?f=Z%3D%5Cleft%20%5C%7B%20%7B%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi45%7D%7D%20%7D%20%5Catop%20%7B%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20e%5E%7Bi-45%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%7D%20%5Csqrt%5B4%5D%7Be%5E%7Bi-45%7D%7D%20%7D%7D%20%5Cright.)
The possible angle respond to:

Been "RAng" the resultant angle, "Ang" the original angle, "n" the degree of the root and "i" a value between 1 and "n"
In this case n=4 with 2 different angles: Ang = 45º and Ang = 315º
Obtaining 8 different angles, therefore 8 different solutions.
D for the same reason percents
Answer:
400
Step-by-step explanation:
Im pretty sure that it is the first answer