Find the median of the first and third quartiles81, 69 ,90, 79, 59, 125, 95, 116, 183, 97, 92, 50, 122, 55, 9250, 55, 59, 69, 79, 81, 90, 92, 92, 95, 97, 116, 122, 125, 183.The first quartile: 50, 55, 59, 69, 79, 81, 90: median of the first quartile is 69The third quartile: 92, 95, 97, 116, 122, 125, 183: median of the third quartile is 116.The interquartile range is the difference between the median of both the first and third quartile: which is 116 - 69 = 47.There are no outliers in this set of data

4 0

2 years ago

4x^2-16=x^2+32<br> Using reverse PEMDAS

telo118 [61]

Answer:

x = ±4 (x = 4 and x = -4)

Step-by-step explanation:

4x²-16 = x²+32

3x²-16 = 32

3x² = 48

x² = 16

x = ±4

Therefore, x = 4 and x = -4

5 0

3 years ago

Help me plz I need it fast

e-lub [12.9K]

1524096 CM is the answer

8 0

3 years ago