Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
<u>Solution:
</u>
Need to calculate ![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
and then multiply the result by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
Let us first evaluate
![\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%5B3%5D%7B1728%7D%3D%5Csqrt%5B3%5D%7B12%20%5Ctimes%2012%20%5Ctimes%2012%7D%3D12)
As need to multiply 12 by
![\Rightarrow 12 \times \sqrt[3]{14903}](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D)
On solving , we get
![\sqrt[3]{14903}=24.608](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D%3D24.608)
![\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D%3D12%20%5Ctimes%2024.608%3D295.306)
Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
I got A I hope this helps
Step two is the answer i hope that helps
If we rearrange the given data set from lowest to highest, we have
The minimum is 3.
The first quartile is the average of 6 and 7, which is 6.5
The median is the average of 8 and 10 which is 9.
The third quartile is the average of 12 and 15 which is 13.5.
The maximum is 18.
