Answer:
The answer is (d) ⇒ ![pq^{2}r\sqrt[3]{pr^{2}}](https://tex.z-dn.net/?f=pq%5E%7B2%7Dr%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D)
Step-by-step explanation:
* To simplify the cube roots:
If its number then the number must be written in the form x³
then we divide the power by 3 to cancel the radical
If its variable we divide its power by 3 to cancel the radical
∵ ![\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bp%5E%7B4%7Dq%5E%7B6%7Dr%5E%7B5%7D%7D%3Dp%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Dq%5E%7B%5Cfrac%7B6%7D%7B3%7D%7Dr%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
∴ 
∵ ![p^{\frac{1}{3}}=\sqrt[3]{p}](https://tex.z-dn.net/?f=p%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Bp%7D)
∵ ![r^{\frac{2}{3}}=\sqrt[3]{r^{2}}](https://tex.z-dn.net/?f=r%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D)
∴ ![p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})](https://tex.z-dn.net/?f=p%28p%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7Dq%5E%7B2%7Dr%28r%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3Dp%28%5Csqrt%5B3%5D%7Bp%7D%29q%5E%7B2%7Dr%28%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D%29)
∴ ![prq^{2}\sqrt[3]{pr^{2}}}](https://tex.z-dn.net/?f=prq%5E%7B2%7D%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D%7D)
∴ The answer is (d)
Answer:
The mean.
Step-by-step explanation:
I would use the mean as the measure of central tendency.
It is the best representation of the data as the variability isn't high; I'd only use the median if outliers are present because outliers don't affect it.
Because there are no outliers here, the mean is better.
Hopefully I helped :)
Answer:
334
Step-by-step explanation:
20+3(7+4)+5+267+9
=20+(3)(11)+5+267+9
=20+33+5+267+9
=53+5+267+9
=58+267+9
=325+9
