Answer:
We conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Step-by-step explanation:
Given the expression
![\sqrt[3]{200k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
so the expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
first solving
![\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
Apply radical rule: ![\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![\sqrt[3]{k^{15}}=k^{\frac{15}{3}}=k^5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D%3Dk%5E%7B%5Cfrac%7B15%7D%7B3%7D%7D%3Dk%5E5)
then solving
![\sqrt[3]{200}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200%7D)
prime factorization: 200: 2³ · 5²
![=\sqrt[3]{2^3\cdot \:5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%5C%3A5%5E2%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
![=\sqrt[3]{2^3}\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%7D%5Csqrt%5B3%5D%7B5%5E2%7D)
Apply radical rule:
![\sqrt[n]{a^n}=a,\:\quad \:a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%2C%5C%3A%5Cquad%20%5C%3Aa%5Cge%200)
so
![=2\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D2%5Csqrt%5B3%5D%7B5%5E2%7D)
Thus, the main expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
![=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Therefore, we conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Answer:
I think the answer would be C.)
Answer:
See explanation below.
Step-by-step explanation:
Note that in △RST and △UVW
- m∠T=180°-m∠R-m∠S;
- m∠W=180°-m∠U-m∠V.
Since ∠R≅∠U and ∠S≅∠V, then ∠T≅∠W.
In ΔRST and ΔUVW:
- ∠S≅∠V (given);
- ∠T≅∠W (proved);
- ST≅VW (given).
ASA theorem that states that if two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.
By ASA theorem ΔRST≅ΔUVW.
So, 
Simplified as mixed number equals to

(
Keep in mind that the LCM was 8.)
Answer:
A) Yes
B) No
C) No
D) No
Step-by-step explanation:
A) Yes, every telephone number that could occur in that community will have an equal chance of being generated because the telephone numbers were generated randomly.
B) No, this method of generating telephone numbers would not result in a simple random sample (SRS) of local residences because the first three digits were generated in a different random way than the last four digits.
C) No, this method would not generate an SRS of local voters because not everybody will be pick up when they call because they may not be at home or they may be too busy to pick up.
D) No, this method is not unbiased in generating samples of households because there will be nonresponse bias since not everyone will pick up the call