Answer:
y 
1/2x + 2 :)
Step-by-step explanation:
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
Margin of error, e = Z*SD/Sqrt (N), where N = Sample population
Assuming a 95% confidence interval and substituting all the values;
At 95% confidence, Z = 1.96
Therefore,
0.23 = 1.96*1.9/Sqrt (N)
Sqrt (N) = 1.96*1.9/0.23
N = (1.96*1.9/0.23)^2 = 262.16 ≈ 263
Minimum sample size required is 263 students.
They both cost 10 dollars
Answer:
The answer to your question is:
Fraction he has review = 17/36
Fraction he have to study = 19/36
Step-by-step explanation:
Data
Monday studied = 2/9
Tuesday = 1/4
Fraction he has review = ?
Fraction he has left = ?
Process
As the denominator is 9, consider the whole is 9.
Then: Fraction left on Monday = 9/9 - 2/9
= 7 / 9
Fraction left on Tuesday = 7/9 - 1/4
= (28 - 9) / 36
= 19 / 36
Fraction he has review = 17/36
Fraction he have to study = 19/36
