<span>The ratio of 2 similar areas is the square of the ratio of their RADIUS or (sides)
R²=(2/3)² = 4/9 This should be the answer
Whereas R³ = (2/3)³ = 8/27 is the ratio of their volume & not area
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Answer:
The simplified version of ![\sqrt[3]{135}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D)
is ![3\sqrt[3]{5}](https://tex.z-dn.net/?f=3%5Csqrt%5B3%5D%7B5%7D)
.
Step-by-step explanation:
The given expression is
According to the property of radical expression.
![\sqrt[n]{x}=(x)^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%3D%28x%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Using this property we get
![\sqrt[3]{135}=(135)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%28135%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\sqrt[3]{135}=(27\times 5)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%2827%5Ctimes%205%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\sqrt[3]{135}=(3^3\times 5)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%283%5E3%5Ctimes%205%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![\sqrt[3]{135}=(3^3)^{\frac{1}{3}}\times (5)^{\frac{1}{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D%283%5E3%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%20%285%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
![[\because (ab)^x=a^xb^x]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28ab%29%5Ex%3Da%5Exb%5Ex%5D)
![\sqrt[3]{135}=3\times \sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D3%5Ctimes%20%5Csqrt%5B3%5D%7B5%7D)
![[\because \sqrt[n]{x}=(x)^{\frac{1}{n}}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5Bn%5D%7Bx%7D%3D%28x%29%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5D)
![\sqrt[3]{135}=3\sqrt[3]{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B135%7D%3D3%5Csqrt%5B3%5D%7B5%7D)
Therefore the simplified version of is .
The extraneous root is
x = 2
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
