Answer:
x,y will be in the ratio 2:3
Step-by-step explanation:
Given that a biased coin has probability 0.6 of turning up heads
In other words, number of heads we obtain by tossing the coin would be binomial with p = 0.6 and q =1-p =0.4
If you toss one time, the distribution of U, no of heads would be
U 0 1
p 0.4 0.6
Winning amount would be 0.6x and losing amount would be 0.4y
If these two are equal then we have

Answer:
1-negative 2-non 3-positive 4-positve 5-non linear 6-negative
I think this might be it. Hope it helps..
See the attached figure
See the attached figure.
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The first equation is
4x + 2x²(3x-5) = 4x + 6x³ - 10x² = 6x³ - 10x² + 4x
So, The degree of the function = 3 , and the number of terms = 3
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The second equation is
(-3x⁴ + 5x³ - 12 ) + ( 7x³ - x⁵ + 6 ) = -x⁵ -3x⁴ +12x³ - 6
So, The degree of the function = 5 , and the number of terms = 4
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The third equation is
(3x² - 3)( 3x² + 3) = 9x⁴ - 9
So, The degree of the function = 4 , and the number of terms = 2
Rationalizing the denominator of ![\frac{\sqrt[3]{2z} }{\sqrt[3]{z^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B2z%7D%20%7D%7B%5Csqrt%5B3%5D%7Bz%5E2%7D%20%7D)
we get
.
Option D is correct.
Step-by-step explanation:
We need to rationalize the denominator: ![\frac{\sqrt[3]{2z} }{\sqrt[3]{z^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B2z%7D%20%7D%7B%5Csqrt%5B3%5D%7Bz%5E2%7D%20%7D)
Solving:
![\frac{\sqrt[3]{2z} }{\sqrt[3]{z^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B2z%7D%20%7D%7B%5Csqrt%5B3%5D%7Bz%5E2%7D%20%7D)
using Radical rule: ![\frac{\sqrt[n]{x}}{\sqrt[n]{y}}=\sqrt[n]{\frac{x}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5Bn%5D%7Bx%7D%7D%7B%5Csqrt%5Bn%5D%7By%7D%7D%3D%5Csqrt%5Bn%5D%7B%5Cfrac%7Bx%7D%7By%7D%20%7D)
![=\sqrt[3]{\frac{2z}{z^2}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2z%7D%7Bz%5E2%7D%7D)
![=\sqrt[3]{\frac{2}{z^{2-1}}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7Bz%5E%7B2-1%7D%7D%7D)
![=\sqrt[3]{\frac{2}{z^{1}}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7Bz%5E%7B1%7D%7D%7D)
![=\sqrt[3]{\frac{2}{z}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7Bz%7D%7D)
We can write it as:
![\frac{\sqrt[3]{2} }{\sqrt[3]{z} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%20%7D%7B%5Csqrt%5B3%5D%7Bz%7D%20%7D)
So, rationalizing the denominator of ![\frac{\sqrt[3]{2z} }{\sqrt[3]{z^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B2z%7D%20%7D%7B%5Csqrt%5B3%5D%7Bz%5E2%7D%20%7D)
we get
.
Option D is correct.
Keywords: Radical Expression
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