Answer:
0.8
Step-by-step explanation:
Answer:
Step-by-step explanation:
When you take the n-th root of a number, you can rewrite the expression by taking it to the 1/n-th power. For example:
![\sqrt[n]{x} =x^\frac{1}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%3Dx%5E%5Cfrac%7B1%7D%7Bn%7D)
For the first expression, we can use this proprtery to get:
![\sqrt[5]{a^x} =(a^x)^\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5Ex%7D%20%3D%28a%5Ex%29%5E%5Cfrac%7B1%7D%7B5%7D)
Using exponent rules, you can combine the exponents by simply multiplying them to get:
![a^\frac{x}{5}](https://tex.z-dn.net/?f=a%5E%5Cfrac%7Bx%7D%7B5%7D)
Moving on to the second expression. It is now the square root, or equivalently a 1/2 power. If we break up the terms under the radical into powers of 2, we can cancel a lot of the terms:
![\frac{\sqrt{81a^3b^1^0} }{\sqrt{3}a } =\frac{\sqrt{81a^2*a*b^1^0} }{\sqrt{3}a }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B81a%5E3b%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%20%3D%5Cfrac%7B%5Csqrt%7B81a%5E2%2Aa%2Ab%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
The a^2 and b^10 can be taken out of the radical because they have perfect roots:
![\frac{\sqrt{81a^2*a*b^1^0} }{\sqrt{3}a }=ab^5\frac{\sqrt{81a} }{\sqrt{3}a }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B81a%5E2%2Aa%2Ab%5E1%5E0%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%3Dab%5E5%5Cfrac%7B%5Csqrt%7B81a%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
The square root of 81 has a perfect root of 9. We have:
![ab^5\frac{\sqrt{81a} }{\sqrt{3}a }=9ab^5\frac{\sqrt{a} }{\sqrt{3}a }](https://tex.z-dn.net/?f=ab%5E5%5Cfrac%7B%5Csqrt%7B81a%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D%3D9ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7B%5Csqrt%7B3%7Da%20%7D)
You can divide 9 and the square root of 3 by breaking up 9 into a product:
![\frac{(3*3)ab^5}{\sqrt{3} } \frac{\sqrt{a} }{a }=3\sqrt{3} ab^5\frac{\sqrt{a} }{a }](https://tex.z-dn.net/?f=%5Cfrac%7B%283%2A3%29ab%5E5%7D%7B%5Csqrt%7B3%7D%20%7D%20%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D%3D3%5Csqrt%7B3%7D%20ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D)
Simply by cancelling the 'a' terms to get:
![3\sqrt{3} ab^5\frac{\sqrt{a} }{a }={3\sqrt{3}}\sqrt{a}b^5=3b^5\sqrt{3a}](https://tex.z-dn.net/?f=3%5Csqrt%7B3%7D%20ab%5E5%5Cfrac%7B%5Csqrt%7Ba%7D%20%7D%7Ba%20%7D%3D%7B3%5Csqrt%7B3%7D%7D%5Csqrt%7Ba%7Db%5E5%3D3b%5E5%5Csqrt%7B3a%7D)
The most frequently occurring element in a set of values is called the mode.
Hope that helped =)
Answer:
when n=7, x=2, ![p=\frac{1}{2}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
Given : Information n=7, x=2, ![p=\frac{1}{2}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7B2%7D)
To find : Evaluate ![C(n,x)\ p^x\ q^{n-x}](https://tex.z-dn.net/?f=C%28n%2Cx%29%5C%20p%5Ex%5C%20q%5E%7Bn-x%7D)
Solution :
The formula given is a binomial distribution,
can be written as ![^nC_x\ p^x\ q^{n-x}](https://tex.z-dn.net/?f=%5EnC_x%5C%20p%5Ex%5C%20q%5E%7Bn-x%7D)
Here, n=7 x=2 and ![p=\frac{1}{2}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7B2%7D)
![q=1-p=1-\frac{1}{2}=\frac{1}{2}](https://tex.z-dn.net/?f=q%3D1-p%3D1-%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Substitute all values in the formula,
![=^7C_2\ (\frac{1}{2})^2\ (\frac{1}{2})^{7-2}](https://tex.z-dn.net/?f=%3D%5E7C_2%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E2%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B7-2%7D)
![=\frac{7!}{2!(7-2)!}\ (\frac{1}{2})^2\ (\frac{1}{2})^{5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B7%21%7D%7B2%21%287-2%29%21%7D%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E2%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B5%7D)
![=\frac{7!}{2!5!}\ (\frac{1}{2})^{2+5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B7%21%7D%7B2%215%21%7D%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B2%2B5%7D)
![=\frac{7\times 6\times 5!}{2\times 5!}\ (\frac{1}{2})^{7}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B7%5Ctimes%206%5Ctimes%205%21%7D%7B2%5Ctimes%205%21%7D%5C%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B7%7D)
![=21\times\frac{1}{128}](https://tex.z-dn.net/?f=%3D21%5Ctimes%5Cfrac%7B1%7D%7B128%7D)
![=\frac{21}{128}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B21%7D%7B128%7D)
Therefore,
when n=7, x=2, ![p=\frac{1}{2}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7B1%7D%7B2%7D)