Answer: ![3ab\sqrt[3]{b^4}](https://tex.z-dn.net/?f=3ab%5Csqrt%5B3%5D%7Bb%5E4%7D)
Step-by-step explanation:
Given the following expression:
![\sqrt[3]{27a^3b^7}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27a%5E3b%5E7%7D)
You need to apply the Product of powers property, which states that:

Then, you can rewrite the expression as following:
![=\sqrt[3]{27a^3b^4b^3}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B27a%5E3b%5E4b%5E3%7D)
The next step is to descompose 27 into its prime factors:

Now you must substitute
inside the given root. Then:
![=\sqrt[3]{3^3a^3b^4b^3}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B3%5E3a%5E3b%5E4b%5E3%7D)
You need to remember that, according to Radicals properties:
![\sqrt[n]{a^n}=a^{\frac{n}{n}}=a^1=a](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%5E%7B%5Cfrac%7Bn%7D%7Bn%7D%7D%3Da%5E1%3Da)
Therefore, the final step is to apply this property in order to finally get the expression is its simplest form. This is:
![=3^{\frac{3}{3}}a^{\frac{3}{3}}b^{\frac{4}{3}}b^{\frac{3}{3}}=3ab^{\frac{4}{3}}b=3ab\sqrt[3]{b^4}](https://tex.z-dn.net/?f=%3D3%5E%7B%5Cfrac%7B3%7D%7B3%7D%7Da%5E%7B%5Cfrac%7B3%7D%7B3%7D%7Db%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Db%5E%7B%5Cfrac%7B3%7D%7B3%7D%7D%3D3ab%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Db%3D3ab%5Csqrt%5B3%5D%7Bb%5E4%7D)
Answer:
The no. of possible handshakes takes place are 45.
Step-by-step explanation:
Given : There are 10 people in the party .
To Find: Assuming all 10 people at the party each shake hands with every other person (but not themselves, obviously) exactly once, how many handshakes take place?
Solution:
We are given that there are 10 people in the party
No. of people involved in one handshake = 2
To find the no. of possible handshakes between 10 people we will use combination over here
Formula : 
n = 10
r= 2
Substitute the values in the formula





No. of possible handshakes are 45
Hence The no. of possible handshakes takes place are 45.
Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>
<span>In part IV we are asked:
</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>
</span> , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>
,</span>(M, E), (M, D), (M, A), (M, S)
<span>
, </span>(E, D), (E, A), (E, S) <span>
, </span>(D, A), (D, S)
, (A, S).}
We can check that the number of the elements of the sample space, n(S) is
1+2+3+4+5+6+7=28.
This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates?
(Answer: 28.)
II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.
III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally?
The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.
Answers:
I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)
{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)
, (C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
, (G, M), (G, E), (G, D), (G, A), (G, S)
,(M, E), (M, D), (M, A), (M, S)
, (E, D), (E, A), (E, S)
, (D, A), (D, S)
, (A, S).}