Answer:
the minimum records to be retrieved by using Chebysher - one sided inequality is 17.
Step-by-step explanation:
Let assume that n should represent the number of the students
SO,
can now be the sample mean of number of students in GPA's
To obtain n such that 
⇒ 
However ;

![E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20D%5Cint%5Climits%5E4_2%20%282%2Be%5E%7B-x%7D%29dx%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7BD%7D%7B3%7D%5Be%5E%7B-4%7D%20%282e%5Ex%20x%5E3%20-3x%5E2%20-6x%20-6%29%5D%5E4__2%7D%7D%3D%2038.21%20%5C%20D)
Similarly;

⇒ 
⇒ 
⇒ 

∴ 
Now; 
Using Chebysher one sided inequality ; we have:

So; 
⇒ 
∴ 
To determine n; such that ;

⇒ 

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.
Answer:
25%- 1875 Votes
<em>7500 x 0.55 = 4125 (</em><em>first candite valid votes</em><em>)</em>
<em>7500 x 0.20 = 1500 (</em><em>invalid</em><em>)</em>
<em>first candite valid votes</em><em> </em><em>(</em><em>55</em><em>) + invalid votes (</em><em>20</em><em>) = 75% of total votes </em>
<em>first candite valid votes</em><em> </em><em>(</em><em>4125</em><em>) + invalid votes (</em><em>1500 </em><em>) = 5625 of total votes </em>
<em />
<em>100 (</em><em>total</em><em> </em><em>percent of votes</em><em>) - 75 (</em><em>total percent of votes</em><em>) = 25% Votes Left</em>
<em>7500 (</em><em>total</em><em> </em><em>number of votes</em><em>) - 5625 (</em><em>total number of votes</em><em>) = 1875 Votes Left</em>
<em>7500 x 0.25 = 1875 (</em><em>valid votes for the other candite</em><em>)</em>
<em> </em>
Answer:
3/8
Step-by-step explanation:
Rearrange, find common denominator.

Answer:
B
Step-by-step explanation: