Solve the attached problem...

6 years agoPossible answers are .0106 .2721 .0047 .1209 .7558

6 years agoYou can use the Pascal's triangle to solve this or Binomial coefficient which is \[ (x+a)^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right)x ^{n}+\left(\begin{matrix}n \\ 1\end{matrix}\right)x ^{n-1}a ^{1}\]

6 years agoHow would I use pascals triangle for this problem?

6 years agoThe pascal's triangle go like this http://en.wikipedia.org/wiki/File:PascalTriangleAnimated2.gif

6 years agoyou count the number 1 outside for the power that you have. For example in the picture if you count the number 1 on the outside there are 5. That's mean the problem is \[(x+a)^{5}\]

6 years agowhat would the x and a be for my question though?

6 years agoyour x = 3/5 and your a would be 2/5

6 years agolet's set up the Pascal triangle first.

6 years agobut then i only got 1, which is not one of my choices

6 years agooh... so i am not necessarily doing it to the fifth power

6 years agoYes 1 is incorrect since the question asks you to Expand you cannot add them together.

6 years agoYes your is to the 10th power

6 years agoso it would be ((3/5)+(2/5))^10?

6 years agoLook at the equation (3/5 + 2/5) ^10. The other stuffs are just there to explain the "history" of the problem. They are not important right now.

6 years agoCorrect!

6 years agobut that still only gets me 1 because (3/5)+(2/5)= 1 and 1^10=1

6 years agoNo you don't add them because the question ask to expand.

6 years agoTry to draw the pascal triangle

6 years ago|dw:1338071183404:dw| Can you do 5 more rows? from here?

6 years ago