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17!

(17-3)! 3!                X .15^3 X .85^17-3

 Oct 8, 2014

Best Answer 

 #2
avatar+33661 
+10

I think this might be asking for the value of the 14th term of the expansion (p + q)17, where p = 0.15 and q = 0.85.

$${\left({\frac{{\mathtt{17}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{17}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.15}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.85}}}^{\left({\mathtt{17}}{\mathtt{\,-\,}}{\mathtt{3}}\right)} = {\mathtt{0.235\: \!856\: \!391\: \!573\: \!379\: \!6}}$$

 Oct 9, 2014
 #1
avatar+23254 
0

I'm not sure what the question is.

 Oct 9, 2014
 #2
avatar+33661 
+10
Best Answer

I think this might be asking for the value of the 14th term of the expansion (p + q)17, where p = 0.15 and q = 0.85.

$${\left({\frac{{\mathtt{17}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{17}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.15}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{{\mathtt{0.85}}}^{\left({\mathtt{17}}{\mathtt{\,-\,}}{\mathtt{3}}\right)} = {\mathtt{0.235\: \!856\: \!391\: \!573\: \!379\: \!6}}$$

Alan Oct 9, 2014
 #3
avatar+130513 
0

Thanks, Alan....I couldn't make heads or tails of that one......!!!!

 

 Oct 9, 2014
 #4
avatar+33661 
0

I was puzzled at first, but then realised the clue was in the title!

 Oct 9, 2014

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