Question 47662
Consider the result after one year (t=1)

(a) 9 1/4% per year, compounded semiannually

A=P(1+0.0925/(2t)^(2t)
A=P(1+0.04625)^2
A=P(1.09463906...

(b) 9% per year, compounded continously
A=Pe^(0.09t)
A=Pe^0.09
A=P(1.09417428...)

So,for one year the 9 % compounded continuously is growing faster.

You would have to look at other values of t (maybe by graphing)
to see which is better in the long run.  I suspect the 
continuous compounding would be the winner.
To do this, let P be $1.00 and let t=x.

Cheers,
Stan H.