Calculate A(N) / B(N) with C(N) digits Consider three number sequences, \$A\$, \$B\$ and \$C\$:

\$A\$: A sequence based on recurrence relations, \$f(n) = f(n-1)+f(n-2)\$, starting with \$f(1) = 3, f(2) = 4\$. So, the sequence begins like this: \$3, 4, 7, 11, 18, 29, 47, 76, ...\$
\$B\$: The composite numbers, that is all integers that aren't primes (or 1): \$4, 6, 8, 9, 10, 12, 14, 15, 16, ...\$
\$C\$: The digits of \$\pi\$: \$3, 1, 4, 1, 5, 9, 2, 6, 5, ...\$

Given a positive integer \$N &lt; 50\$, either as function argument or STDIN, return the decimal value of the fraction \$\frac{A(N)}{B(N)}\$ with \$C(N)\$ digits after the decimal point. Normal rules for rounding applies (round up if the \$N+1\$'th digit is 5 or higher). If the Nth digit of \$\pi\$ is zero, an integer should be printed. scientific notation / Standard form is accepted for numbers higher than 1000.
This is code golf, so the shortest answer in bytes wins.
Some examples:
N = 1: 0.750
N = 2: 0.7
N = 3: 0.8750
N = 4: 1.2
N = 6: 2.416666667
N = 10: 11.056
N = 20: 764.8750

Of course, standard code golf rules applies.
The function must terminate in less than two minutes on any modern laptop.
Pyth, 60 57 58 bytes

.[K`.Rchu,eGsGQjT7e.ftPZQ)Je/u+/*GHhyHy^TQr^T3ZZT\0+xK\.hJ


Test harness

It's pretty straightforward - calculate pi, the fibonacci series and the composites, round to C(n) digits, pad to C(n) digits plus the location of the decimal point digits, done.

A(n): hu,eGsGQjT7

B(n): e.ftPZQ)

C(n): e/u+/*GHhyHy^TQr99ZZT

60 -> 57: Cleaned up the n = 1 special case in the pi calculation.

57 -> 58: Wasn't using high enough precsion for pi for the entire input range - increased 99 iterations to 1000 iterations.

Note on rounding: This uses Python's "nearest even" rounding system, rather than the OP specified "towards infinity" system. However, the difference only matters if the digits immediately following the rounding point are 5000..., e.g. 1.25 rounded to 1 digit. I checked the input range, and this never happens, so the correct result is alway returned.