The perimeter of parallelogram $ABCD$ is 40, and its altitudes are 4 and 7.  Compute $\sin A.$
Label the parallelogram so that the distance between sides $\overline{BC}$ and $\overline{AD}$ is 4, and the distance between sides $\overline{AB}$ and $\overline{CD}$ is 7.  Then $AB = \frac{4}{\sin A}$ and $AD = \frac{7}{\sin A}.$

[asy]
unitsize(1.5 cm);

pair A, B, C, D, P, Q;

A = (0,0);
B = 2*dir(60);
C = B + (3,0);
D = (3,0);
P = (B + reflect(A,D)*(B))/2;
Q = (D + reflect(A,B)*(D))/2;

draw(A--B--C--D--cycle);
draw(B--P,dashed);
draw(D--Q,dashed);

label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, NE);
label("$D$", D, S);
label("$4$", interp(B,P,0.8), E, red);
label("$7$", interp(D,Q,0.5), NE, red);
label("$\frac{4}{\sin A}$", (A + B)/2, NW, red);
label("$\frac{7}{\sin A}$", (A + D)/2, S, red);
[/asy]

Therefore, the perimeter of $ABCD$ is
\[\frac{4}{\sin A} + \frac{7}{\sin A} + \frac{4}{\sin A} + \frac{7}{\sin A} = \frac{22}{\sin A} = 40.\]Hence, $\sin A = \boxed{\frac{11}{20}}.$