In the figure shown, a perpendicular segment is drawn from B in rectangle ABCD to meet diagonal AC at point X.  Side AB is 6 cm and diagonal AC is 10 cm.  How many centimeters away is point X from the midpoint M of the diagonal AC?  Express your answer as a decimal to the nearest tenth.

[asy]
draw((0,0)--(25,0)--(25,15)--(0,15)--cycle,linewidth(1));
draw((0,0)--(25,15),linewidth(1));
draw((7.2,4.3)--(0,15),linewidth(1));
dot((12.5,7.5));
label("$M$",(12.5,7.5),N);
label("$A$",(0,0),S);
label("$B$",(0,15),N);
label("$C$",(25,15),N);
label("$D$",(25,0),S);
label("$X$",(7.2,4.3),SE);
[/asy]
We see that $\triangle ABX \sim \triangle ACB$ since $\angle BXA = \angle ABC$ and $\angle A$ is shared by both triangles. From this similarity, we have the equation $\frac{AB}{AC}=\frac{AX}{AB}$. Plugging in the known values of the sides and solving for $AX$, we have $\frac{6}{10}=\frac{AX}{6}\Rightarrow AX=3.6$. Since $AM=10/2=5$, $XM=5-3.6=\boxed{1.4}$ centimeters.