Graph the ellipse and locate the foci.
\[
16 x^{2}+9 y^{2}=144
\]

Step 1 ：Given the equation of the ellipse is \(16x^2 + 9y^2 = 144\).

Step 2 ：The standard form of an ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where a and b are the semi-major and semi-minor axes respectively.

Step 3 ：Rewrite the given equation in the standard form, we get \(\frac{x^2}{9} + \frac{y^2}{16} = 1\).

Step 4 ：From the standard form, we can identify the values of a and b. Here, \(a = 4\) and \(b = 3\).

Step 5 ：The foci of the ellipse are located at a distance of \(\sqrt{a^2 - b^2}\) from the center along the major axis.

Step 6 ：Calculate the distance to the foci, we get \(\sqrt{4^2 - 3^2} = 2.6457513110645907\).

Step 7 ：Final Answer: The foci of the ellipse are located at a distance of approximately \(\boxed{2.6457513110645907}\) units from the center along the major axis.