\(Given the equation y = Ae^-kx2, taking the natural logarithm of both sides gives:\)

\(\ln y = \ln A - kx^2\)

This is in the form of a straight line \(y = mx + c\) with \(\ln y\) as the dependent variable and \(x^2\) as the independent variable. The slope \(m = -k\) and the intercept \(c = \ln A\).

Using the points (0.64, 0.76) and (1.69, 0.32), calculate the slope:

\(m = \frac{0.32 - 0.76}{1.69 - 0.64} = \frac{-0.44}{1.05} = -0.419\)

Thus, \(k = 0.42\).

Substitute one of the points into the equation \(\ln y = \ln A - kx^2\) to find \(\ln A\):

Using (0.64, 0.76):

\(0.76 = \ln A - 0.42 \times 0.64^2\)

\(0.76 = \ln A - 0.42 \times 0.4096\)

\(0.76 = \ln A - 0.172032\)

\(\ln A = 0.76 + 0.172032 = 0.932032\)

Therefore, \(A = e^{0.932032} \approx 2.80\).