1. The problem is to understand and extract the formula for $m(t)$ given the integral expressions involving $\hat{f}(x)$ and $\hat{F}(x)$.\n\n2. The first expression is \n$$m(t) = \frac{\int x \hat{f}(x) \, dx}{\int \hat{f}(x) \, dx} - t$$\nThis represents the mean of $x$ weighted by $\hat{f}(x)$ minus $t$.\n\n3. The second expression is \n$$m(t) = \frac{\int \hat{F}(x) \, dx}{\int \hat{f}(x) \, dx}$$\nHere, $\hat{F}(x)$ is presumably the cumulative distribution function related to $\hat{f}(x)$.\n\n4. The third expression is \n$$m(t) = \frac{\int \hat{F}(x) \, dx}{\int \hat{f}(x) \, dx} - \hat{F}(t)$$\nThis shows $m(t)$ as the ratio of integrals minus the value of $\hat{F}$ at $t$.\n\n5. These expressions are equivalent forms of $m(t)$, showing relationships between the weighted mean, cumulative distribution, and their integrals.\n\nFinal extracted formula: \n$$m(t) = \frac{\int \hat{F}(x) \, dx}{\int \hat{f}(x) \, dx} - \hat{F}(t)$$