Consider a data consisting of 10 observations $$x_1,x_2,\dots,x_{10}$$, whose mean is $$5$$ and variance is $$7$$. If the mean and the variance of the first 8 observations $$x_1,x_2,\dots,x_8$$ are $$4$$ and $$3.5$$, respectively, and $$x_9 < x_{10}$$, then the value of $$3x_9 + 2x_{10}$$ is ___________.

For $$N = 10$$ observations:

$$\bar{x}_{10} = 5 \implies \sum_{i=1}^{10} x_i = 10 \times 5 = 50$$

$$\sigma_{10}^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - (\bar{x}_{10})^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - 25 = 7 \implies \sum_{i=1}^{10} x_i^2 = 320$$

For the first 8 observations ($$N = 8$$):

$$\bar{x}_{8} = 4 \implies \sum_{i=1}^{8} x_i = 8 \times 4 = 32$$

$$\sigma_{8}^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - (\bar{x}_{8})^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - 16 = 3.5 \implies \sum_{i=1}^{8} x_i^2 = 156$$

Calculating relations for $$x_9$$ and $$x_{10}$$:

$$\sum_{i=1}^{10} x_i - \sum_{i=1}^{8} x_i = x_9 + x_{10} \implies 50 - 32 = x_9 + x_{10} \implies x_9 + x_{10} = 18 \quad \text{--- (1)}$$

$$\sum_{i=1}^{10} x_i^2 - \sum_{i=1}^{8} x_i^2 = x_9^2 + x_{10}^2 \implies 320 - 156 = x_9^2 + x_{10}^2 \implies x_9^2 + x_{10}^2 = 164 \quad \text{--- (2)}$$

$$18^2 = 164 + 2x_9x_{10} \implies 324 = 164 + 2x_9x_{10} \implies x_9x_{10} = 80 \quad \text{--- (3)}$$

Solving the quadratic equation $$t^2 - (x_9+x_{10})t + x_9x_{10} = 0$$:

$$t^2 - 18t + 80 = 0 \implies (t-8)(t-10) = 0 \implies t = 8, 10$$

Since $$x_9 < x_{10}$$: $$x_9 = 8, \quad x_{10} = 10$$

$$3x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44$$