If the noise sources are uncorrelated, then the power (variance) of its sum equals the sum of their powers (variances), because all the co-variances between them are zero, and only the variances remain.
An example with two noise sources (with zero mean):
\$\sigma_N^2 = E[N^2] = E[(X+Y)^2] = E[X^2]+E[Y^2]+2E[XY] = \sigma_X^2 + \sigma_Y^2 +2\sigma_{XY}\$
If \$X\$ and \$Y\$ are uncorrelated then \$\sigma_{XY}=0\$ and \$\sigma_N^2 = \sigma_X^2 + \sigma_Y^2 \$, thus:
\$\sigma_N = \sqrt{\sigma_X^2 + \sigma_Y^2} \$, where \$\sigma_N\$, \$\sigma_X\$ and \$\sigma_Y\$ are the RMS values of \$N\$, \$X\$ and \$Y\$.
It can be demonstrated for any number of sources, if required. The result ends up being the same: power of the sum equals the sum of the powers.