Assume we have $$N_1=5 \text{ dB}$$ $$N_2= - 110 \text{ dB}$$ Then we have $$Y=N_1+N_2=-105 \text{ dB} $$ If I convert to scalar then $$10 \log(X) = -105 \rightarrow X=10^{-10.5}$$
Let me start the problem differently
$$N_1'=5 \text{ dB}= 10^{5/10}$$ $$N_2'= - 110 \text{ dB}= 10^{-11}$$ Then $$Y'=N_1'+N_2'= 10^{5/10}+10^{-11} $$
Why isn't $$Y'=Y$$
$10 \log (X_1) = Y_1$ [in db]
$10 \log (X_2) = Y_2$ [in db]
Y’s can be added to a new total (but not X’s; and they should be multiplied instead. [see below.])
After adding, $10 \log (X_1) + 10 \log (X_2) = Y_1 + Y_2$ [in db]
$10[\log (x_1) + \log (x_2)] = Y_1 + Y_2$ [in db]
$10 \log (X_1 \cdot X_2) = Y_1 + Y_2$ [in db]