Sentence examples for before we assumed from inspiring English sources

For the farms that were infected before, we assumed that a new infection in 2008 will have negligible effect based on the observation that farms infected in 2006 had negligible health problems due to BTV-8 during the epidemic of 2007.

As before, we assumed ten initial cases, and (as in Figure 6) that 30 ring vaccinations were possible per day; then we simulated 100 epidemics assuming one day to find a household contact (and 2 days to find a workplace/social contact).

While, as discussed before, we assumed that most individuals with B-notification that have active disease upon entry to the U.S. are found to be smear-negative, we assume that some fraction of smear-negative individuals will progress to smear-positive if mortality or diagnosis does not occur.

In the sections before we assumed that the entry time T is fixed and entry is taken for granted (Section 2).

As discussed before, we assume that S∗>0.

As before, we assume the validity of condition (2.18).

Similarly as before, we assume that E w w H = N o ⋅ I K.

Similar to before, we assume that (frac {1}{sqrt {K}} widehat {mathbf {H}}), w ℓ, and (mathbf {P}^{frac {1}{2}}) are available in advance and the Hermitian operation is "free".

As before, we assume an even number N of observations forming a complete, undirected, weighted graph, G. Rather than find a minimum-weight non-bipartite matching, we find a minimum-weight r-regular spanning subgraph of G, where 1≤r≤N-2, denoted G r *.

Thus, it follows that lim M → ∞ C s BC* log ( log M ) ≥ ε p. Combining the upper and lower bounds, we have the desired asymptotic throughput scaling result that lim M → ∞ C s BC* log ( log M ) = ε p. As discussed before, we assume that S∗,st>0.

Theorem 3.2 Let 1 < p ≤ 2, 0 < q ≤ ∞, 0 ≤ α < 1 p ′ and s ∈ R. Then the Fourier transform is bounded from the Herz type Besov space K ˙ p α, p B q s to the Herz type amalgams space ( K ˙ 2 − α, p ′, l q ( 〈 z 〉 s − n ( 1 q − 1 p ′ ) + ) ). Proof of Theorem 3.2 As before, we assume s = 0. Let | z | ∞ = max { | z 1 |, …, | z n | }.