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It is not difficult to compute u from X.
By a similar argument, it is not difficult to compute that for each 0 < c ≤ 1, there exists B c ≥ c such that ϕ − 1 ( c s ) ≤ B c ϕ − 1 ( s ), ∀ s ∈ ( 0, 1 κ ).
For all (S geq 1), (H(S) geq H(1)=F(1 - mu_{1 - mu_{)=(mu_{1}+gamma)(sigma_{0}-1)>0) because (H(S)) is monotonically increasing on the interval ([0,+infty)) and (sigma_{0}>1). Therefore (H(S)=0) has exactly one root (S^in(0,1)). It is not difficult to comu_{e the expressions (I^) and (R^) from system (4) at the endemic equilibrium (E_). □.
For each 0 < c ≤ 1, there exists B ˆ c ≤ c such that ϕ ( c s ) = c s 1 + κ c 2 s 2 ≥ B ˆ c s 1 + κ s 2, ∀ s ≥ 0 ; and for each c ≥ 1, there exists A ˆ c ≥ c such that ϕ ( c s ) = c s 1 + κ c 2 s 2 ≤ A ˆ c s 1 + κ s 2, ∀ s ≥ 0. (iii) By a similar argument, it is not difficult to compute that for each 0 < c ≤ 1, there exists B c ≥ c such that ϕ − 1 ( c s ) ≤ B c ϕ − 1 ( s ), ∀ s ∈ ( 0, 1 κ ).
For several levels of exposure or disease, measures are available which correspond to odds ratios, risk ratios and risk differences [ 17], and it is not difficult to compute sample size formulas for these.
Similar(55)
Above we present a relatively simple joint indicator, RTB/HIV (1), which is not difficult to compute and is easily understood.
In recent computers, it is not difficult to employ not only single or double precision arithmetic but also arbitrary-precision arithmetic or symbolic computing.
"It is of course not difficult to compute," says Anurag Acharya, a co-founder of Google Scholar who leads its development.
But it isn't difficult.
It isn't difficult.
It's not difficult".
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.
Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com