Answer t Which notation shows that P is a function of x?his question…
You need to specify what the following are if you want to make it possible to answer your question.
Which one is correct dG(x-0)-dG(x+0)=-1/p(t) or 1/p(t)
Which one is correct dG(x-0)-dG(x+0)=-1/p(t) or 1/p(t)
The equation for this exponential growth function is: P(t) = 76 * 4^t, where P(t) is the population at time t and 4 represents the quadrupling factor. The initial population at time t=0 is 76.
Answer is C. P&T committee
template<typename T> std::vector<T>* create_new_vector (const size_t M) { std::vector<T>* p = nullptr; try { if (p = new std::vector<T>) { p->resize (M); p->shrink_to_fit (); } } catch (std::exception& e) { throw e; } return p; }
A possible exponential decay function for this scenario would be P(t) = P0 * (0.5)^(t/50), where P(t) is the amount remaining after time t, P0 is the initial amount, and t is the time passed in years. This formula represents the decay of a substance with a half-life of 50 years.
p --> q and q --> p are not equivalent p --> q and q --> (not)p are equivalent The truth table shows this. pq p --> q q -->(not)p f f t t f t t t t f f f t t t t
p > q~qTherefore, ~p| p | q | p > q | ~q | ~p || t | t | t | f | f || t | f | t | t | f || f | t | t | f | t || f | f | t | t | t |
4t - 2p
[(1 - p)/(1 - pet)]r for t < -ln(p) where p = probability of success in each trial, r = number of failures before success.