A - makes sure the results are close to the actual value
Answer:
The correct answer is E. none of the above. The population will drops below 100 when t ≥ 38.
Explanation:
Given A= A0 e^kt. The population 10 years ago is A0, the population today is A(10), and we have to find the value of "k" and then the time when population drops below 100.
So, A(t) = 1700 e^kt ⇒ A(10) = 1700 e^k(10) ⇒ 800 = 1700 e^k(10) ⇒
800/1700 = e^k(10) ⇒ln (800/1700) = k(10) ln e ⇒ -0.754/10 = k ⇒
k = -0.0754.
Now you have all the parameters, so you can find the time at which the population drops below 100.
A(t) = 1700 e^kt ⇒ 100 = 1700 e^(-0.0754)t ⇒100/1700 = e^(-0.0754)t ⇒
ln(100/1700) = (-0.0754)t ln e ⇒ [ln(100/1700)]/(-0.0754) = t ⇒
t = 38.
So, the population will drops below 100 when t ≥ 38.
Answer and Explanation:
<u>Cruza</u>: persona sana sin mutaciones con persona portadora
El parental sano aporta alelos normales del gen a la descendencia
El parental portador aporta un alelo normal y un alelo alterado. Tiene el 50% de probabilidad de transmitir la enfermedad a la progenie.
Parental) AA x Aa
Gametas) A A A a
Cuadro de Punnet) A A
A AA AA
a Aa Aa
F1) 2/4 = 1/2 = 50% de la progenie seran individuos sanos AA
2/4 = 1/2 = 50% de la progenie seran individuos portadores Aa
Como uno de los progenitores es portador, cada individuo tiene un 50% de posibilidades heredar el gen alterado.
Answer:
D. To kill insects that commonly spread diseases to people
Reproduction of new cells because if you were seriously burned the cell tissue would be dead and new cells need to grow to take the place of old cells
