Question

The population of mice in Alfred is given by P(t)=2654e7t, where t is in years since 1986. The rate of change of the population is given by the formula
(2654)(7e^(7t)) mice/yr. In year 1991 the population changes by approximately
2.946495792E19 mice. In 1991 8064483601014110592 mice died, which means that how much mice were born that year?

Answer 1

First, we determine t.

t = 1991 - 1986 = 5 years

We use t=5 to find the population in 1986 and the rate of change of population.

Population in 1986:

$P = 2654 e^{7(5)} =4.21 E 10^{18}$

Rate of change of population

$2654(7 e^{7(5)} )=2.95E 10^{19}$

Next, we determine the population in 1991. 

$1991\ population =1986 \ population+5(rate \ of \ change\ of \ population)$

$1991 \ population = 4.21 E 10^{18} +5(2.95E 10^{19})$

$1991 population = 1.52E 10^{20}$

Finally, we subtract the number of deaths from the total population to obtain the number of births:

$1.52E 10^{20}-8064483601014110592 = 1.436455164E 10^{20}$

Therefore, a total estimate of 143,645,516,400,000,000,000 mice were born in 1991.

Answer 2

Answer:  $2.1400474\times 10^{19}$      

Step-by-step explanation:

Since, According to the question,

Total mice die in 1991 = 8064483601014110592

Total changes in the population of mice in 1991 = 2.946495792e 19

Since,

The total changes in the population of the mice in 1991= difference between the population of total mice born and the total mice die.

⇒ 2.946495792e 19 = total mice born on 1991 - 8064483601014110592

⇒ total mice born on 1991 = 2.946495792e 19 + 8064483601014110592

⇒  total mice born on 1991 = 3.7529442e+19

Thus, On 1991 $3.7529442\times 10^{19}$  mice were born.

Note: Here, we assume that the mice born is more than mice dying because the population of mice is increasing every year exponentially.