The ancient fish fossils were dated using C-14. If 1/16 the original amount of C-14 remains in the skeletons, approximately how old are the fish. The half-life of C-14 is 5730. a) 11,460 years
b) 17,190 years
c) 22,920 years
d) 28,650 years​

c) 22,920 years

Explanation:

The half-life of C-14 is 5730 so in 5730 years , amount of C-14 is reduced to half.

In 5730 x 2  years , the amount of C-14 is reduced to 1/4 .

In 5730 x 3 years ,  the amount of C-14 is reduced to 1/8 .

In 5730 x 4 years ,  the amount of C-14 is reduced to 1/16 .

In 22920 years ,  the amount of C-14 is reduced to 1/16 .

So the answer is 22920 years.

it is the second option.

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.