Question

According to the National Center for Health​ Statistics, in​ 1990, 28 % of babies in the United States were born to parents who were not married. Throughout the​ 1990s, this increased by approximately 0.6 % per year. If this trend​ continues, in which year will 76 % of babies be born out of​ wedlock?

Answer 1

76 % of babies will be born out of​ wedlock in the
`81^(st)` year

Explanation:

Given: In​ 1990, 28 % of babies in the United States were born to parents who were not married. This increased by approximately 0.6 % per year.

To find: year when 76 % of babies born out of​ wedlock

Solution:

A sequence is said to be in arithmetic progression if the difference between the terms is same.

For a sequence with first term as 'a' and common difference as 'd', the nth term is given by
`a_n=a+(n-1)d`

The given situation also forms an arithmetic progression with
`a=28\,,\,d=0.6`

Also, put
`a_n=76`

So,

`a_n=a+(n-1)d\\76=28+(n-1)(0.6)\\76-28=0.6(n-1)\\48=0.6(n-1)\\(48)/(0.6)=n-1\\\\80=n-1\\n=80+1\\n=81`

So, 76 % of babies will be born out of​ wedlock in the
`81^(st)` year