**Solutions **

To solve this problem we have to use the quadratic function. An equation where the highest exponent is (usually "x") is a square (2).

Quadratic function = f(x)=ax²+bx+c

Three points have been given to us

⇒ (0 , 77.6)

⇒ (5 , 78)

⇒ (10 , 78.6)

Calculations

(0 , 77.6)

x=0

f(x)=77.6

a(0)²+b(0)+c=77.6

c=77.6

(5 , 78)

a(5)²+b(5)+77.6=78

25a+5b=78-77.6

25a+5b=0.4.............. (1)

(10 , 78.6)

a(10)²+b(10)+77.6=78.6

100a+10b=78.6-77.6

100a+10b=1.............. (2)

As you see equations number (1) and (2) have a system

100a+10b=1

25a + 5b=0.4

We solve this system we have to use elimination. In the elimination method you can add or subtract the equations to get an equation into one variable.

100a+10b=1

-4(25a+5b=0.4)

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

-10b=-0.6

⇒ b=-0.6/-10=0.06

100a+10b=1

-2(25a+5b=0.4)

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

50a = 0.2

⇒ a=0.2/50=0.004

Now we have a, b and c:

α = 0.004

β = 0.06

c = 77.6

Therefore, the quadratic function is:

f(x)=0.004x²+0.06x+77.6

x=year of the beginning of the interval - 1980

The life expectancy for females born between 1995 and 2000 is when

x=1995-1980=15

Therefore we do

f(15)=0.004(15)²+0.06(15)+77.6

f(15)=0.004(225)+0.06(15)+77.6

f(15)=79.4

Females between 2000 and 2005

x=2000-1980=20

Now we do

f(20)=0.004(20)²+0.06(20)+77.6

f(20)=1.6+1.2+77.6

f(20)=80.4

**The life expectancy for females born between 1995 and 2000 = 79.4 years.**

**The life expectancy for females born between 2000 and 2005 = 80.4 years.**