Question

I need help with this question.

Answer 1

do you know any specific methods? there's many ways to do it.

Answer 2

(74.4-73.1 / no.of years) * t +expectancy in 1960.
that would be the equation.

Answer 3

I don't get it..

Answer 4

\[\frac{ Expectancy 1970 -Expectancy 1960 }{ Number of years. } \times T + Expectancy of 1960\]

Answer 5

Okay, now what is the next step?

Answer 6

ummm plug in the values from the table

Answer 7

Can you do just the first one for me? So I can follow the example afterwards

Answer 8

well look on table.. tell me the vallue of expectancy of 1960 and 1970

Answer 9

1960 - 73.1
1970 - 74.4

Answer 10

now put these both values in the formula i wrote above.

Answer 11

But what would I put for number of years, T and Expectancy of 1960?

Answer 12

okay, by linearity the slope remains constant in a straight line, pick any time (t) and expentency E(t) on the line. take any two points in ur given data and find slope, dE(t)/dt e.g 74.4-73.1/1970-1960=0.13 so this remmains contant for any two points, now for any unknown point (t,E(t)) you use slope that is E(t)-1960/t-73.1=0.13, cross multiply and take 1960 on other side u have funtion E(t) in terms of t.

Answer 13

no of years is 1970-1960 bcos u r taking exxpectancy for these years right?
you dont have to put anything for T and expectancy of 1960.. u just told me above ;)

Answer 14

i made a mistake i was E(t)-73.1/t-1960=0.13