In 1995 the life expectancy of a male in a certain country was 70.98 years. In 2002 it was 73.4 years. Let E represent the life expectancy in year t and let t represent the number of years since 1995.
The liner function R(t) that fits the data E(t)= _t+_
round to the nearet 10th
Use the function to predict life expectancy in 2005
R(10)
round to the nearst 10th
From the last line of your question I conclude that the variable of t is to represent the number of years since 1995
So you basically have two ordered pairs,
(0,70.89) and (7,73.4)
the slope would be (73.4-70.89)/7 = .35857
so R(t) = .35857t + b , to match y = mx + b
but (0,70.89) is the "y-intercept" , so ...
R(t) = .35857t + 70.89
then 2005 would be t= 10
R10) = .35857(10) + 70.89
= .....
(To ask for the answer to the nearest 10th is rather silly, since there is only one decimal accuracy given in the data value of 73.4)