Question

In 1995 the life expectancy of a male in a certain country was 63.6 years In 1999 it was 67.4 years Let E represent the life expectancy in year t and let t represent the number of years since 1995
The linear function E(t) that fits the data is
E(t)= t +
(round to the nearest tenth)
Use the function to predict the life expectancy of males in 2006
E(11)=
(round to the nearest tenth)

Answer 1

i love predicting past events

Answer 2

are you sure it says
\[E(t)=t+...\]

Answer 3

just think of finding the equation for the line through the points
\[(0,63.6),(4,67.4)\]

Answer 4

e(t)=

Answer 5

the slope of the line is
\[\frac{67.4-63.6}{4}=\frac{3.8}{4}=.95\]

Answer 6

which is another way of saying the linear model says in 4 years it increased by 3.8 so average in crease is .95 per year

Answer 7

so
\[E(t)=.95t+63.6\] since you started counting at 63.6

Answer 8

2006 is 11 years after 1995 so find
\[E(11)=.95\times 11+63.6\]

Answer 9

can you round that to the nearest 10th please?