Asked by need help please
                ok i don't get this one 
in 1920 the record for a certain race was 45.9 sec in 1980 it was 44.1 sec let r (T) = the record in the race and t= the number of years since 1920
(a) find a linear fuction that fits the data
(b) use the funcation in (a) to perdict the record for 2003 and 2006
(c) find t he year when the record will be 43.20 sec..
help please thatnks in advance
            
        in 1920 the record for a certain race was 45.9 sec in 1980 it was 44.1 sec let r (T) = the record in the race and t= the number of years since 1920
(a) find a linear fuction that fits the data
(b) use the funcation in (a) to perdict the record for 2003 and 2006
(c) find t he year when the record will be 43.20 sec..
help please thatnks in advance
Answers
                    Answered by
            Damon
            
    r = m (t) + b
when year = 1920, t = 0
where t = year - 1920
45.9 = m (0) + b
so b = 45.9
44.1 = m (1980-1920) + 45.9
44.1 = 60 m + 45.9
60 m = -1.8
m = -.03
so
r =-.03 t + 45.9
when year = 2003
t = 2003 - 1920 = 83
r = -.03 (83) + 45.9
r = 43.41
that should get you started
    
when year = 1920, t = 0
where t = year - 1920
45.9 = m (0) + b
so b = 45.9
44.1 = m (1980-1920) + 45.9
44.1 = 60 m + 45.9
60 m = -1.8
m = -.03
so
r =-.03 t + 45.9
when year = 2003
t = 2003 - 1920 = 83
r = -.03 (83) + 45.9
r = 43.41
that should get you started
                    Answered by
            need help please
            
    so what yr will it be 43.20
    
                    Answered by
            need help please
            
    so r(t)=?
    
                    Answered by
            Damon
            
    as I said
r =-.03 t + 45.9
then
43.2 = -.03 t + 45.9
.03 t = 2.7
t = 90
1920 + 90 = 2010
    
r =-.03 t + 45.9
then
43.2 = -.03 t + 45.9
.03 t = 2.7
t = 90
1920 + 90 = 2010
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