1) when t=(1995-1995)=0, E(0)=64.8 and t=6, E(6)=67.5
E(t) = mt +c
first instance,
E(0)=m(0)+c ==> c=64.8
second instance
E(6)=m(6)+64.8 ==> m=(67.5-64.8)/6
m=0.45
E(t)=0.45t+64.8
Was trying to help my sister with her math homework, but its been so long since I've done this stuff that I am kinda stuck. Any help would be appreciated.
1. In 1995, the life expectancy of males in a certain country was 64.8 years. In 2001, it was 67.5 years. Let E represent the life expectancy in year t and let t represent the number of years since 1995
1. The linear function E(t) that fits the data is
E(t)=(?)t+(?)
2.In 1920 the record for a certain race was 45.8 secs. In 1930, it was 45.6 sec. Let R(t)= the record in the race and t=the number of years since 1920.
a)Find a Linear Function that fits the data.
b)Use the function in (a) to predict the record in 2003 and in 2006
c) Find the year when the record will be 44.06 sec.
1) when t=(1995-1995)=0, E(0)=64.8 and t=6, E(6)=67.5
E(t) = mt +c
first instance,
E(0)=m(0)+c ==> c=64.8
second instance
E(6)=m(6)+64.8 ==> m=(67.5-64.8)/6
m=0.45
E(t)=0.45t+64.8
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Algebra 1