Question 1125291

Year Population (in millions)

{{{x}}}|{{{y}}}
{{{1950}}}|{{{ 6.79}}}
{{{1960}}}|{{{ 27.81}}}
{{{1970}}}|{{{ 45.59}}}
{{{1980}}}|{{{ 69.01}}}
{{{1990}}}|{{{ 84.39}}}
{{{2000}}}|{{{ 110.21}}}

A: 
use {{{x=0 }}} for {{{1950}}}, {{{x=10}}} for {{{1960}}},....
{{{x}}}|{{{y}}}
{{{0}}}|{{{ 6.79}}}
{{{10}}}|{{{ 27.81}}}
{{{20}}}|{{{ 45.59}}}
{{{30}}}|{{{ 69.01}}}
{{{40}}}|{{{ 84.39}}}
{{{50}}}|{{{ 110.21}}}


The linear regression equation that best fits this data in the form {{{y=ax+b}}} is: 

{{{y=ax+b}}} use {{{x=0}}}|{{{ y=6.79}}}

{{{6.79=a*0+b}}}
{{{b=6.79}}}

{{{y=ax+b}}} use {{{x=10}}}|{{{ y=27.81}}} and {{{b=6.79}}}

{{{27.81=10a+6.79}}}
{{{27.81-6.79=10a}}}
{{{21.02=10a}}}
{{{a=21.02/10}}}
{{{a=2.102}}}
 
_____ {{{y=2.102x+6.79}}}  _________. Round the constants to the nearest thousandth.



B: Use your regression equation to predict the population if {{{x=5.15}}}

{{{y=2.102*5.15+6.79}}}
{{{y=10.8253+6.79}}}
{{{y=17.615}}}
c: Approximate in what year the population reaches {{{38.05}}} million. 

{{{38.05=2.102x+6.79}}}
{{{38.05 -6.79  =2.102x}}}
{{{31.26  =2.102x}}}
{{{31.26 /2.102=x}}}
{{{x=14.9}}}....Approximately

since {{{x=10}}} represents year {{{1960}}}, {{{x=14.9=10+4.9}}} will represent {{{1960+4.9=1964.9}}} year which means in  year {{{1964}}}, and almost {{{11}}} months which means in November, {{{1964}}}