Question 1148044
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Year 1958 is t=0; so year 2009 is t=51.<br>
The price in 1958 (t=0) was .04 dollars (4 cents):
{{{ab^0 = 0.04}}}
{{{a(1) = 0.04}}}
{{{a = 0.04}}}<br>
The price in 2009 (t=51) was .44 dollars (44 cents):
{{{0.04*b^51 = 0.44}}}
{{{b^51 = 0.44/0.04 = 11}}}
{{{b = 11^(1/51)}}}<br>
The rest I leave to you....<br>
(1) Use a calculator to evaluate b
(2) Form your equation y = ab^t
(3) Check your answer by verifying that ab^51 to the nearest cent is 44 cents
(4) For year 2020 (t=62), evaluate ab^62