# SOLUTION: The Function N(x)= -0.0534x^2 + 0.337x + 0.969 gives the number of active-duty military personnel in the United States Army (in millions) for the years 1965-1972, where x=0 corres

Algebra ->  Algebra  -> Functions -> SOLUTION: The Function N(x)= -0.0534x^2 + 0.337x + 0.969 gives the number of active-duty military personnel in the United States Army (in millions) for the years 1965-1972, where x=0 corres      Log On

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 Question 162201: The Function N(x)= -0.0534x^2 + 0.337x + 0.969 gives the number of active-duty military personnel in the United States Army (in millions) for the years 1965-1972, where x=0 corresponds to 1965, x = 1 corresponds to 1966 and so on. For this period, when was the Army's personnel strength level at its highest, and what was it? I put it into the equation x = -b/2a which I got -3.155 rounded off =3 i then put the -3 into the original equation and I got the answer of -.5226 so I am guessing I did something wrong! Any help? Please and thank you! :)Answer by stanbon(57387)   (Show Source): You can put this solution on YOUR website!The Function N(x)= -0.0534x^2 + 0.337x + 0.969 gives the number of active-duty military personnel in the United States Army (in millions) for the years 1965-1972, where x=0 corresponds to 1965, x = 1 corresponds to 1966 and so on. For this period, when was the Army's personnel strength level at its highest, and what was it? I put it into the equation x = -b/2a which I got -3.155 rounded off =3 i then put the -3 into the original equation and I got the answer of -.5226 so I am guessing I did something wrong! ---------------------------------------------------- For x = -b/2a you should get x = +3.155 which corresponds to the year 1965 + 3.155 which occurs in the year 1968.155 or in 1969. -------- Then N(3.1554307...) = 1.5007 million =============== Cheers, Stan H.