Solve the problem.
-The polynomial $0.0035 x^{3}+0.0053 x^{2}+0.18 x+1.07$ gives the approximate total earnings of a company, in millions of dollars, where $x=0$ corresponds to 1996, $x=1$ corresponds to 1997, and so on. This model is valid for the years from 1996 to 2000. Determine the earnings for 1999.
A)$\$ 2.1$ million
B) $\$ 1.66$ million
C) $\$ 1.48$ million
D) $\$ 1.75$ million

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The Answer of Solve the problem.
-The polynomial $0.0035 x^{3}+0.0053 x^{2}+0.18 x+1.07$ gives the approximate total earnings of a company, in millions of dollars, where $x=0$ corresponds to 1996, $x=1$ corresponds to 1997, and so on. This model is valid for the years from 1996 to 2000. Determine the earnings for 1999.
A)$\$ 2.1$ million
B) $\$ 1.66$ million
C) $\$ 1.48$ million
D) $\$ 1.75$ million

Solve the Problem 0.0035x3+0.0053x2+0.18x+1.070.0035 x^{3}+0.0053 x^{2}+0.18 x+1.070.0035x3+0.0053x2+0.18x+1.07 Gives the Approximate Total Earnings of a Company, in Millions

Solve the Problem $0.0035 x^{3}+0.0053 x^{2}+0.18 x+1.07$ Gives the Approximate Total Earnings of a Company, in Millions