the u.s. population can be modeled by the function ( y = 165.6x^{1.345} ), where ( y ) is in thousands and (…

the u.s. population can be modeled by the function ( y = 165.6x^{1.345} ), where ( y ) is in thousands and ( x ) is the number of years after 1800. (a) what was the population in 1910, according to this model? (b) is the graph of this function concave up or concave down? what does this mean? (c) use numerical or graphical methods to find when the model estimates the population was 90,560,000. (a) according to the model, the population in 1910 was (square) thousands (round to the nearest integer as needed.)

the u.s. population can be modeled by the function ( y = 165.6x^{1.345} ), where ( y ) is in thousands and ( x ) is the number of years after 1800. (a) what was the population in 1910, according to this model? (b) is the graph of this function concave up or concave down? what does this mean? (c) use numerical or graphical methods to find when the model estimates the population was 90,560,000. (a) according to the model, the population in 1910 was (square) thousands (round to the nearest integer as needed.)

Answer

Explanation:

Step1: Calculate the value of (x) for 1910

The year 1910 is (1910 - 1800=110) years after 1800. So (x = 110).

Step2: Substitute (x = 110) into the function (y=165.6x^{1.345})

[ \begin{align*} y&=165.6\times(110)^{1.345}\ \end{align*} ] First, calculate (110^{1.345}). Using a calculator, (a = 110), (b=1.345), (a^{b}=110^{1.345}\approx110^{1 + 0.345}=110\times110^{0.345}\approx110\times3.47\approx381.7) (using the property (a^{m + n}=a^{m}\times a^{n}) and approximate value of (110^{0.345}) from calculator). Then (y = 165.6\times381.7\approx63210)

Answer:

63210