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 92,201 thousands. (round to the nearest integer as needed.) (b) is the graph of this function concave up or concave down? concave up concave down what does this mean? it means that the function is increasing at an increasing rate. (c) the model estimates that the population was 90,560,000 in the year

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 92,201 thousands. (round to the nearest integer as needed.) (b) is the graph of this function concave up or concave down? concave up concave down what does this mean? it means that the function is increasing at an increasing rate. (c) the model estimates that the population was 90,560,000 in the year

Answer

Explanation:

Step1: Substitute the population value into the function

Given (y = 165.6x^{1.345}), and (y=90560) (since (y) is in thousands, (90560000\div1000 = 90560)). So we have the equation (90560=165.6x^{1.345}). First, solve for (x^{1.345}): (\frac{90560}{165.6}=x^{1.345}), which gives (x^{1.345}\approx546.86).

Step2: Solve for (x)

Take both sides to the power of (\frac{1}{1.345}). So (x=(546.86)^{\frac{1}{1.345}}). Using a calculator, (x\approx100).

Answer:

Since (x) is the number of years after 1800, the year is (1800 + 100=1900).