Please answer number 2, please show work. Show the formula used, etc etc. Marvin was doing a study comparing the population growth of his hometown in the early twentieth century with its population growth in the early 1980s.When he came across the two graphs shown here, Marvin noticed that the first graph appeared to be steeper than the second. He concluded that the population was growing faster at the beginning of the century than it was in the early 1980s.**Graph Descriptions:**1. **First Graph (1900 - 1912):** - X-axis: Year (spanning from 1900 to 1912) - Y-axis: Population (in thousands, ranging from 75 to 78) - The graph shows a curve with an upward trajectory indicating a steady population increase over the years.2. **Second Graph (1980 - 1983):** - X-axis: Year (spanning from 1980 to 1983) - Y-axis: Population (in thousands, ranging from 86 to 92) - This graph also shows an upward trend but is less steep than the first graph, suggesting a slower rate of population growth.**Tasks:**1. **Use the first graph to calculate the average increase in population per year for the interval from 1900 to 1912.**2. **Use the second graph to calculate the average increase in population per year for the interval from 1980 to 1983.**

From the provided second graph, the population for the initial period (i.e., for the year 1980) is inferred to be 86.1 thousand approximately. Similarly, the population for the year 1983 is observed to be 88 thousand. For initial population size P0, the population after t years (i.e., the population for the period Pt) is modeled as: Pt=P0ert Where, r is the average rate of growth (in percentage). From the observed values of P0=86.1, the population for the period 1983 (i.e., t=1983-1980=3), i.e., after 3 years, P3 is modeled as: 88=86.1×e3r⇒e3r=8886.1≈1.022067⇒lne3r=ln1.022067 by taking natural logarithm on both sides⇒3r lne≈0.02183⇒3r=0.02183 ∵lne=1⇒r=0.021833⇒r≈0.0073 or 0.73% So the average annual growth (or increase) rate from the year 1980 to 1983 is approximately 0.73%.