5)
Before we see which model best fits the given data, let's plot the points into a graph, that's fundamental! Always plot the points for better visualization, only by plotting the points we can see what kind of model we will use, then, plotting these points we will get:
Here, we are using the y-axis as the life expectancy and the x-axis as the years after 1950. Only looking at the graph we can see that the points follows something like a line, therefore, we are looking for a linear model, like
[tex]y=ax+b[/tex]But what value of "a" and "b" best fits the points? We can find it, but it will take a lot of time and calculus, then, let's look at the graph and see which equation in the option is correct. We only have two options for linear models:
[tex]\begin{gathered} y=0.183x+67.895 \\ \\ y=67.895x+0.183 \end{gathered}[/tex]To see which one is correct we just look at the y-intercept (number without "x"), looking at our graph the y-intercept is something like 68.2, then, the correct answer is
[tex]y=0.183x+67.895[/tex]The final answer for 5) is the letter C.
[tex]y=0.183x+67.895[/tex]6)
Now we can use the equation of 5) to find it, let's find the value of x that we will input into the function, 1995 is 45 years after 1950, then, we will input x = 45.
[tex]\begin{gathered} y=0.183x+67.895 \\ \\ y=0.183\cdot(45)+67.895 \\ \\ y=8.235+67.895 \\ \\ y=76.13\text{ years old} \end{gathered}[/tex]The predicted life expectancy for 1995 is 76.13 years old