Since we will be using powers of 10 and logs, it is best to make the years smaller numbers (imagine taking 10 to the 1940th power). so for the x list, use 40, 45, 50, 55, etc to represent the years after 1900.

a) what kind of model does the scatterplot suggest? (find out from your instructor how to submit any kind of hand made drawings)
b) use logarithms to transform the data (fill in the table below)?

year vehicles log (vehicles)
1940 32.4
1945 31
1950 49.2
1955 62.7
1960 73.9
1965 90.4
1970 108.4
1975 132.9
1980 155.8
1985 171.1

c) identify the point that may be an outlier? provide a reason as to why it may be an outlying point.
d) delete the outlier and write the linear equation (it should be log y-hat = ).
e) write the exponential form of the regression (in the formy hat equals a time b to the power of x).

It might be B, I'm not suree yet.

Answer:     the interval on which the curve is concave up is  (-∞, -1/2)======    we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x    let the function f denote be the anti-derivative of the inside expression, i.e. a function such that .then use the fundamental theorem of calculus:     differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function    find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x    where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation    1 + x + x² = 0results in a discriminant of      meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get      so the since the 2nd derivative of the curve is positive on the interval  (-∞, -1/2), the  curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is  (-∞, -1/2)

The answer is minor seventh