This list shows concentrations of carbon dioxide (in parts per million) in the earth's atmosphere for the years 1950, 1955, 1960, 1965, 1970, 1975, 1980, and 1985. find the mathematical model that best fits the data, and use it to predict the global mean temperature in 2015. code 1950 as 1, 1955 as 2, and so on, in five-year segments.

311 314 317 320 326 331 339 346

a. linear model; 373

b. quadratic model; 412

c. exponential model; 374

d. power model; 346

Choice B

The mathematical model that best fits the data is a polynomial of order 2 as shown on the attachment.

The predicted global mean temperature in 2015 is; 412.24

Step-by-step explanation:

The first step is to obtain a scatter plot of the data and then fit a trend line. A quadratic polynomial fits the data well considering the large value of the coefficient of determination of 0.9984.

To determine the global mean temperature in 2015 we substitute x with 14 in the quadratic model;

y = 0.4643(14)^2 + 0.7976 (14) + 310.07

  = 412.24

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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