The function w(x) = – 0.01x^2 + 0.27x + 8.60 can be used to estimate the number of selfemployed

workers in the U.S., in millions, x years after 1980. How many self-employed workers
were there in the year 1995?

A. 9.8 million
B. 10.4 million
C. 11.15 million
D. 14.9 million

I think the answer is B 10.4 million, but not 100%. Can someone please explain? Thank you

x years after 1980 means that for 1981, x=1, 1982, x=2,.... for 1990, x=10, and 1995, x=15.

If you substitute 15 for x in the function w(x), you will find the population in millions.
Post your choice of answer for verification if you wish.

Thanks I chose B 10.4 Million. I am confident that this is the correct answer. Am I correct?

Well done, that's correct!

To find the number of self-employed workers in the year 1995, we need to substitute x = 1995 - 1980 = 15 into the given function w(x) = -0.01x^2 + 0.27x + 8.60.

Plugging in x = 15, we have:

w(15) = -0.01(15)^2 + 0.27(15) + 8.60
= -0.01(225) + 0.27(15) + 8.60
= -2.25 + 4.05 + 8.60
= 10.40

The estimated number of self-employed workers in the year 1995, according to the given function, is 10.40 million. Therefore, the correct answer is B. 10.4 million.

Keep in mind that this is an estimation provided by the function, and the actual number of self-employed workers in 1995 may differ.