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homework help please!! help needed asap!! 10 points to best answer by end of today?
ok so i have a pre-calculus project due and i need as much help as i can get. i have a hard time with this class and i'm really trying to bring up my grade. i'm not asking for answers because i already did it. i just wanna make sure that my answers are correct.
so here's the question:
the simplest mathematical model to use to represent a set of real-life data is the line. the number of married couples in the u.s. in 1995 was 53.9 million. in 2000, the number of married couples grew to 55.3 million.
a. write a linear equation that gives the number of married couples in terms of the year. let x represent the year with x=5 corresponding to 1995.
-the equation i got was y=0.28x+52.5
i did it on a calculator but i need an explanation of how to get it done by hand and showing work.
b. what is the slope of the line?
- i got 0.28
c. what does the slope of the line tell you about the number of married couples in the u.s.?
- this question i need your help with please
d. sketch a graph of the line.
-i know how to do that.
part 2:
1. find x-intercept. does an x-intercept make sense of the context of the data? explain.
- i don't get this question.
2. make table of values for data.
- i already did that
3. how do the model values compare with the actual values?
- i don't get this one either.
4. i get it
5. do you think a linear model is a good fit for the data from 1991 to 2002? explain.
-i don't get this one either.
please help me. im really trying to get an A on this project.
i'm not in college. i'm in high school. i really need help a.s.a.p.
i'm not in college. i'm in high school. i really need help a.s.a.p.
i'm not in college. i'm in high school. i really need help a.s.a.p.
3 Answers
- sunflowersLv 71 decade agoFavorite Answer
Is there a free math tutor room at your college? It might be a good time to find out where and when it meets.
- 1 decade ago
Part 1c:
The slope of 0.28 means that, from 1995 to 2000, the number of married couples increased by 0.28 million couples per year, on average.
Part 2:1:
The x intercept would be the number of years before 1990 when the number ov married couples was zero, if the number of married couples had increased by 0.28 million couples per year since then.
it would be 52.9 million divided by 0.28 million per year, before 1990.
1 990 - (52.5 / 0.28) = 1 802.59
That would mean that there were no married couples in the year 1802.
It doesn't make sense in the context of the data because marriage numbers are based on population, which grows geometrically, not linearly. And it is also influenced by other, more complicated factors, like costs of setting up a household, etc
Part 1.3. I'm not sure what "the actual values" are. Did the teacher give you marriage numbers for other years? If so, you can calculate the expected number of couples for a year by taking
y = mx + b
expected number of couples = 0.28 * (year-1990) + 52.5
and comparing to the actual number of couples given by the teacher.
If the actual numbers of couples for 1991 and 2002, in your opinion, are close to the numbers you get from the equation (0.28 * (year-1990) + 52.5), then give that opinion.
You could also plot the number of couples versus the year on the chart you made, with the line of best fit, and see if the dots fall close to the line. "Close", of course, is a judgement call, so use your judgement.
