The table shows the approximate population of a city, in thousands, for the years 2009–2013. Which is the best prediction for the city’s approximate population in 2030? A. 153,000 B. 167,000 C. 182,000 D. 195,000

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The table shows the approximate population of a city, in thousands, for the years 2009–2013. Which is the best prediction for the city’s approximate population in 2030? A. 153,000 B. 167,000 C. 182,000 D. 195,000

Mathematics
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well, I would treat years as the x variable and population as the y variable, then write an equation using two points from the table I'll pick the first and last point (2009,78) and (2013,93) then write an equation in y = mx + b form
OK, about to it work out

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I got 1920 and -1936 and when I didvided got -1.00833so on. I don't think I did it right
remember, slope = (y2-y1)/(x2-x1) (93-78) = ?
I was doing 93-73 okay I'm about to do it all over
I mean I was doing 93 minus 2013 okay about to work it all out again
okay I got 3.75 for slope, than I have to work out to get y intercept
ok my slope is 3.75, then I said y-y1=m(x-x1) so it was y-78=3.75-2009 which i got + 78 to -2009= 1931
I'm kinda lost now
wait I think I figured it out and got 2029
ah, I think you've mixed up the equation a bit, let me lend you a hand ^ ^ y-78 = 3.75(x-2009) start by distributing the 3.75
y-78=3.75x-7533.75 y-78+78=3.75x +78-7533.75 y=3.75x-7455.75 3.75x/-7455.75
you can stop here y=3.75x-7455.75 this is the equation we need, now plug in x = 2030
ok so 3.75 times 2030. if so that =7612.5
uh, not quite... we're looking for the value of y 3.75*(2030) - 7455.75
= ?
2030-7455.75
I meant 7612.5-7455.75
ok, now subtract those two values
156.75
good, now we multiply by 1000 to get 156,750 which means that A is the answer that best approximates the population in 2030
Thanks, I don't know to say. You've always been there to help me. Thanks

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