Question

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

Answer 1

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

Answer 2

OK, about to it work out

Answer 3

I got 1920 and -1936 and when I didvided got -1.00833so on. I don't think I did it right

Answer 4

remember, slope = (y2-y1)/(x2-x1)

(93-78) = ?

Answer 5

I was doing 93-73 okay I'm about to do it all over

Answer 6

I mean I was doing 93 minus 2013 okay about to work it all out again

Answer 7

okay I got 3.75 for slope, than I have to work out to get y intercept

Answer 8

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

Answer 9

I'm kinda lost now

Answer 10

wait I think I figured it out and got 2029

Answer 11

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

Answer 12

y-78=3.75x-7533.75
y-78+78=3.75x +78-7533.75
y=3.75x-7455.75
3.75x/-7455.75

Answer 13

you can stop here

y=3.75x-7455.75

this is the equation we need, now plug in x = 2030

Answer 14

ok so 3.75 times 2030. if so that =7612.5

Answer 15

uh, not quite... we're looking for the value of y

3.75*(2030) - 7455.75

Answer 16

= ?

Answer 17

2030-7455.75

Answer 18

I meant 7612.5-7455.75

Answer 19

ok, now subtract those two values

Answer 20

156.75

Answer 21

good, now we multiply by 1000 to get

156,750

which means that A is the answer that best approximates the population in 2030

Answer 22

Thanks, I don't know to say. You've always been there to help me. Thanks