from 1960 to 1990 the average annual consumption of cigarettes per American C can be modeled by the equation C=4024.5+51.4t-3.1t^2 , where t is the number of years since 1960. when did annual cigarette consumption reach 4100??
thanx
2007-01-03 12:02:12 · 7 answers · asked by pearl 2 in Science & Mathematics ➔ Mathematics
Set C to 4100
So:
4100 = 4024.5 + 51.4t - 3.1t^2
Subtract 4100 from both sides:
0 = -75.5 + 51.4t - 3.1t^2
To make things nicer, multiply both sides by -1, and reorder:
0 = 3.1t^2 - 51.4t + 75.5
This takes a form of a quadratic equation and can be solved using the quadratic formula:
t = [-b +/- square-root(b^2 - 4ac)] / (2a)
where:
a = 3.1
b = -51.4
c = 75.5
thus:
t = [-(-51.5) +/- sqrt((-51.4)^2 - 4*3.1*75.5)] / (2*3.1)
t = [51.5 +/- sqrt(2641.96 - 936.2)] / 6.2
t = [51.5 +/- sqrt(1705.76)] / 6.2
t = (51.5 +/- 41.3) / 6.2
which gives:
t = 10.2 / 6.2 = 1.65
or
t = 92.8 / 6.2 = 14.97
without all the rounding above, though, the answers for t are:
1.63 and 14.95
So, according to the model the consumption was FIRST reached in 1961.63 (1961, mid August) then again in 1974.95 (1974, mid December).
Hope this helps.
2007-01-03 12:19:54 · answer #1 · answered by wxchemgeek 2 · 0⤊ 0⤋
4100 = 4024.5+51.4t-3.1t^2
3.1t^2-51.4t+75.5=0
solve for t; then add t to 1960; you get the year annual cigarette consumption reached 4100.
2007-01-03 20:16:20 · answer #2 · answered by Johnny Handsome 2 · 0⤊ 0⤋
C=4024.5+51.4t-3.1t^2=4100
subtract 4100 from each side
-3.1t^2+51.4t-76=0 multiply by -1
3.1t^2-51.4t+76=0
t=(51.4+/-â(51.4^2-4*3.1*76))/6.2
t=(51.4+/-â(2641.96-942.4))/6.2
t=(51.4+/-â1699.56)/6.2
t=(51.4+/-41.2)/6.2
t=10.2/6.2, 92.6/6.2
t=1.6, 14.9
according to this model, consumption climbed to 4100 in the middle of 1961 & kept climbing. It then leveled off & started declining back down to 4100 at the end of 1975 & continued to decline after that.
2007-01-03 20:16:36 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
4100 = 4024.5+51.4t-3.1t^2
0= -75.5 + 51.4t - 3.1t^2
.. then plug into quadratic equation with a= -3.1 b=51.4 c= -75.5
2007-01-03 20:05:47 · answer #4 · answered by arturo r 2 · 0⤊ 0⤋
arturo r explained how to do the algebra (assuming you know the quadratic formula), but once you solve for t you have to figure out what to do with it. I'm guessing that 1960 corresponds to t = 0, so you simply add t to 1960 to get the year you're looking for.
2007-01-03 20:13:57 · answer #5 · answered by actuator 5 · 0⤊ 0⤋
4100 = 4024.5 + 51.4t - 3.1t^2
3.1t^2 - 51.4 t + 75.5 = 0
t = [51.4 +/- sqrt(51.4^2 - 4*3.1*75.5)]/2*3.1
t = [51.4 +/- 41.3]/6.2
t = (51.4+41.3)/6.2 = 14.95, or
t = (51.4 -41.3)/6.2 = 1.63
So it looks like it reached 4100 around June of 1961 and then hit it again inlate December of 1974.
2007-01-03 20:56:36 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
can u give me the formulas of that question...? I'll try to help u
2007-01-03 22:08:00 · answer #7 · answered by algebric 1 · 0⤊ 0⤋