SOLUTION: At​ 12:00 noon, there were 200 bacteria present. By​ 4:00 pm, there were 379 bacteria. If the bacteria grows​ continuously, how many will there be at​ 7:30 pm? There w

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: At​ 12:00 noon, there were 200 bacteria present. By​ 4:00 pm, there were 379 bacteria. If the bacteria grows​ continuously, how many will there be at​ 7:30 pm? There w      Log On

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Question 1144916: At​ 12:00 noon, there were 200 bacteria present. By​ 4:00 pm, there were 379 bacteria. If the bacteria grows​ continuously, how many will there be at​ 7:30 pm?
There will be_bacteria
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I think "continuously" means a straight line
relation. You are given 2 points on a straight line.
( 0, 200 )
( 4, 379 )
Let +N+ = number of bacteria at time +t+,
so the given points are:
( t, N )
For noon, I set +t+=+0+, so at 7:30, +t+=+7.5+
--------------------------------
The general equation is:
+N+=+m%2At+%2B+b+ where +m+ is the slope and
+b+ is the vertical intercept, +N+=+200+ and +t+=+0+
+N+=+m%2At+%2B+200+
and
+m+=+%28+379+-+200+%29+%2F+%28+4+-+0+%29+
+m+=+179+%2F+4+
so
+N+=+%28+179%2F4+%29%2At+%2B+200+
--------------------------------
At 7:30 PM, ++t+=+7.5+
and
+N+=+%28+179%2F4+%29%2A7.5+%2B+200+
You can finish: can't find calculator