SOLUTION: A pound of U.S. mud contains an average of 5.50 x10^(11) bacteria. There are 60.0 drops in one teaspoon, 6.0 teaspoons in an ounce, and 16 ounces in a pound. How many b

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: A pound of U.S. mud contains an average of 5.50 x10^(11) bacteria. There are 60.0 drops in one teaspoon, 6.0 teaspoons in an ounce, and 16 ounces in a pound. How many b      Log On

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Question 762209: A pound of U.S. mud contains an average of 5.50 x10^(11)
bacteria. There are 60.0 drops in one teaspoon, 6.0
teaspoons in an ounce, and 16 ounces in a pound. How many
bacteria live in a drop of U.S. mud?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You have to convert pounds to drops
In terms of units:
( pounds ) x ( ounces / pound ) x ( teaspoons / ounces ) x ( drops / teaspoon )
If you do all the cancellations, you end up with ( drops )
Plugging in numbers:
+%28+1+%29%2A%28+16%2F1+%29%2A%28+6%2F1+%29%2A%28+60%2F1+%29+=+5760+
There are 5,760 props in a pound
( number of bacteria / pound ) x ( pounds / drop ) = ( number of bacteria / drop )
+5.5%2A10%5E11+%2A+%28+1+%2F+5760+%29+=+55000%2A10%5E7+%2F+5760+
+55000%2A10%5E7+%2F+5760+=+9.5486111%2A10%5E7+
There are 95,486,111 bacteria in a drop of mud