I need help on a math question.

A study found that 27% of the fish in a river are trout, 36% are salmon, and 18% are sturgeon. If a large fish trap is set it captures 216 fish, how many of those fish are likely to be fish types other than the ones mentioned?

I'm almost failing in math and I really need help, please? :(

27+36+18=81. The remaining 19% are the "other types" of fish. This is what we want to find.

SO, .19 (19%) times 216 equals the amount that should represent the "other types" that the question is looking for.

.19x216= 41.04

41 fish

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Answer

Probability question, this is how you solve it:

P(trout,salmon,sturgeon) = 0.27 + 0.36 + 0.18 = 0.81
P(other fish) = 1 - 0.81 = 0.19

So 19% of all the fish are other fish types than the ones mentioned.

n = 216
p = 0.19
np = 216(0.19) = 41.04

Therefore, 41 of those fish are other types of fish.

well if your asking what percent will of the fish that are listed out of those how many fish will not be trout, salmon, and sturgeon. then 19% would be a different type of fish. if you asking out of the 216 fish how many will be a diffent type then trout, salmon, and sturgeon then it would be 41 fish would not be trout, salmon, and sturgeon. you just add 27 + 36 +18 and minus that number from 100 to get 19 and to find the number of fish you just mutilply 216 by .27 or .36 or .18 or .19 to get the number of trout, salmon, sturgeon, and a fish not listed.

other fish% = 100 - ( 27 + 36 + 18 ) = 19%

therefore the answer to your question is 216 fish times 19% and is equal to 41 fish.. DONT FAIL MATH and GOOD LUCK!

81% of the fish are those three types. That leaves 19% of other types:

0.19 * 216 = 41.04 which rounds down to 41 other types of fish