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zheka24 [161]
3 years ago
7

Mark u as branliest but answer right plz

Mathematics
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

Step-by-step explanation:

55+4x

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3 years ago
In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)
dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about  44622  humpback whales in the year 2018

5 0
4 years ago
I will be back to help you guys
Luda [366]

Answer:

thx u xox ur so sweet

Step-by-step explanation:

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