Mathematics, 17.04.2020 04:54 daebreonnakelly

The fish population of a lake is decreasing each year. A study is conducted. But, unfortunately some data was lost. The researcher found in her notes that in year one the fish population was 18000 fish and in year three the fish population was 8000 fish. Assume a constant rate of decay. Find a formula F

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Answer from: badpotterchris

$F(t) = 18000(0.6666)^{t}$

Step-by-step explanation:

The fish population after t years can be modeled by the following equation:

$F(t) = F(0)(1-r)^{t}$

In which F(0) is the initial population and r is the constant rate of decay.

Year one the fish population was 18000

This means that F(0) = 18000

In year three the fish population was 8000 fish.

Two years later, so F(2) = 8000

$F(t) = F(0)(1-r)^{t}$

$8000 = 18000(1-r)^{2}$

$(1-r)^{2} = 0.4444$

$\sqrt{(1-r)^{2}} = \sqrt{0.4444}$

1 - r = 0.6666

r = 0.3334

$F(t) = 18000(0.6666)^{t}$

Answer from: Quest

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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