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Nikolay [14]
3 years ago
14

In the year 2005, a total of 750 fish were introduced into a man made lake. The fish population was expected to grow at a rate o

f 8.3% each year. What will the fish population be in 2050?
Mathematics
1 answer:
Butoxors [25]3 years ago
3 0

Answer:

In 2050 fish population is 27000

Step-by-step explanation:

Exponential change can be modeled by,

F=F(0)+x^{n}

Where F(0) is the initial value of 750, x is the rate of change. It is greater then 1 when there is growth and less then 1 when there is decay. and n is the number of time periods.

The rate of change in percentage is 8.3%

That means, after 1 year fish will become 108.3 for 100 fish.

Then x=\frac{108.3}{100}

x=1.083

Difference of years is, n=2050-2005=45

Then, F=750+1.083^{45}

F=27000

So, the fish population in 2050 is 27000

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6 0
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HELP!!! 50 POINTS!! I WILL REPORT IMPROPER ANSWERS!!!
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Answer & Explanation:

1.) Which function is a linear function?

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3 years ago
When the x is slove and you have to slove for the y i got how to do it but i don't know which one am i sloving for in the equati
harina [27]
Since x=8y, subsitute 8y for x in other equation

y=6x-11
y=6(8y)-11
y=48y-11
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x=8y
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x=\frac{88}{47}



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y=\frac{11}{47}
in (x,y) form
(x,y)
(\frac{88}{47},\frac{11}{47})
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