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Stels [109]
3 years ago
13

The number of a particular fish in thousands x years after 1972 can be modeled by f(x)=230(0.886)x. Estimate the year when the n

umber of fish reached 99 thousand.
Mathematics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

The year is 1979.

Step-by-step explanation:

Given:

The function relating fish population and number of years since 1972 is:

f(x)=230(0.886)^x\\Where,\ f(x)\rightarrow \textrm{fish population}\\x\rightarrow \textrm{number of years passed after 1972}

Now, when the fish population reaches 99 thousand, it means that f(x)=99. Now, plugging in f(x)=99 in the above equation and solving for x. This gives,

99=230(0.886)^x\\\frac{99}{230}=0.886^x\\\\\textrm{Taking natural log on both sides}\\\ln(0.886)^x=\ln(\frac{99}{230})\\\\\textrm{Using log property: }\log a^b=b\log a\\x\ln (0.886)=\ln(0.43)\\x\times (-0.12)=-0.84\\x=\frac{-0.84}{-0.12}=7\ years

Therefore, the year after 7 years passed since 1972 is given as:

1972 + 7 = 1979.

So, the year when the number of fish reaches 99 thousand is 1979.

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2 years ago
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Is the number of tosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?

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a) Data and Calculations:

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