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denpristay [2]
3 years ago
10

Each of 16 students will give a 1 1/5 minute speech in English class.

Mathematics
1 answer:
Mice21 [21]3 years ago
8 0

Answer:

19 1/5

Step-by-step explanation:

1 1/5 * 16

6/5 * 16

96/5

19 1/5

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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
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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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