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EleoNora [17]
2 years ago
9

15,000 blue trout were released into the Meherrin River for a scientific study. The function, f(x) = 15,000 (98)x , represents t

he number of blue trout after x years. In 5 years, what will happen to the population?
Mathematics
1 answer:
Norma-Jean [14]2 years ago
0 0

Answer:

I'm guessing you mean f(x)=15,000(9/8)^x. If this is what you mean, the population would increase by about 12,000 (12030.4870605 to be exact).

Step-by-step explanation:

Starting equation: f(x)=15,000(9/8)^x

You can clean up the 9/8 to be 1.125

Now what you want to do is find the answer to (9/8)^5 which is 1.8020324707

Next multiply 1.8020324707 by 15,000 and you get 27030.4870605

Finally 27,030.4870605 - 15,000 gives you 12030.4870605. Which means that the population increased by about 12,000.

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A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o
Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

\dfrac{dA}{dt}=R_{in}-R_{out}

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}

Change In Volume of the tank, \frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}

Therefore:

\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3

This is a linear differential equation in standard form, therefore the integrating factor:

e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2

Multiplying the DE by the integrating factor, we have:

(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000

Therefore, the amount of salt in the tank at any time t is:

A(t)=(50+t)-75000(50+t)^{-2}

After 15 minutes, the amount of salt in the tank is:

A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds

8 0
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Answer: 14.97 per pair of earrings

Step-by-step explanation:

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2 years ago
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kvv77 [185]

Answer:

Option D.

Step-by-step explanation:

In the given graph x-axis represents the number of owners and y-axis represents yearly cost of pets, in hundreds.

It is given that the graph passes through the points (1,15), (3,7) and (10,0).

We need to check whether (4.5, 6) is a realistic solution for the function or not. (4.5,6) point represents

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It is realistic that owners spend $600 a year on their pet(s).

Number of owners can not be a fraction value. So, it is not realistic to have 4.5 owners.

Therefore, the correct option is D.

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ivann1987 [24]

Answer:

See below.

Step-by-step explanation:

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Now, we can write our function. Let x equal the amount of boxes.

The table is a set weight, so that would be our constant.

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The 70 represents the unchanging weight of the table.

In terms of W(x), it will be:

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A lightbulb has a normally distributed light output with mean 5000 end foot-candles and standard deviation of 50 end foot-candle
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To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
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8 0
2 years ago
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