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Kay [80]
4 years ago
6

A pond had an initial population of 120 fish. The number of fish is exponentially decreasing by one-fourth each year. If y repre

sents the number of fish in the pond after x years, determine the graph of the solution set for this situation and the equation modeled by the graph.

Mathematics
1 answer:
Kruka [31]4 years ago
7 0
<span>1) Number of fish in the pond: y
Number of years: x

</span>A pond had an initial population of 120 fish. Then when x=0, y=120. The graph must begin at point (x,y)=(0,120). Only the graph above at right and below at left begin in this point (0,120).

<span>The number of fish is exponentially decreasing by one-fourth each year. Then the first year the number of fish must decrease 120*(1/4)=120/4=30, and the number of fish after the first year must be:
</span>120-30=90=120(1-1/4)=120(4-1)/4=120(3/4). Then when x=1, y=90. The point is (x,y)=(1,90)
In the graph above at right when x=1, y is between 24 and 36. y=90 is not in this interval. then this graph is not the correct.
In the graph below at left when x=1, y is between 84 and 96. y=90 is in this inverval. Then this is the correct graph.

Answer: T<span>he graph of the solution set for this situation is the graph below at left.

2) The equation has the form:
y=y0(r)^x
Where y0 is the initial population and r is the rate of reduction. In this case:
y0=120 and
r=1-1/4=(4-1)/4→r=3/4
Then the equation modeled by the graph is:
y=120(3/4)^x

Answer: The equation modeled by the graph is that above at right:
y=120(3/4)^x</span>
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If $1,000 is invested at 4% simple interest, how much will the investment be worth after 2 years? Please explain how compound an
Novay_Z [31]
<h3>Answer:</h3>
  • simple interest: $1080.00
  • compounded annually: $1081.60
<h3>Step-by-step explanation:</h3>

<em>Simple Interest</em>

Simple interest is computed on the principal amount only. Each year, 4% of the principal is added to the balance. So, at the end of 2 years, the balance is ...

... $1000 + 0.04×$1000 + 0.04×$1000

... = $1000×(1 + 0.04×2) = $1000×1.08

... = $1080.00

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<em>Comment on the computation</em>

The added interest is the rate (per year) multiplied by the number of years. Here, that is 0.04×2×(principal amount). The formula for the simple interest earned is often seen as ...

... I = Prt . . . . . where I is the amount of interest, P is the principal amount, r is the interest rate for the time period, t is the number of time periods.

The account balance (A) with interest added is ...

... A = P + I = P + Prt

... A = P(1 +rt)

Here, the time period is years, and the rate given is an annual rate.

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<em>Compound Interest</em>

Compound interest is computed on the <em>account balance</em> at the beginning of the period, not just the <em>principal</em> amount. After the first period, the account balance includes interest earned so far. So, the interest is earning interest. That is why it is called compounded interest.

Here, the balance at the end of the first year is the principal amount plus the interest that has earned:

... $1000 + 0.04×$1000 = $1000×1.04 = $1040.00

The balance at the end of the second year when the interest is compounded is this account balance plus the interest it earns:

... $1040 + 0.04×$1040 = $1040×1.04 = $1081.60

You may notice that the intial principal, $1000, has been multiplied by the factor 1.04 twice. Using exponents, the multiplier for a period of 2 years is 1.04×1.04 = 1.04².

_____

<em>Comment on the computation</em>

The multiplier of the account balance each year is raised to a power that is the number of years. Here, the account balance at the end of 2 years is (1+0.04)² times the principal amount. A formula that is seen for this is ...

... A = P(1 +r)^t . . . . . where A is the final account balance, P is the principal amount, r is the interest rate for the time period, and t is the number of time periods.

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