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IRINA_888 [86]
4 years ago
8

How much would $125 invested at 8% interest compounded continuously be worth after 16 years? Round your answer to the nearest ce

nt. Do not include units in your answer.
Mathematics
1 answer:
Black_prince [1.1K]4 years ago
6 0
The formula is
A=p e^rt
A future value?
P present value 125
E constant
R interest rate 0.08
T time 16 years
A=125×e^(0.08×16)
A=449.58 round your answer to get
A=450
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Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p
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The question is an illustration of composite functions.

  • Functions c(n) and h(n) are \mathbf{c(n) = 5000 + 250n} and \mathbf{n(h) = 5h}
  • The composite function c(n(h)) is \mathbf{c(n(h)) = 5000 + 1250h}
  • The value of c(n(100)) is \mathbf{c(n(100)) = 130000}
  • The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

  • $5000 in fixed costs plus an additional $250
  • 5 systems in one hour of production

<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

\mathbf{c(n) = Fixed + Additional \times n}

This gives

\mathbf{c(n) = 5000 + 250 \times n}

\mathbf{c(n) = 5000 + 250n}

The function for number of systems is:

\mathbf{n(h) = 5 \times h}

\mathbf{n(h) = 5h}

<u>(b) Function c(n(h))</u>

In (a), we have:

\mathbf{c(n) = 5000 + 250n}

\mathbf{n(h) = 5h}

Substitute n(h) for n in \mathbf{c(n) = 5000 + 250n}

\mathbf{c(n(h)) = 5000 + 250n(h)}

Substitute \mathbf{n(h) = 5h}

\mathbf{c(n(h)) = 5000 + 250 \times 5h}

\mathbf{c(n(h)) = 5000 + 1250h}

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

\mathbf{c(n(100)) = 5000 + 1250 \times 100}

\mathbf{c(n(100)) = 5000 + 125000}

\mathbf{c(n(100)) = 130000}

<u>(d) Interpret (c)</u>

In (c), we have: \mathbf{c(n(100)) = 130000}

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

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