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Free_Kalibri [48]
3 years ago
11

Ava opens an account with $400. She deposits $200/month for 20 years. The interest rate is 2.15%. How much interest will she hav

e earned after 20 years?
a. $61,243.59




b. $48,800




c. 12,443.59




d. None of the above.
Mathematics
1 answer:
artcher [175]3 years ago
3 0

Answer:

  d.  None of the above

Step-by-step explanation:

We assume the sequence of deposits is ...

  month 0: $400

  month 1: $200

  month 2: $200

...

  month 240: $200 . . . . accumulated interest is determined at this point

That is, no interest is earned on the last deposit.

_____

The value of the initial $400 deposit after 20 years at 2.15% interest compounded monthly is ...

  $400×(1 +.0215/12)^(12×20) = $400×1.536666 ≈ $614.67

The value of the $200 annuity at the same interest rate is ...

  $200((1 +.0215/12)^(12×20) -1)/(.0215/12) = $200×299.534612 ≈ $59,906.92

So, the total account value is ...

  $614.67 +59,906.92 = $60521.59

The total amount deposited was ...

  $400 +$200×240 = $48,400

The interest earned is the difference between the account value and the total of deposits:

  $60,521.59 -48,400 = $12,121.59 . . . . interest earned

This value does not match any numerical answer choice, so we conclude the appropriate answer is ...

   None of the above

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
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Answer:

Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

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Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

Assume that the sled and its passenger has reached a height of h meters relative to the base of the slope.

Gain in gravitational potential energy:

\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

Solve for h:

(137 + 68.1)\, h = 63.

\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}.

Therefore, the maximum height that this sled would reach would be approximately 0.31\; \rm m.

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Step-by-step explanation:

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