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garri49 [273]
3 years ago
14

Lena has just become eligible to participate in her​ company's retirement plan. Her company does not match​ contributions, but t

he plan does average an annual return of ​%. Lena is 40 and plans to work to age 65. If she contributes ​$ per​ month, how much will she have in her plan at​ retirement?
Business
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

Throughout her retirement plan, Lena will still have $206,673.13.

Explanation:

The given values are:

Annual Interest Rate

= 12.00%

Monthly Deposit

= $110

Period

= 25 years i.e., 300 months

Monthly Interest Rate

= 1.00%

Now,

The Accumulated Deposits will be:

⇒  110\times 1.01^{299} + 110\times 1.01^{298}+....+ $110\times 1.01 + $110

⇒  \frac{110\times  (1.01^{300} - 1)}{0.01}

⇒  110\times 1,878.846626

⇒  206,673.13  ($)

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A summary should condense a report or article by what percentage is 85 to 90 percent.

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2 years ago
What features should you present when demonstrating 2022 murano’s confident cornering?
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2 years ago
Consider a numerical example using the Solow growth model. Suppose that F(K,N) = K^ 0.5N^ 0.5, with d = 0.1, s = 0.2, n = 0.01,
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Find the solution in the attachments

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4 years ago
Suppose the world population today is 7 billion, and suppose this population grows at a constant rate of 3% per year from now on
oksano4ka [1.4K]

Answer:

a) P(t=100) = 7 e^{0.03*100}=140.599 billion

b) P(t=0) = 7 e^{0.03*0}=7 billion

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c) Figure attached

d) Figure attached

Explanation:

The proportional model on this case would be given by:

\frac{dP}{dt} = kP

Where P is the population size, t the time on years and k a constant.

We can reorder this expression like this:

\frac{dP}{P} = k dt

If we integrate both sides we got:

ln|P| = kt + C

And using exponentials on both sides we got:

P(t) = e^{kt} e^C = P_o e^{kt}

Where P_o=7 billion  represent the initial amount for the starting year t=0.

The rate on this case is given r =3\% = 0.03, so then our model would be given by:

P(t) = 7 e^{0.03t}

Part a

For this case we just need to replace t=100 and we got:

P(t=100) = 7 e^{0.03*100}=140.599 billion

Part b

For this case we have the following:

P(t=0) = 7 e^{0.03*0}=7 billion

P(t=1) = 7 e^{0.03*1}=7.21 billion

P(t=2) = 7 e^{0.03*2}=7.43 billion

P(t=10) = 7 e^{0.03*10}=9.45 billion

P(t=25) = 7 e^{0.03*25}=14.82 billion

P(t=50) = 7 e^{0.03*50}=31.37 billion

Part c

The graph is on the first figure attached.

Part d

If we take a log-log scale we have the following values

We need to exclude the point t=0 since the natural log for 0 is not defined.

ln 1 =0 , ln 2= 0.693, ln 10=2.30, ln 25 =3.22, ln 50= 3.91

The result would be the figure 2 attached. And we see a better result for the graph.

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