Mathematics, 24.07.2019 23:30 genyjoannerubiera

1. suppose you invest $5010 at an annual interest rate of 6.4% compounded continuously. a.) write an equation to model this. b) how much money will you have in the account after 2 years? 2. suppose you invest $100 at 6% annual interest. calculate the amount of money you would have after 1 year if the interest is compounded: a.) quarterly. b.) monthly. c.) daily. 3. suppose you deposit $100,000 in an account today that pays 6% interest compounded annually. write an equation to determine how long it will take before the amount in your account is $500,000 (note: you only have to write the equation). i need because i'm stuck.

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Answer from: ianmorales4894

To solve our problems, we are going to use the formula for compounded interest: $A=P(1+ \frac{r}{n} )^{nt}$
where

is the final amount after

years

is the initial amount

is the interest rate in decimal form

is the number of times the interest is compounded per year

is the time in years

1. A) We know for our problem that the initial investment is $5010, so P=5000

. We also know that the interest rate is 6.4%. To express the interest in decimal form, we are going to divide it by 100%: $r= \frac{6.4}{100} =0.064$ . Since the interest is compounded continuously, it is compounded 365 times per year; therefore, n=365

. Lets replace those values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=5010(1+ \frac{0.064}{365})^{365t}$

We can conclude that the equation that model this situation is $A=5010(1+ \frac{0.064}{365})^{365t}$

B. To find the amount of money you will have after 2 years, we are going to replace

with 2 in the equation from point A:
$A=5010(1+ \frac{0.064}{365})^{365t}$
$A=5010(1+ \frac{0.064}{365})^{(365)(2)}$
A=5694.6

We can conclude that after 2 years you will have $5694.06 in your account.

2. We know for our problem that P=100

, $r= \frac{6}{100} =0.06$ , and t=1

.
A. Since the interest is compounded quarterly, it is compounded 4 times per year; therefore, n=4

. Lets replace the values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=100(1+ \frac{0.06}{4} )^{(4)(1)$
A=106.14

We can conclude that after a year you will have $106.14 in your account.
B. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12

. Lets replace the values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=100(1+ \frac{0.06}{12})^{(12)(1)$
A=106.17

We can conclude that after a year you will have $106.17 in your account.
C. Since the interest is compounded daily, it is compounded 365 times per year; therefore, n=365

. Lets replace the values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=100(1+ \frac{0.06}{365})^{(365)(1)}$
A=106.18

We can conclude that after a year you will have $106.18 in your account.

3. We know for our problem that the initial investment is $100,000, so P=100000

. We also know that the final amount will be $500,000, so A=500000

. The interest rate is 6%, so $r= \frac{6}{100} =0.06$ . Since the interest rate is compunded anually, it is compounded 1 time per year; therefore, n=1

. Lets replace the values in our formula and solve for

:
$A=P(1+ \frac{r}{n} )^{nt}$
$500000=100000(1+ \frac{0.06}{1})^{(1)(t)}$
500000=100000(1+0.06)^t

$\frac{500000}{100000} =1.06^t$
1.06^t=5

$t= \frac{ln(5)}{ln(1.06)}$

We can conclude that $t= \frac{ln(5)}{ln(1.06)}$ is the equation to determine how long it will take before the amount in your account is $500,000

As a bonus:
$t= \frac{ln(5)}{ln(1.06)}$
t=27.6

We can conclude that after 27.6 years you will have $500,000 in your account.

Answer from: Quest

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

Answer from: Quest

the correct answer is b. negative exponential growth.

step-by-step explanation:

you can tell it's an exponential function because it decreases rapidly then levels out and never touches the x-axis.

it is negative because the y-axis decreases as the x-axis increases.

therefore it is a negative exponential growth graph.

hope i !

Categorize the graph as linear increasing, linear decreasing, exponential growth, or negative expone

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