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Y_Kistochka [10]

3 years ago

12

Money is invested into an account earning 4.25% interest compounded annually. If the accumulated value after 18 years will be $2

5,000, approximately how much money is presently in the account?
a.
$5,875
b.
$11,820
c.
$19,125
d.
$23,960

2 answers:

Lunna [17]3 years ago

7 0

We are given with the following:
Effective annual interest, i = 0.0425
Future worth, F = $25000
Number of years, n = 18

We use the formula to solve for the present worth of the money:
P = F / (1 + i)^n
P = 25000 / (1 + 0.0425)^18
P = 11818.73

The closest answer is:
<span>b.
$11,820</span>

yarga [219]3 years ago

5 0

Answer:

B. b. $11,820

Explanation:

I got the right answer on edge/e2020, and you need to

<u><em>STUDY ON YOUR MATH !!!</em></u>

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Joe's annual income has been increasing each year by the same dollar amount. The first year his income was $17 comma 90017,900

Vedmedyk [2.9K]

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points $(x_1,y_1)$ and $(x_2,y_2)$ , then the equation of line is

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of line is

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$y-17900=\frac{2400}{3}(x-1)$

$y-17900=800(x-1)$

$y-17900=800x-800$

Add 17900 on both sides.

$y=800x-800+17900$

$y=800x+17100$

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

$30700=800x+17100$

Subtract 17100 from both sides.

$30700-17100=800x$

$13600=800x$

Divide both sides by 800.

$\frac{13600}{800}=x$

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Therefore, in 17th year his income was $30,700.

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devlian [24]

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