Ryan invested \$4,800$4,800 in an account in the year 1990, and the value has been growing exponentially at a constant rate. The value of the account reached \$6,300$6,300 in the year 1998. Determine the value of the account, to the nearest dollar, in the year 2007.
Well, well, well, looks like Ryan's account is having quite the growth spurt! Let's do some clown-calculations to figure out the value of the account in 2007.
First, we need to figure out the growth rate. To do this, we can divide the final value by the initial value: 6300/4800 = 1.3125.
Now, for the fun part! We can use the growth rate to find the value in 2007. From 1990 to 2007, there are 17 years, right? So we apply the growth rate to the initial value for 17 years: 4800 * (1.3125)^17 = $16,478.11 (rounded to the nearest dollar).
Voila! According to my clown calculations, the value of the account in 2007 would be approximately \$16,478. So keep juggling those investments, Ryan!
To determine the value of the account in the year 2007, we need to find the growth rate over the given time period.
The formula for exponential growth is given by:
A = P(1 + r)^n
Where:
A = Final value of the account
P = Initial investment
r = Growth rate
n = Number of years
We are given:
P = $4,800
A = $6,300
n = 1998 - 1990 = 8 years
Plugging in the values, we can solve for the growth rate:
6,300 = 4,800(1 + r)^8
Divide both sides by 4,800:
6,300/4,800 = (1 + r)^8
1.3125 ≈ (1 + r)^8
Taking the eighth root of both sides:
(1 + r) ≈ (1.3125)^(1/8)
Using a calculator, we find:
(1 + r) ≈ 1.0441
Subtract 1 from both sides:
r ≈ 0.0441
Now, we can find the value of the account in the year 2007 by using the same formula:
A = P(1 + r)^n
Where:
P = $6,300
r = 0.0441
n = 2007 - 1998 = 9 years
Plugging in the values, we get:
A = 6,300(1 + 0.0441)^9
A ≈ $8,721
Therefore, the value of the account in the year 2007 is approximately $8,721.
To solve this problem, we need to use the formula for compound interest. The formula is:
\[A = P \times (1 + r)^n\]
Where:
A is the final amount
P is the initial amount or principal
r is the interest rate per compounding period
n is the number of compounding periods
In this case, the initial amount P is $4,800, and the final amount A is $6,300. The number of years between 1990 and 1998 is 8 years. Plugging in these values, we can solve for the interest rate per compounding period.
\[6,300 = 4,800 \times (1 + r)^8\]
To determine the interest rate per compounding period, we can rearrange the equation:
\[(1 + r)^8 = \frac{6,300}{4,800}\]
Now we can take the eighth root of both sides to solve for (1 + r):
\[(1 + r) = \left(\frac{6,300}{4,800}\right)^{\frac{1}{8}}\]
Now, we can subtract 1 from both sides to solve for r:
\[r = \left(\frac{6,300}{4,800}\right)^{\frac{1}{8}} - 1\]
Once we have the value of r, we can use it to find the value of the account in 2007. The number of years between 1998 and 2007 is 9 years. We can use the formula:
\[A = P \times (1 + r)^n\]
Where P is the final amount in 1998, which is $6,300. The number of years n is 9. Plugging in these values, we can find the value of the account in 2007.
why the backslash and duplication? Just type what you want, rather than attempting copy/paste!
It grew by a factor of 63/48 in 8 years. So the annual interest rate is
r = (63/48)^(1/12)
Thus the amount after t years is
A = 4800(1+r)^t