doris browning deposited $8000 into a savings account paying 5.25% interest. how long will it take for her investment to grow to $10,100. The answer i got was 4yr 11mo 30 days but that is wrong someone please help
I am assuming you are using the standard compound interest method, simple interest is seldom used for periods longer than a year
8000(1.0525)^n = 10100
1.0525^n = 1.2625
n log 1.0525 = log 1.2625
n = 4.5554 years , assuming the rate is 5.25 per annum compounded annually
at continuous compounding
8000 e^(.0525n) = 10100
e^(.0525n) = 1.2625
take ln of both sides
.0525n = ln 1.2625
n = 4.4399 years
btw:
at simple interest
8000 + 8000(.0525)n = 10100
n = 5 years
none of the 3 ways I used produces your answer, I have no idea how you got that.
I belive its 5 years.( asuming the interest rate is anual)
8000x .0525 = 420
(Check[8000/420*5=100])
so 10100-8000 = 2100
Finally 2100/420 = 5
Hope that's right.
To calculate how long it will take for Doris Browning's investment to grow to $10,100, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment ($10,100)
P = the principal amount ($8000)
r = the annual interest rate (5.25% or 0.0525 in decimal form)
n = the number of times interest is compounded per year (assuming it's compounded annually, n = 1)
t = the time in years
Plugging in the values, the equation becomes:
$10,100 = $8000(1 + 0.0525/1)^(1*t)
Dividing both sides by $8000, we get:
1.2625 = 1.0525^t
Now, take the natural logarithm of both sides to solve for t:
ln(1.2625) = ln(1.0525^t)
Using a calculator:
t = ln(1.2625) / ln(1.0525)
t ≈ 0.1638 / 0.0512
t ≈ 3.2 years
Therefore, it will take approximately 3.2 years for Doris Browning's investment to grow to $10,100.
To determine the time it takes for an investment to grow to a certain amount, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount (initial deposit)
r = annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years
In this case, Doris Browning deposited $8000 into a savings account with an interest rate of 5.25% (0.0525 as a decimal). Let's assume the interest is compounded annually, so n = 1.
We want to find out how long it takes for the investment to grow to $10,100, so A = $10,100.
Now we can plug in the values into the formula and solve for t:
$10,100 = $8000(1 + 0.0525/1)^(1t)
Divide both sides by $8000:
1.2625 = (1.0525)^t
To solve for t, we need to take the logarithm of both sides. Let's use the natural logarithm (ln):
ln(1.2625) = ln(1.0525)^t
Using the property of logarithms (ln(a^b) = b * ln(a)):
ln(1.2625) = t * ln(1.0525)
Now, divide both sides by ln(1.0525):
t = ln(1.2625) / ln(1.0525)
Using a calculator, we can find that t ≈ 4.8177 years.
Since the time needs to be in months and days, let's convert the decimal part of the years into the corresponding fraction of a year:
0.8177 years = 0.8177 * 12 months = 9.8124 months
So, the total time it takes for Doris' investment to grow to $10,100 is approximately 4 years, 9 months, and 10 days (assuming the interest is compounded annually).