(7.RP.3) When invested at an annual interest rate of 6% an account earned $180.00 of simple interest in one year. How much money was originally invested in account? (1 pt) *
a) $10.80
b) $108
c) $648
d) $30
e) $300
f) $3,000
3000*.06=180
or more to the point, 180/.06=3000
I = P*r*t = 180.
P*0.06 *1 = 180,
P = $3,000.
To find the amount of money originally invested in the account, we can use the formula for calculating simple interest:
Simple Interest = (Principal) x (Interest Rate) x (Time)
Given that the account earned $180.00 in simple interest, and the annual interest rate is 6% (which can be expressed as 0.06), we can plug in the values into the formula.
$180.00 = (Principal) x 0.06 x 1
Simplifying the equation, we have:
180 = 0.06P
To isolate the principal (P), divide both sides of the equation by 0.06:
180 / 0.06 = P
The original amount invested in the account is therefore:
P = $3000
So, the answer is f) $3,000.
To find the original amount of money invested, we need to use the formula for simple interest:
I = P * r * t
where:
I = interest earned
P = principal (initial amount of money invested)
r = interest rate (as a decimal)
t = time (in years)
In this case, we are given the following information:
I = $180
r = 6% = 0.06 (as a decimal)
t = 1 year
Substituting these values into the formula, we have:
$180 = P * 0.06 * 1
Simplifying the equation, we get:
$180 = 0.06P
To solve for P, we can divide both sides of the equation by 0.06:
$180 / 0.06 = P
Simplifying further:
$3000 = P
Therefore, the amount originally invested in the account was $3,000.
The answer is f) $3,000.