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Javier deposited $250 into a savings account with an interest rate of 1.6%. He made no deposits or withdrawals for 9 months. If interest was compounded annually, which
equation represents the balance in the account after 9 months?
A = 250(1 -0.016)^9
A = 250(1 +0.0016)^9
A = 250(1 +0.016)^0.75
A = 250(1 -0.016)^0.75

Answer 1

`A = 250(1+0.016)^(0.75)`

Explanation:

The equation that represents the balance is expressed using the formula

`A = P(1+r)^(n) \\`

A is the amount after 9 months

P is the principal (amount deposited)

r is the interest rate

n is the time

Given P = $250, r = 1.6% = 0.0016, t =
`(9)/(12)\ years` (months converted to years)

Substituting the values into the formula we have;

`A = 250(1+0.016)^{(9)/(12) }\\A = 250(1+0.016)^(0.75)`

This gives the required answer

Answer 2

The answer is the third equation. A = 250*(1 +0.016)^(0.75)

Explanation:

Since Javier deposited $250 into an account with annual interest rate, then as the years passes his account will grow in the manner shown below:

account(0) = 250

account(1) = account(0)*(1 + 1.6/100) = account(0)*(1 + 0.016) = account(0)*1.016

account(2) = account(1)*1.016 = account(0)*1.016*1.016 = account(0)*(1.016)²

account(3) = account(2)*1.016 = account(0)*(1.016)²*1.016 = account(0)*(1.016)³

account(n) = account(0)*(1.016)^n

Where n is the number of years, account(0) is the initial amount. In this case only 9 months have passed, so we need to convert this value to years, dividing it by 12, which is 9/12 = 0.75. The initial amount was 250, so the equation is:

A = 250*(1.016)^(0.75)

The answer is the third equation.