situation: tina wants to save money for school. tina invests $1,000 in an account that pays an interest rate…

situation: tina wants to save money for school. tina invests $1,000 in an account that pays an interest rate of 6.75%. how many years will it take for the account to reach $18,600? round your answer to the nearest hundredth. enter the correct answer. a = p(1 + r)^t p = amount of money invested r = interest rate percentage in decimal form t = time in years a = total money in the account at time

situation: tina wants to save money for school. tina invests $1,000 in an account that pays an interest rate of 6.75%. how many years will it take for the account to reach $18,600? round your answer to the nearest hundredth. enter the correct answer. a = p(1 + r)^t p = amount of money invested r = interest rate percentage in decimal form t = time in years a = total money in the account at time

Answer

Explanation:

Step1: Identify given values

$P = 1000$, $r=0.0675$, $A = 18600$

Step2: Substitute into formula

$18600=1000(1 + 0.0675)^t$

Step3: Simplify the equation

Divide both sides by 1000: $\frac{18600}{1000}=(1.0675)^t$, so $18.6=(1.0675)^t$

Step4: Take the natural - logarithm of both sides

$\ln(18.6)=\ln(1.0675^t)$

Step5: Use logarithm property

$\ln(18.6)=t\ln(1.0675)$

Step6: Solve for t

$t=\frac{\ln(18.6)}{\ln(1.0675)}$ $t\approx\frac{2.9228}{0.0653}\approx44.76$

Answer:

$44.76$