Question

Suppose a deposit of $ 2 , 000 in a savings account that paid an annual interest rate r (compounded yearly) is worth $ 2 , 209 after 2 years. using the formula a = p ( 1 + r ) t , we have 2 , 209 = 2 , 000 ( 1 + r ) 2 solve for r to find the annual interest rate (to the nearest tenth).

Answer 1

Answer:

5.1% (nearest tenth)

Step-by-step explanation:

Annual Compound Interest Formula

$\large \sf A=P\left(1+r\right)^{t}$

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• t = time (in years)

Given:

• A = $2,209
• P = $2,000
• t = 2 years

Substitute the given values into the formula and solve for r:

$\implies \sf 2209=2000\left(1+r\right)^{2}$

$\implies \sf \dfrac{2209}{2000}=(1+r)^2$

$\implies \sf 1.1045=(1+r)^2$

$\implies \sf \sqrt{1.1045}=1+r$

$\implies \sf r = \sqrt{1.1045}-1$

$\implies \sf r = 0.05095194942...$

$\implies \sf r = 5.095194942...\%$

Therefore, the annual interest rate is 5.1% (nearest tenth)