Question

The balance of an investment is increasing according to the rule: Next year’s balance = 1.08 x current balance What is the constant percent increase per year? If the original balance of the investment was $2000 find a formula that gives the balance B (in dollars) after T years.

Answer 1

Answer:

The constant percent increase per year is 8%.

$B(T) = 2000(1.08)^{T}{/tex]Step-by-step explanation:The amount in the balance after T years is given by the following equation:[tex]B(T) = B(0)(1+r)^{T}$

In which B(0) is the initial amount and r is the growth rate, as a decimal.

Next year’s balance = 1.08 x current balance

So

$B(1) = (1.08)B(0)$

This means that:

$1 + r = 1.08$

$r = 1.08 - 1$

$r = 0.08$

The constant percent increase per year is 8%

If the original balance of the investment was $2000 find a formula that gives the balance B (in dollars) after T years.

This means that $B(0) = 2000$

So

$B(T) = B(0)(1+r)^{T}$

$B(T) = 2000(1+0.08)^{T}$

[tex]B(T) = 2000(1.08)^{T}{/tex]

Answer 2

Answer:

$\large \boxed{1. 8 \, \%; \, 2. \, B_{T} = 2000(1.08)^{T}}$

Step-by-step explanation:

1. Rate of increase

 Let B = this year's balance. Then

1.08B = next year's balance

1.08B = B + 0.08B = B + 8 % of B

$\text{The rate of increase of B is \large \boxed{\mathbf{8 \%}} per year.}$

2. The formula

Each year, the balance is multiplied by 1.08.

The end of a year is the same as the end of the previous year.  

If T = the number of years, then the balance at the beginning of Year 1 is B₀.

$B_{T} = B_{0}(1.08)^{T}$

If B₀ = $2000, the formula becomes

$\large \boxed{\mathbf{B_{T} = 2000(1.08)^{T}}}$