Maite has money in an interest-bearing account. The table shows how much money is in the account at the end of each year.
N
Year Amount
1 $1,000.00
$1,030.00
3 $1,060.90
4 $1,092.73
5 $1,125.51
This situation represents
sequence.
The common
is
At the end of the seventh year, Maite will have $
in the account

Respuesta :

Determining how much money will be in the account of Maite at the end of each year, we use an exponential growth factor, since this is a geometric sequence.

1. This situation represents a geometric  sequence.

A geometric sequence increases by a common exponential growth factor.

2. The common exponential factor is 1.03 (which gives a growth rate of 3% annually).  See how this factor is determined below.

3. At the end of the seventh year, Maite will have $1,194.05 in the account.  See the calculation below.

Data and Calculations:

Year    Amount

1       $1,000.00

2      $1,030.00

3      $1,060.90

4      $1,092.73

5       $1,125.51

6      $1,159.27 ($1,125.51 * 1.03)

7      $1,194.05 ($1,159.27 * 1.03)

The common exponential factor = 1.03 (1 + 0.03)

To obtain the common exponential factor, subtract Year 2 account balance from Year 1 account balance.  Divide the result by Year 1 account balance. This operation can also be carried out with Year 2 and Year 3 balances or Year 4 and Year 5 balances.

To determine how much money will be in the account of Maite at the end of Year 6, using Year 5 as a base = (Year 5 account balance * Exponential Factor)

= $1,159.27 ($1,125.51 * 1.03)

To determine Year 7 account balance, we use Year 6 above as the base

= $1.194.05 ($1,159.27 * 1.03)

Learn more about geometric sequence and exponential growth here: https://brainly.com/question/7154553

Answer:

This situation represents a geometric sequence.

The common ratio is 1.03

At the end of the seventh year, Maite will have $1,194.06 in the account.

Step-by-step explanation:

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