   Chapter 2.2, Problem 88E Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

Investments Refer to Exercise 86. In August of which year will an investment of $4,000 made in August 2013 first exceed$4,400? [HINT: See Example 5; you saw this exercise before in Section 1.2.]

To determine

The time when investment of $4000 exceeds$4400 if Ally Bank was giving 0.61% interest on online saving accounts reinvested daily.

Explanation

Given Information:

Ally Bank was giving 0.61% interest on online saving accounts reinvested daily.

The amount of $4000 deposited in August 2013. Convert the rate of interest in the decimal form. r=0.61%=0.61100=0.0061 Since the interest is reinvested daily therefore the value of m is, m=365 Here leap year is avoided Consider the formula of compound interest is, A(t)=P(1+rm)mt Substitute P=$4000, r=0.0061 and m=365 in the expression A(t)=P(1+rm)mt.

A(t)=$4000(1+0.0061365)365t=$4000(1.0000167)365t=$4000((1.0000167)365)t=$4000(1.00612)t

Substitute t=8 in the expression A(t)=\$4000(1

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