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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Finding Account Balances In Exercises 19-22, complete the table to determine the balance A for P dollars invested at rate r for t years, compounded n times per year.

n 1 2 4 12 365 Continuous compounding
A

P = $ 3000 ,   r = 3.5 % ,   t = 10  years

To determine

To calculate: The amount A for corresponding value of n that is, number of times compounded per year in the following table when $3000 is invested at the rate of 3.5% for 10 years,

n 1 2 4 12 365 Continuous compounding
A
Explanation

Given Information:

The amount of $3000 is invested at the rate of 3.5% for 10 years and the table is as follows,

n 1 2 4 12 365 Continuous compounding
A

Formula used:

The formula to compute the balance amount after t years when interest is compounded n times per year is,

A=P(1+rn)nt

Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals, t is the number of years and n is the number of times compounded per year.

The formula to compute the balance amount after t years when interest is compounded continuously is,

A=Pert

Where, P is the deposited amount, A is the amount after t years, r is the interest rate in decimals and t is the number of years.

Calculation:

Consider that the amount of $3000 is invested at the rate of 3.5% for 10 years.

Here, P=$3000, r=3.5% and t=10 years.

Simplify the rate as,

r=3.5%=3.5100=0.035

When n=1,

Substitute 1 for number of times compounded per year, 0.035 for interest rate, $3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as,

A=P(1+rn)nt=$3000(1+0.0351)(1)(10)

Solve further as,

A=$3000(1.035)10$30001.41060$4231.80

When n=2,

Substitute 2 for number of times compounded per year, 0.035 for interest rate, $3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as,

A=P(1+rn)nt=$3000(1+0.0352)(2)(10)

Solve further as,

A=$3000(1.0175)20$30001.414778$4244.33

When n=4,

Substitute 4 for number of times compounded per year, 0.035 for interest rate, $3000 for the amount deposited and 10 years for the time in the formula, A=P(1+rn)nt as,

A=P(1+rn)nt=$3000(1+0

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P-17RECh-4 P-18RECh-4 P-19RECh-4 P-20RECh-4 P-21RECh-4 P-22RECh-4 P-23RECh-4 P-24RECh-4 P-25RECh-4 P-26RECh-4 P-27RECh-4 P-28RECh-4 P-29RECh-4 P-30RECh-4 P-31RECh-4 P-32RECh-4 P-33RECh-4 P-34RECh-4 P-35RECh-4 P-36RECh-4 P-37RECh-4 P-38RECh-4 P-39RECh-4 P-40RECh-4 P-41RECh-4 P-42RECh-4 P-43RECh-4 P-44RECh-4 P-45RECh-4 P-46RECh-4 P-47RECh-4 P-48RECh-4 P-49RECh-4 P-50RECh-4 P-51RECh-4 P-52RECh-4 P-53RECh-4 P-54RECh-4 P-55RECh-4 P-56RECh-4 P-57RECh-4 P-58RECh-4 P-59RECh-4 P-60RECh-4 P-61RECh-4 P-62RECh-4 P-63RECh-4 P-64RECh-4 P-65RECh-4 P-66RECh-4 P-67RECh-4 P-68RECh-4 P-69RECh-4 P-70RECh-4 P-71RECh-4 P-72RECh-4 P-73RECh-4 P-74RECh-4 P-75RECh-4 P-76RECh-4 P-77RECh-4 P-78RECh-4 P-79RECh-4 P-80RECh-4 P-81RECh-4 P-82RECh-4 P-83RECh-4 P-84RECh-4 P-85RECh-4 P-86RECh-4 P-87RECh-4 P-88RECh-4 P-89RECh-4 P-90RECh-4 P-91RECh-4 P-92RECh-4 P-93RECh-4 P-94RECh-4 P-95RECh-4 P-96RECh-4 P-97RECh-4 P-98RECh-4 P-99RECh-4 P-100RECh-4 P-101RECh-4 P-102RECh-4 P-103RECh-4 P-104RECh-4 P-105RECh-4 P-106RECh-4 P-107RECh-4 P-108RECh-4 P-109RECh-4 P-110RECh-4 P-111RECh-4 P-112RECh-4 P-113RECh-4 P-114RECh-4 P-115RECh-4 P-116RECh-4 P-117RECh-4 P-118RECh-4 P-119RECh-4 P-120RECh-4 P-121RECh-4 P-122RECh-4 P-123RECh-4 P-124RECh-4 P-125RECh-4 P-126RECh-4 P-127RECh-4 P-128RECh-4 P-129RECh-4 P-130RECh-4 P-131RECh-4 P-132RECh-4 P-1TYSCh-4 P-2TYSCh-4 P-3TYSCh-4 P-4TYSCh-4 P-5TYSCh-4 P-6TYSCh-4 P-7TYSCh-4 P-8TYSCh-4 P-9TYSCh-4 P-10TYSCh-4 P-11TYSCh-4 P-12TYSCh-4 P-13TYSCh-4 P-14TYSCh-4 P-15TYSCh-4 P-16TYSCh-4 P-17TYSCh-4 P-18TYSCh-4 P-19TYSCh-4 P-20TYSCh-4 P-21TYSCh-4 P-22TYSCh-4 P-23TYSCh-4 P-24TYSCh-4 P-25TYSCh-4 P-26TYSCh-4 P-27TYSCh-4 P-28TYSCh-4 P-29TYS

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