Step 1

a) Simple interest for an amount P, at an interest rate of r (in decimals) per annum and for T years is given by:

Plugging the values:

\(I=PTr\)

\(I=(10000)(4)(0.055)\)

\(I=2200\)

Step 2

b) Amount after continuous compounding for T years of principle P at an interest rate of r is given by:

Plugging the values:

So the interest is given by:

\(A=Pe^{rT}\)

\(A=(10000)e^{(0.05)(4)}\)

\(A=(10000)e^{0.2}\)

\(A=(10000)(1.221402)\)

\(A=12214.03\)

\(I=12214.03-10000=2214.03\)

Step 3

c) Hence, simple interest results in less total interest.

a) Simple interest for an amount P, at an interest rate of r (in decimals) per annum and for T years is given by:

Plugging the values:

\(I=PTr\)

\(I=(10000)(4)(0.055)\)

\(I=2200\)

Step 2

b) Amount after continuous compounding for T years of principle P at an interest rate of r is given by:

Plugging the values:

So the interest is given by:

\(A=Pe^{rT}\)

\(A=(10000)e^{(0.05)(4)}\)

\(A=(10000)e^{0.2}\)

\(A=(10000)(1.221402)\)

\(A=12214.03\)

\(I=12214.03-10000=2214.03\)

Step 3

c) Hence, simple interest results in less total interest.