Occasionally a savings account may actually pay interest compounded continuously. For each deposit, find the interest earned if interest is compounded (a) semiannually, (b) quarterly, (c) monthly, (d) daily, and (e) continuously. Use 1 year=365 days

1. Occasionally a savings account may actually pay interest compounded continuously. For each deposit, find the interest earned if interest is compounded (a) semiannually, (b) quarterly, (c) monthly, (d) daily, and (e) continuously. Use 1 year = 365 days.

Principal $1031

Rate 1.4%

Time 3 years

a) $ 44.07

b) $ 44.15

c) $ 44.20

d) $ 44.22

e) $ 44.22

Step-by-step explanation:

The formula to find the total amount earned using compound interest is given as:

A = P(1 + r/n)^nt

Where A = Total amount earned after time t

P = Principal = $1031

r = Interest rate = 1.4%

n = compounding frequency

t = Time in years = 3 years

For each deposit, find the interest earned if interest is compounded

(a) semiannually

This means the interest is compounded 2 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/2) ^2 × 3

A = 1031 (1 + 0.007)^6

A = $ 1,075.07

A = P + I where

I = A - P

I = $1075.07 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.07

(b) quarterly

This means the interest is compounded 4 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/4) ^4 × 3

A = 1031 (1 + 0.014/4)^12

A = $ 1,075.15

I = A - P

I = $1075.15 - $1031

A = P + I where

P (principal) = $ 1,031.00

I (interest) = $ 44.15

(c) monthly,

This means the interest is compounded 12 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/12) ^12 × 3

A = 1031 (1 + 0.014/12)^36

A = $ 1,075.20

A = P + I where

I = A - P

I = $1075.20 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.20

(d) daily,Use 1 year = 365 days

This means the interest is compounded 365 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/365) ^2 × 3

A = 1031 (1 + 0.00365)^365 × 3

A = $ 1,075.22

A = P + I where

I = A - P

I = $1075.22 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.22

(e) continuously. .

This means the interest is compounded 2 times in a year

Hence:

A = Pe^rt

A = 1031 × e ^0.014 × 3

A = $ 1,075.22

A = P + I where

I = A - P

I = $1075.22 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.22

rick has $1400 and sally has $1500

step-by-step explanation:

to find amount of money rick has after 10 years :

principal = $1000 , time = 10 years , rate = 4% annually

to find amount of money sally has after 10 years :

her friend gives to pay her $50 per year, therefore money she has after 10 years = 50 × 10 = $500

also, sally gets $1000 after 10 years,

therefore, total amount of money sally has after 10 years = 500 + 1000 = $1500

so, after 10 years : rick has $1400 and sally has $1500