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Klio2033 [76]
3 years ago
5

Help me with this problem

Mathematics
2 answers:
Paha777 [63]3 years ago
8 0

The answer to your question is C because they are related ints


siniylev [52]3 years ago
5 0
The answer of your problem is C. Because 25 and 100 are related integers.
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Occasionally a savings account may actually pay interest compounded continuously. For each deposit, find the interest earned if
Ugo [173]

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

Answer:

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

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