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slega [8]
3 years ago
7

Cindy invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. approximately how much does she

have in her account after two years?
Business
1 answer:
stiks02 [169]3 years ago
6 0

Annual Compound Formula is:

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

Where:

A is the future value of the investment

P is the principal investment

r is the annual interest rate

<span>n is the number of  interest compounded per year</span>

t is the number of years the money is invested


So for the given problem:

P = $10,000

r = 0.0396

n = 2 since it is semi-annual

t = 2 years

 

Solution:

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

A = $10,000 ( 1 + 0.0396/2) ^ (2)(2)

A = $10000 (1.00815834432633616)

A = $10,815.83 is the amount after two years

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Which of the following is an example of an illiquid asset?
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Read 2 more answers
Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows, leather seats and cruise control.
Law Incorporation [45]

This question is incomplete! here is the complete question with answer!

Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows, leather seats and cruise control.

Quantity A : Number of different combinations possible

Quantity B : 25

A The quantity in Column A is greater

B The quantity in Column B is greater

C The two quantities are equal.

D The relationship cannot be determined from the information given.

Answer:

A The quantity in Column A is greater

Explanation:

Lets first analyze the number of different combinations possible!

To find out the combinations we use the following formula:

The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter.

nCr=\frac{n!}{r!(n-r)!}

Car X can have these 5  features: sunroof, stereo, tinted windows, leather seats and cruise control.

Car X has 2 options for each feature: Included or Not included

If car X has no feature at all then:

5C0=\frac{5!}{0!(5-0)!}=\frac{5!}{0!*5!}=1

If car X has 1 feature included then:

5C1=\frac{5!}{1!(5-1)!}=\frac{5!}{1!*4!}=5

If car X has 2 features included then:

5C2=\frac{5!}{2!(5-2)!}=\frac{5!}{2!*3!}=10

If car X has 3 features included then:

5C3=\frac{5!}{3!(5-3)!}=\frac{5!}{3!*2!}=10

If car X has 4 features included then:

5C4=\frac{5!}{4!(5-4)!}=\frac{5!}{4!*1!}=5

If car X has all of the 5 features included then:

5C5=\frac{5!}{5!(5-5)!}=\frac{5!}{5!*0!}=1

Now sum all the possible combinations: 1+5+10+10+5+1=32

So, quantity A is 32 and quantity B is 25

Therefore, we can conclude that the quantity in column A is greater

8 0
3 years ago
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