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Vlad1618 [11]
3 years ago
7

Skylar plans to use $3400 to open a savings account with an annual interest rate of 1.15%. How much more interest will he earn o

ver 13 years if he chooses a compound interest account that compounds interest quarterly instead annually? Round your answer to the nearest cent.
interest compounded annually: A = P (1 + r)t
interest compounded quarterly: A = P (1 +r/4)4t
 
Mathematics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

$ 2.5

Step-by-step explanation:

Given that

Total Amount to be invested = P = $3400

Total Time of investment = t = 13 years

Rate of interest = r = 1.15 %

Lets calculate Simple interest first

As we know formula for simple interest is

A = P (1 +r)^{t}

Substituting the values

A = 3400(1+ \frac{1.15}{100} )^{13}

A =  $ 3944.9

Lets now calculate the compound interest

As we know formula for simple interest is

A = P (1 + \frac{r}{n})^{nt}

 As compound interest applies every 3 months so n =3

So,

A = 3400 (1 + \frac{1.15}{100*4} )^{13*4}

A  = $ 3947.4

Now Lets calculate how much more he will earn using compound interest.

It can be found by taking difference of compound interest and simple interest

Amount = 3947.4 - 3944.9 = $ 2.5

So Skyler must have earned 2.5 dollar more using the compound interest.

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(a) P(x\le 3) = 0.75

(b) P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

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Step-by-step explanation:

Given

\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

Solving (b): Probability of at most 3 CDs

Here, we consider:

\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}

This probability is calculated as:

P(x\le 3) = P(1) + P(2) + P(3)

This gives:

P(x\le 3) = 0.30 + 0.25 + 0.20

P(x\le 3) = 0.75

<em>(b) is the same as (a)</em>

<em />

Solving (c): Probability of 5 or more CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x \ge 5) = P(5) + P(6\ or\ more)

This gives:

P(x\ge 5) = 0.05 + 0.05

P(x \ge 5) = 0.10

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}

This probability is calculated as:

P(x = 1\ or\ 2) = P(1) + P(2)

This gives:

P(x = 1\ or\ 2) = 0.30 + 0.25

P(x = 1\ or\ 2) = 0.55

Solving (e): Probability of more than 2 CDs

Here, we consider:

\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}

This probability is calculated as:

P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)

This gives:

P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05

P(x > 2) = 0.45

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