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Aleonysh [2.5K]
3 years ago
10

Ralph Chase plans to sell a piece of property for ​$150000. He wants the money to be paid off in two ways minus−a ​short-term no

te at 12​% interest and a​ long-term note at 9​% interest. Find the amount of each note if the total annual interest paid is ​$15900.
The amount of the 12​% note is ​$___. and the amount of the 9​% note is ​$___
Mathematics
1 answer:
Nezavi [6.7K]3 years ago
6 0

Answer:

The amount of 12% note is $90,000

The amount of 9% note is $80,000

Step-by-step explanation:

Missing word "The amount of the 12% note is $ and the amount of the 9% note is $"

Selling price = $170,000

Let x be the amount of money paid short term, therefore the amount of money paid long term is (170000 - x)

Given total annual interest = $18,000

Total annual interest = Interest due to short term (x) + interest due to long term (170000 - x)

18,000 = 0.12x + 0.09*(170000 - x)

18,000 = 0.12x + 15,300 - 0.09x

18,000 = 0.03x + 15,300

0.03x = 18,000 - 15,300

0.03x = 2,700

x = 2,700/0.03

x = 90,000

So, the amount of 12% note is $90,000

So long term paid money = $170,000 - $90,000 = $80,000

So, the amount of 9% note is $80,000

plz mark me as barinly

Step-by-step explanation:

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The number of ways of the books can be arranged are illustrations of permutations.

  • When the books are arranged in any order, the number of arrangements is 3628800
  • When the mathematics book must not be together, the number of arrangements is 2903040
  • When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
  • When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

\mathbf{Novels = 3}

\mathbf{Mathematics = 2}

\mathbf{Chemistry = 5}

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<u>(a) The books in any order</u>

First, we calculate the total number of books

\mathbf{n = Novels + Mathematics + Chemistry}

\mathbf{n = 3 + 2 +  5}

\mathbf{n = 10}

The number of arrangement is n!:

So, we have:

\mathbf{n! = 10!}

\mathbf{n! = 3628800}

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

\mathbf{Maths\ together = 2 \times 9!}

Using the complement rule, we have:

\mathbf{Maths\ not\ together = Total - Maths\ together}

This gives

\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}

\mathbf{Maths\ not\ together = 2903040}

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the novels in:

\mathbf{Novels = 3!\ ways}

Next, arrange the chemistry books in:

\mathbf{Chemistry = 5!\ ways}

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

\mathbf{n =Mathematics + 1 + 1}

\mathbf{n =2 + 1 + 1}

\mathbf{n =4}

So, the number of arrangements is:

\mathbf{Arrangements = n! \times 3! \times 5!}

\mathbf{Arrangements = 4! \times 3! \times 5!}

\mathbf{Arrangements = 17280}

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

\mathbf{Mathematics = 2}

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the mathematics in:

\mathbf{Mathematics = 2!}

Literally, the number of chemistry and mathematics now is:

\mathbf{n =Chemistry + 1}

\mathbf{n =5 + 1}

\mathbf{n =6}

So, the number of arrangements of these books is:

\mathbf{Arrangements = n! \times 2!}

\mathbf{Arrangements = 6! \times 2!}

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

\mathbf{Arrangements = ^7P_3}

So, the total arrangement is:

\mathbf{Total = 6! \times 2!\times ^7P_3}

\mathbf{Total = 6! \times 2!\times 210}

\mathbf{Total = 302400}

Read more about permutations at:

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