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WITCHER [35]
3 years ago
7

Clabber company has bonds outstanding with a par value of $100,000 and a carrying value of $97,300. if the company calls these b

onds at a price of $95,000, the gain or loss on retirement is
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Gain or loss = book value of the bonds- the amount paid to the bond hold

Gain or loss = $97,300 - $ 95,000 = $ 2300

The book value of the bond = face value of the bond+ Unamortized premium

                                    =$100,000+$x= $2300
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Solve for x: 1 1/5 x -2 1/3 < 1/7 x+ 1 1/2
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Solving inequalities is similar to solving a regular equation. The only thing you need to worry about with inequalities is that when you are dividing or multiplying by a negative number, you must flip the inequality sign. You won't have to worry about that here if the coefficient in front of x is positive when you divide!

Also remember:
- To add/subtract fractions, you have to turn all mixed numbers into improper fractions (if needed), <span>find a common denominator on both fractions (if needed), add/subtract the numerators, put the sum/difference over the common denominator, and simplify if needed. 

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Back to the problem:
</span>You are told that 1 \frac{1}{5} x - 2 \frac{1}{3} \ \textless \ \frac{1}{7} x + 1 \frac{1}{2} and you have to solve for x.
<span>
1) </span>Using the info for converting mixed numbers to improper fractions from above, you know that 1 \frac{1}{2} =  \frac{3}{2} and 2 \frac{1}{3} =  \frac{7}{3} and 1 \frac{1}{5} x =  \frac{6}{5} x. Now add/subtract using the info above, isolating the variable x by adding 2 \frac{1}{3} to both sides and subtracting \frac{1}{7} x from both sides. 
1 \frac{1}{5} x - 2 \frac{1}{3} \ \textless \ \frac{1}{7} x + 1 \frac{1}{2}\\&#10;\frac{6}{5} x -  \frac{7}{3} \ \textless \ \frac{1}{7} x + \frac{3}{2}\\&#10;(\frac{6}{5} x  - \frac{1}{7} x) \ \textless \ (\frac{3}{2} + \frac{7}{3})\\&#10;(\frac{42}{35} x  - \frac{5}{35} x) \ \textless \ (\frac{9}{6} + \frac{14}{6})\\&#10; \frac{37}{35} x \ \textless \   \frac{23}{6}

2) Divide both sides by \frac{37}{35} to get the inequality for x. Remember the info for dividing from above:
\frac{37}{35} x \ \textless \ \frac{23}{6}\\&#10;x \ \textless \  \frac{23}{6} \div \frac{37}{35}  \\&#10;x \ \textless \  \frac{23}{6} \times \frac{35}{37} \\&#10;x \ \textless \  \frac{805}{222}

Your final answer is x < \frac{805}{222} or x < 3 \frac{139}{222}.
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