anonymous
  • anonymous
a company had a total debt of $320,000 in 1980. between 1980 and 1987 it was able to reduce its debt 15% each year. approximate the companys debt in 1987 to the nearest 1000
Mathematics
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anonymous
  • anonymous
a company had a total debt of $320,000 in 1980. between 1980 and 1987 it was able to reduce its debt 15% each year. approximate the companys debt in 1987 to the nearest 1000
Mathematics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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gw2011
  • gw2011
The answer to the nearest 1000 is $103,000 which is for the end of 1987 . Here is how you can calculate it: (D)(0.85)^7 where D = the total debt (0.85)^7 = the amount of debt remaining which is taken to the power of 7 since the debt is reduced each year beginning in 1981 and going to the end of 1987 which is 7 years. Therefore, the formula can now be used as follows: (D)(0.85)^7 = ($320,000)(0.85)^7 = $102,584.67 and to the nearest 1000 the answer is $103,000. I've assumed that the total debt at the end of 1980 is $320,000 and that the debt is to be calculated for the end of 1987.

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