An investment of 1 will double in 27.72 years at a force of interest, δ. An investment of 1 will increase to 7.04 in n years at a nominal rate of interest numerically equal to δ and convertible once every two years. Calculate n.

Answers

Answer 1
Answer:

Answer:

80

Explanation:

According to the given situation, the computation of n is shown below:-

EXP[27.72δ]=2

δ =0.025

m = 1 ÷ 2

(1 + 0.025 ÷ (1 ÷ 2))^n ÷ 2 = 7.04

n ÷ 2 × ln(1.05)=ln(7.04)

n ÷ 2=40

n = 80

Therefore for computing the n we simply applied the above formula i.e. by considering all the information given in the question

Hence,the n is 80

Answer 2
Answer:

To find the number of years it takes for an investment of $1 to increase to $7.04 at a nominal rate of interest numerically equal to δ and convertible once every two years, we can use the formula A = P(1 + r/m)^mt. Using this formula, we can solve for t by substituting the given values into the equation and solving for t using logarithms.

To find n, the number of years it takes for an investment of $1 to increase to $7.04 at a nominal rate of interest numerically equal to δ and convertible once every two years, we can use the formula:



A = P(1 + r/m)mt



Where A is the final amount, P is the initial investment, r is the nominal rate of interest, m is the number of times interest is compounded per year, and t is the number of years.



In this case, A = $7.04, P = $1, r = δ, and m = 2 (since it is convertible once every two years). Using this information, we can solve for t:



$7.04 = $1(1 + δ/2)2t



Divide both sides by $1:



7.04 = (1 + δ/2)2t



Take the logarithm of both sides:



log(7.04) = log((1 + δ/2)2t)



Apply the power rule of logarithms:



log(7.04) = 2t * log(1 + δ/2)



Divide both sides by 2 * log(1 + δ/2):



t = log(7.04) / (2 * log(1 + δ/2))



Plug in the value of δ to find the value of t.

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Answers

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

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Answers

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Answers

Answer:

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Answers

Answer:

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Explanation:

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Answers

Answer:

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