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71000

(1+ 0.07) 30 (1)

360(15)

360

71000 1 0 07 30 1 360 15 360

Answer :

Answer :

  • P = 25,898.04

Explanation :

we are given that the future value of an investment is 74000 with an interest rate of 7% and compounding periods equal to 360 over a period of 15 years and asked to find the initial principal amount (P).

we can work out the value of P using the formula ,

  • A = P(1 + r/n)^n*t

where,

  • A = future value of the investment made ( 74000 )
  • P = Initial principal amount
  • r = Interest rate (7% = 0.07)
  • n = compounding periods
  • t = time period (15 years )

plugging in the values,

  • 74000 = P(1 + 0.07/360)^(360(15))
  • P = (74000)/(1 + 0.07/360)^(360(15))
  • P = (74000)/(2.857359....)
  • P = 25,898.04 ( 2 d.p. )

thus, the initial principal amount is 25,898.04.

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Rewritten by : Barada

Answer:

[tex]P \approx 25898 [/tex]

Step-by-step explanation:

To simplify the given expression, we follow the order of operations, starting from the innermost parentheses and working outward.

Given:

[tex] P = \dfrac{74000}{\left(1 + \dfrac{0.07}{360}\right)^{360 \times 15}} [/tex]

First, let's simplify the expression inside the parentheses:

[tex] 1 + \dfrac{0.07}{360} [/tex]

To add 1 and [tex] \dfrac{0.07}{360} [/tex], we need to make the denominators the same:

[tex] 1 + \dfrac{0.07}{360} = \dfrac{360}{360} + \dfrac{0.07}{360} = \dfrac{360 + 0.07}{360} [/tex]

[tex] 1 + \dfrac{0.07}{360} = \dfrac{360.07}{360} [/tex]

Now, the expression becomes:

[tex] P = \dfrac{74000}{\left(\dfrac{360.07}{360}\right)^{360 \times 15}} [/tex]

We can simplify this further by dividing 360.07 by 360:

[tex] P = \dfrac{74000}{\left(\dfrac{360.07}{360}\right)^{360 \times 15}} \\\\= \dfrac{74000}{\left(1.000194444\right)^{360 \times 15}} [/tex]

Now, we raise [tex]1.000194444[/tex] to the power of [tex]360 \times 15[/tex]:

[tex] P = \dfrac{74000}{(1.000194444)^{5400}} [/tex]

Now, we compute the final result:

[tex] P \approx \dfrac{74000}{2.857359452 } [/tex]

[tex] P \approx 25898.03671 [/tex]

[tex] P \approx 25898 \textsf{(in nearest whole number)}[/tex]

So, the simplified expression is approximately:

[tex]\sf P \approx 25898 [/tex].