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s2008m [1.1K]
3 years ago
10

If $64,000 is invested in an IRA account with an annual interest rate of 8% compounded once a year, what is the value of the acc

ount after 6 years? Round your answer to two decimal places.
Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer: the value of the account after 6 years is $101559.96

Step-by-step explanation:

If $64,000 is invested in an IRA account, then

Principal = $64,000

So P = 64,000

The rate at which $64000 was compounded is 8%

So r = 8/100 = 0.08

If it is compounded once in a year, this means that it is compounded annually (and not semi annually, quarterly or others). So

n = 1

We want to determine the value of the account after 6 years, this means

time, t = 6

Applying the compound interest formula,

A = P(1 + r/n)^nt

A = amount after n number of years

A = 64000( 1 + 0.08/1)^1×6

A = 64000(1.08)^6

A= 64000×1.58687432294

A= 101559.956668416

Approximately $101559.96 to 2 decimal places

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given first random sample size n₁ = 500</em>

Given  Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

<em>First sample proportion </em>

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<em>Level of significance = α = 0.05</em>

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<u><em>Step(ii)</em></u><em>:-</em>

<em>Null hypothesis : H₀: There  is no significance difference between these proportions</em>

<em>Alternative Hypothesis :H₁: There  is significance difference between these proportions</em>

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<em>         </em>P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2}  }{n_{1}+ n_{2}  } = \frac{500 X 0.13+700 X0.19  }{500 + 700 } = 0.165<em></em>

<em>        Q = 1 - P = 1 - 0.165 = 0.835</em>

<em></em>Z = \frac{0.13-0.19  }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }<em></em>

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<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>Alternative hypothesis is accepted at 0.05 level of significance</em>

<u><em>Conclusion:</em></u><em>-</em>

<em>There is there is a difference between these proportions at α = 0.05</em>

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