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WINSTONCH [101]
2 years ago
11

On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any

additional deposits or withdrawals since then, and if the interest rate has stayed the same, in what year did his balance hit $2300, according to the rule of 72?
Mathematics
2 answers:
dalvyx [7]2 years ago
7 0
According to the 72 rule
72/rate=time
72÷9.6=7.5 years

Another way to solve by using the main equation
2300=1150(1+0.096/4)^4t
Solve for t
t=((log(2,300÷1,150)÷log(1+(0.096÷4))÷4))=7.31years

Hope it helps :-)
Harman [31]2 years ago
7 0

Answer: 7.5 years.

Step-by-step explanation:

Here the given amount = $1150

Amount after getting compound interest in this amount = $ 2300

Since, $2300 is double to the amount $1150

According to the rule of 72, an amount is doubled if the product of annual rate and period (in years) is equal to 72. Also, with the help of this rule we get the approximate value of rate or time.

Since, Here the annual rate of interest = 9.6%

Let the given amount $1150 is doubled ($2300)  in t years.

The, 9.6 × t = 72

⇒ t = 72/9.6 = 7.5

Thus, given amount $1150 is doubled in 7.5 years(approx).

Verification : Finding the number of year with help of general formula of compound interest.

2300=1150(1+\frac{9.6}{100})^{t}

2=(1+\frac{9.6}{100})^{t}

2=(1+\frac{9.6}{100})^{t}

2^{1/t}=(1+\frac{9.6}{100})

⇒ t = 7.562 years.

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Step-by-step explanation:

-3x(4x - 5xy + 8y)

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A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample
8090 [49]

Answer:

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>Given second sample size n₂ = 700</em>

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

<em>critical value = 1.96</em>

<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>

<em>Test statistic </em>

<em></em>Z = \frac{p_{1} ^{-}-p^{-} _{2}  }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }<em></em>

<em>where </em>

<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>

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<em>Z =  -2.76</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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Rounding up

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