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Aleks [24]
3 years ago
13

At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr

om the account.Which explicit formula can be used to find the account’s balance at the beginning of year 14? What is the balance?
A.A(n) = (250)(n – 1)(0.03); $97.50 B.A(n) = 250 + (n)(0.03 • 250); $355.00 C.A(n) = 250 + (n – 1)(0.03); $250.39 D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50
Mathematics
2 answers:
olganol [36]3 years ago
8 0

Answer:

the correct answer is D

Step-by-step explanation:

i just did it and it was right

Dennis_Churaev [7]3 years ago
4 0
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
</span>
Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250

At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
</span>
A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5

So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50 </span>
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Answer:

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And the confidence interval is given by:

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And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

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p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

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And the confidence interval is given by:

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And for this case the interval contains the value 0.16, so then we can conclude at 5% of significance that the true proportion is not different from 0.16

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Please help with c and i. and ii.
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