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2 years ago

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Mike's checking account pays 0.5% interest monthly and has a $7.50 service fee. If Mike's balance last month was $1423, find the

amount of credit or debit to his account after interest and service fee are both applied.

2 answers:

BARSIC [14]2 years ago

7 0

(1,423×0.005)−7.50
=−0.385 debit

marissa [1.9K]2 years ago

3 0

Answer:

The balance in the account when debit and credit is $1422.615.

Step-by-step explanation:

As given

Mike's checking account pays 0.5% interest monthly and has a $7.50 service fee.

If Mike's balance last month was $1423.

Than

0.5% is written in the decimal form.

$= \frac{0.5}{100}$

= 0.005

Thus

Monthly interest price = 0.005 × 1423

= $7.115

As given

service fee = $7.50

As the service tax is greater than the monthly interest price.

Thus

Debit = $7.50 - $7.115

= $0.385

The balance in the account when debit and credit = $1423 - $0.385

The balance in the account when debit and credit = $1422.615

Therefore the balance in the account when debit and credit is $1422.615 .

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Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

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And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

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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".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

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Using the normal standard table, excel or a calculator we see that:

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