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Sauron [17]
2 years ago
13

The account balance on April 1st is $50.51. On April 15th, a payment of $15.00 is made. On April 25th, a purchase of $19.27 is m

ade. What is the finance charge if the annual rate is 18%? What is the new account balance?
Mathematics
2 answers:
Ilia_Sergeevich [38]2 years ago
4 0

Answer:

Step-by-step explanation:

Combine the information in the problem and the chart using the average balance method to solve the problem.

The account balance on April 1st is $50.51. On April 15th, a payment of $15.00 is made. What is the average daily balance for the account? What is the finance charge if the annual rate is 18%? What is the new account balance?

matrenka [14]2 years ago
3 0
Hi there!

The true answer is as follows:

Finance charge = $0.76
To get this you need to take the account balance on April 1st and multiply by the annual rate of 18%, which looks like this:

$50.51 × 0.18 = 9.0918
Then you divide your answer by 12 for the months of the year to get:

9.0918 ÷ 12 = 0.75765
Which rounds to $0.76 for you finance charge

New balance = $55.54
To get this you take the account balance on April 1st and subtract the payment of $15.00 and then add the purchase of $19.27 and the finance charge of $0.76, like so:

$50.51 - $15.00 = $35.51 + $19.27 + $0.76 = $55.54

Your friend, ASIAX
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Answer:

(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}} = 10.862

(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}} = 33.138

And the confidence interval for the difference of means is given by:

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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".

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\bar X_{A}=242 represent the mean for the sample A

\bar X_{B}=220 represent the mean for the sample B

s_{A}=20 represent the sample standard deviation for the sample A

s_{B}=31 represent the sample standard deviation for the sample B

n_{A}=47 sample size selected A

n_{B}=42 sample size selected B

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For this case the confidence interval for the difference of means \mu_A -\mu_B is given by:

(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}

The degrees of freedom are given by:

df = n_A +n_B -2= 47+42-2=87

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,87)".And we see that t_{\alpha/2}=1.988

And the confidence interval would be given by

(242-220) -1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}} = 10.862

(242-220) +1.988 \sqrt{\frac{20^2}{47} +\frac{31^2}{42}} = 33.138

And the confidence interval for the difference of means is given by:

10.862 \leq \mu_A -\mu_B \leq 33.138

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