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harina [27]
1 year ago
9

Calculate the unpaid balance, finance charge, and new balance using the unpaid balance method. Note: interest rate is given as a

monthly rate.
Previous balance = $283.63
Payments/credits = $60.00

Unpaid balance = $ a. 343.63 b. 223.63

Monthly rate = 1.5%

Finance charge = $ a. 3.35 b. 4.25
New purchases = $51.36
New balance = $ a. 278.34 b. 223.63 c. 3.35
Mathematics
1 answer:
inn [45]1 year ago
5 0
Unpaid balance 283.63-60=223.63
Finance charge
223.63*(0.015/12)=3.35
New balance
223.63+3.35+51.36==278.34
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Which is a stretch of an exponential growth function? f(x) = Two-thirds (two-thirds) Superscript x f(x) = Three-halves (two-thir
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Answer:

f(x) = Three-halves (three-halves) Superscript x

f(x) = Two-thirds (three-halves) Superscript x

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An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
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Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

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\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

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Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

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Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

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So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

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2 years ago
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