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Karo-lina-s [1.5K]
1 year ago
15

jesse had a balance of 1200 on a credit card with an APR of 18.7% compounded monthly. About how much will he save in interest ov

er the course of a year if he transfers his balance to a credit card with an APR of 12.5% compounded monthly?
Mathematics
1 answer:
igor_vitrenko [27]1 year ago
5 0

Jesse saved about $85.77 in interest over the course of a year if he transfer from one credit card to another.

<h3>What is the interest accrued on savings?</h3>

The interest accrued on saving is the interest gained on saving or investment over a period of time.

ar

The amount of money saved in interest by Jesse over a period of 12 months provided that the transferred his balance to a credit card is calculated as follows:

Amount saved = \mathbf{P_x(1 + \dfrac{r}{P_x})^{12}-P_y(1 + \dfrac{r}{P_y})^{12}}

Amount saved = \mathbf{1200(1 + \dfrac{18.7}{1200})^{12} - 1200(1 + \dfrac{12.5}{1200})^{12}}

Amount saved = $85.768

Amount saved = $85.77

Learn more about interest accrued on a savings here:

brainly.com/question/1434515

#SPJ1

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tia_tia [17]

Answer:

a) 47.917-1.796\frac{2.811}{\sqrt{12}}=46.460    

47.917+1.796\frac{2.811}{\sqrt{12}}=49.374    

b) Null hypothesis: \mu =50

Alternative hypothesis: \mu \neq 50

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.

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.  

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We have the following data: 44 49 45 51 49 53 49 44 47 50 46 48

And we can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}

And replacing we got:

\bar X= 47.917, s= 2.811

\mu population mean (variable of interest)  

n=12 represent the sample size  

Part a: Calculate the confidence interval

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)  

The degrees of freedom are given:

df= n-1= 12-1=11

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,11)".And we see that t_{\alpha/2}=1.796

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

47.917-1.796\frac{2.811}{\sqrt{12}}=46.460    

47.917+1.796\frac{2.811}{\sqrt{12}}=49.374    

Part b

For this case the system of hypothesis are:

Null hypothesis: \mu =50

Alternative hypothesis: \mu \neq 50

And after analyze the interval obtained the upper limit not contains the value of 50, so then we can conclude that the claim is not satisfied for this case, and the population mean seems to be lower than 50 at 10% of significance.

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Answer:

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Step-by-step explanation:

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Solving the equation to find the value of x.

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cancelling out -2x and 2x

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Answer:

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x^2 = 19^2 - 12^2

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x^2 = 217

x = \sqrt{217}

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