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Kisachek [45]
3 years ago
13

what is the monthly payment for a loan if the amount to finance is 12,385, the APR is 6.9 percent for 5 years and the monthly pa

yment per $100 is $2.24
Mathematics
2 answers:
jeyben [28]3 years ago
8 0

Answer:$277.42

Step-by-step explanation:

12,385 divided by 100 = 123.85 multiply by 2.24 = 277.42 that is your answer.

dimulka [17.4K]3 years ago
4 0
<span>$12,385 APR = 6.9% 5 years= $864.565 monthly payment per $100 is $2.44 $12,385 + $864.565 =$13,222.565/60 =$220.37+ $2.44(2) =$225.25 Monthly</span>
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Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal
Usimov [2.4K]

Answer:

a) We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

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And we got for this case z_{crit}=2.33

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b) We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

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And we got for this case z_{crit}=\pm 1.96

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Step-by-step explanation:

Part a

We have that the significance is given by \alpha =0.01 and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case z_{crit}=2.33

So then the rejection region would be z>2.33

Part b

We have that the significance is given by \alpha =0.05, \alpha/2 =0.025 and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case z_{crit}=\pm 1.96

So then the rejection region would be z>1.96 \cup z

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