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Mnenie [13.5K]
2 years ago
9

Stefani is purchasing a house and finances $410,350 of the purchasing price. The mortgage is a 30 year 7/3 ARM at 6.5% with a 4/

13 cap structure. What will the remaining balance be after the first seven years? (show work)
A= $371,023.06
B= $217,870.06
C= $357,112.36
D= $192,479.94

Mathematics
1 answer:
poizon [28]2 years ago
6 0
I used an online mortgage calculator.

Inputted the following information:
Loan amount: 410,350
loan rate: 6.5%
loan term: 30 years

It gave me the following results:
monthly payment: 2,593.69
annual payment: 31,124.29

31,124.29 * 7 years = 217,870.03 total payment after 7 years.

Based on the Yearly Amortization Table, the remaining balance after the first seven years is <span>$373,327.70

The figure is nearest to Choice A. 371, 023.06</span>

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3 years ago
Helppppopleaseee????
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Answer:

LQ = 5

Median = 6

Step-by-step explanation:

\frac{n+1}{2} gives us 5.5, so we do the average of position 5 and 6. This gives us 6, so the median is 6.

Now we find the median of the bottom half of the numbers, which is 5, so the LQ is 5.

Now we find the median of the upper half of the numbers, which is 7, so the UQ is 7.

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2 years ago
What is the math problem answer?
azamat

To find the mid point add both X coordinates together and divide by 2 and then do the same with the y coordinates.

X coordinate: 4 + -6 = 4-6 = -2 / 2 = -1

Y coordinate: -2 + 3 = 1 / 2 = 1/2


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3 years ago
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

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Answer:

D

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