Answer:
D. $31,337.27
Step-by-step explanation:
We have that the initial amount of the loan is $5500.
Miranda took the loan for 4 years. So, the total present value is $5500×4 = $22,000.
The rate of interest on the loan is 7.5% i.e. 0.075 and it was for the duration of 10 years.
Also, it is given that the loan was compounded annually.
We have the formula as,
i.e. ![PV=\frac{P\times [1-(1+\frac{r}{n})^{-t\times n}]}{\frac{r}{n}}](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B%5Cfrac%7Br%7D%7Bn%7D%29%5E%7B-t%5Ctimes%20n%7D%5D%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D)
Substituting the values, we get,
i.e. ![PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B%5Cfrac%7B0.075%7D%7B12%7D%29%5E%7B-10%5Ctimes%2012%7D%5D%7D%7B%5Cfrac%7B0.075%7D%7B12%7D%7D)
i.e. ![22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281%2B0.00625%29%5E%7B-120%7D%5D%7D%7B0.00625%7D)
i.e. ![22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-%281.00625%29%5E%7B-120%7D%5D%7D%7B0.00625%7D)
i.e. ![22000=\frac{P\times [1-0.4735]}{0.00625}](https://tex.z-dn.net/?f=22000%3D%5Cfrac%7BP%5Ctimes%20%5B1-0.4735%5D%7D%7B0.00625%7D)
i.e. 
i.e. 
i.e. 
i.e. 
Thus, the total lifetime cost to pay of the loans compounded annually = 261.16 × 120 = $31,339.2
Hence, the total cost close to the answer is $31,337.27
Answer:
c
Step-by-step explanation:
Yesssssssssssssssssssssssssss
Answer:
Y = 5.5x
The constant is 5.5
since no matter what y and x are there will always be a 5.5 difference
Also mark brainliest please
Average rate of change can be calculated by determining the
rate of change at x = a, and at x = b
f’(x) =2 (3^x) ln(3)
f’(0) = 2 ln(3)
f’(1) = 6 ln(3)
f’(2) = 18 ln(3)
f’(3) = 54 ln(3)
Average:
at section A = [6 ln(3) – 2 ln(3)]/1 = 4 ln(3)
at section B = [54 ln(3) – 18 ln(3)]/1 = 36 ln(3)
section B is 9 times larger.
Based from the f’(x), f’(x) varies as the power of x. so the
greater of value of x, the greater the rate of change.
