Answer:
Considering is a compunded increase rate of the yearly car insurance fee and x is the initial value of the insurance fee then
x×(1.05)^12=$86.82
×=$86.82/(1.05)^12
I got 441 times using this formula 7x3^2=x
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
C, there are 32 more students who want a beef meal than want a chicken meal.
200/25 = 8
12 x 8 = 96
8 x 8 = 64
5 x 8 = 40
& 96 - 64 = 32
B , I remember having this and it was B , but your answers could be in different spots then mine but I say B