Answer:
After 11 weeks, Darnell′s savings account will have a total of $8,360.
Step-by-step explanation:
The data provided is as follows:
n: 1 2 3 4
f (n): 260 360 460 560
Consider the data for f (n).
The series f (n) follows an arithmetic sequence with a common difference of 100 and first term as 260.
The nth term of an arithmetic sequence is:
![a_{n}=\frac{n}{2}[2a+(n-1)d]](https://tex.z-dn.net/?f=a_%7Bn%7D%3D%5Cfrac%7Bn%7D%7B2%7D%5B2a%2B%28n-1%29d%5D)
Compute the value of f (11) as follows:
![f(11)=\frac{11}{2}[(2\times260)+(11-1)\times 100]](https://tex.z-dn.net/?f=f%2811%29%3D%5Cfrac%7B11%7D%7B2%7D%5B%282%5Ctimes260%29%2B%2811-1%29%5Ctimes%20100%5D)
![=5.5\times[520+1000]\\\\=5.5\times 1520\\\\=8360](https://tex.z-dn.net/?f=%3D5.5%5Ctimes%5B520%2B1000%5D%5C%5C%5C%5C%3D5.5%5Ctimes%201520%5C%5C%5C%5C%3D8360)
Thus, after 11 weeks, Darnell′s savings account will have a total of $8,360.
Answer:

Step-by-step explanation:
Salvage value=$1000
Purchased value=$11,000
In order to find the balance in accumulated depreciation at december 31,2015 using the units of activity we will use the following formula:

In the above equation $10000 came from Purchased value - salvage Value
Answer: X=-2 1/5, X=-1
Step-by-step explanation:
The distance from CF is 9 units, 1/5 that = 1 4/5, add that to -4 is -2 1/5.
The distance from BD is 6 units, 2/3 that =4, add that to -5 is -1.
Hope this helps!
Attached the solution and work.