Answer:
the fourth one
Step-by-step explanation:
hope it helped!!!
Here is the answer to the given problem above. Given that the business owes $8000 on a loan and every month, it pays 1212. So for the first month, the remaining is 6,788. On the second month is 5,576, and on the third month, it is 4,364. In the fourth month, the remaining would be 3,152, and in the fifth month is 1,940. Therefore, on the sixth month, if the business pays 1,212 still, the remaining would be $728 only. Hope this answer helps.
Answer:
P = 0.0909
Step-by-step explanation:
To know the number of ways or combinations in which we can select x elements from a group of n elements, we can use the following equation:
![nCx=\frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=nCx%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
So, if you sat down at your computer and randomly loaded 4 of the 12 problems, there are 495 different possibilities and it is calculated as:
![12C4=\frac{12!}{4!(12-4)!}=495](https://tex.z-dn.net/?f=12C4%3D%5Cfrac%7B12%21%7D%7B4%21%2812-4%29%21%7D%3D495)
Then, from 495 different possibilities, there are 45 possibilities that both this problem and Richard Rusczyk's problem were among the four you loaded. This 45 possibilities are calculated as:
![(1C1)*(1C1)*(10C2)=(\frac{1!}{1!(1-1)!})*(\frac{1!}{1!(1-1)!})*(\frac{10!}{2!(10-2)!})=45](https://tex.z-dn.net/?f=%281C1%29%2A%281C1%29%2A%2810C2%29%3D%28%5Cfrac%7B1%21%7D%7B1%21%281-1%29%21%7D%29%2A%28%5Cfrac%7B1%21%7D%7B1%21%281-1%29%21%7D%29%2A%28%5Cfrac%7B10%21%7D%7B2%21%2810-2%29%21%7D%29%3D45)
Because you need to select: this problem and there is only one, the problem that Richard Rusczyk wrote and there is only one, and 2 problems from the other 10.
Finally, the probability that both this problem and Richard Rusczyk's problem were among the four you loaded is equal to:
![P=\frac{45}{495}=0.0909](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B45%7D%7B495%7D%3D0.0909)
x÷-9≥3 then you multiply by -9 so after it will be x≥-27