Answer:
Step-by-step explanation:
Let's solve for 
in the first equation and then solve for 
in the second equation.
I will then use the following identity to get right of the parameter, 
:
(Pythagorean Identity).
Let's begin with .
Subtract 2 on both sides:
Divide both sides by -3:
Now time for the second equation, .
Subtract 1 on both sides:
Divide both sides by 4:
Now let's plug it into our Pythagorean Identity:
now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
Answer:
The probability of both points falling in the same row or column is 7/19, or approximately 37%
Step-by-step explanation:
The easiest way to solve this is to think of it rephrased as "what is the probability that your second point will be in the same row or column as your first point". With that frame of reference, you can simply consider how many other points are left that do or do not fall in line with the selected one.
After selecting one, there are 19 points left.
The row that the first one falls in will have 3 remaining empty points.
The column will have 4 remaining empty points.
Add those up and you have 7 possible points that meet the conditions being checked.
So the probability of both points falling in the same row or column is 7/19, or approximately 37%
50 degrees is one while the other two are 130 degrees
