5/9 = 0.56
11/20 = 55/100 = 55%
So that means 5/9ths is bigger than 11/20ths
Hope this helped, trust me i know best:)
![\bf \stackrel{\textit{multiplying both sides by LCD of 3}}{3(y+5)=3\left[ \cfrac{5}{3}(x-3) \right]}\implies 3y+15=5(x-3) \\\\\\ 3y+15=5x-15\implies -5x+3y=-30\implies \stackrel{\textit{multiplying by -1}}{5x-3y=30}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20LCD%20of%203%7D%7D%7B3%28y%2B5%29%3D3%5Cleft%5B%20%5Ccfrac%7B5%7D%7B3%7D%28x-3%29%20%5Cright%5D%7D%5Cimplies%203y%2B15%3D5%28x-3%29%0A%5C%5C%5C%5C%5C%5C%0A3y%2B15%3D5x-15%5Cimplies%20-5x%2B3y%3D-30%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20by%20-1%7D%7D%7B5x-3y%3D30%7D)
bearing in mind the standard form uses all integers, and the x-variable cannot have a negative coefficient.
Answer: D
Suppose that 35\%35%35, percent of students in a school can sing their national anthem correctly. The music teacher takes an SRS of 303030 students from the population of 750750750 students at the school and finds that 40\%40%40, percent of students sampled can sing their national anthem correctly. The teacher plans to take more samples like this.
Step-by-step explanation: Khan Academy Explanation
The key is to find the first term a(1) and the difference d.
in an arithmetic sequence, the nth term is the first term +(n-1)d
the firs three terms: a(1), a(1)+d, a(1)+2d
the next three terms: a(1)+3d, a(1)+4d, a(1)+5d,
a(1) + a(1)+d +a(1)+2d=108
a(1)+3d + a(1)+4d + a(1)+5d=183
subtract the first equation from the second equation: 9d=75, d=75/9=25/3
Plug d=25/3 in the first equation to find a(1): a(1)=83/3
the 11th term is: a(1)+(25/3)(11-1)=83/3 +250/3=111
Please double check my calculation. <span />
In simplest form: 2 10101/62500
