Answer:20
Step-by-step explanation:
Answer:
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Step-by-step explanation:
Given that;
the frequencies of there alternatives are;
Frequency A = 60
Frequency B = 12
Frequency C = 48
Total = 60 + 12 + 48 = 120
Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;
Relative Frequency A = Frequency A / total = 60 / 120 = 0.5
Relative Frequency B = Frequency B / total = 12 / 120 = 0.1
Relative Frequency C = Frequency C / total = 48 / 120 = 0.4
therefore;
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Check the picture below.
since in a rhombus the diagonals bisect each other, thus EC = EA.
now, the rhombus is simply 4 congruent triangles, we know the base and height of one of them, thus
![\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=8\\ h=15 \end{cases}\implies A=\cfrac{1}{2}(8)(15)\implies A=60 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of all 4 triangles}}{4(60)\implies 240}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20triangle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dbh~~%20%5Cbegin%7Bcases%7D%20b%3D8%5C%5C%20h%3D15%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%288%29%2815%29%5Cimplies%20A%3D60%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20all%204%20triangles%7D%7D%7B4%2860%29%5Cimplies%20240%7D)
9y^3+8y=4y+18y^2
9y^3-18y^2+4y=0
*Divide the whole thing by y
9y^2-18y+4=0
*use quadratic formula to solve
