How do I find the Q1 and Q3?<br><br>
0,0,1,2,2,3,4,4,4,4,5,6,6,7,7
Angelina_Jolie [31]
Answer:
Q1 = 2
Q3 = 6
Step-by-step explanation:
Mathematically, we have
Q1 = (n + 1)/4 th term
where n is the number of terms
By the count, we have n as 15
Q1 = (15 + 1)/4
Q1 = 4th term
Looking at the arrangement, the 4th term is 2
For Q3
Q3 = 3(n + 1)/4 th term
n = 15
Q3 = 3 * 4 = 12th term
The 12th term is 6
So that is the 3rd quartile
It’s D
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\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
Answer:
x + y ≤
3x + 5y ≥ 1100
Step-by-step explanation:
Given:
Seating capacity of theater = 250
Cost of each child ticket = $3
Cost of each adult ticket = $5
Cost per performance = $1100 at least
Find:
System of inequalities
Computation:
Let;
x = Number of children's tickets
y = Number of adult tickets
So
x + y ≤
3x + 5y ≥ 1100
