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
keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%20a%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%28r%5E%7B-7%7Db%5E%7B-8%7D%29%5E0%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7D%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20t%5E4%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20%5Ccfrac%7Bt%5E4%7D%7Bw%7D%20)
Answer:
To my knowledge of congruency, I would say the answer is A,
Step-by-step explanation:
Hope this helped!!!
Answer:
hello :
If f(x)=x-2 and g(x)=1/2x,
g(f(x)) = g ( x - 2 )
= 1/2(x-2)
g(f(x)) = (1/2)x - 1
-33=-11*3
-8=-11+3
so
(x-11)(x+3)
