well, looking at the picture of this vertically opening parabola, it has a vertex at 0,0 and it passes through 2,1 hmm ok
![~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = a(x-0)^2+0\qquad \stackrel{\textit{we also know that}}{x=2\qquad y = 1}\qquad \implies 1=a(2-0)^2+0 \\\\\\ 1=4a\implies \cfrac{1}{4}=a~\hspace{10em} \boxed{y=\cfrac{1}{4}x^2}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20y%20%3D%20a%28x-0%29%5E2%2B0%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bwe%20also%20know%20that%7D%7D%7Bx%3D2%5Cqquad%20y%20%3D%201%7D%5Cqquad%20%5Cimplies%201%3Da%282-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%201%3D4a%5Cimplies%20%5Ccfrac%7B1%7D%7B4%7D%3Da~%5Chspace%7B10em%7D%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B4%7Dx%5E2%7D)
No sé lo siento esqje necesito mas puntos lol hahaha
Let the consecutive even numbers be n and n+2.
From given condition,
n + n+2 = 66
2n = 64
∴ n = 32.
Therefore, the two consecutive even numbers are.
n = 32 and n+2 = 34.
Use copy dot flip to solve.
= 3 5/7 / 2/1
= 26/7 * 1/2 (Turned into improper fraction)
= 26/14 (Improper Fraction Answer)
= 1 12/14 (Mixed Number Answer Simplified)
= 1 6/7 (Final Answer Simplified)
Hope This Helped! Good Luck!
Answer:
Margin of Error = 5.4088 ;
Confidence interval = (30.1 ; 40.9)
Interval estimate are almost the same
Step-by-step explanation:
Given that :
Population standard deviation, σ = 9.3
Sample size, n = 8
Xbar = 35.5
Confidence level = 90%
The confidence interval:
Xbar ± Margin of error
Margin of Error = Zcritical * σ/sqrt(n)
Zcritical at 90% = 1.645
Margin of Error = 1.645 * 9.3/sqrt(8) = 5.4088
Confidence interval :
Xbar ± Margin of error
35.5 ± 5.4088
Lower boundary = (35.5 - 5.4088) = 30.0912 = 30.1
Upper boundary = (35.5 + 5.4088) = 40.9088 = 40.9
(30.1 ; 40.9)
T distribution =. (30.5 ; 40.5)
Normal distribution = (30.1, 40.9)