Answer: The lower bound of confidence interval would be 0.116.
Step-by-step explanation:
Since we have given that
p = 13.2%= 0.132
n = 1105
At 90% confidence,
z = 1.645
So, Margin of error would be
So, the lower bound of the confidence interval would be
Hence, the lower bound of confidence interval would be 0.116.
Answer:
x = (-29)/5
Step-by-step explanation:
Solve for x:
(2 (x - 5))/3 = (3 (x + 1))/2
Multiply both sides by 6:
(6×2 (x - 5))/3 = (6×3 (x + 1))/2
6/3 = (3×2)/3 = 2:
2×2 (x - 5) = (6×3 (x + 1))/2
6/2 = (2×3)/2 = 3:
2×2 (x - 5) = 3×3 (x + 1)
2×2 = 4:
4 (x - 5) = 3×3 (x + 1)
3×3 = 9:
4 (x - 5) = 9 (x + 1)
Expand out terms of the left hand side:
4 x - 20 = 9 (x + 1)
Expand out terms of the right hand side:
4 x - 20 = 9 x + 9
Subtract 9 x from both sides:
(4 x - 9 x) - 20 = (9 x - 9 x) + 9
4 x - 9 x = -5 x:
-5 x - 20 = (9 x - 9 x) + 9
9 x - 9 x = 0:
-5 x - 20 = 9
Add 20 to both sides:
(20 - 20) - 5 x = 20 + 9
20 - 20 = 0:
-5 x = 9 + 20
9 + 20 = 29:
-5 x = 29
Divide both sides of -5 x = 29 by -5:
(-5 x)/(-5) = 29/(-5)
(-5)/(-5) = 1:
x = 29/(-5)
Multiply numerator and denominator of 29/(-5) by -1:
Answer: x = (-29)/5
let's say the point dividing JK is say point P, so the JK segment gets split into two pieces, JP and PK
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ J(-25,10)\qquad K(5,-20)\qquad \qquad \stackrel{\textit{ratio from J to K}}{7:3} \\\\\\ \cfrac{J~~\begin{matrix} P \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} P \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~K} = \cfrac{7}{3}\implies \cfrac{J}{K} = \cfrac{7}{3}\implies3J=7K\implies 3(-25,10)=7(5,-20)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%20%5C%5C%5C%5C%5C%5C%20J%28-25%2C10%29%5Cqquad%20K%285%2C-20%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20J%20to%20K%7D%7D%7B7%3A3%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7BJ~~%5Cbegin%7Bmatrix%7D%20P%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B~~%5Cbegin%7Bmatrix%7D%20P%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~K%7D%20%3D%20%5Ccfrac%7B7%7D%7B3%7D%5Cimplies%20%5Ccfrac%7BJ%7D%7BK%7D%20%3D%20%5Ccfrac%7B7%7D%7B3%7D%5Cimplies3J%3D7K%5Cimplies%203%28-25%2C10%29%3D7%285%2C-20%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf P=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill\\\\ P=\left(\cfrac{(3\cdot -25)+(7\cdot 5)}{7+3}\quad ,\quad \stackrel{\textit{y-coordinate}}{\cfrac{(3\cdot 10)+(7\cdot -20)}{7+3}}\right) \\\\\\ P=\left( \qquad ,\quad \cfrac{30-140}{10} \right)\implies P=\left(\qquad ,~~\cfrac{-110}{10} \right)\implies P=(\qquad ,\quad -11)](https://tex.z-dn.net/?f=%5Cbf%20P%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20P%3D%5Cleft%28%5Ccfrac%7B%283%5Ccdot%20-25%29%2B%287%5Ccdot%205%29%7D%7B7%2B3%7D%5Cquad%20%2C%5Cquad%20%5Cstackrel%7B%5Ctextit%7By-coordinate%7D%7D%7B%5Ccfrac%7B%283%5Ccdot%2010%29%2B%287%5Ccdot%20-20%29%7D%7B7%2B3%7D%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20P%3D%5Cleft%28%20%5Cqquad%20%2C%5Cquad%20%5Ccfrac%7B30-140%7D%7B10%7D%20%5Cright%29%5Cimplies%20P%3D%5Cleft%28%5Cqquad%20%2C~~%5Ccfrac%7B-110%7D%7B10%7D%20%5Cright%29%5Cimplies%20P%3D%28%5Cqquad%20%2C%5Cquad%20-11%29)
