Answer:
Step-by-step explanation:
We can calculate this confidence interval using the population proportion calculation. To do this we must find p' and q'
Where p' = 14/100= 0.14 (no of left handed sample promotion)
q' = 1-p' = 1-0.14= 0.86
Since the requested confidence level is CL = 0.98, then α = 1 – CL = 1 – 0.98 = 0.02/2= 0.01, z (0.01) = 2.326
Using p' - z alpha √(p'q'/n) for the lower interval - 0.14-2.326√(0.14*0.86/100)
= -2.186√0.00325
= -2.186*0.057
= 12.46%
Using p' + z alpha √(p'q'/n)
0.14+2.326√(0.14*0.86/100)
= 0.466*0.057
= 26.5%
Thus we estimate with 98% confidence that between 12% and 27% of all Americans are left handed.
The correct answer is option c
First you need to multiply by a factor that will givr you the same value in one of the x or y value.
(6x-12y=24)1
+(-x-6y=4)-2
6x-12y=24
=2x+12y=-8
8x=16÷8
x=2
now that you have x, substitute x by 2 in one of the eqauations
-2-6y=4
+2. +2
=-6y=6
÷-6. ÷-6
y=-1
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Step-by-step explanation:
