Answer:
a) CI = ( 5,1 ; 5,7 )
b) SE = 0,1
Step-by-step explanation:
a) Sample random n = 100
Mean = μ = 5,4
Standard deviation s = 1,3
CI = 99 % α = 1 % α = 0,01 α/2 = 0,005
z(c) for 0,005 is from z-table z(c) = 2,575
z(c) = ( X - μ ) /s/√n CI = μ ± z(c) * s/√n
CI = 5,4 ± 2,575* 1,3/10
CI = 5,4 ± 0,334
CI = ( 5,1 ; 5,7 )
b) SE = Standard deviation / √n
SE = 1,3 /10 SE = 0,1
We can support that with 99 % of probability our random variable will be in the CI.
X + k y = 1
k x + y = 1 / * ( - k )
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x + k y = 1
- k² x - k y = - k
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x - k² x = 1 - k
x ( 1 - k² ) = 1 - k
x = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
y = 1 - k( 1 - k )/( 1 - k² )
y = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
a ) For k = - 1 this system has no solution.
b ) For k ≠ - 1 and k ≠ 1, the system has unique solution:
( x , y ) = ( 1/ (1 + k) , 1/( 1 + k ) ).
c ) For k = 1, there are infinitely many solutions.
Answer:
A is correct and .. AAAAA
Step-by-step explanation:
A
