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.
Answer: Option b
Step-by-step explanation:
Linear equations have the following form:
Where the exponents n, m, s and h are always 0 or 1
To know which equations are nonlinear, identify among the options given, those that have exponents other than 1 or 0
Note that in option b) the exponent of the variable x is therefore the equation is nonlinear
Finally the answer is the option b
<span>£1 equals to 1.43 us dollars.
7 </span>× 1.43 = <span>10.01
</span>10.01 ÷ 5 = <span>2.002
</span>
2 fives are equal to <span>£7. </span>
Answer:
Step-by-step explanation:
f(x) = 4-x
g(x) = h+k
g(f(x)) = 2-16x+26
so put f(x) in g(x)
h+k
h((4-x)(4-x) + k
h(-8x+16)+k
if h = 2 , then
2-16x+32 + k
and we want 26 instead of 32 so subtract 6 so K = (-6)
2-16x+32 + (-6)
2-16x+32 - 6
2-16x+26
h=2
k=(-6)