Answer:
The distribution is approximately normal with mean = 2.8 and standard error = 0.4
Step-by-step explanation:
We are given;
Mean; μ = 2.8
Standard deviation; σ = 4
Sample size; n = 100
Now, the central limit theorem states that the sample mean with a sample size(n) from a population mean (μ) and population standard deviation(σ) will, for large value of n, have an approximately normal distribution with mean μ and standard error as (σ/√n)
The sample size is 100 and thus it's very large because it's bigger than minimum of 30 for approximate distribution.
Thus, SE = (σ/√n) = 4/√100 = 4/10 = 0.4
Thus,from the central limit theorem I described, we can say that the distribution is approximately normal with mean = 2.8 and standard error = 0.4
sorry dude I'm stupid I would help but I'm a idot
Y=54x+10
If he paid $30 it would be
30=54x + 10
Answer:
Move all terms to the left side and set equal to zero. Then set each factor equal to zero.Move all terms to the left side and set equal to zero. Then set each factor equal to zero.
Solution
= − 3
11−11+35=3 11+24=3
11+24=311n+24=3n11n+24=3n11+24−24=3−24 1
Answer:
x=-2
Step-by-step explanation:
