Answer:
0.005 `; 0.00499 ;
No, because np < 10 ;
2000
Step-by-step explanation:
Given that:
Number of samples , n = 100
Proportion, p = x / n
p = 1 / 200
= 0.005
p = μ
Standard deviation of sample proportion :
σp = sqrt((p(1 - p)) / n)
σp = sqrt((0.005(1 - 0.005)) / 200)
σp = sqrt((0.005(0.995)) / 200)
σp = sqrt(0.004975 / 200)
σp = sqrt(0.000024875)
σp = 0.0049874
σp = 0.00499
np = 100 * 0.005 = 0.5
n(1 - p) = 100(1-0.05) = 95
Smallest value of n for which sampling distribution is approximately normal
np ≥ 10
0.005n ≥ 10
To obtain the smallest value of n,
0.005n = 10
n = 10 / 0.005
n = 2000
Answer: i can’t read it could try upload it again
Explanation: Henlo :3
Answer:
7. (7^2)(4^2) or 49 * 16
8.18^5
Step-by-step explanation:
7. You can do that or simplify the exponents.
8. When dividing two exponents you can subtract the exponentes. *(note i didn't simplify 18^5 because it said expression, and a simplified number is not an expression.)
