Let n = required random sample size.
Assume that the population standard deviation is known as σ.
Let m = sample mean.
At the 95% confidence level, the expected range is
(m - k(σ/√n), m + k(σ/√n))
where k = 1.96.
Therefore the error margin is 1.96(σ/√n).
Because the error margin is specified as 3% or 0.03, therefore
(1.96σ)/√n = 0.03
√n = (1.96σ)/0.03
n = 128.05σ²
This means that the sample size is about 128 times the population variance.
Answer:
Smallest sample size = 128.05σ², where σ = population standard deviation.
the difference of 5 times the cube of x cubed and the quotient of 4 times x and 3
5*x^3 - (4x/3)
It would take 4750 because 1L is 5000 ml and 5000-250=4750. Use the web to do it in metric measurements.
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NB: values with 10 is to the first power.
