Answer:
hmmmm. probably 2? I think it's two
We can solve this problem by referring to the standard
probability distribution tables for z.
We are required to find for the number of samples given the
proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value
of z equivalent to:
z = 1.96
Since the problem states that it should be within the true
proportion then p = 0.5
Now we can find for the sample size using the formula:
n = (z^2) p q /E^2
where,
<span> p = 0.5</span>
q = 1 – p = 0.5
E = estimate of 5% = 0.05
Substituting:
n = (1.96^2) 0.5 * 0.5 / 0.05^2
n = 384.16
<span>Around 385students are required.</span>
Answer:
Step-by-step explanation:multipy -4x and 4 by 2 were the 8x and the -8x would cancel then add and bring down answer of x=15
