What do you need help with
Answer:
865
Step-by-step explanation:
We have that in 95% confidence level the value of z has a value of 1.96. This can be confirmed in the attached image of the normal distribution.
Now we have the following formula:
n = [z / E] ^ 2 * (p * q)
where n is the sample size, which is what we want to calculate, "E" is the error that is 2% or 0.02. "p" is the probability they give us, 5 out of 50, is the same as 1 out of 10, that is 0.1. "q" is the complement of p, that is, 1 - 0.1 = 0.9, that is, the value of q is 0.9.
Replacing these values we are left with:
n = [1.96 / 0.02] ^ 2 * [(0.1) * (0.9)]
n = 864.36
865 by rounding to the largest number.
1/5 of 152 is 30.4
1/2 of 152 is 76
1/4 of 152 is 38
hope this helped :)
Answer:
40
Step-by-step explanation:
![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] = ad-bc](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20ad-bc)